[{"title": "Boyle\u2019s Law .txt", "text": "So in order to explain exactly how individual gas molecules behave, scientists came up with something called a kinetic molecular theory. And what this theory is is it's basically a bunch of assumptions that they make about gases that helps us understand how individual gas molecules interact. So the kinetic theory is used to explain the behavior of gases on a nanoscale level. Now, in order to look at the macroscopic level or explain gas behavior on a macroscopic level, much larger level, we have to look at something else. Now, scientists came up with different equations and formulas to explain macroscopic gas behavior. The first formula we're going to look at and discuss is called Boils Law."}, {"title": "Boyle\u2019s Law .txt", "text": "Now, in order to look at the macroscopic level or explain gas behavior on a macroscopic level, much larger level, we have to look at something else. Now, scientists came up with different equations and formulas to explain macroscopic gas behavior. The first formula we're going to look at and discuss is called Boils Law. Now, Boils Law works under certain conditions. Now, if we have a constant temperature and constant number of moles or N constant number of molecules, then we can use something called Boils Law. And what Boils Law relates is it relates volume and pressure."}, {"title": "Boyle\u2019s Law .txt", "text": "Now, Boils Law works under certain conditions. Now, if we have a constant temperature and constant number of moles or N constant number of molecules, then we can use something called Boils Law. And what Boils Law relates is it relates volume and pressure. And what it states is that volume is directly proportional to the inverse of one over P. Or said another way, volume is inversely proportional to one over P. And we can represent this as VP equals constant. In other words, if we rearrange this and multiply this by some constant, we get this formula. And what this basically says is that under these conditions of constant temperature and constant number of moles, v times p will always be a constant."}, {"title": "Boyle\u2019s Law .txt", "text": "And what it states is that volume is directly proportional to the inverse of one over P. Or said another way, volume is inversely proportional to one over P. And we can represent this as VP equals constant. In other words, if we rearrange this and multiply this by some constant, we get this formula. And what this basically says is that under these conditions of constant temperature and constant number of moles, v times p will always be a constant. So when B increases, p decreases, or when P increases, V decreases and so on. And our constant depends on the temperature and the number of moles. So if temperature increases or its temperature changes or N changes, this constant will also change."}, {"title": "Boyle\u2019s Law .txt", "text": "So when B increases, p decreases, or when P increases, V decreases and so on. And our constant depends on the temperature and the number of moles. So if temperature increases or its temperature changes or N changes, this constant will also change. In other words, the number that you get when you multiply D times P will also change. Now, suppose we have some gas or some sample of gas. And suppose we have one set of conditions and a second set of conditions."}, {"title": "Boyle\u2019s Law .txt", "text": "In other words, the number that you get when you multiply D times P will also change. Now, suppose we have some gas or some sample of gas. And suppose we have one set of conditions and a second set of conditions. So suppose I have the following. Suppose I have some container with pressure one and volume one. And I have the same container, but with a smaller volume and a different pressure."}, {"title": "Boyle\u2019s Law .txt", "text": "So suppose I have the following. Suppose I have some container with pressure one and volume one. And I have the same container, but with a smaller volume and a different pressure. So one set of conditions and second set of conditions. Now, what this law does is it explains macroscopic phenomenon. Like, for example, why is it that when I take a balloon filled with air and I push it hard enough, it explodes?"}, {"title": "Boyle\u2019s Law .txt", "text": "So one set of conditions and second set of conditions. Now, what this law does is it explains macroscopic phenomenon. Like, for example, why is it that when I take a balloon filled with air and I push it hard enough, it explodes? Well, why did that occur? Well, this can be explained by Boyle's Law and I'll show you in a second. Well, this equation can be rearranged in this format if we're dealing with two different sets of conditions."}, {"title": "Boyle\u2019s Law .txt", "text": "Well, why did that occur? Well, this can be explained by Boyle's Law and I'll show you in a second. Well, this equation can be rearranged in this format if we're dealing with two different sets of conditions. Notice that p times V will always give you a constant when you're talking about the same temperature and the same number of mole. So if I have one set of conditions p one times v one, that will give me a constant. And if I have the second set of conditions p two times v two, it will give me the same constant, right?"}, {"title": "Boyle\u2019s Law .txt", "text": "Notice that p times V will always give you a constant when you're talking about the same temperature and the same number of mole. So if I have one set of conditions p one times v one, that will give me a constant. And if I have the second set of conditions p two times v two, it will give me the same constant, right? So I can set them equal. This guy is equal to the same constant that this number represents. So this is my equation for two sets of data or two sets of conditions."}, {"title": "Boyle\u2019s Law .txt", "text": "So I can set them equal. This guy is equal to the same constant that this number represents. So this is my equation for two sets of data or two sets of conditions. Now let's look at this picture. Well, once again, why is it that a balloon explodes? Well, when the balloon is when you're not compressing the balloon, when you're just dangerous up, it has a certain pressure and a certain volume."}, {"title": "Boyle\u2019s Law .txt", "text": "Now let's look at this picture. Well, once again, why is it that a balloon explodes? Well, when the balloon is when you're not compressing the balloon, when you're just dangerous up, it has a certain pressure and a certain volume. When you take it in your hand and you begin squeezing it, you begin decreasing the volume. Boils law states that if you decrease volume, pressure must increase because our constant remains the same. And that means pressure will begin to increase and the ball or the balloon will pop when the pressure is large enough for it to burst open and pop."}, {"title": "Boyle\u2019s Law .txt", "text": "When you take it in your hand and you begin squeezing it, you begin decreasing the volume. Boils law states that if you decrease volume, pressure must increase because our constant remains the same. And that means pressure will begin to increase and the ball or the balloon will pop when the pressure is large enough for it to burst open and pop. And that's exactly why balloon, when squeezed, will eventually pop. So let's look at ventral sensation. Suppose that this is our balloon and this is our compressed balloon."}, {"title": "Boyle\u2019s Law .txt", "text": "And that's exactly why balloon, when squeezed, will eventually pop. So let's look at ventral sensation. Suppose that this is our balloon and this is our compressed balloon. Well, our gas molecule in this condition are further in part than they are in this condition. And that means if they're further apart here, they will make less collisions than here. And that means that there are less collisions."}, {"title": "Boyle\u2019s Law .txt", "text": "Well, our gas molecule in this condition are further in part than they are in this condition. And that means if they're further apart here, they will make less collisions than here. And that means that there are less collisions. Less of the molecules are colliding with the walls. And so with less collisions, that means we have less pressure. So the bigger the volume, the smaller the pressure."}, {"title": "Boyle\u2019s Law .txt", "text": "Less of the molecules are colliding with the walls. And so with less collisions, that means we have less pressure. So the bigger the volume, the smaller the pressure. So once again, we see that we can use the kinetic theory to explain nanoscopic or nanoscale behavior of these molecules. And once again, the kinetic theory explains boiler's law. A smaller volume means less room to navigate and increase in number of collisions."}, {"title": "Boyle\u2019s Law .txt", "text": "So once again, we see that we can use the kinetic theory to explain nanoscopic or nanoscale behavior of these molecules. And once again, the kinetic theory explains boiler's law. A smaller volume means less room to navigate and increase in number of collisions. This increase in collisions will increase our pressure because by definition, pressure is forced per unit area. And if we have more molecules hitting the walls, we have more force and so a higher pressure. So this is Boyle's Law and Boyle's Law is used to explain macroscopic behavior."}, {"title": "Boyle\u2019s Law .txt", "text": "This increase in collisions will increase our pressure because by definition, pressure is forced per unit area. And if we have more molecules hitting the walls, we have more force and so a higher pressure. So this is Boyle's Law and Boyle's Law is used to explain macroscopic behavior. So let's examine the graphs of Boyle's Law or a graph of Boyle's Law. Now, we can have two graphs. We can graph volume and pressure."}, {"title": "Boyle\u2019s Law .txt", "text": "So let's examine the graphs of Boyle's Law or a graph of Boyle's Law. Now, we can have two graphs. We can graph volume and pressure. Or we can grab volume and one over pressure. So let's graph this guy first. So recall that I said that volume is inversely proportional to one over P. Now mathematically what that means is we have this type of a graph in which as we increase our volume, our pressure decreases."}, {"title": "Boyle\u2019s Law .txt", "text": "Or we can grab volume and one over pressure. So let's graph this guy first. So recall that I said that volume is inversely proportional to one over P. Now mathematically what that means is we have this type of a graph in which as we increase our volume, our pressure decreases. Or if we decrease our volume, decrease that volume in the balloon, our pressure will begin to increase. If we continue to increase or decrease the volume, that pressure will begin to increase exponentially, right? And that's what this represents."}, {"title": "Boyle\u2019s Law .txt", "text": "Or if we decrease our volume, decrease that volume in the balloon, our pressure will begin to increase. If we continue to increase or decrease the volume, that pressure will begin to increase exponentially, right? And that's what this represents. Now instead, suppose that I graph volume over one over P. Well, how would that look? Well, if I grab the volume over one over P, whenever this guy increases, this guy increases by the same ratio amount. And that's because volume times pressure gives you a constant."}, {"title": "Boyle\u2019s Law .txt", "text": "Now instead, suppose that I graph volume over one over P. Well, how would that look? Well, if I grab the volume over one over P, whenever this guy increases, this guy increases by the same ratio amount. And that's because volume times pressure gives you a constant. If this increases by say, two times, then this must decrease by two times. That's why this guy is a straight line, the slope is constant, versus on this graph, the slope varies, it changes. And if you wanted to find the slope, you would have to use calculus and approximate it using lines tangent to any point on the line."}, {"title": "Structure of Atoms .txt", "text": "Now, all matter and mass is composed of very tiny units called atoms. Everything we see, we touch, we feel is composed of atoms. Now, atoms themselves are composed of nucleuses surrounded by electrons. Now, a nucleus is composed of two types of particles called protons and neutrons. Now, protons and neutrons have approximately the same weight. A neutron is a tiny bit heavier than protons, but for all purposes we can approximate that these guys have the same exact mass."}, {"title": "Structure of Atoms .txt", "text": "Now, a nucleus is composed of two types of particles called protons and neutrons. Now, protons and neutrons have approximately the same weight. A neutron is a tiny bit heavier than protons, but for all purposes we can approximate that these guys have the same exact mass. Electrons, however, have a very small mass, much smaller than that of protons or neutrons. In fact, it's 1800 times smaller than a proton or a neutron. Now, if we look at this table and we look at their masses, a proton has one AMU, a neutron has one AMU."}, {"title": "Structure of Atoms .txt", "text": "Electrons, however, have a very small mass, much smaller than that of protons or neutrons. In fact, it's 1800 times smaller than a proton or a neutron. Now, if we look at this table and we look at their masses, a proton has one AMU, a neutron has one AMU. Now, AMU is simply atomic mass unit. We're going to discuss that in detail in another lecture. But an electron has a mass of 5.5 times ten to negative four AMU that's much smaller than that of proton or a neutron."}, {"title": "Structure of Atoms .txt", "text": "Now, AMU is simply atomic mass unit. We're going to discuss that in detail in another lecture. But an electron has a mass of 5.5 times ten to negative four AMU that's much smaller than that of proton or a neutron. The charge, however, of a proton, an electron has the same magnitude 1.6\ntimes ten to negative 19 Coulombs. However, the sign of a proton is positive, while the sign of an electron is negative. A neutron has VR charge."}, {"title": "Structure of Atoms .txt", "text": "The charge, however, of a proton, an electron has the same magnitude 1.6\ntimes ten to negative 19 Coulombs. However, the sign of a proton is positive, while the sign of an electron is negative. A neutron has VR charge. It's a neutral charge. Now let's look at the structure. Now, in the illustration above, we see our atom."}, {"title": "Structure of Atoms .txt", "text": "It's a neutral charge. Now let's look at the structure. Now, in the illustration above, we see our atom. Now, this whole guy is our nucleus. And our nucleus is composed of two particles, protons and neutrons. In this atom we have two protons and two neutrons."}, {"title": "Structure of Atoms .txt", "text": "Now, this whole guy is our nucleus. And our nucleus is composed of two particles, protons and neutrons. In this atom we have two protons and two neutrons. The protons are quantitatively charged, while the neutrons are neutrally charged. Now, the electron is found orbiting our atom, our nucleus. And the distance between our nucleus and the electron is quite large."}, {"title": "Structure of Atoms .txt", "text": "The protons are quantitatively charged, while the neutrons are neutrally charged. Now, the electron is found orbiting our atom, our nucleus. And the distance between our nucleus and the electron is quite large. And in fact, atoms are mostly composed of empty space. And in fact, if our atom with the size of a football field, our nucleus will be the size of a marble. So you can imagine that our entire atom, for the most part, is composed of empty space."}, {"title": "Structure of Atoms .txt", "text": "And in fact, atoms are mostly composed of empty space. And in fact, if our atom with the size of a football field, our nucleus will be the size of a marble. So you can imagine that our entire atom, for the most part, is composed of empty space. And that's because our electrons are very, very small and they orbit our nucleus at a very, very great distance compared to the size of the nucleus itself. So what holds the nucleus or the protons and neutrons together? Well, the force that holds the nucleus together is called a nuclear force."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "Well, it tells us two important things. First, gas molecules have no volume. And second, gas molecules exert no force on one another. That means if we had a system continuing containing two gas molecules, gas molecule A and gas molecule B, that means these two guys have no volume and they exert no force on one another, so they can't communicate. So in many different ways, molecule A is invisible to molecule B and molecule B is invisible to molecule A. And that leads directly into the following results pressure created by one gas molecule is independent of the pressure created by another gas molecule."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "That means if we had a system continuing containing two gas molecules, gas molecule A and gas molecule B, that means these two guys have no volume and they exert no force on one another, so they can't communicate. So in many different ways, molecule A is invisible to molecule B and molecule B is invisible to molecule A. And that leads directly into the following results pressure created by one gas molecule is independent of the pressure created by another gas molecule. In other words, the pressure that this gas molecule exerts on the walls of my system of my container is independent of what this guy exerts on my wall of the container. So to find the total pressure, my system of two gas molecules, I simply have to add up this pressure and this pressure. And that's exactly what B states."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "In other words, the pressure that this gas molecule exerts on the walls of my system of my container is independent of what this guy exerts on my wall of the container. So to find the total pressure, my system of two gas molecules, I simply have to add up this pressure and this pressure. And that's exactly what B states. The total pressure of any system of gas molecules is equal to the pressure exerted by each an individual gas molecule. So p one plus p two plus all the way up to PM, assuming that my system is composed of M gas molecules. So this is simply a mathematical way of representing any series."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "The total pressure of any system of gas molecules is equal to the pressure exerted by each an individual gas molecule. So p one plus p two plus all the way up to PM, assuming that my system is composed of M gas molecules. So this is simply a mathematical way of representing any series. So let's look at this system. So suppose we have a closed system of five gas molecules, two blue and three red. Now, we have two types of molecules and a total of five molecules."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "So let's look at this system. So suppose we have a closed system of five gas molecules, two blue and three red. Now, we have two types of molecules and a total of five molecules. So my one type is blue. My second type is red. To find the total pressure of my system here, I would simply add up all the individual pressures."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "So my one type is blue. My second type is red. To find the total pressure of my system here, I would simply add up all the individual pressures. So five molecules altogether. So I add up to five molecules p blue one plus p blue two plus p red one plus p red two plus p red three. Now, I could also say that my p total is equal to the pressure due to blue gas molecules."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "So five molecules altogether. So I add up to five molecules p blue one plus p blue two plus p red one plus p red two plus p red three. Now, I could also say that my p total is equal to the pressure due to blue gas molecules. This guy plus this guy plus the pressure due to red gas molecule or p one plus p two plus p three. Another way of saying this is the pressure due to my blue gas molecule is the partial pressure of my blue molecule and this guy is the partial pressure of my red molecule. So partial pressure refers to the pressure exerted by one type of gas molecule."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "This guy plus this guy plus the pressure due to red gas molecule or p one plus p two plus p three. Another way of saying this is the pressure due to my blue gas molecule is the partial pressure of my blue molecule and this guy is the partial pressure of my red molecule. So partial pressure refers to the pressure exerted by one type of gas molecule. In this case, a blue gas molecule or a red gas molecule in a mixture of gases. In this case, two types of gas molecules, red and blue. So another way of representing this formula or equation is the following."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "In this case, a blue gas molecule or a red gas molecule in a mixture of gases. In this case, two types of gas molecules, red and blue. So another way of representing this formula or equation is the following. So we can represent the partial pressure of any gas in a mixture of gasses by the following equation. The partial pressure is equal to the mole fraction of that gas in our mixture times the total pressure. In other words, I want to represent this in this following way let's see what we do."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "So we can represent the partial pressure of any gas in a mixture of gasses by the following equation. The partial pressure is equal to the mole fraction of that gas in our mixture times the total pressure. In other words, I want to represent this in this following way let's see what we do. P total is equal to well, how many blue molecules on the system? Two. How many molecules altogether?"}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "P total is equal to well, how many blue molecules on the system? Two. How many molecules altogether? Five. So the mole fraction of my blue guy is two over five times my P total plus the partial pressure of my red molecules. We have three red molecules over five red molecules times the P total."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "Five. So the mole fraction of my blue guy is two over five times my P total plus the partial pressure of my red molecules. We have three red molecules over five red molecules times the P total. So this is the partial pressure due to the blue molecules. These guys are the same plus the partial pressures into the red molecules. These guys are also the same."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "So this is the partial pressure due to the blue molecules. These guys are the same plus the partial pressures into the red molecules. These guys are also the same. Gives us P total. And look, common denominator adds a two and three. I get five and five."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "Gives us P total. And look, common denominator adds a two and three. I get five and five. The five cancel and I simply get P total equals P total. So this formula makes sense. And in fact, this formula is called a Dalton's Law of partial pressures."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "The five cancel and I simply get P total equals P total. So this formula makes sense. And in fact, this formula is called a Dalton's Law of partial pressures. And this law can be derived using the ideal gas law. And let's see how. Well, suppose in part E, we have three types of gas molecules and each type has N number of moles."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "And this law can be derived using the ideal gas law. And let's see how. Well, suppose in part E, we have three types of gas molecules and each type has N number of moles. So n one, N two and n three. So red molecules, blue molecules and purple molecules. Well, what's the total number of molecules or moles in molecules of my system?"}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "So n one, N two and n three. So red molecules, blue molecules and purple molecules. Well, what's the total number of molecules or moles in molecules of my system? Well, in this case, it was two plus three. So we added them. So we do the same thing."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "Well, in this case, it was two plus three. So we added them. So we do the same thing. We add the moles, add the number of molecules. The N total is n plus n one plus n two plus n three is my total. So, if I want to find the total pressure using the ideal gas law, the following has to be done."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "We add the moles, add the number of molecules. The N total is n plus n one plus n two plus n three is my total. So, if I want to find the total pressure using the ideal gas law, the following has to be done. I rearrange it a bit and bring the D over on this side. And again, P total is equal to while my volume is constant, I'm assuming it's constant. My temperature is constant."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "I rearrange it a bit and bring the D over on this side. And again, P total is equal to while my volume is constant, I'm assuming it's constant. My temperature is constant. Also, R is a gas constant. It's always constant. And now I plug in N total or the total number of molecules in my system."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "Also, R is a gas constant. It's always constant. And now I plug in N total or the total number of molecules in my system. Now I can go to the next step. And instead of writing N total, I plug in all these three guides or the addition of these three guides and I get in parentheses n one plus n two plus n three times RT divided by V.\nAnd now I need simple algebra to distribute to each guide. And I get n one R T over V plus n two r T over V plus n three r T over V. And we see using the ideal gas law that we get the same exact thing as up here."}, {"title": "Dalton\u2019s Law on Partial Pressure .txt", "text": "Now I can go to the next step. And instead of writing N total, I plug in all these three guides or the addition of these three guides and I get in parentheses n one plus n two plus n three times RT divided by V.\nAnd now I need simple algebra to distribute to each guide. And I get n one R T over V plus n two r T over V plus n three r T over V. And we see using the ideal gas law that we get the same exact thing as up here. So P total is equal to partial pressure of gas one plus partial pressure of gas two and plus partial pressure of gas three. So once again, we found this formula first using the kinetic theory. And then we confirm that in fact, this works using the ideal gas law."}, {"title": "Oxidation Numbers Example .txt", "text": "So let's begin. In this molecule, we have an N atom at an fatom. An fatom, according to our table, precedes or is more important than an N atom. And that means we first assign our number. Oxidation number to F gets negative one. But since we have three F's, that means this one has a negative three."}, {"title": "Oxidation Numbers Example .txt", "text": "And that means we first assign our number. Oxidation number to F gets negative one. But since we have three F's, that means this one has a negative three. And since our entire molecule is neutral, this one must be plus three. Plus three minus three gives you neutral. So zero charge makes sense."}, {"title": "Oxidation Numbers Example .txt", "text": "And since our entire molecule is neutral, this one must be plus three. Plus three minus three gives you neutral. So zero charge makes sense. Let's look at Ammonium NH Four. NH four has an overall charge of plus one. Now let's look at the atoms."}, {"title": "Oxidation Numbers Example .txt", "text": "Let's look at Ammonium NH Four. NH four has an overall charge of plus one. Now let's look at the atoms. Our H atom is more important than N atom, so we assign to H first. Since H is not attached to a metal atom, we assign H a plus one. So four h is get plus four."}, {"title": "Oxidation Numbers Example .txt", "text": "Our H atom is more important than N atom, so we assign to H first. Since H is not attached to a metal atom, we assign H a plus one. So four h is get plus four. Since we want a plus one overall charge, this guy must be a minus three. Minus three plus four gives you a plus one. Let's look at no three."}, {"title": "Oxidation Numbers Example .txt", "text": "Since we want a plus one overall charge, this guy must be a minus three. Minus three plus four gives you a plus one. Let's look at no three. This guy has negative three as an overall charge. So let's look at the individual atoms. All is more important than N. That means we first assign our oval, so all gets a negative two."}, {"title": "Oxidation Numbers Example .txt", "text": "This guy has negative three as an overall charge. So let's look at the individual atoms. All is more important than N. That means we first assign our oval, so all gets a negative two. Since we have a three zeros, this becomes negative six for three OS. And since we want a negative one overall, our N must be plus five. Notice how in our three examples, our N differed in every example."}, {"title": "Oxidation Numbers Example .txt", "text": "Since we have a three zeros, this becomes negative six for three OS. And since we want a negative one overall, our N must be plus five. Notice how in our three examples, our N differed in every example. In this one, it was a plus three. In this one, there was a negative three, and this one was a plus five. Let's look at four."}, {"title": "Oxidation Numbers Example .txt", "text": "In this one, it was a plus three. In this one, there was a negative three, and this one was a plus five. Let's look at four. In four we have an ionic compound, and that means we have an atom and another molecule. So this atom, since it's in the second group, gets a plus two. So this is a plus two."}, {"title": "Oxidation Numbers Example .txt", "text": "In four we have an ionic compound, and that means we have an atom and another molecule. So this atom, since it's in the second group, gets a plus two. So this is a plus two. And we see from this example that each individual guy gets a minus one. So minus one times two gives us a plus two for this whole guy. Another way of seeing it a minus two, sorry."}, {"title": "Oxidation Numbers Example .txt", "text": "And we see from this example that each individual guy gets a minus one. So minus one times two gives us a plus two for this whole guy. Another way of seeing it a minus two, sorry. Another way of seeing it would have been to realize that this whole compound that is neutral. So if this is plus two, this must be minus two. Now, the individual atoms are the same as this example."}, {"title": "Oxidation Numbers Example .txt", "text": "Another way of seeing it would have been to realize that this whole compound that is neutral. So if this is plus two, this must be minus two. Now, the individual atoms are the same as this example. Let's look at this guy. Well, from this example, we know that each ammonia molecule has a plus one charge. That means two ammonia molecules will have a plus two charge."}, {"title": "Oxidation Numbers Example .txt", "text": "Let's look at this guy. Well, from this example, we know that each ammonia molecule has a plus one charge. That means two ammonia molecules will have a plus two charge. Now, since this entire compound is neutral, that means this guy has to be a minus two. And now let's assign individual oxation numbers to each atom. So o perceives s that means we assign to o first."}, {"title": "Oxidation Numbers Example .txt", "text": "Now, since this entire compound is neutral, that means this guy has to be a minus two. And now let's assign individual oxation numbers to each atom. So o perceives s that means we assign to o first. O gets a negative two. So negative two times four atoms gives you a negative eight for O. And since we want an overall negative two on this molecule, this S must have a plus six."}, {"title": "Oxidation Numbers Example .txt", "text": "O gets a negative two. So negative two times four atoms gives you a negative eight for O. And since we want an overall negative two on this molecule, this S must have a plus six. Plus six minus eight gives you a negative two. Now, we're not going to assign anything to this guy because we already did that in example, too. Let's look at six."}, {"title": "Oxidation Numbers Example .txt", "text": "Plus six minus eight gives you a negative two. Now, we're not going to assign anything to this guy because we already did that in example, too. Let's look at six. B a. So four. So, once again, BA BA, since it's in a second group, must have a plus two."}, {"title": "Oxidation Numbers Example .txt", "text": "B a. So four. So, once again, BA BA, since it's in a second group, must have a plus two. So BA gets a plus two. Let's look at so four. Now, so four from this example, we know has a negative two."}, {"title": "Oxidation Numbers Example .txt", "text": "So BA gets a plus two. Let's look at so four. Now, so four from this example, we know has a negative two. So this guy must be negative two. Another way of seeing it would have been to realize that this whole thing is neutral. So if this is plus two, this might be minus two."}, {"title": "Oxidation Numbers Example .txt", "text": "So this guy must be negative two. Another way of seeing it would have been to realize that this whole thing is neutral. So if this is plus two, this might be minus two. Now, we're not going to assign anything to individual atoms, because we did that in this example. Let's look at seven. Now, in seven, we have copper and S, right?"}, {"title": "Oxidation Numbers Example .txt", "text": "Now, we're not going to assign anything to individual atoms, because we did that in this example. Let's look at seven. Now, in seven, we have copper and S, right? S precedes copper. So S is in the same group as oxygen. So S must have a minus two."}, {"title": "Oxidation Numbers Example .txt", "text": "S precedes copper. So S is in the same group as oxygen. So S must have a minus two. So S as a minus two. Now, this whole compound has a neutral charge. So if this is minus two, our copper must be plus one each, right?"}, {"title": "Oxidation Numbers Example .txt", "text": "So S as a minus two. Now, this whole compound has a neutral charge. So if this is minus two, our copper must be plus one each, right? So this makes our two coppers plus two. Let's look at this guy. Now, we already did ammonia."}, {"title": "Oxidation Numbers Example .txt", "text": "So this makes our two coppers plus two. Let's look at this guy. Now, we already did ammonia. We know that's. Plus one. So this whole thing is plus one."}, {"title": "Oxidation Numbers Example .txt", "text": "We know that's. Plus one. So this whole thing is plus one. That means this whole guy must be plus one. Now this whole guy must be minus one. And we saw that in this example, it was minus one."}, {"title": "Oxidation Numbers Example .txt", "text": "That means this whole guy must be plus one. Now this whole guy must be minus one. And we saw that in this example, it was minus one. So this whole guy is minus one. Now let's look at individual atoms. Individual atoms are the same exact as these two guys."}, {"title": "Oxidation Numbers Example .txt", "text": "So this whole guy is minus one. Now let's look at individual atoms. Individual atoms are the same exact as these two guys. This must be negative two times three, negative six. For three oxygen molecules to create a negative one. Overall, on this molecule, this N must be a plus five."}, {"title": "Oxidation Numbers Example .txt", "text": "This must be negative two times three, negative six. For three oxygen molecules to create a negative one. Overall, on this molecule, this N must be a plus five. Plus five, minus six, negative one. You can check that. And it works for Ammonium, the same thing that we did here."}, {"title": "Oxidation Numbers Example .txt", "text": "Plus five, minus six, negative one. You can check that. And it works for Ammonium, the same thing that we did here. This H is assigned first, so it gets a plus one times four, a plus four. So this guy gets a plus four. And this end, to create a plus one or charge, must be negative three."}, {"title": "Oxidation Numbers Example .txt", "text": "This H is assigned first, so it gets a plus one times four, a plus four. So this guy gets a plus four. And this end, to create a plus one or charge, must be negative three. So negative three plus four gives you positive one works. Let's look at the last one, this guy. Well, we already know from this example, in this example that so four is negative two."}, {"title": "Oxidation Numbers Example .txt", "text": "So negative three plus four gives you positive one works. Let's look at the last one, this guy. Well, we already know from this example, in this example that so four is negative two. So four must be negative two. So our Zn must be plus two because we want a mutual atom. Now let's look at individual guys."}, {"title": "Oxidation Numbers Example .txt", "text": "So four must be negative two. So our Zn must be plus two because we want a mutual atom. Now let's look at individual guys. Well, it's the same thing as we did here. This guy assigned first to oxygen. So oxygen gets a negative two, times four, negative eight."}, {"title": "Cell diagram .txt", "text": "Electrochemical cells can be represented using something called a cell diagram or an electrochemical cell diagram. Now, this is simply a simplification of this drawing here, and it looks like this. So let's examine our full drawing. So within our anode, there's an oxidation reaction. So copper solid is oxidized into aqueous copper and it releases electrons. Electrons travel in this conductor to this metal bar, the cadmium."}, {"title": "Cell diagram .txt", "text": "So within our anode, there's an oxidation reaction. So copper solid is oxidized into aqueous copper and it releases electrons. Electrons travel in this conductor to this metal bar, the cadmium. The cadmium solid. And then the electrons react with our aqueous cadmium, forming our solid cadmium. So reduction occurs in the cathode and oxidation occurs in the anode."}, {"title": "Cell diagram .txt", "text": "The cadmium solid. And then the electrons react with our aqueous cadmium, forming our solid cadmium. So reduction occurs in the cathode and oxidation occurs in the anode. Now, this diagram is very tedious to draw. We can represent this diagram in a simple way using a cell diagram seen on this side. So these guys are equivalent representations."}, {"title": "Cell diagram .txt", "text": "Now, this diagram is very tedious to draw. We can represent this diagram in a simple way using a cell diagram seen on this side. So these guys are equivalent representations. Now, the double bar in the middle of the vertical double bar represents our sold bridge. The single vertical lines represents our separation of phases. For example, we have our solid copper and our aqueous copper."}, {"title": "Cell diagram .txt", "text": "Now, the double bar in the middle of the vertical double bar represents our sold bridge. The single vertical lines represents our separation of phases. For example, we have our solid copper and our aqueous copper. So acreage copper sound and solution and beaker one. And the solid copper is this electrode bar, the same way that we have the electrode bar this side and acreage copper on this side. And this is our separation of phases."}, {"title": "Cell diagram .txt", "text": "So acreage copper sound and solution and beaker one. And the solid copper is this electrode bar, the same way that we have the electrode bar this side and acreage copper on this side. And this is our separation of phases. The same concept on this side. This cadmium bar is separated by phases in the solution. So we have the ions here and the solid cadmium in the bar the same way that this bar here, this vertical line represents these two phases."}, {"title": "Cell diagram .txt", "text": "The same concept on this side. This cadmium bar is separated by phases in the solution. So we have the ions here and the solid cadmium in the bar the same way that this bar here, this vertical line represents these two phases. Now, this is their anode and this is their cathode. So the thing on your left is always the anode. The thing on your right is always the cathode."}, {"title": "Cell diagram .txt", "text": "Now, this is their anode and this is their cathode. So the thing on your left is always the anode. The thing on your right is always the cathode. And electrons travel from this guy to this guy. So the way you read this is that copper solid oxidize releasing two electrons and this ion. Now, these two electrons travel to this side into the cathode."}, {"title": "Half equivalence point .txt", "text": "Now, if you don't know what Titration is and you don't know what an equivalence point Is, then check out the link below. So, once again, we're tightrading an acid with the base. And here's our Titration curve where the y axis is PH and the x axis is the volume of base added. Now, we've defined the equivalence point to be the point at which all the acid has been neutralized. So every single molecule of acid in our buffer system has been neutralized. Now, we can also define something called the half equivalence point."}, {"title": "Half equivalence point .txt", "text": "Now, we've defined the equivalence point to be the point at which all the acid has been neutralized. So every single molecule of acid in our buffer system has been neutralized. Now, we can also define something called the half equivalence point. And the half equivalence point is the point at which exactly half of the acid in our buffer system has been neutralized. Now, suppose we choose our buffer system to consist of acetic acid. Now, acetic acid associates it to acetate ion and an H plus ion."}, {"title": "Half equivalence point .txt", "text": "And the half equivalence point is the point at which exactly half of the acid in our buffer system has been neutralized. Now, suppose we choose our buffer system to consist of acetic acid. Now, acetic acid associates it to acetate ion and an H plus ion. Now, suppose we begin adding some volume of our base. And suppose we choose our base to be ammonia. So we're adding the volume of ammonia into our acetic acid."}, {"title": "Half equivalence point .txt", "text": "Now, suppose we begin adding some volume of our base. And suppose we choose our base to be ammonia. So we're adding the volume of ammonia into our acetic acid. Now, the reaction looks like this. The conjugate acid, acetic acid dissociates into the conjugate base acetate ion. This base, the ammonia gains an H becoming ammonia."}, {"title": "Half equivalence point .txt", "text": "Now, the reaction looks like this. The conjugate acid, acetic acid dissociates into the conjugate base acetate ion. This base, the ammonia gains an H becoming ammonia. Now, so, let's look at this definition again. The half equivalence point is the point at which exactly half of the assets, exactly half of this guy has dissociated into this guy Aka. Has been neutralized."}, {"title": "Half equivalence point .txt", "text": "Now, so, let's look at this definition again. The half equivalence point is the point at which exactly half of the assets, exactly half of this guy has dissociated into this guy Aka. Has been neutralized. So that means we can define the half equivalence point in another way. The half equivalence point is the point at which the concentration of the conjugate acid equals the concentration of the conjugate base. Because half of this is now this."}, {"title": "Half equivalence point .txt", "text": "So that means we can define the half equivalence point in another way. The half equivalence point is the point at which the concentration of the conjugate acid equals the concentration of the conjugate base. Because half of this is now this. So the concentration of this guy equals this guy. Well, why is this definition important? Well, we'll see why in a second."}, {"title": "Half equivalence point .txt", "text": "So the concentration of this guy equals this guy. Well, why is this definition important? Well, we'll see why in a second. Let's look at the Henderson Hasselblack formula or equation. Now, if you don't know what this formula is, check out the link above. So, this equation states that PH is equal to PKA of our acid plus log of this ratio the concentration of the conjugate base over the concentration of the conjugate acid."}, {"title": "Half equivalence point .txt", "text": "Let's look at the Henderson Hasselblack formula or equation. Now, if you don't know what this formula is, check out the link above. So, this equation states that PH is equal to PKA of our acid plus log of this ratio the concentration of the conjugate base over the concentration of the conjugate acid. And this PH is the PH of our buffer system. So notice that since this guy equals this guy, this divided by this is one. So what's inside here is simply one."}, {"title": "Half equivalence point .txt", "text": "And this PH is the PH of our buffer system. So notice that since this guy equals this guy, this divided by this is one. So what's inside here is simply one. So let's rewrite it. PH is equal to PKA plus log of one. But what's log of one?"}, {"title": "Half equivalence point .txt", "text": "So let's rewrite it. PH is equal to PKA plus log of one. But what's log of one? Well, log of one is zero. And that means PH equals PKA. Well, that's nice and all, but why is that important?"}, {"title": "Half equivalence point .txt", "text": "Well, log of one is zero. And that means PH equals PKA. Well, that's nice and all, but why is that important? Where's the significance? Well, this means that now we can choose the PH of our bumper system by simply choosing an acid with PKA that's closest to our desired PH. So suppose, for example, I want my bubble system to have a PH of 4.7."}, {"title": "Half equivalence point .txt", "text": "Where's the significance? Well, this means that now we can choose the PH of our bumper system by simply choosing an acid with PKA that's closest to our desired PH. So suppose, for example, I want my bubble system to have a PH of 4.7. Now, how I find the asset to use is I simply find the acid with the PKA value closest to 4.7. Now, I go online I find my table, I look up an acid with a PH that's a PKA closest to a PH of 4.7, and I find that it's acetic acid. So now I know, using this equation here, that if I choose my buffer sydney system to consist of acetic acetic acid, my PH of my bumper system will be 4.7."}, {"title": "Parts per million Example .txt", "text": "The formula is mass of compound x divided by total mass of solution multiplied by ten to the six or million. Now, since this is a ratio that units cancel and Ppm is unitless, so now it's still an example using parts per million. In this example, we start with 25 bowls of water in a cup. Here's our cup. The blue dots are the water molecules, nothing else exists. You want to find them out in grams of HCL to add to create a 90,000 parts per million solution."}, {"title": "Parts per million Example .txt", "text": "Here's our cup. The blue dots are the water molecules, nothing else exists. You want to find them out in grams of HCL to add to create a 90,000 parts per million solution. So you want to go from this cup to this cup, where this cup contains HCL molecules in the concentration of 90,000 parts per million. We want to find the amount of grams of the red dots we need to add to create such a solution. The first step is to calculate the molecular weight of water."}, {"title": "Parts per million Example .txt", "text": "So you want to go from this cup to this cup, where this cup contains HCL molecules in the concentration of 90,000 parts per million. We want to find the amount of grams of the red dots we need to add to create such a solution. The first step is to calculate the molecular weight of water. To calculate the molecular weight of water, we simply add the atomic weight of oxygen plus two times the atomic weight of H because we have a substrate of two, so we get 16 grams/mol of oxygen plus two times 1 gram/mol of H gives us 18 grams/mol. So the molecular weight of water is 18 grams/mol. Now, to find the amount in grams of water that we have in our initial solution, we need to multiply the molecular weight by the 25 moles of H 20 that we have."}, {"title": "Parts per million Example .txt", "text": "To calculate the molecular weight of water, we simply add the atomic weight of oxygen plus two times the atomic weight of H because we have a substrate of two, so we get 16 grams/mol of oxygen plus two times 1 gram/mol of H gives us 18 grams/mol. So the molecular weight of water is 18 grams/mol. Now, to find the amount in grams of water that we have in our initial solution, we need to multiply the molecular weight by the 25 moles of H 20 that we have. So 18 grams/mol times 25 moles gives you 450. Now, moles cancel, the grams are left, so we have 450 grams of H 20. Now, we want to find the amount of HCL we need to add in terms of grams to create a 90,000 parts per million solution."}, {"title": "Parts per million Example .txt", "text": "So 18 grams/mol times 25 moles gives you 450. Now, moles cancel, the grams are left, so we have 450 grams of H 20. Now, we want to find the amount of HCL we need to add in terms of grams to create a 90,000 parts per million solution. So we simply use the parts per million formula. We say, well, we want to create a 90,000 grams or parts per million solution equals x is the amount of HCL grams we need to add divided by the total amount of grounds we have the solution. So we already have 450 grams of water plus the amount of HCL and grounds we will add."}, {"title": "Parts per million Example .txt", "text": "So we simply use the parts per million formula. We say, well, we want to create a 90,000 grams or parts per million solution equals x is the amount of HCL grams we need to add divided by the total amount of grounds we have the solution. So we already have 450 grams of water plus the amount of HCL and grounds we will add. So plus x times ten to the 6th or 1 million. We do a little bit of simple algebra to solve for x, we divide through by ten to the six we get 90,000 divided by 1 million equals this guy. Now, we multiply through by 450 plus x and we get 0.9, which is simply this guy times 450 plus x equals the x was left over on that side, so equals x."}, {"title": "Solubility Product Constant .txt", "text": "Before we talk about KSP, let's talk about the solubility of ionic compounds. Now, all ionic compounds have the ability to dissociate into their ion form when added into water. For example, let's take ionic compound sodium chloride. When we add sodium chloride chloride into water, it dissociates into two ions, sodium and chloride. Now, this reaction is called the forward reaction or dissolution. The reverse reaction is just as likely to occur, and that's called precipitation, the sufformation of ionic compound from its ion form."}, {"title": "Solubility Product Constant .txt", "text": "When we add sodium chloride chloride into water, it dissociates into two ions, sodium and chloride. Now, this reaction is called the forward reaction or dissolution. The reverse reaction is just as likely to occur, and that's called precipitation, the sufformation of ionic compound from its ion form. Now, initially, when we add photos chloride into water, the forward rate is much higher than the reverse rate. Eventually, however, though dynamic equilibrium is achieved, at this point, the forward rate is equal to the reverse rate. And at this point, the solution is said to be saturated, which basically means that the concentration of the ions or dissolved ions is that it's maximum."}, {"title": "Solubility Product Constant .txt", "text": "Now, initially, when we add photos chloride into water, the forward rate is much higher than the reverse rate. Eventually, however, though dynamic equilibrium is achieved, at this point, the forward rate is equal to the reverse rate. And at this point, the solution is said to be saturated, which basically means that the concentration of the ions or dissolved ions is that it's maximum. So these guys are at the maximum. Now, whenever we talk about normal equations or normal reactions, not fluidation reactions, we talk about equilibrium constants. In the same way, when we talk about salvation reactions, we could talk about something called solubility product constant or KST."}, {"title": "Solubility Product Constant .txt", "text": "So these guys are at the maximum. Now, whenever we talk about normal equations or normal reactions, not fluidation reactions, we talk about equilibrium constants. In the same way, when we talk about salvation reactions, we could talk about something called solubility product constant or KST. Now, when we determine the normal equilibrium constant, we don't include solids and liquids in our calculation. And in the same way, when we talk about salvation or solubility product constant KFC, we don't include solids and liquids. For example, let's take a reaction of solid Orion compound AB that associates in water into A plus B."}, {"title": "Solubility Product Constant .txt", "text": "Now, when we determine the normal equilibrium constant, we don't include solids and liquids in our calculation. And in the same way, when we talk about salvation or solubility product constant KFC, we don't include solids and liquids. For example, let's take a reaction of solid Orion compound AB that associates in water into A plus B. Now, since we don't count the solids, we don't count the liquids, but we do count gases and Aqueous compounds. When we determine the KSP or the solubility of bonus constants, we don't count this guy or the other guy. We only count these two guys."}, {"title": "Solubility Product Constant .txt", "text": "Now, since we don't count the solids, we don't count the liquids, but we do count gases and Aqueous compounds. When we determine the KSP or the solubility of bonus constants, we don't count this guy or the other guy. We only count these two guys. So KSP is equal to the concentration of A times the concentration of B. In this problem, we're given some unknown amount of barium sulfate and some unknown amount of water in a cup. Now, we want to mix the two and wait for dynamic equilibrium to establish."}, {"title": "Solubility Product Constant .txt", "text": "So KSP is equal to the concentration of A times the concentration of B. In this problem, we're given some unknown amount of barium sulfate and some unknown amount of water in a cup. Now, we want to mix the two and wait for dynamic equilibrium to establish. Once equilibrium establishes, we're given that the KSP or the Solubility product is equal to 1.0 times ten to negative ten at 25 degrees Celsius. So we want to find the solubility of barium sulfate. To find the solubility of barium sulfate, we must first write the dissociation reaction for barium sulfate."}, {"title": "Solubility Product Constant .txt", "text": "Once equilibrium establishes, we're given that the KSP or the Solubility product is equal to 1.0 times ten to negative ten at 25 degrees Celsius. So we want to find the solubility of barium sulfate. To find the solubility of barium sulfate, we must first write the dissociation reaction for barium sulfate. Therefore, we get 1 mol of barium sulfate in its solid form, dissociates into 1 mol of barium plus 1 mol of sulfate, and both guys are in the Aqueous form. The first step is to write the KSP equation. To write the KSP equation, we simply realize that this guy is a solid and therefore he doesn't count."}, {"title": "Solubility Product Constant .txt", "text": "Therefore, we get 1 mol of barium sulfate in its solid form, dissociates into 1 mol of barium plus 1 mol of sulfate, and both guys are in the Aqueous form. The first step is to write the KSP equation. To write the KSP equation, we simply realize that this guy is a solid and therefore he doesn't count. In this equation. These guys only count because they're both atreus. Remember, we never count solids and we never count liquids."}, {"title": "Solubility Product Constant .txt", "text": "In this equation. These guys only count because they're both atreus. Remember, we never count solids and we never count liquids. Therefore, KSP is equal to the concentration of barium times the concentration of sulfate ion. Finally, since this guy is 1.0 times cents and negative ten, we say KSP is equal to 1.0 times ten to negative ten equals. Now, since this is X and this is X, you write X is here."}, {"title": "Solubility Product Constant .txt", "text": "Therefore, KSP is equal to the concentration of barium times the concentration of sulfate ion. Finally, since this guy is 1.0 times cents and negative ten, we say KSP is equal to 1.0 times ten to negative ten equals. Now, since this is X and this is X, you write X is here. Now, since barium there's 1 mol of barrium, we put a one in front of the X for barium. And since there's a 1 mol of sulfate, we put the 1 mol in front of the X. So we get one X times one X equals X squared."}, {"title": "Catalysts .txt", "text": "So earlier in another lecture, we spoke about the relationship between temperature and reaction rate. And we said that as we increase our temperature, our reaction rate also increases because on average, more molecules will have enough kinetic energy to overcome the activation energy. Now, today we're going to look at something called catalysts. Now, catalysts are organic or inorganic molecules that also, like temperature, affect our rate of reaction. Now, let's look at the following hypothetical example in which reactants A plus B react to form a product AB. Now, let's suppose that our reaction is reversible, meaning it goes forward and backward."}, {"title": "Catalysts .txt", "text": "Now, catalysts are organic or inorganic molecules that also, like temperature, affect our rate of reaction. Now, let's look at the following hypothetical example in which reactants A plus B react to form a product AB. Now, let's suppose that our reaction is reversible, meaning it goes forward and backward. And that means an equilibrium. Our rate forward will be the same as the rate backwards. Now let's look at the catalyzed reaction."}, {"title": "Catalysts .txt", "text": "And that means an equilibrium. Our rate forward will be the same as the rate backwards. Now let's look at the catalyzed reaction. Suppose we add a catalyst, catalyst C, to our reactants. Now, before we look at the mechanism by which it increases the rate, let's make sure we understand the fact that catalysts are not used up in reaction. In other words, if you add some catalyst to our reactants, you will get that same catalyst back at the end of your reaction."}, {"title": "Catalysts .txt", "text": "Suppose we add a catalyst, catalyst C, to our reactants. Now, before we look at the mechanism by which it increases the rate, let's make sure we understand the fact that catalysts are not used up in reaction. In other words, if you add some catalyst to our reactants, you will get that same catalyst back at the end of your reaction. Now, that catalyst might react somehow with one of the reactants, maybe covalently or non covalent. In other words, it might buy to it and help them for the products. But at the end it will separate and you will be able to get your feed back."}, {"title": "Catalysts .txt", "text": "Now, that catalyst might react somehow with one of the reactants, maybe covalently or non covalent. In other words, it might buy to it and help them for the products. But at the end it will separate and you will be able to get your feed back. All right? So let's look at the mechanism by which these catalysts affect our reaction rates. So, in order to see this, we have to go back to our Iranians equation."}, {"title": "Catalysts .txt", "text": "All right? So let's look at the mechanism by which these catalysts affect our reaction rates. So, in order to see this, we have to go back to our Iranians equation. This equation we spoke about when we spoke about temperature and reaction rate. So K, our reaction constant is equal to z times p. Now, z and p are the scarcity factor and the frequency of collisions. Now, this guy e is what our catalyst affects."}, {"title": "Catalysts .txt", "text": "This equation we spoke about when we spoke about temperature and reaction rate. So K, our reaction constant is equal to z times p. Now, z and p are the scarcity factor and the frequency of collisions. Now, this guy e is what our catalyst affects. Now, catalysts speed up reactions by lowering the activation energy needed to convert the reactants to products. Now, this in turn increases the number of molecules that have enough kinetic energy to climb that activation barrier. In other words, it decreases this activation energy EA, thereby increasing this e component."}, {"title": "Catalysts .txt", "text": "Now, catalysts speed up reactions by lowering the activation energy needed to convert the reactants to products. Now, this in turn increases the number of molecules that have enough kinetic energy to climb that activation barrier. In other words, it decreases this activation energy EA, thereby increasing this e component. And this in turn increases our rate constant, which is directly proportional to rate of reaction. And that's how the rates of reactions are increased by catabalists. Now, let's look at this graph."}, {"title": "Catalysts .txt", "text": "And this in turn increases our rate constant, which is directly proportional to rate of reaction. And that's how the rates of reactions are increased by catabalists. Now, let's look at this graph. It's energy in the Y axis versus time or progress or reaction on the x axis. Now, this black curve is the curve that represents before additional catalysts. Notice activation energy goes all the way up to this blue level."}, {"title": "Catalysts .txt", "text": "It's energy in the Y axis versus time or progress or reaction on the x axis. Now, this black curve is the curve that represents before additional catalysts. Notice activation energy goes all the way up to this blue level. Now, when you add that catalyst, what happens is that activation energy is lowered by this much to this red level. And that means more molecules, on average will have enough kinetic energy to climb this new activation barrier and form the product. And that's exactly what happens when you add a catalyst."}, {"title": "Catalysts .txt", "text": "Now, when you add that catalyst, what happens is that activation energy is lowered by this much to this red level. And that means more molecules, on average will have enough kinetic energy to climb this new activation barrier and form the product. And that's exactly what happens when you add a catalyst. Now, it's very important to understand the following point. Catalysts do not, and I repeat, do not affect the equilibrium of reaction. In other words, what catalysts do is they speed up their forward reaction and reverse reaction."}, {"title": "Catalysts .txt", "text": "Now, it's very important to understand the following point. Catalysts do not, and I repeat, do not affect the equilibrium of reaction. In other words, what catalysts do is they speed up their forward reaction and reverse reaction. But the final concentrations of our product and reactions remain the same. In other words, let's look at this uncannyze and catalyze reaction. Again, suppose that the concentration and equilibrium of our uncatalyzed are as following we have concentration of A, we have concentration of B and construction of our product AB."}, {"title": "Catalysts .txt", "text": "But the final concentrations of our product and reactions remain the same. In other words, let's look at this uncannyze and catalyze reaction. Again, suppose that the concentration and equilibrium of our uncatalyzed are as following we have concentration of A, we have concentration of B and construction of our product AB. Now, for the catalyzed reaction, even though equilibrium will be reached much quicker because of a catalyst, the final concentrations are exactly the same. They have not changed. In other words, catalysts do not touch the equilibrium of our reaction."}, {"title": "Catalysts .txt", "text": "Now, for the catalyzed reaction, even though equilibrium will be reached much quicker because of a catalyst, the final concentrations are exactly the same. They have not changed. In other words, catalysts do not touch the equilibrium of our reaction. They affect the kinetics of our reaction, but they do not affect equilibrium. Now, we're going to examine the two types of catalysts. So we have heterogeneous catalysts are molecules that are in a different state compared to the reactants."}, {"title": "Catalysts .txt", "text": "They affect the kinetics of our reaction, but they do not affect equilibrium. Now, we're going to examine the two types of catalysts. So we have heterogeneous catalysts are molecules that are in a different state compared to the reactants. In other words, if our reactants are in a gas state or liquid state then our catalysts are in a solid state. Now, when we're dealing with heterogeneous catalysts, namely Salad catalysts, this is what happens. Our reactants absorb momentarily or bind to the catalyst which weaken the bonds, decreasing activation energy which in turn increases the reaction rate."}, {"title": "Catalysts .txt", "text": "In other words, if our reactants are in a gas state or liquid state then our catalysts are in a solid state. Now, when we're dealing with heterogeneous catalysts, namely Salad catalysts, this is what happens. Our reactants absorb momentarily or bind to the catalyst which weaken the bonds, decreasing activation energy which in turn increases the reaction rate. So let's look at the following uncategorized reaction. BR two reacts with C two h four to produce C two H two BR two. Now, this by itself is a very slow occurring reaction."}, {"title": "Catalysts .txt", "text": "So let's look at the following uncategorized reaction. BR two reacts with C two h four to produce C two H two BR two. Now, this by itself is a very slow occurring reaction. But if you add a catalyst, a metal catalyst, this reaction will speed up. Let's look at the following illustration. So this is our metal catalyst."}, {"title": "Catalysts .txt", "text": "But if you add a catalyst, a metal catalyst, this reaction will speed up. Let's look at the following illustration. So this is our metal catalyst. What happens is this reaction momentarily binds to the surface of our catalyst and this weakens the double bond. And then this other reactant can come from the top, attacking these carbons, thereby creating our product. Now, this is how metal catalysts act."}, {"title": "Catalysts .txt", "text": "What happens is this reaction momentarily binds to the surface of our catalyst and this weakens the double bond. And then this other reactant can come from the top, attacking these carbons, thereby creating our product. Now, this is how metal catalysts act. An example of such a metal catalyst is, for example, fuel cells. In fuel cells, plant and catalyst acts in the same manner to speed up the reactions the oxidation and reduction reactions in an anode in a cathode. Now, if you want to learn more about fuel cells, check out the link above."}, {"title": "Catalysts .txt", "text": "An example of such a metal catalyst is, for example, fuel cells. In fuel cells, plant and catalyst acts in the same manner to speed up the reactions the oxidation and reduction reactions in an anode in a cathode. Now, if you want to learn more about fuel cells, check out the link above. So now let's look at homogeneous catalysts. Now, homogeneous catalysts are catalysts that are in the same state as our reactant, usually liquid or gas. A great and common example of a homogeneous catalyst are acids."}, {"title": "Catalysts .txt", "text": "So now let's look at homogeneous catalysts. Now, homogeneous catalysts are catalysts that are in the same state as our reactant, usually liquid or gas. A great and common example of a homogeneous catalyst are acids. Now, these guys weaken bonds by adding an H plus ion to one of the reactants, thereby lowering the activation energy and speeding up our reaction. For example, let's look at the following reaction. Now, this actually involves a bit of organic chemistry but bear with me and I'll try to explain it."}, {"title": "Catalysts .txt", "text": "Now, these guys weaken bonds by adding an H plus ion to one of the reactants, thereby lowering the activation energy and speeding up our reaction. For example, let's look at the following reaction. Now, this actually involves a bit of organic chemistry but bear with me and I'll try to explain it. What happens is one of the H molecules, one of the H ions is added to this age group, to this oxygen group and this weakens this bond here. So then the hydroxide form act as a base or a nucleophile attacking this carbon bond, thereby displacing this weaker bond. And it was weakened by the h group, remember?"}, {"title": "Catalysts .txt", "text": "What happens is one of the H molecules, one of the H ions is added to this age group, to this oxygen group and this weakens this bond here. So then the hydroxide form act as a base or a nucleophile attacking this carbon bond, thereby displacing this weaker bond. And it was weakened by the h group, remember? So displacing. It forming our product. Now we have the oh group instead of the Och three group."}, {"title": "Catalysts .txt", "text": "So displacing. It forming our product. Now we have the oh group instead of the Och three group. And this is exactly how homogeneous catalysts act. In other words, they momentarily bind with our reactants, help them out, and then at the end, after a reaction is finished, they've move away, and you can isolate the catalyst at the end of your reaction. Now, a great example of biological catalysts are enzymes."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "The first experiment, which we'll talk about in great detail in another lecture was called a photoelectric experiment or simply the Photo Electric effect. And this experiment was conducted by Einstein. And what Einstein showed was that light, an electromagnetic phenomenon, had both particle like properties as well as wavelike properties. In other words, light has the following property called wave particle duality. And what this property shows or tells us is that whenever it's convenient, light can act as a wave. And whenever it's convenient, light will act as a particle."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "In other words, light has the following property called wave particle duality. And what this property shows or tells us is that whenever it's convenient, light can act as a wave. And whenever it's convenient, light will act as a particle. Now, following this experiment, another experiment was conducted known as the BROGLEY Experiment. And what that experiment showed was that not only light has its property but other subatomic particles, like electrons also have this duality property or the wave particle duality property. Now, these two experiments led directly to the following result the Uncertainty Principle, or the Heisenberg Uncertainty Principle, named after the guy who came up with the principle, Heisenberg."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "Now, following this experiment, another experiment was conducted known as the BROGLEY Experiment. And what that experiment showed was that not only light has its property but other subatomic particles, like electrons also have this duality property or the wave particle duality property. Now, these two experiments led directly to the following result the Uncertainty Principle, or the Heisenberg Uncertainty Principle, named after the guy who came up with the principle, Heisenberg. Now, what this principle showed was that it showed that as you move downward in size from something large to the subatomic level the less your objects act like particles and the more they act as a wave. In other words, if you get down to the subatomic level to the electrons and protons and neutrons the less your objects act as solid spheres and the more your objects act as waves. Now, to demonstrate what this uncertainty principle states, I'll use the following example."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "Now, what this principle showed was that it showed that as you move downward in size from something large to the subatomic level the less your objects act like particles and the more they act as a wave. In other words, if you get down to the subatomic level to the electrons and protons and neutrons the less your objects act as solid spheres and the more your objects act as waves. Now, to demonstrate what this uncertainty principle states, I'll use the following example. Suppose I have this relatively large ball which, from where you're sitting you can probably tell where the ball is and you can tell if the ball isn't moving so you could tell its velocity. Now, suppose I go smaller. Suppose I hold up this ball."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "Suppose I have this relatively large ball which, from where you're sitting you can probably tell where the ball is and you can tell if the ball isn't moving so you could tell its velocity. Now, suppose I go smaller. Suppose I hold up this ball. Now, once again, this is a relatively large ball. And from where you're sitting, you could probably tell that the ball isn't moving and you could tell where the ball is. Now, suppose I go even smaller."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "Now, once again, this is a relatively large ball. And from where you're sitting, you could probably tell that the ball isn't moving and you could tell where the ball is. Now, suppose I go even smaller. Suppose I go down to this really tiny marble which you probably can't see from where you're sitting. But I'll move it closer. There's my particle."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "Suppose I go down to this really tiny marble which you probably can't see from where you're sitting. But I'll move it closer. There's my particle. There's my solid sphere. Now, now that you saw the sphere, you could probably see that. You could probably see it from where you're sitting."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "There's my solid sphere. Now, now that you saw the sphere, you could probably see that. You could probably see it from where you're sitting. But suppose now, I walk a mile away or a kilometer away and suppose I hold this ball. Now, now, this ball becomes a spec. You could still see it, but it's much, much smaller."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "But suppose now, I walk a mile away or a kilometer away and suppose I hold this ball. Now, now, this ball becomes a spec. You could still see it, but it's much, much smaller. Now, suppose I walk a mile away and I hold this ball up. Now, this ball you probably won't see really. Well, you might see it if you have really good vision."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "Now, suppose I walk a mile away and I hold this ball up. Now, this ball you probably won't see really. Well, you might see it if you have really good vision. But I don't think I'll see it a mile away. Now, suppose I hold this really tiny marble, the solid sphere, from a mile away you definitely won't see this one. So in other words, the smaller you go, the less you see its position and the less you see its velocity."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "But I don't think I'll see it a mile away. Now, suppose I hold this really tiny marble, the solid sphere, from a mile away you definitely won't see this one. So in other words, the smaller you go, the less you see its position and the less you see its velocity. If I walk 5 miles away and I hold either of these balls, you won't see any ball and you won't be able to tell where the ball is and with what speed or with what velocity it's moving. The point is, and what this uncertainty principles show, is that as you shrink down to the atom and then to the sub atom, to the electron, you no longer are dealing with solid spheres. They're no longer solid spheres, and they act more as waves."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "If I walk 5 miles away and I hold either of these balls, you won't see any ball and you won't be able to tell where the ball is and with what speed or with what velocity it's moving. The point is, and what this uncertainty principles show, is that as you shrink down to the atom and then to the sub atom, to the electron, you no longer are dealing with solid spheres. They're no longer solid spheres, and they act more as waves. In other words, they have both wavelike properties and solid properties. And that means, because elementary particles are no longer solid spheres, there is no way to know its position and at the same time, its velocity with complete certainty. So the formula or the equation for this uncertainty principle is the following plaques constant a very, very small number divided by two is always less than our change in x or the uncertainty of our position."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "In other words, they have both wavelike properties and solid properties. And that means, because elementary particles are no longer solid spheres, there is no way to know its position and at the same time, its velocity with complete certainty. So the formula or the equation for this uncertainty principle is the following plaques constant a very, very small number divided by two is always less than our change in x or the uncertainty of our position. Change in position times mass times change in velocity. Now remember, mass times velocity is momentum. So this guy is change in momentum."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "Change in position times mass times change in velocity. Now remember, mass times velocity is momentum. So this guy is change in momentum. In other words, this is the uncertainty of our position and this is the uncertainty of our momentum or velocity. And what this equation basically says is the following the less our change in axis, if this guy is very small, that means we know more information about our position, where our electron is located. And that means if this guy decreases and this is a constant, this guy must increase, the smaller our change in excess, the more we know about our position, the greater our change in b is, the less we know about our velocity."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "In other words, this is the uncertainty of our position and this is the uncertainty of our momentum or velocity. And what this equation basically says is the following the less our change in axis, if this guy is very small, that means we know more information about our position, where our electron is located. And that means if this guy decreases and this is a constant, this guy must increase, the smaller our change in excess, the more we know about our position, the greater our change in b is, the less we know about our velocity. And likewise, the same holds the more we know about our velocity change in velocity. The less our change in velocity is. And the less we know about our change in x, the less we know about our position."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "And likewise, the same holds the more we know about our velocity change in velocity. The less our change in velocity is. And the less we know about our change in x, the less we know about our position. In other words, we can't be very certain about our position and at the same time about our velocity. That's what the uncertainty principle tells us. And this has to do with the duality nature of subatomic particles, electrons and protons, as well as a duality of light."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "In other words, we can't be very certain about our position and at the same time about our velocity. That's what the uncertainty principle tells us. And this has to do with the duality nature of subatomic particles, electrons and protons, as well as a duality of light. In other words, when you go from a large ball from this ball to a subatomic particle, our particle loses its solid sphere like properties. It stops acting like a solid sphere and starts acting more like a wave. And therefore, we can no longer pinpoint exactly where our object is and at the same time, what its velocity is, what its momentum."}, {"title": "Heisenberg\u2019s Uncertainty Principle .txt", "text": "In other words, when you go from a large ball from this ball to a subatomic particle, our particle loses its solid sphere like properties. It stops acting like a solid sphere and starts acting more like a wave. And therefore, we can no longer pinpoint exactly where our object is and at the same time, what its velocity is, what its momentum. The last thing I want to mention is the following this principle has nothing to do with how inaccurate or how accurate our instrument is, or how inaccurate or accurate our methods or experimental methods are. In other words, even if we have the perfect instrument and our methods were the perfect methods, we still would not be able to pinpoint exactly where our object is, our electron is, and exactly with what velocity and in which direction our electron is traveling. This principle has nothing to do with our instruments."}, {"title": "Fuel Cells .txt", "text": "Now, fuel cells are very commonly used on spacecraft. They provide electricity to the various supply and system spacecrafts. So let's look at oxidation and reduction oxygen reactions found in a fuel cell. So our oxidation is as follows. A diatomic hydrogen is oxidized and it releases two H plus ions and two electrons. Our reduction reaction is as follows."}, {"title": "Fuel Cells .txt", "text": "So our oxidation is as follows. A diatomic hydrogen is oxidized and it releases two H plus ions and two electrons. Our reduction reaction is as follows. A diatomic oxygen molecule takes up those two electrons and also takes up the two H plus ions forming water in a liquid state. Now, our neck reduction reaction is found by simply adding up these guys. We see that the H two plus ions cancel, the electrons cancel, and we have the following redox reaction."}, {"title": "Fuel Cells .txt", "text": "A diatomic oxygen molecule takes up those two electrons and also takes up the two H plus ions forming water in a liquid state. Now, our neck reduction reaction is found by simply adding up these guys. We see that the H two plus ions cancel, the electrons cancel, and we have the following redox reaction. Now, our e is 0.7. Our cell potential for our fuel cell is zero 7 volts. It's positive."}, {"title": "Fuel Cells .txt", "text": "Now, our e is 0.7. Our cell potential for our fuel cell is zero 7 volts. It's positive. Now let's look at the layout of a fuel cell. A fuel cell, like any other electrochemical cell, has an anode and a cathode. It has a conductor that carries electrons from the anode to the cathode."}, {"title": "Fuel Cells .txt", "text": "Now let's look at the layout of a fuel cell. A fuel cell, like any other electrochemical cell, has an anode and a cathode. It has a conductor that carries electrons from the anode to the cathode. And this is our outside system that receives electricity in the form of moving electrons. Now, like always, like most cases, our anode is negatively charged and out cathode is positively charged. And that's why electrons travel from the negative charge to the positive charge."}, {"title": "Fuel Cells .txt", "text": "And this is our outside system that receives electricity in the form of moving electrons. Now, like always, like most cases, our anode is negatively charged and out cathode is positively charged. And that's why electrons travel from the negative charge to the positive charge. Now, inside our anode, we need to allow H two molecules in the gas state in. And that's why we have an outside power source that allows those H two irons or H two molecules inside our anode. And to make sure our pressure is not increasing, make sure there's no build up in pressure, this needs to be released back into some outside location."}, {"title": "Fuel Cells .txt", "text": "Now, inside our anode, we need to allow H two molecules in the gas state in. And that's why we have an outside power source that allows those H two irons or H two molecules inside our anode. And to make sure our pressure is not increasing, make sure there's no build up in pressure, this needs to be released back into some outside location. That's why we have this guy on the bottom. So when this H two molecule enters our system, it is oxidized. But how is it oxidized?"}, {"title": "Fuel Cells .txt", "text": "That's why we have this guy on the bottom. So when this H two molecule enters our system, it is oxidized. But how is it oxidized? Well, this brown layer is a platinum catalyst. And this platinum acts to catalyze or speed up that reaction going from our reacting to products. So when this guy in a our anode, it reacts with the platinum catalyst producing two moles of H plus ions and two moles of electrons."}, {"title": "Fuel Cells .txt", "text": "Well, this brown layer is a platinum catalyst. And this platinum acts to catalyze or speed up that reaction going from our reacting to products. So when this guy in a our anode, it reacts with the platinum catalyst producing two moles of H plus ions and two moles of electrons. Now, these two moles of electrons travel via the conductor this way. Notice we have a membrane. And this membrane does not allow our electrons to pass from this anode to capital via this membrane."}, {"title": "Fuel Cells .txt", "text": "Now, these two moles of electrons travel via the conductor this way. Notice we have a membrane. And this membrane does not allow our electrons to pass from this anode to capital via this membrane. This membrane only allows H plus ions to flow or protons to flow. Now, why should we allow protons to flow? Well, we'll talk about that in a bit."}, {"title": "Fuel Cells .txt", "text": "This membrane only allows H plus ions to flow or protons to flow. Now, why should we allow protons to flow? Well, we'll talk about that in a bit. But notice some of the H or diatomic H must leave because we can't have a pressure build up in this system. So now we have the two electrons traveling all the way to this cathode. Now, when it travels through this guy, this guy provides electricity to some outside source."}, {"title": "Fuel Cells .txt", "text": "But notice some of the H or diatomic H must leave because we can't have a pressure build up in this system. So now we have the two electrons traveling all the way to this cathode. Now, when it travels through this guy, this guy provides electricity to some outside source. This is where the electrical work is done. Now, when this electron or two electrons travel all the way down to this cathode. These electrons react with the oxygen molecule, reducing it."}, {"title": "Fuel Cells .txt", "text": "This is where the electrical work is done. Now, when this electron or two electrons travel all the way down to this cathode. These electrons react with the oxygen molecule, reducing it. But notice that in order for this build up of H plus ions not to occur, these H plus ions must pass to this side. So this, in a way, acts as a sole bridge because if this membrane wasn't here, we'd have a build up of positive charge here and a lack of positive charge here. And then that means our electrons will stop flowing."}, {"title": "Fuel Cells .txt", "text": "But notice that in order for this build up of H plus ions not to occur, these H plus ions must pass to this side. So this, in a way, acts as a sole bridge because if this membrane wasn't here, we'd have a build up of positive charge here and a lack of positive charge here. And then that means our electrons will stop flowing. So to close the circuit, we need this membrane. And so these H plus ions travel from the anode to the cathode. And when they reach this position, they react with the oxygen and the electrons forming water."}, {"title": "Fuel Cells .txt", "text": "So to close the circuit, we need this membrane. And so these H plus ions travel from the anode to the cathode. And when they reach this position, they react with the oxygen and the electrons forming water. Now, this water needs to be released somewhere because if the water remains, there's a build up of water and now cell would eventually stop functioning. So this water leaves through some outside pump and is stored somewhere else. Now, notice, the same way we need to allow H two molecules inside our ano, we need to allow o two molecules inside our capital."}, {"title": "Fuel Cells .txt", "text": "Now, this water needs to be released somewhere because if the water remains, there's a build up of water and now cell would eventually stop functioning. So this water leaves through some outside pump and is stored somewhere else. Now, notice, the same way we need to allow H two molecules inside our ano, we need to allow o two molecules inside our capital. And that's why we have this guy here. When this enters, when this oxygen enters the capital, it reacts with the H plus I the electrons coming in, and it forms our water. And this is a continuous process and it powers some outside source in this area here."}, {"title": "Fuel Cells .txt", "text": "And that's why we have this guy here. When this enters, when this oxygen enters the capital, it reacts with the H plus I the electrons coming in, and it forms our water. And this is a continuous process and it powers some outside source in this area here. So we have a few problems with our fuel cells. The first major problem is diatomic. H two molecule does not occur in nature."}, {"title": "Fuel Cells .txt", "text": "So we have a few problems with our fuel cells. The first major problem is diatomic. H two molecule does not occur in nature. And it's very difficult and takes a lot of energy and money to generate. So it's very, very expensive. And that's why places like NASA use it."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "In this lecture, we're going to look at something called salts. Now, salts are formed whenever acids and bases react. So for example, let's look at a hypothetical reaction of a hypothetical acid and a hypothetical base. So we have HX plus MLH. So these guys associate to form the H ion, xion, Emion and hydroxide ion. So this H plus ion and the hydroxide ion will react to form water and this Xion and the Mion will react to form our salt."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "So we have HX plus MLH. So these guys associate to form the H ion, xion, Emion and hydroxide ion. So this H plus ion and the hydroxide ion will react to form water and this Xion and the Mion will react to form our salt. Now, different types of salts exist. Let's see what types of salts are formed when strong acids react with strong bases. So to illustrate this, let's see an example."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "Now, different types of salts exist. Let's see what types of salts are formed when strong acids react with strong bases. So to illustrate this, let's see an example. So a strong acid, hydrochloric acid and a strong base, sodium hydroxide dissociate to form an H plus ion, a chloride ion, sodium ion and hydroxide ion. Now, in the same way that these guys react to form water, these two guys will react to form our water, while these two guys will react to former salt. Now, notice I wrote mutual salt."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "So a strong acid, hydrochloric acid and a strong base, sodium hydroxide dissociate to form an H plus ion, a chloride ion, sodium ion and hydroxide ion. Now, in the same way that these guys react to form water, these two guys will react to form our water, while these two guys will react to former salt. Now, notice I wrote mutual salt. And in fact, strong acids and strong bases react to form Mutual Salt. And that's because our Final Solution has no presence of acids or bases. And let's see why."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "And in fact, strong acids and strong bases react to form Mutual Salt. And that's because our Final Solution has no presence of acids or bases. And let's see why. Well, this guy has a high Ka value because it's a strong acid. And this guy has a high KB value also because it's a strong base. And that means our equilibrium will lie all the way to the right."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "Well, this guy has a high Ka value because it's a strong acid. And this guy has a high KB value also because it's a strong base. And that means our equilibrium will lie all the way to the right. So in equilibrium, we're not going to have any of these guys present and we're not going to have any of this guy or this guy present due to these two assets and bases. Now, odor minization of water will still occur, but we're going to have the same concentration of this as this. So in our Final Solution, we're only going to have water and the neutral salt present, or just the salt present."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "So in equilibrium, we're not going to have any of these guys present and we're not going to have any of this guy or this guy present due to these two assets and bases. Now, odor minization of water will still occur, but we're going to have the same concentration of this as this. So in our Final Solution, we're only going to have water and the neutral salt present, or just the salt present. And because our concentrations of H plus and oh minus are equal, this is a neutral salt. So combining these two guys to form this guy and this guy is the same thing as taking a cup of water and adding some salt inside. The result is the same as if you would take some hydrochloric acid and some sodium hydroxide, mix them and get this, the two results are the same."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "And because our concentrations of H plus and oh minus are equal, this is a neutral salt. So combining these two guys to form this guy and this guy is the same thing as taking a cup of water and adding some salt inside. The result is the same as if you would take some hydrochloric acid and some sodium hydroxide, mix them and get this, the two results are the same. So let's see what types of salts are formed when strong bases and weak acids react. So once again, let's illustrate using an example. So acetic acid, a weak acid and sodium hydroxide, a strong base, react and dissociate into acetate ion, h plus ion, sodium ion, and hydroxide ion."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "So let's see what types of salts are formed when strong bases and weak acids react. So once again, let's illustrate using an example. So acetic acid, a weak acid and sodium hydroxide, a strong base, react and dissociate into acetate ion, h plus ion, sodium ion, and hydroxide ion. Now, in the same way that these two guys form water, and these two guys from water, this guy and this guy will also form out of water. But now this acetate ion and this sodium will form a basic salt. So we see that strong bases and weak acids produce basic salts."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "Now, in the same way that these two guys form water, and these two guys from water, this guy and this guy will also form out of water. But now this acetate ion and this sodium will form a basic salt. So we see that strong bases and weak acids produce basic salts. Well, this has to do because of a hydrolysis reaction. And let's see what happens so this is a weak acid and that means its Ka will be low. So equilibrium for this guy will lie all the way to the left, not the right as in this case."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "Well, this has to do because of a hydrolysis reaction. And let's see what happens so this is a weak acid and that means its Ka will be low. So equilibrium for this guy will lie all the way to the left, not the right as in this case. That means at our solution, when our solution is formed, equilibrium, we're going to have some of these guys present, right? We're going to have a bunch of these guys present. And that means at equilibrium, we're not only going to have water and salt, we're also going to have this guy present."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "That means at our solution, when our solution is formed, equilibrium, we're going to have some of these guys present, right? We're going to have a bunch of these guys present. And that means at equilibrium, we're not only going to have water and salt, we're also going to have this guy present. So we're going to have this ion, acetate ion, and our water molecule from here. And these guys will now react because this will act as an acid and this will act as a base, right? Because this is a conjugate base of this acid."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "So we're going to have this ion, acetate ion, and our water molecule from here. And these guys will now react because this will act as an acid and this will act as a base, right? Because this is a conjugate base of this acid. And because this conjugate acid is weak, this conjugate base is strong. And that means it will react with water to form back deciding acid and a hydroxide ion. And this hydroxide ion is what creates the basic solution."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "And because this conjugate acid is weak, this conjugate base is strong. And that means it will react with water to form back deciding acid and a hydroxide ion. And this hydroxide ion is what creates the basic solution. And because we have a basic solution, we're going to have the basic salt. So a times strong bases react with weak acids, we produce basic salts. Now, how basic our salt is depends on the KB value of our reaction."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "And because we have a basic solution, we're going to have the basic salt. So a times strong bases react with weak acids, we produce basic salts. Now, how basic our salt is depends on the KB value of our reaction. The higher the KB value, the stronger this base. And that means the more basic our salt. So let's look at what types of salts are formed when strong acids react with weak bases."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "The higher the KB value, the stronger this base. And that means the more basic our salt. So let's look at what types of salts are formed when strong acids react with weak bases. So let's examine the following example. Hydrochloric acid, a strong acid reacts with ammonia, a weak base, and in the presence of water, that they associate into H plus ion, a chloride ion. And our ammonia."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "So let's examine the following example. Hydrochloric acid, a strong acid reacts with ammonia, a weak base, and in the presence of water, that they associate into H plus ion, a chloride ion. And our ammonia. Now ammonia reacts with H to create Ammonium, a positively charged ion. Now this positively charged ion then reacts with the chloride to neutralize the charge, creating an acidic salt. Now, since we begin with water, we also have water at the end result."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "Now ammonia reacts with H to create Ammonium, a positively charged ion. Now this positively charged ion then reacts with the chloride to neutralize the charge, creating an acidic salt. Now, since we begin with water, we also have water at the end result. Now let's examine why we have an acidic salt. Remember, we begin with a weak acid that has a low KB value. And that means equilibrium will be far to the left."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "Now let's examine why we have an acidic salt. Remember, we begin with a weak acid that has a low KB value. And that means equilibrium will be far to the left. Our reactants will be favored, or at least our ammonia will be favored. So that means this weak base has a strong or relatively strong conjugate acid. So what happens when our conjugate acid is formed?"}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "Our reactants will be favored, or at least our ammonia will be favored. So that means this weak base has a strong or relatively strong conjugate acid. So what happens when our conjugate acid is formed? What happens when Ammonium is formed, when this guy combines with this guy? Well, we get ammonia and Ammonium will not want to exist in this state because it's a relatively strong conjugate acid. So it will want to dissociate back into our ammonia."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "What happens when Ammonium is formed, when this guy combines with this guy? Well, we get ammonia and Ammonium will not want to exist in this state because it's a relatively strong conjugate acid. So it will want to dissociate back into our ammonia. But now it dissociates back into ammonia in the presence of water. And when water is in the mixture, what happens? Well, this guy is an acid, so this guy must be a base."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "But now it dissociates back into ammonia in the presence of water. And when water is in the mixture, what happens? Well, this guy is an acid, so this guy must be a base. And our base will accept the H ion forming back our ammonia and creating hydronium. So now, equilibrium, we're going to have more hydronium. And that means our acidity of our solution will increase."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "And our base will accept the H ion forming back our ammonia and creating hydronium. So now, equilibrium, we're going to have more hydronium. And that means our acidity of our solution will increase. So our salt will be acidic. Therefore, we see that conjugate acid determines the PH of our solution. And the higher the K. A, the stronger our conjugate acid and the more acidic our salt or the more acidic our solution."}, {"title": "Acidic Basic and Neutral Salts .txt", "text": "So our salt will be acidic. Therefore, we see that conjugate acid determines the PH of our solution. And the higher the K. A, the stronger our conjugate acid and the more acidic our salt or the more acidic our solution. Now, that means whenever we mix strong acids and weak bases, we create acidic salt. So let's look at what types of salts are created when weak bases and weak acids are reactive. So there is actually a competition between the conjugate acid and the conjugate base."}, {"title": "Intro to Quantum Mechanics .txt", "text": "Now, in this lecture, I'm going to give a very, very brief introduction to quantum mechanics. The reason this branch of physics is important in chemistry is it becomes important when we talk about quantum numbers and the photoelectric effect. Now, quantum mechanics is the study of microscopic phenomena such as electromagnetic magnetic waves and the study of subatomic particles such as electrons, protons, neutrons, et cetera. Now, the main idea, the main conclusion that we get from quantum mechanics is the following tiny subatomic particles called electrons and protons and neutrons and other particles. Now, subatomic simply means smaller than an atom gain or lose energy in discrete amount called photons. Now, we're going to talk more about photons and what they are when we'll talk about the photoelectric effect."}, {"title": "Intro to Quantum Mechanics .txt", "text": "Now, the main idea, the main conclusion that we get from quantum mechanics is the following tiny subatomic particles called electrons and protons and neutrons and other particles. Now, subatomic simply means smaller than an atom gain or lose energy in discrete amount called photons. Now, we're going to talk more about photons and what they are when we'll talk about the photoelectric effect. Now, this statement is analogous to the following everyday situation. Suppose we have a vending machine. And this vending machine only accepts coins that are $0.25."}, {"title": "Intro to Quantum Mechanics .txt", "text": "Now, this statement is analogous to the following everyday situation. Suppose we have a vending machine. And this vending machine only accepts coins that are $0.25. So it won't accept two coins. It won't accept five cent coins. It won't accept dollar bills."}, {"title": "Intro to Quantum Mechanics .txt", "text": "So it won't accept two coins. It won't accept five cent coins. It won't accept dollar bills. It only accepts a single coin known as the quarter. Now, in the same analogous way, electrons and protons and neutrons and other subatomic particles accept energy bursts this feet amount called a photon. Now, one other result from quantum mechanics is the following."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "So we already spoke about the root mean square velocity of any set of points. Now, we're going to look specifically at the root mean square velocity for gases. Now, in gases, root mean square velocity is given by the following formula, which can be derived using calculus. Now, I will spare you the calculus and simply give you the formula. But if you're curious about where this formula comes from, leave A comment and I'll Show you. So, VRMs is equal to the square root of three times R times T divided by M, where M is our molar mass of gas, t is our temperature in Kelvin, and R is the molar gas constant."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "Now, I will spare you the calculus and simply give you the formula. But if you're curious about where this formula comes from, leave A comment and I'll Show you. So, VRMs is equal to the square root of three times R times T divided by M, where M is our molar mass of gas, t is our temperature in Kelvin, and R is the molar gas constant. Now, my goal is I want to use this VRMs to find the average kinetic energy. Because remember, kinetic energy of anything is given by one half times mass times V squared. Now, to find the kinetic average, I basically plug in my Dr meh into my D, and that will give me the average, because remember, this guy is the average or the quadratic average of my sets of points."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "Now, my goal is I want to use this VRMs to find the average kinetic energy. Because remember, kinetic energy of anything is given by one half times mass times V squared. Now, to find the kinetic average, I basically plug in my Dr meh into my D, and that will give me the average, because remember, this guy is the average or the quadratic average of my sets of points. So average kinetic energy is equal to one half times and times VR mass squared equals one half times mass. Now, I take my formula for BRMs, for Dances, and plug into my equation. The radical will disappear, because the radical has an exponent of a half a half times two is one."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "So average kinetic energy is equal to one half times and times VR mass squared equals one half times mass. Now, I take my formula for BRMs, for Dances, and plug into my equation. The radical will disappear, because the radical has an exponent of a half a half times two is one. So I simply get one half times mass over molam mass times R and T. Now, remember, molar mass has units of mass divided by moles. So the mass will cancel, and moles will go on top, and we'll get one half the mass will cancel. So this M and this M will cancel."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "So I simply get one half times mass over molam mass times R and T. Now, remember, molar mass has units of mass divided by moles. So the mass will cancel, and moles will go on top, and we'll get one half the mass will cancel. So this M and this M will cancel. The moles will go on top, and we'll have R times T. So my file final representation of my kinetic or average kinetic energy is three times N times R times T divided by two, where N is our number of moles. So Suppose we have 1 mol of any gas. Well, since we have 1 Mol, our N will be one."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "The moles will go on top, and we'll have R times T. So my file final representation of my kinetic or average kinetic energy is three times N times R times T divided by two, where N is our number of moles. So Suppose we have 1 mol of any gas. Well, since we have 1 Mol, our N will be one. And for 1 Mol, our formula becomes average kinetic energy of 1 mol of gas. That's equal to three times R times T divided by two. So the only non constant in this equation is T. So we see kinetic energy depends strictly on temperature, or 1 mol of gas depends strictly on the temperature."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "And for 1 Mol, our formula becomes average kinetic energy of 1 mol of gas. That's equal to three times R times T divided by two. So the only non constant in this equation is T. So we see kinetic energy depends strictly on temperature, or 1 mol of gas depends strictly on the temperature. Now, suppose instead I want to find one molecule. What's the kinetic average of one molecule of gas? Not 1 mol."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "Now, suppose instead I want to find one molecule. What's the kinetic average of one molecule of gas? Not 1 mol. Well, remember, a mole has an avocado's number of molecules. So I simply take my formula and divide it by N. So three times R times two divided by two N. The reason I singled out the R and the N is because this guy is known as a Boltzmann constant. In other words, scientists use some value k that equals r divided by n. So we replace this R divided by n with k called the Bolton constant, and our equation becomes three times k times t over two."}, {"title": "Average Kinetic Energy and Root Mean Square Velocity .txt", "text": "Well, remember, a mole has an avocado's number of molecules. So I simply take my formula and divide it by N. So three times R times two divided by two N. The reason I singled out the R and the N is because this guy is known as a Boltzmann constant. In other words, scientists use some value k that equals r divided by n. So we replace this R divided by n with k called the Bolton constant, and our equation becomes three times k times t over two. Which is the same thing as saying three times R times t divided by two times n.\nSo once again, for one molecule, a single molecule, to find the kinetic average of that molecule, we simply use this formula where k is at boltsman constant. Now, to find the average kinetic energy of a mole of molecules, we have to use this formula. Now note that at any given time, our molecule can have any kinetic energy."}, {"title": "Colloids .txt", "text": "That means that they are nonhomogeneous solutions. Recall that a homogeneous solution is a solution in which the compounds that compose the mixture are in the same state while colloids are composed of compounds in different states. Now, they also contain larger solid particles. For example, blood, which is a colloid, is composed mostly of water and hemoglobin. Hemoglobin is the larger particle, the solute, while water is a solvent, okay? Now, the larger particles are also called colloidal particles."}, {"title": "Colloids .txt", "text": "For example, blood, which is a colloid, is composed mostly of water and hemoglobin. Hemoglobin is the larger particle, the solute, while water is a solvent, okay? Now, the larger particles are also called colloidal particles. So hemoglobin is the colloidal particle. Now, other examples of colloids exist. Fog."}, {"title": "Colloids .txt", "text": "So hemoglobin is the colloidal particle. Now, other examples of colloids exist. Fog. Fog is composed of liquid particles found in the gas state. Smoke is composed of solid particles or carbon, found in the gas state. Wood cream is composed of gas particles found in the liquid state."}, {"title": "Colloids .txt", "text": "Fog is composed of liquid particles found in the gas state. Smoke is composed of solid particles or carbon, found in the gas state. Wood cream is composed of gas particles found in the liquid state. And paint is composed of solid particles found in the liquid state. Now, carbon systems also experience something called a Tyndall effect. The tyle effect is a scattering of light due to the presence of large particles."}, {"title": "Colloids .txt", "text": "And paint is composed of solid particles found in the liquid state. Now, carbon systems also experience something called a Tyndall effect. The tyle effect is a scattering of light due to the presence of large particles. Now, if you look at this system here, pretend that this is a cup and inside is a colloidal system, okay? And these are the colloidal particles or the large particles. Now, when light will enter it's going to bounce back and forth between these particles and it's going to lose energy."}, {"title": "Colloids .txt", "text": "Now, if you look at this system here, pretend that this is a cup and inside is a colloidal system, okay? And these are the colloidal particles or the large particles. Now, when light will enter it's going to bounce back and forth between these particles and it's going to lose energy. Eventually, when it comes out, it's going to come out with less energy because remember, light carries energy. Light is composed of photons. That means that the system, the colloidal system will be translucent so you won't be able to see through it very clearly."}, {"title": "Colloids .txt", "text": "Eventually, when it comes out, it's going to come out with less energy because remember, light carries energy. Light is composed of photons. That means that the system, the colloidal system will be translucent so you won't be able to see through it very clearly. The solid portion of the colloid is called a continuous phase. The solid portion of the colloid is called dispersed phase. Now, when water is a solvent, classification occurs in two ways."}, {"title": "Colloids .txt", "text": "The solid portion of the colloid is called a continuous phase. The solid portion of the colloid is called dispersed phase. Now, when water is a solvent, classification occurs in two ways. Lyophilicaloids are those colloids that have strong attractions between continuous and dispersed phases. Examples include protein water, which is found in blood. Protein is a dispersed phase."}, {"title": "Colloids .txt", "text": "Lyophilicaloids are those colloids that have strong attractions between continuous and dispersed phases. Examples include protein water, which is found in blood. Protein is a dispersed phase. Water is a continuous phase and they mix very well. Biophobic colloids are those colloids that lack attraction between continuous and dispersed phases. And examples include fat and water mixtures."}, {"title": "Colloids .txt", "text": "Water is a continuous phase and they mix very well. Biophobic colloids are those colloids that lack attraction between continuous and dispersed phases. And examples include fat and water mixtures. We know that fat, the dispersed phase and water continuous days don't mix very well. Now, there are two ways to filter colloids. The first way is by heating and then filtration."}, {"title": "Colloids .txt", "text": "We know that fat, the dispersed phase and water continuous days don't mix very well. Now, there are two ways to filter colloids. The first way is by heating and then filtration. If you heat a colloid, this will cause the salute to coagulate and then you could filter it by simple filtration. The second way is via dialysis. Dialysis is a separation using a semi permeable membrane."}, {"title": "Colloids .txt", "text": "If you heat a colloid, this will cause the salute to coagulate and then you could filter it by simple filtration. The second way is via dialysis. Dialysis is a separation using a semi permeable membrane. For example, you have a semi permeable membrane here. Red blood cells mixed with water. The red blood cells are the dispersed phase."}, {"title": "Cell voltage example .txt", "text": "So we begin with a certain electrochemical cell that's composed of cadmium and zinc. Now, we are given the cell voltage or the electromotive force around cells, and it's given to be zero point 36 volts. Under standard conditions, that means 1 bar pressure and one molar concentration. Now, this guy is also at 25 degrees Celsius. So we are also given the cell voltage for one of the two half reactions, one of the two half cells. And the zinc half reaction is given negative zero point 76 volts."}, {"title": "Cell voltage example .txt", "text": "Now, this guy is also at 25 degrees Celsius. So we are also given the cell voltage for one of the two half reactions, one of the two half cells. And the zinc half reaction is given negative zero point 76 volts. We want to find our goal is to find the cell voltage for the other half cell for the other half reaction. So, in the first step, we begin our problem by writing their reduction reaction for the entire equation for the entire process. So, solid zinc reacts with cadmium ions to produce aqueous zinc and solid cadmium."}, {"title": "Cell voltage example .txt", "text": "We want to find our goal is to find the cell voltage for the other half cell for the other half reaction. So, in the first step, we begin our problem by writing their reduction reaction for the entire equation for the entire process. So, solid zinc reacts with cadmium ions to produce aqueous zinc and solid cadmium. So let's figure out which one is oxidized and which one is reduced. This will help us place the metal that belongs to the anode and the metal that belongs to the cathode. So, zinc solid goes from a neutral charge to a plus two charge."}, {"title": "Cell voltage example .txt", "text": "So let's figure out which one is oxidized and which one is reduced. This will help us place the metal that belongs to the anode and the metal that belongs to the cathode. So, zinc solid goes from a neutral charge to a plus two charge. That means it loses electrons. So that means it is oxidized. It's a reducing agent."}, {"title": "Cell voltage example .txt", "text": "That means it loses electrons. So that means it is oxidized. It's a reducing agent. Now, this guy cadmium ion goes from an ion to a neutral charge, and that means this guy is reduced. So it's the oxidizing agent. It gains those electrons that are released by this zinc solid metal."}, {"title": "Cell voltage example .txt", "text": "Now, this guy cadmium ion goes from an ion to a neutral charge, and that means this guy is reduced. So it's the oxidizing agent. It gains those electrons that are released by this zinc solid metal. So that means in step two, when we draw our electrochemical cell, the first half cell will contain our zinc solid. And that's because zinc is oxidized, and the anode always contains the oxidation reaction. So this is our zinc metal."}, {"title": "Cell voltage example .txt", "text": "So that means in step two, when we draw our electrochemical cell, the first half cell will contain our zinc solid. And that's because zinc is oxidized, and the anode always contains the oxidation reaction. So this is our zinc metal. That means this must be our cadmium metal. And so electrons will travel via the conductor, via this volt meter and into this cathode, into this electrode. The volt meter, by the way, is the thing that reads zero point 36 volts."}, {"title": "Cell voltage example .txt", "text": "That means this must be our cadmium metal. And so electrons will travel via the conductor, via this volt meter and into this cathode, into this electrode. The volt meter, by the way, is the thing that reads zero point 36 volts. So this sold bridge is placed here because it plays the role of closing the circuit. Without the sole bridge, this guy would not function. Electrons would not flow, and it's very important."}, {"title": "Cell voltage example .txt", "text": "So this sold bridge is placed here because it plays the role of closing the circuit. Without the sole bridge, this guy would not function. Electrons would not flow, and it's very important. So, once again, our anode, our place where oxidation occurs, and our cathode, the place where reduction occurs. So electrons travel from this way, from this electrode to this electrode. So when they leave this cell, this solid zinc releases electrons, right?"}, {"title": "Cell voltage example .txt", "text": "So, once again, our anode, our place where oxidation occurs, and our cathode, the place where reduction occurs. So electrons travel from this way, from this electrode to this electrode. So when they leave this cell, this solid zinc releases electrons, right? Electrons begin to flow here. It also releases a zinc ion into this solution. And when the electrons travel this way, they combine."}, {"title": "Cell voltage example .txt", "text": "Electrons begin to flow here. It also releases a zinc ion into this solution. And when the electrons travel this way, they combine. When they reach this metal, they combine with the cadmium ions forming our cadmium solid. So, in the third step, we basically want to use this cell voltage formula to find our cell voltage for the cathode. Remember?"}, {"title": "Cell voltage example .txt", "text": "When they reach this metal, they combine with the cadmium ions forming our cadmium solid. So, in the third step, we basically want to use this cell voltage formula to find our cell voltage for the cathode. Remember? Now, we know that this guy, this value of negative zero point 76 represents the value for the anode. So we use this formula to solve for our cathode. So we know our 0.36 volts, which is told by the voltmeter."}, {"title": "Cell voltage example .txt", "text": "Now, we know that this guy, this value of negative zero point 76 represents the value for the anode. So we use this formula to solve for our cathode. So we know our 0.36 volts, which is told by the voltmeter. Now, this we don't know. So we let a DX minus now in parentheses we have a negative 0.76 volts, right? So negative and negative becomes a positive."}, {"title": "Charles Law .txt", "text": "And what this theory did for us is it helped us gain more intuition about how individual gas molecules interact on a very small microscopic level. Now we also spoke about a law called Boils Law. And what Boils Law did for us is it helps help us gain more intuition about the macroscopic behavior of gases. In other words, what happens at constant temperature when we take a balloon filled with air and squeeze it. Well, we said that, and Boyle's Law explained this, that when we squeeze the balloon, we decrease volume, increase pressure, and eventually our balloon will pop. Now we're going to look at another law called Charles Law which also helps us explain the macroscopic large scale behavior of many gas molecules and how they interact with one another and with our system."}, {"title": "Charles Law .txt", "text": "In other words, what happens at constant temperature when we take a balloon filled with air and squeeze it. Well, we said that, and Boyle's Law explained this, that when we squeeze the balloon, we decrease volume, increase pressure, and eventually our balloon will pop. Now we're going to look at another law called Charles Law which also helps us explain the macroscopic large scale behavior of many gas molecules and how they interact with one another and with our system. So for Charles Law to work, two conditions must hold. We must have constant pressure and we must have constant number of molecules. So our N number of moles stays the same."}, {"title": "Charles Law .txt", "text": "So for Charles Law to work, two conditions must hold. We must have constant pressure and we must have constant number of molecules. So our N number of moles stays the same. And what Charles Law does is it relates volume and temperature. Remember Boyle's Law related volume and pressure? Well, Charles Law relates temperature and volume when pressure is constant."}, {"title": "Charles Law .txt", "text": "And what Charles Law does is it relates volume and temperature. Remember Boyle's Law related volume and pressure? Well, Charles Law relates temperature and volume when pressure is constant. And what it says is that volume is directly proportional to temperature. And this means if we bring the T over or if we multiply a constant by T and we bring T over, what we get is the following v divided by T is equal to a constant. Remember, in Boyle's Law we saw that P times V gave us a constant and that if we increase pressure, our volume must decrease while keeping the constant the same."}, {"title": "Charles Law .txt", "text": "And what it says is that volume is directly proportional to temperature. And this means if we bring the T over or if we multiply a constant by T and we bring T over, what we get is the following v divided by T is equal to a constant. Remember, in Boyle's Law we saw that P times V gave us a constant and that if we increase pressure, our volume must decrease while keeping the constant the same. Well, in this case, we have the same kind of situation, except now we have volume divided by temperature. In other words. Now for this constant to remain a constant and not change if we increase volume, say by two, our temperature also must increase by the same amount by two."}, {"title": "Charles Law .txt", "text": "Well, in this case, we have the same kind of situation, except now we have volume divided by temperature. In other words. Now for this constant to remain a constant and not change if we increase volume, say by two, our temperature also must increase by the same amount by two. So whenever we increase volume, we increase temperature. Or if we decrease volume, we must decrease the temperature for this guy to remain constant. Now notice this constant remains or depends on two things on the pressure and on the number of molecules."}, {"title": "Charles Law .txt", "text": "So whenever we increase volume, we increase temperature. Or if we decrease volume, we must decrease the temperature for this guy to remain constant. Now notice this constant remains or depends on two things on the pressure and on the number of molecules. For example, if we have more molecules, our constant will be higher. And we'll learn more about that when we talk about the ideal gas law. For now, it's sufficient to say that our cost depends on both of these guys."}, {"title": "Charles Law .txt", "text": "For example, if we have more molecules, our constant will be higher. And we'll learn more about that when we talk about the ideal gas law. For now, it's sufficient to say that our cost depends on both of these guys. So the same way we did for Boyle's Law, for Charles Law, we can rewrite this relation or this relation in the following manner. Now, suppose I have some system, some gas system and I have two sets of different conditions for this gas system. Well, I can relate them in the following manner."}, {"title": "Charles Law .txt", "text": "So the same way we did for Boyle's Law, for Charles Law, we can rewrite this relation or this relation in the following manner. Now, suppose I have some system, some gas system and I have two sets of different conditions for this gas system. Well, I can relate them in the following manner. Assuming that these guys are both constant. In other words, note that the constant always stays the same when our pressure number of moles stays the same. That means if we have some conditions, v one and T one and a second condition of V two and T two."}, {"title": "Charles Law .txt", "text": "Assuming that these guys are both constant. In other words, note that the constant always stays the same when our pressure number of moles stays the same. That means if we have some conditions, v one and T one and a second condition of V two and T two. And these two conditions are under the same pressure and number of moles, that means our constant will be the same. And so I can say V one over T one equals V two over t two equals that same constant. So once again, for a gas sample with two different sets of T's and Vs for two different conditions, this is our law."}, {"title": "Charles Law .txt", "text": "And these two conditions are under the same pressure and number of moles, that means our constant will be the same. And so I can say V one over T one equals V two over t two equals that same constant. So once again, for a gas sample with two different sets of T's and Vs for two different conditions, this is our law. This is our Charles Law. Now, whenever we talk about gases, and we talk about these different laws that revolve around gases, our temperature is never in Celsius, it's only in Kelvin. And that means we have to convert Celsius to Kelvin."}, {"title": "Charles Law .txt", "text": "This is our Charles Law. Now, whenever we talk about gases, and we talk about these different laws that revolve around gases, our temperature is never in Celsius, it's only in Kelvin. And that means we have to convert Celsius to Kelvin. And this is how you do it. To get the Kelvin temperature, you take the Celsius temperature and you add 273.15 to it. For example, if our Celsius temperature is ten degrees Celsius, I simply add 273.15 and I get 283.15 Kelvin."}, {"title": "Charles Law .txt", "text": "And this is how you do it. To get the Kelvin temperature, you take the Celsius temperature and you add 273.15 to it. For example, if our Celsius temperature is ten degrees Celsius, I simply add 273.15 and I get 283.15 Kelvin. That's our temperature in Kelvin. So let's think of an example where Charles Law is evident. So suppose it's someone's birthday and it's winter."}, {"title": "Charles Law .txt", "text": "That's our temperature in Kelvin. So let's think of an example where Charles Law is evident. So suppose it's someone's birthday and it's winter. So it's cold outside and you need to go and buy somebody a present. So he decides to buy a balloon. So you go inside the store, you fill up the balloon with some helium."}, {"title": "Charles Law .txt", "text": "So it's cold outside and you need to go and buy somebody a present. So he decides to buy a balloon. So you go inside the store, you fill up the balloon with some helium. Now suppose once you fill the balloon up, there are no holes in the balloon. So the number of moles or number of molecules inside our balloon remains constant. Also, let's suppose that our pressure inside the store and outside is one ATM atmospheric pressure."}, {"title": "Charles Law .txt", "text": "Now suppose once you fill the balloon up, there are no holes in the balloon. So the number of moles or number of molecules inside our balloon remains constant. Also, let's suppose that our pressure inside the store and outside is one ATM atmospheric pressure. So let's suppose that our pressure is also constant. Now, obviously, if it's winter, it's much colder outside than inside. So suppose I fill up my balloon with N number of moles of helium, and suppose now I go outside with that balloon."}, {"title": "Charles Law .txt", "text": "So let's suppose that our pressure is also constant. Now, obviously, if it's winter, it's much colder outside than inside. So suppose I fill up my balloon with N number of moles of helium, and suppose now I go outside with that balloon. Well, on the inside, inside the store, we were at one condition. We had some D two and t two, right? Once we step outside, our temperature drops."}, {"title": "Charles Law .txt", "text": "Well, on the inside, inside the store, we were at one condition. We had some D two and t two, right? Once we step outside, our temperature drops. So what happens to volume? Well, according to Charles Law, this guy states that if we go outside and my temperature drops, then my volume must drop as well to make sure that our constant stays the same. That means if I go outside with my balloon, my balloon will shrivel, it will become smaller, and that's because of Charles Law."}, {"title": "Charles Law .txt", "text": "So what happens to volume? Well, according to Charles Law, this guy states that if we go outside and my temperature drops, then my volume must drop as well to make sure that our constant stays the same. That means if I go outside with my balloon, my balloon will shrivel, it will become smaller, and that's because of Charles Law. So once again, we see how macro scale events are explained by Charles Law, just like they were explained by Boyle's Law. So we have yet another law that helps us explain how gas molecules behave when we have a lot of them together, not simply individual gas molecules. Now, we can, of course, explain how Charles Law functions on an individual level using the kinetic theory."}, {"title": "Charles Law .txt", "text": "So once again, we see how macro scale events are explained by Charles Law, just like they were explained by Boyle's Law. So we have yet another law that helps us explain how gas molecules behave when we have a lot of them together, not simply individual gas molecules. Now, we can, of course, explain how Charles Law functions on an individual level using the kinetic theory. Now, kinetic theory explains Charles Law on a microscopic level at constant pressure. So we have constant pressure, right? Our pressure doesn't change."}, {"title": "Charles Law .txt", "text": "Now, kinetic theory explains Charles Law on a microscopic level at constant pressure. So we have constant pressure, right? Our pressure doesn't change. So if our temperature increases, what happens to the molecules? The molecules gain more kinetic energy. And if they gain more kinetic energy, they gain more speed."}, {"title": "Charles Law .txt", "text": "So if our temperature increases, what happens to the molecules? The molecules gain more kinetic energy. And if they gain more kinetic energy, they gain more speed. They become quicker. And that means the only way that our pressure stays constant with higher kinetic energy is if the volume expands. So that's exactly what happens."}, {"title": "Charles Law .txt", "text": "They become quicker. And that means the only way that our pressure stays constant with higher kinetic energy is if the volume expands. So that's exactly what happens. In order to ensure that there is constant pressure. An increase in temperature means there's an increase in kinetic energy. And this increase in kinetic energy means that there are more molecules, or the molecules are pushing against the wall of the container with greater force."}, {"title": "Charles Law .txt", "text": "In order to ensure that there is constant pressure. An increase in temperature means there's an increase in kinetic energy. And this increase in kinetic energy means that there are more molecules, or the molecules are pushing against the wall of the container with greater force. And so they must expand container, increase the volume. So let's look at the graph of volume times our temperature. Now."}, {"title": "Charles Law .txt", "text": "And so they must expand container, increase the volume. So let's look at the graph of volume times our temperature. Now. Temperature here is in Kelvin. So our origin is at zero. Zero?"}, {"title": "Charles Law .txt", "text": "Temperature here is in Kelvin. So our origin is at zero. Zero? Remember, we can't have negative volume. Our small volume of zero is zero, so can't go below this. Now our temperature in Kelvin is zero."}, {"title": "Charles Law .txt", "text": "Remember, we can't have negative volume. Our small volume of zero is zero, so can't go below this. Now our temperature in Kelvin is zero. So we saw it as zero. Zero? At zero volume."}, {"title": "Charles Law .txt", "text": "So we saw it as zero. Zero? At zero volume. We have zero Kelvin. But remember, zero Kelvin is unattainable. Everything has volume."}, {"title": "Charles Law .txt", "text": "We have zero Kelvin. But remember, zero Kelvin is unattainable. Everything has volume. And that means our temperature must be somewhere above zero. Also notice that v one over t one means that our t one can't be zero. Otherwise, we get an undefined number, a number that has infinity as its answer."}, {"title": "Charles Law .txt", "text": "And that means our temperature must be somewhere above zero. Also notice that v one over t one means that our t one can't be zero. Otherwise, we get an undefined number, a number that has infinity as its answer. Because any number over zero is infinity. So let's look at this law. Why is it that we have a linear function?"}, {"title": "Charles Law .txt", "text": "Because any number over zero is infinity. So let's look at this law. Why is it that we have a linear function? Linear line? Well, that's because this is our slope. A constant slope."}, {"title": "Charles Law .txt", "text": "Linear line? Well, that's because this is our slope. A constant slope. So this guy's constant. Whenever D increases, team must increase by the same amount. If this is doubled, this is doubled."}, {"title": "Sp Hybridization.txt", "text": "So let's begin with the following depiction. In our first case, let's suppose we have two different atoms. The first atom donates an S orbital, and the second atom donates a p orbital. So these two orbitals will combine in a way to form the following molecular orbitals. So, because we input two, we should get back to orbitals. And that's exactly what we get."}, {"title": "Sp Hybridization.txt", "text": "So these two orbitals will combine in a way to form the following molecular orbitals. So, because we input two, we should get back to orbitals. And that's exactly what we get. The first molecular orbital is known as the bonding molecular orbital. It's lower in energy. And the second one is known as the anti bonding molecular orbital."}, {"title": "Sp Hybridization.txt", "text": "The first molecular orbital is known as the bonding molecular orbital. It's lower in energy. And the second one is known as the anti bonding molecular orbital. It's the one higher in energy. So that's the first case. That's the normal case that we're used to seeing."}, {"title": "Sp Hybridization.txt", "text": "It's the one higher in energy. So that's the first case. That's the normal case that we're used to seeing. So let's suppose we try a different thing. Now, let's suppose we have a single atom. And that single atom has both an S orbital as well as a p orbital."}, {"title": "Sp Hybridization.txt", "text": "So let's suppose we try a different thing. Now, let's suppose we have a single atom. And that single atom has both an S orbital as well as a p orbital. What happens is, within that single atom, these two orbitals can combine in such a way to produce something that we know as hybridized orbitals. In other words, we have a single atom. Within that single atom, an S orbital interacts with a p orbital to produce two hybridized orbitals."}, {"title": "Sp Hybridization.txt", "text": "What happens is, within that single atom, these two orbitals can combine in such a way to produce something that we know as hybridized orbitals. In other words, we have a single atom. Within that single atom, an S orbital interacts with a p orbital to produce two hybridized orbitals. Now, once again, we input two orbitals. So we should get back to hybridized orbitals. And that's exactly what we see happen here."}, {"title": "Sp Hybridization.txt", "text": "Now, once again, we input two orbitals. So we should get back to hybridized orbitals. And that's exactly what we see happen here. Now, when this S combines with this positive P, we get the following hybridized orbitals. In other words, this positive region simply combines with this positive region, and this becomes smaller. So a positive S orbital combines with a positive p orbital."}, {"title": "Sp Hybridization.txt", "text": "Now, when this S combines with this positive P, we get the following hybridized orbitals. In other words, this positive region simply combines with this positive region, and this becomes smaller. So a positive S orbital combines with a positive p orbital. The two greens combine, the blue becomes smaller to form an enlarged positive green lobe and a smaller or thinner negative blue lobe. And the same happens when this part is negative. We get the following because two negative lobes combine to form this enlarged negative section, enlarged negative lobe, and the smaller positive green lobe."}, {"title": "Sp Hybridization.txt", "text": "The two greens combine, the blue becomes smaller to form an enlarged positive green lobe and a smaller or thinner negative blue lobe. And the same happens when this part is negative. We get the following because two negative lobes combine to form this enlarged negative section, enlarged negative lobe, and the smaller positive green lobe. Now, in this lecture, we're going to only talk about SP hypersized orbitals. In future lectures, we're also going to talk about SP two and SP three hyperze orbitals. So, what is an SP hyperized orbital?"}, {"title": "Sp Hybridization.txt", "text": "Now, in this lecture, we're going to only talk about SP hypersized orbitals. In future lectures, we're also going to talk about SP two and SP three hyperze orbitals. So, what is an SP hyperized orbital? Well, this is simply an orbital produced using 50% S orbitals and 50% P orbitals. In other words, when we're combining our orbitals within that given atom, 50% comes from S and 50% comes from P. And this is known as an SP hyper dice orbital. That's exactly what we have in this situation here."}, {"title": "Sp Hybridization.txt", "text": "Well, this is simply an orbital produced using 50% S orbitals and 50% P orbitals. In other words, when we're combining our orbitals within that given atom, 50% comes from S and 50% comes from P. And this is known as an SP hyper dice orbital. That's exactly what we have in this situation here. So let's look at an example in nature. So where is this evidence? So let's look at one particular example in which a Beryllium atom combines with two H atoms."}, {"title": "Sp Hybridization.txt", "text": "So let's look at an example in nature. So where is this evidence? So let's look at one particular example in which a Beryllium atom combines with two H atoms. So let's examine the electron configuration of Beryllium. So, Beryllium, in its neutral state, has four electrons, four protons, four neutrons. So the electron configuration goes like this."}, {"title": "Sp Hybridization.txt", "text": "So let's examine the electron configuration of Beryllium. So, Beryllium, in its neutral state, has four electrons, four protons, four neutrons. So the electron configuration goes like this. We have two electrons that go into our one S, and we have two electrons that go into the two S. Now, we also have the two p orbitals. But since we have no more electrons, there are zero electrons in the two P orbital. So we can either represent it this way, or we can simply remove the two P. Now, for my purposes, I'm going to leave it in this way, and we'll see why."}, {"title": "Sp Hybridization.txt", "text": "We have two electrons that go into our one S, and we have two electrons that go into the two S. Now, we also have the two p orbitals. But since we have no more electrons, there are zero electrons in the two P orbital. So we can either represent it this way, or we can simply remove the two P. Now, for my purposes, I'm going to leave it in this way, and we'll see why. So my question is the following will this Be donate a two P orbital to bind with the H, or will it donate a hybrid orbital? In other words, which situation is more stable? So let's examine it this way."}, {"title": "Sp Hybridization.txt", "text": "So my question is the following will this Be donate a two P orbital to bind with the H, or will it donate a hybrid orbital? In other words, which situation is more stable? So let's examine it this way. Let's draw out our pictures. Let's suppose that Be forms this hybridized orbital, and then this hybridized orbital interacts with the H atom to form our covalent bond. And let's also suppose that we have a Beryllium atom that donates a simple two P orbital to interact with that H atom."}, {"title": "Sp Hybridization.txt", "text": "Let's draw out our pictures. Let's suppose that Be forms this hybridized orbital, and then this hybridized orbital interacts with the H atom to form our covalent bond. And let's also suppose that we have a Beryllium atom that donates a simple two P orbital to interact with that H atom. Let's see which one is more stable. Well, recall that whenever bonds are formed, bonds or covalent bonds are formed by the overlap of atomic orbitals, as we see here. And we know that the better the overlap, the larger the lobes, the more stable our compound is, the more stable our bond is."}, {"title": "Sp Hybridization.txt", "text": "Let's see which one is more stable. Well, recall that whenever bonds are formed, bonds or covalent bonds are formed by the overlap of atomic orbitals, as we see here. And we know that the better the overlap, the larger the lobes, the more stable our compound is, the more stable our bond is. So in which situation do we have a more stabilized overlap? A larger overlap? Well, clearly this case has a much bigger lobe."}, {"title": "Sp Hybridization.txt", "text": "So in which situation do we have a more stabilized overlap? A larger overlap? Well, clearly this case has a much bigger lobe. And that means the interaction will be much better in this hybridized interaction. In other words, this hybridized lobe creates a larger lobe. And that means, because we have a larger lobe, we have a better overlap."}, {"title": "Sp Hybridization.txt", "text": "And that means the interaction will be much better in this hybridized interaction. In other words, this hybridized lobe creates a larger lobe. And that means, because we have a larger lobe, we have a better overlap. And so that means this is much more stable. And so this will not occur. We're going to have this type of bond."}, {"title": "Sp Hybridization.txt", "text": "And so that means this is much more stable. And so this will not occur. We're going to have this type of bond. In other words, within this benh, when Be bonds to H, it creates a hybridized orbital, which then bonds to the one S of the H. And let's see exactly that in this energy diagram. So we can imagine this being the energy diagram. So the higher up we go, the more energy we have."}, {"title": "Sp Hybridization.txt", "text": "In other words, within this benh, when Be bonds to H, it creates a hybridized orbital, which then bonds to the one S of the H. And let's see exactly that in this energy diagram. So we can imagine this being the energy diagram. So the higher up we go, the more energy we have. The lower we go, the less energy we have. What happens is the following. The Beryllium creates this hybridized orbital SD hybridized which comes from one S and one P an St hybridized orbital."}, {"title": "Sp Hybridization.txt", "text": "The lower we go, the less energy we have. What happens is the following. The Beryllium creates this hybridized orbital SD hybridized which comes from one S and one P an St hybridized orbital. And that orbital, which is a bit slightly higher in energy than our one S of the H atom. So this is the H atom, and this is the one S orbital. And this is the SP hybridized orbital of our Beryllium."}, {"title": "Sp Hybridization.txt", "text": "And that orbital, which is a bit slightly higher in energy than our one S of the H atom. So this is the H atom, and this is the one S orbital. And this is the SP hybridized orbital of our Beryllium. They will interact. So we input two atomic orbitals, and we get two molecular orbitals. So we get the one lower in energy and the one higher in energy."}, {"title": "Sp Hybridization.txt", "text": "They will interact. So we input two atomic orbitals, and we get two molecular orbitals. So we get the one lower in energy and the one higher in energy. So, once again, let's recap. So hybridization is simply a process that occurs within an atom. Within an atom, the orbitals can interact in a way to produce these hybridized orbitals that contain larger sections and smaller sections."}, {"title": "Mass percent example.txt", "text": "Mass percentage is simply another way of finding the concentration of the solution. Mass percentage is equal to mass of some compound x divided by the total mass of the solution times 100. The 100 gives you the percentage. This is a fraction, so you divide mass by mass, so the units cancel out. So mass percent is unitless. Now let's do a problem with mass percentage."}, {"title": "Mass percent example.txt", "text": "This is a fraction, so you divide mass by mass, so the units cancel out. So mass percent is unitless. Now let's do a problem with mass percentage. The question tells us that we have 49 grams of gold, 25 grams of carbon, .5 water. We need to find the mass percent of carbon, water and gold in our solution. The first step is to find the mass percentage of gold."}, {"title": "Mass percent example.txt", "text": "The question tells us that we have 49 grams of gold, 25 grams of carbon, .5 water. We need to find the mass percent of carbon, water and gold in our solution. The first step is to find the mass percentage of gold. To find the mass percentage of gold, we simply use the formula. So 49 grams of gold divided by the total mass of the solution, 49 grams plus 25 grams plus now we can't add kilograms to grams. So the first step is to convert this to grams."}, {"title": "Mass percent example.txt", "text": "To find the mass percentage of gold, we simply use the formula. So 49 grams of gold divided by the total mass of the solution, 49 grams plus 25 grams plus now we can't add kilograms to grams. So the first step is to convert this to grams. We get plus 500 grams. Now we multiply the whole thing by 100 to find the percentage, and we get 8.5%. The mass perceptive of gold is 8.5."}, {"title": "Mass percent example.txt", "text": "We get plus 500 grams. Now we multiply the whole thing by 100 to find the percentage, and we get 8.5%. The mass perceptive of gold is 8.5. To find the mass perceptive carbon, we follow the same exact formula. 25 grams of carbon divided by the total grams of the solution multiplied by 100 gives us 4.4%. So the max percent of carbon is 4.4."}, {"title": "Mass percent example.txt", "text": "To find the mass perceptive carbon, we follow the same exact formula. 25 grams of carbon divided by the total grams of the solution multiplied by 100 gives us 4.4%. So the max percent of carbon is 4.4. The last step could be done in two ways. One way, you simply use the formula. You plug things in, you find the result."}, {"title": "Mass percent example.txt", "text": "The last step could be done in two ways. One way, you simply use the formula. You plug things in, you find the result. A quicker way would be simply to realize that if you add these guys up and subtracted from 100, we get the mass percent of the final thing final compound within our solution. Namely water. So 100 -8.5 plus 4.5 gives you 87."}, {"title": "Second Law of Thermodynamics .txt", "text": "From the side ramp here of the heat engine, we can see that that's the case. The energy that comes from the hot body, some of that energy goes into doing work, expanding the piston increase, increasing the volume. And some of that goes into the cold body, decreasing the temperature as the piston moves back into its original position, thereby keeping the temperature constant. Okay, we know by conservation of energy that the input energy equals the output energy. That means QH, which means the energy input is equal to QC, the energy transferred into the cold body plus the work done by the system or by the molecules within the system. And this directly correlates the first law of thermodynamics."}, {"title": "Second Law of Thermodynamics .txt", "text": "Okay, we know by conservation of energy that the input energy equals the output energy. That means QH, which means the energy input is equal to QC, the energy transferred into the cold body plus the work done by the system or by the molecules within the system. And this directly correlates the first law of thermodynamics. And in fact, it's the same thing. It's basically this. Okay, so we basically are saying that engines, heat engines aren't completely 100% efficient in converting heat into work."}, {"title": "Second Law of Thermodynamics .txt", "text": "And in fact, it's the same thing. It's basically this. Okay, so we basically are saying that engines, heat engines aren't completely 100% efficient in converting heat into work. So how efficient are they? Well, this formula here where E stands for efficiency or engine efficiency, can basically tell you how efficient an engine is. If you know the temperature of the cold body and the temperature of the hot body, you can find the efficiency."}, {"title": "Second Law of Thermodynamics .txt", "text": "So how efficient are they? Well, this formula here where E stands for efficiency or engine efficiency, can basically tell you how efficient an engine is. If you know the temperature of the cold body and the temperature of the hot body, you can find the efficiency. And this also shows you can see from algebra and basic calculus that ends, this becomes zero or tenths of zero. That is, as TC decreases and Th increases or the difference between these two guys increases, the efficiency also increases. Okay?"}, {"title": "Second Law of Thermodynamics .txt", "text": "And this also shows you can see from algebra and basic calculus that ends, this becomes zero or tenths of zero. That is, as TC decreases and Th increases or the difference between these two guys increases, the efficiency also increases. Okay? You can pluck some values in and you'll see that as this becomes smaller and this becomes larger, that E becomes more efficient. Okay, finally, let's talk about refrigerators and air conditioners. So refrigerators and air conditioners are basically reverse heat engines."}, {"title": "Second Law of Thermodynamics .txt", "text": "You can pluck some values in and you'll see that as this becomes smaller and this becomes larger, that E becomes more efficient. Okay, finally, let's talk about refrigerators and air conditioners. So refrigerators and air conditioners are basically reverse heat engines. What actually happens is work is inputted so that heat can be transferred from a cold body to a hot body or energy can be transferred from a cold body to a hot body. This decreases the temperature of the system but increases the temperature of the outside. For example, in this room in the summer, if I have an air conditioner and I plug it into the outlet, the energy that goes into the air conditioner basically does work on the inside room."}, {"title": "Homo-Lumo interactions.txt", "text": "In this lecture, I'd like to examine the homoluma interaction between compounds. So let's look at the following example. Let's suppose we have compound one and alkane reacting with compound two, our hydrochloric acid. So this is a simple additional reaction. So in this reaction, this alkin actively lowers based base. This acts as a lewis acid."}, {"title": "Homo-Lumo interactions.txt", "text": "So this is a simple additional reaction. So in this reaction, this alkin actively lowers based base. This acts as a lewis acid. So this donates a pair of electrons. This accepts a pair of electrons. So our intermediate reactants are the intermediate carbocation that has a positive charge on this carbon and has an extra h that it got from the hydrochloric acid."}, {"title": "Homo-Lumo interactions.txt", "text": "So this donates a pair of electrons. This accepts a pair of electrons. So our intermediate reactants are the intermediate carbocation that has a positive charge on this carbon and has an extra h that it got from the hydrochloric acid. Now, this chlorine, or chloride atom now has an extra pair of non bonding electrons, and so it develops a negative charge. In the second step of this addiction reaction, we have the chloride ion donating a pair of non bombing electrons. So this is our lewis base and our lewis acid."}, {"title": "Homo-Lumo interactions.txt", "text": "Now, this chlorine, or chloride atom now has an extra pair of non bonding electrons, and so it develops a negative charge. In the second step of this addiction reaction, we have the chloride ion donating a pair of non bombing electrons. So this is our lewis base and our lewis acid. And so we form the following final product. So, let's examine this picture more closely using molecular and atomic orbitals. So let's draw our molecular orbitals or atomic orbitals for this reaction."}, {"title": "Homo-Lumo interactions.txt", "text": "And so we form the following final product. So, let's examine this picture more closely using molecular and atomic orbitals. So let's draw our molecular orbitals or atomic orbitals for this reaction. So, here's our alkane. So our sigma bond and our pi bond, creating the double bond. What happens is this pair of electrons."}, {"title": "Homo-Lumo interactions.txt", "text": "So, here's our alkane. So our sigma bond and our pi bond, creating the double bond. What happens is this pair of electrons. So, this bond is composed of a pair of electrons, one electron in this two p orbital and the second electron in this two p orbital. So these two electrons attack this h atom, taking that h atom away from this chlorine atom. And we develop the following diagram."}, {"title": "Homo-Lumo interactions.txt", "text": "So, this bond is composed of a pair of electrons, one electron in this two p orbital and the second electron in this two p orbital. So these two electrons attack this h atom, taking that h atom away from this chlorine atom. And we develop the following diagram. So, this is our intermediate cargo cation. So, this bond has been formed, this cobalt sigma ch bond. And now we have a positive charge on this twopie orbital because we have 1233 electrons."}, {"title": "Homo-Lumo interactions.txt", "text": "So, this is our intermediate cargo cation. So, this bond has been formed, this cobalt sigma ch bond. And now we have a positive charge on this twopie orbital because we have 1233 electrons. And that means we have a positive charge on the two p orbital. So, once again, this is SP two hybridized, and this is a planar molecule. So what happens next?"}, {"title": "Homo-Lumo interactions.txt", "text": "And that means we have a positive charge on the two p orbital. So, once again, this is SP two hybridized, and this is a planar molecule. So what happens next? Well, next we have this lewis base. We have our chloride atom. And this non bonding pair of electrons attacks or attaches overlaps with this two p orbital, forming our spinal product."}, {"title": "Homo-Lumo interactions.txt", "text": "Well, next we have this lewis base. We have our chloride atom. And this non bonding pair of electrons attacks or attaches overlaps with this two p orbital, forming our spinal product. So, in the first step, this pi bond acted as a lewis base, donating this pair of electrons and this h atom on this compound on the hydrochloric acid active as a lewis acid, donating that h, donating that empty one s orbital. And likewise, here, this is the lewis acid because it has an empty two p orbital. And this is the lewis base because it has a pair of non bonding electrons."}, {"title": "Homo-Lumo interactions.txt", "text": "So, in the first step, this pi bond acted as a lewis base, donating this pair of electrons and this h atom on this compound on the hydrochloric acid active as a lewis acid, donating that h, donating that empty one s orbital. And likewise, here, this is the lewis acid because it has an empty two p orbital. And this is the lewis base because it has a pair of non bonding electrons. So, what exactly is a lewis athens based reaction? So, a lewis athens based reaction is the interaction between a filled molecular orbital, as we saw here, and an antimolecular orbital. And this is known as a homolumo interaction."}, {"title": "Homo-Lumo interactions.txt", "text": "So, what exactly is a lewis athens based reaction? So, a lewis athens based reaction is the interaction between a filled molecular orbital, as we saw here, and an antimolecular orbital. And this is known as a homolumo interaction. Homo simply meaning highest occupied molecular orbital, and lumo, meaning lowest unoccupied molecular orbital. So if we go back to the first step in this additional reaction, we see that our homo, the highest occupied molecular orbital, is the pi bond. So this is our lowest base."}, {"title": "Homo-Lumo interactions.txt", "text": "Homo simply meaning highest occupied molecular orbital, and lumo, meaning lowest unoccupied molecular orbital. So if we go back to the first step in this additional reaction, we see that our homo, the highest occupied molecular orbital, is the pi bond. So this is our lowest base. And in this case, our lowest unoccupied molecular orbital is the antibonding Sigma Bond. Remember, the bonding Sigma Bond is completely filled. The antibonding has no electrons, and so that means it must be the lowest unoccupied molecular orbital."}, {"title": "Homo-Lumo interactions.txt", "text": "And in this case, our lowest unoccupied molecular orbital is the antibonding Sigma Bond. Remember, the bonding Sigma Bond is completely filled. The antibonding has no electrons, and so that means it must be the lowest unoccupied molecular orbital. Likewise, in this step, this was our lumo lowest unoccupied molecular orbital. And this was our homo, the Lewis base. So let's look at the second more closely in terms of energy."}, {"title": "Homo-Lumo interactions.txt", "text": "Likewise, in this step, this was our lumo lowest unoccupied molecular orbital. And this was our homo, the Lewis base. So let's look at the second more closely in terms of energy. So, our homo is the lowest base. It's the highest occupied molecular orbital that has the non bonding pair of electrons. And this is our lumo lowest unoccupied molecular orbital."}, {"title": "Homo-Lumo interactions.txt", "text": "So, our homo is the lowest base. It's the highest occupied molecular orbital that has the non bonding pair of electrons. And this is our lumo lowest unoccupied molecular orbital. It's the two p orbital, and it's a bit higher in energy than our homo. So they interact. They overlap to form our molecular Sigma Bond."}, {"title": "Homo-Lumo interactions.txt", "text": "It's the two p orbital, and it's a bit higher in energy than our homo. So they interact. They overlap to form our molecular Sigma Bond. And the two electrons go into this bonding molecular orbital. And no electrons go into the antibinding molecular orbital because it's higher in energy. So, once again, as an overview, a Lewis acidbased reaction is simply a reaction between a filled orbital of one compound and an empty orbital of second compound."}, {"title": "Electrolytic cells .txt", "text": "Now, electrolytic cells are electrochemical cells that are supplied with an outside source of electrons, which allows reactant favorite redox reactions to occur. Now, recall that voltaic cells convert chemical energy into electrical work via the process of moving electrons in a spontaneous product favorite reaction. So, unlike voltaic cells, electrolytic cells do the opposite. They use up electrical work to power reactant favorite non spontaneous reactions. Now, let's look at an example. Let's look at the decomposition of molten sodium chloride."}, {"title": "Electrolytic cells .txt", "text": "They use up electrical work to power reactant favorite non spontaneous reactions. Now, let's look at an example. Let's look at the decomposition of molten sodium chloride. Now, what molten means is that it's heated to a certain temperature so that it goes from a solid state to a liquid state. This is not an aqueous state. It's a liquid state of sodium chloride, meaning these guys dissociate, but there is no water in our mixture."}, {"title": "Electrolytic cells .txt", "text": "Now, what molten means is that it's heated to a certain temperature so that it goes from a solid state to a liquid state. This is not an aqueous state. It's a liquid state of sodium chloride, meaning these guys dissociate, but there is no water in our mixture. There's no solvent. So let's look at our electrolytic electrochemical cell. So it's composed of not two half cells, but one half cell."}, {"title": "Electrolytic cells .txt", "text": "There's no solvent. So let's look at our electrolytic electrochemical cell. So it's composed of not two half cells, but one half cell. So one beaker. Now, within this beaker, we have melted or liquid sodium chloride. So we have a bunch of sodium molecules or sodium ions chloride ions."}, {"title": "Electrolytic cells .txt", "text": "So one beaker. Now, within this beaker, we have melted or liquid sodium chloride. So we have a bunch of sodium molecules or sodium ions chloride ions. So these two electrodes are in there, so they're made from the same exact material. And what happens is we connect these guards to an outside power source, like a battery or a voltage cell. Now, what happens is this battery powers."}, {"title": "Electrolytic cells .txt", "text": "So these two electrodes are in there, so they're made from the same exact material. And what happens is we connect these guards to an outside power source, like a battery or a voltage cell. Now, what happens is this battery powers. It allows electrons to transfer in this direction. So if they transfer this way, that means this metal obtains these electrons. So this metal or electrode forms a negative charge, while this electrode forms a positive charge, because electrons will be taken away from this electrode."}, {"title": "Electrolytic cells .txt", "text": "It allows electrons to transfer in this direction. So if they transfer this way, that means this metal obtains these electrons. So this metal or electrode forms a negative charge, while this electrode forms a positive charge, because electrons will be taken away from this electrode. So, since this develops a negative charge, let's see what happens with the portion that's immersed into our liquid. Well, we said some of the sodium molecules will be moving around, and since they are positively charged, they will be attracted to this negatively charged electrode. Likewise, these chloride atoms are negatively charged, so they will be attracted to this positively charged electrode."}, {"title": "Electrolytic cells .txt", "text": "So, since this develops a negative charge, let's see what happens with the portion that's immersed into our liquid. Well, we said some of the sodium molecules will be moving around, and since they are positively charged, they will be attracted to this negatively charged electrode. Likewise, these chloride atoms are negatively charged, so they will be attracted to this positively charged electrode. So we'll have a separation of sodium and chloride in our liquid. Now, what happens when our sodium positively charged ion hits this negatively charged electron? Well, some of the electrons will transfer into our sodium molecule, and that means our sodium will be reduced."}, {"title": "Electrolytic cells .txt", "text": "So we'll have a separation of sodium and chloride in our liquid. Now, what happens when our sodium positively charged ion hits this negatively charged electron? Well, some of the electrons will transfer into our sodium molecule, and that means our sodium will be reduced. So this section is where reduction occurs. And that means by definition, it must be our cathode. Now, likewise, when these molecules or ions hit this little electrode, they give off some of these electrons, because electrons want to move from a negative charge to a positive charge."}, {"title": "Electrolytic cells .txt", "text": "So this section is where reduction occurs. And that means by definition, it must be our cathode. Now, likewise, when these molecules or ions hit this little electrode, they give off some of these electrons, because electrons want to move from a negative charge to a positive charge. So when this hits it, electrons travel inside this electrode, and they enter our circuit and travel all the way down here. So what happens when electrons leave? Well, this guy is oxidized into diatomic gas, and so it evaporates into our environment."}, {"title": "Electrolytic cells .txt", "text": "So when this hits it, electrons travel inside this electrode, and they enter our circuit and travel all the way down here. So what happens when electrons leave? Well, this guy is oxidized into diatomic gas, and so it evaporates into our environment. And this is where oxidation takes place. And so, by definition, this guy is our anode. So notice two important differences between voltaic cells and electrolytic cells."}, {"title": "Electrolytic cells .txt", "text": "And this is where oxidation takes place. And so, by definition, this guy is our anode. So notice two important differences between voltaic cells and electrolytic cells. Our cathode in this situation is negative, and our anode is positive. But in voltaic cells, it's reversed. Our cathode is positive and our Amote is negative."}, {"title": "Electrolytic cells .txt", "text": "Our cathode in this situation is negative, and our anode is positive. But in voltaic cells, it's reversed. Our cathode is positive and our Amote is negative. And that's because this electron doesn't travel this way like it does in voltaic cells, but it travels this way due to this outside battery source. Another important difference, obviously, is the fact that in electrolytic cells, we have outside battery source. We have an outside power source."}, {"title": "Electrolytic cells .txt", "text": "And that's because this electron doesn't travel this way like it does in voltaic cells, but it travels this way due to this outside battery source. Another important difference, obviously, is the fact that in electrolytic cells, we have outside battery source. We have an outside power source. But in this will take cells. We don't have it. Now, so let's look at the oxidation reaction that occurs in our anode."}, {"title": "Electrolytic cells .txt", "text": "But in this will take cells. We don't have it. Now, so let's look at the oxidation reaction that occurs in our anode. So two of these molecules, two of these ions give off those two electrons, forming our diatomic gas molecule, and the diatomic gas molecule evaporates into our environment. Now, let's look at a reduction reaction. This reduction reaction occurs in the following manner."}, {"title": "Electrolytic cells .txt", "text": "So two of these molecules, two of these ions give off those two electrons, forming our diatomic gas molecule, and the diatomic gas molecule evaporates into our environment. Now, let's look at a reduction reaction. This reduction reaction occurs in the following manner. Two sodium ions react with two electrons. When they hit this metal, they take up those two electrons, forming two sodium solid molecules or two moles of sodium solid molecules. Now, our net reaction is just an addition of this guy to this guy."}, {"title": "Electrolytic cells .txt", "text": "Two sodium ions react with two electrons. When they hit this metal, they take up those two electrons, forming two sodium solid molecules or two moles of sodium solid molecules. Now, our net reaction is just an addition of this guy to this guy. Notice that electrons cancel, and we simply get the following net reduction reaction. Now, if we were to look up the electron potentials or the cell potentials for this reaction and this reaction, we would get the following voltages. Now, to find the net or the final cell voltage, we simply add these guys up, and we get negative 4.0\n72 volts."}, {"title": "Electrolytic cells .txt", "text": "Notice that electrons cancel, and we simply get the following net reduction reaction. Now, if we were to look up the electron potentials or the cell potentials for this reaction and this reaction, we would get the following voltages. Now, to find the net or the final cell voltage, we simply add these guys up, and we get negative 4.0\n72 volts. So that means that this much voltage must be supplied to our electrolytic cell by this battery to power this reaction. So decomposition of this guy required energy. Now, other decomposition reactions are very popular."}, {"title": "Neuron Cells Part I .txt", "text": "Well, there are many different examples of concentration cells. Today we're going to look at a very important biological example of a concentration cell called a neuron cell. Now, neuron cells are simply specialized concentration cells found within our nervous system, within our body that communicate with one another via changes in ion concentration. Now, these changes in ion concentration create a difference in voltage or something called a cell voltage. And this difference in voltage creates electrical signals. Now, these electrical signals travel from one cell to another, and this is how cells communicate."}, {"title": "Neuron Cells Part I .txt", "text": "Now, these changes in ion concentration create a difference in voltage or something called a cell voltage. And this difference in voltage creates electrical signals. Now, these electrical signals travel from one cell to another, and this is how cells communicate. Now let's look at something called a resting electrical potential resell. Now, our cells within our body, specifically neuron cells, establish electrical potentials or cell voltages at rest. And what this guy simply means, it's the cell voltage produced by our cell when no signals are being transducted or conducted from one cell to another."}, {"title": "Neuron Cells Part I .txt", "text": "Now let's look at something called a resting electrical potential resell. Now, our cells within our body, specifically neuron cells, establish electrical potentials or cell voltages at rest. And what this guy simply means, it's the cell voltage produced by our cell when no signals are being transducted or conducted from one cell to another. Now let's look at a portion of our cell membrane found at the exxon hillock. The exxon hillock is simply the portion of the neuron where our signal is generated. But remember, we're talking about the resting potential."}, {"title": "Neuron Cells Part I .txt", "text": "Now let's look at a portion of our cell membrane found at the exxon hillock. The exxon hillock is simply the portion of the neuron where our signal is generated. But remember, we're talking about the resting potential. That means no signals are being generated just yet. So let's examine the different types of ions that are present within our body, within our cells. So we see that we have calcium, we have potassium, we have sodium, and we have chloride."}, {"title": "Neuron Cells Part I .txt", "text": "That means no signals are being generated just yet. So let's examine the different types of ions that are present within our body, within our cells. So we see that we have calcium, we have potassium, we have sodium, and we have chloride. Now, when we're at our resting potential, we have a lower concentration of calcium, sodium chloride inside the cell. This is the inside than the outside. On the contrary, we have a higher concentration of potassium ions on the inside than the outside."}, {"title": "Neuron Cells Part I .txt", "text": "Now, when we're at our resting potential, we have a lower concentration of calcium, sodium chloride inside the cell. This is the inside than the outside. On the contrary, we have a higher concentration of potassium ions on the inside than the outside. Now notice we have a semipermandal membrane. So we have hydrophilic heads and hydrophobic tails. These guys are transport membranes or transport proteins."}, {"title": "Neuron Cells Part I .txt", "text": "Now notice we have a semipermandal membrane. So we have hydrophilic heads and hydrophobic tails. These guys are transport membranes or transport proteins. And these proteins allow ions to flow in or as a cell, the active transport or passive transport. Now, so today we're only going to look at this guy here, potassium ion. But notice that our resting electrical potential, the cell or our cell voltage, is generated by simply adding up the cell voltages of all these four ions."}, {"title": "Neuron Cells Part I .txt", "text": "And these proteins allow ions to flow in or as a cell, the active transport or passive transport. Now, so today we're only going to look at this guy here, potassium ion. But notice that our resting electrical potential, the cell or our cell voltage, is generated by simply adding up the cell voltages of all these four ions. When we add all these guys up, we get our final cell voltage or the resting electrical potential. Now, today, to save time, I'm only going to show you for this potassium ion. You can do these on your own."}, {"title": "Neuron Cells Part I .txt", "text": "When we add all these guys up, we get our final cell voltage or the resting electrical potential. Now, today, to save time, I'm only going to show you for this potassium ion. You can do these on your own. So let's take this potassium ion and let's create a concentration cell or a specialized concentration cell called a neuron cell. So this is our electrochemical concentration cell for potassium. This is our negatively charged anode and our positively charged cathode."}, {"title": "Neuron Cells Part I .txt", "text": "So let's take this potassium ion and let's create a concentration cell or a specialized concentration cell called a neuron cell. So this is our electrochemical concentration cell for potassium. This is our negatively charged anode and our positively charged cathode. Now, this is where oxidation of potassium takes place. And this is where reduction of potassium takes place. You could think of this conductor, that electrodes and a sulfbridge, as representing the cell membrane."}, {"title": "Neuron Cells Part I .txt", "text": "Now, this is where oxidation of potassium takes place. And this is where reduction of potassium takes place. You could think of this conductor, that electrodes and a sulfbridge, as representing the cell membrane. And the solution on this in this anode is the outside solution and the cathode is the inside solution. And that's because initially in a concentration cell, this guy is more dilute. That means it's the outside, because remember, we have less potassium on the outside than on the inside."}, {"title": "Neuron Cells Part I .txt", "text": "And the solution on this in this anode is the outside solution and the cathode is the inside solution. And that's because initially in a concentration cell, this guy is more dilute. That means it's the outside, because remember, we have less potassium on the outside than on the inside. So this guy must be the inside. So now let's see what happens. Well, electrons leave this potassium ion or leave the potassium solid ion this electrode and travel via the conductor, via the cell membrane onto this electrode."}, {"title": "Neuron Cells Part I .txt", "text": "So this guy must be the inside. So now let's see what happens. Well, electrons leave this potassium ion or leave the potassium solid ion this electrode and travel via the conductor, via the cell membrane onto this electrode. At the same time, they release these potassium ions into our solution. So the concentration of this guy on the outside becomes greater. Likewise, these guys on the inside here are taken up because they react with the electrons to form our K solid."}, {"title": "Neuron Cells Part I .txt", "text": "At the same time, they release these potassium ions into our solution. So the concentration of this guy on the outside becomes greater. Likewise, these guys on the inside here are taken up because they react with the electrons to form our K solid. And this changes the concentration of our inside and outside. So this is our oxidation reaction, where our potassium is oxidized, and our reduction reaction, where the potassium accepts the electrons form of the solid. Now, if we if we want to find the net rebus reaction, we simply add these two guys up the east cancel, the K solids cancel, and we are left with K plus inside and K plus outside."}, {"title": "Neuron Cells Part I .txt", "text": "And this changes the concentration of our inside and outside. So this is our oxidation reaction, where our potassium is oxidized, and our reduction reaction, where the potassium accepts the electrons form of the solid. Now, if we if we want to find the net rebus reaction, we simply add these two guys up the east cancel, the K solids cancel, and we are left with K plus inside and K plus outside. Notice the reactants is the inside. This is where we begin, and this guy is the outside. This is where we end because electrons travel this way, but the ions want to travel this way because we have more ions on the outside on the inside than the outside."}, {"title": "Neuron Cells Part I .txt", "text": "Notice the reactants is the inside. This is where we begin, and this guy is the outside. This is where we end because electrons travel this way, but the ions want to travel this way because we have more ions on the outside on the inside than the outside. So, once again, in the electrochemical cell setup, electrons travel this way, but C plus atoms travel this way because this concentration increases while this guy decreases. Now, if we look up the cell voltage of this oxidation reaction and this reduction reaction, we find that they're the same magnitude but different signs. So if we add them up, that means they will be zero."}, {"title": "Calculating the equivalence point .txt", "text": "Now, if you don't know what the equivalence point is, check out the link below. So in the beginning, we have a buffer solution of some known asset. So if we know the asset, that means we know the asset amortization constant. We can simply look that up. So in my first step, I basically want to find a KB. And I want to find a KB using this formula here."}, {"title": "Calculating the equivalence point .txt", "text": "We can simply look that up. So in my first step, I basically want to find a KB. And I want to find a KB using this formula here. Now, if you don't know what this formula is or where it comes from, check out the link above and I'll tell you why in a second. We want to find the KB. Well, kw, something we know at some given temperature at a 25 degree celsius, kw is tens of negative 14."}, {"title": "Calculating the equivalence point .txt", "text": "Now, if you don't know what this formula is or where it comes from, check out the link above and I'll tell you why in a second. We want to find the KB. Well, kw, something we know at some given temperature at a 25 degree celsius, kw is tens of negative 14. It's the ionization constant of water. Now, this guy equals ka, something we know times KB. So we find KB by simply dividing kw by ka."}, {"title": "Calculating the equivalence point .txt", "text": "It's the ionization constant of water. Now, this guy equals ka, something we know times KB. So we find KB by simply dividing kw by ka. Now, why do we need the KB? Well, remember what the equivalent point is. It's the point at which all the asset has been neutralized by some base, right?"}, {"title": "Calculating the equivalence point .txt", "text": "Now, why do we need the KB? Well, remember what the equivalent point is. It's the point at which all the asset has been neutralized by some base, right? So I can use the KB to find the amount of base needed to neutralize our acid completely. And then if I know my concentration of base, I could find the poh. And using the poh, I can find the PH."}, {"title": "Calculating the equivalence point .txt", "text": "So I can use the KB to find the amount of base needed to neutralize our acid completely. And then if I know my concentration of base, I could find the poh. And using the poh, I can find the PH. And that's exactly what we do. So in my second step, I basically use the KB or the base annette and constant. I equate that to my equilibrium expression, which states that the concentration of hydroxide, what I'm looking for, equals the concentration of the conjugate acid over the concentration of the conjugate base."}, {"title": "Calculating the equivalence point .txt", "text": "And that's exactly what we do. So in my second step, I basically use the KB or the base annette and constant. I equate that to my equilibrium expression, which states that the concentration of hydroxide, what I'm looking for, equals the concentration of the conjugate acid over the concentration of the conjugate base. Now, I could get this guy in this side and divide by this guy and get the concentration that I'm looking at equals a known constant, unknown amount, and a known amount. Now I solve and I find my concentration. Next, I find my poh by using the formula which is negative lot of the concentration found here in step three."}, {"title": "Calculating the equivalence point .txt", "text": "Now, I could get this guy in this side and divide by this guy and get the concentration that I'm looking at equals a known constant, unknown amount, and a known amount. Now I solve and I find my concentration. Next, I find my poh by using the formula which is negative lot of the concentration found here in step three. And finally, in the final step, I solve for that PH by using the formula 14 equals poh plus PH. Now, if you don't know where this formula comes from, check out the link right there. So basically rearrange and find my PH."}, {"title": "The Cage Effect of Solvents .txt", "text": "On average, molecules found in the liquid state collide 100 times more frequently than the same molecules found in the gas state. That means we should be able to assume that reactions occur quicker in the liquid state than in the gas state because reactions require collisions. Now, this is not actually the case, and this is because of an effect called the effect of solvent molecules or Cage effect of solvent molecules. Now let's look at the following reaction. Suppose a red molecule must react with an orange molecule to produce a red orange product. Now, suppose we dissolve these guys in a liquid."}, {"title": "The Cage Effect of Solvents .txt", "text": "Now let's look at the following reaction. Suppose a red molecule must react with an orange molecule to produce a red orange product. Now, suppose we dissolve these guys in a liquid. So in liquid reactants are dissolved in a solvent, like, for example, water, which ends up predominating the solution. Therefore, most of the collisions in liquids occur between solvent and solute. And this means that even though the collisions occur more frequently in liquid than acoustic solutions, a lot of those collisions are between solvent and soluble molecules."}, {"title": "The Cage Effect of Solvents .txt", "text": "So in liquid reactants are dissolved in a solvent, like, for example, water, which ends up predominating the solution. Therefore, most of the collisions in liquids occur between solvent and solute. And this means that even though the collisions occur more frequently in liquid than acoustic solutions, a lot of those collisions are between solvent and soluble molecules. And the only way you react molecules is if the reactive molecules react. So, because most of the collisions occur between solid molecules and solid molecules, that means, on average, reactions in the gas state and liquid state will be approximately the same. Now let's look at this Cage effect."}, {"title": "The Cage Effect of Solvents .txt", "text": "And the only way you react molecules is if the reactive molecules react. So, because most of the collisions occur between solid molecules and solid molecules, that means, on average, reactions in the gas state and liquid state will be approximately the same. Now let's look at this Cage effect. Suppose we have our system where the red molecules are water solid molecules and the red and orange molecules or the molecules spoken about here, are the reactants. So notice that the red molecule is in a cage of solid molecules and before it leaves, they make many collisions with the water cage. Eventually, however, it will bounce out."}, {"title": "The Cage Effect of Solvents .txt", "text": "Suppose we have our system where the red molecules are water solid molecules and the red and orange molecules or the molecules spoken about here, are the reactants. So notice that the red molecule is in a cage of solid molecules and before it leaves, they make many collisions with the water cage. Eventually, however, it will bounce out. And if it bounces into another cage where this origin molecule is present, then it will react to former products. But otherwise, if it jumps into another cage that doesn't have another reactor molecule, it will continue balancing. And this greatly slows down our reaction in liquid and increased space."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "Second the X axis is temperature. Now two main differences exist between the two phase diagrams. First at one atmospheric pressure. Water exists in all three phases. However, for the carbon dioxide diagram, we see that carbon dioxide exists only in the solid phase and in the gas days. Let's look at the water diagram first."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "Water exists in all three phases. However, for the carbon dioxide diagram, we see that carbon dioxide exists only in the solid phase and in the gas days. Let's look at the water diagram first. So at low temperatures, from here to here, we can find water in the solid state. At medium temperatures, from here to here, we can find water in the liquid state. And finally, at high temperatures above this temperature, we can find water in the gas state."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "So at low temperatures, from here to here, we can find water in the solid state. At medium temperatures, from here to here, we can find water in the liquid state. And finally, at high temperatures above this temperature, we can find water in the gas state. For carbon dioxide. However, at one ATM, at low temperatures, below this temperature, we find that in the solid base and then we see that above this temperature, our solid sublons directly into the gas state. And so we could only find carbon dioxide in a solid and gas state at 180 m. The only way we could get into the liquid state is if we increase pressure and then increase temperature."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "For carbon dioxide. However, at one ATM, at low temperatures, below this temperature, we find that in the solid base and then we see that above this temperature, our solid sublons directly into the gas state. And so we could only find carbon dioxide in a solid and gas state at 180 m. The only way we could get into the liquid state is if we increase pressure and then increase temperature. The second main difference, and perhaps the more important difference, is the following. For the phase diagram of water. The boundary between the solid and the liquid line."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "The second main difference, and perhaps the more important difference, is the following. For the phase diagram of water. The boundary between the solid and the liquid line. This line has a negative slope. While for the carbon dioxide phase diagram, the boundary between the solid and liquid is positive. It has a positive slope."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "This line has a negative slope. While for the carbon dioxide phase diagram, the boundary between the solid and liquid is positive. It has a positive slope. So it's increasing here. And it's decreasing here. And this happens because water has special properties."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "So it's increasing here. And it's decreasing here. And this happens because water has special properties. As a solid. The molecules in the solid states are very far apart or further apart than they are in the liquid water states. And that means liquid."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "As a solid. The molecules in the solid states are very far apart or further apart than they are in the liquid water states. And that means liquid. The molecules are closer. So for some given volume water, liquid or liquid water is more dense than solid water. And that's why ice states a float."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "The molecules are closer. So for some given volume water, liquid or liquid water is more dense than solid water. And that's why ice states a float. Because ice is less dense than water. So for this guy, however, the solid is more dense than the liquid, and therefore, this slope is positive. So if you place a solid carbon dioxide into the liquid, it's going to sink straight down."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "Because ice is less dense than water. So for this guy, however, the solid is more dense than the liquid, and therefore, this slope is positive. So if you place a solid carbon dioxide into the liquid, it's going to sink straight down. Now, one more effect because of this negative slope is the following Because the slope is negative. If we keep our temperature constant, say, somewhere right here. So if we keep this temperature constant, we see that we can actually make the solid become a liquid by simply increasing our pressure."}, {"title": "Carbon Dioxide vs Water Phase Diagrams .txt", "text": "Now, one more effect because of this negative slope is the following Because the slope is negative. If we keep our temperature constant, say, somewhere right here. So if we keep this temperature constant, we see that we can actually make the solid become a liquid by simply increasing our pressure. So at constant temperature, we can make a solid become a liquid by simply compressing it, increasing the pressure. But for this situation, we can't. The only way we get a solid to become a liquid is if we increase temperature."}, {"title": "Phase Change of Water .txt", "text": "In this lecture, we're going to look at the phase change of water at constant pressure at one ATM when it goes from a very low temperature to a very high temperature. So the Y axis is temperature in Celsius. The X axis is energy input. And Joules, let's begin at negative 60 Celsius. So we want to go from negative negative 60 celsius to zero celsius. And we do so by heating our system."}, {"title": "Phase Change of Water .txt", "text": "And Joules, let's begin at negative 60 Celsius. So we want to go from negative negative 60 celsius to zero celsius. And we do so by heating our system. So what happens on a microscopic level when we heat the system? Well, on a microscopic level, we increase the kinetic energy of our molecules because kinetic energy is what's responsible for increasing temperature. So our potential energy stays the same from this point to this point."}, {"title": "Phase Change of Water .txt", "text": "So what happens on a microscopic level when we heat the system? Well, on a microscopic level, we increase the kinetic energy of our molecules because kinetic energy is what's responsible for increasing temperature. So our potential energy stays the same from this point to this point. But our kinetic energy increases. Now, when we get to zero degrees Celsius, a phase change occurs. So solid becomes a liquid and the change in temperature notice is zero."}, {"title": "Phase Change of Water .txt", "text": "But our kinetic energy increases. Now, when we get to zero degrees Celsius, a phase change occurs. So solid becomes a liquid and the change in temperature notice is zero. So the slope is zero. So from this point to this point, the temperature is the same. It's zero degrees Celsius."}, {"title": "Phase Change of Water .txt", "text": "So the slope is zero. So from this point to this point, the temperature is the same. It's zero degrees Celsius. And that's because all the energy input goes into increasing potential energy of our substance. And potential energy increase is what causes the change in phase. When we finish the phase change, we want to increase temperature once again."}, {"title": "Phase Change of Water .txt", "text": "And that's because all the energy input goes into increasing potential energy of our substance. And potential energy increase is what causes the change in phase. When we finish the phase change, we want to increase temperature once again. So once again, all the energy input goes into increasing our kinetic energy of our molecules. Because kinetic energy is what's responsible for increasing temperature. So when we get to 100 degrees Celsius, once again, our slope is zero."}, {"title": "Phase Change of Water .txt", "text": "So once again, all the energy input goes into increasing our kinetic energy of our molecules. Because kinetic energy is what's responsible for increasing temperature. So when we get to 100 degrees Celsius, once again, our slope is zero. That means our temperature change is zero. All the energy goes into increasing potential energy of our bonds found within the liquid and gas phase. So eventually all the liquid becomes gas."}, {"title": "Phase Change of Water .txt", "text": "That means our temperature change is zero. All the energy goes into increasing potential energy of our bonds found within the liquid and gas phase. So eventually all the liquid becomes gas. And once again, we follow a linear graph here. So any increase in energy will increase kinetic energy of our molecules. So, once again, let's review."}, {"title": "Phase Change of Water .txt", "text": "And once again, we follow a linear graph here. So any increase in energy will increase kinetic energy of our molecules. So, once again, let's review. So when the slope is zero, we deal with phase changes. And here all energy input goes into increasing potential energy of the system. No change in kinetic energy is observed."}, {"title": "Phase Change of Water .txt", "text": "So when the slope is zero, we deal with phase changes. And here all energy input goes into increasing potential energy of the system. No change in kinetic energy is observed. And so the change in temperature in both cases is zero. When we talk about going from this guy to this guy or from this guy to this guy, the intermediate phases between the phase changes, we talk about energy input that goes into increasing kinetic energy of the system because kinetic energy of the system is what increases temperature. Because we want to go from zero to 100 and from negative 60 to zero."}, {"title": "Phase Change of Water .txt", "text": "And so the change in temperature in both cases is zero. When we talk about going from this guy to this guy or from this guy to this guy, the intermediate phases between the phase changes, we talk about energy input that goes into increasing kinetic energy of the system because kinetic energy of the system is what increases temperature. Because we want to go from zero to 100 and from negative 60 to zero. We want to increase temperature and not the potential energy of the system. So let's talk about one last thing. So when we go from solid to liquid, that's called melting."}, {"title": "Phase Change of Water .txt", "text": "We want to increase temperature and not the potential energy of the system. So let's talk about one last thing. So when we go from solid to liquid, that's called melting. Okay? And melting according to this graph is endothermic. And that's because our final energy level is somewhere here."}, {"title": "Phase Change of Water .txt", "text": "Okay? And melting according to this graph is endothermic. And that's because our final energy level is somewhere here. Our initial energy level is somewhere here. So a larger amount of energy minus a smaller amount of energy gives us a positive number. So the change in enthalpy of fusion is positive."}, {"title": "Phase Change of Water .txt", "text": "Our initial energy level is somewhere here. So a larger amount of energy minus a smaller amount of energy gives us a positive number. So the change in enthalpy of fusion is positive. Melting is endothermic. Likewise, going backward or freezing is exothermic because we're taking this number and we're subtracting this number from this number. A smaller number minus a larger number gives you a negative number, and that means freezing."}, {"title": "Phase Change of Water .txt", "text": "Melting is endothermic. Likewise, going backward or freezing is exothermic because we're taking this number and we're subtracting this number from this number. A smaller number minus a larger number gives you a negative number, and that means freezing. Going this way is exothermic. And so it releases energy into the environment, heating the environment and cooling our system. So let's look at vaporization."}, {"title": "Phase Change of Water .txt", "text": "Going this way is exothermic. And so it releases energy into the environment, heating the environment and cooling our system. So let's look at vaporization. Vaporization is the process by which liquid molecules go into the gas state. So it's going from here to here. Once again, just like the change in enthalpy of fusion, change in enthalpy of vaporization is also positive."}, {"title": "Phase Change of Water .txt", "text": "Vaporization is the process by which liquid molecules go into the gas state. So it's going from here to here. Once again, just like the change in enthalpy of fusion, change in enthalpy of vaporization is also positive. It's endothermic because we take a large number and we subtract a small number from the large number and we get a positive number. And once again, going from this stage to this stage, or from a gas to a liquid, which is called condensation, well, that's exothermic. That's negative, because we take this number, subtracted from this number, and we get a negative number."}, {"title": "Acid Ionization Constant .txt", "text": "Remember, water molecules can act as both acids and bases. And in fact, if you add two water molecules together, one will act as an acid and the second one will act as a base, creating a conjugate acid and a conjugate base. In other words, if this is our acid, it will donate the H, creating an oh ion. And if this is our base, it will accept that H creating a hydronium ion. So now, if we wanted to, we can also write the equilibrium equation for this reaction. In other words, kw, our ionization constant for water is equal to hydronium concentration times the hydroxide concentration."}, {"title": "Acid Ionization Constant .txt", "text": "And if this is our base, it will accept that H creating a hydronium ion. So now, if we wanted to, we can also write the equilibrium equation for this reaction. In other words, kw, our ionization constant for water is equal to hydronium concentration times the hydroxide concentration. And at 25 degrees Celsius, this equals 10 times ten to negative 14. Now, this process is called the autoimilization reaction. And if you want to learn more about this reaction, check out the link below."}, {"title": "Acid Ionization Constant .txt", "text": "And at 25 degrees Celsius, this equals 10 times ten to negative 14. Now, this process is called the autoimilization reaction. And if you want to learn more about this reaction, check out the link below. Now, in the same way that we talk about ionization constants of water, we can also talk about ionization constants of acids. Except now they're not Kw, they're ka, where A is for our acid. So let's suppose we have a hypothetical acid Ha reacting with a water molecule in a liquid state."}, {"title": "Acid Ionization Constant .txt", "text": "Now, in the same way that we talk about ionization constants of water, we can also talk about ionization constants of acids. Except now they're not Kw, they're ka, where A is for our acid. So let's suppose we have a hypothetical acid Ha reacting with a water molecule in a liquid state. Now, what will happen? Well, this acid will donate the H, releasing the H, while the base will accept that age, creating a hydronium ion and a conjugate base. So this is our conjugate acid, our conjugate base and our conjugate base and our conjugate acid."}, {"title": "Acid Ionization Constant .txt", "text": "Now, what will happen? Well, this acid will donate the H, releasing the H, while the base will accept that age, creating a hydronium ion and a conjugate base. So this is our conjugate acid, our conjugate base and our conjugate base and our conjugate acid. Now, let's write the equilibrium constant stress in the same way we did for water for this acid. So ka, our acid annetzation constant is equal to our concentration of our hydronium times, the concentration of the conjugate base. Now, in this case, our conjugate base is simply this guy here."}, {"title": "Acid Ionization Constant .txt", "text": "Now, let's write the equilibrium constant stress in the same way we did for water for this acid. So ka, our acid annetzation constant is equal to our concentration of our hydronium times, the concentration of the conjugate base. Now, in this case, our conjugate base is simply this guy here. Now, both guys are included in our numerator because both guys are in aqueous form. Remember, liquids and solids are not included. And that's why we didn't include these two water molecules."}, {"title": "Acid Ionization Constant .txt", "text": "Now, both guys are included in our numerator because both guys are in aqueous form. Remember, liquids and solids are not included. And that's why we didn't include these two water molecules. Now, on the bottom, since we have an Ha in the Aqueous state, our conjugate acid, we must include the conjugate acid as well. So Ha gets incorporated into our equilibrium constant expression. Now, what is a k value or Ka value?"}, {"title": "Acid Ionization Constant .txt", "text": "Now, on the bottom, since we have an Ha in the Aqueous state, our conjugate acid, we must include the conjugate acid as well. So Ha gets incorporated into our equilibrium constant expression. Now, what is a k value or Ka value? Well, the ionization constant is simply a ratio from a mathematical perspective. And what the Ka is, it's the amount of product formed over the amount of reactants left over. So if this number is very large, then that means a lot of product was formed and very little reactants left over."}, {"title": "Acid Ionization Constant .txt", "text": "Well, the ionization constant is simply a ratio from a mathematical perspective. And what the Ka is, it's the amount of product formed over the amount of reactants left over. So if this number is very large, then that means a lot of product was formed and very little reactants left over. So what does that tell us about the Ha? Well, if this reaction is favored this way, if a lot of product is formed and very little reactor is left over, that means this acid is very good at giving off that age. So it must be a very good acid by definition of an acid."}, {"title": "Acid Ionization Constant .txt", "text": "So what does that tell us about the Ha? Well, if this reaction is favored this way, if a lot of product is formed and very little reactor is left over, that means this acid is very good at giving off that age. So it must be a very good acid by definition of an acid. So that means if our Ka is large, we have a good acid. Likewise, if our Ka is small, that means very little product is formed and a lot of reactors left over. That means this asset is very bad at releasing that age, so it's a bad asset."}, {"title": "Acid Ionization Constant .txt", "text": "So that means if our Ka is large, we have a good acid. Likewise, if our Ka is small, that means very little product is formed and a lot of reactors left over. That means this asset is very bad at releasing that age, so it's a bad asset. Now, we can deduce that if the Ka is greater than one, then that means it's a strong acid. And if the Ka is less than one, that means it's a weak acid. Now let's look at one more thing."}, {"title": "Acid Ionization Constant .txt", "text": "Now, we can deduce that if the Ka is greater than one, then that means it's a strong acid. And if the Ka is less than one, that means it's a weak acid. Now let's look at one more thing. So what happens if our Ka is less than one? Then this guy must be a weak acid. But that means our conjugate base must be a good base."}, {"title": "Acid Ionization Constant .txt", "text": "So what happens if our Ka is less than one? Then this guy must be a weak acid. But that means our conjugate base must be a good base. Likewise, if this was a good acid and our Ka was above one, that means this was a bad conjugate base. Now let's look at a few examples. Nitric acid in aqueous state could act with water to produce hydronium ion plus nitrate ion."}, {"title": "Acid Ionization Constant .txt", "text": "Likewise, if this was a good acid and our Ka was above one, that means this was a bad conjugate base. Now let's look at a few examples. Nitric acid in aqueous state could act with water to produce hydronium ion plus nitrate ion. Now, the ka for this reaction for this acid is 20. And that means, according to this theory, it's a very good acid. And indeed it is."}, {"title": "Acid Ionization Constant .txt", "text": "Now, the ka for this reaction for this acid is 20. And that means, according to this theory, it's a very good acid. And indeed it is. Now let's look at hydrofluoric acid. So hydrofluoric acid in the equated states react with water in the liquid state to produce hydronium plus the fion. Now, the Ka for this reaction for this particular acid is very low."}, {"title": "Acid Ionization Constant .txt", "text": "Now let's look at hydrofluoric acid. So hydrofluoric acid in the equated states react with water in the liquid state to produce hydronium plus the fion. Now, the Ka for this reaction for this particular acid is very low. It's 7.2 times ten to negative four. And that means, according to this theory, it's a bad acid. And in fact, it is a bad acid."}, {"title": "Acid Ionization Constant .txt", "text": "It's 7.2 times ten to negative four. And that means, according to this theory, it's a bad acid. And in fact, it is a bad acid. It's a weak acid. So now we can use this ka value to determine whether or not an acid is a good acid or a bad asset. Now, before we have to look at the polarity of the bond, we have to look at the bond strength and we have to look at the conjugate base."}, {"title": "Ion Pairing .txt", "text": "Now, ion Pairing is A Momentary aggregation of Electrically charged ions found In A Concentrated solution Of Two Or More compounds. Now to really grasp what we mean by Ion pairing, let's create a solution and see what happens with the natural solution. So let's mix liquid water and solid sodium chloride and see what happens. Well, once we mix them, the sodium chloride dissociates and forms ions. So we have a bunch of polar water molecules separated by ions. At any given time, two of these ions might be separated by a solvent water molecule, and at this point, they can't interact with one another because of this separation."}, {"title": "Ion Pairing .txt", "text": "Well, once we mix them, the sodium chloride dissociates and forms ions. So we have a bunch of polar water molecules separated by ions. At any given time, two of these ions might be separated by a solvent water molecule, and at this point, they can't interact with one another because of this separation. However, at another place, the solid molecule, the water molecule, might not separate them. And at this point, because of their proximity, the ions will form an ion pair. And this is only for the moment."}, {"title": "Ion Pairing .txt", "text": "However, at another place, the solid molecule, the water molecule, might not separate them. And at this point, because of their proximity, the ions will form an ion pair. And this is only for the moment. Imagine two ions floating or flying around in the liquid, right? Eventually they will get close to each other. And when they get close to each other, they will form that momentary bond."}, {"title": "Ion Pairing .txt", "text": "Imagine two ions floating or flying around in the liquid, right? Eventually they will get close to each other. And when they get close to each other, they will form that momentary bond. But it's only for the moment. Because if they're traveling, they will attract and will continue to travel. So eventually they will break."}, {"title": "Ion Pairing .txt", "text": "But it's only for the moment. Because if they're traveling, they will attract and will continue to travel. So eventually they will break. So they travel, attract and break. And it's only for the moment. And that's the difference between an Ion pair and an ionic bond."}, {"title": "Ion Pairing .txt", "text": "So they travel, attract and break. And it's only for the moment. And that's the difference between an Ion pair and an ionic bond. An ionic bond is not for the moment. It Stays. It Exists."}, {"title": "Ion Pairing .txt", "text": "An ionic bond is not for the moment. It Stays. It Exists. If this sodium chloride is left untouched, it will continue. The ionic bond will continue to exist. But in this case, the Ion pair only exists for a moment."}, {"title": "Ion Pairing .txt", "text": "If this sodium chloride is left untouched, it will continue. The ionic bond will continue to exist. But in this case, the Ion pair only exists for a moment. Now Ion Pairing does not exist in an ideal solution. And that's because in an ideal solution, by definition, all the ions are separated from one another by solvent molecules. So no attraction between two ions exists."}, {"title": "Ion Pairing .txt", "text": "Now Ion Pairing does not exist in an ideal solution. And that's because in an ideal solution, by definition, all the ions are separated from one another by solvent molecules. So no attraction between two ions exists. So Ion pairing only exists in nonideal conditions. Now, if you want to see why Ion Pairing is important and where it's applied, check out the video below. In that video, I talk about something called boiling point elevation and melting point depression."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Well, Gibbs free energy, just like enthalpy, is a manmade concept, which means it cannot be measured. So you cannot use an instrument to measure Gibbs free energy of some object. Gibbs free energy could only measure expansion fundamentally. Now, that's because Gibbs free energy is defined by a formula. And this formula only holds under certain conditions. If these conditions aren't met, Gibbs free energy breaks down."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Now, that's because Gibbs free energy is defined by a formula. And this formula only holds under certain conditions. If these conditions aren't met, Gibbs free energy breaks down. Now, let's look at these conditions. These conditions are constant temperature and pressure. The reaction must be reversible, and there is no mechanical work done."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Now, let's look at these conditions. These conditions are constant temperature and pressure. The reaction must be reversible, and there is no mechanical work done. Only TV work is allowed to be done. Now, let's see the result of constant temperature and pressure. In an isolated system, the number of moles stays the same."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Only TV work is allowed to be done. Now, let's see the result of constant temperature and pressure. In an isolated system, the number of moles stays the same. And that's because there is no exchange in mass. So, if the number of moles stay the same, if the temperature is constant and pressure is constant, and according to the ideal gas law, the volume remains constant, so there is no volume change. So, according to the formula for change in enthalpy, if there's no volume change, then this guy, the PV work done, is zero."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "And that's because there is no exchange in mass. So, if the number of moles stay the same, if the temperature is constant and pressure is constant, and according to the ideal gas law, the volume remains constant, so there is no volume change. So, according to the formula for change in enthalpy, if there's no volume change, then this guy, the PV work done, is zero. And so we can approximate the change in enthalpy to simply equaling change in internal energy or change in energy or heat. Now, before we jump into the formula, let's look at entropy and entropy. So what is entropy?"}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "And so we can approximate the change in enthalpy to simply equaling change in internal energy or change in energy or heat. Now, before we jump into the formula, let's look at entropy and entropy. So what is entropy? Well, entropy tells us if the reaction is exothermic or endothermic. It does not tell us if the reaction is spontaneous. Okay?"}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Well, entropy tells us if the reaction is exothermic or endothermic. It does not tell us if the reaction is spontaneous. Okay? That's important. Now let's look at entropy. Now, from another video, we learned that entropy is nature's tendency to create the most probable system, okay?"}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "That's important. Now let's look at entropy. Now, from another video, we learned that entropy is nature's tendency to create the most probable system, okay? So, for example, if we have this isolated system, we have four molecules on this side. Entropy, by definition, will tell us that two of the molecules will want to go here and will want to create this system here. In other words, this system is more probable."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "So, for example, if we have this isolated system, we have four molecules on this side. Entropy, by definition, will tell us that two of the molecules will want to go here and will want to create this system here. In other words, this system is more probable. Okay? So we see that entropy and not enthalpy tells us if the reaction is spontaneous. Therefore, entropy determines spontaneity and not enthalpy."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Okay? So we see that entropy and not enthalpy tells us if the reaction is spontaneous. Therefore, entropy determines spontaneity and not enthalpy. Okay? Now let's look at Gibbs free energy. So, gifts free energy combines entropy and enthalpy the same way that enthalpy combines internal energy and change in volume."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Okay? Now let's look at Gibbs free energy. So, gifts free energy combines entropy and enthalpy the same way that enthalpy combines internal energy and change in volume. The formula for Gibbs free energy is as follows. The change in Gibbs free energy is equal to the change in entropy minus temperature times change in entropy. Now, when the change in Gibbs free energy is zero, the reaction is said to be at equilibrium."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "The formula for Gibbs free energy is as follows. The change in Gibbs free energy is equal to the change in entropy minus temperature times change in entropy. Now, when the change in Gibbs free energy is zero, the reaction is said to be at equilibrium. Okay? The rate of the full reaction is the same as the rate of the reverse reaction. Now, when GIBS free energy is negative, the reaction is said to be spontaneous."}, {"title": "Introduction to Gibbs Free Energy .txt", "text": "Okay? The rate of the full reaction is the same as the rate of the reverse reaction. Now, when GIBS free energy is negative, the reaction is said to be spontaneous. If it's positive, it's said to be non spontaneous or non spontaneous. Okay? Now, what git speed energy basically tells us is that even an exothermic reaction can be non spontaneous if this portion, if the change in entropy is negative, enough."}, {"title": "Half Reactions .txt", "text": "So recall that redox reactions are chemical reactions that involve the transfer of electrons from one atom to another atom. Now, this means one atom is oxidized and one atom is reduced. So that means any redox reaction can be broken down into two types of reactions an oxidation reactions, in which an atom loses electrons and a reduction reaction, in which atoms gain electrons. So let's look at a very simple redox reaction. So, zinc metal reacts with aqueous copper to produce aqueous zinc and metal copper. So let's break this reaction into an oxidation and a reduction reaction."}, {"title": "Half Reactions .txt", "text": "So let's look at a very simple redox reaction. So, zinc metal reacts with aqueous copper to produce aqueous zinc and metal copper. So let's break this reaction into an oxidation and a reduction reaction. So, what gets oxidized? Well, our zinc solid goes from a neutral charge to a plus two charge. That means our zinc loses two electrons."}, {"title": "Half Reactions .txt", "text": "So, what gets oxidized? Well, our zinc solid goes from a neutral charge to a plus two charge. That means our zinc loses two electrons. So let's write the equation for that. So, let's write the oxidation equation. So, zinc solid becomes zinc plus two because it loses two electrons."}, {"title": "Half Reactions .txt", "text": "So let's write the equation for that. So, let's write the oxidation equation. So, zinc solid becomes zinc plus two because it loses two electrons. So our oxidation reaction involves the loss of electrons. Now let's write the reduction reaction. In our reduction reaction, our copper gets reduced because it gains those two electrons that are lost by the zinc."}, {"title": "Half Reactions .txt", "text": "So our oxidation reaction involves the loss of electrons. Now let's write the reduction reaction. In our reduction reaction, our copper gets reduced because it gains those two electrons that are lost by the zinc. So our copper plus two gains two electrons plus two electrons producing a neutral copper solid molecule. So this is our oxidation and this is our reduction reactions. Now, these two reactions are each called the half reaction."}, {"title": "Half Reactions .txt", "text": "So our copper plus two gains two electrons plus two electrons producing a neutral copper solid molecule. So this is our oxidation and this is our reduction reactions. Now, these two reactions are each called the half reaction. And the half reaction is simply a way for us to visualize more clearly the transfer of electrons from one atom to another. Because notice in this net reaction, there was no transfer of electrons. We couldn't visualize the transfer electrons."}, {"title": "Half Reactions .txt", "text": "And the half reaction is simply a way for us to visualize more clearly the transfer of electrons from one atom to another. Because notice in this net reaction, there was no transfer of electrons. We couldn't visualize the transfer electrons. But in these half reactions, we have the plus two E and plus two e.\nSo this is simply a better way for us to see the movement of electrons. Now, to go back to our net reaction net reduce reaction. What we simply do is we add up the two half reactions."}, {"title": "Half Reactions .txt", "text": "But in these half reactions, we have the plus two E and plus two e.\nSo this is simply a better way for us to see the movement of electrons. Now, to go back to our net reaction net reduce reaction. What we simply do is we add up the two half reactions. So we add up the two half reactions by first adding up all the molecules on this side and then add up all the molecules on this side. So zinc solid plus acreage copper plus two electrons gives us this produces everything on this side. So zinc or acreage zinc plus copper solid plus our two electrons."}, {"title": "Half Reactions .txt", "text": "So we add up the two half reactions by first adding up all the molecules on this side and then add up all the molecules on this side. So zinc solid plus acreage copper plus two electrons gives us this produces everything on this side. So zinc or acreage zinc plus copper solid plus our two electrons. Now, notice one thing. That two electrons up here on this side and on this side. And that means by using simple algebra we just cross these guys out, we subtract two e, and what we get is our final net reaction."}, {"title": "Half Reactions .txt", "text": "Now, notice one thing. That two electrons up here on this side and on this side. And that means by using simple algebra we just cross these guys out, we subtract two e, and what we get is our final net reaction. So that means no electrons appear in the net reaction ever. And that's because the number of electrons released by our oxidation reaction is equal to the number gained by our reduction reaction. So whatever is gained must be lost somewhere."}, {"title": "Freezing Point depression Example .txt", "text": "Now, our initial freezing point is six degrees Celsius. Our final freezing point is 2.7 degrees Celsius. Our constant for freezing is 20 Celsius times kilograms per mole. Now, so we basically start with 100 grams of cyclohexane which freezes at six degrees Celsius. We add 2 grams of unknown compound to our beaker of 100 grams of cyclohexane and that solution freezing point drops to 2.7 degrees Celsius. Now, our goal is to use the freezing point depression formula to find the Molar mass."}, {"title": "Freezing Point depression Example .txt", "text": "Now, so we basically start with 100 grams of cyclohexane which freezes at six degrees Celsius. We add 2 grams of unknown compound to our beaker of 100 grams of cyclohexane and that solution freezing point drops to 2.7 degrees Celsius. Now, our goal is to use the freezing point depression formula to find the Molar mass. In the first step we write our freezing depression formula and that basically state that's the change in temperature equal there constant times molarity times I I is a bond half factor and in this case it's one. Because our unknown compound does not associate into anything, it stays the way it is. So you plug in our values and we get six -20 or 2.7 degrees Celsius."}, {"title": "Freezing Point depression Example .txt", "text": "In the first step we write our freezing depression formula and that basically state that's the change in temperature equal there constant times molarity times I I is a bond half factor and in this case it's one. Because our unknown compound does not associate into anything, it stays the way it is. So you plug in our values and we get six -20 or 2.7 degrees Celsius. Divided by 20 our constant Celsius cancels moles goes on top and we get 0.165\nmoles per kilogram. And this is our Molarity. Now, in the second step, let's look at the formula for molality."}, {"title": "Freezing Point depression Example .txt", "text": "Divided by 20 our constant Celsius cancels moles goes on top and we get 0.165\nmoles per kilogram. And this is our Molarity. Now, in the second step, let's look at the formula for molality. Molarity equals moles of compound divided by a kilogram of solids. Now we're given kilogram of solids, so we know that we also know molality. Now let's change our moles of compound formula to something else."}, {"title": "Freezing Point depression Example .txt", "text": "Molarity equals moles of compound divided by a kilogram of solids. Now we're given kilogram of solids, so we know that we also know molality. Now let's change our moles of compound formula to something else. Let's see, how else can we represent the moles of compound? Now, whenever you want to find the moles of compound, we basically take our amount in grams divided by the molecular weight of that same compound and we get our moles. So another way of finding or representing moles of compound is simply grams of compound divided by molecular weight."}, {"title": "Freezing Point depression Example .txt", "text": "Let's see, how else can we represent the moles of compound? Now, whenever you want to find the moles of compound, we basically take our amount in grams divided by the molecular weight of that same compound and we get our moles. So another way of finding or representing moles of compound is simply grams of compound divided by molecular weight. But molecular weight and molar mass mean the same thing. So we simply write molar mass and now we have this, we have Molarity and all we need is to find the Molar mass. Remember, the grams of compound was given initially 2 grams of unknown compound and we can check using our units that this makes sense, grams divided by grams per mole."}, {"title": "Freezing Point depression Example .txt", "text": "But molecular weight and molar mass mean the same thing. So we simply write molar mass and now we have this, we have Molarity and all we need is to find the Molar mass. Remember, the grams of compound was given initially 2 grams of unknown compound and we can check using our units that this makes sense, grams divided by grams per mole. The grams cancel moles goes on top and we're left with moles per kilogram. And that's exactly what molality is. Now, our third step."}, {"title": "Freezing Point depression Example .txt", "text": "The grams cancel moles goes on top and we're left with moles per kilogram. And that's exactly what molality is. Now, our third step. In our last step, we basically plug in our values. So for molality, which we got from the first step is zero point 65 equals two divided by x. Our unknown."}, {"title": "Freezing Point depression Example .txt", "text": "In our last step, we basically plug in our values. So for molality, which we got from the first step is zero point 65 equals two divided by x. Our unknown. The molar mass entire thing divided by 0.10.1\ncomes from 100 grams. Remember, we want to deal with kilograms of our solid and we're given 100 grams. So divide this by 1000, we get 0.1\nkg, bring this over, multiply this out, then bring it back and we at two divided by 0.165 equals x."}, {"title": "Balancing Redox reactions example .txt", "text": "We want to follow seven steps that I've outlined in another lecture and want to get a final net balanced redox reaction. So let's follow these seven steps. In the first step, we want to find or determine the oxidized atom and the reduced atom. So let's find a reduced atom first. So, which atom gains electrons? So I've written out the oxidation states for each atom."}, {"title": "Balancing Redox reactions example .txt", "text": "So let's find a reduced atom first. So, which atom gains electrons? So I've written out the oxidation states for each atom. So let's see which atom becomes more negative. So, our Mm, or permanganate atom, goes from a plus seven to a plus two. That means it gains five electrons."}, {"title": "Balancing Redox reactions example .txt", "text": "So let's see which atom becomes more negative. So, our Mm, or permanganate atom, goes from a plus seven to a plus two. That means it gains five electrons. So that means our Mm is reduced. Now, likewise, let's find an atom that loses those electrons. Well, our carbon atom is plus three on this side and plus four on that side."}, {"title": "Balancing Redox reactions example .txt", "text": "So that means our Mm is reduced. Now, likewise, let's find an atom that loses those electrons. Well, our carbon atom is plus three on this side and plus four on that side. And that means when it goes from here to here, it is oxidized. It loses those electrons gained by this Mm. So our carbon is oxidized."}, {"title": "Balancing Redox reactions example .txt", "text": "And that means when it goes from here to here, it is oxidized. It loses those electrons gained by this Mm. So our carbon is oxidized. We're double step one. Let's go to step two. And step two."}, {"title": "Balancing Redox reactions example .txt", "text": "We're double step one. Let's go to step two. And step two. We want to write out the two half reactions for our unbalanced equations. So we want to write out the oxidation reaction and the reduction reaction. So, oxidation basically states that this guy goes from this guy to this guy in here."}, {"title": "Balancing Redox reactions example .txt", "text": "We want to write out the two half reactions for our unbalanced equations. So we want to write out the oxidation reaction and the reduction reaction. So, oxidation basically states that this guy goes from this guy to this guy in here. And then the reduction reaction, our ion, goes to this ion. So this is our reduction reaction. Now, I combine steps three and four in steps."}, {"title": "Balancing Redox reactions example .txt", "text": "And then the reduction reaction, our ion, goes to this ion. So this is our reduction reaction. Now, I combine steps three and four in steps. In step three, I basically want to balance out the atoms that are not oxygen atoms and not hydrogen atoms. So let's start with oxidation reaction. So, an oxidation health reaction."}, {"title": "Balancing Redox reactions example .txt", "text": "In step three, I basically want to balance out the atoms that are not oxygen atoms and not hydrogen atoms. So let's start with oxidation reaction. So, an oxidation health reaction. Let's balance out the carbon atoms. So, to balance this carbon atom out, there are two atoms on this side and only one atom on this side. That means I want to multiply this guy by two."}, {"title": "Balancing Redox reactions example .txt", "text": "Let's balance out the carbon atoms. So, to balance this carbon atom out, there are two atoms on this side and only one atom on this side. That means I want to multiply this guy by two. And that's exactly what I do here. Now, I have two carbon atoms on this side and two carbon atoms on this side. Next, I want to balance out the carbon or the MN atoms on this side in this reduction reaction."}, {"title": "Balancing Redox reactions example .txt", "text": "And that's exactly what I do here. Now, I have two carbon atoms on this side and two carbon atoms on this side. Next, I want to balance out the carbon or the MN atoms on this side in this reduction reaction. So, since I have one MN here and one Amen here, it's already balanced. So, in the fourth step, let's balance out the oxygen molecules by adding water molecules. And let's balance the h atoms by adding H plus ions."}, {"title": "Balancing Redox reactions example .txt", "text": "So, since I have one MN here and one Amen here, it's already balanced. So, in the fourth step, let's balance out the oxygen molecules by adding water molecules. And let's balance the h atoms by adding H plus ions. So let's see our oxidation reaction. So, we have four O atoms on this side and only two atoms on this side. But I already multiply this carbon dioxide by two."}, {"title": "Balancing Redox reactions example .txt", "text": "So let's see our oxidation reaction. So, we have four O atoms on this side and only two atoms on this side. But I already multiply this carbon dioxide by two. That means our oxygens already bounced out. So I have four here and four here. Two times two."}, {"title": "Balancing Redox reactions example .txt", "text": "That means our oxygens already bounced out. So I have four here and four here. Two times two. So the oxygen is bounced out. Let's balance out the h atoms. I have two atoms on this side and no h atoms on that side, right?"}, {"title": "Balancing Redox reactions example .txt", "text": "So the oxygen is bounced out. Let's balance out the h atoms. I have two atoms on this side and no h atoms on that side, right? So in order to balance this, I have to add two h atoms on this side. That's exactly what I do. So I'm done with my oxidation half reaction."}, {"title": "Balancing Redox reactions example .txt", "text": "So in order to balance this, I have to add two h atoms on this side. That's exactly what I do. So I'm done with my oxidation half reaction. Let's look at the reduction half reaction. Remember, the MNS are already balanced out, so let's balance out oxygen. I have four oxygen this side, so that means you have to add four water molecules on this side."}, {"title": "Balancing Redox reactions example .txt", "text": "Let's look at the reduction half reaction. Remember, the MNS are already balanced out, so let's balance out oxygen. I have four oxygen this side, so that means you have to add four water molecules on this side. If I add four water molecules on this side, I balance out the O. But now I have to balance out the H, because now I have four times two, eight H ions on this side or atoms on this side. That means I balance this guy out by adding eight H plus ions to this side, and I get this following reaction."}, {"title": "Balancing Redox reactions example .txt", "text": "If I add four water molecules on this side, I balance out the O. But now I have to balance out the H, because now I have four times two, eight H ions on this side or atoms on this side. That means I balance this guy out by adding eight H plus ions to this side, and I get this following reaction. So I'm done with steps three and four. Now let's go to step five. In step five, you basically want to balance out the entire charge down on this side and this side, and then the entire charge down on this side and this side."}, {"title": "Balancing Redox reactions example .txt", "text": "So I'm done with steps three and four. Now let's go to step five. In step five, you basically want to balance out the entire charge down on this side and this side, and then the entire charge down on this side and this side. In other words, on this side of the oxidation half reaction, our charge is neutral. On this side, it's not neutral because we have two positive charges. So this side is neutral."}, {"title": "Balancing Redox reactions example .txt", "text": "In other words, on this side of the oxidation half reaction, our charge is neutral. On this side, it's not neutral because we have two positive charges. So this side is neutral. This side is plus two. So to make this side zero, I have to add two electrons. That's exactly what I do."}, {"title": "Balancing Redox reactions example .txt", "text": "This side is plus two. So to make this side zero, I have to add two electrons. That's exactly what I do. And now my charge is neutral and neutral because these two pluses cancel these two minuses. Now let's follow the same steps for this reduction reaction. So on this side, I have a minus one and eight pluses."}, {"title": "Balancing Redox reactions example .txt", "text": "And now my charge is neutral and neutral because these two pluses cancel these two minuses. Now let's follow the same steps for this reduction reaction. So on this side, I have a minus one and eight pluses. That means our charge on this side is plus seven. Now on this side, my charge is plus two. So I want to neutralize my charges."}, {"title": "Balancing Redox reactions example .txt", "text": "That means our charge on this side is plus seven. Now on this side, my charge is plus two. So I want to neutralize my charges. That means if I have plus seven here, I have to add seven electrons, and that's exactly what I do. Now, on this side, to neutralize this charge, I have to add two electrons. That's exactly what I do."}, {"title": "Balancing Redox reactions example .txt", "text": "That means if I have plus seven here, I have to add seven electrons, and that's exactly what I do. Now, on this side, to neutralize this charge, I have to add two electrons. That's exactly what I do. I add two electrons. But now notice that I have two electrons here and seven electrons here. That means I could subtract two electrons from both sides."}, {"title": "Balancing Redox reactions example .txt", "text": "I add two electrons. But now notice that I have two electrons here and seven electrons here. That means I could subtract two electrons from both sides. And what happens is this guy cancels out and this guy becomes a five, because seven minus five is seven minus two is five. So I'm done. Step five."}, {"title": "Balancing Redox reactions example .txt", "text": "And what happens is this guy cancels out and this guy becomes a five, because seven minus five is seven minus two is five. So I'm done. Step five. In step six, I want to consider the fact that energy or charge is concerned, like mass is concerned. And that means I have to whatever amount of electrons lost by one atom or by one side has to be gained by another side. So notice on this side, I have two electrons, on this side, I have five electrons."}, {"title": "Balancing Redox reactions example .txt", "text": "In step six, I want to consider the fact that energy or charge is concerned, like mass is concerned. And that means I have to whatever amount of electrons lost by one atom or by one side has to be gained by another side. So notice on this side, I have two electrons, on this side, I have five electrons. That means I want to find a factor or a common factor of five and two. And this factor is ten, because two goes into ten five times, and five goes into ten two times. So I basically multiply this whole guy out by five, and I multiply this whole guy out by two, and I get the following."}, {"title": "Balancing Redox reactions example .txt", "text": "That means I want to find a factor or a common factor of five and two. And this factor is ten, because two goes into ten five times, and five goes into ten two times. So I basically multiply this whole guy out by five, and I multiply this whole guy out by two, and I get the following. So five times this guy gives me this, five times this guy gives me this guy five times, two H gives me ten H, and five times this guy gives me ten electrons. I multiply this whole guy by two. I get two of these guys plus 16 of H pluses plus ten of east equals two of these guys and eight of these guys."}, {"title": "Balancing Redox reactions example .txt", "text": "So five times this guy gives me this, five times this guy gives me this guy five times, two H gives me ten H, and five times this guy gives me ten electrons. I multiply this whole guy by two. I get two of these guys plus 16 of H pluses plus ten of east equals two of these guys and eight of these guys. Now, when I go from step six to seven, I basically want to add these equations up, right? I want to add my two half reactions. So I add all the guys on the left side, and then I add all the guys on the right side."}, {"title": "Balancing Redox reactions example .txt", "text": "Now, when I go from step six to seven, I basically want to add these equations up, right? I want to add my two half reactions. So I add all the guys on the left side, and then I add all the guys on the right side. So I get this whole guy plus nothing more on this side. So I go down to my other reactions, plus two of these guys, plus 16 of these guys, plus ten electrons. And now I'm done with everything on the left side."}, {"title": "Balancing Redox reactions example .txt", "text": "So I get this whole guy plus nothing more on this side. So I go down to my other reactions, plus two of these guys, plus 16 of these guys, plus ten electrons. And now I'm done with everything on the left side. Now I move on to the right side. This guy equals ten carbon dioxide molecules plus two Mm, two plus molecules, right? Plus ten H plus molecules, plus eight water molecules plus ten electrons."}, {"title": "Balancing Redox reactions example .txt", "text": "Now I move on to the right side. This guy equals ten carbon dioxide molecules plus two Mm, two plus molecules, right? Plus ten H plus molecules, plus eight water molecules plus ten electrons. So this is my equation. Now, my next goal before I get to my finalized answer, I have to cross out some things. The first thing I crossed out is a ten electron."}, {"title": "Balancing Redox reactions example .txt", "text": "So this is my equation. Now, my next goal before I get to my finalized answer, I have to cross out some things. The first thing I crossed out is a ten electron. So we have ten electrons here and seven electrons here. That cancels out. Next, notice that I have 16 H plus ions on this side and ten plus H ions on this side."}, {"title": "Balancing Redox reactions example .txt", "text": "So we have ten electrons here and seven electrons here. That cancels out. Next, notice that I have 16 H plus ions on this side and ten plus H ions on this side. That means I could cancel out this guy by subtracting ten H plus from this side and ten H plus from this side. So I canceled out, and 16 minus ten is six. So my final equation, my final balanced redox equation, is five H, two C, two four plus two M and minus one four plus six of these guys."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "Today we're going to talk about roll slaw for ideal fluids. So what is a pure liquid? A pure liquid is simply a liquid that is not contaminated by any other compound or molecule. For example, suppose we have a closed system, a closed container, and inside this container we have pure water. But that simply means that the only types of molecules found within our system or water molecules. Now, what will happen to our system if it's left untouched?"}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "For example, suppose we have a closed system, a closed container, and inside this container we have pure water. But that simply means that the only types of molecules found within our system or water molecules. Now, what will happen to our system if it's left untouched? Well, eventually, some of the water molecules found on the surface of the liquid will escape into the gas state and will become gas molecules. And this is called evaporation. Now, on the rate of evaporation and condensation are equal."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "Well, eventually, some of the water molecules found on the surface of the liquid will escape into the gas state and will become gas molecules. And this is called evaporation. Now, on the rate of evaporation and condensation are equal. The system is said to be in dynamic equilibrium. At this point, we could measure something called vapor pressure. Vapor pressure is the pressure exerted by evaporating molecules or gas molecules found in dynamic equilibrium with the pure liquid molecules."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "The system is said to be in dynamic equilibrium. At this point, we could measure something called vapor pressure. Vapor pressure is the pressure exerted by evaporating molecules or gas molecules found in dynamic equilibrium with the pure liquid molecules. Now, what will happen to our system if we add a non volatile compound to our pure substance? Remember, a non volatile compound is a compound that will not evaporate. And remember, evaporation only occurs on the surface area of the liquid."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "Now, what will happen to our system if we add a non volatile compound to our pure substance? Remember, a non volatile compound is a compound that will not evaporate. And remember, evaporation only occurs on the surface area of the liquid. So therefore, let's look at what happens to our surface area after we add this substance. The surface area of the liquid is constant, it remains the same. And that's because our system, our container, remains constant."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "So therefore, let's look at what happens to our surface area after we add this substance. The surface area of the liquid is constant, it remains the same. And that's because our system, our container, remains constant. It will not change in shape or size. So if you take the crosssectional before addition and after addition, the crosssectional area will remain the same. What will change however, is the number of pure liquid molecules found on the surface area and this number will decrease."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "It will not change in shape or size. So if you take the crosssectional before addition and after addition, the crosssectional area will remain the same. What will change however, is the number of pure liquid molecules found on the surface area and this number will decrease. And this is because of the presence of nonvolatile compounds. Now for example, let's look at the before picture. Before addition, we have water molecules found on the surface area."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "And this is because of the presence of nonvolatile compounds. Now for example, let's look at the before picture. Before addition, we have water molecules found on the surface area. And these water molecules will escape and they're going to condense back into the liquid state. Now, after the addition, we're going to have some non volatile compounds replacing these molecules. And that means less molecules or less pure molecules present on the surface."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "And these water molecules will escape and they're going to condense back into the liquid state. Now, after the addition, we're going to have some non volatile compounds replacing these molecules. And that means less molecules or less pure molecules present on the surface. And remember, evaporation occurs on the surface of the liquid. So there are less volatile molecules, less molecules that evaporate, less gas molecules will be present at equilibrium. And this means the pressure of the vapor pressure is less."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "And remember, evaporation occurs on the surface of the liquid. So there are less volatile molecules, less molecules that evaporate, less gas molecules will be present at equilibrium. And this means the pressure of the vapor pressure is less. And this is Rolls law. Rolt's law gives us the following equation. The vapor pressure after the addition is equal to the mole fraction of the pure substance times the vapor pressure before addition."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "And this is Rolls law. Rolt's law gives us the following equation. The vapor pressure after the addition is equal to the mole fraction of the pure substance times the vapor pressure before addition. And this is known as Rolls law. Now let's look at the addition of a volatile compound to a pure mixture. A volatile compound is simply a compound that will evaporate."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "And this is known as Rolls law. Now let's look at the addition of a volatile compound to a pure mixture. A volatile compound is simply a compound that will evaporate. So what happens to our surface area? Well, the surface area remains the same, and that's because once again, our container does not change in shape or size. Its cross sectional area remains constant."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "So what happens to our surface area? Well, the surface area remains the same, and that's because once again, our container does not change in shape or size. Its cross sectional area remains constant. And once again, we have a less pure molecules found on the surface. And that's because they're replaced by the new volatile content. But now there's a major difference."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "And once again, we have a less pure molecules found on the surface. And that's because they're replaced by the new volatile content. But now there's a major difference. Now we're dealing with a content that is allowed to evaporate. So let's look at the difference. The before picture, before addition, we only have water molecules or pure molecules found on a surface."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "Now we're dealing with a content that is allowed to evaporate. So let's look at the difference. The before picture, before addition, we only have water molecules or pure molecules found on a surface. And these guys will evaporate and will create a certain vapor pressure. The vapor pressure is this the after picture? After we add the volatile compound, we're going to have a mixture of the new compound and the old compound found on the surface area."}, {"title": "Raoul\u2019s Law for Ideal Fluids .txt", "text": "And these guys will evaporate and will create a certain vapor pressure. The vapor pressure is this the after picture? After we add the volatile compound, we're going to have a mixture of the new compound and the old compound found on the surface area. But now the red guys, the new compound, are allowed to evaporate. So now the vapor pressure will be due to these molecules and these molecules. So Roth's law will tell us that the final vapor pressure of our system will be the vapor pressure due to the new guys, plus the vapor pressure due to the old guys."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "So if you're not sure about entropy, check out the link below. So one of the most basic definitions of entropy is that entropy is the tendency of a system to even out. For example, let's look at system. This system is composed of two sections connected by a bridge and we have two molecules on this side and eight molecules on this side. What entropy tells us is that eventually three of these molecules will end up on this side and that this system will be the most probable system. So we want a system that's even not uneven or balanced, not on balance, in which we have five molecules here and five molecules here."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "This system is composed of two sections connected by a bridge and we have two molecules on this side and eight molecules on this side. What entropy tells us is that eventually three of these molecules will end up on this side and that this system will be the most probable system. So we want a system that's even not uneven or balanced, not on balance, in which we have five molecules here and five molecules here. Let's look at system C.\nNow, in system C we basically have a cell. And this cell is surrounded by semipermeable membrane that allows water molecules in and out but does not allow any size molecules to go in or out. Now, in this cell, however, we have a small hole in the membrane and this hole is large enough to allow the molecules in and out."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "Let's look at system C.\nNow, in system C we basically have a cell. And this cell is surrounded by semipermeable membrane that allows water molecules in and out but does not allow any size molecules to go in or out. Now, in this cell, however, we have a small hole in the membrane and this hole is large enough to allow the molecules in and out. So water is going to travel in and out and so will the molecules. So eventually, according to what entropy tells us, we're going to want to go from system C to system D. That is, we want to travel from this system to this even system in which we have five molecules on this side on the inside and five molecules on the outside. Now let's look at a third system."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "So water is going to travel in and out and so will the molecules. So eventually, according to what entropy tells us, we're going to want to go from system C to system D. That is, we want to travel from this system to this even system in which we have five molecules on this side on the inside and five molecules on the outside. Now let's look at a third system. Let's look at system E. Now, in system E, we basically have the same system as above, except now our hole is filled in. So we have a continuous circular membrane. So now we can't talk about the side molecules going in and out because side molecules can't go in and out."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "Let's look at system E. Now, in system E, we basically have the same system as above, except now our hole is filled in. So we have a continuous circular membrane. So now we can't talk about the side molecules going in and out because side molecules can't go in and out. And that's because of the membrane. The membrane is a barrier to the side molecules. So we must talk about something else."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "And that's because of the membrane. The membrane is a barrier to the side molecules. So we must talk about something else. We must talk about the amount of solid molecules per some given volume. And in fact, that's concentration. So our concentration on the outside, because we have more molecules, is higher than the concentration on the inside."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "We must talk about the amount of solid molecules per some given volume. And in fact, that's concentration. So our concentration on the outside, because we have more molecules, is higher than the concentration on the inside. So what will happen? Well, entropy tells us that in this case, the only thing that can move, which is a solvent and the water will move and water will travel from the inside to the outside. So our cell will shrink in size and our concentration will increase in size because now we have less volume."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "So what will happen? Well, entropy tells us that in this case, the only thing that can move, which is a solvent and the water will move and water will travel from the inside to the outside. So our cell will shrink in size and our concentration will increase in size because now we have less volume. But the same amount of solution, now the concentration on the outside will decrease because now we have the same amount, but a larger volume, a larger amount of water. And that means eventually our concentrations will equal out. So now let's define osmosis."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "But the same amount of solution, now the concentration on the outside will decrease because now we have the same amount, but a larger volume, a larger amount of water. And that means eventually our concentrations will equal out. So now let's define osmosis. Osmosis is the movement of solvent. In our case, water from an area of a lower side concentration to an area of a higher sized concentration. So in the system above, osmosis occurred from the inside the cell into the outside, because water moves from a lower concentration to a higher concentration."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "Osmosis is the movement of solvent. In our case, water from an area of a lower side concentration to an area of a higher sized concentration. So in the system above, osmosis occurred from the inside the cell into the outside, because water moves from a lower concentration to a higher concentration. Okay? Now, we can also talk about something called osmotic pressure. Osmotic pressure is the pressure required to stop osmosis."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "Okay? Now, we can also talk about something called osmotic pressure. Osmotic pressure is the pressure required to stop osmosis. So, for example, suppose we apply some pressure to our cell membrane, and if this pressure equals asthmatic pressure, well, then no osmosis will occur, no movement of water will occur. And this becomes very important when you're talking about hydrostatic pressure and asthmatic pressure in the capillaries of our body. So the last thing I want to talk about is a formula."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "So, for example, suppose we apply some pressure to our cell membrane, and if this pressure equals asthmatic pressure, well, then no osmosis will occur, no movement of water will occur. And this becomes very important when you're talking about hydrostatic pressure and asthmatic pressure in the capillaries of our body. So the last thing I want to talk about is a formula. And this formula could be only used when we talk about ideal conditions. So ideal solutions and solutions that have very low or small concentrations. Now, if you want to look at a problem using this formula, check out the link below."}, {"title": "Osmosis and Osmotic Pressure.txt", "text": "And this formula could be only used when we talk about ideal conditions. So ideal solutions and solutions that have very low or small concentrations. Now, if you want to look at a problem using this formula, check out the link below. Now let's look at the formula. The formula basically states that osmotic pressure is equal to molarity of solution times our gas constants r times our temperature in Kelvin times I. Now, I is called a vant half factor."}, {"title": "Methyl Compounds .txt", "text": "So let's discuss methyl compounds. Now, methyl compounds are simply compounds that have a side chain or side group ch three. And that ch three is attached covalently to some other atomyl chloride compound given here, x. Now, this x could be anything, and here are a few examples. We have methyl chloride, methyl alcohol, or methanol. We have methylamine, and we have methylcyanide."}, {"title": "Methyl Compounds .txt", "text": "Now, this x could be anything, and here are a few examples. We have methyl chloride, methyl alcohol, or methanol. We have methylamine, and we have methylcyanide. So let's compare methyl compounds to something that we already spoke about methane. In fact, methane is methyl compound where the x has been replaced with an H. Now, methane is the simplest alkane, and methane has symmetry. And that's because the carbon, the central carbon, is attached to four identical H atoms, 1234."}, {"title": "Methyl Compounds .txt", "text": "So let's compare methyl compounds to something that we already spoke about methane. In fact, methane is methyl compound where the x has been replaced with an H. Now, methane is the simplest alkane, and methane has symmetry. And that's because the carbon, the central carbon, is attached to four identical H atoms, 1234. And that means all the ch bonds will be exactly the same as the other. All these ch bonds will be SP three hybridized, and the angles between a two bonds will be 109.5\ndegrees. So, once again, each bond is identical because we have a single carbon atom attached to four identical H atoms."}, {"title": "Methyl Compounds .txt", "text": "And that means all the ch bonds will be exactly the same as the other. All these ch bonds will be SP three hybridized, and the angles between a two bonds will be 109.5\ndegrees. So, once again, each bond is identical because we have a single carbon atom attached to four identical H atoms. Hence, our bonds are all SP three hybridized. So let's take this methane and compare it to a methylcogddam. So here we have a methylcogdam, where we replace the H with an x."}, {"title": "Methyl Compounds .txt", "text": "Hence, our bonds are all SP three hybridized. So let's take this methane and compare it to a methylcogddam. So here we have a methylcogdam, where we replace the H with an x. This X could be any atom. Now, we have bonds that are not all identical. In other words, we still have three of these ch bonds, but now we have a different CX bond."}, {"title": "Methyl Compounds .txt", "text": "This X could be any atom. Now, we have bonds that are not all identical. In other words, we still have three of these ch bonds, but now we have a different CX bond. For example, if I replace this x with a chloride atom, so this is a chloride. That means the chloride, since the chloride is more electronegative than either the carbon or the H atoms, that means that chloride will pull electrons more strongly than either of the H atom. And so there will be an unequal electron density in this bond."}, {"title": "Methyl Compounds .txt", "text": "For example, if I replace this x with a chloride atom, so this is a chloride. That means the chloride, since the chloride is more electronegative than either the carbon or the H atoms, that means that chloride will pull electrons more strongly than either of the H atom. And so there will be an unequal electron density in this bond. Electrons will be closer to this x atom, to this CL atom than to the carbon atom. And that means that this will be a slightly asymmetrical molecule, asymmetrical compound. And so it will slightly deviate from this methane compound."}, {"title": "Class Clapeyron Example .txt", "text": "So we are given alcohol, and we know that our alcohol at 20 degrees Celsius has a vapor pressure of 40 mercury. And at 61 degrees Celsius, it has a vapor pressure of 360 millimeters of mercury. And our goal is to find a change in entropy of vaporization of 1 mol of alcohol. So to find that, we have to use something called a class here to clay for an equation. Now, if you want to learn more about this equation, where it comes from, and why it's important, check out the link below. So let's look at our equation."}, {"title": "Class Clapeyron Example .txt", "text": "So to find that, we have to use something called a class here to clay for an equation. Now, if you want to learn more about this equation, where it comes from, and why it's important, check out the link below. So let's look at our equation. Notice that in this equation we have three knowns and two unknowns. So we have a constant R, it's a gas constant. We know that."}, {"title": "Class Clapeyron Example .txt", "text": "Notice that in this equation we have three knowns and two unknowns. So we have a constant R, it's a gas constant. We know that. Now we have the pressure and we have a temperature. What we don't have is this C constant and this entropy of vaporization. In fact, that's exactly what we want to find."}, {"title": "Class Clapeyron Example .txt", "text": "Now we have the pressure and we have a temperature. What we don't have is this C constant and this entropy of vaporization. In fact, that's exactly what we want to find. So if we somehow know this, we can find that. But an even better tactic would be to get rid of this. So notice we have a single equation and two unknowns."}, {"title": "Class Clapeyron Example .txt", "text": "So if we somehow know this, we can find that. But an even better tactic would be to get rid of this. So notice we have a single equation and two unknowns. So mathematically we can't solve this. But if we come up with a system of equations, two equations and two unknowns, we could somehow manipulate the two equations, get rid of that C, and solve for our unknown. And in fact, that's exactly what we're going to do."}, {"title": "Class Clapeyron Example .txt", "text": "So mathematically we can't solve this. But if we come up with a system of equations, two equations and two unknowns, we could somehow manipulate the two equations, get rid of that C, and solve for our unknown. And in fact, that's exactly what we're going to do. Notice we have initial conditions and final conditions. So why not come up with two of these equations, one for the initial condition and one for the final condition. That's exactly what we do here."}, {"title": "Class Clapeyron Example .txt", "text": "Notice we have initial conditions and final conditions. So why not come up with two of these equations, one for the initial condition and one for the final condition. That's exactly what we do here. This one is for our final conditions, where PF and TF are P final and T final. And this guy is for our initial conditions, p initial and T initial. And in order to get rid of these two C's, let's subtract this guy from this guy."}, {"title": "Class Clapeyron Example .txt", "text": "This one is for our final conditions, where PF and TF are P final and T final. And this guy is for our initial conditions, p initial and T initial. And in order to get rid of these two C's, let's subtract this guy from this guy. Okay, that's exactly what we do right here. So this guy minus this guy. Now notice we have the equal sign here."}, {"title": "Class Clapeyron Example .txt", "text": "Okay, that's exactly what we do right here. So this guy minus this guy. Now notice we have the equal sign here. We're going to subtract everything on this side first, then everything on this side next. So natural log of P final, minus natural log of P initial, and we get exactly this. Next, we subtract this whole section from this whole section."}, {"title": "Class Clapeyron Example .txt", "text": "We're going to subtract everything on this side first, then everything on this side next. So natural log of P final, minus natural log of P initial, and we get exactly this. Next, we subtract this whole section from this whole section. So first we take this and we put it here. That's exactly what we did. Next, we subtract this guy from this guy."}, {"title": "Class Clapeyron Example .txt", "text": "So first we take this and we put it here. That's exactly what we did. Next, we subtract this guy from this guy. But notice we have a negative sign here, and that means we have to distribute this negative sign to here and here. So this guy becomes positive. So we get plus this guy, and this negative makes this positive a negative."}, {"title": "Class Clapeyron Example .txt", "text": "But notice we have a negative sign here, and that means we have to distribute this negative sign to here and here. So this guy becomes positive. So we get plus this guy, and this negative makes this positive a negative. So this guy becomes a negative. And now notice we have a plus C and a minus C. So the C's cancel and we get just this guy plus this guy. And that's exactly what we have right here."}, {"title": "Class Clapeyron Example .txt", "text": "So this guy becomes a negative. And now notice we have a plus C and a minus C. So the C's cancel and we get just this guy plus this guy. And that's exactly what we have right here. Now our next step is to basically equate this guy, to simply this guy and rewrite the equations to a better looking formula so this guy, using the logs of logs, we can rewrite in this format. So natural log of P final divided by P initial equals. Now, notice on this side, we have two common terms or one common term."}, {"title": "Class Clapeyron Example .txt", "text": "Now our next step is to basically equate this guy, to simply this guy and rewrite the equations to a better looking formula so this guy, using the logs of logs, we can rewrite in this format. So natural log of P final divided by P initial equals. Now, notice on this side, we have two common terms or one common term. This guy and this guy are two of the same term. It's the same term. So basically, you want to take this guy out of our equation and leave this guy and this guy alone."}, {"title": "Class Clapeyron Example .txt", "text": "This guy and this guy are two of the same term. It's the same term. So basically, you want to take this guy out of our equation and leave this guy and this guy alone. So negative change in anthropy vaporization over r, our common term, times one over t, final minus one over T initial. Notice that here this was a positive. But since we're taking a negative out, this becomes a negative."}, {"title": "Class Clapeyron Example .txt", "text": "So negative change in anthropy vaporization over r, our common term, times one over t, final minus one over T initial. Notice that here this was a positive. But since we're taking a negative out, this becomes a negative. And to check that, we multiply this out and we should get this form. And, in fact, we do. Now, our final step before we plug in chug is simply to rewrite this so that we have this thing on one side and everything else, all the knowns on the other side."}, {"title": "Class Clapeyron Example .txt", "text": "And to check that, we multiply this out and we should get this form. And, in fact, we do. Now, our final step before we plug in chug is simply to rewrite this so that we have this thing on one side and everything else, all the knowns on the other side. So you want the unknown on one side and the knowns on the other side. And this is what we get. So what we do is we find the common denominator here, multiply this by Ti and this by TF, bring everything over, bring the R over, and then bring the negative over, and we get this."}, {"title": "Class Clapeyron Example .txt", "text": "So you want the unknown on one side and the knowns on the other side. And this is what we get. So what we do is we find the common denominator here, multiply this by Ti and this by TF, bring everything over, bring the R over, and then bring the negative over, and we get this. Finally, we plug in all our information. So our R, it's 8.31\njoules per mole times Kelvin, times natural log of 360 or 40, which is simply nine times. Now, we have to use our temperature in Kelvin, and that means we have to convert from celebrates to Kelvin by simply adding 273 to each temperature."}, {"title": "Class Clapeyron Example .txt", "text": "Finally, we plug in all our information. So our R, it's 8.31\njoules per mole times Kelvin, times natural log of 360 or 40, which is simply nine times. Now, we have to use our temperature in Kelvin, and that means we have to convert from celebrates to Kelvin by simply adding 273 to each temperature. So this is what we get. We basically plug this into our calculator. We solve, and we get approximately 44 kilojoules per mole."}, {"title": "Alkenes and Double Bonds .txt", "text": "But unlike alkanes, alkines contain a double bond. And in this lecture, we're going to examine exactly what a double bond is. So let's begin by looking at the simplest alkhine, known as ethylene, also known as ethylene. Now, ethylene is composed of two carbons connected by a double bond and two H atoms found on both sides of those carbons. So let's begin by building or creating this ethylene molecule. So let's create it using a methylratical."}, {"title": "Alkenes and Double Bonds .txt", "text": "Now, ethylene is composed of two carbons connected by a double bond and two H atoms found on both sides of those carbons. So let's begin by building or creating this ethylene molecule. So let's create it using a methylratical. Recall that a methylradical is simply a carbon atom attached to three H bonds or three H atoms via SP two hybridized orbitals. So each of these sigma bonds, covalent sigma bonds, are SP two hybridized. And we also have a pure two p orbital that contains a single electron within that two p orbital."}, {"title": "Alkenes and Double Bonds .txt", "text": "Recall that a methylradical is simply a carbon atom attached to three H bonds or three H atoms via SP two hybridized orbitals. So each of these sigma bonds, covalent sigma bonds, are SP two hybridized. And we also have a pure two p orbital that contains a single electron within that two p orbital. So, to build this ethylene, let's simply replace the H atom here with a methylene atom with a ch two atom, or a ch two molecule. Sorry. What do we get?"}, {"title": "Alkenes and Double Bonds .txt", "text": "So, to build this ethylene, let's simply replace the H atom here with a methylene atom with a ch two atom, or a ch two molecule. Sorry. What do we get? Well, if we say if we replace this with a ch two molecule, we get the following picture. So, we get a carbon carbon bond. We get four PH bonds, two on each side."}, {"title": "Alkenes and Double Bonds .txt", "text": "Well, if we say if we replace this with a ch two molecule, we get the following picture. So, we get a carbon carbon bond. We get four PH bonds, two on each side. And we have a two p orbital on both of these carbons that has an electron in each orbital. So first, we let's examine what type of bond this carbon carbon bond is. So this carbon donates an SP two hybridized orbital."}, {"title": "Alkenes and Double Bonds .txt", "text": "And we have a two p orbital on both of these carbons that has an electron in each orbital. So first, we let's examine what type of bond this carbon carbon bond is. So this carbon donates an SP two hybridized orbital. Remember, these guys are SP two hybridized. And this carbon also donates an SP two hybridized orbital. So when we combine two atomic orbitals, we must form two molecular orbitals according to quantum mechanics."}, {"title": "Alkenes and Double Bonds .txt", "text": "Remember, these guys are SP two hybridized. And this carbon also donates an SP two hybridized orbital. So when we combine two atomic orbitals, we must form two molecular orbitals according to quantum mechanics. And so our lower end energy more stable molecular orbital will be due to the overlap of these green regions. And we will get the following SP two, SP two sigma bonding molecular orbital, or simply mo. So the two electrons, one electron in each of this carbon in each of these SP two hybridized orbitals will go into this lower in energy bonding molecular orbital."}, {"title": "Alkenes and Double Bonds .txt", "text": "And so our lower end energy more stable molecular orbital will be due to the overlap of these green regions. And we will get the following SP two, SP two sigma bonding molecular orbital, or simply mo. So the two electrons, one electron in each of this carbon in each of these SP two hybridized orbitals will go into this lower in energy bonding molecular orbital. Now, we're also going to have this antibonding molecular orbital. But since it's high in energy and it's morphed and it's less stable, the electrons will not go into that orbital. So both electrons will be in this molecular orbital."}, {"title": "Alkenes and Double Bonds .txt", "text": "Now, we're also going to have this antibonding molecular orbital. But since it's high in energy and it's morphed and it's less stable, the electrons will not go into that orbital. So both electrons will be in this molecular orbital. And so this covalent bond is SP two SP two hybridized sigma bonding molecular orbital. And now notice one more thing. Notice we have these two two p orbitals, and they're both parallel to one another."}, {"title": "Alkenes and Double Bonds .txt", "text": "And so this covalent bond is SP two SP two hybridized sigma bonding molecular orbital. And now notice one more thing. Notice we have these two two p orbitals, and they're both parallel to one another. In other words, these two guides are parallel to one another, and they're perpendicular to either of these ch bonds. And so, because these guides are parallel and because they have the same exact energy as one another, they will create an overlapping condition. So, once again, just like we have an overlap here, we're going to have an overlap here."}, {"title": "Alkenes and Double Bonds .txt", "text": "In other words, these two guides are parallel to one another, and they're perpendicular to either of these ch bonds. And so, because these guides are parallel and because they have the same exact energy as one another, they will create an overlapping condition. So, once again, just like we have an overlap here, we're going to have an overlap here. So let's combine these two p orbitals. So here we have a two p orbital from this carbon combined with a two P orbital from this carbon. Once again, we're combining two atomic orbitals to form two different molecular orbitals."}, {"title": "Alkenes and Double Bonds .txt", "text": "So let's combine these two p orbitals. So here we have a two p orbital from this carbon combined with a two P orbital from this carbon. Once again, we're combining two atomic orbitals to form two different molecular orbitals. However, now they're no longer Sigma. They're called pi. Okay?"}, {"title": "Alkenes and Double Bonds .txt", "text": "However, now they're no longer Sigma. They're called pi. Okay? So one is a pi, or two P, two pi bonding molecular orbital. And the second one is a two P two pi antibonding molecular orbital. So, there will be one known between these two orbitals."}, {"title": "Alkenes and Double Bonds .txt", "text": "So one is a pi, or two P, two pi bonding molecular orbital. And the second one is a two P two pi antibonding molecular orbital. So, there will be one known between these two orbitals. And that means this guy will be higher in energy and less stable. And so electrons will tend to go into the lower in energy more stable bond. So this pi bond here."}, {"title": "Alkenes and Double Bonds .txt", "text": "And that means this guy will be higher in energy and less stable. And so electrons will tend to go into the lower in energy more stable bond. So this pi bond here. Now, notice one difference between our Sigma and our Pi bonds. Both of these SP two hybridized orbitals contain 33% or 33.3% as character, while these two p orbitals contain no as character. So that means, because these guys contain the more stable S character, these are more stable."}, {"title": "Alkenes and Double Bonds .txt", "text": "Now, notice one difference between our Sigma and our Pi bonds. Both of these SP two hybridized orbitals contain 33% or 33.3% as character, while these two p orbitals contain no as character. So that means, because these guys contain the more stable S character, these are more stable. So that means they're lower in energy than these two P orbitals. So when these two orbitals, when these two p orbitals combine to form a pi orbital, this pi orbital is higher in energy than this SP two SP two Sigma bonding molecular orbital. And that means this will be more stable than our Pi bond."}, {"title": "Alkenes and Double Bonds .txt", "text": "So that means they're lower in energy than these two P orbitals. So when these two orbitals, when these two p orbitals combine to form a pi orbital, this pi orbital is higher in energy than this SP two SP two Sigma bonding molecular orbital. And that means this will be more stable than our Pi bond. So now let's redraw our diagram for this silent molecule. So here we have our two carbons, and they create an SP two hybridized molecular orbital given here. And it also creates this interaction here between our pure two P orbitals."}, {"title": "Alkenes and Double Bonds .txt", "text": "So now let's redraw our diagram for this silent molecule. So here we have our two carbons, and they create an SP two hybridized molecular orbital given here. And it also creates this interaction here between our pure two P orbitals. Remember, this electron is found at the same time in this region, as well as in this region. So that means there will be interaction between these two lobes here. And so this is known as our Pi bond, and this is known as our Sigma Bond."}, {"title": "Alkenes and Double Bonds .txt", "text": "Remember, this electron is found at the same time in this region, as well as in this region. So that means there will be interaction between these two lobes here. And so this is known as our Pi bond, and this is known as our Sigma Bond. And once again, this pi bond will be higher in energy than this Sigma Bond. Another way of drawing this a more simpler way is simply with two bonds here, two dashes here. Now, notice that this is a Sigma bond, and the top one is a Pi bond."}, {"title": "Alkenes and Double Bonds .txt", "text": "And once again, this pi bond will be higher in energy than this Sigma Bond. Another way of drawing this a more simpler way is simply with two bonds here, two dashes here. Now, notice that this is a Sigma bond, and the top one is a Pi bond. So let's review. The Sigma bond contains more S character and is therefore lower in energy and more stable or stronger. Because remember, the more stable something is, the more stronger it is then the Pi bond."}, {"title": "Alkenes and Double Bonds .txt", "text": "So let's review. The Sigma bond contains more S character and is therefore lower in energy and more stable or stronger. Because remember, the more stable something is, the more stronger it is then the Pi bond. Therefore, the Pi bond is less stable because it's higher in energy, and therefore it's more reactive. Now, whenever we input energy to break our Pi bond or to break our double bond, our pi bonds break first. So let's look at one more important detail about our double bonds."}, {"title": "Alkenes and Double Bonds .txt", "text": "Therefore, the Pi bond is less stable because it's higher in energy, and therefore it's more reactive. Now, whenever we input energy to break our Pi bond or to break our double bond, our pi bonds break first. So let's look at one more important detail about our double bonds. So, this is a single bond. This is an ethylene molecule. So notice that in an ethylene molecule, we have one Sigma bond, and the Sigma bond is able to rotate."}, {"title": "Alkenes and Double Bonds .txt", "text": "So, this is a single bond. This is an ethylene molecule. So notice that in an ethylene molecule, we have one Sigma bond, and the Sigma bond is able to rotate. So we create confirmations or confirmations of ethane. So we could have an eclipse confirm, and we could have a staggered confirm. Now, notice what happens in our double bond molecule."}, {"title": "Alkenes and Double Bonds .txt", "text": "So we create confirmations or confirmations of ethane. So we could have an eclipse confirm, and we could have a staggered confirm. Now, notice what happens in our double bond molecule. So here we have the following ethylene lean. So, notice that these ch bonds on Ethylene lean are on the same plane. And these two orbitals here, this Pi orbital is created by an overlap of two P orbitals down perpendicular to either of BCH bonds."}, {"title": "Alkenes and Double Bonds .txt", "text": "So here we have the following ethylene lean. So, notice that these ch bonds on Ethylene lean are on the same plane. And these two orbitals here, this Pi orbital is created by an overlap of two P orbitals down perpendicular to either of BCH bonds. And notice what happens. Notice now there is no rotation. And that's because if there was rotation, these two P orbitals would no longer be in parallel."}, {"title": "Alkenes and Double Bonds .txt", "text": "And notice what happens. Notice now there is no rotation. And that's because if there was rotation, these two P orbitals would no longer be in parallel. When I rotate these, these guys would lose that overlap and therefore would destabilize the molecule. So that means, because these two P orbitals like to stay in parallel to one another, they like to stay in the same plane. This double bond will not allow rotation."}, {"title": "Kinetic Molecular Theory .txt", "text": "And they call these assumptions the kinetic molecular theory. Now, from the math, a Niagara point of view, this isn't really a theory because under real conditions these assumptions don't hold. Yes, these assumptions are important to make because they allow us to come up with concrete conclusions about the behavior of gas molecules. So let's begin. The first assumption is the fact that volume of gas molecules is zero. So where does this assumption come from?"}, {"title": "Kinetic Molecular Theory .txt", "text": "So let's begin. The first assumption is the fact that volume of gas molecules is zero. So where does this assumption come from? Well, it comes from the observation that gases are easily compressed and mixed very well. And this is because the distance between the molecules is much larger than the size of the molecules. Now let's look at our inflated ball."}, {"title": "Kinetic Molecular Theory .txt", "text": "Well, it comes from the observation that gases are easily compressed and mixed very well. And this is because the distance between the molecules is much larger than the size of the molecules. Now let's look at our inflated ball. Within our ball, we have lots of different molecules. But the difference between any two molecules is much greater than the size of the molecule itself. And that's why we can compress it because when we compress it, there's lots of space for the molecules to move."}, {"title": "Kinetic Molecular Theory .txt", "text": "Within our ball, we have lots of different molecules. But the difference between any two molecules is much greater than the size of the molecule itself. And that's why we can compress it because when we compress it, there's lots of space for the molecules to move. On the contrary, on solids and liquids the density is much higher and so there is not too much space for them to move. And that's why we can't compress them easily. And that's exactly why when we take this inflated ball, we see that we can easily compress it because there's lots of space for the molecules to move."}, {"title": "Kinetic Molecular Theory .txt", "text": "On the contrary, on solids and liquids the density is much higher and so there is not too much space for them to move. And that's why we can't compress them easily. And that's exactly why when we take this inflated ball, we see that we can easily compress it because there's lots of space for the molecules to move. But if this a ball was filled with solid or liquid, I would not be able to compress it without changing its shape or volume. Let's look at the second assumption. Gases move at high velocities in all different directions."}, {"title": "Kinetic Molecular Theory .txt", "text": "But if this a ball was filled with solid or liquid, I would not be able to compress it without changing its shape or volume. Let's look at the second assumption. Gases move at high velocities in all different directions. So what's the observation or what's the experience from everyday life that tells us that gas is, in fact move at high velocities? Well, for example, if you forgot to wash your feet or you've been wearing your shoes for way too long, you know that if you take off your shoes and there's a girl or a voice sitting across the room, they will definitely smell you instantaneously. That's why you better keep your shoes on."}, {"title": "Kinetic Molecular Theory .txt", "text": "So what's the observation or what's the experience from everyday life that tells us that gas is, in fact move at high velocities? Well, for example, if you forgot to wash your feet or you've been wearing your shoes for way too long, you know that if you take off your shoes and there's a girl or a voice sitting across the room, they will definitely smell you instantaneously. That's why you better keep your shoes on. That's because when you take off your shoes the molecules of air trap in your socks and in your shoes escape and move at very high speeds in all different directions. So the person sitting across the room from you will definitely smell you. So you better keep those shoes on."}, {"title": "Kinetic Molecular Theory .txt", "text": "That's because when you take off your shoes the molecules of air trap in your socks and in your shoes escape and move at very high speeds in all different directions. So the person sitting across the room from you will definitely smell you. So you better keep those shoes on. So the second assumption about high velocities also accounts for the fact that gases will expand into any container quickly and completely. Let's look at our third assumption. So, gas molecules exert no forces on one another due to mass and charge."}, {"title": "Kinetic Molecular Theory .txt", "text": "So the second assumption about high velocities also accounts for the fact that gases will expand into any container quickly and completely. Let's look at our third assumption. So, gas molecules exert no forces on one another due to mass and charge. From everyday experience we know that if we take an object and drop it, it will slide down. Well, why does it slide down? Because the Earth, a much greater mass, pulls the object and this object pulls the Earth as well."}, {"title": "Kinetic Molecular Theory .txt", "text": "From everyday experience we know that if we take an object and drop it, it will slide down. Well, why does it slide down? Because the Earth, a much greater mass, pulls the object and this object pulls the Earth as well. But the Earth is so large, it doesn't really move too much. And in fact, any two objects that have mass will exert a pulling force. Now, the same way charge also exerts a pulling and an attraction force."}, {"title": "Kinetic Molecular Theory .txt", "text": "But the Earth is so large, it doesn't really move too much. And in fact, any two objects that have mass will exert a pulling force. Now, the same way charge also exerts a pulling and an attraction force. Now, all these pulling attraction forces can be neglected in a gas system. Well, this fifth part is not really an assumption. It's more of a conclusion."}, {"title": "Kinetic Molecular Theory .txt", "text": "Now, all these pulling attraction forces can be neglected in a gas system. Well, this fifth part is not really an assumption. It's more of a conclusion. Now, average kinetic energy of molecules is proportional to the temperature. And that simply means if we increase our temperature, we have more kinetic energy. And one observation regarding this assumption is that reactions occur quicker when our temperatures are higher."}, {"title": "Kinetic Molecular Theory .txt", "text": "Now, average kinetic energy of molecules is proportional to the temperature. And that simply means if we increase our temperature, we have more kinetic energy. And one observation regarding this assumption is that reactions occur quicker when our temperatures are higher. And that's because there are more collisions between any molecules. And so these colliding molecules are allowed to react, and so they create products. And that's why our rates are higher."}, {"title": "Kinetic Molecular Theory .txt", "text": "And that's because there are more collisions between any molecules. And so these colliding molecules are allowed to react, and so they create products. And that's why our rates are higher. Now, I want to mention one more thing. Now, recall that kinetic energy is equal to one half mass times velocity squared. So suppose I have two molecules, one heavy molecule and one light molecule with the same kinetic energy."}, {"title": "Kinetic Molecular Theory .txt", "text": "Now, I want to mention one more thing. Now, recall that kinetic energy is equal to one half mass times velocity squared. So suppose I have two molecules, one heavy molecule and one light molecule with the same kinetic energy. Well, according to this formula, if the kinetic energies are the same, then the higher or the heavier molecule will have a lower velocity, while the lighter molecule will have a higher velocity. We could also talk about the average velocities of the molecules, and that's simply average of all the molecules found in our system. So, on average, if you pull out a molecule from our system, it will have an average speed."}, {"title": "Elements and Isotopes.txt", "text": "Now, over 100 different types of atoms exist and each atom is called an element. Now each element found on the periodic table of elements is represented in the following way where this x is the symbol of our atom. Now in this case it's just x. It's a hypothetical symbol. But for example, carbon has the letter C and oxygen has the letter O. Now this A and this C are usually numbers but in this case we're going to use letters."}, {"title": "Elements and Isotopes.txt", "text": "It's a hypothetical symbol. But for example, carbon has the letter C and oxygen has the letter O. Now this A and this C are usually numbers but in this case we're going to use letters. The A is the atomic mass and the Z is the atomic number of our element. Now the atomic mass is the mass of that element. It's the number of protons and the number of neutrons."}, {"title": "Elements and Isotopes.txt", "text": "The A is the atomic mass and the Z is the atomic number of our element. Now the atomic mass is the mass of that element. It's the number of protons and the number of neutrons. Now note that electrons are not counted Naratomic mass because their mass is much smaller than that the proton or the neutron. Now the atomic number is the number of protons of our element. Now that number, the atomic number is the identity number of that element."}, {"title": "Elements and Isotopes.txt", "text": "Now note that electrons are not counted Naratomic mass because their mass is much smaller than that the proton or the neutron. Now the atomic number is the number of protons of our element. Now that number, the atomic number is the identity number of that element. It's used to identify our element. It's the fingerprint of that element. And that's because any element can have different number of electrons or neutrons but it will always have the same number of protons."}, {"title": "Elements and Isotopes.txt", "text": "It's used to identify our element. It's the fingerprint of that element. And that's because any element can have different number of electrons or neutrons but it will always have the same number of protons. And that's why you could use the atomic number to identify our x, our element. The second that the number of protons changes that means our element also changes. Now let's go into something called isotopes of elements."}, {"title": "Elements and Isotopes.txt", "text": "And that's why you could use the atomic number to identify our x, our element. The second that the number of protons changes that means our element also changes. Now let's go into something called isotopes of elements. Now, two or more atoms that contain the same number of protons mean they're the same elements but have different number of neutrons, are called isotopes of that same. Let's look at a very common example of carbon. Now carbon has three isotopes."}, {"title": "Elements and Isotopes.txt", "text": "Now, two or more atoms that contain the same number of protons mean they're the same elements but have different number of neutrons, are called isotopes of that same. Let's look at a very common example of carbon. Now carbon has three isotopes. Now in each case because this is a carbon atom it must have the same number of Z. The atomic number must be the same. In other words, all three have six protons."}, {"title": "Elements and Isotopes.txt", "text": "Now in each case because this is a carbon atom it must have the same number of Z. The atomic number must be the same. In other words, all three have six protons. But notice carbon twelve has six neutrons. Carbon 13 has seven neutrons and carbon 14 has eight neutrons thereby giving the atomic mass six plus 612, six plus 713 and six plus eight is 14. So that makes sense because to get the atomic mass to get the a you have to add up the protons and neutrons."}, {"title": "Elements and Isotopes.txt", "text": "But notice carbon twelve has six neutrons. Carbon 13 has seven neutrons and carbon 14 has eight neutrons thereby giving the atomic mass six plus 612, six plus 713 and six plus eight is 14. So that makes sense because to get the atomic mass to get the a you have to add up the protons and neutrons. Now there is a unit that scientists created to deal with very, very small amounts of atoms. Now the AMU, or atomic mass unit is the unit of mass used for elements and compounds. Now by definition we define that one atom, a single atom of carbon."}, {"title": "Elements and Isotopes.txt", "text": "Now there is a unit that scientists created to deal with very, very small amounts of atoms. Now the AMU, or atomic mass unit is the unit of mass used for elements and compounds. Now by definition we define that one atom, a single atom of carbon. Twelve is composed of twelve AMU and everything else is relative to this amount. For example, an H atom is one AMU and it's relative to this. In other words, this is twelve times that's heavier in terms of AMU than an H atom."}, {"title": "Elements and Isotopes.txt", "text": "Twelve is composed of twelve AMU and everything else is relative to this amount. For example, an H atom is one AMU and it's relative to this. In other words, this is twelve times that's heavier in terms of AMU than an H atom. A single h atom. Now, note that this is one atom of carbon. That's a very tiny amount."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "So, from experiencing the world around us, we know that solids come in many different forms, sizes and shapes. Now, in this lecture, we're going to look at exactly that. We're going to examine the different type of structures formed by solids. So two main types of structures exist, and we'll talk about Crystalline Solids or simply crystals and amorphous solids. So let's begin with the crystalline solids. So all Crystalline Solids or simply crystals have a very well ordered shape or structure."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "So two main types of structures exist, and we'll talk about Crystalline Solids or simply crystals and amorphous solids. So let's begin with the crystalline solids. So all Crystalline Solids or simply crystals have a very well ordered shape or structure. And because of that, they have a very sharp melting point. What that means is that the melting point's range is very small. It melts over a very, very small range of temperature."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "And because of that, they have a very sharp melting point. What that means is that the melting point's range is very small. It melts over a very, very small range of temperature. Now. Four main types of Crystalline solids or simply crystals exist. Ionic Crystalline solids."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Now. Four main types of Crystalline solids or simply crystals exist. Ionic Crystalline solids. Metallic Crystalline solids. Molecular Crystalline solids and network covalent crystalline solids. Now, let's begin with our ionic crystals."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Metallic Crystalline solids. Molecular Crystalline solids and network covalent crystalline solids. Now, let's begin with our ionic crystals. Now, these crystals consist of ions that are held together by electrostatic forces. Electrostatic forces are forces between positively charged ions and negatively charged ions of different atoms. Now, let's see in some examples."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Now, these crystals consist of ions that are held together by electrostatic forces. Electrostatic forces are forces between positively charged ions and negatively charged ions of different atoms. Now, let's see in some examples. Sodium chloride is an ionic crystal. Lithium chloride is also an ionic crystal. Calcium chloride or calcium dichloride is also an ionic crystal."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Sodium chloride is an ionic crystal. Lithium chloride is also an ionic crystal. Calcium chloride or calcium dichloride is also an ionic crystal. And basically, whenever an alkaline metal or an alkaline earth metal reacts with a halogen, these guys will always produce, or will most of the time actually will always produce ionic crystals. Now let's look at metallic crystals. Now, these guys consist of single metal molecules that are held together by a sea of electrons."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "And basically, whenever an alkaline metal or an alkaline earth metal reacts with a halogen, these guys will always produce, or will most of the time actually will always produce ionic crystals. Now let's look at metallic crystals. Now, these guys consist of single metal molecules that are held together by a sea of electrons. And examples include any type of alkali metal or alkaline earth metal. For example, a composite which actually isn't an alkaline or alkali metal, but is a transition metal, but still it's a metal that has a metallic solid structure. Other examples are sodium metal or potassium metal or leafy metal or calcium metal."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "And examples include any type of alkali metal or alkaline earth metal. For example, a composite which actually isn't an alkaline or alkali metal, but is a transition metal, but still it's a metal that has a metallic solid structure. Other examples are sodium metal or potassium metal or leafy metal or calcium metal. Any of these guys have a metallic like structure or metallic crystal structure. So now let's look at the third type of Crystalline Solids known as molecular solids. Now, these crystals, or these Crystalline solids consist of molecules held together by intermolecular forces called down their balls forces."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Any of these guys have a metallic like structure or metallic crystal structure. So now let's look at the third type of Crystalline Solids known as molecular solids. Now, these crystals, or these Crystalline solids consist of molecules held together by intermolecular forces called down their balls forces. And examples include ice. In other words, when you freeze ice, when you take energy away from water and form ice, the water molecules form a very structured formation. And this creates what we know as ice."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "And examples include ice. In other words, when you freeze ice, when you take energy away from water and form ice, the water molecules form a very structured formation. And this creates what we know as ice. And ice are called molecular crystals or molecular solids. The fourth and final type of Crystalline Solids we're going to look at are network covalent crystals. Now, these consist of network of atoms or molecules held together by covalent bonds."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "And ice are called molecular crystals or molecular solids. The fourth and final type of Crystalline Solids we're going to look at are network covalent crystals. Now, these consist of network of atoms or molecules held together by covalent bonds. These covalent bonds can be both non polar and polar covalent. Now, an example is diamonds. So what's the structure of a diamond?"}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "These covalent bonds can be both non polar and polar covalent. Now, an example is diamonds. So what's the structure of a diamond? Diamond consists of solely carbon atoms held together by covalent bonds. These Sigma covalent bonds are very strong, and that's exactly why our diamonds are so strong. It's very hard to break diamonds."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Diamond consists of solely carbon atoms held together by covalent bonds. These Sigma covalent bonds are very strong, and that's exactly why our diamonds are so strong. It's very hard to break diamonds. Now let's look at the second type of structures of solids. And these guys are known as amorphous solids. Now, these guys don't really have a well structured shape and because of this, they melt over a very wide range of temperatures."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Now let's look at the second type of structures of solids. And these guys are known as amorphous solids. Now, these guys don't really have a well structured shape and because of this, they melt over a very wide range of temperatures. Examples of this include some plastics and some glass. Now, we should also mention that there is another type of solid or another type of formations that solid solids form. And this is called polymers."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Examples of this include some plastics and some glass. Now, we should also mention that there is another type of solid or another type of formations that solid solids form. And this is called polymers. Now, we can have both polymers of amorphous solids and polymers of crystalline solids. Now, when we melt polymer solids very quickly, we get amorphous solids. When we melt polymers very slowly over a very long range of time, we get crystalline solids."}, {"title": "Crystalline Solids and Amorphous Solids .txt", "text": "Now, we can have both polymers of amorphous solids and polymers of crystalline solids. Now, when we melt polymer solids very quickly, we get amorphous solids. When we melt polymers very slowly over a very long range of time, we get crystalline solids. Now, some examples of biopolymers biological polymers include DNA proteins which are basically composed of many amino acids. Macromolecules such as carbohydrates glycogen, for example, starch. All these guys are examples of biological polymers."}, {"title": "Electromotive force.txt", "text": "So that means in electrochemical cells called voltaic cells, electrons flow from the Higher electric potential electrode, the lower electric potential electrode. So that means they travel from the anode electrode to the cathode electrode. So we can define something called the electromotive force or simply EMF as the difference in this electric potential between the anode and the cathode. So to really understand what the Electric Motor Force is, we have to explore a concept called Electric Potential Energy. So electric potential energy, or simply electrical work is equal to the charge of an object times the change in electrical potential. Now, we just said the change in electrical potential between the animal and the capital is simply EMF."}, {"title": "Electromotive force.txt", "text": "So to really understand what the Electric Motor Force is, we have to explore a concept called Electric Potential Energy. So electric potential energy, or simply electrical work is equal to the charge of an object times the change in electrical potential. Now, we just said the change in electrical potential between the animal and the capital is simply EMF. So this is our EMF. So in order to understand what the EMF is, we have to really understand what charges and what electrical work or electrical potential energy is. So let's look at electrical work first."}, {"title": "Electromotive force.txt", "text": "So this is our EMF. So in order to understand what the EMF is, we have to really understand what charges and what electrical work or electrical potential energy is. So let's look at electrical work first. So electrical potential energy is similar to gravitational potential energy. So we know that any two objects say this marker and this marker will pull each other and this pull will be due to their masses. And the pull or the force is given by the gravitational constant."}, {"title": "Electromotive force.txt", "text": "So electrical potential energy is similar to gravitational potential energy. So we know that any two objects say this marker and this marker will pull each other and this pull will be due to their masses. And the pull or the force is given by the gravitational constant. Time mass one times mass two divided by the distance between them. Now, the electrical potential energy is similar to this concept, except now they will pull each other not due to mass, but due to charge. So if this object has a charge one and this object has charge two, and their charges are opposite, then they will pull each other."}, {"title": "Electromotive force.txt", "text": "Time mass one times mass two divided by the distance between them. Now, the electrical potential energy is similar to this concept, except now they will pull each other not due to mass, but due to charge. So if this object has a charge one and this object has charge two, and their charges are opposite, then they will pull each other. Now, if the charge are the same charge, they will push away. And that's the difference between electrical potential energy and gravitational potential energy. So let's look at what charge is."}, {"title": "Electromotive force.txt", "text": "Now, if the charge are the same charge, they will push away. And that's the difference between electrical potential energy and gravitational potential energy. So let's look at what charge is. Charge is simply the amount of electrons found in some object. And charge is measured in units called coulombs. And one electron."}, {"title": "Electromotive force.txt", "text": "Charge is simply the amount of electrons found in some object. And charge is measured in units called coulombs. And one electron. One electron has a charge of 1.622 times ten to the negative. 19 coulombs. So that means two electrons will be two times this amount, three electrons, three times this amount, and so on."}, {"title": "Electromotive force.txt", "text": "One electron has a charge of 1.622 times ten to the negative. 19 coulombs. So that means two electrons will be two times this amount, three electrons, three times this amount, and so on. Now, my question is, how many electrons are found in one coulomb of charge? Well, to find that answer, we must divide one coulomb by 1.622\ntimes ten to 19 coulombs per electron. And that will give us 6.24 times 18 electrons."}, {"title": "Electromotive force.txt", "text": "Now, my question is, how many electrons are found in one coulomb of charge? Well, to find that answer, we must divide one coulomb by 1.622\ntimes ten to 19 coulombs per electron. And that will give us 6.24 times 18 electrons. So this many electrons will be found in one coulomb of charge. That means a coulomb is a pretty big amount of charge. So whenever someone says one coulem of charge moves from this position to this position, that means 6.24\ntimes ten to the 18 electrons move from this position to this position."}, {"title": "Electromotive force.txt", "text": "So this many electrons will be found in one coulomb of charge. That means a coulomb is a pretty big amount of charge. So whenever someone says one coulem of charge moves from this position to this position, that means 6.24\ntimes ten to the 18 electrons move from this position to this position. That's what a coulomb is. That's what charges it's. The movement of electrons from some point A to some point B."}, {"title": "Electromotive force.txt", "text": "That's what a coulomb is. That's what charges it's. The movement of electrons from some point A to some point B. So, now that we know what electrical work is and what charges, we can rearrange this formula to give us the change in electric potential. So, by bringing this guy over to this side, we get electrical work divided by charge. And by the way, electrical work, like any work, has the units of Joules."}, {"title": "Electromotive force.txt", "text": "So, now that we know what electrical work is and what charges, we can rearrange this formula to give us the change in electric potential. So, by bringing this guy over to this side, we get electrical work divided by charge. And by the way, electrical work, like any work, has the units of Joules. So the units of electrical work divided by charge is Joules divided by Coulomb. And this equals our changing electrical potential, which is also the voltage difference. And which is also what we said before is the electromotive force or EMF."}, {"title": "Electromotive force.txt", "text": "So the units of electrical work divided by charge is Joules divided by Coulomb. And this equals our changing electrical potential, which is also the voltage difference. And which is also what we said before is the electromotive force or EMF. Now, whenever we talk about the EMF of an electrochemical cell, we could also call our EMF the cell voltage because we're talking about an electrochemical cell. And the cell voltage shows how much work can be done for every coulomb produced by a redox reaction in an electrochemical cell. So that basically says that when in an electrochemical cell, one coolant of charge moves from the anode to the cathode, x amount of work can be done."}, {"title": "Electromotive force.txt", "text": "Now, whenever we talk about the EMF of an electrochemical cell, we could also call our EMF the cell voltage because we're talking about an electrochemical cell. And the cell voltage shows how much work can be done for every coulomb produced by a redox reaction in an electrochemical cell. So that basically says that when in an electrochemical cell, one coolant of charge moves from the anode to the cathode, x amount of work can be done. So let's now look at the difference between this battery or this electric chemical cell and this smaller electrochemical cell. So let us examine the difference between this D battery and the AAA battery. Well, according to this label, it says that our electromotive force of this deep battery is exactly 1.5 volts."}, {"title": "Electromotive force.txt", "text": "So let's now look at the difference between this battery or this electric chemical cell and this smaller electrochemical cell. So let us examine the difference between this D battery and the AAA battery. Well, according to this label, it says that our electromotive force of this deep battery is exactly 1.5 volts. So EMF of D is 1.5\nvolts. What is the EMF of the AAA battery? Well, the EMF of this guy is also 1.5 volts."}, {"title": "Electromotive force.txt", "text": "So EMF of D is 1.5\nvolts. What is the EMF of the AAA battery? Well, the EMF of this guy is also 1.5 volts. So EMF is 1.5 volts. That's weird. How come this more expensive larger battery has the same EMF as the smaller and cheaper battery?"}, {"title": "Electromotive force.txt", "text": "So EMF is 1.5 volts. That's weird. How come this more expensive larger battery has the same EMF as the smaller and cheaper battery? Well, let's examine exactly what EMF is. Remember, EMF is the amount of energy produced when warm Coulomb travels from this anode to this cathode. Or said another way, when 6.24\ntimes ten to the 18 electrons travel from this point to this point, they produce 1.5 joy or joules of work."}, {"title": "Electromotive force.txt", "text": "Well, let's examine exactly what EMF is. Remember, EMF is the amount of energy produced when warm Coulomb travels from this anode to this cathode. Or said another way, when 6.24\ntimes ten to the 18 electrons travel from this point to this point, they produce 1.5 joy or joules of work. So what this means is that in this battery and in this battery, when one Coulomb charge travels from the animals to the cathode, both batteries produce the same amount of work. The same energy is released to do work. So what's the difference between the D and the AAA?"}, {"title": "Electromotive force.txt", "text": "So what this means is that in this battery and in this battery, when one Coulomb charge travels from the animals to the cathode, both batteries produce the same amount of work. The same energy is released to do work. So what's the difference between the D and the AAA? Why is this more expensive? Well, it's more expensive because it's bigger. It could hold more charge."}, {"title": "Electromotive force.txt", "text": "Why is this more expensive? Well, it's more expensive because it's bigger. It could hold more charge. That's the difference. This guy holds much more charge and eventually the charge here will run out. But this guy will still have enough charge."}, {"title": "Electromotive force.txt", "text": "That's the difference. This guy holds much more charge and eventually the charge here will run out. But this guy will still have enough charge. So the charge and electrons will continue traveling from here to here. So this guy literally houses more electrons and that's why it's more expensive. So, for example, this battery will be able to run a light bulb for say, 15 minutes, while this guy will run a light bulb for say, 4 hours."}, {"title": "Henderson Hasselbalch example .txt", "text": "In this example, we begin with 0.05 molar of Peruvic acid and 0.07 molar of sodium Peruvine. Our ka for our acid is 3.1 times ten to negative three. We want to find the PH of our buffer solution. Once we mix these two guys, there are two methods we can use use to find the PH of the buffer solution. Our first method involves the Henderson Hasselblack formula. And if you haven't seen this formula before or you don't know where it comes from, check out the link below."}, {"title": "Henderson Hasselbalch example .txt", "text": "Once we mix these two guys, there are two methods we can use use to find the PH of the buffer solution. Our first method involves the Henderson Hasselblack formula. And if you haven't seen this formula before or you don't know where it comes from, check out the link below. The second method is to simply use the Ka. So let's do the first method first. So our PH is equal to PKA plus or log of concentration of conjugate base divided by the concentration of conjugate acid."}, {"title": "Henderson Hasselbalch example .txt", "text": "The second method is to simply use the Ka. So let's do the first method first. So our PH is equal to PKA plus or log of concentration of conjugate base divided by the concentration of conjugate acid. This equals, remember, PKA is simply negative log of Ka. So this equals negative log of Ka. And our Ka is 3.1\ntimes tens of eight."}, {"title": "Henderson Hasselbalch example .txt", "text": "This equals, remember, PKA is simply negative log of Ka. So this equals negative log of Ka. And our Ka is 3.1\ntimes tens of eight. So we plug it into here plus log of this guy over this guy we get 0.07 over 0.05\napproximately equals we plug this into our calculator and we get 2.65. So that's our PH. Now let's find using the same PH using the Ka."}, {"title": "Henderson Hasselbalch example .txt", "text": "So we plug it into here plus log of this guy over this guy we get 0.07 over 0.05\napproximately equals we plug this into our calculator and we get 2.65. So that's our PH. Now let's find using the same PH using the Ka. So remember, our equation for conjugate acid and conjugate base is conjugate acid plus our water gives us conjugate base plus our hydronium ion. So let's write the equilibrium equation for this guy. So, Ka, our acid ionization constant is equal to the concentration of hydronium times the concentration of Pyruvate divided by the concentration of pyruvic acid."}, {"title": "Henderson Hasselbalch example .txt", "text": "So remember, our equation for conjugate acid and conjugate base is conjugate acid plus our water gives us conjugate base plus our hydronium ion. So let's write the equilibrium equation for this guy. So, Ka, our acid ionization constant is equal to the concentration of hydronium times the concentration of Pyruvate divided by the concentration of pyruvic acid. Now, initially we begin with some amount of this guy and some amount of this guy, right? We don't have any of this guy. Or we approximate this guy to be zero."}, {"title": "Henderson Hasselbalch example .txt", "text": "Now, initially we begin with some amount of this guy and some amount of this guy, right? We don't have any of this guy. Or we approximate this guy to be zero. At the end of our equilibrium, this guy is x. And since this and this is a ratio of one to one, this guy must be x two. So the concentration of this guy increases by x."}, {"title": "Henderson Hasselbalch example .txt", "text": "At the end of our equilibrium, this guy is x. And since this and this is a ratio of one to one, this guy must be x two. So the concentration of this guy increases by x. The concentration of this guy increases by x, and the concentration of this guy decreases by x because this guy dissociates into this guy. So we get our hydronium concentration, we represent as x. Our pyruvate concentration we present as x because we begin with this concentration divide by since we begin with this amount of Peruvic acid and then some dissociates into the conjugate base, we say that it's 0.5 minus x."}, {"title": "Henderson Hasselbalch example .txt", "text": "The concentration of this guy increases by x, and the concentration of this guy decreases by x because this guy dissociates into this guy. So we get our hydronium concentration, we represent as x. Our pyruvate concentration we present as x because we begin with this concentration divide by since we begin with this amount of Peruvic acid and then some dissociates into the conjugate base, we say that it's 0.5 minus x. Now we approximate because the x is much smaller than 0.5 or 0.7 or 0.5. We approximate this to be 0.7. X divided by 0.5 equals our Ka."}, {"title": "Henderson Hasselbalch example .txt", "text": "Now we approximate because the x is much smaller than 0.5 or 0.7 or 0.5. We approximate this to be 0.7. X divided by 0.5 equals our Ka. So equals 3.1 times ten negative three. We bring the 0.5\nover. Then we divide by 0.7 and we get x equals 0.221."}, {"title": "Henderson Hasselbalch example .txt", "text": "So equals 3.1 times ten negative three. We bring the 0.5\nover. Then we divide by 0.7 and we get x equals 0.221. And notice that this is indeed much smaller than 0.5 or zero point 75. So our approximation was accurate. Now, to find a PH, we basically plug in the concentration into our PH into our negative log."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "Today we're going to look at a very simple concept but one that's confused very often. We're going to look at avocado's number and moles. Now, avogadjo's number, just like any other number, is simply a number but it's a very large number. More specifically, it's 6.022 times ten to 23. So it's a very, very big number. Now, avogatro number is usually used in relation with moles."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "More specifically, it's 6.022 times ten to 23. So it's a very, very big number. Now, avogatro number is usually used in relation with moles. Now, a mole which is represented by lowercase N is a group that consists of an avogadro's number of anything. Now, in the same way that a dozen represents a group of twelve, a mole represents a group of this many things. Now, we could have a dozen roses, a dozen eggs, a dozen chickens, a dozen chairs, cars, books, people."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "Now, a mole which is represented by lowercase N is a group that consists of an avogadro's number of anything. Now, in the same way that a dozen represents a group of twelve, a mole represents a group of this many things. Now, we could have a dozen roses, a dozen eggs, a dozen chickens, a dozen chairs, cars, books, people. In the same way we can have a mole of anything. Now, the only thing is a mole is usually used for very small things. For example, we can talk about a mole of protons, a mole of electrons, a mole of atoms, a mole of molecules."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "In the same way we can have a mole of anything. Now, the only thing is a mole is usually used for very small things. For example, we can talk about a mole of protons, a mole of electrons, a mole of atoms, a mole of molecules. Because these things are very very small it usually doesn't make sense to say a mole of people because that's impossible. We can't have a mole of people because we only have 7 billion people in the world. So moles are used for very small quantities of something."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "Because these things are very very small it usually doesn't make sense to say a mole of people because that's impossible. We can't have a mole of people because we only have 7 billion people in the world. So moles are used for very small quantities of something. For example, we can talk about a mole of atoms which is 6.22 times ten to 23 atoms. But we can't really or it doesn't make sense to talk about a mole of books or a mole of x because I don't think we have this many eggs in the world. Maybe we do, but it's just a very large number."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "For example, we can talk about a mole of atoms which is 6.22 times ten to 23 atoms. But we can't really or it doesn't make sense to talk about a mole of books or a mole of x because I don't think we have this many eggs in the world. Maybe we do, but it's just a very large number. Now, moles and a bogondo's number can be used to answer the following questions. For example how many molecules of water are in 20 grams of water? Now, to answer this question we have to first figure out how many moles are in 20 grams of water."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "Now, moles and a bogondo's number can be used to answer the following questions. For example how many molecules of water are in 20 grams of water? Now, to answer this question we have to first figure out how many moles are in 20 grams of water. And then we multiply that by our avocado's number remembering that 1 mol of anything is this many atoms or molecules in our case. Now, let's take our amount in grams of water and divide it by our molecular weight of water which we can find on the periodic table which is 18 grams/mol. The way we found that is we simply added up the atomic weight of oxygen plus two atomic weights of H because we have two H atoms."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "And then we multiply that by our avocado's number remembering that 1 mol of anything is this many atoms or molecules in our case. Now, let's take our amount in grams of water and divide it by our molecular weight of water which we can find on the periodic table which is 18 grams/mol. The way we found that is we simply added up the atomic weight of oxygen plus two atomic weights of H because we have two H atoms. So 16 for oxygen and two times one for H two gives us 18 grams/mol for a single water molecule. So 20 grams of water divided by 18 grams/mol we see that the grams cancel, the moles go on top and we get one point moles of water. So this means within our 20 grams of water there are one point eleven moles of water."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "So 16 for oxygen and two times one for H two gives us 18 grams/mol for a single water molecule. So 20 grams of water divided by 18 grams/mol we see that the grams cancel, the moles go on top and we get one point moles of water. So this means within our 20 grams of water there are one point eleven moles of water. Now, we know that in 1 mol of anything there are this many molecules. So to find the number of molecules in one point eleven moles of water we simply multiply this number by Avogastro's number. So we get 6.22 times ten to the 23 atoms."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "Now, we know that in 1 mol of anything there are this many molecules. So to find the number of molecules in one point eleven moles of water we simply multiply this number by Avogastro's number. So we get 6.22 times ten to the 23 atoms. Or actually, in our case, this should be molecules. Molecules per mole, times one point eleven moles of water. So the moles cancel and we're left with 6.68 times tens of 23 molecules of water."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "Or actually, in our case, this should be molecules. Molecules per mole, times one point eleven moles of water. So the moles cancel and we're left with 6.68 times tens of 23 molecules of water. So this is how many molecules are found in 20 grams of water. Now, suppose I asked the question, how many atoms are found in 20 grams of water? Well, that means we take this number and multiply it by three."}, {"title": "Avogadro\u2019s Number and Moles .txt", "text": "So this is how many molecules are found in 20 grams of water. Now, suppose I asked the question, how many atoms are found in 20 grams of water? Well, that means we take this number and multiply it by three. Why? Well, this is a molecule, but an atom represents a single atom within this molecule. So we have one atom, two three atoms, one oxygen and two H atoms."}, {"title": "Resonance Forms Example .txt", "text": "Now, this lecture we're going to do three examples shown here, and our goal will be to find as many resident forms as possible. So remember, whenever we're trying to do resonance and we're trying to find resonant forms, we're only moving electrons. The atoms never actually move. So let's redraw this molecule with the atoms being in the same exact place as shown here. So we have the H atom, the C, the H atom here. We have our N atom and the two H's here."}, {"title": "Resonance Forms Example .txt", "text": "So let's redraw this molecule with the atoms being in the same exact place as shown here. So we have the H atom, the C, the H atom here. We have our N atom and the two H's here. So notice our atoms haven't actually moved. But what has happened? Well, what can we do here?"}, {"title": "Resonance Forms Example .txt", "text": "So notice our atoms haven't actually moved. But what has happened? Well, what can we do here? Well, look at this plus charge here. We don't like having charges. Charges means destabilization."}, {"title": "Resonance Forms Example .txt", "text": "Well, look at this plus charge here. We don't like having charges. Charges means destabilization. We want to stabilize this lowest dot structure by removing this plus one charge. The way we can do that is basically take this double bond here, or take these two electrons, a pair of electrons, and use arrow formulasm to move these electrons here. Remember, an arrow, a double headed arrow simply means that a pair of electrons is being moved."}, {"title": "Resonance Forms Example .txt", "text": "We want to stabilize this lowest dot structure by removing this plus one charge. The way we can do that is basically take this double bond here, or take these two electrons, a pair of electrons, and use arrow formulasm to move these electrons here. Remember, an arrow, a double headed arrow simply means that a pair of electrons is being moved. So these two electrons are being moved from here on to here. And so let's draw our two electrons here. Now, what has happened is this N now has five electrons, and that means this has been neutralized to a neutral charge."}, {"title": "Resonance Forms Example .txt", "text": "So these two electrons are being moved from here on to here. And so let's draw our two electrons here. Now, what has happened is this N now has five electrons, and that means this has been neutralized to a neutral charge. But now this carbon has a positive charge. And so let's draw a plus charge here. And this concludes our resonance forms."}, {"title": "Resonance Forms Example .txt", "text": "But now this carbon has a positive charge. And so let's draw a plus charge here. And this concludes our resonance forms. For this molecule, we have two major resonance forms. So let's go to part two. In part two, we have the following molecule, where we have two oxygen, carbon, carbon, and three HS."}, {"title": "Resonance Forms Example .txt", "text": "For this molecule, we have two major resonance forms. So let's go to part two. In part two, we have the following molecule, where we have two oxygen, carbon, carbon, and three HS. So our goal will be to draw as many resonance forms or as many major resin forms as possible. So let's begin by moving electrons. Well, what's one way that we can move electrons here?"}, {"title": "Resonance Forms Example .txt", "text": "So our goal will be to draw as many resonance forms or as many major resin forms as possible. So let's begin by moving electrons. Well, what's one way that we can move electrons here? Well, we can basically take a pair of electrons here. We can create a double bond here, and these two electrons can be moved onto this oxygen, and we will create the following molecule or compound. Remember, our atoms have not actually moved, so our atoms remain the same."}, {"title": "Resonance Forms Example .txt", "text": "Well, we can basically take a pair of electrons here. We can create a double bond here, and these two electrons can be moved onto this oxygen, and we will create the following molecule or compound. Remember, our atoms have not actually moved, so our atoms remain the same. What does move are electrons. So now this has essentially flipped. Now we have a negative charge on this upper oxygen and a neutral charge on the bottom."}, {"title": "Resonance Forms Example .txt", "text": "What does move are electrons. So now this has essentially flipped. Now we have a negative charge on this upper oxygen and a neutral charge on the bottom. Well, what's another way that we can rearrange things? Well, we can surely take this bond here, the double bond, the pair of electrons, and move it onto this oxygen here. So let's do that."}, {"title": "Resonance Forms Example .txt", "text": "Well, what's another way that we can rearrange things? Well, we can surely take this bond here, the double bond, the pair of electrons, and move it onto this oxygen here. So let's do that. So once again, keeping in mind that the resonant form represents an arrow that looks like this. And once again, no atoms are moved. Now, we have the following picture."}, {"title": "Resonance Forms Example .txt", "text": "So once again, keeping in mind that the resonant form represents an arrow that looks like this. And once again, no atoms are moved. Now, we have the following picture. In this picture, we have now developed a minus on top, a minus on the bottom, and we have a plus one on the carbon. So now we have the following species. Now, notice here we have only one negative charge, one negative charge."}, {"title": "Resonance Forms Example .txt", "text": "In this picture, we have now developed a minus on top, a minus on the bottom, and we have a plus one on the carbon. So now we have the following species. Now, notice here we have only one negative charge, one negative charge. But in this rest and form, we have two negative charges and a positive charge. Remember, whenever we have a lot of charge, a lot of charge stabilizes the structure. And that basically means that these two resonant forms will be more important than this resin form."}, {"title": "Resonance Forms Example .txt", "text": "But in this rest and form, we have two negative charges and a positive charge. Remember, whenever we have a lot of charge, a lot of charge stabilizes the structure. And that basically means that these two resonant forms will be more important than this resin form. So these are our major resonant forms. Now let's jump to part three. In part three, we basically have a very similar structure to this, except now we have an H bonded to our oxygen, so we have a neutral atom."}, {"title": "Resonance Forms Example .txt", "text": "So these are our major resonant forms. Now let's jump to part three. In part three, we basically have a very similar structure to this, except now we have an H bonded to our oxygen, so we have a neutral atom. So let's continue and let's draw what's one other way that we can draw this in terms of resonance? Well, we can surely take this pair of electron, place it here, and these electrons will move on to this oxygen, and we will develop the following lewis dot structure or resin four. Remember, our atoms don't actually move."}, {"title": "Resonance Forms Example .txt", "text": "So let's continue and let's draw what's one other way that we can draw this in terms of resonance? Well, we can surely take this pair of electron, place it here, and these electrons will move on to this oxygen, and we will develop the following lewis dot structure or resin four. Remember, our atoms don't actually move. Atoms stay the same. But now we have the following picture. So here we have an H two."}, {"title": "Resonance Forms Example .txt", "text": "Atoms stay the same. But now we have the following picture. So here we have an H two. Now, notice what happens. This has 12345. So this develops a plus one charge."}, {"title": "Resonance Forms Example .txt", "text": "Now, notice what happens. This has 12345. So this develops a plus one charge. This develops a negative one charge. And so an overall net charge on this higher molecule is still zero, just like it is here. So let's draw one other resin form for this molecule or compound."}, {"title": "Resonance Forms Example .txt", "text": "This develops a negative one charge. And so an overall net charge on this higher molecule is still zero, just like it is here. So let's draw one other resin form for this molecule or compound. What's another way we can draw it? Well, what if we just simply take this bond and move it back here? Well, let's try that."}, {"title": "Resonance Forms Example .txt", "text": "What's another way we can draw it? Well, what if we just simply take this bond and move it back here? Well, let's try that. So we will have the following picture. Now, this carbon will develop a plus one charge. This will be neutral."}, {"title": "Resonance Forms Example .txt", "text": "So we will have the following picture. Now, this carbon will develop a plus one charge. This will be neutral. Let's build our electrons here. And this guy here will have a negative charge. So we're going to have an overall net charge of zero."}, {"title": "Resonance Forms Example .txt", "text": "Let's build our electrons here. And this guy here will have a negative charge. So we're going to have an overall net charge of zero. So what happened here? Well, here we have electrons that went onto here. So this concludes our major resonance structures."}, {"title": "The pH scale .txt", "text": "And that's because pure water dissociates into H plus and hydroxide. So at any given time, we can measure the concentration of of this guy in terms of molar amount. Now, if you want to learn more about how H plus relates to acids and oh, Myers relates to bases, check out the link below. So what happens when we add some substance to our pure water, our PR substance? Well, for example, suppose we add HDL to our water. What happens?"}, {"title": "The pH scale .txt", "text": "So what happens when we add some substance to our pure water, our PR substance? Well, for example, suppose we add HDL to our water. What happens? Well, Htl dissociates its a Hydride ion and a chloride ion. So our concentration or amount in molar of H plus will increase because we have the H coming from water and the H coming from HDL. And so our concentration of our solution will increase."}, {"title": "The pH scale .txt", "text": "Well, Htl dissociates its a Hydride ion and a chloride ion. So our concentration or amount in molar of H plus will increase because we have the H coming from water and the H coming from HDL. And so our concentration of our solution will increase. Now we can measure the concentrations in terms of molar amount, but the values we get can range from anywhere from ten to the negative 14 molar. That's a very small number. That's one divided by 100 trillion."}, {"title": "The pH scale .txt", "text": "Now we can measure the concentrations in terms of molar amount, but the values we get can range from anywhere from ten to the negative 14 molar. That's a very small number. That's one divided by 100 trillion. And they could become as large as ten molar. Okay, so this becomes a problem. This is very inconvenient because say if you want to graph concentration in molar versus temperature, a graph would not be possible because our y scale would be just too big."}, {"title": "The pH scale .txt", "text": "And they could become as large as ten molar. Okay, so this becomes a problem. This is very inconvenient because say if you want to graph concentration in molar versus temperature, a graph would not be possible because our y scale would be just too big. The range would be too big. So scientists came up with a way or a convenient way of expressing the concentration of our hydrogen ions in solution. And this convenient way is called a PH scale."}, {"title": "The pH scale .txt", "text": "The range would be too big. So scientists came up with a way or a convenient way of expressing the concentration of our hydrogen ions in solution. And this convenient way is called a PH scale. The PH scale is defined by the following formula. PH of our solution is equal to negative log of the concentration of hydrogen ions. And that's equivalent to saying negative log of the concentration of hydronium ions."}, {"title": "The pH scale .txt", "text": "The PH scale is defined by the following formula. PH of our solution is equal to negative log of the concentration of hydrogen ions. And that's equivalent to saying negative log of the concentration of hydronium ions. Because this guy and this guy are really one and the same. So let's see why logs are convenient. But first, let's see what logs are."}, {"title": "The pH scale .txt", "text": "Because this guy and this guy are really one and the same. So let's see why logs are convenient. But first, let's see what logs are. Logs are simply another way of representing exponents. And if you want to find an exponent, you use logs. For example, suppose we're given this equation here."}, {"title": "The pH scale .txt", "text": "Logs are simply another way of representing exponents. And if you want to find an exponent, you use logs. For example, suppose we're given this equation here. So ten to the negative four is equal to one over 10,000, which is another way of saying 0.1. Okay? So on a lock scale, this can be represented in the following way."}, {"title": "The pH scale .txt", "text": "So ten to the negative four is equal to one over 10,000, which is another way of saying 0.1. Okay? So on a lock scale, this can be represented in the following way. Where this is our base, our result and our exponent. We get log base ten, log base ten. In the inside we get our result."}, {"title": "The pH scale .txt", "text": "Where this is our base, our result and our exponent. We get log base ten, log base ten. In the inside we get our result. So 0.0001\nand that equals to our exponent negative four. Here, to make this positive, I simply multiply both sides by negative one and I brought the negative to this side. And that's why this guy is positive."}, {"title": "The pH scale .txt", "text": "So 0.0001\nand that equals to our exponent negative four. Here, to make this positive, I simply multiply both sides by negative one and I brought the negative to this side. And that's why this guy is positive. So for example, suppose if I had the ten and I had this value, but this was my unknown, I could simply use this to find my exponent. That's why logs are convenient whenever you don't know our exponent, but you know the result and you know the base. You can use longs to find the exponent."}, {"title": "The pH scale .txt", "text": "So for example, suppose if I had the ten and I had this value, but this was my unknown, I could simply use this to find my exponent. That's why logs are convenient whenever you don't know our exponent, but you know the result and you know the base. You can use longs to find the exponent. That's why logs are used. So let's represent this guy as well. So this guy ten to negative five is equal to one over 100,000, which is 0.1\nand equivalently on the log scale if you're presented in terms of negative log of base ten of 0.0001, and that equals to five."}, {"title": "The pH scale .txt", "text": "That's why logs are used. So let's represent this guy as well. So this guy ten to negative five is equal to one over 100,000, which is 0.1\nand equivalently on the log scale if you're presented in terms of negative log of base ten of 0.0001, and that equals to five. Now notice what happens. A decrease of tenfold going from this guy to this guy increases by increment of only one, going from four to five. Now, let's represent ten to negative six and ten to negative seven as well."}, {"title": "The pH scale .txt", "text": "Now notice what happens. A decrease of tenfold going from this guy to this guy increases by increment of only one, going from four to five. Now, let's represent ten to negative six and ten to negative seven as well. Well, we get this result and we get six and seven. Now notice this. Going from tens of negative four to ten to negative six is a decrease in 100 fold."}, {"title": "The pH scale .txt", "text": "Well, we get this result and we get six and seven. Now notice this. Going from tens of negative four to ten to negative six is a decrease in 100 fold. And going from 10th to negative four to ten to negative seven is a decrease in 1000 fold. But on a PH scale, on a log scale, it only decreases by an increment of two and an increment of three. And that means now, we can graph these guys."}, {"title": "The pH scale .txt", "text": "And going from 10th to negative four to ten to negative seven is a decrease in 1000 fold. But on a PH scale, on a log scale, it only decreases by an increment of two and an increment of three. And that means now, we can graph these guys. We can graph PH versus, say, temperature, right? And this will create a very good graph. That's exactly why we use PH in the PH scale, just simply because it's a convenient way of converting inconvenient Molar numbers to convenient numbers."}, {"title": "The pH scale .txt", "text": "We can graph PH versus, say, temperature, right? And this will create a very good graph. That's exactly why we use PH in the PH scale, just simply because it's a convenient way of converting inconvenient Molar numbers to convenient numbers. Okay? Now, one last note about this is that our range for PH can range anywhere from zero to 14, where zero is the smallest possible PH and 14 is the largest possible PH. So let's look at this size."}, {"title": "The pH scale .txt", "text": "Okay? Now, one last note about this is that our range for PH can range anywhere from zero to 14, where zero is the smallest possible PH and 14 is the largest possible PH. So let's look at this size. So at 25 degrees Celsius, we say that a PH of seven is neutral. And that's because PH of seven is smack in the middle. PH of water is seven because we have just as many H ions as we have hydroxide ions."}, {"title": "The pH scale .txt", "text": "So at 25 degrees Celsius, we say that a PH of seven is neutral. And that's because PH of seven is smack in the middle. PH of water is seven because we have just as many H ions as we have hydroxide ions. So our PH is right in the middle. Remember, this represents acidity, and this represents how basic something is. Now, a PH of less than seven represents something that's acidic."}, {"title": "The pH scale .txt", "text": "So our PH is right in the middle. Remember, this represents acidity, and this represents how basic something is. Now, a PH of less than seven represents something that's acidic. And that's because as our PH decreases, our concentration of H ions increases. Because, look, as we go from seven to four, we go from ten to negative seven to ten to negative four Molar amounts of H plus in the same way that a PH of seven is basic. And that's because as we go down or as we go up on the PH scale, we go down, we decrease on the Molar scale."}, {"title": "Triple bond of Acetylene.txt", "text": "So this compound is composed of two carbon atoms and two H atoms. The bonds between the carbon and the H are one SSP hybridized. And the bond, a sigma bond between the two carbon atoms is SPSP hybridized. Now, these two other bonds are the two Pi bonds, as we'll see in just a second. So let's try to build our molecular orbital diagram for this compound using the atomic orbitals of the individual atoms. So once again, we have our two H atoms."}, {"title": "Triple bond of Acetylene.txt", "text": "Now, these two other bonds are the two Pi bonds, as we'll see in just a second. So let's try to build our molecular orbital diagram for this compound using the atomic orbitals of the individual atoms. So once again, we have our two H atoms. They're in the neutral state. So that means they each have one electron each in the one S orbital. So here we have the one S orbital of the first H and the one S orbital of the second H. Now, this black dot is simply our electron, the balanced electron, to be specific."}, {"title": "Triple bond of Acetylene.txt", "text": "They're in the neutral state. So that means they each have one electron each in the one S orbital. So here we have the one S orbital of the first H and the one S orbital of the second H. Now, this black dot is simply our electron, the balanced electron, to be specific. So now let's take our two carbon atoms. So, once again, these guys are all separated. They haven't yet combined to form our ethylene compound."}, {"title": "Triple bond of Acetylene.txt", "text": "So now let's take our two carbon atoms. So, once again, these guys are all separated. They haven't yet combined to form our ethylene compound. So here we have one carbon, and here we have the second carbon. So two of the orbitals for this carbon are SP hybridized. And the same thing goes for this carbon."}, {"title": "Triple bond of Acetylene.txt", "text": "So here we have one carbon, and here we have the second carbon. So two of the orbitals for this carbon are SP hybridized. And the same thing goes for this carbon. These two orbitals are SP hybridized, shown here as well. Now, these two orbitals are pure two p orbitals, and we have one electron in each. So altogether, each carbon donates four balance electrons."}, {"title": "Triple bond of Acetylene.txt", "text": "These two orbitals are SP hybridized, shown here as well. Now, these two orbitals are pure two p orbitals, and we have one electron in each. So altogether, each carbon donates four balance electrons. So we have one balanced electron each here and four balance electrons here. Each carbon also has two electrons in the one S orbital. But that's not shown because the one S orbital does not react."}, {"title": "Triple bond of Acetylene.txt", "text": "So we have one balanced electron each here and four balance electrons here. Each carbon also has two electrons in the one S orbital. But that's not shown because the one S orbital does not react. So now these two carbons can overlap. And to be specific, these SP hybridized orbitals will overlap, and they will each share an electron. Likewise, the one S orbital of this H will overlap with the SP orbital of this carbon."}, {"title": "Triple bond of Acetylene.txt", "text": "So now these two carbons can overlap. And to be specific, these SP hybridized orbitals will overlap, and they will each share an electron. Likewise, the one S orbital of this H will overlap with the SP orbital of this carbon. And likewise, the one orbital here will interact with the SP orbital of the second carbon. And we will form our ethnic compound, shown here. So this is a molecular orbital diagram of our cyclist alkyne."}, {"title": "Triple bond of Acetylene.txt", "text": "And likewise, the one orbital here will interact with the SP orbital of the second carbon. And we will form our ethnic compound, shown here. So this is a molecular orbital diagram of our cyclist alkyne. So here we have the SPST hybridized orbital formed from the overlap of each individual SP hybridized orbital. And here we have our one SSP on both sides bonds created by the overlap of the one S and SP hybridized orbital of the carbon. And notice we still have these pure p orbitals."}, {"title": "Triple bond of Acetylene.txt", "text": "So here we have the SPST hybridized orbital formed from the overlap of each individual SP hybridized orbital. And here we have our one SSP on both sides bonds created by the overlap of the one S and SP hybridized orbital of the carbon. And notice we still have these pure p orbitals. So we have one pure P orbital in the Y direction, y being up, and we have one pure P orbital in the z direction C being coming out of the board. And what will happen is there will be an overlap between our two p orbitals that are parallel in the same plane. So these two guys, these two orbitals, will interact, and these two orbitals will interact."}, {"title": "Triple bond of Acetylene.txt", "text": "So we have one pure P orbital in the Y direction, y being up, and we have one pure P orbital in the z direction C being coming out of the board. And what will happen is there will be an overlap between our two p orbitals that are parallel in the same plane. So these two guys, these two orbitals, will interact, and these two orbitals will interact. Notice that since these two orbitals are along the y axis and these two orbitals are along the z axis, they're perpendicular to one another. And in fact, both of these two appear to the orbitals are perpendicular to these orbitals to the one s SP and to the SPSP. In other words, the SP and the SP and the one s and the SP are along the X axis, while these orbitals are along the Y axis and these orbitals are along the Z axis."}, {"title": "Triple bond of Acetylene.txt", "text": "Notice that since these two orbitals are along the y axis and these two orbitals are along the z axis, they're perpendicular to one another. And in fact, both of these two appear to the orbitals are perpendicular to these orbitals to the one s SP and to the SPSP. In other words, the SP and the SP and the one s and the SP are along the X axis, while these orbitals are along the Y axis and these orbitals are along the Z axis. So all of these guys are perpendicular to one another. So now let's look at the energy diagram for all these interactions. So the One s interacts with the SP to form the one s SP?"}, {"title": "Triple bond of Acetylene.txt", "text": "So all of these guys are perpendicular to one another. So now let's look at the energy diagram for all these interactions. So the One s interacts with the SP to form the one s SP? Molecular orbital. In other words, notice that the one s is lower in energy than the SP because the one s has more s character than the SP does. And so when they interact, they form two types of molecular orbitals bonding and the antibonding molecular orbitals."}, {"title": "Triple bond of Acetylene.txt", "text": "Molecular orbital. In other words, notice that the one s is lower in energy than the SP because the one s has more s character than the SP does. And so when they interact, they form two types of molecular orbitals bonding and the antibonding molecular orbitals. So since this is higher in energy, electrons will not be found in this rather destabilizing molecular orbital. So let's look at the SP and SP interaction. So when the carbon and the carbon interact to form our bond, our sigma bond, we have an SP interaction within the atomic orbitals to form our molecular orbitals."}, {"title": "Triple bond of Acetylene.txt", "text": "So since this is higher in energy, electrons will not be found in this rather destabilizing molecular orbital. So let's look at the SP and SP interaction. So when the carbon and the carbon interact to form our bond, our sigma bond, we have an SP interaction within the atomic orbitals to form our molecular orbitals. Or bonding sigma bonding. Molecular orbital. Likewise."}, {"title": "Triple bond of Acetylene.txt", "text": "Or bonding sigma bonding. Molecular orbital. Likewise. Since we input two atomic orbitals, we form two molecular orbitals. So we also form the antibiotic SP SP sigma molecular orbital. And since this is higher energy no electrons go into here."}, {"title": "Triple bond of Acetylene.txt", "text": "Since we input two atomic orbitals, we form two molecular orbitals. So we also form the antibiotic SP SP sigma molecular orbital. And since this is higher energy no electrons go into here. Electrons are only found in the lower stabilizing orbital. Now let's look at the two p. Two p interaction. In other words, the Pi bonding."}, {"title": "Triple bond of Acetylene.txt", "text": "Electrons are only found in the lower stabilizing orbital. Now let's look at the two p. Two p interaction. In other words, the Pi bonding. Remember, these bonds here are all known as Sigma bonds. While these bonds are known, the ones in red are known as pi bonds. So when the two p and two p interact, which are, by the way, higher in energy than the SP, they form the two p two p pi bonding molecular orbital."}, {"title": "Triple bond of Acetylene.txt", "text": "Remember, these bonds here are all known as Sigma bonds. While these bonds are known, the ones in red are known as pi bonds. So when the two p and two p interact, which are, by the way, higher in energy than the SP, they form the two p two p pi bonding molecular orbital. And the two p two pi antibonding molecular orbital. And once again, both electrons go into the lower in energy pi bonding molecular orbital. And no electrons go into the two p two pi antibonding molecular orbital."}, {"title": "Acid Ionization Constant Example .txt", "text": "In this example, we begin with a 0.03 molar of vitonic acid and a PH of 3.5 or 25 degrees Celsius. We want to find the hydroxide concentration as well as the PKA of our botanic acid. Now, we'll find this guy first. So, in step one and three, I find that PKA. And step two, I find the hydroxide filtration. So let's skip to step two."}, {"title": "Acid Ionization Constant Example .txt", "text": "So, in step one and three, I find that PKA. And step two, I find the hydroxide filtration. So let's skip to step two. Now, the first thing I do is I use the formula PH plus poh equals 14. Now, if you're not certain about this formula and you don't know where this formula comes from, check out the link below. So, I simply take our PH, plug it in into this number, and we get 14 equals 3.5 plus poh subtract by 3.5\nand I get poh is equal to 10.5."}, {"title": "Acid Ionization Constant Example .txt", "text": "Now, the first thing I do is I use the formula PH plus poh equals 14. Now, if you're not certain about this formula and you don't know where this formula comes from, check out the link below. So, I simply take our PH, plug it in into this number, and we get 14 equals 3.5 plus poh subtract by 3.5\nand I get poh is equal to 10.5. And now I simply apply the formula for poh. So, Poh is equal to 10.5,\nwhich is equal to negative log of the concentration of oh. Now, I convert this log to exponents."}, {"title": "Acid Ionization Constant Example .txt", "text": "And now I simply apply the formula for poh. So, Poh is equal to 10.5,\nwhich is equal to negative log of the concentration of oh. Now, I convert this log to exponents. By first bringing the negative sign to this side, I see that my base is ten. And I raise that base ten to the negative 10.5 exponent. So, my concentration of hydroxide is equal to ten to the negative 10.5."}, {"title": "Acid Ionization Constant Example .txt", "text": "By first bringing the negative sign to this side, I see that my base is ten. And I raise that base ten to the negative 10.5 exponent. So, my concentration of hydroxide is equal to ten to the negative 10.5. Plug this into my calculator and get 3.16 times ten to the negative eleven molar. So, this is my concentration of hydroxide. So, I found the first part."}, {"title": "Acid Ionization Constant Example .txt", "text": "Plug this into my calculator and get 3.16 times ten to the negative eleven molar. So, this is my concentration of hydroxide. So, I found the first part. Now let's look at a PKA. So, before anything else, we must write the dissociation reaction for our acid. So, our acid in the aqueous state reacts with water and liquid state to form a conjugate base and a conjugate acid."}, {"title": "Acid Ionization Constant Example .txt", "text": "Now let's look at a PKA. So, before anything else, we must write the dissociation reaction for our acid. So, our acid in the aqueous state reacts with water and liquid state to form a conjugate base and a conjugate acid. Now, our goal is to find the Ka. And then we can find the PKA by using the lock formula. So, what is Ka?"}, {"title": "Acid Ionization Constant Example .txt", "text": "Now, our goal is to find the Ka. And then we can find the PKA by using the lock formula. So, what is Ka? Well, Ka is the ratio of these guys to this guy. Remember, we don't use the liquid component, so we don't really care about this guy. So we can forget about this guy for now."}, {"title": "Acid Ionization Constant Example .txt", "text": "Well, Ka is the ratio of these guys to this guy. Remember, we don't use the liquid component, so we don't really care about this guy. So we can forget about this guy for now. What we want to find is that equilibrium concentration of this guy, this guy and this guy. Then we plug that in into our equilibrium expression and find our Ka. And then we find the PKA by using logs."}, {"title": "Acid Ionization Constant Example .txt", "text": "What we want to find is that equilibrium concentration of this guy, this guy and this guy. Then we plug that in into our equilibrium expression and find our Ka. And then we find the PKA by using logs. So we know that we begin. Our initial condition is 0.3 molar of Betonic acid. So our initial condition is we have 0.3 of this guy, we don't care about this guy, we have none of this guy."}, {"title": "Acid Ionization Constant Example .txt", "text": "So we know that we begin. Our initial condition is 0.3 molar of Betonic acid. So our initial condition is we have 0.3 of this guy, we don't care about this guy, we have none of this guy. And we have a very small amount of this guy. Because remember, in pure water, some water dissociates into HBO plus. And if you're not sure about what I'm talking about, check out the link below."}, {"title": "Acid Ionization Constant Example .txt", "text": "And we have a very small amount of this guy. Because remember, in pure water, some water dissociates into HBO plus. And if you're not sure about what I'm talking about, check out the link below. But this amount is so small, it tends to the negative seven molar. And that's very small. And so we could approximate this to be zero."}, {"title": "Acid Ionization Constant Example .txt", "text": "But this amount is so small, it tends to the negative seven molar. And that's very small. And so we could approximate this to be zero. So initially, we have zero of this, zero of this and 0.3\nmolar of our acid. We want to find the equilibrium concentration. So in equilibrium, our PH is 3.5."}, {"title": "Acid Ionization Constant Example .txt", "text": "So initially, we have zero of this, zero of this and 0.3\nmolar of our acid. We want to find the equilibrium concentration. So in equilibrium, our PH is 3.5. That means we could use the lock formula to find the concentration of this guy. Now, after we find the concentration of this guy, the concentration of this guy is the same thing. That's because 1 mol and 1 mol, our ratio of this guy to this guy is one to one."}, {"title": "Acid Ionization Constant Example .txt", "text": "That means we could use the lock formula to find the concentration of this guy. Now, after we find the concentration of this guy, the concentration of this guy is the same thing. That's because 1 mol and 1 mol, our ratio of this guy to this guy is one to one. So that means whatever the amount and molar of this guy is, this guy is the same. So let's find the concentration of hydronium. So, PH equals 3.5\nto my given information is equal to negative log of the concentration of hydrogen."}, {"title": "Acid Ionization Constant Example .txt", "text": "So that means whatever the amount and molar of this guy is, this guy is the same. So let's find the concentration of hydronium. So, PH equals 3.5\nto my given information is equal to negative log of the concentration of hydrogen. And that means we have to convert this guy to exponents. So our base is ten. We bring a negative over."}, {"title": "Acid Ionization Constant Example .txt", "text": "And that means we have to convert this guy to exponents. So our base is ten. We bring a negative over. We get base ten to the negative 3.5 because this is my exponent. So we get our concentration is equal to ten to negative 3.5. And we get 3.16 times ten to negative four molar."}, {"title": "Acid Ionization Constant Example .txt", "text": "We get base ten to the negative 3.5 because this is my exponent. So we get our concentration is equal to ten to negative 3.5. And we get 3.16 times ten to negative four molar. So this is my concentration of hydronium as well as my concentration of this conjugate base. And that's because there's a one to one ratio. And this was a two and this was a one."}, {"title": "Acid Ionization Constant Example .txt", "text": "So this is my concentration of hydronium as well as my concentration of this conjugate base. And that's because there's a one to one ratio. And this was a two and this was a one. Then I would have to multiply this guy by two to find this. But since it's one to one, they're equal. Okay, so now we have my equilibrium concentration of my conjugate base and my conjugate acid."}, {"title": "Acid Ionization Constant Example .txt", "text": "Then I would have to multiply this guy by two to find this. But since it's one to one, they're equal. Okay, so now we have my equilibrium concentration of my conjugate base and my conjugate acid. And I have to find my equilibrium concentration of my initial acid. So initially at 0.3 and at equilibrium, that means I have to have a little bit less than 0.03\nbecause a little bit of it is dissociated into this guy. But how much less is exactly?"}, {"title": "Acid Ionization Constant Example .txt", "text": "And I have to find my equilibrium concentration of my initial acid. So initially at 0.3 and at equilibrium, that means I have to have a little bit less than 0.03\nbecause a little bit of it is dissociated into this guy. But how much less is exactly? Well, the initial amount of vitonic acid minus final amount of conjugate base. So this guy in molar minus this guy in molar will give me the amount of vitamin acid left over. So 0.3 the amount I began with minus 3.116 times ten to negative four."}, {"title": "Acid Ionization Constant Example .txt", "text": "Well, the initial amount of vitonic acid minus final amount of conjugate base. So this guy in molar minus this guy in molar will give me the amount of vitamin acid left over. So 0.3 the amount I began with minus 3.116 times ten to negative four. This guy gives me how much of my vitonic acid is left over at equilibrium or 0.2968 molar. So now I have all my equilibrium concentrations. So I simply write my equilibrium expression."}, {"title": "Acid Ionization Constant Example .txt", "text": "This guy gives me how much of my vitonic acid is left over at equilibrium or 0.2968 molar. So now I have all my equilibrium concentrations. So I simply write my equilibrium expression. So Ka is equal to the concentration of this guy times the concentration of this guy divided by the concentration of this guy. So, 3.16\ntimes ten to the negative four times 3.16 times ten to negative four divided by 0.2968 gives me 3.36\ntimes ten to negative six. So that's our Ka."}, {"title": "Acid Ionization Constant Example .txt", "text": "So Ka is equal to the concentration of this guy times the concentration of this guy divided by the concentration of this guy. So, 3.16\ntimes ten to the negative four times 3.16 times ten to negative four divided by 0.2968 gives me 3.36\ntimes ten to negative six. So that's our Ka. But we're not done because we want to find the PKA. So I simply use the log formula. So, PKA gives us negative log of this guy produces 5.47."}, {"title": "Degree of Unsaturation.txt", "text": "Now, if the compound is relatively simple, you can figure out what the molecular structure of that compound is using your molecular formula. But if the compound is relatively complicated and the molecular formula contains lots of different atoms, then it becomes very difficult at determining what the molecular structure is using only your molecular formula. Now, one tool that we can use to figure out what the molecular structure is is known as degree of unsaturation. So degree of unsaturation, given by the Greek letter omega, gives us the total number of pi bonds and or ring structures for that given molecular formula. And the formula to find degree of unsaturation is given by this guy here. So two N plus two minus x, and the whole thing divided by two, where N represents your number of carbons and x represents your number of hydrogen atoms."}, {"title": "Degree of Unsaturation.txt", "text": "So degree of unsaturation, given by the Greek letter omega, gives us the total number of pi bonds and or ring structures for that given molecular formula. And the formula to find degree of unsaturation is given by this guy here. So two N plus two minus x, and the whole thing divided by two, where N represents your number of carbons and x represents your number of hydrogen atoms. Now, whenever we're using this formula and we're trying to determine degree bond saturation, we have to follow three important rules. Rule number one, replace all halogens with hydrogens. Rule number two, whenever you're given a molecule or compound with oxygens or sulfur, omit the oxygens and sulfur in your formula."}, {"title": "Degree of Unsaturation.txt", "text": "Now, whenever we're using this formula and we're trying to determine degree bond saturation, we have to follow three important rules. Rule number one, replace all halogens with hydrogens. Rule number two, whenever you're given a molecule or compound with oxygens or sulfur, omit the oxygens and sulfur in your formula. And rule number three, remove nitrogens along with one hydrogen per nitrogen atom. So let's look at how we can use this rule and how we can use or how we can use this formula and these three rules. So let's begin with example one."}, {"title": "Degree of Unsaturation.txt", "text": "And rule number three, remove nitrogens along with one hydrogen per nitrogen atom. So let's look at how we can use this rule and how we can use or how we can use this formula and these three rules. So let's begin with example one. So, in example one, according to our molecular formula, we have six carbons. So our M is six, and we have twelve H atoms. So our x is twelve."}, {"title": "Degree of Unsaturation.txt", "text": "So, in example one, according to our molecular formula, we have six carbons. So our M is six, and we have twelve H atoms. So our x is twelve. So let's use our formula to find degree of unsaturation for compound one. So we have two times six plus two. Put that in parentheses."}, {"title": "Degree of Unsaturation.txt", "text": "So let's use our formula to find degree of unsaturation for compound one. So we have two times six plus two. Put that in parentheses. Minus twelve divided by two equals. So, once again, our N is six. Our x is twelve."}, {"title": "Degree of Unsaturation.txt", "text": "Minus twelve divided by two equals. So, once again, our N is six. Our x is twelve. So we get two times 612 plus 214 minus twelve, that's two divided by two. And that gives us one, because two divided by two gives us one. So our degree of unsaturation for compound one is one."}, {"title": "Degree of Unsaturation.txt", "text": "So we get two times 612 plus 214 minus twelve, that's two divided by two. And that gives us one, because two divided by two gives us one. So our degree of unsaturation for compound one is one. That means that we either have one pi bond or we have one ring structure in this molecular compound. So let's look at example two. Here we have five carbons and six H atoms."}, {"title": "Degree of Unsaturation.txt", "text": "That means that we either have one pi bond or we have one ring structure in this molecular compound. So let's look at example two. Here we have five carbons and six H atoms. So, once again, our M is five and x is six. So two times five plus two. So that's twelve minus six, that's six divided by two gives us, well, six divided by two gives us three."}, {"title": "Degree of Unsaturation.txt", "text": "So, once again, our M is five and x is six. So two times five plus two. So that's twelve minus six, that's six divided by two gives us, well, six divided by two gives us three. So, once again, our number of carbons, N is five. Two times 510 plus 212 minus six inch atoms. So that six on top and two on the bottom."}, {"title": "Degree of Unsaturation.txt", "text": "So, once again, our number of carbons, N is five. Two times 510 plus 212 minus six inch atoms. So that six on top and two on the bottom. So we have three degrees of unsaturation. So that means we can either have three pi bonds or we can have two pi bonds and one ring structure. Or two ring structures and Y pi bond and one Pi bond or three ring structures and zero Pi bonds."}, {"title": "Degree of Unsaturation.txt", "text": "So we have three degrees of unsaturation. So that means we can either have three pi bonds or we can have two pi bonds and one ring structure. Or two ring structures and Y pi bond and one Pi bond or three ring structures and zero Pi bonds. So let's look at example number three. So here we have five carbons. So our N is five h or eight h atoms."}, {"title": "Degree of Unsaturation.txt", "text": "So let's look at example number three. So here we have five carbons. So our N is five h or eight h atoms. But now we have bromine, we have two BRS, we have a halogen. And that means we have to refer to rule number one, which states replace halogens with hydrogens. So since we have two halogens, that means we have two more H atoms."}, {"title": "Degree of Unsaturation.txt", "text": "But now we have bromine, we have two BRS, we have a halogen. And that means we have to refer to rule number one, which states replace halogens with hydrogens. So since we have two halogens, that means we have two more H atoms. So we have eight plus two, a total of ten H atoms. So our x is ten. So two times five plus two, that's twelve minus ten, that's two divided by two gives us or two divided by two gives us one."}, {"title": "Degree of Unsaturation.txt", "text": "So we have eight plus two, a total of ten H atoms. So our x is ten. So two times five plus two, that's twelve minus ten, that's two divided by two gives us or two divided by two gives us one. So this compound either has one Pi bond or one ring structure. So let's look at compound number four. For example four."}, {"title": "Degree of Unsaturation.txt", "text": "So this compound either has one Pi bond or one ring structure. So let's look at compound number four. For example four. So here we have five carbons. Our N is five. We have eleven H atoms, but we also have N atoms."}, {"title": "Degree of Unsaturation.txt", "text": "So here we have five carbons. Our N is five. We have eleven H atoms, but we also have N atoms. So let's go back to our rules. According to rule number three, remove nitrogens along with one H.\nSo every time we remove a nitrogen, we remove one H. So that means we no longer have eleven HS, but we have eleven minus three HS. Because if we remove three ends, we have to remove three H's."}, {"title": "Degree of Unsaturation.txt", "text": "So let's go back to our rules. According to rule number three, remove nitrogens along with one H.\nSo every time we remove a nitrogen, we remove one H. So that means we no longer have eleven HS, but we have eleven minus three HS. Because if we remove three ends, we have to remove three H's. So we have a total of eight H atoms. So our x is eight. So our two times five plus two."}, {"title": "Degree of Unsaturation.txt", "text": "So we have a total of eight H atoms. So our x is eight. So our two times five plus two. So that gives us twelve minus eight. That gives us four divided by two. So four divided by two gives us two."}, {"title": "Degree of Unsaturation.txt", "text": "So that gives us twelve minus eight. That gives us four divided by two. So four divided by two gives us two. So that means we either have two ring structures, zero Pi bonds, one ring structure, or one Pi bond, or two Pi bonds and zero ring structures. Let's put an equal sign here. Example number five."}, {"title": "Degree of Unsaturation.txt", "text": "So that means we either have two ring structures, zero Pi bonds, one ring structure, or one Pi bond, or two Pi bonds and zero ring structures. Let's put an equal sign here. Example number five. So in this compound we now have six N. So our carbon number is six. We have eight HS, but we also have two BRS and three oxygens. Recall that whenever we see an oxygen or a sulfur, according to rule number two, we do not count them."}, {"title": "Degree of Unsaturation.txt", "text": "So in this compound we now have six N. So our carbon number is six. We have eight HS, but we also have two BRS and three oxygens. Recall that whenever we see an oxygen or a sulfur, according to rule number two, we do not count them. We omit them in our calculation. But since rule number one states that halogens count as hydrogens, we have not eight, but eight plus two. So ten h atoms."}, {"title": "Degree of Unsaturation.txt", "text": "We omit them in our calculation. But since rule number one states that halogens count as hydrogens, we have not eight, but eight plus two. So ten h atoms. So our x in this case is ten. So two times 612 plus two gives us 14 minus ten, that gives us four divided by two. So four divided by two gives us two."}, {"title": "Degree of Unsaturation.txt", "text": "So our x in this case is ten. So two times 612 plus two gives us 14 minus ten, that gives us four divided by two. So four divided by two gives us two. So once again we have six carbons. So our N is six. Two times 612 plus 214 minus ten, because eight plus two is ten."}, {"title": "Degree of Unsaturation.txt", "text": "So once again we have six carbons. So our N is six. Two times 612 plus 214 minus ten, because eight plus two is ten. So 14 minus ten is four divided by two is two. So that means in this compound we either have two Pi bonds, zero ring structures, one ring structure and y pi bonds, or two ring structures and zero pi bonds. So let's look at example number six, our last example."}, {"title": "Degree of Unsaturation.txt", "text": "So 14 minus ten is four divided by two is two. So that means in this compound we either have two Pi bonds, zero ring structures, one ring structure and y pi bonds, or two ring structures and zero pi bonds. So let's look at example number six, our last example. So now we have seven carbons. So our N is seven. We have eight H atoms and we have one BR."}, {"title": "Degree of Unsaturation.txt", "text": "So now we have seven carbons. So our N is seven. We have eight H atoms and we have one BR. So that means we have nine H atoms, but we also have three nitrogens. So we have to remove those three nitrogens. So nine minus three is six."}, {"title": "Degree of Unsaturation.txt", "text": "So that means we have nine H atoms, but we also have three nitrogens. So we have to remove those three nitrogens. So nine minus three is six. Oxygens don't count, so that means our x is six. So we have two multiplied by seven, which is 14 plus two. That gives us 16 minus we said that it was six for x."}, {"title": "Degree of Unsaturation.txt", "text": "Oxygens don't count, so that means our x is six. So we have two multiplied by seven, which is 14 plus two. That gives us 16 minus we said that it was six for x. So we get 16 minus six divided by two gives us five. So 16 minus six of ten divided by two gives us five. So this means we have some combination of pi bonds and ring structures, such that when I add up my ring structures and pi bonds, I get five."}, {"title": "Work.txt", "text": "In the previous video, we discussed one type of energy transfers, namely heat transfers. We said that energy could travel from a hot object to a cold object and that three types of heat transfers exist convection, conduction, and radiation. Another form of energy transfer is work, and there are only Two types of energy transfers heat and work. So if it's not heat, it must be work and vice versa. So in this lecture, we're going to focus primarily on work. So from a physics perspective, what is work?"}, {"title": "Work.txt", "text": "So if it's not heat, it must be work and vice versa. So in this lecture, we're going to focus primarily on work. So from a physics perspective, what is work? From a physics perspective, if you want to move a box a distance D, or from a point X to a point Y, you have to exert A force on that box over a distance D.\nAnd assuming the force is constant, you can find the work done by this formula. Work equals force times distance travel. If the force isn't constant, you have to do a little bit of calculus, and you have to use the integral."}, {"title": "Work.txt", "text": "From a physics perspective, if you want to move a box a distance D, or from a point X to a point Y, you have to exert A force on that box over a distance D.\nAnd assuming the force is constant, you can find the work done by this formula. Work equals force times distance travel. If the force isn't constant, you have to do a little bit of calculus, and you have to use the integral. Okay, now, from a chemistry perspective, what is work? Well, when we're in chemistry, we basically want to work with atoms and molecules, and we say that a single molecule or an atom has some kinetic energy, translational energy. And collectively, the translational or kinetic energy of the entire system can do work."}, {"title": "Work.txt", "text": "Okay, now, from a chemistry perspective, what is work? Well, when we're in chemistry, we basically want to work with atoms and molecules, and we say that a single molecule or an atom has some kinetic energy, translational energy. And collectively, the translational or kinetic energy of the entire system can do work. But now it doesn't do work by acting over some distance. It does work by expansion. So expansion work is called PD work, and you find the work done by multiplying the pressure when it's constant by the change in volume."}, {"title": "Work.txt", "text": "But now it doesn't do work by acting over some distance. It does work by expansion. So expansion work is called PD work, and you find the work done by multiplying the pressure when it's constant by the change in volume. So this equation only holds when the pressure is constant. When the pressure isn't constant, we have to use the integral. Okay, so there are many different examples that could be used to describe a chemical work, but for the most part, people use two examples, one with balloons and the other one with pistons."}, {"title": "Work.txt", "text": "So this equation only holds when the pressure is constant. When the pressure isn't constant, we have to use the integral. Okay, so there are many different examples that could be used to describe a chemical work, but for the most part, people use two examples, one with balloons and the other one with pistons. Okay, so how balloons expand? Balloons expand because you blow in air. In other words, you blow in oxygen molecules, water vapor, you blow in carbon dioxide, you blow in nitrogen molecules, and so on."}, {"title": "Work.txt", "text": "Okay, so how balloons expand? Balloons expand because you blow in air. In other words, you blow in oxygen molecules, water vapor, you blow in carbon dioxide, you blow in nitrogen molecules, and so on. And the increase in the number of molecules increases the system's kinetic energy, collective kinetic energy. And that, in turn, pushes on the walls of the balloon. And this push exerts a pressure or A force on the outside molecules, the surrounding molecules in the air."}, {"title": "Work.txt", "text": "And the increase in the number of molecules increases the system's kinetic energy, collective kinetic energy. And that, in turn, pushes on the walls of the balloon. And this push exerts a pressure or A force on the outside molecules, the surrounding molecules in the air. And this does work on those molecules, expanding the balloon and moving those molecules away. Okay? So collectively, the molecules or the atoms within the system, within a stationary system expand by doing work on the surrounding molecules."}, {"title": "Work.txt", "text": "And this does work on those molecules, expanding the balloon and moving those molecules away. Okay? So collectively, the molecules or the atoms within the system, within a stationary system expand by doing work on the surrounding molecules. Okay? The same concept works for a piston. In a piston, at some constant pressure, you have some molecules floating around within this area, okay?"}, {"title": "Work.txt", "text": "Okay? The same concept works for a piston. In a piston, at some constant pressure, you have some molecules floating around within this area, okay? And the atmosphere exerts a force on the piston downward. Now, what happens if, for example, you heat the system? If you heat the system, you increase the kinetic energy of the molecules inside here by increasing their speed."}, {"title": "Work.txt", "text": "And the atmosphere exerts a force on the piston downward. Now, what happens if, for example, you heat the system? If you heat the system, you increase the kinetic energy of the molecules inside here by increasing their speed. And this, in turn, will push against the piston. It will push against those molecules found in the air here and will do work on them. And it will expand the entire system."}, {"title": "Work.txt", "text": "And this, in turn, will push against the piston. It will push against those molecules found in the air here and will do work on them. And it will expand the entire system. So, once again, the collective kinetic energy of the system inside does work on the outside molecules, the surrounding molecules. Okay? And that is how it expands."}, {"title": "Work.txt", "text": "So, once again, the collective kinetic energy of the system inside does work on the outside molecules, the surrounding molecules. Okay? And that is how it expands. And when force or when pressure is constant, we can find the equation. We can find the work done on the surroundings by simply using the equation. Work equals volume or change in volume times pressure."}, {"title": "Work.txt", "text": "And when force or when pressure is constant, we can find the equation. We can find the work done on the surroundings by simply using the equation. Work equals volume or change in volume times pressure. Okay? Now, how do we come up with this equation? Let's circle this equation that we keep on mentioning."}, {"title": "Work.txt", "text": "Okay? Now, how do we come up with this equation? Let's circle this equation that we keep on mentioning. Okay, so work equals pressure times change in volume, right? So we start with pressure. When we talk about chemistry and chemical systems, we always talk about pressure."}, {"title": "Work.txt", "text": "Okay, so work equals pressure times change in volume, right? So we start with pressure. When we talk about chemistry and chemical systems, we always talk about pressure. And pressure is force per unit area, okay? Now what happens if we just multiply this side by A and this side by A? Well, if you multiply this side by A, the A's cancel out."}, {"title": "Work.txt", "text": "And pressure is force per unit area, okay? Now what happens if we just multiply this side by A and this side by A? Well, if you multiply this side by A, the A's cancel out. We get S. If you multiply this side by A, we get PA. Okay? So force is equal to pressure times area."}, {"title": "Work.txt", "text": "We get S. If you multiply this side by A, we get PA. Okay? So force is equal to pressure times area. Okay? Now what happens if we multiply by distance, both sides by distance? You multiply this side by distance, we get force times distance."}, {"title": "Work.txt", "text": "Okay? Now what happens if we multiply by distance, both sides by distance? You multiply this side by distance, we get force times distance. If you multiply this side by distance, we get pressure times area times distance. But remember, that area times distance is simply the volume. That is, if we go back to our Piston example and the Piston moves a certain distance, say, D, then this area A times distance D will give us this whole volume here or the change in volume."}, {"title": "Work.txt", "text": "If you multiply this side by distance, we get pressure times area times distance. But remember, that area times distance is simply the volume. That is, if we go back to our Piston example and the Piston moves a certain distance, say, D, then this area A times distance D will give us this whole volume here or the change in volume. Therefore, pressure times change in volume is equal to work. And that's how we derive this equation, from pressure. Okay?"}, {"title": "Work.txt", "text": "Therefore, pressure times change in volume is equal to work. And that's how we derive this equation, from pressure. Okay? And that's what work is in a chemical perspective or from a chemical perspective. Now let's look at a graph of pressure versus volume, when pressure is constant, okay? What we see is pressure is the y axis, the vertical axis."}, {"title": "Work.txt", "text": "And that's what work is in a chemical perspective or from a chemical perspective. Now let's look at a graph of pressure versus volume, when pressure is constant, okay? What we see is pressure is the y axis, the vertical axis. It's constant. So it's the same throughout the entire process. The volume, however, changes."}, {"title": "Work.txt", "text": "It's constant. So it's the same throughout the entire process. The volume, however, changes. It goes from some lower volume to some higher volume. Okay? What the work is, is this."}, {"title": "Work.txt", "text": "It goes from some lower volume to some higher volume. Okay? What the work is, is this. The pressure which is constant. The pressure is the vertical side, this side. So that's pressure times the change in volume v two or V final minus V one, B initial."}, {"title": "Work.txt", "text": "The pressure which is constant. The pressure is the vertical side, this side. So that's pressure times the change in volume v two or V final minus V one, B initial. So this side, this is change in volume. So this side times this side gives us the work. So since this is a rectangle and the area of a rectangle is is side times width, what we see or side times width, what we see is that work is this shade regions here."}, {"title": "Work.txt", "text": "So this side, this is change in volume. So this side times this side gives us the work. So since this is a rectangle and the area of a rectangle is is side times width, what we see or side times width, what we see is that work is this shade regions here. And this is actually work. Pressure times change in volume. Okay?"}, {"title": "Work.txt", "text": "And this is actually work. Pressure times change in volume. Okay? Now, if we want to find a work when pressure is in constant, we could still do it. But then we have to use the integral over some area, okay? We could no longer use this equation because this equation assumes that pressure is constant throughout the experiment."}, {"title": "Internal Energy of Matter .txt", "text": "Well, internal energy is the collective energy of all the different types of molecules found within a system. Now, a system can be composed of billions of different molecules. For simplification purposes, we're going to look at a system with only two diatomic oxygen molecules, moles. Now, let's look at the types of energies molecules could have. Well, they can have either kinetic energy or potential energy. Kinetic energy is the energy due to motion, and it could be subdivided into three different types."}, {"title": "Internal Energy of Matter .txt", "text": "Now, let's look at the types of energies molecules could have. Well, they can have either kinetic energy or potential energy. Kinetic energy is the energy due to motion, and it could be subdivided into three different types. Translational energy is the energy due to velocity. Now, when a mass moves with a certain velocity, it carries a certain kinetic energy. And this becomes important in gasses and liquids."}, {"title": "Internal Energy of Matter .txt", "text": "Translational energy is the energy due to velocity. Now, when a mass moves with a certain velocity, it carries a certain kinetic energy. And this becomes important in gasses and liquids. In solids, vibrational energy takes over. In solids, the molecules don't really move too much. They vibrate."}, {"title": "Internal Energy of Matter .txt", "text": "In solids, vibrational energy takes over. In solids, the molecules don't really move too much. They vibrate. And this vibration is created due to repulsion and attraction of protons and electrons. Now, if we look at this diatomic oxygen molecule, we'll see that this is the case. This diatomic oxygen molecule is composed of two nuclei."}, {"title": "Internal Energy of Matter .txt", "text": "And this vibration is created due to repulsion and attraction of protons and electrons. Now, if we look at this diatomic oxygen molecule, we'll see that this is the case. This diatomic oxygen molecule is composed of two nuclei. These two nuclei repel each other because they are positively charged. So they want to move away from each other. But notice that each one has an electron cloud around it."}, {"title": "Internal Energy of Matter .txt", "text": "These two nuclei repel each other because they are positively charged. So they want to move away from each other. But notice that each one has an electron cloud around it. And this guy will attract these guys, and this guy will attract these guys. So this will create motion, a reverse motion. Instead of repulsing, they want to attract, okay?"}, {"title": "Internal Energy of Matter .txt", "text": "And this guy will attract these guys, and this guy will attract these guys. So this will create motion, a reverse motion. Instead of repulsing, they want to attract, okay? So this creates repulsion, but they want to attract due to the attraction between this and this and that and that. Okay, so they create this harmonic vibrational force. And this is important in solid, okay, rotational."}, {"title": "Internal Energy of Matter .txt", "text": "So this creates repulsion, but they want to attract due to the attraction between this and this and that and that. Okay, so they create this harmonic vibrational force. And this is important in solid, okay, rotational. The first type of kinetic energy, rotational energy is the energy due to torque. Now, when molecules, for example, the atomic oxygen moves, it moves and rotates. It rotates this way."}, {"title": "Internal Energy of Matter .txt", "text": "The first type of kinetic energy, rotational energy is the energy due to torque. Now, when molecules, for example, the atomic oxygen moves, it moves and rotates. It rotates this way. The same way when you throw a Frisbee, the Frisbee rotates and it moves. And this rotation creates a curved path, okay? And this rotation also creates kinetic energy."}, {"title": "Internal Energy of Matter .txt", "text": "The same way when you throw a Frisbee, the Frisbee rotates and it moves. And this rotation creates a curved path, okay? And this rotation also creates kinetic energy. Now let's look at the different types of potential energies that exist. Potential energies are energies due to positional placement in space, three types exist. Rough mass is the energy due to a stationary mass."}, {"title": "Internal Energy of Matter .txt", "text": "Now let's look at the different types of potential energies that exist. Potential energies are energies due to positional placement in space, three types exist. Rough mass is the energy due to a stationary mass. Now, any mass has energy, and Einstein showed this with the equation E equals MC squared. The larger the mass, the larger the energy. Electrostatic energy is the energy due to the attraction and repulsion of the protons in the nuclei or the nucleus and the electrons in the electron cloud, okay?"}, {"title": "Internal Energy of Matter .txt", "text": "Now, any mass has energy, and Einstein showed this with the equation E equals MC squared. The larger the mass, the larger the energy. Electrostatic energy is the energy due to the attraction and repulsion of the protons in the nuclei or the nucleus and the electrons in the electron cloud, okay? This creates an electrostatic potential energy. Intermolecular energy is the energy due to neighboring molecules. If we go back to our system that's composed of two diatomic oxygen molecules, we see that the protons found on this nucleus will attract the electrons of the neighboring atom."}, {"title": "Internal Energy of Matter .txt", "text": "This creates an electrostatic potential energy. Intermolecular energy is the energy due to neighboring molecules. If we go back to our system that's composed of two diatomic oxygen molecules, we see that the protons found on this nucleus will attract the electrons of the neighboring atom. And the same thing for this one. The proton here will attract electrons here, and they will create potential energy, okay? Now, to find an internal energy of the system, I would literally have to look at every single one of these for this molecule."}, {"title": "Internal Energy of Matter .txt", "text": "And the same thing for this one. The proton here will attract electrons here, and they will create potential energy, okay? Now, to find an internal energy of the system, I would literally have to look at every single one of these for this molecule. Add that up. I would have to look at the same situation here, take all these, add them up for this one and then I would sum them together. And that would be my total or my internal energy."}, {"title": "Internal Energy of Matter .txt", "text": "Add that up. I would have to look at the same situation here, take all these, add them up for this one and then I would sum them together. And that would be my total or my internal energy. Now, if I had billions of molecules, I would have to do it for every single molecule. Now, one last thing I want to mention is that internal energy is a state function. And what that means is simply that eternal energy does not depend on the process or the pathway taken to get to this system."}, {"title": "Internal Energy of Matter .txt", "text": "Now, if I had billions of molecules, I would have to do it for every single molecule. Now, one last thing I want to mention is that internal energy is a state function. And what that means is simply that eternal energy does not depend on the process or the pathway taken to get to this system. What it does depend on is the current system at hand and it's also an external property. And that simply means that with increase in size of the system, internal energy of the system will also increase. And that's simply because if you increase the number of molecules in a system, there are more kinetic energies and potential energies to sum up and so that internal energy will increase."}, {"title": "Autoionization Example .txt", "text": "Our second solution is a seven molar sodium hydroxide solution given also at 25 degrees Celsius. We want to find two things. First, you want to find the concentration of hydroxide. Now, seven molar nitric acid solution. And second, we want to find the concentration of hydronium in our seven molar sodium hydroxide solution. So, in the first step, we're going to do Part A."}, {"title": "Autoionization Example .txt", "text": "Now, seven molar nitric acid solution. And second, we want to find the concentration of hydronium in our seven molar sodium hydroxide solution. So, in the first step, we're going to do Part A. So, first we must write the odor ionization reaction for water. So two moles of water react to produce 1 mol of conjugate acid and 1 mol of conjugate base. Now, let's write the equilibrium constants equation or expression for our odorionization reaction."}, {"title": "Autoionization Example .txt", "text": "So, first we must write the odor ionization reaction for water. So two moles of water react to produce 1 mol of conjugate acid and 1 mol of conjugate base. Now, let's write the equilibrium constants equation or expression for our odorionization reaction. So, Kw is equal to concentration of hydronium times the concentration of hydroxide. So we know our Kw. That's a constant at a 25 degree Celsius, it's ten to negative 14."}, {"title": "Autoionization Example .txt", "text": "So, Kw is equal to concentration of hydronium times the concentration of hydroxide. So we know our Kw. That's a constant at a 25 degree Celsius, it's ten to negative 14. Now, we also know this guy, that's seven molar because we're dealing with an acid. This is seven molar of acid. And this acid associates into H plus and some other ion."}, {"title": "Autoionization Example .txt", "text": "Now, we also know this guy, that's seven molar because we're dealing with an acid. This is seven molar of acid. And this acid associates into H plus and some other ion. And this H plus increases the concentration of both this guy and this guy. And in fact, H and H 30 plus are one and the same. They're meant to represent the same thing."}, {"title": "Autoionization Example .txt", "text": "And this H plus increases the concentration of both this guy and this guy. And in fact, H and H 30 plus are one and the same. They're meant to represent the same thing. So our concentration of hydronium in our solution is seven molar. So we know Kw and we know this guy. Now, we could plug it in and find our result."}, {"title": "Autoionization Example .txt", "text": "So our concentration of hydronium in our solution is seven molar. So we know Kw and we know this guy. Now, we could plug it in and find our result. Now, by the way, if you're confused at this part, or if you're confused about the autoimization of water, check out the link below. So we basically take our numbers, we plug them in, and we find that 10th of the negative 14 divided by seven gives you 1.43 times ten to negative 15 molar of this guy of seven molar nitric acid solution. So this means that our concentration of our base, of our hydroxide is very, very small."}, {"title": "Autoionization Example .txt", "text": "Now, by the way, if you're confused at this part, or if you're confused about the autoimization of water, check out the link below. So we basically take our numbers, we plug them in, and we find that 10th of the negative 14 divided by seven gives you 1.43 times ten to negative 15 molar of this guy of seven molar nitric acid solution. So this means that our concentration of our base, of our hydroxide is very, very small. And this means this must be a very strong acid or a very acidic solution. So let's do part two, part B in this section. So, our sodium hydroxide dissociates into sodium and hydroxide."}, {"title": "Autoionization Example .txt", "text": "And this means this must be a very strong acid or a very acidic solution. So let's do part two, part B in this section. So, our sodium hydroxide dissociates into sodium and hydroxide. So this must be our base. So our seven molar concentration now refers to the concentration of hydroxide. So we simply repeat the step kw 10th to negative 14 equal it unknown times seven."}, {"title": "Autoionization Example .txt", "text": "So this must be our base. So our seven molar concentration now refers to the concentration of hydroxide. So we simply repeat the step kw 10th to negative 14 equal it unknown times seven. We bring the seven over and we get divided. And we get 4.1.43 times ten to negative 15 molar of this guy of seven molar sodium hydroxide. So these two numbers have the same magnitude, but they mean two different things."}, {"title": "Autoionization Example .txt", "text": "We bring the seven over and we get divided. And we get 4.1.43 times ten to negative 15 molar of this guy of seven molar sodium hydroxide. So these two numbers have the same magnitude, but they mean two different things. In this case, this means the concentration of hydroxide. That means it's a very small concentration of base. So this is an acidic solution."}, {"title": "Buffer Systems .txt", "text": "Now to begin, let's suppose we have a system of one liter of pure water that has a PH of seven. Let's examine what happens to our PH when we add a small amount of acid or base to our system. Well, let's begin with the acid. Suppose we add zero one mo of HCL to our one liter system. Let's see the new PH. Well, the PH is equal to negative log zero one, and that gives us a PH of two."}, {"title": "Buffer Systems .txt", "text": "Suppose we add zero one mo of HCL to our one liter system. Let's see the new PH. Well, the PH is equal to negative log zero one, and that gives us a PH of two. That means if we add this little hydrochloric acid, our PH drops by five increments. That's equivalent to a 100,000 fold increase in the hydronium concentration of our mixture. So now let's add the same amount of sodium hydroxide, a base to our system."}, {"title": "Buffer Systems .txt", "text": "That means if we add this little hydrochloric acid, our PH drops by five increments. That's equivalent to a 100,000 fold increase in the hydronium concentration of our mixture. So now let's add the same amount of sodium hydroxide, a base to our system. What will be the new PH? Well, first we calculate the Poh. And the Poh is equal to negative log of 0.1,\ngives it two."}, {"title": "Buffer Systems .txt", "text": "What will be the new PH? Well, first we calculate the Poh. And the Poh is equal to negative log of 0.1,\ngives it two. Now we subtract two from 14 and we get a PH of twelve. That means our PH increases by five increments. That's equivalent to a 100,000 fold increase in the hydroxide concentration."}, {"title": "Buffer Systems .txt", "text": "Now we subtract two from 14 and we get a PH of twelve. That means our PH increases by five increments. That's equivalent to a 100,000 fold increase in the hydroxide concentration. That's a very big increase in PH. So the takeaway from this is that adding a small amount of acid or base to pure water will drastically change the PH of water. Now, if you're confused about how we got this part or this part, check out the link below."}, {"title": "Buffer Systems .txt", "text": "That's a very big increase in PH. So the takeaway from this is that adding a small amount of acid or base to pure water will drastically change the PH of water. Now, if you're confused about how we got this part or this part, check out the link below. Now Aqueous solutions, unlike pure water, solutions resist changes in PH when we add acid or base. And that's because aqueous solutions have buffer systems. And a buffer is simply a chemical system which resists a PH change."}, {"title": "Buffer Systems .txt", "text": "Now Aqueous solutions, unlike pure water, solutions resist changes in PH when we add acid or base. And that's because aqueous solutions have buffer systems. And a buffer is simply a chemical system which resists a PH change. So for example, suppose we had a system of 0.5 molar of acetic acid mixed with 0.5\nmolar of sodium acetate in one liter of water, and this PH and this system's PH was 4.74. So now what happens to our PH if we add 0.1 molar of hydrochloric acid, as we did in part A? Well, now our PH will only decrease from 4.74\nto 4.72."}, {"title": "Buffer Systems .txt", "text": "So for example, suppose we had a system of 0.5 molar of acetic acid mixed with 0.5\nmolar of sodium acetate in one liter of water, and this PH and this system's PH was 4.74. So now what happens to our PH if we add 0.1 molar of hydrochloric acid, as we did in part A? Well, now our PH will only decrease from 4.74\nto 4.72. That's a change of 0.2. That's a very small change. Well, that's because this system has a buffer system."}, {"title": "Buffer Systems .txt", "text": "That's a change of 0.2. That's a very small change. Well, that's because this system has a buffer system. And buffer systems are really important. For example, our blood is a buffer system. And if our PH of our blood decreases even slightly, we will suffocate and die."}, {"title": "Buffer Systems .txt", "text": "And buffer systems are really important. For example, our blood is a buffer system. And if our PH of our blood decreases even slightly, we will suffocate and die. So these guys are very important. Now let's see why this happens. Now, before we look at how they work, let's look at why they work."}, {"title": "Buffer Systems .txt", "text": "So these guys are very important. Now let's see why this happens. Now, before we look at how they work, let's look at why they work. What are the few requirements of buffer systems? Well, first, we have to have a weak acid and a weak base. And second, the weak acid cannot react with our weak base, because if they did, our buffer system would be neutralized and we wouldn't have a buffer to work with."}, {"title": "Buffer Systems .txt", "text": "What are the few requirements of buffer systems? Well, first, we have to have a weak acid and a weak base. And second, the weak acid cannot react with our weak base, because if they did, our buffer system would be neutralized and we wouldn't have a buffer to work with. Now, what's one thing that satisfies these two requirements? Well, that's a conjugate acid base pair. Whenever a conjugate acid reacts with a conjugate base, it produces another conjugate acid and base pair for example, acetate ion and acetic acid react to produce acetate ion and acetic acid."}, {"title": "Buffer Systems .txt", "text": "Now, what's one thing that satisfies these two requirements? Well, that's a conjugate acid base pair. Whenever a conjugate acid reacts with a conjugate base, it produces another conjugate acid and base pair for example, acetate ion and acetic acid react to produce acetate ion and acetic acid. So a conjugate base pair reacts to produce another conjugate base pair. So nothing is neutralized. And there are buffer remains unchanged."}, {"title": "Buffer Systems .txt", "text": "So a conjugate base pair reacts to produce another conjugate base pair. So nothing is neutralized. And there are buffer remains unchanged. So normally, buffers contain equal amounts of conjugate acid as conjugate base. For example, in this buffer system, we have the same amount of acetic acid as the acetate ion. Now, some exceptions do exist."}, {"title": "Buffer Systems .txt", "text": "So normally, buffers contain equal amounts of conjugate acid as conjugate base. For example, in this buffer system, we have the same amount of acetic acid as the acetate ion. Now, some exceptions do exist. For example, our blood, our blood has much more conjugate base than conjugate acid. But that's because our body produces many more acidic byproducts than basic byproducts. And so we need more base to neutralize our acid."}, {"title": "Buffer Systems .txt", "text": "For example, our blood, our blood has much more conjugate base than conjugate acid. But that's because our body produces many more acidic byproducts than basic byproducts. And so we need more base to neutralize our acid. Now, let's see how these buffers work. So, for example, suppose we have the buffer system above, composed of acetic acid methodate ion. Now, suppose we add a strong base such as sodium hydroxide to our system."}, {"title": "Buffer Systems .txt", "text": "Now, let's see how these buffers work. So, for example, suppose we have the buffer system above, composed of acetic acid methodate ion. Now, suppose we add a strong base such as sodium hydroxide to our system. What will happen? Well, this base reacts with our conjugate acid to produce back the conjugate base and water. So, before this base can affect our system, it's neutralized into a water molecule."}, {"title": "Buffer Systems .txt", "text": "What will happen? Well, this base reacts with our conjugate acid to produce back the conjugate base and water. So, before this base can affect our system, it's neutralized into a water molecule. So our PH only changes slightly. Likewise, let's see what happens when we add acid to our buffer system. Well, hydrochloric acid first reacts with water, producing hydronium ion and the CL ion."}, {"title": "Atomic Orbitals .txt", "text": "So organic chemistry is essentially the study of covalent bonds. And covalent bonds are formed by the overlap of atomic orbitals. And that means in order to understand what covalent bonds are, we must first understand what atomic orbitals are. So let's begin. So here we have so here we have Boris model. Now, Boris'model is essentially a depiction of the nucleus, the protons found in the nucleus and the electrons found orbiting our nucleus."}, {"title": "Atomic Orbitals .txt", "text": "So let's begin. So here we have so here we have Boris model. Now, Boris'model is essentially a depiction of the nucleus, the protons found in the nucleus and the electrons found orbiting our nucleus. Now, according to the Boris model, and for this particular atom, our electrons are orbiting in a perfect circle. So for this atom, we have two electrons found in this circle and two electrons found in the outer circular orbit. Now, as you may or may not know, Bored model is actually an inaccurate depiction of our atomic nucleus and electrons."}, {"title": "Atomic Orbitals .txt", "text": "Now, according to the Boris model, and for this particular atom, our electrons are orbiting in a perfect circle. So for this atom, we have two electrons found in this circle and two electrons found in the outer circular orbit. Now, as you may or may not know, Bored model is actually an inaccurate depiction of our atomic nucleus and electrons. And that's because electrons do not actually occupy these perfect or circular and spherical orbits. Now, Schrodinger described a pathway that our electrons follow using wave equations. So in other words, our electrons follow certain orbits, certain pathways that are not circular."}, {"title": "Atomic Orbitals .txt", "text": "And that's because electrons do not actually occupy these perfect or circular and spherical orbits. Now, Schrodinger described a pathway that our electrons follow using wave equations. So in other words, our electrons follow certain orbits, certain pathways that are not circular. And what this person did is he described the pathways that they take using wave equations. Now, wave equations are simply mathematical representations of the pathways that our electrons do take. And just like any simple equation, we can also solve wave equations for solutions."}, {"title": "Atomic Orbitals .txt", "text": "And what this person did is he described the pathways that they take using wave equations. Now, wave equations are simply mathematical representations of the pathways that our electrons do take. And just like any simple equation, we can also solve wave equations for solutions. And the solutions to these wave equations are called wave functions. Now, orbitals are the same thing as wave functions. So orbitals are wave functions."}, {"title": "Atomic Orbitals .txt", "text": "And the solutions to these wave equations are called wave functions. Now, orbitals are the same thing as wave functions. So orbitals are wave functions. So orbitals are solutions to these wave equations. And since wave equations are simply mathematical representations of the pathway that electrons take, if we solve these wave equations, we can find the probability of an electron being at a certain region in a certain volume. And these probabilities are given by orbitals."}, {"title": "Atomic Orbitals .txt", "text": "So orbitals are solutions to these wave equations. And since wave equations are simply mathematical representations of the pathway that electrons take, if we solve these wave equations, we can find the probability of an electron being at a certain region in a certain volume. And these probabilities are given by orbitals. So orbitals represent certain shapes or volumes within which our electrons are most likely in. The reason I say most likely is because orbitals are probabilities. Now, before we talk more about orbitals, let's recall what quantum numbers are."}, {"title": "Atomic Orbitals .txt", "text": "So orbitals represent certain shapes or volumes within which our electrons are most likely in. The reason I say most likely is because orbitals are probabilities. Now, before we talk more about orbitals, let's recall what quantum numbers are. Quantum numbers are simply the idea of our electrons. So if we have a unique electron in a given atom, that electron has four unique quantum numbers that are unique to that electron. So we have the principal quantum number, we have the Zimmerfo quantum number, and we have two more quantum numbers."}, {"title": "Atomic Orbitals .txt", "text": "Quantum numbers are simply the idea of our electrons. So if we have a unique electron in a given atom, that electron has four unique quantum numbers that are unique to that electron. So we have the principal quantum number, we have the Zimmerfo quantum number, and we have two more quantum numbers. Now, the principal quantum number gives the energy level of that electron. The second quantum number, known as the Zimmerfa quantum number gives or designates the shape of the orbital, and it's represented by the letter L, and it could be 00:12 and so on, zero being the S shape, one being the P shape, two being the D shape. The third quantum number specifies exactly which orbital that our electron is in."}, {"title": "Atomic Orbitals .txt", "text": "Now, the principal quantum number gives the energy level of that electron. The second quantum number, known as the Zimmerfa quantum number gives or designates the shape of the orbital, and it's represented by the letter L, and it could be 00:12 and so on, zero being the S shape, one being the P shape, two being the D shape. The third quantum number specifies exactly which orbital that our electron is in. And the fourth quantum number gives the spin electron spin of our electron. So we could have either plus one half spin or minus one half spin. So in this lecture, we're only going to deal with the S or the P orbital."}, {"title": "Atomic Orbitals .txt", "text": "And the fourth quantum number gives the spin electron spin of our electron. So we could have either plus one half spin or minus one half spin. So in this lecture, we're only going to deal with the S or the P orbital. So let's begin with the s orbital. So the s orbital, which is one of the solutions to the wave equations, is given by the spherical shape. So this sphere is the s orbital."}, {"title": "Atomic Orbitals .txt", "text": "So let's begin with the s orbital. So the s orbital, which is one of the solutions to the wave equations, is given by the spherical shape. So this sphere is the s orbital. And what it basically states is that our electron is most likely in this sphere here. Now, of course, as we're talking about probabilities, there is still a probability that our electron will be found outside this spherical shape. But it's very unlikely and that's why we say it's most likely in this orbital."}, {"title": "Atomic Orbitals .txt", "text": "And what it basically states is that our electron is most likely in this sphere here. Now, of course, as we're talking about probabilities, there is still a probability that our electron will be found outside this spherical shape. But it's very unlikely and that's why we say it's most likely in this orbital. So the p orbital, unlike the s orbital, have a dumbbell like shape or sideways eight. Now we have the PX orbital, we have the PY orbital and the PZ orbital. In other words, if we label this as the x axis, this as the y axis and this as the Z axis."}, {"title": "Atomic Orbitals .txt", "text": "So the p orbital, unlike the s orbital, have a dumbbell like shape or sideways eight. Now we have the PX orbital, we have the PY orbital and the PZ orbital. In other words, if we label this as the x axis, this as the y axis and this as the Z axis. So Z axis is coming out of the board or going into the board. Then we have the following three orbitals. Now, if we take these guys and put them together, we get the overall p orbital and it's given by the following picture, which kind of looks like a flower, a three dimensional flower."}, {"title": "Atomic Orbitals .txt", "text": "So Z axis is coming out of the board or going into the board. Then we have the following three orbitals. Now, if we take these guys and put them together, we get the overall p orbital and it's given by the following picture, which kind of looks like a flower, a three dimensional flower. Now we have the X orbital. So this guy here, we have the y orbital. So this guy here and we have this z orbital, the PV orbital which is coming out of the board."}, {"title": "Atomic Orbitals .txt", "text": "Now we have the X orbital. So this guy here, we have the y orbital. So this guy here and we have this z orbital, the PV orbital which is coming out of the board. Now, just like on the XYZ axis, we have the positive side. So Y going this way is positive, x going this way is positive and D going out of the board is positive. We also have the positive sides or positive probabilities of the orbitals."}, {"title": "Atomic Orbitals .txt", "text": "Now, just like on the XYZ axis, we have the positive side. So Y going this way is positive, x going this way is positive and D going out of the board is positive. We also have the positive sides or positive probabilities of the orbitals. So this green part is the positive and the blue part is the negative. Now, because we're dealing with waves and waves have nodes and anti nodes. These guys will also have nodes and anti nodes."}, {"title": "Atomic Orbitals .txt", "text": "So this green part is the positive and the blue part is the negative. Now, because we're dealing with waves and waves have nodes and anti nodes. These guys will also have nodes and anti nodes. Now the nodes are these guys here. So if you could think of the eight and the eight intersects at this point. This point is the node."}, {"title": "Atomic Orbitals .txt", "text": "Now the nodes are these guys here. So if you could think of the eight and the eight intersects at this point. This point is the node. And what it basically states is that our electron has a zero probability of being in this place. So the node means zero probability of finding electron at this place. That means we're never going to find an electron here, here or here, or in the cumulative picture."}, {"title": "Atomic Orbitals .txt", "text": "And what it basically states is that our electron has a zero probability of being in this place. So the node means zero probability of finding electron at this place. That means we're never going to find an electron here, here or here, or in the cumulative picture. We're never going to find the electron at the origin at the point on this XYZ axis. So why are these guys important? How can we use these guys to represent pictures of our atoms?"}, {"title": "Atomic Orbitals .txt", "text": "We're never going to find the electron at the origin at the point on this XYZ axis. So why are these guys important? How can we use these guys to represent pictures of our atoms? Okay, so let's take an example. Let's take the carbon atom. So carbon, a neutral carbon, has six protons."}, {"title": "Atomic Orbitals .txt", "text": "Okay, so let's take an example. Let's take the carbon atom. So carbon, a neutral carbon, has six protons. Hence this subscript six and six electrons. So that means if we were to draw our electron configuration, we would get this depiction. So two electrons go into this one s, two electrons go into this two s, and two electrons go into this picture here."}, {"title": "Atomic Orbitals .txt", "text": "Hence this subscript six and six electrons. So that means if we were to draw our electron configuration, we would get this depiction. So two electrons go into this one s, two electrons go into this two s, and two electrons go into this picture here. But remember, we have to follow the poly exclusion principle which basically states that a maximum of two electrons can be placed into any orbital. So two electrons can be placed into the s.\nTwo electrons each can be placed into the PX, PY and PZ. So cumulatively, we're going to have a total, a maximum of six electrons that can be placed into this flower shaped p orbital, because this actually includes three separate orbitals."}, {"title": "Atomic Orbitals .txt", "text": "But remember, we have to follow the poly exclusion principle which basically states that a maximum of two electrons can be placed into any orbital. So two electrons can be placed into the s.\nTwo electrons each can be placed into the PX, PY and PZ. So cumulatively, we're going to have a total, a maximum of six electrons that can be placed into this flower shaped p orbital, because this actually includes three separate orbitals. So two can be placed from here into here and into here. Now, also recall Honduras. Hans Rule basically states that before we begin completely filling these orbitals, we first have to place one electron here, one electron here, and one electron here."}, {"title": "Atomic Orbitals .txt", "text": "So two can be placed from here into here and into here. Now, also recall Honduras. Hans Rule basically states that before we begin completely filling these orbitals, we first have to place one electron here, one electron here, and one electron here. So we have to go in order. And that's exactly what we do here. One electron is placed into the PX."}, {"title": "Atomic Orbitals .txt", "text": "So we have to go in order. And that's exactly what we do here. One electron is placed into the PX. And one electron is placed into the PY. And that's exactly what we do here. So let's look at what happened."}, {"title": "Atomic Orbitals .txt", "text": "And one electron is placed into the PY. And that's exactly what we do here. So let's look at what happened. So the electrons are the brown dots. I used brown because we already used blue, and I don't want to confuse you guys further. So the blue or the brown are our electrons."}, {"title": "Atomic Orbitals .txt", "text": "So the electrons are the brown dots. I used brown because we already used blue, and I don't want to confuse you guys further. So the blue or the brown are our electrons. So the one that's orbital I did not depict. And that's because the one that's orbital is simply a smaller black sphere found within this two s sphere. This black sphere."}, {"title": "Atomic Orbitals .txt", "text": "So the one that's orbital I did not depict. And that's because the one that's orbital is simply a smaller black sphere found within this two s sphere. This black sphere. Here is the two s sphere. So let's imagine that we took our two electrons and placed it into our one s. And now we're taking out two electrons and place it into the two s. The two s is this black sphere here. And I took two electrons and placed it into the black sphere, as shown here."}, {"title": "Atomic Orbitals .txt", "text": "Here is the two s sphere. So let's imagine that we took our two electrons and placed it into our one s. And now we're taking out two electrons and place it into the two s. The two s is this black sphere here. And I took two electrons and placed it into the black sphere, as shown here. Now, I have one electron that I place into the X. So the green region and one electron that I placed into the Y region. So the Y orbital or the green part of the Y orbital?"}, {"title": "Atomic Orbitals .txt", "text": "Now, I have one electron that I place into the X. So the green region and one electron that I placed into the Y region. So the Y orbital or the green part of the Y orbital? And that's the picture. Or the picture using atomic orbitals of our carbon. Now."}, {"title": "Atomic Orbitals .txt", "text": "And that's the picture. Or the picture using atomic orbitals of our carbon. Now. We'll see in later lectures how when these guys interact with other orbitals, with other atoms, they form something called covalent bonds. Now let's look at neon. Now."}, {"title": "Atomic Orbitals .txt", "text": "We'll see in later lectures how when these guys interact with other orbitals, with other atoms, they form something called covalent bonds. Now let's look at neon. Now. Neon has ten protons and ten electrons. In fact, it has a perfect electron configuration. It's a noble gas."}, {"title": "Atomic Orbitals .txt", "text": "Neon has ten protons and ten electrons. In fact, it has a perfect electron configuration. It's a noble gas. So all the electrons, all the atomic orbitals should be filled. So let's look at our electron configuration. So we have one s, two, two s, two, two PX, two, two p y two and two PZ, two."}, {"title": "Atomic Orbitals .txt", "text": "So all the electrons, all the atomic orbitals should be filled. So let's look at our electron configuration. So we have one s, two, two s, two, two PX, two, two p y two and two PZ, two. So once again, two electrons go into the one s which isn't shown. Two electrons go into the two s, which is shown. And here we have two electrons."}, {"title": "Atomic Orbitals .txt", "text": "So once again, two electrons go into the one s which isn't shown. Two electrons go into the two s, which is shown. And here we have two electrons. Two electrons go into our x orbital, two electrons go into the two p y orbital and two electrons go into the two PZ orbital. So all the Orbitals all the Green Orbitals are filled. And so this is our atomic orbital representation of our neon, in which it has a perfect election configuration."}, {"title": "Introduction to Resonance Forms .txt", "text": "A lewis structure is simply an electronic configuration of our atoms, our molecules and compounds. So let's begin by using formaldehyde. So we're going to have this compound that's compared composed of one carbon, one oxygen and two h atoms. Now we're going to draw the lewis down structure for our formaldehyde. So our first step is to count the number of balanced electrons. Balanced electrons, once again, of those electrons that come from the outermost shells of our atoms."}, {"title": "Introduction to Resonance Forms .txt", "text": "Now we're going to draw the lewis down structure for our formaldehyde. So our first step is to count the number of balanced electrons. Balanced electrons, once again, of those electrons that come from the outermost shells of our atoms. So oxygen has six valence electrons, carbon has four valence electrons, and h has one each. We have two h, and so two valence electrons come from our two HS. So we have a total of twelve balanced electrons."}, {"title": "Introduction to Resonance Forms .txt", "text": "So oxygen has six valence electrons, carbon has four valence electrons, and h has one each. We have two h, and so two valence electrons come from our two HS. So we have a total of twelve balanced electrons. So let's begin drawing our loose dot structure. So we have carbon, two h's and 10. So let's begin by first drawing our sigma, or Covalent bonds."}, {"title": "Introduction to Resonance Forms .txt", "text": "So let's begin drawing our loose dot structure. So we have carbon, two h's and 10. So let's begin by first drawing our sigma, or Covalent bonds. So we have two bonds between the chas, and we have one bond between the oxygen. So here's our Covalent sigma bond, covalent sigma bond, and Covalent sigma bond. So so far, we have used up six balance electrons."}, {"title": "Introduction to Resonance Forms .txt", "text": "So we have two bonds between the chas, and we have one bond between the oxygen. So here's our Covalent sigma bond, covalent sigma bond, and Covalent sigma bond. So so far, we have used up six balance electrons. We have six more balanced electrons that we can use. Notice that carbon and oxygen both can develop double bonds. So let's create a double bond between carbon and oxygen."}, {"title": "Introduction to Resonance Forms .txt", "text": "We have six more balanced electrons that we can use. Notice that carbon and oxygen both can develop double bonds. So let's create a double bond between carbon and oxygen. So we place two electrons and we create a pi bond. So now we have four more balanced electrons we can place, and we place them around the oxygen like so. Now, because oxygen has 123456, carbon has 1234, and h has one each, this is a neutral lewis structure, lewis form."}, {"title": "Introduction to Resonance Forms .txt", "text": "So we place two electrons and we create a pi bond. So now we have four more balanced electrons we can place, and we place them around the oxygen like so. Now, because oxygen has 123456, carbon has 1234, and h has one each, this is a neutral lewis structure, lewis form. Now, the following problem arises. Now, this is not the only lewis structure that exists. Others exist."}, {"title": "Introduction to Resonance Forms .txt", "text": "Now, the following problem arises. Now, this is not the only lewis structure that exists. Others exist. And in fact, here's one other one. Instead of creating that pi bond, by placing those two balanced electrons into our pipeline here, we could have simply placed those two electrons on oxygen, and that would create another lewis dot structure. However, this structure has a negative charge on oxygen and a plus charge on the carbon, because we only have one, two, three bonds here, and we have 123-4567 electrons on the oxygen."}, {"title": "Introduction to Resonance Forms .txt", "text": "And in fact, here's one other one. Instead of creating that pi bond, by placing those two balanced electrons into our pipeline here, we could have simply placed those two electrons on oxygen, and that would create another lewis dot structure. However, this structure has a negative charge on oxygen and a plus charge on the carbon, because we only have one, two, three bonds here, and we have 123-4567 electrons on the oxygen. So notice the following. We have two different lewis structures for formaldehyde, and in fact, this idea, this concept is called resonance. And these structures, lewis dot structures, are called resonant forms."}, {"title": "Introduction to Resonance Forms .txt", "text": "So notice the following. We have two different lewis structures for formaldehyde, and in fact, this idea, this concept is called resonance. And these structures, lewis dot structures, are called resonant forms. So let's define resonant forms. Resonant forms are the different combination of the possible lewis dot structures for our compounds. Now, whenever we draw resonant forms, the following two things have to always be kept in mind."}, {"title": "Introduction to Resonance Forms .txt", "text": "So let's define resonant forms. Resonant forms are the different combination of the possible lewis dot structures for our compounds. Now, whenever we draw resonant forms, the following two things have to always be kept in mind. The first one is, since lewis structures are electronic configurations, we only move electrons and we never move any atoms. Now, notice what this arrow represents. This arrow is known as arrow formulasm."}, {"title": "Introduction to Resonance Forms .txt", "text": "The first one is, since lewis structures are electronic configurations, we only move electrons and we never move any atoms. Now, notice what this arrow represents. This arrow is known as arrow formulasm. A double headed arrow simply means two electrons are being moved. So in this case, I have two electrons moving from my pi bond, and they move on to one of the orbitals on the oxygen. And now I no longer have this because these electrons have moved here."}, {"title": "Introduction to Resonance Forms .txt", "text": "A double headed arrow simply means two electrons are being moved. So in this case, I have two electrons moving from my pi bond, and they move on to one of the orbitals on the oxygen. And now I no longer have this because these electrons have moved here. So this is, once again, this arrow signifies movement of electrons and not movement of atoms. In other words, if I draw this molecule or this compound where I took this atom and placed the atom onto one of the orbitals here, so now there's a bond, a sigma bond between oxygen, carbon. This is not a resonant form."}, {"title": "Introduction to Resonance Forms .txt", "text": "So this is, once again, this arrow signifies movement of electrons and not movement of atoms. In other words, if I draw this molecule or this compound where I took this atom and placed the atom onto one of the orbitals here, so now there's a bond, a sigma bond between oxygen, carbon. This is not a resonant form. This is not a lewis dot structure for formaldehyde. In fact, this is not even formaldehyde. It's another molecule."}, {"title": "Introduction to Resonance Forms .txt", "text": "This is not a lewis dot structure for formaldehyde. In fact, this is not even formaldehyde. It's another molecule. It's another compound altogether. So, once again, in resonant forms, there's only movement of electrons, never movement of atoms. Notice another important point that we'll talk about in detail."}, {"title": "Introduction to Resonance Forms .txt", "text": "It's another compound altogether. So, once again, in resonant forms, there's only movement of electrons, never movement of atoms. Notice another important point that we'll talk about in detail. In just a moment. There is a double headed arrow, like so, and this represents resonant forms, okay? And we'll see why this is different than equilibrium arrows."}, {"title": "Introduction to Resonance Forms .txt", "text": "In just a moment. There is a double headed arrow, like so, and this represents resonant forms, okay? And we'll see why this is different than equilibrium arrows. In just a second, let's look at Nitromethane. We're going to do a second example in which we're going to draw the resonant forms. So Nitromethane has one N, two O's, one C and three H's."}, {"title": "Introduction to Resonance Forms .txt", "text": "In just a second, let's look at Nitromethane. We're going to do a second example in which we're going to draw the resonant forms. So Nitromethane has one N, two O's, one C and three H's. So in order to draw our loose dot structures, let's count the balanced electrons. So we have three balanced electrons from H.\nWe have four balanced electrons from our carbon. So we have and that means we have five balanced electrons coming from N, and we have two OS."}, {"title": "Introduction to Resonance Forms .txt", "text": "So in order to draw our loose dot structures, let's count the balanced electrons. So we have three balanced electrons from H.\nWe have four balanced electrons from our carbon. So we have and that means we have five balanced electrons coming from N, and we have two OS. That means we have twelve electrons, balanced electrons coming from oxygen. So altogether, we should have a total of 24 balance electrons, right? Twelve plus five plus four plus three should give us twelve."}, {"title": "Introduction to Resonance Forms .txt", "text": "That means we have twelve electrons, balanced electrons coming from oxygen. So altogether, we should have a total of 24 balance electrons, right? Twelve plus five plus four plus three should give us twelve. So let's begin. We draw our carbon, our H atoms, three H atoms around carbon. Then we draw our end right next to our carbon."}, {"title": "Introduction to Resonance Forms .txt", "text": "So let's begin. We draw our carbon, our H atoms, three H atoms around carbon. Then we draw our end right next to our carbon. And then two o's, like so. We start by creating sigma bonds or covalent bonds. One bond here, a bond here, a bond here between the HS, a fourth bond between the carbon and end."}, {"title": "Introduction to Resonance Forms .txt", "text": "And then two o's, like so. We start by creating sigma bonds or covalent bonds. One bond here, a bond here, a bond here between the HS, a fourth bond between the carbon and end. So all the orbitals here are filled. Now we create a bond between sigma bond between o and this o here. Now we have 123-45-6789 711, twelve valve electrons left over."}, {"title": "Introduction to Resonance Forms .txt", "text": "So all the orbitals here are filled. Now we create a bond between sigma bond between o and this o here. Now we have 123-45-6789 711, twelve valve electrons left over. So we create this pi bond here. We place two electrons onto our oxygen, like we did here, and three pairs of electrons here. So that means this guy has a negative charge, like this one has here."}, {"title": "Introduction to Resonance Forms .txt", "text": "So we create this pi bond here. We place two electrons onto our oxygen, like we did here, and three pairs of electrons here. So that means this guy has a negative charge, like this one has here. This N has a positive charge because N likes to have five electrons. It only has four electrons here, and this has six. So it's neutral."}, {"title": "Introduction to Resonance Forms .txt", "text": "This N has a positive charge because N likes to have five electrons. It only has four electrons here, and this has six. So it's neutral. So a plus charge and a minus charge creates a net charge of zero. And this does have a net charge of zero. Now, this isn't the only lewis dot structure."}, {"title": "Introduction to Resonance Forms .txt", "text": "So a plus charge and a minus charge creates a net charge of zero. And this does have a net charge of zero. Now, this isn't the only lewis dot structure. If we move two electrons here and we take these two electrons and create a double bond here, we have the following a distant lewis structure. Basically, these guys flip. Now we have a negative charge here."}, {"title": "Introduction to Resonance Forms .txt", "text": "If we move two electrons here and we take these two electrons and create a double bond here, we have the following a distant lewis structure. Basically, these guys flip. Now we have a negative charge here. We still have a positive charge here, and we have a neutral charge here. So once again, we have a combination of Lewis structures. And these guys are known as resonance forms."}, {"title": "Introduction to Resonance Forms .txt", "text": "We still have a positive charge here, and we have a neutral charge here. So once again, we have a combination of Lewis structures. And these guys are known as resonance forms. And the entire concept is known as resonance. Now notice other structures exist. We could have simply taken this double bond, placed in here and created a plus two charge, a minus two charge, and a minus two charge."}, {"title": "Introduction to Resonance Forms .txt", "text": "And the entire concept is known as resonance. Now notice other structures exist. We could have simply taken this double bond, placed in here and created a plus two charge, a minus two charge, and a minus two charge. So more Lewis structures do exist. And now we come to the most important point about resonant forms nitromethane. This compound does not spend half of its time as one resonant form and half of its time as the other."}, {"title": "Introduction to Resonance Forms .txt", "text": "So more Lewis structures do exist. And now we come to the most important point about resonant forms nitromethane. This compound does not spend half of its time as one resonant form and half of its time as the other. It is a combination of the two. In other words, this arrow does not mean it's an equilibrium. In other words, our nitromethane doesn't spend half the time as this compound, and then it converts to this compound."}, {"title": "Introduction to Resonance Forms .txt", "text": "It is a combination of the two. In other words, this arrow does not mean it's an equilibrium. In other words, our nitromethane doesn't spend half the time as this compound, and then it converts to this compound. The entire nitromethane is a combination of these two molecules. Its actual structure is somewhere in the middle of these two resonant forms. And let's look at the following important observation."}, {"title": "Introduction to Resonance Forms .txt", "text": "The entire nitromethane is a combination of these two molecules. Its actual structure is somewhere in the middle of these two resonant forms. And let's look at the following important observation. So let's suppose we have some compound X and it occur and it reacts in some way, and it converts to a completely different molecule, different compound where the atoms have moved. And this is why. Now let's wait until equilibrium has been achieved."}, {"title": "Introduction to Resonance Forms .txt", "text": "So let's suppose we have some compound X and it occur and it reacts in some way, and it converts to a completely different molecule, different compound where the atoms have moved. And this is why. Now let's wait until equilibrium has been achieved. So the arrows, the rates going forward and reverse are the same. Now notice that these two arrows are different than this arrow. Now let's suppose we have some compound A and B, which are resonance forms."}, {"title": "Introduction to Resonance Forms .txt", "text": "So the arrows, the rates going forward and reverse are the same. Now notice that these two arrows are different than this arrow. Now let's suppose we have some compound A and B, which are resonance forms. Now, this once again, does not mean that A converts to B and then B converts to A. Right? What this means is that the actual form of this molecule is a combination of the two."}, {"title": "Introduction to Resonance Forms .txt", "text": "Now, this once again, does not mean that A converts to B and then B converts to A. Right? What this means is that the actual form of this molecule is a combination of the two. It's somewhere in between. Our actual molecule that this resin forms represents is an A, nor is it D. It's somewhere in between. And let's call it C. Okay, so this is equivalent to some molecule C, which is a combination of these two molecules."}, {"title": "Introduction to Resonance Forms .txt", "text": "It's somewhere in between. Our actual molecule that this resin forms represents is an A, nor is it D. It's somewhere in between. And let's call it C. Okay, so this is equivalent to some molecule C, which is a combination of these two molecules. And this arrow does not mean equilibrium. Our molecules here aren't at equilibrium. They're not converting from this form to that form at a dad form to this form."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So thus far we have combined a one S orbital and a one S orbital. And we've also combined a one S with a two P. In this lecture, we're going to combine a two p with a two P to four molecular orbitals. Now, the first thing we have to realize is that there are at least two different ways that we can orient our two p orbitals. In part A, we have a parallel orientation. In other words, our two p orbitals are simply parallel to one another. In Part B, we have an orthogonal or perpendicular orientation in which our one two p orbital is perpendicular to our second two p orbital."}, {"title": "Molecular Orbital Formation Example .txt", "text": "In part A, we have a parallel orientation. In other words, our two p orbitals are simply parallel to one another. In Part B, we have an orthogonal or perpendicular orientation in which our one two p orbital is perpendicular to our second two p orbital. So let's see which one of these forms molecular orbitals. So let's begin with part B. In Part B, we have two ways that we can orient or combine them."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So let's see which one of these forms molecular orbitals. So let's begin with part B. In Part B, we have two ways that we can orient or combine them. We can either combine the positive of this with the positive of this, or we can combine the positive orbital here and the negative orbital here. Negative simply means we flip the signs. In other words, this green is the positive, this green is the negative."}, {"title": "Molecular Orbital Formation Example .txt", "text": "We can either combine the positive of this with the positive of this, or we can combine the positive orbital here and the negative orbital here. Negative simply means we flip the signs. In other words, this green is the positive, this green is the negative. So let's begin by adding or by combining the positive two p and the positive to P. We get the following orientation. Now notice in this orientation we have the positive interacting in a bonding way with the positive and so that forms a bond. And here we have an anti bonding interaction because the negative of this two p orbital interacts with the positive section of this two p orbital."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So let's begin by adding or by combining the positive two p and the positive to P. We get the following orientation. Now notice in this orientation we have the positive interacting in a bonding way with the positive and so that forms a bond. And here we have an anti bonding interaction because the negative of this two p orbital interacts with the positive section of this two p orbital. So we have bonding and antibounding. So let's look at the negative. Or simply we're combining two p positive and two p negative."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So we have bonding and antibounding. So let's look at the negative. Or simply we're combining two p positive and two p negative. That means this becomes blue and this becomes green. So we have this orientation. Once again we have a negative interacting with a negative."}, {"title": "Molecular Orbital Formation Example .txt", "text": "That means this becomes blue and this becomes green. So we have this orientation. Once again we have a negative interacting with a negative. So that means we get a bonding orientation and we have positive interacting with a negative to produce antibonding. So that means because we have bonding and antibonding, we have no net interaction. In other words, no net interaction because the bonding and the antibonding exactly cancel out."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So that means we get a bonding orientation and we have positive interacting with a negative to produce antibonding. So that means because we have bonding and antibonding, we have no net interaction. In other words, no net interaction because the bonding and the antibonding exactly cancel out. So this type of orthogonal orientation does not work. It does not create molecular orbitals. So let's go to the parallel."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So this type of orthogonal orientation does not work. It does not create molecular orbitals. So let's go to the parallel. So once again, in the parallel combination we have two different waves that we can combine. Remember, we're inputting two atomic orbitals, so we should get back two molecular orbitals. So let's go this way first."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So once again, in the parallel combination we have two different waves that we can combine. Remember, we're inputting two atomic orbitals, so we should get back two molecular orbitals. So let's go this way first. In other words, we're combining a two p positive and a two p positive. So we get this orientation. Notice we have positive interacting with a positive and negative interacting with a negative."}, {"title": "Molecular Orbital Formation Example .txt", "text": "In other words, we're combining a two p positive and a two p positive. So we get this orientation. Notice we have positive interacting with a positive and negative interacting with a negative. So we have bonding interactions. Likewise, if we combine positive two p and a negative two p, we switch these guys. So this becomes a negative blue and this becomes a positive green and we have this orientation."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So we have bonding interactions. Likewise, if we combine positive two p and a negative two p, we switch these guys. So this becomes a negative blue and this becomes a positive green and we have this orientation. In this orientation we have positive and negative. So we have antiboming and antiboding. So there exists a notal plane smacked between these two guys."}, {"title": "Molecular Orbital Formation Example .txt", "text": "In this orientation we have positive and negative. So we have antiboming and antiboding. So there exists a notal plane smacked between these two guys. Right in the middle. So that means there's no electron density here. So electrons can't be found on this node or nodal plane."}, {"title": "Molecular Orbital Formation Example .txt", "text": "Right in the middle. So that means there's no electron density here. So electrons can't be found on this node or nodal plane. So this, in fact, creates this type of parallel orientation creates two molecular orbitals. One bonding and one anti bonding. So let's draw our energy diagram once again."}, {"title": "Molecular Orbital Formation Example .txt", "text": "So this, in fact, creates this type of parallel orientation creates two molecular orbitals. One bonding and one anti bonding. So let's draw our energy diagram once again. Going up. Our energy increases. Going down."}, {"title": "Molecular Orbital Formation Example .txt", "text": "Going up. Our energy increases. Going down. Our energy decreases. So since our two p orbitals are exactly identical, that means they're on the same energy level. So they're on the same level here."}, {"title": "Molecular Orbital Formation Example .txt", "text": "Our energy decreases. So since our two p orbitals are exactly identical, that means they're on the same energy level. So they're on the same level here. Likewise, because they're identical. They each have one electron. Now these two electrons will want to go in this orbital."}, {"title": "Molecular Orbital Formation Example .txt", "text": "Likewise, because they're identical. They each have one electron. Now these two electrons will want to go in this orbital. Or in that orbital. Well, probably this one. And that's because this one is more stabilizing."}, {"title": "Molecular Orbital Formation Example .txt", "text": "Or in that orbital. Well, probably this one. And that's because this one is more stabilizing. It's stabilizing because it's lower in energy. And nature likes that. Nature creates a system in which the energy is lower than before."}, {"title": "Molecular Orbital Formation Example .txt", "text": "It's stabilizing because it's lower in energy. And nature likes that. Nature creates a system in which the energy is lower than before. And so our electrons will combine to form our bonding molecular orbital. And notice that the spins are different. They're Opposites."}, {"title": "Molecular Orbital Formation Example .txt", "text": "And so our electrons will combine to form our bonding molecular orbital. And notice that the spins are different. They're Opposites. And this fact is due the pole exclusion principle that states that any orbital has the maximum two electrons and these two electrons have opposite spins. Now notice that electrons can still go into this orbital, right? But they don't want to."}, {"title": "Molecular Orbital Formation Example .txt", "text": "And this fact is due the pole exclusion principle that states that any orbital has the maximum two electrons and these two electrons have opposite spins. Now notice that electrons can still go into this orbital, right? But they don't want to. And so that's why you won't find electrons there. If the electrons do somehow end up in this orbital, this orbital will create a destabilizing effect. And that means the nuclei will repel one another and will try to break that covalent bond."}, {"title": "Introduction to Entropy.txt", "text": "I will talk to you about the concept of entropy. So what is entropy? Well, the most basic definition of entropy is that entropy is the measure of a disorder. Every system the best and a very good definition because it's kind of vague and you can't quantify with that definition. You can't use numbers. So let's explore the concept of entropy using public probability, and maybe we can come up with a better definition using probability."}, {"title": "Introduction to Entropy.txt", "text": "Every system the best and a very good definition because it's kind of vague and you can't quantify with that definition. You can't use numbers. So let's explore the concept of entropy using public probability, and maybe we can come up with a better definition using probability. Okay, so let's look at this system here. This system is composed of two containers connected by a bridge, and within the container, there are four different molecules. And these molecules are allowed to diffuse from one side to the other side."}, {"title": "Introduction to Entropy.txt", "text": "Okay, so let's look at this system here. This system is composed of two containers connected by a bridge, and within the container, there are four different molecules. And these molecules are allowed to diffuse from one side to the other side. So let's see what's the most probable situation that we can get. Okay, so what's the likelihood that we're going to get four different molecules all on the left side? Well, this thing could only occur one time, or there's one way that this can occur."}, {"title": "Introduction to Entropy.txt", "text": "So let's see what's the most probable situation that we can get. Okay, so what's the likelihood that we're going to get four different molecules all on the left side? Well, this thing could only occur one time, or there's one way that this can occur. So let's look at this side. What's the likelihood that you get two different molecules on the left side of the container or on the left container? Okay, well, there are six different ways that this can occur."}, {"title": "Introduction to Entropy.txt", "text": "So let's look at this side. What's the likelihood that you get two different molecules on the left side of the container or on the left container? Okay, well, there are six different ways that this can occur. And this means this type of situation is six times as probable. That basically means that if you take a snapshot at any given time of this system, that this snapshot of this picture is more or six times more likely to occur. So now we can come up with a better definition of entropy using probability."}, {"title": "Introduction to Entropy.txt", "text": "And this means this type of situation is six times as probable. That basically means that if you take a snapshot at any given time of this system, that this snapshot of this picture is more or six times more likely to occur. So now we can come up with a better definition of entropy using probability. Entropy is the tendency of a system to take its most probable form. So in the situation or in the system we have here, what's the most probable form? Well, it's clear that this one."}, {"title": "Introduction to Entropy.txt", "text": "Entropy is the tendency of a system to take its most probable form. So in the situation or in the system we have here, what's the most probable form? Well, it's clear that this one. It must be this one. Okay, now you could imagine this is only with four molecules. You could imagine how unlikely this becomes when we get millions and billions of different molecules."}, {"title": "Introduction to Entropy.txt", "text": "It must be this one. Okay, now you could imagine this is only with four molecules. You could imagine how unlikely this becomes when we get millions and billions of different molecules. Okay, this becomes much more likely with greater amount of molecules. Now let's explore the relationship between the second law of thermodynamics and entropy. In another video recently, the second law of thermodynamics basically states that heat cannot be completely converted into work."}, {"title": "Introduction to Entropy.txt", "text": "Okay, this becomes much more likely with greater amount of molecules. Now let's explore the relationship between the second law of thermodynamics and entropy. In another video recently, the second law of thermodynamics basically states that heat cannot be completely converted into work. Here we will see a slightly different definition of the second law of thermodynamics. So let's explore these two isolated systems that are the same size, that have the same number of molecules and the same type of molecules. Which one do you think is more probable to occur?"}, {"title": "Introduction to Entropy.txt", "text": "Here we will see a slightly different definition of the second law of thermodynamics. So let's explore these two isolated systems that are the same size, that have the same number of molecules and the same type of molecules. Which one do you think is more probable to occur? Well, let's find out. Let's use entropy to find out. Okay, well, in this system, the molecules seem to be scattered as far away from each other as possible."}, {"title": "Introduction to Entropy.txt", "text": "Well, let's find out. Let's use entropy to find out. Okay, well, in this system, the molecules seem to be scattered as far away from each other as possible. In this system, they're very structured. They're in one ball. Okay, so let's see which ones are more probable."}, {"title": "Introduction to Entropy.txt", "text": "In this system, they're very structured. They're in one ball. Okay, so let's see which ones are more probable. So in this system, you have bunch of nuclei, positively charged nuclei close to each other. Now, positive charges repel. And so these guys are going to want to naturally move away from each other, as far away from each other as possible."}, {"title": "Introduction to Entropy.txt", "text": "So in this system, you have bunch of nuclei, positively charged nuclei close to each other. Now, positive charges repel. And so these guys are going to want to naturally move away from each other, as far away from each other as possible. They, in fact, would want to form this structure here. Okay? So this more structured system will want to turn into this less structured system."}, {"title": "Introduction to Entropy.txt", "text": "They, in fact, would want to form this structure here. Okay? So this more structured system will want to turn into this less structured system. Okay, so let's go back to our definition, or definitions of entropy. One definition states that entropy is the measure of this order of a system. So since this is more structured, there's more order, so it's less disordered."}, {"title": "Introduction to Entropy.txt", "text": "Okay, so let's go back to our definition, or definitions of entropy. One definition states that entropy is the measure of this order of a system. So since this is more structured, there's more order, so it's less disordered. Okay? This means there is a lower entropy or low entropy because it's more structured, more ordered. Okay?"}, {"title": "Introduction to Entropy.txt", "text": "Okay? This means there is a lower entropy or low entropy because it's more structured, more ordered. Okay? Now, here it's the opposite. Since here there's less structure and less order, it's more disordered, so there is a higher entropy. Now let's go to our second definition of entropy, which basically states that entropy is the tendency of a system to take its most probable form."}, {"title": "Introduction to Entropy.txt", "text": "Now, here it's the opposite. Since here there's less structure and less order, it's more disordered, so there is a higher entropy. Now let's go to our second definition of entropy, which basically states that entropy is the tendency of a system to take its most probable form. So, once again, which one is more probable? Well, this one is more probable. Therefore, there's a higher entropy, and this guy is less probable, so there's a lower entropy."}, {"title": "Introduction to Entropy.txt", "text": "So, once again, which one is more probable? Well, this one is more probable. Therefore, there's a higher entropy, and this guy is less probable, so there's a lower entropy. Okay, so we can refine the second law of thermodynamics into the following. The second law of thermodynamics states that the entropy of an isolated system will never decrease. It will either stay the same, or it will increase."}, {"title": "Hund\u2019s Rule .txt", "text": "And they do so according to Coulomb's law, which is given by the following formula the force that at either charge field due to the other charges given by constant k times charge of one times charge of two divided by distance between them squared. So if I take two electrons, q one and q two a distance R apart this electron, this charge q two, will feel a force due to this charge q one, the same charge, and this force will be in this direction. And this force is given by this law. Likewise, this charge Q one will also feel a force due to this charge q two. And the force will be in the opposite direction with the same magnitude. And it's also given by this equation, Coulomb's law."}, {"title": "Hund\u2019s Rule .txt", "text": "Likewise, this charge Q one will also feel a force due to this charge q two. And the force will be in the opposite direction with the same magnitude. And it's also given by this equation, Coulomb's law. So what Coulomb's Law says is the following if we place two electrons next to each other, they will repel. They will create repulsion forces. So this leads into the following fact placing electrons into the same orbital in a subshell will create repulsion forces."}, {"title": "Hund\u2019s Rule .txt", "text": "So what Coulomb's Law says is the following if we place two electrons next to each other, they will repel. They will create repulsion forces. So this leads into the following fact placing electrons into the same orbital in a subshell will create repulsion forces. And this explains two ideas. This is why a maximum of two electrons can go into an orbital. Because if we place three, four or five electrons into the same orbital, this will increase our force dramatically, creating a lot of repulsion forces, creating a lot of repulsion."}, {"title": "Hund\u2019s Rule .txt", "text": "And this explains two ideas. This is why a maximum of two electrons can go into an orbital. Because if we place three, four or five electrons into the same orbital, this will increase our force dramatically, creating a lot of repulsion forces, creating a lot of repulsion. And that means this idea explains the Poly Exclusion Principle, which states that a maximum two electrons can be placed into any given orbital. Now, this principle also explains Honduras. And what Hungry states is the following."}, {"title": "Hund\u2019s Rule .txt", "text": "And that means this idea explains the Poly Exclusion Principle, which states that a maximum two electrons can be placed into any given orbital. Now, this principle also explains Honduras. And what Hungry states is the following. It says that electrons will not go into an occupied orbital occupied by some electron until all the orbitals within that subshell are already filled. So, for example, let's look at the electron configuration of nitrogen. And it's the following two electrons are placed into the one s orbital."}, {"title": "Hund\u2019s Rule .txt", "text": "It says that electrons will not go into an occupied orbital occupied by some electron until all the orbitals within that subshell are already filled. So, for example, let's look at the electron configuration of nitrogen. And it's the following two electrons are placed into the one s orbital. Two electrons are placed into the two s orbital, right? We're not placing three electrons into our s orbital or four electrons, because only a maximum of two electrons can go into any given orbital because of this principle that we mentioned above. Now, let's look at our p orbitals."}, {"title": "Hund\u2019s Rule .txt", "text": "Two electrons are placed into the two s orbital, right? We're not placing three electrons into our s orbital or four electrons, because only a maximum of two electrons can go into any given orbital because of this principle that we mentioned above. Now, let's look at our p orbitals. Remember, there are three p orbitals. And what Hungry tells us is that before we add two electrons into an orbital, first all the orbitals must be filled with at least one electron. And that's exactly why we first add one electron to the PX orbital."}, {"title": "Hund\u2019s Rule .txt", "text": "Remember, there are three p orbitals. And what Hungry tells us is that before we add two electrons into an orbital, first all the orbitals must be filled with at least one electron. And that's exactly why we first add one electron to the PX orbital. Then we add the second electron to the PY orbital. And then we're adding the third electron into the PV orbital to give us a total of two plus two, four plus three seven electrons. This nitrogen has seven protons and seven electrons in its neutral state."}, {"title": "Hund\u2019s Rule .txt", "text": "Then we add the second electron to the PY orbital. And then we're adding the third electron into the PV orbital to give us a total of two plus two, four plus three seven electrons. This nitrogen has seven protons and seven electrons in its neutral state. Now let's look at oxygen. Oxygen has eight protons, so it has eight electrons. So let's draw the electron configuration according to Hungry."}, {"title": "Hund\u2019s Rule .txt", "text": "Now let's look at oxygen. Oxygen has eight protons, so it has eight electrons. So let's draw the electron configuration according to Hungry. So two electrons are placed into s, and two electrons are placed into the two s. Right. So that's because of the poly exclusion principle. Once again it states a maximum."}, {"title": "Hund\u2019s Rule .txt", "text": "So two electrons are placed into s, and two electrons are placed into the two s. Right. So that's because of the poly exclusion principle. Once again it states a maximum. Two electrons will go into an orbital. Next, we begin filling our p orbitals. We have three p orbitals, and now we have four electrons."}, {"title": "Hund\u2019s Rule .txt", "text": "Two electrons will go into an orbital. Next, we begin filling our p orbitals. We have three p orbitals, and now we have four electrons. So first, we distribute the three electrons the following way. We place one into P, one into Y, and then one into D. And now, since all of them are filled, my fourth electron will go into filling completely this orbital. This PX orbital."}, {"title": "Hund\u2019s Rule .txt", "text": "So first, we distribute the three electrons the following way. We place one into P, one into Y, and then one into D. And now, since all of them are filled, my fourth electron will go into filling completely this orbital. This PX orbital. For example, if I dealt with the next atom, if I had one more electron, I place it into this y and if I had one more electron, I place it into my Z and I get a noble gas configuration. So Hans Rule can be represented in the following graphic way. So let's look at nitrogen."}, {"title": "Hund\u2019s Rule .txt", "text": "For example, if I dealt with the next atom, if I had one more electron, I place it into this y and if I had one more electron, I place it into my Z and I get a noble gas configuration. So Hans Rule can be represented in the following graphic way. So let's look at nitrogen. So here's my energy axis. And here's just my X axis. Now, this bar represents my one s orbital."}, {"title": "Hund\u2019s Rule .txt", "text": "So here's my energy axis. And here's just my X axis. Now, this bar represents my one s orbital. This black bar represents my two S orbital. The reason this one is lower than this one is because one s orbital is at a lower state energy state than the two s orbitals. And so the two s is a bit higher."}, {"title": "Hund\u2019s Rule .txt", "text": "This black bar represents my two S orbital. The reason this one is lower than this one is because one s orbital is at a lower state energy state than the two s orbitals. And so the two s is a bit higher. Likewise, the two PX and the two PX. Two PY and two PZ are higher than either this guy and this guy. That means they will be higher."}, {"title": "Hund\u2019s Rule .txt", "text": "Likewise, the two PX and the two PX. Two PY and two PZ are higher than either this guy and this guy. That means they will be higher. And these guys aren't the same level. So that means they will be at the same level. So now, when I place electrons, I place them in the following way."}, {"title": "Hund\u2019s Rule .txt", "text": "And these guys aren't the same level. So that means they will be at the same level. So now, when I place electrons, I place them in the following way. The upward arrow represents the electron spin of plus one half. The downward arrow represents the electron spin of negative one half. So first I draw my up arrow, my down arrow, and I finished with the one s.\nNext, I put two electrons into my two s. Upward arrow, downward arrow."}, {"title": "Hund\u2019s Rule .txt", "text": "The upward arrow represents the electron spin of plus one half. The downward arrow represents the electron spin of negative one half. So first I draw my up arrow, my down arrow, and I finished with the one s.\nNext, I put two electrons into my two s. Upward arrow, downward arrow. And finally, I put one electron in each. So I begin with the plus one. Half."}, {"title": "Hund\u2019s Rule .txt", "text": "And finally, I put one electron in each. So I begin with the plus one. Half. So upward, upward and upward. And now I'm done. This is my graphic representation of Hans rule for nitrogen."}, {"title": "Hund\u2019s Rule .txt", "text": "So upward, upward and upward. And now I'm done. This is my graphic representation of Hans rule for nitrogen. Now let's look at the graphic representation for Hans Rule for oxygen. So we start by drawing the same bars. And now we start filling our orbitals."}, {"title": "Hund\u2019s Rule .txt", "text": "Now let's look at the graphic representation for Hans Rule for oxygen. So we start by drawing the same bars. And now we start filling our orbitals. So up down. Up down, up up. And then I take my Fourth Electron and Final Electron and put it into my two X."}, {"title": "Hund\u2019s Rule .txt", "text": "So up down. Up down, up up. And then I take my Fourth Electron and Final Electron and put it into my two X. So I draw it down one. Because according to our rules, we can't have electrons that have the same spin in the same orbital. If we put two electrons in the same orbital, they must always have opposite spin."}, {"title": "Acidity of Hydrides .txt", "text": "Now, any compound that's composed of two elements in which one element is a hydrogen is called a Hydride. Now, Hydrides have varying levels of acidity. They could be acidic, neutral, or basic. Now, if we look at our emeritus table, we see a trend. We see that as we go across the period from left to right, the acidity of Hydrides increases. And as we go down a group, the acidity of Hydride also increases."}, {"title": "Acidity of Hydrides .txt", "text": "Now, if we look at our emeritus table, we see a trend. We see that as we go across the period from left to right, the acidity of Hydrides increases. And as we go down a group, the acidity of Hydride also increases. So that means atoms that exist on the left side of the period that form Hydrides are basic atoms that exist in the middle, that transition atoms or transition metals that form Hydrides with H form either mutual or basic Hydrides. And for the most part, atoms that I found on this part on the right side form either weak acids, strong acids, or neutral Hydrides. So for example, let's look at sodium and lithium and potassium."}, {"title": "Acidity of Hydrides .txt", "text": "So that means atoms that exist on the left side of the period that form Hydrides are basic atoms that exist in the middle, that transition atoms or transition metals that form Hydrides with H form either mutual or basic Hydrides. And for the most part, atoms that I found on this part on the right side form either weak acids, strong acids, or neutral Hydrides. So for example, let's look at sodium and lithium and potassium. All these guys are found on the left side. That means they will form basic Hydrides. And in fact, all metal Hydrides are either basic or neutral Hydrides."}, {"title": "Acidity of Hydrides .txt", "text": "All these guys are found on the left side. That means they will form basic Hydrides. And in fact, all metal Hydrides are either basic or neutral Hydrides. Now, we could also say, with the exception of one molecule, all nonmetal Hydrides are either neutral or acidic. The exception is ammonia. Ammonia is the only nonmetal that forms a weak base."}, {"title": "Acidity of Hydrides .txt", "text": "Now, we could also say, with the exception of one molecule, all nonmetal Hydrides are either neutral or acidic. The exception is ammonia. Ammonia is the only nonmetal that forms a weak base. Now, let's examine the right side of the periodic table. On the right side, all the way on the right side, and all the way down, we see that we have strong acids. So HCL and HBR are both strong acids."}, {"title": "Acidity of Hydrides .txt", "text": "Now, let's examine the right side of the periodic table. On the right side, all the way on the right side, and all the way down, we see that we have strong acids. So HCL and HBR are both strong acids. Now, because fluorine is found on the top, it's a weak acid. Remember, as we go up a group, the acid strength decreases. So this guy is a weak acid."}, {"title": "Acidity of Hydrides .txt", "text": "Now, because fluorine is found on the top, it's a weak acid. Remember, as we go up a group, the acid strength decreases. So this guy is a weak acid. Water is neutral, so we leave that alone. Now, these two guys are weak acids as well, because remember, we move back a group, that means our acid strength decreased. So these guys are weak acids."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "Now, previously, we spoke about atomic orbitals and covalent bonds. In this lecture, we'll see how atomic orbitals of atoms combine to form covalent bonds, which are really molecular orbitals. So let's begin with the following simple example. So let's say we want to combine two neutral H atoms in a way to form a diatomic H two molecule. So we want to form a covalent bond. So what must happen first?"}, {"title": "Introduction to Molecular Orbitals .txt", "text": "So let's say we want to combine two neutral H atoms in a way to form a diatomic H two molecule. So we want to form a covalent bond. So what must happen first? Well, initially, these two mutual H atoms are very far apart. And since they're neutral, they each have a proton, a nucleus, and an electron surrounding that nucleus. So what happens as I begin to bring these molecules closer and closer and closer?"}, {"title": "Introduction to Molecular Orbitals .txt", "text": "Well, initially, these two mutual H atoms are very far apart. And since they're neutral, they each have a proton, a nucleus, and an electron surrounding that nucleus. So what happens as I begin to bring these molecules closer and closer and closer? Well, as I begin moving them closer, they begin to feel electrostatic force Coulomb's law. Now, as I move them closer and closer, eventually I will bring them to a point when the electrostatic repulsion of the nuclei of the protons down the nuclei will balance out the electrostatic attraction between the electrons and the protons in the nuclei. In other words, let's look at the following example."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "Well, as I begin moving them closer, they begin to feel electrostatic force Coulomb's law. Now, as I move them closer and closer, eventually I will bring them to a point when the electrostatic repulsion of the nuclei of the protons down the nuclei will balance out the electrostatic attraction between the electrons and the protons in the nuclei. In other words, let's look at the following example. So, when I have these two protons a certain distance apart, the repulsion between these two protons going this way will equal the attraction of these electrons. In other words, this proton of H atom one will attract the electron of H atom two. And likewise, H atom two will attract the electron of H atom one."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "So, when I have these two protons a certain distance apart, the repulsion between these two protons going this way will equal the attraction of these electrons. In other words, this proton of H atom one will attract the electron of H atom two. And likewise, H atom two will attract the electron of H atom one. And in fact, when there are 0.7\nangstromes, angstrom simply means one times ten to the negative 10 meters apart. When they're this distance apart, a bond will form, a covalent bond will form. And in fact, as you move the two H atoms closer and closer from a far distance apart, energy begins to decrease until we reach this point, until we reach 0.7\nangstromes away."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "And in fact, when there are 0.7\nangstromes, angstrom simply means one times ten to the negative 10 meters apart. When they're this distance apart, a bond will form, a covalent bond will form. And in fact, as you move the two H atoms closer and closer from a far distance apart, energy begins to decrease until we reach this point, until we reach 0.7\nangstromes away. So if we graph energy versus our distance between them, we will see that a distance far apart, somewhere right here, we're going to have some energy. And as we begin moving them closer and closer, our energy will begin to decrease until we reach this point. And at this point, we have the minimum amount of energy."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "So if we graph energy versus our distance between them, we will see that a distance far apart, somewhere right here, we're going to have some energy. And as we begin moving them closer and closer, our energy will begin to decrease until we reach this point. And at this point, we have the minimum amount of energy. In other words, nature likes to minimize energy. The more destabilizing it is, the more energy we have. The less energy we have, the more stabilizing our compound is."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "In other words, nature likes to minimize energy. The more destabilizing it is, the more energy we have. The less energy we have, the more stabilizing our compound is. So, in other words, the beginning condition, our initial atoms are at a higher energy than our final molecule. Now, what happens when I continue pushing them closer and closer? Well, as that begins pushing them closer and closer and closer, the repulsion forces begin to increase dramatically."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "So, in other words, the beginning condition, our initial atoms are at a higher energy than our final molecule. Now, what happens when I continue pushing them closer and closer? Well, as that begins pushing them closer and closer and closer, the repulsion forces begin to increase dramatically. And that's exactly why we see that as we go past 0.7 atroms, our energy begins to increase. So as you bring atoms closer and closer, repulsion of the positively charged nuclei causes a sharp increase in energy, as we see here. So as they're moving closer past the distance, the electrons, as well, as protons begin to repel one another, and the energy dramatically increases once again."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "And that's exactly why we see that as we go past 0.7 atroms, our energy begins to increase. So as you bring atoms closer and closer, repulsion of the positively charged nuclei causes a sharp increase in energy, as we see here. So as they're moving closer past the distance, the electrons, as well, as protons begin to repel one another, and the energy dramatically increases once again. To recap, nature likes to form stabilizing structures. Nature will not form a structure that is higher in energy. In other words, if this was higher in energy than this, this molecule would not form."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "To recap, nature likes to form stabilizing structures. Nature will not form a structure that is higher in energy. In other words, if this was higher in energy than this, this molecule would not form. The reason this form spontaneously is because our energy of initial molecules or atoms is lower than the final. So now, let's examine atomic orbitals. So, what is the atomic orbital that our electron is in?"}, {"title": "Introduction to Molecular Orbitals .txt", "text": "The reason this form spontaneously is because our energy of initial molecules or atoms is lower than the final. So now, let's examine atomic orbitals. So, what is the atomic orbital that our electron is in? Well, it's the one S orbital. So let's say we have this atom. Let's call it ha subscript A."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "Well, it's the one S orbital. So let's say we have this atom. Let's call it ha subscript A. And let's call this guy H subscript B. So this here, this PSI, Greek letter PSI, simply represents the orbital or wave function. So let's say PSI subscript haply means that this is the one S orbital of our ha atom."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "And let's call this guy H subscript B. So this here, this PSI, Greek letter PSI, simply represents the orbital or wave function. So let's say PSI subscript haply means that this is the one S orbital of our ha atom. And this is the one s orbital of our HB atom. So when these orbitals atomic orbitals are very far apart, nothing really happens. But as I move them closer and closer, eventually, when I get to this point, these atomic orbitals will overlap, and they will create something known as the molecular orbital or molecular bonding orbital."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "And this is the one s orbital of our HB atom. So when these orbitals atomic orbitals are very far apart, nothing really happens. But as I move them closer and closer, eventually, when I get to this point, these atomic orbitals will overlap, and they will create something known as the molecular orbital or molecular bonding orbital. Now, this guy is represented by PSI phi. So phi bonding is our molecular orbital. So when we see this symbol, we usually think atomic orbitals."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "Now, this guy is represented by PSI phi. So phi bonding is our molecular orbital. So when we see this symbol, we usually think atomic orbitals. And this is five. When we see this symbol five, we think about molecular orbitals. So, once again, atomic orbitals will combine two form molecular orbitals, or also known as covalent bonds."}, {"title": "Introduction to Molecular Orbitals .txt", "text": "And this is five. When we see this symbol five, we think about molecular orbitals. So, once again, atomic orbitals will combine two form molecular orbitals, or also known as covalent bonds. Now, from quantum mechanics, we know that whatever number of atomic orbitals that combine, they will form the same amount of molecular orbitals. In other words, there's a conservation number that we have to take into consideration. So, because two atomic orbitals combined, we should form two molecular orbitals."}, {"title": "Acid Strength.txt", "text": "So before we talk about acid and basis, we must define what acid and bases are. Well, if you want to learn more about the various types of definitions that exist between acid and bases, check out the link below. In this lecture we're going to focus on the bronzed lottery acid based concept. So according to that definition, acids are defined by their ability to donate an H plus ion, while bases are defined by their ability to accept an H plus ion. So what makes a compound X a better acid than compound Y? Well, stronger acids are better at donating than H plus ion than weaker acids are."}, {"title": "Acid Strength.txt", "text": "So according to that definition, acids are defined by their ability to donate an H plus ion, while bases are defined by their ability to accept an H plus ion. So what makes a compound X a better acid than compound Y? Well, stronger acids are better at donating than H plus ion than weaker acids are. And stronger bases are better at accepting that H plus ion than weaker bases. That means the reason that compound X is a better asset than compound Y well, is because compound X releases that H ion with greater ease compared to compound Y. Now, three things make up a good asset bond strength, polarity of bond, and stability of conjugate base."}, {"title": "Acid Strength.txt", "text": "And stronger bases are better at accepting that H plus ion than weaker bases. That means the reason that compound X is a better asset than compound Y well, is because compound X releases that H ion with greater ease compared to compound Y. Now, three things make up a good asset bond strength, polarity of bond, and stability of conjugate base. So how strong is the bond connecting the H ion and the atom in the compound? Well, suppose you're holding somebody's hand and you're holding their hand tightly with a good grip. Well, then the other person will not be able to let go of their hand that easily."}, {"title": "Acid Strength.txt", "text": "So how strong is the bond connecting the H ion and the atom in the compound? Well, suppose you're holding somebody's hand and you're holding their hand tightly with a good grip. Well, then the other person will not be able to let go of their hand that easily. However, if your grip is weak and you're not holding it tightly, then that person will be able to let go of their hand the same way that weak bonds will release the H plus ion with greater ease. So whenever you have a base that comes around, that base will be able to take away that H plus ion if the bond is weak. So weaker bonds equals better acids."}, {"title": "Acid Strength.txt", "text": "However, if your grip is weak and you're not holding it tightly, then that person will be able to let go of their hand the same way that weak bonds will release the H plus ion with greater ease. So whenever you have a base that comes around, that base will be able to take away that H plus ion if the bond is weak. So weaker bonds equals better acids. So let's look at the polarity of our bond. So how polar is our bond? Let's examine HCH bond in a methane and an HCL bond in a hydrochloric acid or HCL."}, {"title": "Acid Strength.txt", "text": "So let's look at the polarity of our bond. So how polar is our bond? Let's examine HCH bond in a methane and an HCL bond in a hydrochloric acid or HCL. So the strength of these bonds is relatively the same. So if we strictly look at part C, the bond strength, we will determine that our acid strength is the same. But that's not the case."}, {"title": "Acid Strength.txt", "text": "So the strength of these bonds is relatively the same. So if we strictly look at part C, the bond strength, we will determine that our acid strength is the same. But that's not the case. This is a much better asset than our methane molecule. So why is that? Well, we have to look at the polarity."}, {"title": "Acid Strength.txt", "text": "This is a much better asset than our methane molecule. So why is that? Well, we have to look at the polarity. Remember, polarity comes from electronegativity. And the greater the difference in electronegativity, the more polar our bond is. So let's look at this guy."}, {"title": "Acid Strength.txt", "text": "Remember, polarity comes from electronegativity. And the greater the difference in electronegativity, the more polar our bond is. So let's look at this guy. The difference between electronegativity between the C atom and the H atom is smaller than the difference between a CL atom and the H atom. And that's because the CL atom is much more electronic negative. That means it's going to pull the electrons toward itself."}, {"title": "Acid Strength.txt", "text": "The difference between electronegativity between the C atom and the H atom is smaller than the difference between a CL atom and the H atom. And that's because the CL atom is much more electronic negative. That means it's going to pull the electrons toward itself. So the density will not be equal, whereas here it will be equal, or pretty much equal. Therefore, this section of the bond will be weak. And so when a base comes around, it will be able to pull away this H atom with greater ease."}, {"title": "Acid Strength.txt", "text": "So the density will not be equal, whereas here it will be equal, or pretty much equal. Therefore, this section of the bond will be weak. And so when a base comes around, it will be able to pull away this H atom with greater ease. Therefore, the more polar our bond, the more likely that it will break and release that H plus ion. Finally, let's look at the stability of conjugate bases. Now, if you forgot what a conjugate acid in base is, check out the link below."}, {"title": "Acid Strength.txt", "text": "Therefore, the more polar our bond, the more likely that it will break and release that H plus ion. Finally, let's look at the stability of conjugate bases. Now, if you forgot what a conjugate acid in base is, check out the link below. Now, we're going to explore the difference between chloric acid and hypochlorous acid. So if we look at the polarity and the bond strength, we see that this ho bond and this ho bond are identical. And that means, according to polarity and bond strength, these assets should have the same strain."}, {"title": "Acid Strength.txt", "text": "Now, we're going to explore the difference between chloric acid and hypochlorous acid. So if we look at the polarity and the bond strength, we see that this ho bond and this ho bond are identical. And that means, according to polarity and bond strength, these assets should have the same strain. But experimentally, we know that chlorot acid is a better asset than the Hypochlorous acid. So let's examine why this has to do with conjugate basis. Let's look at the conjugate base of Hypochlorous acid."}, {"title": "Acid Strength.txt", "text": "But experimentally, we know that chlorot acid is a better asset than the Hypochlorous acid. So let's examine why this has to do with conjugate basis. Let's look at the conjugate base of Hypochlorous acid. While when this H associates, it creates a negative charge on the O atom, creating this guy here. Now, this guy can be resonant stabilized by the formation of a double bond, and this creates a negative charge on the seal atom. So we have two resonance stabilized states."}, {"title": "Acid Strength.txt", "text": "While when this H associates, it creates a negative charge on the O atom, creating this guy here. Now, this guy can be resonant stabilized by the formation of a double bond, and this creates a negative charge on the seal atom. So we have two resonance stabilized states. Now, let's examine the conjugate base of our chloric acid. Well, this guy is resonant stabilized by three states, in fact, four states. I'll show you the last one in a bit."}, {"title": "Acid Strength.txt", "text": "Now, let's examine the conjugate base of our chloric acid. Well, this guy is resonant stabilized by three states, in fact, four states. I'll show you the last one in a bit. So this negative atom can be distributed to this oxygen and then this oxygen. And that happens when this guy forms a double bond, displacing this bond, forming a negative bond here. And then this lone pair forms a double bond with this bond, displacing this lone pair, creating a negative charge here."}, {"title": "Acid Strength.txt", "text": "So this negative atom can be distributed to this oxygen and then this oxygen. And that happens when this guy forms a double bond, displacing this bond, forming a negative bond here. And then this lone pair forms a double bond with this bond, displacing this lone pair, creating a negative charge here. In fact, a fourth resonance stabilized state exists in which three double bonds exist, and the CL atom has a negative charge. So we see that resonance stabilization is good. Whenever we have more resonance stabilization, that means we have a more stable conjugate base."}, {"title": "Acid Strength.txt", "text": "In fact, a fourth resonance stabilized state exists in which three double bonds exist, and the CL atom has a negative charge. So we see that resonance stabilization is good. Whenever we have more resonance stabilization, that means we have a more stable conjugate base. So this guy will exist, and it will be more likely that it will exist than this guy. And therefore, chloride acid will be more likely to go this way to lose and dissociate and to form this state than this guy. This guy will be less likely to dissociate because it's only stabilized by two resonance stabilized states, or its conjugate base is only stabilized by two states versus four states in this case."}, {"title": "Acid Strength.txt", "text": "So this guy will exist, and it will be more likely that it will exist than this guy. And therefore, chloride acid will be more likely to go this way to lose and dissociate and to form this state than this guy. This guy will be less likely to dissociate because it's only stabilized by two resonance stabilized states, or its conjugate base is only stabilized by two states versus four states in this case. So, once again, the more stable our conjugate base is, the stronger our asset. That also means as acid strain decreases, say, from going this guy to going to this guy, the strength of our conjugate base increases. In other words, this guy will be more likely to accept an atom and go to this guy than this guy because this guy exists by itself in a more stable state."}, {"title": "Cell Voltage Equation .txt", "text": "So let's begin by first writing out the two half reactions of this reduction reaction. So let's see which guy is oxidized and which guy is reduced. Well, our iron atom goes from a neutral charge to a plus two charge, while our cavmium atom goes from a plus two charge to a neutral charge. That means this atom loses two electrons and this atom gains those same two electrons. So this is our oxidized atom and our reduced atom or our reducing agent and oxidizing agent. So let's go to step one and let's see our two half reactions."}, {"title": "Cell Voltage Equation .txt", "text": "That means this atom loses two electrons and this atom gains those same two electrons. So this is our oxidized atom and our reduced atom or our reducing agent and oxidizing agent. So let's go to step one and let's see our two half reactions. So our oxidation half reaction is the following. Our solid iron becomes a positively charged molecule plus two electrons because it releases those two electrons while our cadmium aqueous atom gains those two electrons forming our cadmium solid. So this is our reduction reaction and oxidation reaction."}, {"title": "Cell Voltage Equation .txt", "text": "So our oxidation half reaction is the following. Our solid iron becomes a positively charged molecule plus two electrons because it releases those two electrons while our cadmium aqueous atom gains those two electrons forming our cadmium solid. So this is our reduction reaction and oxidation reaction. So let's look at the cell diagram for this electrochemical cell. So remember, these two vertical lines represent the sole bridge and these guys simply represent separations of phases. So this and this are in different phases and these guys are in different phases also."}, {"title": "Cell Voltage Equation .txt", "text": "So let's look at the cell diagram for this electrochemical cell. So remember, these two vertical lines represent the sole bridge and these guys simply represent separations of phases. So this and this are in different phases and these guys are in different phases also. So this is our anode and this is our cathode. So what happens is two electrons leave this atom forming our aqueous iron atom and these two electrons travel via the conductor to this guy reacting with this positively charged atom forming our solid cadmium. So let's go to step two."}, {"title": "Cell Voltage Equation .txt", "text": "So this is our anode and this is our cathode. So what happens is two electrons leave this atom forming our aqueous iron atom and these two electrons travel via the conductor to this guy reacting with this positively charged atom forming our solid cadmium. So let's go to step two. Now, in step two and three, what we want to do or do is find a cell voltage of our electrochemical cell and then use the cell voltage to find our equilibrium constant KC. So let's go to step two. Now, this is our formula that we want to use to find the cell voltage where this is the cell voltage of the reduction reaction and the cell voltage of the oxidation reaction."}, {"title": "Cell Voltage Equation .txt", "text": "Now, in step two and three, what we want to do or do is find a cell voltage of our electrochemical cell and then use the cell voltage to find our equilibrium constant KC. So let's go to step two. Now, this is our formula that we want to use to find the cell voltage where this is the cell voltage of the reduction reaction and the cell voltage of the oxidation reaction. Now, we basically look these guys up on our table for reduction half reactions on the statement conditions and we find that our reduction cell voltage is zero point 43 negative, while our oxidation half reaction is negative zero point 44. Well, actually our reduction going this way because only reduction half reactions are listed. So we have to look at the guy going this way."}, {"title": "Cell Voltage Equation .txt", "text": "Now, we basically look these guys up on our table for reduction half reactions on the statement conditions and we find that our reduction cell voltage is zero point 43 negative, while our oxidation half reaction is negative zero point 44. Well, actually our reduction going this way because only reduction half reactions are listed. So we have to look at the guy going this way. So that is negative zero point 44. Now, I put this negative here in here because we want to convert this to an oxidation because in this anode oxidation that reduction occurs and that's why we have the negative sign here. So what we get is these negatives become a positive and we basically add this guy to this guy and we get zero point 37 volts."}, {"title": "Cell Voltage Equation .txt", "text": "So that is negative zero point 44. Now, I put this negative here in here because we want to convert this to an oxidation because in this anode oxidation that reduction occurs and that's why we have the negative sign here. So what we get is these negatives become a positive and we basically add this guy to this guy and we get zero point 37 volts. This is our cell voltage of our electrochemical cell. Now, in the previous lecture we learned that there's a relationship between our cell voltage and our equilibrium constant, namely this equation here. Now, we also saw in that same lecture that we can convert this formula at 25 degrees Celsius to the following formula."}, {"title": "Cell Voltage Equation .txt", "text": "This is our cell voltage of our electrochemical cell. Now, in the previous lecture we learned that there's a relationship between our cell voltage and our equilibrium constant, namely this equation here. Now, we also saw in that same lecture that we can convert this formula at 25 degrees Celsius to the following formula. Log k equals number of moles times our cell voltage divided by this number here. Zero point 52. Now, this number comes from the fact that both r and F are constants."}, {"title": "Cell Voltage Equation .txt", "text": "Log k equals number of moles times our cell voltage divided by this number here. Zero point 52. Now, this number comes from the fact that both r and F are constants. And at 25 degrees Celsius, t is also constant. Now t is in Kelvin. And we also basically converted our natural log to log of base ten."}, {"title": "Cell Voltage Equation .txt", "text": "And at 25 degrees Celsius, t is also constant. Now t is in Kelvin. And we also basically converted our natural log to log of base ten. Now, let's plug our guides in. So our E is from here, and n, we look at this equation. We see that N represents two moles of electrons."}, {"title": "Cell Voltage Equation .txt", "text": "Now, let's plug our guides in. So our E is from here, and n, we look at this equation. We see that N represents two moles of electrons. So n is two. That's what we get. Two times zero point 37 results from this guy divided by zero point 52, and we get 1.25 equals log k.\nNow, we change this entire thing to exponents, and we get ten to the 1.5."}, {"title": "Cell Voltage Equation .txt", "text": "So n is two. That's what we get. Two times zero point 37 results from this guy divided by zero point 52, and we get 1.25 equals log k.\nNow, we change this entire thing to exponents, and we get ten to the 1.5. You plug that into the calculator, and it's approximately 17.8. So our K is 17.8. And what does that mean?"}, {"title": "Cell Voltage Equation .txt", "text": "You plug that into the calculator, and it's approximately 17.8. So our K is 17.8. And what does that mean? Well, remember we said if our K is above one, that means our reaction is product favorite. It's spontaneous. So this guy, the equilibrium lies on the right."}, {"title": "Cell Voltage Equation .txt", "text": "Well, remember we said if our K is above one, that means our reaction is product favorite. It's spontaneous. So this guy, the equilibrium lies on the right. That means almost all of these guys are converted to our products. And this is the same thing as our e. Remember, our E says what our E gives us a positive value for cell voltage and what the a positive value for cell voltage mean? Remember, a positive value for cell voltage means our reaction is product favored, and a negative value means it's reacting favored."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "So here we have the nonpolar covalent bonds, and here we have the polar covalent bond. Now, notice right off the bat the similarity. Both guys, both bonds are covalent bonds. And what, what that simply means is that there is a sharing of electrons. In other words, one atom donates an electron and a second atom also donates an electron. The difference lies is a polarity and we'll see what that means in just a second."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "And what, what that simply means is that there is a sharing of electrons. In other words, one atom donates an electron and a second atom also donates an electron. The difference lies is a polarity and we'll see what that means in just a second. So let's begin with the non polar covalent bond. So let's suppose we have two atoms. So two atoms that are exactly the same mean they have the same amount of protons in the nucleus and the same amount of electrons."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "So let's begin with the non polar covalent bond. So let's suppose we have two atoms. So two atoms that are exactly the same mean they have the same amount of protons in the nucleus and the same amount of electrons. So here we have our picture where we have our nucleus one with some amount of electrons and nucleus two with the same amount with the same number of electrons. And each nucleus or each atom donates an electron. So one coming from this atom and the second coming from this atom."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "So here we have our picture where we have our nucleus one with some amount of electrons and nucleus two with the same amount with the same number of electrons. And each nucleus or each atom donates an electron. So one coming from this atom and the second coming from this atom. So let's look at Coulomb's law for a second. Coulomb's law gives us the force felt by two charges some distance apart. What it states is the following."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "So let's look at Coulomb's law for a second. Coulomb's law gives us the force felt by two charges some distance apart. What it states is the following. If we have two charges, q one and Q two, if we multiply them together and multiplied by the constant k and divided by the distance between their center of charges squared, we get the force that each charge feels due to the other charge. Remember, plus charges repel and plus and minus attract. So notice what we have here."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "If we have two charges, q one and Q two, if we multiply them together and multiplied by the constant k and divided by the distance between their center of charges squared, we get the force that each charge feels due to the other charge. Remember, plus charges repel and plus and minus attract. So notice what we have here. The charge in this nucleus is the same as the charge in this nucleus because we have the same amount of protons in those nuclei. And we both have one electrons, one electron here and one electron here. And they also have the same amount of charge."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "The charge in this nucleus is the same as the charge in this nucleus because we have the same amount of protons in those nuclei. And we both have one electrons, one electron here and one electron here. And they also have the same amount of charge. So that means that this nuclei will exert a force on this electron and this force will be equal to the force that this nuclei nucleus exerts on this electron. So this electron will pull this or this proton will pull this electron with the same amount of force that this proton will pull on this electron. So there will be an equal distribution of charge."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "So that means that this nuclei will exert a force on this electron and this force will be equal to the force that this nuclei nucleus exerts on this electron. So this electron will pull this or this proton will pull this electron with the same amount of force that this proton will pull on this electron. So there will be an equal distribution of charge. These electrons will be found equidistant between these two nuclei. Distance will be the same exact. And this only occurs when we have the same amount of protons found in the nucleus."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "These electrons will be found equidistant between these two nuclei. Distance will be the same exact. And this only occurs when we have the same amount of protons found in the nucleus. So for example, if we have an H and an H, both nuclei have one protons each. If we have an F and an F, both nuclei have nine protons each and so on. Basically, when we have two of the same atoms, we're going to have a non polar covalent bond as we have here and as we have here."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "So for example, if we have an H and an H, both nuclei have one protons each. If we have an F and an F, both nuclei have nine protons each and so on. Basically, when we have two of the same atoms, we're going to have a non polar covalent bond as we have here and as we have here. Now, notice we have a double bond here. But it doesn't change the fact that this is a nonpolar covalent bond because we have an oxygen with some amount of protons and a second oxygen with the same amount of protons. Now, another way to look at it is via electronegativity."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Now, notice we have a double bond here. But it doesn't change the fact that this is a nonpolar covalent bond because we have an oxygen with some amount of protons and a second oxygen with the same amount of protons. Now, another way to look at it is via electronegativity. Now, both atoms have the same amount of electronegativity and that simply means they will attract electrons with the same affinity. And that basically means that our electrons will be found smack in the middle. They'll be equidistant between our two atoms."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Now, both atoms have the same amount of electronegativity and that simply means they will attract electrons with the same affinity. And that basically means that our electrons will be found smack in the middle. They'll be equidistant between our two atoms. And so we're going to have symmetry. In other words, if we take a line and cross it this way, this section will be symmetrical to this section. And that means we're going to have a non polar covalent bond."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "And so we're going to have symmetry. In other words, if we take a line and cross it this way, this section will be symmetrical to this section. And that means we're going to have a non polar covalent bond. Now let's look at polar covalent bonds. Polar covalent bond simply means there will be an unequal distribution of charge. And let's see why."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Now let's look at polar covalent bonds. Polar covalent bond simply means there will be an unequal distribution of charge. And let's see why. Well, suppose we have an atom and a second atom that have different number of protons in their nuclei. Suppose we have a larger atom with a larger number of protons than this atom. And what that basically means, because of coulomb law, because the charge will be greater for this nucleus than for this nucleus, the force that these electrons feel due to this nucleus will be larger than due to this nucleus."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Well, suppose we have an atom and a second atom that have different number of protons in their nuclei. Suppose we have a larger atom with a larger number of protons than this atom. And what that basically means, because of coulomb law, because the charge will be greater for this nucleus than for this nucleus, the force that these electrons feel due to this nucleus will be larger than due to this nucleus. And so there will be an unequal sharing or an unequal distribution of electrons. So there will be an unequal distribution of charge between these two atoms. And that means we're going to have a polar covalent bond."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "And so there will be an unequal sharing or an unequal distribution of electrons. So there will be an unequal distribution of charge between these two atoms. And that means we're going to have a polar covalent bond. Now, another way we're presenting this is by the following depiction. So because our electrons will be closer to the larger atom, we're going to develop a partial, not a full, but a partial negative charge. This simply means partial negative."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Now, another way we're presenting this is by the following depiction. So because our electrons will be closer to the larger atom, we're going to develop a partial, not a full, but a partial negative charge. This simply means partial negative. Now, there will be a partial positive charge because electrons will be shifted this way. There will be a partial positive charge on this smaller atom. Now, examples include HF, hohc and many, many more examples."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Now, there will be a partial positive charge because electrons will be shifted this way. There will be a partial positive charge on this smaller atom. Now, examples include HF, hohc and many, many more examples. Basically, whenever you have two different atoms, such as here HF, we're going to have an equal distribution of charge. Electrons are going to be closer to the larger atom because this f, for example, has nine protons, while this h has only one proton. So our nucleus will pull these electrons stronger than the h. So our electrons will be closer this way."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Basically, whenever you have two different atoms, such as here HF, we're going to have an equal distribution of charge. Electrons are going to be closer to the larger atom because this f, for example, has nine protons, while this h has only one proton. So our nucleus will pull these electrons stronger than the h. So our electrons will be closer this way. Now, another way of representing this unequal distribution is by using this arrow. So we draw an arrow towards where our electrons are being pulled and our electrons are being pulled towards the larger nucleus. So towards the f.\nAnd we draw kind of a plus sign on the end where there's a partial positive charge."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "Now, another way of representing this unequal distribution is by using this arrow. So we draw an arrow towards where our electrons are being pulled and our electrons are being pulled towards the larger nucleus. So towards the f.\nAnd we draw kind of a plus sign on the end where there's a partial positive charge. So the same thing goes for h and o, we're going to have an arrow this way and we're going to have an arrow this way. Now, another way of looking at this is via electronegativity. The atom that has a larger electronegativity, it will pull or attract electrons more strongly."}, {"title": "Polar and Nonpolar Covalent Bonds .txt", "text": "So the same thing goes for h and o, we're going to have an arrow this way and we're going to have an arrow this way. Now, another way of looking at this is via electronegativity. The atom that has a larger electronegativity, it will pull or attract electrons more strongly. That means we're going to have because this one is more electronegative than this atom. And this one is more electronegative than this and this is more electronegative than this. We're going to have our error pointing in this direction and we're going to have a polar covalent bond."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "So for any given alkene, may different types of isomers can exist. In this lecture we're going to compare the sits or the Z isomers with the trans or the E isomers. And we're going to discuss which ones are more stable. So let's suppose we're working with three hexane. Now three hexane has two types of isomers. It has the trans three hexen isomer or the e three hexenisomer."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "So let's suppose we're working with three hexane. Now three hexane has two types of isomers. It has the trans three hexen isomer or the e three hexenisomer. And there's also the S three hexenisomer or the Z three hexane isomer. Now, trans simply means the smaller H groups are on opposite sides of the double bond, while the E means that the two higher priority groups are in opposite sides of our double bond. Likewise, sin simply means that the two H groups are on the same side of the double bond, while the Z means that the two priority, the higher priority groups are the same side of the double bond."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "And there's also the S three hexenisomer or the Z three hexane isomer. Now, trans simply means the smaller H groups are on opposite sides of the double bond, while the E means that the two higher priority groups are in opposite sides of our double bond. Likewise, sin simply means that the two H groups are on the same side of the double bond, while the Z means that the two priority, the higher priority groups are the same side of the double bond. So we have trans and we have the CIS three hexane. So let's examine our three dimensional model of this three hexane. So here we have our CIS three hexane."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "So we have trans and we have the CIS three hexane. So let's examine our three dimensional model of this three hexane. So here we have our CIS three hexane. So here's our double bond and here our single bonds. So we have the methyl group, the methyl group and our two HS here. Now, one important detail that you must remember about double bonds versus single bonds."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "So here's our double bond and here our single bonds. So we have the methyl group, the methyl group and our two HS here. Now, one important detail that you must remember about double bonds versus single bonds. Double bonds do not rotate their rigid while single bonds, single covalent bonds are able to rotate. So this bond here will rotate and this bond here will also rotate. So let's suppose that this bond rotates and this bond also rotates."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "Double bonds do not rotate their rigid while single bonds, single covalent bonds are able to rotate. So this bond here will rotate and this bond here will also rotate. So let's suppose that this bond rotates and this bond also rotates. What will happen then? Well, if these two bonds rotate, look at what happens when they rotate. They will bump and this bumping will cause steric hinderance."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "What will happen then? Well, if these two bonds rotate, look at what happens when they rotate. They will bump and this bumping will cause steric hinderance. This will interfere and destabilize this CIS three hexane. In other words, there is this bumping effect when these two single bonds rotate. And this means or this creates a high energy destabilizing interaction in the CIS compound."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "This will interfere and destabilize this CIS three hexane. In other words, there is this bumping effect when these two single bonds rotate. And this means or this creates a high energy destabilizing interaction in the CIS compound. What about our trans compound? Let's suppose we move this ethyl group into or onto the bottom and now we have our transexane. So now no matter how much they rotate, there's no interaction between these destabilizing ethyl groups between these large ethyl groups."}, {"title": "Cis-Trans Z-e Isomer Stability .txt", "text": "What about our trans compound? Let's suppose we move this ethyl group into or onto the bottom and now we have our transexane. So now no matter how much they rotate, there's no interaction between these destabilizing ethyl groups between these large ethyl groups. And that means the trans isomer is the more stable isomer because it does not have the destabilizing interaction between the two ethyl groups like in the CIS compound. Therefore, it has a lower or more negative enthalpy of formation. There is a difference of about one kilogal per mole of energy between this guy and this compound."}, {"title": "Using a Barometer .txt", "text": "Now, you take the cup, you fill it up with mercury. You take the tube, you fill that up to the rib of mercury. You take the tube, flip it upside down, and place it into the cup. Now, once you place it to the cup, a vacuum is created in this section. And that's because the force of gravity pulls down on the liquid. But it will pull down to a certain extent."}, {"title": "Using a Barometer .txt", "text": "Now, once you place it to the cup, a vacuum is created in this section. And that's because the force of gravity pulls down on the liquid. But it will pull down to a certain extent. And that's because the air molecules found on the outside, such as oxygen molecules, carbon dioxide molecules, nitrogen molecules, exert a certain force on these two sections here on this liquid. And when the force of these molecules equals the force of the gravity on the liquid, when these two forces equal, it will stop pulling down, and it will level off at some height H. So then you can calculate that height, H, from this point to this point. And you can use this formula to find the pressure of the atmosphere where pressure is equal to density of liquid times gravitational constant times height H.\nSo let's see what happens when the atmosphere pressure increases."}, {"title": "Using a Barometer .txt", "text": "And that's because the air molecules found on the outside, such as oxygen molecules, carbon dioxide molecules, nitrogen molecules, exert a certain force on these two sections here on this liquid. And when the force of these molecules equals the force of the gravity on the liquid, when these two forces equal, it will stop pulling down, and it will level off at some height H. So then you can calculate that height, H, from this point to this point. And you can use this formula to find the pressure of the atmosphere where pressure is equal to density of liquid times gravitational constant times height H.\nSo let's see what happens when the atmosphere pressure increases. Suppose you decrease the altitude where you are measuring the pressure, so the pressure increases. If the pressure increases, then the force will increase. That's pushing down this liquid."}, {"title": "Using a Barometer .txt", "text": "Suppose you decrease the altitude where you are measuring the pressure, so the pressure increases. If the pressure increases, then the force will increase. That's pushing down this liquid. It will push the liquid down lower and will raise this higher. So h will increase. And according to the formula, we see just that."}, {"title": "Using a Barometer .txt", "text": "It will push the liquid down lower and will raise this higher. So h will increase. And according to the formula, we see just that. If the pressure has increased while these two guys are held constant, h will increase. The same is true for a lower pressure. Suppose we go higher up to some mountain top."}, {"title": "Using a Barometer .txt", "text": "If the pressure has increased while these two guys are held constant, h will increase. The same is true for a lower pressure. Suppose we go higher up to some mountain top. At this mountaintop, the pressure is lower. So H will be lower as well, because these two guys are constant. And that's because if the pressure is lower, the force of gravity will be higher."}, {"title": "Using a Barometer .txt", "text": "At this mountaintop, the pressure is lower. So H will be lower as well, because these two guys are constant. And that's because if the pressure is lower, the force of gravity will be higher. And so it will push this level up, and it will lower this level. Okay? So the height, the total height, will be lower."}, {"title": "Using a Barometer .txt", "text": "And so it will push this level up, and it will lower this level. Okay? So the height, the total height, will be lower. Now, last thing I want to talk about is this little vacuum here. Now, remember, this liquid we chose, mercury, will evaporate some of its molecules. So technically, this isn't a vacuum."}, {"title": "Using a Barometer .txt", "text": "Now, last thing I want to talk about is this little vacuum here. Now, remember, this liquid we chose, mercury, will evaporate some of its molecules. So technically, this isn't a vacuum. And there are some vapor molecules flowing around or flying around in this area in the space here. Now, what happens if we increase the vapor pressure here? If we increase the vapor pressure, the pressure due to the molecules in the air will push on this section here."}, {"title": "Using a Barometer .txt", "text": "And there are some vapor molecules flowing around or flying around in this area in the space here. Now, what happens if we increase the vapor pressure here? If we increase the vapor pressure, the pressure due to the molecules in the air will push on this section here. It will decrease this level and decrease the height. So there will be some discrepancy in the atmospheric pressure if there are molecules found within this section. So the more volatile the liquid that's used in this section, the higher the bigger pressure is."}, {"title": "Using a Barometer .txt", "text": "It will decrease this level and decrease the height. So there will be some discrepancy in the atmospheric pressure if there are molecules found within this section. So the more volatile the liquid that's used in this section, the higher the bigger pressure is. And that means the larger the difference from here to here, the lower this section will be. Now, the same way that you could measure the temperature of the atmosphere using a thermometer, you can also measure atmospheric pressure using a barometer. A barometer can be semblasing three things a long cylindrical tube open at one end and then select usually mercury and a cup."}, {"title": "Using a Barometer .txt", "text": "And that means the larger the difference from here to here, the lower this section will be. Now, the same way that you could measure the temperature of the atmosphere using a thermometer, you can also measure atmospheric pressure using a barometer. A barometer can be semblasing three things a long cylindrical tube open at one end and then select usually mercury and a cup. Now, you take the cup, you fill it up with mercury. You take the tube, you fill that up to the rib of mercury. You take the tube, flip it upside down, and place it into the cup."}, {"title": "Using a Barometer .txt", "text": "Now, you take the cup, you fill it up with mercury. You take the tube, you fill that up to the rib of mercury. You take the tube, flip it upside down, and place it into the cup. Now, once you place it to the cup, a vacuum is created in this section. And that's because the force of gravity pulls down on the liquid. But it will pull down to a certain extent."}, {"title": "Using a Barometer .txt", "text": "Now, once you place it to the cup, a vacuum is created in this section. And that's because the force of gravity pulls down on the liquid. But it will pull down to a certain extent. And that's because the air molecules found on the outside, such as oxygen molecules, carbon dioxide molecules, nitrogen molecules, exert a certain force on these two sections here on this liquid. And when the force of these molecules equals the force of the gravity on the liquid, when these two forces equal, it will stop pulling down, and it will level off at some height H. So then you can calculate that height H from this point to this point. And you can use this formula to find the pressure of the atmosphere where pressure is equal to density of liquid times gravitational constant times height H.\nSo let's see what happens when the atmosphere pressure increases."}, {"title": "Using a Barometer .txt", "text": "And that's because the air molecules found on the outside, such as oxygen molecules, carbon dioxide molecules, nitrogen molecules, exert a certain force on these two sections here on this liquid. And when the force of these molecules equals the force of the gravity on the liquid, when these two forces equal, it will stop pulling down, and it will level off at some height H. So then you can calculate that height H from this point to this point. And you can use this formula to find the pressure of the atmosphere where pressure is equal to density of liquid times gravitational constant times height H.\nSo let's see what happens when the atmosphere pressure increases. Suppose you decrease the altitude where you are measuring the pressure, so the pressure increases. If the pressure increases, then the force will increase. That's pushing down this liquid."}, {"title": "Using a Barometer .txt", "text": "Suppose you decrease the altitude where you are measuring the pressure, so the pressure increases. If the pressure increases, then the force will increase. That's pushing down this liquid. It will push the liquid down lower and will raise this higher. So h will increase. And according to the formula, we see just that."}, {"title": "Using a Barometer .txt", "text": "It will push the liquid down lower and will raise this higher. So h will increase. And according to the formula, we see just that. If the pressure has increased while these two guys are held constant, h will increase. The same is true for a lower pressure. Supposedly, go higher up to some mountain top."}, {"title": "Using a Barometer .txt", "text": "If the pressure has increased while these two guys are held constant, h will increase. The same is true for a lower pressure. Supposedly, go higher up to some mountain top. At this mountaintop, the pressure is lower. So H will be lower as well, because these two guys are constant. And that's because if the pressure is lower, the force of gravity will be higher."}, {"title": "Using a Barometer .txt", "text": "At this mountaintop, the pressure is lower. So H will be lower as well, because these two guys are constant. And that's because if the pressure is lower, the force of gravity will be higher. And so it will push this level up, and it will lower this level. Okay? So the height, the total height, will be lower."}, {"title": "Using a Barometer .txt", "text": "And so it will push this level up, and it will lower this level. Okay? So the height, the total height, will be lower. Now, last thing I want to talk about is this little vacuum here. Now, remember, this liquid we chose, mercury, will evaporate some of its molecules. So, technically, this isn't a vacuum."}, {"title": "Using a Barometer .txt", "text": "Now, last thing I want to talk about is this little vacuum here. Now, remember, this liquid we chose, mercury, will evaporate some of its molecules. So, technically, this isn't a vacuum. And there are some vapor molecules flowing around or flying around in this area in the space here. Now, what happens if we increase the vapor pressure here? If we increase the vapor pressure, the pressure due to the molecules in the air will push on this section here."}, {"title": "Definition of Temperature .txt", "text": "In this lecture, I will talk to you about the concept of temperature. Now, temperature can be defined in many different ways. In this video, I've outlined three major definitions of temperature. The first definition talks about energy transfer. The second definition talks about the ideal gas law. And the third definition couples thermal energy or kinetic energy with temperature."}, {"title": "Definition of Temperature .txt", "text": "The first definition talks about energy transfer. The second definition talks about the ideal gas law. And the third definition couples thermal energy or kinetic energy with temperature. Now let's go to the first definition. Now let's remember what heat is. Heat is a transfer of energy from a cold body to a hot body, which means that an energy transfer only occurs when there is a difference in temperature."}, {"title": "Definition of Temperature .txt", "text": "Now let's go to the first definition. Now let's remember what heat is. Heat is a transfer of energy from a cold body to a hot body, which means that an energy transfer only occurs when there is a difference in temperature. So one way to define temperature is to say that temperature is a property of matter that determines if energy transfer can occur. That is, if there is no difference in temperature. If the temperature of one object and a second object are the same, no energy transfer due to heat can occur."}, {"title": "Definition of Temperature .txt", "text": "So one way to define temperature is to say that temperature is a property of matter that determines if energy transfer can occur. That is, if there is no difference in temperature. If the temperature of one object and a second object are the same, no energy transfer due to heat can occur. Okay? And that's pretty intuitive. Let's look at the second definition, the ideal gas law."}, {"title": "Definition of Temperature .txt", "text": "Okay? And that's pretty intuitive. Let's look at the second definition, the ideal gas law. Well, the ideal gas law tells us that pressure times volume equals number of moles times a constant R times temperature t.\nFrom that, when we keep pressure constant and the number of moles constant, we get Charles Law or v one over t one equals V two over t two. And when we graph this, we get the following. Suppose we graph it at one ATM, we get a linear relationship."}, {"title": "Definition of Temperature .txt", "text": "Well, the ideal gas law tells us that pressure times volume equals number of moles times a constant R times temperature t.\nFrom that, when we keep pressure constant and the number of moles constant, we get Charles Law or v one over t one equals V two over t two. And when we graph this, we get the following. Suppose we graph it at one ATM, we get a linear relationship. If we graph it at three ATM, we also get a linear relationship. If we graph it at five at M, we also get a linear relationship. Now, at any other temperature, it would also show a linear relationship."}, {"title": "Definition of Temperature .txt", "text": "If we graph it at three ATM, we also get a linear relationship. If we graph it at five at M, we also get a linear relationship. Now, at any other temperature, it would also show a linear relationship. Now, one thing can be seen from this graph is that at every single temperature, the graph is a set at exactly one point. And that one point is right here on the x axis. And we can define temperature to be zero Kelvin at this point."}, {"title": "Definition of Temperature .txt", "text": "Now, one thing can be seen from this graph is that at every single temperature, the graph is a set at exactly one point. And that one point is right here on the x axis. And we can define temperature to be zero Kelvin at this point. That is below this, right? It cannot exist. No temperature below this can't exist."}, {"title": "Definition of Temperature .txt", "text": "That is below this, right? It cannot exist. No temperature below this can't exist. There are only temperatures above absolute zero. And that's another way to define temperature, using graphs and using equations. So this isn't that intuitive."}, {"title": "Definition of Temperature .txt", "text": "There are only temperatures above absolute zero. And that's another way to define temperature, using graphs and using equations. So this isn't that intuitive. This is more intuitive. Now, the third definition is also intuitive. The third definition talks about thermal energy or a kinetic energy of a system."}, {"title": "Definition of Temperature .txt", "text": "This is more intuitive. Now, the third definition is also intuitive. The third definition talks about thermal energy or a kinetic energy of a system. Now, remember, the kinetic energy is the translational energies plus the rotational energies plus the vibrational energies of a system, right? So when we talk about fluids, we talk about mainly translational and rotational energies. And we can formulate this formula here, which basically says translational energy plus rotational energy equals three over two times the constant K times temperature."}, {"title": "Definition of Temperature .txt", "text": "Now, remember, the kinetic energy is the translational energies plus the rotational energies plus the vibrational energies of a system, right? So when we talk about fluids, we talk about mainly translational and rotational energies. And we can formulate this formula here, which basically says translational energy plus rotational energy equals three over two times the constant K times temperature. So from this equation, we see that if we increase temperature, we increase translational energy or the speed, and we increase rotational energy or the torque for solace. However, there's really no translational energies or rotational energies at low temperature. So we can't really have a relation between temperature and kinetic energy at low temperatures."}, {"title": "Definition of Temperature .txt", "text": "So from this equation, we see that if we increase temperature, we increase translational energy or the speed, and we increase rotational energy or the torque for solace. However, there's really no translational energies or rotational energies at low temperature. So we can't really have a relation between temperature and kinetic energy at low temperatures. Now, at high temperatures, solids actually begin to translate and rotate slightly. So at high temperatures, we can't use this formula for solids. But what we could say is that for high temperatures, as temperature increases, so does the kinetic energy."}, {"title": "Definition of Temperature .txt", "text": "Now, at high temperatures, solids actually begin to translate and rotate slightly. So at high temperatures, we can't use this formula for solids. But what we could say is that for high temperatures, as temperature increases, so does the kinetic energy. Now, one last thing I want to mention is from this definition, I said that temperature is a property. Now, more specifically, temperature is an intensive property. Now remember, when you divide an extensive property by another extensive property, you get an intensive property."}, {"title": "Definition of Temperature .txt", "text": "Now, one last thing I want to mention is from this definition, I said that temperature is a property. Now, more specifically, temperature is an intensive property. Now remember, when you divide an extensive property by another extensive property, you get an intensive property. And I mentioned this in my previous video, okay? And we see this from this fact here. Kinetic energy, which is an extensive property."}, {"title": "Definition of Temperature .txt", "text": "And I mentioned this in my previous video, okay? And we see this from this fact here. Kinetic energy, which is an extensive property. If you divide it by the number of moles, which is also an extensive property, you get temperature, right? And so extensive property divided by an extensive property gives you an intensive property. And so that's why temperature must be an intensive property."}, {"title": "Solution formation and heat of solution .txt", "text": "Intermolecular bonds are non COBALTING forces that hold two or more different molecules in place. Suppose we have a single molecule. What holds that molecule in place? Well, covalent bonds or the shared electrons between different atoms of that same molecule hold that molecule in place. What holds a bunch of different molecules in place? Well, noncovalent forces such as dipole forces or lumpin forces hold these guys in place."}, {"title": "Solution formation and heat of solution .txt", "text": "Well, covalent bonds or the shared electrons between different atoms of that same molecule hold that molecule in place. What holds a bunch of different molecules in place? Well, noncovalent forces such as dipole forces or lumpin forces hold these guys in place. And these forces are called intermolecular forces. So intermocular forces are simply non covalent bonds between different molecules. They hold different molecules in place."}, {"title": "Solution formation and heat of solution .txt", "text": "And these forces are called intermolecular forces. So intermocular forces are simply non covalent bonds between different molecules. They hold different molecules in place. Now what are solutions? Solutions are mixtures of at least two compounds. Suppose we had a Beaker A with compound X and Baker B with compound Y?"}, {"title": "Solution formation and heat of solution .txt", "text": "Now what are solutions? Solutions are mixtures of at least two compounds. Suppose we had a Beaker A with compound X and Baker B with compound Y? Now suppose we wanted to make a solution out of compound X and Y. So we mix them. What would have to happen before a solution is formed?"}, {"title": "Solution formation and heat of solution .txt", "text": "Now suppose we wanted to make a solution out of compound X and Y. So we mix them. What would have to happen before a solution is formed? So by definition, we said solution formation is a formation of intermolecular bonds between X and Y. Well, before a bond is formed between X and Y, the bonds between the XS and the YS must break. So before X and Y forms a solution, intermolecular bonds between compound X and intermolecular bonds between compound Y must break."}, {"title": "Solution formation and heat of solution .txt", "text": "So by definition, we said solution formation is a formation of intermolecular bonds between X and Y. Well, before a bond is formed between X and Y, the bonds between the XS and the YS must break. So before X and Y forms a solution, intermolecular bonds between compound X and intermolecular bonds between compound Y must break. Once all these intermocular bonds break, then intermolecular bonds between compounds X and Y must form. Now remember, the breaking of a bond is endothermic. So these guys are endothermic."}, {"title": "Solution formation and heat of solution .txt", "text": "Once all these intermocular bonds break, then intermolecular bonds between compounds X and Y must form. Now remember, the breaking of a bond is endothermic. So these guys are endothermic. The formation of bonds is exothermic or energy is released. Now, from another lecture we saw that change in enthalpy is equal to a change in internal energy plus PD work done or the work done by the system on the environment to create that system. We also saw that if pressure is held constant and the number of moles is held constant, the change in volume is zero."}, {"title": "Solution formation and heat of solution .txt", "text": "The formation of bonds is exothermic or energy is released. Now, from another lecture we saw that change in enthalpy is equal to a change in internal energy plus PD work done or the work done by the system on the environment to create that system. We also saw that if pressure is held constant and the number of moles is held constant, the change in volume is zero. If the change in volume is zero, this term becomes zero. So we can say change in enthalpy is simply equal to change in internal energy. Remember, when bonds are broken, reaction is endothermic."}, {"title": "Solution formation and heat of solution .txt", "text": "If the change in volume is zero, this term becomes zero. So we can say change in enthalpy is simply equal to change in internal energy. Remember, when bonds are broken, reaction is endothermic. Okay? Energy is required to break a bond. So the enthalpy of this system, of this guy is positive."}, {"title": "Solution formation and heat of solution .txt", "text": "Okay? Energy is required to break a bond. So the enthalpy of this system, of this guy is positive. Enthalpy change of this guy is also positive. Now, when bonds are formed, energy is released, they're exothermic. So the formation of Ynex is negative."}, {"title": "Solution formation and heat of solution .txt", "text": "Enthalpy change of this guy is also positive. Now, when bonds are formed, energy is released, they're exothermic. So the formation of Ynex is negative. Now if we add all these guys up, we get something called heat of solution. And heat of solution can tell you if the solution formation is exothermic or endothermic. Now if this guy is negative, it's exothermic."}, {"title": "Solution formation and heat of solution .txt", "text": "Now if we add all these guys up, we get something called heat of solution. And heat of solution can tell you if the solution formation is exothermic or endothermic. Now if this guy is negative, it's exothermic. And bonds formed are stronger or more stable than the bonds broken. So this bond between X and Y is greater or is stronger and more stable than that bond or this bond. If the heat of solution is positive, the bonds formed are weaker than the bonds broken."}, {"title": "Solution formation and heat of solution .txt", "text": "And bonds formed are stronger or more stable than the bonds broken. So this bond between X and Y is greater or is stronger and more stable than that bond or this bond. If the heat of solution is positive, the bonds formed are weaker than the bonds broken. So the bond here is weaker and less stable than the bond here. Or here. Now, entropy will never tell you if a solution is spontaneous."}, {"title": "Solution formation and heat of solution .txt", "text": "So the bond here is weaker and less stable than the bond here. Or here. Now, entropy will never tell you if a solution is spontaneous. Likewise, heat of solution does not tell you if a reaction is spontaneous. Now, remember, only entropy dictates spontaneity. Luckily, solutions usually increase in entropy, and entropy determine spontaneity."}, {"title": "Solution formation and heat of solution .txt", "text": "Likewise, heat of solution does not tell you if a reaction is spontaneous. Now, remember, only entropy dictates spontaneity. Luckily, solutions usually increase in entropy, and entropy determine spontaneity. Now, under right conditions, solutions are usually spontaneous. And that's because entropy is increased. It's not because the reaction is exothermic."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "The force on either of the proton or the electron is given by k, a constant times q one, the q, or the charge of the protons times q two, the charge, the electrons divided by the distance between them squared. And what Coulomb's law does, it describes the force and electron experiences due to the pull of the proton nucleus. So let's look at two atoms, and let's compare these guys. First, let's look at hydrogen, and then let's look at helium. Well, recall that hydrogen has exactly one proton and one neutron in its neutral state. And that means we can find the electrostatic force or the pull on this electron due to this proton by simply applying our coulombs equation."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "First, let's look at hydrogen, and then let's look at helium. Well, recall that hydrogen has exactly one proton and one neutron in its neutral state. And that means we can find the electrostatic force or the pull on this electron due to this proton by simply applying our coulombs equation. And this guy states that our force due to this electron or on this electron due to this proton is given by k, a constant times the charge of our proton, times the charge of our electron divided by our distance r between them squared. And this is the force that this guy feels due to this guy and also the force that this proton feels due to this electron. Now, note that their forces are negative, but they have the same magnitude."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "And this guy states that our force due to this electron or on this electron due to this proton is given by k, a constant times the charge of our proton, times the charge of our electron divided by our distance r between them squared. And this is the force that this guy feels due to this guy and also the force that this proton feels due to this electron. Now, note that their forces are negative, but they have the same magnitude. So now let's look at helium. Remember, the outermost electrons are the electrons that will be pulled away. The inner electrons found on lower energy levels will not be pulled away first."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So now let's look at helium. Remember, the outermost electrons are the electrons that will be pulled away. The inner electrons found on lower energy levels will not be pulled away first. So in helium, we now have two protons and two electrons. So our nucleus is composed of two protons, and our shells are composed of one electron each. So now we have an inner shell and an outermost shell."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So in helium, we now have two protons and two electrons. So our nucleus is composed of two protons, and our shells are composed of one electron each. So now we have an inner shell and an outermost shell. And recall that electrons on the outermost shell will be pulled away. So this is the electron we have to worry about. This is the electron that will be pulled away."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "And recall that electrons on the outermost shell will be pulled away. So this is the electron we have to worry about. This is the electron that will be pulled away. So if we calculate the force that this guy experiences due to this proton nucleus, we'll see a discrepancy. And this discrepancy comes from something called the shielding effect. Now, this innermost atom or this innermost electron creates a shielding effect."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So if we calculate the force that this guy experiences due to this proton nucleus, we'll see a discrepancy. And this discrepancy comes from something called the shielding effect. Now, this innermost atom or this innermost electron creates a shielding effect. In other words, inner electrons surrounding our nucleus shield the outermost electron from some of that charge due to our protons decreasing the net pull of the nucleus on our outermost electron. Now, the final amount of charge felt by this outermost electron due to our proton nucleus is called the effective nuclear charge. In other words, because there is a second electron found even closer to our nucleus, that means the charge that this guy will feel due to this proton nucleus will be less than what it would feel if this electron wasn't here."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "In other words, inner electrons surrounding our nucleus shield the outermost electron from some of that charge due to our protons decreasing the net pull of the nucleus on our outermost electron. Now, the final amount of charge felt by this outermost electron due to our proton nucleus is called the effective nuclear charge. In other words, because there is a second electron found even closer to our nucleus, that means the charge that this guy will feel due to this proton nucleus will be less than what it would feel if this electron wasn't here. That means if we use the same exact formula for Coulomb's law for this guy, we'll get this result. In other words, the force that this guy feels this charge due to our proton nucleus. This guy divided by R squared times K will be greater than the actual force that this guy feels."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "That means if we use the same exact formula for Coulomb's law for this guy, we'll get this result. In other words, the force that this guy feels this charge due to our proton nucleus. This guy divided by R squared times K will be greater than the actual force that this guy feels. Why? Well, because some of its positive charge will be dissipated to this electron. So that means our final effective nuclear charge will be less."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "Why? Well, because some of its positive charge will be dissipated to this electron. So that means our final effective nuclear charge will be less. So the way we find our effective nuclear charge is using the following formula. Our effective nuclear charge given by ZF equals to the nuclear charge or the actual nuclear charge of our proton nucleus minus the average number of electrons separating our outermost electron and our proton nucleus. And we see that for helium, if our nuclear charge on our proton nucleus is two, then this guy actually feels a charge of 1.69, because we're subtracting the charge that this guy experiences."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So the way we find our effective nuclear charge is using the following formula. Our effective nuclear charge given by ZF equals to the nuclear charge or the actual nuclear charge of our proton nucleus minus the average number of electrons separating our outermost electron and our proton nucleus. And we see that for helium, if our nuclear charge on our proton nucleus is two, then this guy actually feels a charge of 1.69, because we're subtracting the charge that this guy experiences. So at the end, this atomos electron will experiences on average less pole. So let's compare the effective nuclear charge of the following two atoms. Let's look at lithium and Beryllium."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So at the end, this atomos electron will experiences on average less pole. So let's compare the effective nuclear charge of the following two atoms. Let's look at lithium and Beryllium. Now, notice that Beryllium is one guy over to the right on their periodic table. That means it has four protons and four electrons, while lithium has three protons and three electrons. Let's compare the atomic structure."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "Now, notice that Beryllium is one guy over to the right on their periodic table. That means it has four protons and four electrons, while lithium has three protons and three electrons. Let's compare the atomic structure. So, the atomic structure of lithium is the following. Two of its electrons are found in the innermost shell, the one s shell. And one electron, because it only has three electrons, is found in the outermost two s shell."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So, the atomic structure of lithium is the following. Two of its electrons are found in the innermost shell, the one s shell. And one electron, because it only has three electrons, is found in the outermost two s shell. Now, let's look at the comic structure of Beryllium. This guy has one extra electron or one more electron than lithium. Now, that means that it also just like lithium has two electrons on the one s in it orbital."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "Now, let's look at the comic structure of Beryllium. This guy has one extra electron or one more electron than lithium. Now, that means that it also just like lithium has two electrons on the one s in it orbital. But now, because it has an extra electron, it has two electrons in the outermost two s orbital. So the ratio in lithium of inner electrons to outer electrons is two to one, while the ratio of inner to outer electrons in Beryllium is two to two. So the ratio is greater in this guy."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "But now, because it has an extra electron, it has two electrons in the outermost two s orbital. So the ratio in lithium of inner electrons to outer electrons is two to one, while the ratio of inner to outer electrons in Beryllium is two to two. So the ratio is greater in this guy. The significance of this is the following. Because our ratio of inner to outer is greater, that means the inner electrons will take away or shield more charge. So lithium will experience a strong shielding effect due to these inner electrons."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "The significance of this is the following. Because our ratio of inner to outer is greater, that means the inner electrons will take away or shield more charge. So lithium will experience a strong shielding effect due to these inner electrons. So this single hour electron will not experience as much charge as it would if these guys weren't here. Now, let's compare this guy. Notice that in this guy we have one more electron in our outer shell."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So this single hour electron will not experience as much charge as it would if these guys weren't here. Now, let's compare this guy. Notice that in this guy we have one more electron in our outer shell. And what that basically means is that because our ratio of eight to outer is smaller, that means this shielding effect will not be as great, because now we have an extra electron floating around our shell. So these guys will experience more charge than Dixon electron on the lithium. And what that means is that these two electrons will be pulled closer to the nucleus than here."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "And what that basically means is that because our ratio of eight to outer is smaller, that means this shielding effect will not be as great, because now we have an extra electron floating around our shell. So these guys will experience more charge than Dixon electron on the lithium. And what that means is that these two electrons will be pulled closer to the nucleus than here. And therefore the radius of this atom will be smaller than the radius of this atom. And that's because of the following. They both have a one s and a two s orbital."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "And therefore the radius of this atom will be smaller than the radius of this atom. And that's because of the following. They both have a one s and a two s orbital. So technically, their radius should be the same. But it's not. It's smaller in this guy because it has a weaker shielding effect."}, {"title": "Effective Nuclear Charge and the Shielding Effect .txt", "text": "So technically, their radius should be the same. But it's not. It's smaller in this guy because it has a weaker shielding effect. In other words, the two electrons found on the outermost atom on the outermost shell experience more charge and so therefore are pulled closer. So, once again, the shielding effect is not as great in Beryllium as it is in lithium because the extra electron in Beryllium is added to the same energy level. The same two S orbital."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "In this lecture, I'd like to talk about the concept of electron configuration. But before we get into that, it's really important to talk about the following principle known as the AFP principle. Now, this principle will help us explain electron configuration. Now, what this principle states is the following. Whenever an atom adds a new proton to create a new atom element, it must also add another electron to neutralize that extra charge that comes from that extra proton. So the problem is the following whenever we add a proton, we know exactly where our proton goes."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, what this principle states is the following. Whenever an atom adds a new proton to create a new atom element, it must also add another electron to neutralize that extra charge that comes from that extra proton. So the problem is the following whenever we add a proton, we know exactly where our proton goes. That proton goes into the nucleus along with all the other protons. What about our electrons? We have options as to where to place those electrons."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "That proton goes into the nucleus along with all the other protons. What about our electrons? We have options as to where to place those electrons. Because we have many different shells, we have many different subshells and we have made different orbitals within our subshells. So. Luckily, nature always tends to form the lowest possible energy state."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Because we have many different shells, we have many different subshells and we have made different orbitals within our subshells. So. Luckily, nature always tends to form the lowest possible energy state. Therefore, whenever we add electrons to our atom, that electron will go into the lowest possible subshell. Lowest possible energy subshell. So let's examine the following concepts, and I state the following as electrons move further from the nucleus, our energy level will increase."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Therefore, whenever we add electrons to our atom, that electron will go into the lowest possible subshell. Lowest possible energy subshell. So let's examine the following concepts, and I state the following as electrons move further from the nucleus, our energy level will increase. And this is true. Now let's examine why it's true. Well, let's look at the following illustration."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "And this is true. Now let's examine why it's true. Well, let's look at the following illustration. Suppose we have a proton now, a nucleus. And this is an electron. A distance."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Suppose we have a proton now, a nucleus. And this is an electron. A distance. R away from it. Now Coulomb's law will give us some force that this guy feels due to this electron. And then it will also give us the same force, except in negative direction, that this guy feels due to this proton."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "R away from it. Now Coulomb's law will give us some force that this guy feels due to this electron. And then it will also give us the same force, except in negative direction, that this guy feels due to this proton. Now, my question is, how do we get this electron not a distance our way, but a distance to our way? In other words, how do we move this electron from this guy from this distance to this distance? Well, remember, if this is a positive force and this is a negative force, these guys are tracting."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, my question is, how do we get this electron not a distance our way, but a distance to our way? In other words, how do we move this electron from this guy from this distance to this distance? Well, remember, if this is a positive force and this is a negative force, these guys are tracting. They want to come close. So in order for me to move this guy a distance R here a distance two R from our nucleus. That means I have to do work on that electron."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "They want to come close. So in order for me to move this guy a distance R here a distance two R from our nucleus. That means I have to do work on that electron. So if We Consider Our Proton And our electron a system, that means I have to do work on my system to move this electron a distance r away from this electron, right? So that means work must be done on our system. And because work is a transfer of energy, energy must be transferred into our system."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So if We Consider Our Proton And our electron a system, that means I have to do work on my system to move this electron a distance r away from this electron, right? So that means work must be done on our system. And because work is a transfer of energy, energy must be transferred into our system. So our overall energy of our system increases as the electron moves away from our nucleus. And that's exactly why placing electrons further away from our nucleus will increase the energy of our system. That's exactly why nature tends to place electrons as close to our atom, as close to our nucleus as possible."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So our overall energy of our system increases as the electron moves away from our nucleus. And that's exactly why placing electrons further away from our nucleus will increase the energy of our system. That's exactly why nature tends to place electrons as close to our atom, as close to our nucleus as possible. Because placing it further away increases the energy of our system. And nature tends to take the lowest energy state. So that's exactly why when we look at the principal quantum numbers as our principal quantum numbers increase as we go from n equals one to n equals two to n equals three, or from Orbital S P to D. Our energy will increase as we go down this table."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Because placing it further away increases the energy of our system. And nature tends to take the lowest energy state. So that's exactly why when we look at the principal quantum numbers as our principal quantum numbers increase as we go from n equals one to n equals two to n equals three, or from Orbital S P to D. Our energy will increase as we go down this table. So now let's talk about electron configuration. Electron configuration is simply a systematic approach to representing and showing exactly where our electrons are placed within any given atom. In other words, in which shells or in which subshells are our electrons down?"}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So now let's talk about electron configuration. Electron configuration is simply a systematic approach to representing and showing exactly where our electrons are placed within any given atom. In other words, in which shells or in which subshells are our electrons down? So let's look at the simplest atom. The h atom. The H Atom has one proton and one electron."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So let's look at the simplest atom. The h atom. The H Atom has one proton and one electron. The proton is the nucleus. The electron is down, orbiting our nucleus. Now, what are the Quantum numbers of this electron?"}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "The proton is the nucleus. The electron is down, orbiting our nucleus. Now, what are the Quantum numbers of this electron? Well, this guy has the first principal level. The first principal quantum number, or N equals one. And that means if N equals one, our second quantum number, our L must be zero."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Well, this guy has the first principal level. The first principal quantum number, or N equals one. And that means if N equals one, our second quantum number, our L must be zero. And if l is zero, that means our third quantum number, the orbital in which our electron is located is the s orbital. So to represent this in an electron configuration way, we simply do the following. We put a coefficient in front of our S. So the Wand represents our shell, our principal quantum number."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "And if l is zero, that means our third quantum number, the orbital in which our electron is located is the s orbital. So to represent this in an electron configuration way, we simply do the following. We put a coefficient in front of our S. So the Wand represents our shell, our principal quantum number. Our S represents our subshell. And at the same time, it also represents our orbital in which our electron is in. The SuperScript of one represents the number of electrons."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Our S represents our subshell. And at the same time, it also represents our orbital in which our electron is in. The SuperScript of one represents the number of electrons. So we have one electron. So we have a SuperScript of one. So this Is The Electron Configuration for this atom."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So we have one electron. So we have a SuperScript of one. So this Is The Electron Configuration for this atom. For the h atom. Now, any atom or any element on the periodic table has an electron configuration. So let's look at another one."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "For the h atom. Now, any atom or any element on the periodic table has an electron configuration. So let's look at another one. Let's look at oxygen. Oxygen has an atomic number of eight. And that means in its neutral state, it has eight electrons."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Let's look at oxygen. Oxygen has an atomic number of eight. And that means in its neutral state, it has eight electrons. So let's look at its electron configuration. Now, notice that n equals One. It could fit two electrons."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So let's look at its electron configuration. Now, notice that n equals One. It could fit two electrons. Because when n equals one, our l equals zero. And we only have one orbital, the S orbital. So we can place the maximum of two electrons into any orbital."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Because when n equals one, our l equals zero. And we only have one orbital, the S orbital. So we can place the maximum of two electrons into any orbital. So we can place two electrons into shell level. N equals one. How about N equals two?"}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So we can place two electrons into shell level. N equals one. How about N equals two? Well, n equals two has a maximum of two to the two. So maximum of four orbitals. That means if we have four orbitals, I have an s orbital and three p orbitals."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Well, n equals two has a maximum of two to the two. So maximum of four orbitals. That means if we have four orbitals, I have an s orbital and three p orbitals. Right? So I could put a maximum of eight electrons into my N equals two energy level into my N equals two shell. So I place two electrons into my s orbital."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Right? So I could put a maximum of eight electrons into my N equals two energy level into my N equals two shell. So I place two electrons into my s orbital. This guy's my s orbital. And then I could put six electrons into my three p orbitals. So 1234, I put Four, because I only have four left over."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "This guy's my s orbital. And then I could put six electrons into my three p orbitals. So 1234, I put Four, because I only have four left over. So now let's look at the electron configuration for our oxygen. In our principal quantum number one. N equals one."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So now let's look at the electron configuration for our oxygen. In our principal quantum number one. N equals one. We have only the X orbital. So we place two electrons here. So one s, two."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "We have only the X orbital. So we place two electrons here. So one s, two. And now N equals two. We have an S and the three P's. And that means we first place two electrons into our S.\nSo two F, two."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "And now N equals two. We have an S and the three P's. And that means we first place two electrons into our S.\nSo two F, two. And then we distribute electrons into our piece. And we have three P's. So we start by putting one in each."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "And then we distribute electrons into our piece. And we have three P's. So we start by putting one in each. And I'll explain in a little bit why we put one here, one here, one here, and then one here. And not two here and two here and none here. I'll explain why in a second."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "And I'll explain in a little bit why we put one here, one here, one here, and then one here. And not two here and two here and none here. I'll explain why in a second. Now, so I place one here, one here, one here, and then I place my fourth one into my X. So we can also represent this guy in the following way. We simply erase all these X's as ZS and simply say two P and we place a four on top."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, so I place one here, one here, one here, and then I place my fourth one into my X. So we can also represent this guy in the following way. We simply erase all these X's as ZS and simply say two P and we place a four on top. Now, whenever we do this, we make the assumption that you understand the fact that this is not a single orbital but this is actually three orbitals. Because if this was a single orbital, the four would not make sense because you could only place the maximum of two electrons into any orbital according to the Pole Exclusion Principle. So that's the electron configuration."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, whenever we do this, we make the assumption that you understand the fact that this is not a single orbital but this is actually three orbitals. Because if this was a single orbital, the four would not make sense because you could only place the maximum of two electrons into any orbital according to the Pole Exclusion Principle. So that's the electron configuration. One. F two. Two."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "One. F two. Two. F two. Two. P four."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "F two. Two. P four. Since oxygen has a maximum of eight electrons, that means two plus two plus four eight electrons. So this makes sense. So every atom on the table, every element has its own electron configuration."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Since oxygen has a maximum of eight electrons, that means two plus two plus four eight electrons. So this makes sense. So every atom on the table, every element has its own electron configuration. Now, let's look at sodium. Sodium has the following electron configuration, right? It has eleven electrons in its neutral state."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, let's look at sodium. Sodium has the following electron configuration, right? It has eleven electrons in its neutral state. How about neon? Neon has ten electrons. It has a perfect configuration."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "How about neon? Neon has ten electrons. It has a perfect configuration. Every single orbital is completely filled. And remember, every atom on a table wants to become a noble gas. So every atom wants to become or wants to take the electron configuration of a noble gas."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Every single orbital is completely filled. And remember, every atom on a table wants to become a noble gas. So every atom wants to become or wants to take the electron configuration of a noble gas. And Neon is, in fact, a noble gas. Another way representing electro configurations are using noble gas configurations. Notice that one S, two, two S two and two P, six is identical to one S Two, two S two and two P six of Neon."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "And Neon is, in fact, a noble gas. Another way representing electro configurations are using noble gas configurations. Notice that one S, two, two S two and two P, six is identical to one S Two, two S two and two P six of Neon. So we can replace this whole guy simply with neon. And these brackets simply mean the electron configuration of Neon. And then we add this guy three S one."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So we can replace this whole guy simply with neon. And these brackets simply mean the electron configuration of Neon. And then we add this guy three S one. And this means this is the electron configuration of our sodium, right? Because sodium has one more electron than neon. That electron goes into the three S one energy level."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "And this means this is the electron configuration of our sodium, right? Because sodium has one more electron than neon. That electron goes into the three S one energy level. Now, the ground state of any atom represents its lowest energy state. This is the ground state of sodium. This is the ground state of hydrogen, and this is the ground state of oxygen."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, the ground state of any atom represents its lowest energy state. This is the ground state of sodium. This is the ground state of hydrogen, and this is the ground state of oxygen. Now, when atoms go from a ground state to an excited state, what that means is they form ions or they form excited state atoms. And that simply means electrons jump to a higher state. And we'll talk more about that when we'll talk about the photoelectric effect."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, when atoms go from a ground state to an excited state, what that means is they form ions or they form excited state atoms. And that simply means electrons jump to a higher state. And we'll talk more about that when we'll talk about the photoelectric effect. So one important aspect of electron configuration must be understood. Electron configuration doesn't necessarily have to order things from lowest energy to highest energy. Although that's usually the case, that doesn't have to happen."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So one important aspect of electron configuration must be understood. Electron configuration doesn't necessarily have to order things from lowest energy to highest energy. Although that's usually the case, that doesn't have to happen. In other words, let's look at the following important facts. So four S that energy level is at a lower energy than three D, and five S is at a lower energy level than four D.\nAnd this holds for six S and 5D as well. Now, these things you just have to simply remember."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "In other words, let's look at the following important facts. So four S that energy level is at a lower energy than three D, and five S is at a lower energy level than four D.\nAnd this holds for six S and 5D as well. Now, these things you just have to simply remember. Now, if we look at the electron configuration for Bromine, which has 35 35 protons and 35 electrons, we see the following electron configuration. Now, this guy makes sense because one S comes before two S and two S comes before two P and two P comes before three S, and three S comes before three P. In other words, this goes from lowest energy level to highest energy level. And since we just said that four S is at a lower energy level than 3D, it would make sense to place the four S two before the this would be the correct election configuration if you're actually ordering from lowest energy level to highest energy level."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Now, if we look at the electron configuration for Bromine, which has 35 35 protons and 35 electrons, we see the following electron configuration. Now, this guy makes sense because one S comes before two S and two S comes before two P and two P comes before three S, and three S comes before three P. In other words, this goes from lowest energy level to highest energy level. And since we just said that four S is at a lower energy level than 3D, it would make sense to place the four S two before the this would be the correct election configuration if you're actually ordering from lowest energy level to highest energy level. But this doesn't have to be the case. In other words, if I flip these guys, that isn't wrong. That's allowed."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "But this doesn't have to be the case. In other words, if I flip these guys, that isn't wrong. That's allowed. That's allowed to happen. And that's because electron configuration doesn't necessitate that you have to order them from lowest energy to highest energy. Although, if in a question and asks you for the proper electron configuration that order things from lowest to highest, then that has to happen."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "That's allowed to happen. And that's because electron configuration doesn't necessitate that you have to order them from lowest energy to highest energy. Although, if in a question and asks you for the proper electron configuration that order things from lowest to highest, then that has to happen. Then this is the proper electron configuration. So ions also can be represented using electron configuration. For example, let's look at sodium plus has eleven protons and ten elections."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "Then this is the proper electron configuration. So ions also can be represented using electron configuration. For example, let's look at sodium plus has eleven protons and ten elections. So in order to write the electric configuration for this guy, we simply take away one electron from the highest energy level. So in this case, it would be one S, two two S, two two P, six. And we took out one electron from the three S, one orbital, so that that guy disappeared."}, {"title": "Aufbau Principle and Electron Configuration .txt", "text": "So in order to write the electric configuration for this guy, we simply take away one electron from the highest energy level. So in this case, it would be one S, two two S, two two P, six. And we took out one electron from the three S, one orbital, so that that guy disappeared. Now, if we make this guy into an ion, right? Let's say we make it into BR plus ion, we take one electron away. That means we take it away from the highest energy level."}, {"title": "Real Gases and van der waals equation .txt", "text": "So far we have only spoken about the way ideal gas molecules behave. We still haven't spoken about how actual or real gas molecules behave. Now, recall that any ideal gas molecule will behave accordingly with the kinetic molecular theory which makes a few important assumptions. Now, in this lecture, we're going to look at the way real gas gas molecules behave and which assumptions are broken down and under which conditions. So on the conditions of standard temperature of zero Celsius and standard pressure of one ATM we can use the ideal gas law to solve problems and to figure out how gas molecules behave. Now, under conditions of high pressure of about 1000 ATM and low temperatures the ideal gas law breaks down meaning our gas molecules no longer behave accordingly with this law."}, {"title": "Real Gases and van der waals equation .txt", "text": "Now, in this lecture, we're going to look at the way real gas gas molecules behave and which assumptions are broken down and under which conditions. So on the conditions of standard temperature of zero Celsius and standard pressure of one ATM we can use the ideal gas law to solve problems and to figure out how gas molecules behave. Now, under conditions of high pressure of about 1000 ATM and low temperatures the ideal gas law breaks down meaning our gas molecules no longer behave accordingly with this law. And we can't use this law to solve any form of problem. Now, let's see why this happens and let's see what breaks down. So suppose we have this system in which we have nine molecules and this system is under constant temperature but our pressure is one ATM."}, {"title": "Real Gases and van der waals equation .txt", "text": "And we can't use this law to solve any form of problem. Now, let's see why this happens and let's see what breaks down. So suppose we have this system in which we have nine molecules and this system is under constant temperature but our pressure is one ATM. Now, suppose we go from this system to a much smaller system. So we decrease volume and we increase pressure but our temperature remains constant. So let's go back to this system, this system which was at standard temperature and pressure, these two values our distance between any two molecules was much larger than the distance or the size of the actual molecule itself."}, {"title": "Real Gases and van der waals equation .txt", "text": "Now, suppose we go from this system to a much smaller system. So we decrease volume and we increase pressure but our temperature remains constant. So let's go back to this system, this system which was at standard temperature and pressure, these two values our distance between any two molecules was much larger than the distance or the size of the actual molecule itself. So that means we can use the kinetic theory to approximate how this behaves. Because if the distance is far then we can assume the volume of the molecules to be very, very small. And since they're far away, we can assume they don't attract or repel each other."}, {"title": "Real Gases and van der waals equation .txt", "text": "So that means we can use the kinetic theory to approximate how this behaves. Because if the distance is far then we can assume the volume of the molecules to be very, very small. And since they're far away, we can assume they don't attract or repel each other. However, when we get to this state what happens here at high temperatures? Suppose this is 1000 ATM and still zero Celsius. So our volume has decreased tremendously and our pressure has increased but temperature remains the same."}, {"title": "Real Gases and van der waals equation .txt", "text": "However, when we get to this state what happens here at high temperatures? Suppose this is 1000 ATM and still zero Celsius. So our volume has decreased tremendously and our pressure has increased but temperature remains the same. So at very high pressures the distance between any two molecules is approximately the same as the size of the molecule itself. Now, and I claim under this condition the kinetic theory breaks down. So let's see, what about the kinetic theory that breaks down in this system?"}, {"title": "Real Gases and van der waals equation .txt", "text": "So at very high pressures the distance between any two molecules is approximately the same as the size of the molecule itself. Now, and I claim under this condition the kinetic theory breaks down. So let's see, what about the kinetic theory that breaks down in this system? Well, now, the molecules are very close to each other. They're so close, in fact, that they will attract each other and repel each other. So the forces that we spoke about before that we neglected."}, {"title": "Real Gases and van der waals equation .txt", "text": "Well, now, the molecules are very close to each other. They're so close, in fact, that they will attract each other and repel each other. So the forces that we spoke about before that we neglected. Now we have to take them into consideration. And these forces follow. Coulomb's law means that if our distance decreases, our force increases."}, {"title": "Real Gases and van der waals equation .txt", "text": "Now we have to take them into consideration. And these forces follow. Coulomb's law means that if our distance decreases, our force increases. So if this one's positive and this one's negative they will attract each other according to Coulomb's law. So now one of the assumptions of the kinetic theory breaks down namely, the electrostatic forces. Electrostatic forces cannot be neglected when pressures are very high."}, {"title": "Real Gases and van der waals equation .txt", "text": "So if this one's positive and this one's negative they will attract each other according to Coulomb's law. So now one of the assumptions of the kinetic theory breaks down namely, the electrostatic forces. Electrostatic forces cannot be neglected when pressures are very high. Now, let's look at the volume. Let's take this picture and zoom in. This is what we get."}, {"title": "Real Gases and van der waals equation .txt", "text": "Now, let's look at the volume. Let's take this picture and zoom in. This is what we get. Notice that the space in between the molecules is approximately the same or has the same volume as the molecules themselves. Now, the volume can no longer be neglected. So we can say the volume is zero because now the molecules actually take up a lot of space much more than in this picture."}, {"title": "Real Gases and van der waals equation .txt", "text": "Notice that the space in between the molecules is approximately the same or has the same volume as the molecules themselves. Now, the volume can no longer be neglected. So we can say the volume is zero because now the molecules actually take up a lot of space much more than in this picture. And that means the second assumption in our kinetic theory also breaks down. Namely, volume is no longer zero. So we see that in extreme conditions of high pressure our ideal gas law no longer holds."}, {"title": "Real Gases and van der waals equation .txt", "text": "And that means the second assumption in our kinetic theory also breaks down. Namely, volume is no longer zero. So we see that in extreme conditions of high pressure our ideal gas law no longer holds. Now let's see why. Under low temperatures the ideal gas law also breaks down. Well, if we're at low temperatures that means each molecule has a smaller kinetic energy and therefore it's traveling with a smaller velocity."}, {"title": "Real Gases and van der waals equation .txt", "text": "Now let's see why. Under low temperatures the ideal gas law also breaks down. Well, if we're at low temperatures that means each molecule has a smaller kinetic energy and therefore it's traveling with a smaller velocity. And therefore, they will all drop to the bottom of the container and they will collect and get very close. And if they're close, that means they're feeling electrostatic forces. And so our kinetic theory also breaks down on the low temperatures."}, {"title": "Real Gases and van der waals equation .txt", "text": "And therefore, they will all drop to the bottom of the container and they will collect and get very close. And if they're close, that means they're feeling electrostatic forces. And so our kinetic theory also breaks down on the low temperatures. Now let's compare the pressure of ideal and real systems. Now, for the pressure of an ideal system, remember they're not feeling electrostatic forces and that means they hit the wall of the container and they're not attracted or repelled by other molecules. For real situations, for real gases, the pressure is less."}, {"title": "Real Gases and van der waals equation .txt", "text": "Now let's compare the pressure of ideal and real systems. Now, for the pressure of an ideal system, remember they're not feeling electrostatic forces and that means they hit the wall of the container and they're not attracted or repelled by other molecules. For real situations, for real gases, the pressure is less. Well, why is it less? Well, when the molecule in the real gas travels it's attracted by other molecules. And that means if it's attracted by other molecules if it's pulled by the other molecules it will hit the wall with less force and therefore, a smaller pressure will result."}, {"title": "Real Gases and van der waals equation .txt", "text": "Well, why is it less? Well, when the molecule in the real gas travels it's attracted by other molecules. And that means if it's attracted by other molecules if it's pulled by the other molecules it will hit the wall with less force and therefore, a smaller pressure will result. That means for ideal pressures, ideal pressures are higher than real pressures. Likewise, let's examine the conditions for volume of ideal versus volume of real. Now, the volume of ideal is less than volume of real because when you're taken into consideration the volume of real gases you're taking into consideration the volume of the molecules."}, {"title": "Real Gases and van der waals equation .txt", "text": "That means for ideal pressures, ideal pressures are higher than real pressures. Likewise, let's examine the conditions for volume of ideal versus volume of real. Now, the volume of ideal is less than volume of real because when you're taken into consideration the volume of real gases you're taking into consideration the volume of the molecules. And that means the volume will be plus the volume of the molecules. And so the volume of real gases will be greater. So now let's see what the gas law is for real gases."}, {"title": "Real Gases and van der waals equation .txt", "text": "And that means the volume will be plus the volume of the molecules. And so the volume of real gases will be greater. So now let's see what the gas law is for real gases. So recall that the ideal gas law states that pressure of the ideal system times volume of ideal system gives you NRT. And now notice that in a real system and an ideal system this NRT remains the same. This remains a constant."}, {"title": "Real Gases and van der waals equation .txt", "text": "So recall that the ideal gas law states that pressure of the ideal system times volume of ideal system gives you NRT. And now notice that in a real system and an ideal system this NRT remains the same. This remains a constant. Because if we're talking about the same temperature our T in both ideal and non ideal conditions stays the same. Our R remains a constant and our number of moles does not change. So in both ideal and non ideal systems this guy remains constant."}, {"title": "Real Gases and van der waals equation .txt", "text": "Because if we're talking about the same temperature our T in both ideal and non ideal conditions stays the same. Our R remains a constant and our number of moles does not change. So in both ideal and non ideal systems this guy remains constant. The only thing that changes is pressure and volume. And so suppose we have P ideal and V ideal. Now, from this information we know that our P real is smaller than our P ideal."}, {"title": "Real Gases and van der waals equation .txt", "text": "The only thing that changes is pressure and volume. And so suppose we have P ideal and V ideal. Now, from this information we know that our P real is smaller than our P ideal. And that means we have to add some term to our P real to equate that to our P. Likewise, our Vreal is larger than the ideal and that's why we have to subtract some term from it, some term Y to get the ideal. And so, from experiments, scientists found out what this X and what this Y was. This X is N squared times A over V. Two this Y is N times B."}, {"title": "Real Gases and van der waals equation .txt", "text": "And that means we have to add some term to our P real to equate that to our P. Likewise, our Vreal is larger than the ideal and that's why we have to subtract some term from it, some term Y to get the ideal. And so, from experiments, scientists found out what this X and what this Y was. This X is N squared times A over V. Two this Y is N times B. Now, an is simply number of moles, and B is volume a and B are constants that depend on the gas being used. And so they're different for different gases. Now, once again, the reason we have this guy, the reason we're adding this guy to P real is because P real is smaller than P ideal."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "In this lecture we're going to talk about the concept of conjugate acidbased pairs. So whenever we have an acid that donates an H plus ion to a base, a new base and a new acid are formed. So, for example, let's look at the reaction of acetic acid and water. So in this case, our bronze and Lyley acid is our acetic acid and our bronchylori base is our water. And that's because this guy has an extra H ion. It donates that H plus ion."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "So in this case, our bronze and Lyley acid is our acetic acid and our bronchylori base is our water. And that's because this guy has an extra H ion. It donates that H plus ion. And this guy has a lone pair of electrons on its oxygen. And it has the potential to gain a proton. So this is our brocholyari base."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "And this guy has a lone pair of electrons on its oxygen. And it has the potential to gain a proton. So this is our brocholyari base. Now, when they react this guy, our CBIC acid loses NH ion, while this guy, our water, gains NH ion. And that means if we look at these guys, this becomes our new base, our new bronze of Larry base. Because it now has the potential to gain an H ion because it has that electron pair on the O atom."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "Now, when they react this guy, our CBIC acid loses NH ion, while this guy, our water, gains NH ion. And that means if we look at these guys, this becomes our new base, our new bronze of Larry base. Because it now has the potential to gain an H ion because it has that electron pair on the O atom. This guy now has an extra H ion. So it has the potential to donate one. So this guy becomes our bronchit Larry acid."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "This guy now has an extra H ion. So it has the potential to donate one. So this guy becomes our bronchit Larry acid. So we see that this statement holds true whenever we have an acid that reacts with a base, a new base and a new acid are formed. So let's define conjugate acid base pairs. So a pair of molecules or ions related to each other by the loss or gain of a single H plus ion are called conjugate acid base pairs."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "So we see that this statement holds true whenever we have an acid that reacts with a base, a new base and a new acid are formed. So let's define conjugate acid base pairs. So a pair of molecules or ions related to each other by the loss or gain of a single H plus ion are called conjugate acid base pairs. So let's go back to our above system. So we have an acid loser than H ion becoming a base. So this acid and this base are the conjugate acid base pairs."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "So let's go back to our above system. So we have an acid loser than H ion becoming a base. So this acid and this base are the conjugate acid base pairs. Likewise, this base gains an H becomes a new acid. So this base and this acid are the conjugate acid base pairs. So whenever we have a bronched Larry acid reacting with a bronzedidlory base, we will always form a conjugate Bronxed Larry acid and a conjugate bronson Larry base."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "Likewise, this base gains an H becomes a new acid. So this base and this acid are the conjugate acid base pairs. So whenever we have a bronched Larry acid reacting with a bronzedidlory base, we will always form a conjugate Bronxed Larry acid and a conjugate bronson Larry base. For example, let's look at another reaction in which an acid, a bronsolaric acid reacts with a base water. What will happen is this has the potential to gain an H. So it will give off the H.\nAnd the lone pair of electrons on the base will take that H forming our new conjugate acid plus our new conjugate base. So this acid and this base are conjugate acid base pairs."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "For example, let's look at another reaction in which an acid, a bronsolaric acid reacts with a base water. What will happen is this has the potential to gain an H. So it will give off the H.\nAnd the lone pair of electrons on the base will take that H forming our new conjugate acid plus our new conjugate base. So this acid and this base are conjugate acid base pairs. And this base and this acid are conjugate acid base pairs. So notice that one member of a conjugate acid base pair is always found on one side, while the second member of the conjugate base pair is always found on the other side of the equation. You'll never find both members on one side."}, {"title": "Conjugate Acid-Base Pairs .txt", "text": "And this base and this acid are conjugate acid base pairs. So notice that one member of a conjugate acid base pair is always found on one side, while the second member of the conjugate base pair is always found on the other side of the equation. You'll never find both members on one side. For example, this guy and this guy are found on different sides. And this guy and this guy are found different sides, just like this acid and this new base are found on different sides. And this base and this acid are found on different sides, so this always holds true."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "In this lecture, we're going to talk about four important periodic trends atomic radius, ionization energy, electronegativity, and electron affinity of atoms. Now, let's begin with atomic radius. So what is an atomic radius? Well, it's exactly what you think it is. If you think about atom as being a sphere, then our radius begins at the center of our nucleus and ends at the outermost electron shell. So, for this atom, our radius is the black line."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Well, it's exactly what you think it is. If you think about atom as being a sphere, then our radius begins at the center of our nucleus and ends at the outermost electron shell. So, for this atom, our radius is the black line. So I want to ask the question what happens to our italic radius as we go from left to right across the period on our periodic table? For example, let's take the following period. Let's begin with lithium and go all the way up to four."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So I want to ask the question what happens to our italic radius as we go from left to right across the period on our periodic table? For example, let's take the following period. Let's begin with lithium and go all the way up to four. What happens to our atomic radius? Well, we see that atomic radius decreases as we go from lithium to fluorine. Why is that?"}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "What happens to our atomic radius? Well, we see that atomic radius decreases as we go from lithium to fluorine. Why is that? Well, it's because of two things. First, the number of protons or number of protons found in our nucleus increases as we go from lithium to fluorine. And second, the number of electrons found on our most electron shell also increases."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Well, it's because of two things. First, the number of protons or number of protons found in our nucleus increases as we go from lithium to fluorine. And second, the number of electrons found on our most electron shell also increases. And this means, according to Coulomb's law, the force also increases. In other words, the force with which the protons pull the outermost electrons increases. And this means that our effective nuclear charge on our atom increases."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And this means, according to Coulomb's law, the force also increases. In other words, the force with which the protons pull the outermost electrons increases. And this means that our effective nuclear charge on our atom increases. And if the force is stronger, so the protons are pulling our outermost electrons with a greater force, that means our radius will decrease. The difference between the center, the nucleus and the outermost electron will decrease as we go across the period. So that means our lithium will have the highest radius, the largest radius, and the smallest effective nuclear charge, while our fluoride will have the highest effective nuclear charge and the smallest atomic radius."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And if the force is stronger, so the protons are pulling our outermost electrons with a greater force, that means our radius will decrease. The difference between the center, the nucleus and the outermost electron will decrease as we go across the period. So that means our lithium will have the highest radius, the largest radius, and the smallest effective nuclear charge, while our fluoride will have the highest effective nuclear charge and the smallest atomic radius. So now let's talk about a group. What happens as we go from top of the group to the bottom of the group? So let's look at the following group."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So now let's talk about a group. What happens as we go from top of the group to the bottom of the group? So let's look at the following group. Let's begin with lithium and go to sodium, then potassium, and so on. Well, as we go down the group, our atomic radius tends to increase. And this is because with which atom we add a new energy shell."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Let's begin with lithium and go to sodium, then potassium, and so on. Well, as we go down the group, our atomic radius tends to increase. And this is because with which atom we add a new energy shell. So let's look at the following two atoms. Let's look at lithium and let's look at sodium. Sodium is right below lithium on the same group on the periodic table."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So let's look at the following two atoms. Let's look at lithium and let's look at sodium. Sodium is right below lithium on the same group on the periodic table. So notice that we have two energy levels, one S and two S for the lithium, while the sodium has not two, but three energy levels. One S, two F and three S.\nNow, this addition of the three S means that our atom will grow in size, will enlarge. Where this guy, the atomos guy, is the three F. Shell."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So notice that we have two energy levels, one S and two S for the lithium, while the sodium has not two, but three energy levels. One S, two F and three S.\nNow, this addition of the three S means that our atom will grow in size, will enlarge. Where this guy, the atomos guy, is the three F. Shell. So that means when we move one down to potassium, potassium will have a four S. So potassium will be even larger than sodium and definitely larger than lithium. And that's exactly what we see. In other words, as we go down a group, our atomic radius tends to grow in size, while as we go across the period, our atomic radius tends to decrease because our effective nuclear charge of our atom tends to increase."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So that means when we move one down to potassium, potassium will have a four S. So potassium will be even larger than sodium and definitely larger than lithium. And that's exactly what we see. In other words, as we go down a group, our atomic radius tends to grow in size, while as we go across the period, our atomic radius tends to decrease because our effective nuclear charge of our atom tends to increase. So that's atomic radius. Now let's look at the ionization energy of our atoms. So what is ionization energy?"}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So that's atomic radius. Now let's look at the ionization energy of our atoms. So what is ionization energy? Well, electrons don't simply come off the atoms by themselves. Remember, electrons are held together by electrostatic force that comes from the positively charged nucleus and the negatively charged electrons. So something must pull those electrons away."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Well, electrons don't simply come off the atoms by themselves. Remember, electrons are held together by electrostatic force that comes from the positively charged nucleus and the negatively charged electrons. So something must pull those electrons away. In other words, work or energy must be inputted into our system to pull that electron off. So therefore, we can define our ionization energy to be the energy required to pull off that electron. That outermost electron."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "In other words, work or energy must be inputted into our system to pull that electron off. So therefore, we can define our ionization energy to be the energy required to pull off that electron. That outermost electron. Now more than one electron can be pulled off. For example, calcium. Calcium in its neutral state, can take away two electrons to become calcium plus two."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Now more than one electron can be pulled off. For example, calcium. Calcium in its neutral state, can take away two electrons to become calcium plus two. So that means some atoms can pull away or can give off more than one electron. Now, the energy required to pull away that first electron is known as The First ionization energy, while the energy Required to pull away that second electron is Known as A second ionization energy and So on. Now let's look at the following now, the less likely an atom gives up the electron, the more energy is required to pull that electron off."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So that means some atoms can pull away or can give off more than one electron. Now, the energy required to pull away that first electron is known as The First ionization energy, while the energy Required to pull away that second electron is Known as A second ionization energy and So on. Now let's look at the following now, the less likely an atom gives up the electron, the more energy is required to pull that electron off. And we see that as we go across a period, our ionization energy of our atom tends to increase. And to explain that, let's look at Coulomb's Law. Now, Coulomb's law once again states that the force is equal to a constant k times charge q one times charge q two divided by the distance between them squared, where this guy is a charge due to the nucleus and q two is the charge due to the electrons."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And we see that as we go across a period, our ionization energy of our atom tends to increase. And to explain that, let's look at Coulomb's Law. Now, Coulomb's law once again states that the force is equal to a constant k times charge q one times charge q two divided by the distance between them squared, where this guy is a charge due to the nucleus and q two is the charge due to the electrons. Now what happens as we move, for example, from lithium to fluorine? We already said that our effective nuclear charge tends to increase as we go from left to right. So fluorine has the highest force."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Now what happens as we move, for example, from lithium to fluorine? We already said that our effective nuclear charge tends to increase as we go from left to right. So fluorine has the highest force. In other words, the protons found in the nucleus. Pull those electrons on the outermost electron shell with a lot of force, much more force than lithium or beryllium or boron or carbon. And that means it's going to require much more energy to pull those outermost electrons off."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "In other words, the protons found in the nucleus. Pull those electrons on the outermost electron shell with a lot of force, much more force than lithium or beryllium or boron or carbon. And that means it's going to require much more energy to pull those outermost electrons off. And that's exactly why as we go across the period from lithium to fluorine, our ionization energy tends to increase. Because as we go this way, we have a higher effective nuclear charge, which means we have a greater force. And also take this into account."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And that's exactly why as we go across the period from lithium to fluorine, our ionization energy tends to increase. Because as we go this way, we have a higher effective nuclear charge, which means we have a greater force. And also take this into account. As we go across, our atomic radius decreases, and that means our denominator. Our R also increases. So we see that our cues increase, our Rs decrease."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "As we go across, our atomic radius decreases, and that means our denominator. Our R also increases. So we see that our cues increase, our Rs decrease. And whenever the denominator decreases, that means our force tends to increase. So not only does this increase in charge cause the force to go up, but also the decrease in the r of the atomic radius tends to increase our force. So therefore, as we go from left to right, our ionization energy also increases."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And whenever the denominator decreases, that means our force tends to increase. So not only does this increase in charge cause the force to go up, but also the decrease in the r of the atomic radius tends to increase our force. So therefore, as we go from left to right, our ionization energy also increases. How Bad? When we go down a group. Well, when we go down the group, our atomic radius increases."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "How Bad? When we go down a group. Well, when we go down the group, our atomic radius increases. And that means if we go back to Coulomb's Law, if our atomic radius increases, that means the difference between Q one and Q two or the protons and electrons increases. So our R also increases. And if there are increases, our denominators increase."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And that means if we go back to Coulomb's Law, if our atomic radius increases, that means the difference between Q one and Q two or the protons and electrons increases. So our R also increases. And if there are increases, our denominators increase. And that means our force is less. So as we go down a group, our ionization energy tends to decrease. So to wrap up, basically, the higher your ionization energy is, the less likely you are to give up electrons."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And that means our force is less. So as we go down a group, our ionization energy tends to decrease. So to wrap up, basically, the higher your ionization energy is, the less likely you are to give up electrons. And we'll see that this directly translates into something called electronegativity. So let's look at the third periodic trend called electronegativity. Now, electronegativity is simply the ability of atoms to accept or attract other electrons."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And we'll see that this directly translates into something called electronegativity. So let's look at the third periodic trend called electronegativity. Now, electronegativity is simply the ability of atoms to accept or attract other electrons. And we see that as we go from left to right, across the period, from Lithium to Fluorine, our Electronegativity increases. So let's examine why. Let's look at atomic structure and the electron configuration of lithium and compare to that of fluorine."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And we see that as we go from left to right, across the period, from Lithium to Fluorine, our Electronegativity increases. So let's examine why. Let's look at atomic structure and the electron configuration of lithium and compare to that of fluorine. Now, lithium has three electrons and three protons. So its nucleus is composed of only three protons, while its inner shell is composed of two electrons, and its outer shell is composed only of a single electron. Now let's look at the fluorine."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Now, lithium has three electrons and three protons. So its nucleus is composed of only three protons, while its inner shell is composed of two electrons, and its outer shell is composed only of a single electron. Now let's look at the fluorine. Fluorine has nine protons in its nucleus. While two electrons are found in the inner shell, seven electrons are found on the outer shell. And that means because we have a higher nuclear or effective nuclear charge, we have a higher force."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Fluorine has nine protons in its nucleus. While two electrons are found in the inner shell, seven electrons are found on the outer shell. And that means because we have a higher nuclear or effective nuclear charge, we have a higher force. In other words, because we have nine protons in our nucleus and seven electrons on our outer shell, our force with which our nucleus pulls those electrons is much greater than the force with which these three protons are pulling a single electron. And so that means if we place some arbitrary electron equidistant between these two atoms what? We see that this fluorine will pull this electron with much more force than this guy."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "In other words, because we have nine protons in our nucleus and seven electrons on our outer shell, our force with which our nucleus pulls those electrons is much greater than the force with which these three protons are pulling a single electron. And so that means if we place some arbitrary electron equidistant between these two atoms what? We see that this fluorine will pull this electron with much more force than this guy. And that means as we go from lithium to beryllium, to boron, to a carbon to nitrogen to oxygen, finally, to fluorine our electronegativity increases, and in fact, fluorine is the most electronegative atom and electronegativity is actually measured on a scale called the polling scale, and it's given a highest value of 4.0. Now, as we go across the period from left to right, we see that our electronegativity increases how about as we go from top to bottom? Well, as we go from top to bottom, our atomic radius increases."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And that means as we go from lithium to beryllium, to boron, to a carbon to nitrogen to oxygen, finally, to fluorine our electronegativity increases, and in fact, fluorine is the most electronegative atom and electronegativity is actually measured on a scale called the polling scale, and it's given a highest value of 4.0. Now, as we go across the period from left to right, we see that our electronegativity increases how about as we go from top to bottom? Well, as we go from top to bottom, our atomic radius increases. So our force with which our protons in the nucleus pull those electrons decreases. Because remember, force, according to Coulomb's law, is equal to K times q one times q two over r two. So our denominator increases, decreasing our force."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So our force with which our protons in the nucleus pull those electrons decreases. Because remember, force, according to Coulomb's law, is equal to K times q one times q two over r two. So our denominator increases, decreasing our force. And so, therefore, as we go from top to bottom, our electronegativity decreases. Now for noble gases. Electro."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And so, therefore, as we go from top to bottom, our electronegativity decreases. Now for noble gases. Electro. Negativity is undefined. And that's because in Noble gas structure, the electron configuration is perfect. And what noble gases can't accept any more electrons?"}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Negativity is undefined. And that's because in Noble gas structure, the electron configuration is perfect. And what noble gases can't accept any more electrons? Because notice that this twopie orbital can accept one more electron. And that's exactly why fluorine can accept one more electron. But the next atom, the noble gas after this guy can't accept any more electrons because it has a two p six electron configuration."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Because notice that this twopie orbital can accept one more electron. And that's exactly why fluorine can accept one more electron. But the next atom, the noble gas after this guy can't accept any more electrons because it has a two p six electron configuration. So let's look at our final periodic trend called electron affinity. Now. Electron affinity is the amount of energy released when an atom gains an electron."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "So let's look at our final periodic trend called electron affinity. Now. Electron affinity is the amount of energy released when an atom gains an electron. Remember, the only way to take an electron away from the outer shell of an atom is to apply work, is to input energy. Because work must be done against the force of the protons in the nucleus attracting those electrons, those outer electrons. So that means the reverse must be the following whenever an electron or whenever an atom gains an electron energy must be released."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "Remember, the only way to take an electron away from the outer shell of an atom is to apply work, is to input energy. Because work must be done against the force of the protons in the nucleus attracting those electrons, those outer electrons. So that means the reverse must be the following whenever an electron or whenever an atom gains an electron energy must be released. And that's exactly what happens when Fluorine, for example, gains electrons. When fluorine goes from a neutral molecule, gains an electron to form an anion, it loses energy. The energy level of the outer shell is lower."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And that's exactly what happens when Fluorine, for example, gains electrons. When fluorine goes from a neutral molecule, gains an electron to form an anion, it loses energy. The energy level of the outer shell is lower. And therefore this molecule, this anion is more stable than the neutral counterpart. So this reaction going this way is exothermic. Now."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "And therefore this molecule, this anion is more stable than the neutral counterpart. So this reaction going this way is exothermic. Now. That means whenever we go from Lithium to Beryllium to Boron and so on, whenever we go from left to right, our electron activity increases. And that means this guy, this reaction is more exothermic for fluorine than for lithium. In other words, whenever our fluorine gains electrons, it becomes stable and it loses a lot of energy."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "That means whenever we go from Lithium to Beryllium to Boron and so on, whenever we go from left to right, our electron activity increases. And that means this guy, this reaction is more exothermic for fluorine than for lithium. In other words, whenever our fluorine gains electrons, it becomes stable and it loses a lot of energy. On the contrary, whenever this guy gains electrons, the reaction for this guy is endothermic. Because this guy, the lithium in the neutral state is more stable than the anion than lithium minus one. And that's exactly what we mean by electron affinity."}, {"title": "Atomic Radius, Ionization Energy, Electronegativity and Electron Affinity .txt", "text": "On the contrary, whenever this guy gains electrons, the reaction for this guy is endothermic. Because this guy, the lithium in the neutral state is more stable than the anion than lithium minus one. And that's exactly what we mean by electron affinity. Now. Likewise, as we go from top to bottom on our group in group, on the period our electron affinity decreases. And that's because our atomic radius increases as we go from top to bottom."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "Now, the atom that gains the electrons becomes more negative and is said to be reduced. The atom that loses the electrons is said to be more positive and is said to be oxidized. So let's look at atoms A and B. Suppose atom A loses an electron, while atom B gains that same electron. That means our charge of A goes from A neutral charge to A plus one charge. It loses an electron, while atom B gains an electron."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "Suppose atom A loses an electron, while atom B gains that same electron. That means our charge of A goes from A neutral charge to A plus one charge. It loses an electron, while atom B gains an electron. So Its charge goes from A neutral charge to a negative one charge. So species A or atom A is said to be oxidized, while atom B is set to be reduced. Now, we can also look at it another way."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "So Its charge goes from A neutral charge to a negative one charge. So species A or atom A is said to be oxidized, while atom B is set to be reduced. Now, we can also look at it another way. Atom A is a reducing agent. Why? Well, because it reduces atom B."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "Atom A is a reducing agent. Why? Well, because it reduces atom B. It makes this atom more negative. So we can also look at atom B as an oxidizing agent, because atom D takes away that electron from A, and it oxidizes A, and that's why it's the oxidizing agent. So Oxidation, or the loss of electrons and reduction or the gain of electron always comes in a pair, the same way that acids are always paired with the base."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "It makes this atom more negative. So we can also look at atom B as an oxidizing agent, because atom D takes away that electron from A, and it oxidizes A, and that's why it's the oxidizing agent. So Oxidation, or the loss of electrons and reduction or the gain of electron always comes in a pair, the same way that acids are always paired with the base. So let's look at the most common redox reaction out there two H, two molecules combined with a single O, two molecule forming two molecules of water. So on this side, our H two is in its atomic state. It's in its elemental state, and so is oxygen."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "So let's look at the most common redox reaction out there two H, two molecules combined with a single O, two molecule forming two molecules of water. So on this side, our H two is in its atomic state. It's in its elemental state, and so is oxygen. That means they both have A charge of zero. Now, in this case, our oxygen becomes negative two, and our H becomes a positive two. And two H's cause A positive two charge."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "That means they both have A charge of zero. Now, in this case, our oxygen becomes negative two, and our H becomes a positive two. And two H's cause A positive two charge. So our overall charge is zero. But each atom gains or loses electrons. So let's see what happens."}, {"title": "Oxidation-Reduction Reactions .txt", "text": "So our overall charge is zero. But each atom gains or loses electrons. So let's see what happens. So, our H atom is the reducing agent, and it loses electrons, which means it's oxidized. And that loss of electrons, those electrons are transferred to our oxygen molecule, and that means our oxygen molecule is reduced. So this oxygen molecule is the oxidizing agent because it gains those electrons."}, {"title": "Order of Reactions .txt", "text": "So our reactive x and the gas state reacts in a single step to produce two products, y and z, also in a gas state. Now, we already spoke about what the rate law was. The rate law of any reaction is a mathematical representation of the relationship between the concentration of the reactants and our rate of reaction. And we said that the general form is as follows rate of forward reaction is equal to the constant of our Ford reaction times the concentration of reactants to some power A. Now, this A represents the order of our reactants. And since the only reactant is the x reactant, this A represents the order of the entire reaction."}, {"title": "Order of Reactions .txt", "text": "And we said that the general form is as follows rate of forward reaction is equal to the constant of our Ford reaction times the concentration of reactants to some power A. Now, this A represents the order of our reactants. And since the only reactant is the x reactant, this A represents the order of the entire reaction. Now, my question is how do we determine the order of our reaction, namely A and our rate constant KF? Well, one way to do it, as we saw, is to determine using experimental results. So we find the initial concentrations and the initial rate and we use that information to determine their relationship or the relationship between A and our rate of our reaction."}, {"title": "Order of Reactions .txt", "text": "Now, my question is how do we determine the order of our reaction, namely A and our rate constant KF? Well, one way to do it, as we saw, is to determine using experimental results. So we find the initial concentrations and the initial rate and we use that information to determine their relationship or the relationship between A and our rate of our reaction. And using that this guy and our concentration, we can then find our KAP. Well, that's one way to do it. A second way to do it is with graphing."}, {"title": "Order of Reactions .txt", "text": "And using that this guy and our concentration, we can then find our KAP. Well, that's one way to do it. A second way to do it is with graphing. So we can graph concentration of reactants against time or progress of reaction. Now, there are three main graphs that we can get. Now let's look at the first one."}, {"title": "Order of Reactions .txt", "text": "So we can graph concentration of reactants against time or progress of reaction. Now, there are three main graphs that we can get. Now let's look at the first one. The first one is known as the 0th order of our reaction. And that means our A is equal to zero. So this exponent A is zero and hence 0th order of our react."}, {"title": "Order of Reactions .txt", "text": "The first one is known as the 0th order of our reaction. And that means our A is equal to zero. So this exponent A is zero and hence 0th order of our react. Now let's recall what the rate of reaction is. Well, the rate of any reaction or reactant is given by the following formula. Since we're going from reactants to products, our reactant concentration is decreasing, so the rate is negative."}, {"title": "Order of Reactions .txt", "text": "Now let's recall what the rate of reaction is. Well, the rate of any reaction or reactant is given by the following formula. Since we're going from reactants to products, our reactant concentration is decreasing, so the rate is negative. Change in x of concentration x divided by change in time is equal to our rate of forward. So k times our concentration of x. Now, since we're dealing with A equals zero, the zeros order, we plug in our zero for the exponent A."}, {"title": "Order of Reactions .txt", "text": "Change in x of concentration x divided by change in time is equal to our rate of forward. So k times our concentration of x. Now, since we're dealing with A equals zero, the zeros order, we plug in our zero for the exponent A. That means this guy, the x will go to zero. So let's rearrange this a bit. Let's bring the negative over here."}, {"title": "Order of Reactions .txt", "text": "That means this guy, the x will go to zero. So let's rearrange this a bit. Let's bring the negative over here. Let's bring the T on this side and make this guy one because 80 power is one and we get change in concentration of x equals negative constant k times change in time. Now let's represent this guy and this guy following way. So the final concentration minus the initial concentration equals negative k in parentheses time final minus time initial."}, {"title": "Order of Reactions .txt", "text": "Let's bring the T on this side and make this guy one because 80 power is one and we get change in concentration of x equals negative constant k times change in time. Now let's represent this guy and this guy following way. So the final concentration minus the initial concentration equals negative k in parentheses time final minus time initial. But what is time initial? Time initial is time zero. So this guy is zero."}, {"title": "Order of Reactions .txt", "text": "But what is time initial? Time initial is time zero. So this guy is zero. So let's go here. Now, change in concentration of final minus change in concentration of initial gives you negative k times t final because this guy becomes zero. Now let's bring the initial over to this side and we get concentration of our final equals negative k times t final plus our initial concentration of x."}, {"title": "Order of Reactions .txt", "text": "So let's go here. Now, change in concentration of final minus change in concentration of initial gives you negative k times t final because this guy becomes zero. Now let's bring the initial over to this side and we get concentration of our final equals negative k times t final plus our initial concentration of x. And notice that this equation has the same form as y equals MX plus b. This is our general equation for a line. And that means if we plot this guy, we will get a line."}, {"title": "Order of Reactions .txt", "text": "And notice that this equation has the same form as y equals MX plus b. This is our general equation for a line. And that means if we plot this guy, we will get a line. So this is our y number, this is our Y intercept, this is our x value and this is our slope. Our slope is negative k. So if we plot this where our Y is the concentration of X and our X is the time or the progress of reaction, we get the following negative slope where this point is the .0\ncomma concentration of initial because it's time to zero, we plug in t zero and we get simply this guy equals this guy. That's exactly what this guy is."}, {"title": "Order of Reactions .txt", "text": "So this is our y number, this is our Y intercept, this is our x value and this is our slope. Our slope is negative k. So if we plot this where our Y is the concentration of X and our X is the time or the progress of reaction, we get the following negative slope where this point is the .0\ncomma concentration of initial because it's time to zero, we plug in t zero and we get simply this guy equals this guy. That's exactly what this guy is. This is our initial concentration of our reactor. Now the slope represents negative k, this is our slope. So to find our slope, we simply find the final concentration, subtract that from initial concentration and divide it by the change in time, or M equals change in Y over change in X and we find our K value."}, {"title": "Order of Reactions .txt", "text": "This is our initial concentration of our reactor. Now the slope represents negative k, this is our slope. So to find our slope, we simply find the final concentration, subtract that from initial concentration and divide it by the change in time, or M equals change in Y over change in X and we find our K value. And that's how we find our K.\nNow, in other words, to summarize, if after graphing our concentration of reactants versus time, we get a negative slope like this, we get a graph like this. That means our A must be one. So now let's look at what the graph would look like if our A was actually one, if our order was first order or A equals one."}, {"title": "Order of Reactions .txt", "text": "And that's how we find our K.\nNow, in other words, to summarize, if after graphing our concentration of reactants versus time, we get a negative slope like this, we get a graph like this. That means our A must be one. So now let's look at what the graph would look like if our A was actually one, if our order was first order or A equals one. Now once again we begin the same way. Negative change in concentration of x over change in time is equal to our constant KDF times the concentration of x to the one power. So now this guy remains."}, {"title": "Order of Reactions .txt", "text": "Now once again we begin the same way. Negative change in concentration of x over change in time is equal to our constant KDF times the concentration of x to the one power. So now this guy remains. So let's rearrange them a bit and we get well, we bring that change in concentration of x over this side and we bring our time or change in time to this side. We also bring the negative over to this side and we get change in concentration of x divided by concentration of x equals negative our constant times change in time. Now, in order to get from this guy to this guy, we have to use calculus and integrals."}, {"title": "Order of Reactions .txt", "text": "So let's rearrange them a bit and we get well, we bring that change in concentration of x over this side and we bring our time or change in time to this side. We also bring the negative over to this side and we get change in concentration of x divided by concentration of x equals negative our constant times change in time. Now, in order to get from this guy to this guy, we have to use calculus and integrals. Now I will spare you the calculus and we'll simply jump to this step. If you're interested about what to do from this step to this step, leave a comment and I'll show you. So after we use calculus, we get natural log of the concentration of final minus natural log of concentration of initial equals negative k times time final minus time initial."}, {"title": "Order of Reactions .txt", "text": "Now I will spare you the calculus and we'll simply jump to this step. If you're interested about what to do from this step to this step, leave a comment and I'll show you. So after we use calculus, we get natural log of the concentration of final minus natural log of concentration of initial equals negative k times time final minus time initial. Now this time initial is zero. So that becomes zero and we bring over this guy to the right side and we get natural log of the concentration final equals negative k times t final plus because this guy was brought over natural log of the concentration of an initial. Now, this guy also has the same form as the line as a line y equals MX plus B, where this guy is our Y value this guy."}, {"title": "Order of Reactions .txt", "text": "Now this time initial is zero. So that becomes zero and we bring over this guy to the right side and we get natural log of the concentration final equals negative k times t final plus because this guy was brought over natural log of the concentration of an initial. Now, this guy also has the same form as the line as a line y equals MX plus B, where this guy is our Y value this guy. The b is our Y intercept. Our slope is negative k, and our x value is time. So let's graph this on an XY plane."}, {"title": "Order of Reactions .txt", "text": "The b is our Y intercept. Our slope is negative k, and our x value is time. So let's graph this on an XY plane. So the y axis is our natural log of x of our concentration of reactive and our x axis is time or progress or reaction. And we see that once again our slope is negative k, our Y intercept is this guy. And in order to find our K, we simply use the slope."}, {"title": "Order of Reactions .txt", "text": "So the y axis is our natural log of x of our concentration of reactive and our x axis is time or progress or reaction. And we see that once again our slope is negative k, our Y intercept is this guy. And in order to find our K, we simply use the slope. In other words, change in y over change in x. Now once again, if our A equals one, then when we graph this equation on the XY plane, we should get a negative slope like this a straight line. Now, suppose our A was zero, and we tried graphing this equation the same way."}, {"title": "Order of Reactions .txt", "text": "In other words, change in y over change in x. Now once again, if our A equals one, then when we graph this equation on the XY plane, we should get a negative slope like this a straight line. Now, suppose our A was zero, and we tried graphing this equation the same way. Then since A equals zero, this would not be a straight line, it'd be some other line. And that means we can conclude that our A is not one, our A must be something else. And in order to determine A equals zero, we have to actually graph what we graphed before and after we graph."}, {"title": "Order of Reactions .txt", "text": "Then since A equals zero, this would not be a straight line, it'd be some other line. And that means we can conclude that our A is not one, our A must be something else. And in order to determine A equals zero, we have to actually graph what we graphed before and after we graph. If we got a straight negative line, we can conclude that our A equals zero. So finally, let's look at the graph for A equals two. So what will the graph look like when our exponent is two?"}, {"title": "Order of Reactions .txt", "text": "If we got a straight negative line, we can conclude that our A equals zero. So finally, let's look at the graph for A equals two. So what will the graph look like when our exponent is two? So this is called a second order reaction. And we begin the same way. Negative change in concentration of x divided by changes time."}, {"title": "Order of Reactions .txt", "text": "So this is called a second order reaction. And we begin the same way. Negative change in concentration of x divided by changes time. Our rate equal to this rate KF times the concentration is the second power because A equals two. Now, once again we rearrange. We get the following."}, {"title": "Order of Reactions .txt", "text": "Our rate equal to this rate KF times the concentration is the second power because A equals two. Now, once again we rearrange. We get the following. Then we use calculus and simple algebra to find the following final equation. Now once again, if you're curious about this step, leave a comment and I'll show you. Now this equation also looks like a line."}, {"title": "Order of Reactions .txt", "text": "Then we use calculus and simple algebra to find the following final equation. Now once again, if you're curious about this step, leave a comment and I'll show you. Now this equation also looks like a line. So y equals MX plus B, where our Y is this guy. Our slope is a k, our X is time, and our Y intercept is this guy. So we're going to get a straight line, positive straight line with a positive slope."}, {"title": "Order of Reactions .txt", "text": "So y equals MX plus B, where our Y is this guy. Our slope is a k, our X is time, and our Y intercept is this guy. So we're going to get a straight line, positive straight line with a positive slope. And our intercept is this guy. And now our Y is this guy. One over change in concentration."}, {"title": "Reaction Quotient .txt", "text": "So far we have spoken about the equilibrium constant of our reactions. And we said that the equilibrium constant k allows us to see how far our reaction proceeded at equilibrium. In other words, it only gives us information about our reaction at equilibrium when equilibrium is established. Now, it would be nice is if we could somehow know more information about our reaction or about the progress of our reaction before equilibrium is actually established. And this is exactly what the reaction quotient does. It provides us more information about our reaction before equilibrium is established."}, {"title": "Reaction Quotient .txt", "text": "Now, it would be nice is if we could somehow know more information about our reaction or about the progress of our reaction before equilibrium is actually established. And this is exactly what the reaction quotient does. It provides us more information about our reaction before equilibrium is established. Now, the reaction quotient q is also just like the equilibrium constant k, the ratio of the concentration of products to the concentration of reactants. But unlike our k, q does not refer to the equilibrium concentration. Now, if we look at the following reaction, that is not an equilibrium in which x moles of A react with y moles of B to form our products c and d, both having z and w moles respectively."}, {"title": "Reaction Quotient .txt", "text": "Now, the reaction quotient q is also just like the equilibrium constant k, the ratio of the concentration of products to the concentration of reactants. But unlike our k, q does not refer to the equilibrium concentration. Now, if we look at the following reaction, that is not an equilibrium in which x moles of A react with y moles of B to form our products c and d, both having z and w moles respectively. Now our quotient, our reaction quotient is the same thing as our equilibrium constant in the sense that it's a ratio. It's the ratio of this guy times the ratio. It's the concentration of this guy times the concentration of this guy divided by the concentration of A times the concentration of B."}, {"title": "Reaction Quotient .txt", "text": "Now our quotient, our reaction quotient is the same thing as our equilibrium constant in the sense that it's a ratio. It's the ratio of this guy times the ratio. It's the concentration of this guy times the concentration of this guy divided by the concentration of A times the concentration of B. Now, these exponents come from the coefficients X-Y-Z and W. Now, if our q is equal to our k, that means our reaction is already at equilibrium. Now, if q is greater than k at the beginning of our reaction, that means our reaction is reacting favorite. And that's because let's look at the ratio."}, {"title": "Reaction Quotient .txt", "text": "Now, these exponents come from the coefficients X-Y-Z and W. Now, if our q is equal to our k, that means our reaction is already at equilibrium. Now, if q is greater than k at the beginning of our reaction, that means our reaction is reacting favorite. And that's because let's look at the ratio. If q is greater than k, that means the concentrations of our product c and d is greater than that in equilibrium. And that means our equilibrium will shift this way. In other words, these products will tend to convert to reactants."}, {"title": "Reaction Quotient .txt", "text": "If q is greater than k, that means the concentrations of our product c and d is greater than that in equilibrium. And that means our equilibrium will shift this way. In other words, these products will tend to convert to reactants. And so our reaction is reactant favorite, meaning it's favored. This way our reverse reaction is favored over our four reactions. Now, if k is if q is less than k, that means our reaction is products favored."}, {"title": "Reaction Quotient .txt", "text": "And so our reaction is reactant favorite, meaning it's favored. This way our reverse reaction is favored over our four reactions. Now, if k is if q is less than k, that means our reaction is products favored. So let's go back to the ratio. So q being less than k means that our concentration of and d, our product is less than data equilibrium. And what that means is A and B will tend to react to produce the C and D that is seen at equilibrium."}, {"title": "Concentration Cells .txt", "text": "And this difference in reactions creates something called electric potential, also known as a cell voltage. And we saw that understanding conditions of 1 bar pressure and one molar concentration, we can find find our cell voltage by using this formula where we first find our cell voltage of the anode and then subtract that from the capital. And these values can be looked up on a reduction half reaction table. Now, what happens when our conditions are not standard? Well, then we have to use this formula. And we saw that this formula is called Nurse Formula."}, {"title": "Concentration Cells .txt", "text": "Now, what happens when our conditions are not standard? Well, then we have to use this formula. And we saw that this formula is called Nurse Formula. And what it basically tells us is that the concentration of our solutions in beaker One and Baker Two will make a difference. They will influence the final cell voltage of our cell. So, as of now, we've only really spoken about these types of electrochemical cells in which beaker One contains the oxidation reaction of some metal x."}, {"title": "Concentration Cells .txt", "text": "And what it basically tells us is that the concentration of our solutions in beaker One and Baker Two will make a difference. They will influence the final cell voltage of our cell. So, as of now, we've only really spoken about these types of electrochemical cells in which beaker One contains the oxidation reaction of some metal x. And Baker Two, the half cell two contains the reduction reaction in which some other metal, metal y, is reduced. So let's see what the overall net reduction reaction is. Well, our x metal releases an electron and also releases an x plus ion to our solution, increases the concentration of the x plus ion in Baker One."}, {"title": "Concentration Cells .txt", "text": "And Baker Two, the half cell two contains the reduction reaction in which some other metal, metal y, is reduced. So let's see what the overall net reduction reaction is. Well, our x metal releases an electron and also releases an x plus ion to our solution, increases the concentration of the x plus ion in Baker One. This electron then travels via the conductor into this second electrode, a different metal. Now, inside this electrode, this electron reacts with this y plus ion that is taken up into the electrode and they react to form our Y solid. Now, notice one type of reaction occurred that deals with X in this beaker."}, {"title": "Concentration Cells .txt", "text": "This electron then travels via the conductor into this second electrode, a different metal. Now, inside this electrode, this electron reacts with this y plus ion that is taken up into the electrode and they react to form our Y solid. Now, notice one type of reaction occurred that deals with X in this beaker. And a second type of reaction, a different reaction occurred in beaker Two. And this difference created something called a cell voltage. And we saw that once again, on the standing conditions, we use this formula, and on the nonstanding conditions, we use this formula."}, {"title": "Concentration Cells .txt", "text": "And a second type of reaction, a different reaction occurred in beaker Two. And this difference created something called a cell voltage. And we saw that once again, on the standing conditions, we use this formula, and on the nonstanding conditions, we use this formula. Now, I want to talk about a different type of electrochemical cell called the concentration cell. Now, in this cell, we also have two half cells or two beakers. We also have a salt bridge and a conductor with a bolt meter."}, {"title": "Concentration Cells .txt", "text": "Now, I want to talk about a different type of electrochemical cell called the concentration cell. Now, in this cell, we also have two half cells or two beakers. We also have a salt bridge and a conductor with a bolt meter. The only difference is that now we have the same exact electrode. So this electrode might be composed of some metal x, and this electrode is also composed of that same metal. So now we have the same type of reaction occurring in Baker One and Baker Two."}, {"title": "Concentration Cells .txt", "text": "The only difference is that now we have the same exact electrode. So this electrode might be composed of some metal x, and this electrode is also composed of that same metal. So now we have the same type of reaction occurring in Baker One and Baker Two. The only difference is we have different concentrations of solutions because we learned from this equation that even though this guy might be zero, the concentration difference will still create an electric potential or cell voltage. Now, what is this guy for, this type of electrochemical cell called the concentration cell? Well, let's see."}, {"title": "Concentration Cells .txt", "text": "The only difference is we have different concentrations of solutions because we learned from this equation that even though this guy might be zero, the concentration difference will still create an electric potential or cell voltage. Now, what is this guy for, this type of electrochemical cell called the concentration cell? Well, let's see. Well, our E of the cell or cell voltage of our cell is equal. To notice this guy is now zero, because if we try to find our E on the standard conditions what we will get is that this guy and this guy are actually the same. The value on our table is the same."}, {"title": "Concentration Cells .txt", "text": "Well, our E of the cell or cell voltage of our cell is equal. To notice this guy is now zero, because if we try to find our E on the standard conditions what we will get is that this guy and this guy are actually the same. The value on our table is the same. So we subtract some number x minus x gives us zero so this guy goes to zero and what we are left with is this equation where now our cell voltage will depend strictly on the concentrations of our solutions. So let's look at an example of this reaction in order to really understand what a concentration cell is, let's create one. Suppose we have the following electrochemical setup we have two has cells in hassle number one, oxidation of solid copper takes place so this electrode is solid copper and our initial concentration of aqueous copper in beaker one is zero four five molar."}, {"title": "Concentration Cells .txt", "text": "So we subtract some number x minus x gives us zero so this guy goes to zero and what we are left with is this equation where now our cell voltage will depend strictly on the concentrations of our solutions. So let's look at an example of this reaction in order to really understand what a concentration cell is, let's create one. Suppose we have the following electrochemical setup we have two has cells in hassle number one, oxidation of solid copper takes place so this electrode is solid copper and our initial concentration of aqueous copper in beaker one is zero four five molar. Now in hassle number two, reduction of copper takes place and this copper or aqueous copper becomes copper solid. Now, this electrode is also copper solid but our concentration in beaker one is 0.5 molar so this guy is more concentrated and this guy is more dilute. So let's see what the net rethink reaction for this electrochemical cell is."}, {"title": "Concentration Cells .txt", "text": "Now in hassle number two, reduction of copper takes place and this copper or aqueous copper becomes copper solid. Now, this electrode is also copper solid but our concentration in beaker one is 0.5 molar so this guy is more concentrated and this guy is more dilute. So let's see what the net rethink reaction for this electrochemical cell is. So we add the oxidation reaction and the reduction reaction up and we get our net reactor to be the following notice that this copper solid and this copper solid cancel because they appear on both sides of the equation. Also the electrons cancel because they appear on both sides of the equation. What we are left with is this guy and this guy."}, {"title": "Concentration Cells .txt", "text": "So we add the oxidation reaction and the reduction reaction up and we get our net reactor to be the following notice that this copper solid and this copper solid cancel because they appear on both sides of the equation. Also the electrons cancel because they appear on both sides of the equation. What we are left with is this guy and this guy. Now, this guy actually comes from half cell number two, right? Because when we ask these guys up, this guy from hazel number two will appear on the left side of our net equation while this aqueous copper will appear on the right side of our equation. So we go from a higher concentration from zero five to a lower concentration zero five and that makes sense."}, {"title": "Concentration Cells .txt", "text": "Now, this guy actually comes from half cell number two, right? Because when we ask these guys up, this guy from hazel number two will appear on the left side of our net equation while this aqueous copper will appear on the right side of our equation. So we go from a higher concentration from zero five to a lower concentration zero five and that makes sense. So if we were to write the equilibrium constant expression it would be this concentration divided by this concentration and we'll see why that's important in a second. 1st, let's review exactly what happens in this setup. Well, this is the more dilute solution."}, {"title": "Concentration Cells .txt", "text": "So if we were to write the equilibrium constant expression it would be this concentration divided by this concentration and we'll see why that's important in a second. 1st, let's review exactly what happens in this setup. Well, this is the more dilute solution. Initially, we have very little aqueous cu two plus ions dissolved in our solution but as that reaction continues, this metal bar releases more of the copper ions into solution increasing the solution, making more concentrated. Likewise, this acreage copper is taken up by the metal bar because when electrons travel from this electrode to this electrode they react to form copper solid so this becomes more diluted or less concentrated. Eventually, when the two concentrations equal our reaction will stop because our cell voltage will be zero."}, {"title": "Concentration Cells .txt", "text": "Initially, we have very little aqueous cu two plus ions dissolved in our solution but as that reaction continues, this metal bar releases more of the copper ions into solution increasing the solution, making more concentrated. Likewise, this acreage copper is taken up by the metal bar because when electrons travel from this electrode to this electrode they react to form copper solid so this becomes more diluted or less concentrated. Eventually, when the two concentrations equal our reaction will stop because our cell voltage will be zero. So let's calculate our cell voltage and our initial condition of this concentration and this concentration. So our formula is this one but notice that this guy is zero because if we were to look up the values for this oxidation reaction and this reduction reaction we see that the magnitude is the same but signs are different so if we add these guys up, we will get zero. That's why this guy is zero."}, {"title": "Concentration Cells .txt", "text": "So let's calculate our cell voltage and our initial condition of this concentration and this concentration. So our formula is this one but notice that this guy is zero because if we were to look up the values for this oxidation reaction and this reduction reaction we see that the magnitude is the same but signs are different so if we add these guys up, we will get zero. That's why this guy is zero. So our E is simply this whole guy here. Notice this is for 25 degrees Celsius. Now, why do we say log 0.5 over 0.5?"}, {"title": "Concentration Cells .txt", "text": "So our E is simply this whole guy here. Notice this is for 25 degrees Celsius. Now, why do we say log 0.5 over 0.5? Well, remember, Q is like the equilibrium constant, except it's not an equilibrium. It's also the concentration of products over reactants. And in our net, we do reaction."}, {"title": "Concentration Cells .txt", "text": "Well, remember, Q is like the equilibrium constant, except it's not an equilibrium. It's also the concentration of products over reactants. And in our net, we do reaction. The product is this guy in half cell number one, the concentration of 0.5005. And this is half cell number two, the reactant, which is found in this half cell number two. So our concentration is zero five."}, {"title": "Concentration Cells .txt", "text": "The product is this guy in half cell number one, the concentration of 0.5005. And this is half cell number two, the reactant, which is found in this half cell number two. So our concentration is zero five. And that's why we divide 0.05 by 0.5. Now, our N number of moles of electrons is two moles, right? Two moles up here on this side and two moles up here on this side."}, {"title": "Concentration Cells .txt", "text": "And that's why we divide 0.05 by 0.5. Now, our N number of moles of electrons is two moles, right? Two moles up here on this side and two moles up here on this side. So our N is two. So we take our calculator, we plug it in, we get a negative number. But since we have a negative on the outside, that means our E, or cell voltage of our cell, is positive."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "Now, we're going to specifically focus on melting point and boiling point. Now, before we get to the trends, let's define intramolecular bonds and intermolecular bonds. Now, intramolecular bonds are the bonds between the individual joe atoms within a given compound. The intermolecular bonds are the bonds between different compounds. Now, here we have two alkanes. We have neopentane and pentane."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "The intermolecular bonds are the bonds between different compounds. Now, here we have two alkanes. We have neopentane and pentane. The intramlecular bonds within neopentane are the carbon carbon Covalent bonds and carbon H Covalent bonds. Likewise, the intramolecular bonds in pentane are the carbon carbon Covalent bonds and the carbon H Covalent bonds, which aren't shown. So for both types of alkanes and in fact, for all alkanes, the intramlecular bonds are Covalent bonds."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "The intramlecular bonds within neopentane are the carbon carbon Covalent bonds and carbon H Covalent bonds. Likewise, the intramolecular bonds in pentane are the carbon carbon Covalent bonds and the carbon H Covalent bonds, which aren't shown. So for both types of alkanes and in fact, for all alkanes, the intramlecular bonds are Covalent bonds. What about the intermolecular bonds? Well, if we just examine this molecule by itself, if we simply have one molecule or one compound, that means we have no intermolecular bonds. Because in order to have an inter molecular bond, you have to have a second compound next to that compound."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "What about the intermolecular bonds? Well, if we just examine this molecule by itself, if we simply have one molecule or one compound, that means we have no intermolecular bonds. Because in order to have an inter molecular bond, you have to have a second compound next to that compound. So if this is all by itself, we have no intermolecular bonds. If we place this pentane or some other alkane next to our neopentane, these two compounds will attract one another via intermolecular bonds. And in this case, in the case of alkanes, we have Vanderbilt forces."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "So if this is all by itself, we have no intermolecular bonds. If we place this pentane or some other alkane next to our neopentane, these two compounds will attract one another via intermolecular bonds. And in this case, in the case of alkanes, we have Vanderbilt forces. So the intermolecular bonds of alkanes are Vanderbilt forces. And these are simply instantaneous dipoles created between the protons found in the nuclei and the electrons surrounding or orbiting those nuclei. So these valuable forces exist only for a moment, and that's exactly why they are relatively weak forces compared to bonds like polar bonds."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "So the intermolecular bonds of alkanes are Vanderbilt forces. And these are simply instantaneous dipoles created between the protons found in the nuclei and the electrons surrounding or orbiting those nuclei. So these valuable forces exist only for a moment, and that's exactly why they are relatively weak forces compared to bonds like polar bonds. Now, so let's talk about this blue region here. So this blue region on pentane and neopentane represents the boundarbal forces, the instantaneous dipoles created by these two molecules. Notice that this is relatively symmetrical."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "Now, so let's talk about this blue region here. So this blue region on pentane and neopentane represents the boundarbal forces, the instantaneous dipoles created by these two molecules. Notice that this is relatively symmetrical. It looks like a sphere, while this has an oval shape. And this will become important when we'll talk about trends. And we'll see why in just a second."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "It looks like a sphere, while this has an oval shape. And this will become important when we'll talk about trends. And we'll see why in just a second. So the longer the carbon chain is, the higher the boiling point. So why is that? So what that statement states is that this longer pentane molecule will have a higher boiling point than neopentane, which is shorter and more symmetrical."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "So the longer the carbon chain is, the higher the boiling point. So why is that? So what that statement states is that this longer pentane molecule will have a higher boiling point than neopentane, which is shorter and more symmetrical. Now let's define what a boiling point is. The boiling point is the point at which our pentane goes from liquid or our meal pentane. It goes from liquid to gas."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "Now let's define what a boiling point is. The boiling point is the point at which our pentane goes from liquid or our meal pentane. It goes from liquid to gas. And that means whenever we go from liquid to gas, we're breaking intermolecular bonds. So that means if this has a higher boiling point, that means this guy forms stronger intermolecular bonds. And that's exactly why longer the longer our carbon chain is, the stronger our intermolecular bonds are, because we have more surface area, we have more vandalbile forces than in this symmetrical case."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "And that means whenever we go from liquid to gas, we're breaking intermolecular bonds. So that means if this has a higher boiling point, that means this guy forms stronger intermolecular bonds. And that's exactly why longer the longer our carbon chain is, the stronger our intermolecular bonds are, because we have more surface area, we have more vandalbile forces than in this symmetrical case. Remember, whenever we have a sphere, we have the minimum amount of surface area. So the longer our alkane is, the higher the boiling point is. Now, on the contrary, the more symmetry we have in our alkane, the higher the melting point is."}, {"title": "Boiling and Melting Points of Alkanes .txt", "text": "Remember, whenever we have a sphere, we have the minimum amount of surface area. So the longer our alkane is, the higher the boiling point is. Now, on the contrary, the more symmetry we have in our alkane, the higher the melting point is. Now, why is that? Well, in a solid, we have a lot of symmetry, right? We have a crystal lattice, and we have very well structured molecules."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "So when we talk about systems, we simply talk about objects or things of interest, things that we're studying. And everything outside of the system is called the surrounding. Now, when we talk about entropy, we represent entropy with the letter S and changes in entropy with delta S.\nSo according to the conservation of energy, delta S or the change in entropy of the system plus the change in entropy, the surrounding will give you the change in entropy of the universe. Okay? Now remember, the Universe is an isolated system. And according to the second law of thermodynamics, an isolated system cannot decrease in entropy."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "Okay? Now remember, the Universe is an isolated system. And according to the second law of thermodynamics, an isolated system cannot decrease in entropy. That is, the entropy must either remain the same or it might increase. Okay? So we could represent this formula in another way, basically the same thing, except now we have this zero here, greater than or equal to zero."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "That is, the entropy must either remain the same or it might increase. Okay? So we could represent this formula in another way, basically the same thing, except now we have this zero here, greater than or equal to zero. So the total entropy of the universe can never decrease. It must either remain the same or it must increase. What this also says is that the entropy of the system can decrease as long as there is a greater or equal increase in the entropy of the surrounding."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "So the total entropy of the universe can never decrease. It must either remain the same or it must increase. What this also says is that the entropy of the system can decrease as long as there is a greater or equal increase in the entropy of the surrounding. Suppose that this decreases by ten. So this is negative ten. If this is ten, there's an increase in the universe by ten."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "Suppose that this decreases by ten. So this is negative ten. If this is ten, there's an increase in the universe by ten. Then you get negative ten plus ten will give you zero. So that works. If this decreases by negative ten and this increases by, say, eleven, then negative ten plus eleven will give you one also positive."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "Then you get negative ten plus ten will give you zero. So that works. If this decreases by negative ten and this increases by, say, eleven, then negative ten plus eleven will give you one also positive. So that works. But if we get negative ten, so the system decreases in entropy by negative ten, but the universe increases only by nine, then we get negative ten plus nine will give you negative one, a negative number, a violation of the second law of thermodynamics. And this basically tells us that if the entropy of the system decreases, then the entropy of the surroundings must increase by the same amount or greater."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "So that works. But if we get negative ten, so the system decreases in entropy by negative ten, but the universe increases only by nine, then we get negative ten plus nine will give you negative one, a negative number, a violation of the second law of thermodynamics. And this basically tells us that if the entropy of the system decreases, then the entropy of the surroundings must increase by the same amount or greater. You can never increase by less. Okay, now let's look at these isolated systems. We're going to define entropy in a different way."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "You can never increase by less. Okay, now let's look at these isolated systems. We're going to define entropy in a different way. Now, a third way. Now, in this system, we have six molecules on this side and two molecules on this side. Okay?"}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "Now, a third way. Now, in this system, we have six molecules on this side and two molecules on this side. Okay? So we have more molecules here than here. So according to the probability theory of entropy, or the probability definition of entropy, what will happen? Well, two of these will want to go into here."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "So we have more molecules here than here. So according to the probability theory of entropy, or the probability definition of entropy, what will happen? Well, two of these will want to go into here. That situation will be more probable. So we want to even out. We want our system to be even."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "That situation will be more probable. So we want to even out. We want our system to be even. We want four molecules here and four molecules here. Okay? So according to the probability definition of entropy, this is more likely, more probable."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "We want four molecules here and four molecules here. Okay? So according to the probability definition of entropy, this is more likely, more probable. So another way to define entropy would be as follows. Entropy is nature's way to spread energy evenly throughout a system. So if you have a reaction where the reactants have much more molecules than the product, then the equation or the reaction will tend to go right because there are less molecules on the product."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "So another way to define entropy would be as follows. Entropy is nature's way to spread energy evenly throughout a system. So if you have a reaction where the reactants have much more molecules than the product, then the equation or the reaction will tend to go right because there are less molecules on the product. Side than the reactive side. The last thing that we should talk about when we talk about entropy is the third law of thermodynamics. But before we get into that, let's define entropy mathematically using a formula."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "Side than the reactive side. The last thing that we should talk about when we talk about entropy is the third law of thermodynamics. But before we get into that, let's define entropy mathematically using a formula. The formula for Entropy is as follows. Change in entropy is equal to change in heat over temperature. And it has the units joules per kelvin."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "The formula for Entropy is as follows. Change in entropy is equal to change in heat over temperature. And it has the units joules per kelvin. Now, what the third law does is it assigns a zero entropy value to all elements that are at zero kelvin. But remember, kelvin or zero Kelvin is attainable. And let's t y well, this formula says that the temperature, in fact, was zero."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "Now, what the third law does is it assigns a zero entropy value to all elements that are at zero kelvin. But remember, kelvin or zero Kelvin is attainable. And let's t y well, this formula says that the temperature, in fact, was zero. The zero here would be a denominator. And that's impossible mathematically, because anything over zero is undefined. Okay, now, last two things that we should mention is that entropy is an extensive property."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "The zero here would be a denominator. And that's impossible mathematically, because anything over zero is undefined. Okay, now, last two things that we should mention is that entropy is an extensive property. And what that basically means that if the system grows in size, entropy increases. So if the system grows in volume, the entropy increases because there is more room for the molecules to move around. If temperature increases, the kinetic energy of the molecules increase."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "And what that basically means that if the system grows in size, entropy increases. So if the system grows in volume, the entropy increases because there is more room for the molecules to move around. If temperature increases, the kinetic energy of the molecules increase. The molecules move around more violently, quicker and faster, until the entropy increases. If the number of moles increases, the number of particles increases. There are more particles, so the entropy increases as well."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "The molecules move around more violently, quicker and faster, until the entropy increases. If the number of moles increases, the number of particles increases. There are more particles, so the entropy increases as well. The last thing we should mention is that entropy is a state function. And what that basically means is that change in entropy of the forward reaction is equal to the negative of that reverse reaction. And what that means is as follows."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "The last thing we should mention is that entropy is a state function. And what that basically means is that change in entropy of the forward reaction is equal to the negative of that reverse reaction. And what that means is as follows. If we look at the reaction here, a plus B forms C, we go from two molecules to one molecule. So entropy decreases. The reverse reaction would mean that we go from C to A plus B."}, {"title": "Entropy and Third Law of Thermodynamics .txt", "text": "If we look at the reaction here, a plus B forms C, we go from two molecules to one molecule. So entropy decreases. The reverse reaction would mean that we go from C to A plus B. So we go from C one molecule to two molecules. So, according to this, because entropy is a state function, the reverse reaction is the negative. The Ford reaction."}, {"title": "Vapor pressure.txt", "text": "Whenever we talk about vapor pressure, we talk about a closed system. A closed system is a system in which no mass leaves the system. So no molecules leave the system. Now, we also talk about two phases being present in the system either liquid and gas or solid in gas. Now imagine we had a closed container. In this container."}, {"title": "Vapor pressure.txt", "text": "Now, we also talk about two phases being present in the system either liquid and gas or solid in gas. Now imagine we had a closed container. In this container. We had a batch vacuum, and we could somehow inject a liquid into this vacuum. So we have empty space and we have the liquid on the bottom. Now, any of these molecules at any given time has some kinetic energy."}, {"title": "Vapor pressure.txt", "text": "We had a batch vacuum, and we could somehow inject a liquid into this vacuum. So we have empty space and we have the liquid on the bottom. Now, any of these molecules at any given time has some kinetic energy. Now, if the kinetic energy is high enough, they could escape the liquid phase and become a gas molecule. And this is called evaporation. Condensation is the opposite."}, {"title": "Vapor pressure.txt", "text": "Now, if the kinetic energy is high enough, they could escape the liquid phase and become a gas molecule. And this is called evaporation. Condensation is the opposite. It's the process by which gas molecules hitting the surface of the liquid get stuck in the liquid. Now, when the rates of evaporation equal condensation, the state is an equilibrium. And when the state is in the equilibrium, we can measure the pressure exerted by the gas molecules on the wall of the container."}, {"title": "Vapor pressure.txt", "text": "It's the process by which gas molecules hitting the surface of the liquid get stuck in the liquid. Now, when the rates of evaporation equal condensation, the state is an equilibrium. And when the state is in the equilibrium, we can measure the pressure exerted by the gas molecules on the wall of the container. And this pressure is called vapor pressure. So vapor pressure, by definition, is the pressure exerted by gas molecules that are in equilibrium with their liquid or solid state. And at this point, the rate of molecules leaving the solid or liquid state is equal to the rate at which they enter that state."}, {"title": "Vapor pressure.txt", "text": "And this pressure is called vapor pressure. So vapor pressure, by definition, is the pressure exerted by gas molecules that are in equilibrium with their liquid or solid state. And at this point, the rate of molecules leaving the solid or liquid state is equal to the rate at which they enter that state. Now, from everyday experience, we know a hot liquid will evaporate quicker than a cold liquid. And this is because vapor pressure is related to kinetic energy and temperature. This equation shows exactly that."}, {"title": "Vapor pressure.txt", "text": "Now, from everyday experience, we know a hot liquid will evaporate quicker than a cold liquid. And this is because vapor pressure is related to kinetic energy and temperature. This equation shows exactly that. This equation shows that if we increase temperature, we will increase the partial pressure. And that means we have more kinetic energy. The more kinetic energy we have, the more molecules we have that will leave the liquid state and travel into the gas state."}, {"title": "Vapor pressure.txt", "text": "This equation shows that if we increase temperature, we will increase the partial pressure. And that means we have more kinetic energy. The more kinetic energy we have, the more molecules we have that will leave the liquid state and travel into the gas state. Volatility is the ability of molecules to evaporate. So a volatile substance will evaporate quickly. An example of one is alcohol."}, {"title": "Vapor pressure.txt", "text": "Volatility is the ability of molecules to evaporate. So a volatile substance will evaporate quickly. An example of one is alcohol. Nonvolatile substances are those molecules that will not evaporate quickly. Boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. At this point, more liquid molecules have a higher kinetic energy and therefore more likely to escape into the gas state."}, {"title": "Vapor pressure.txt", "text": "Nonvolatile substances are those molecules that will not evaporate quickly. Boiling point is the temperature at which the vapor pressure of the liquid equals atmospheric pressure. At this point, more liquid molecules have a higher kinetic energy and therefore more likely to escape into the gas state. Melting point is the temperature at which the vapor pressure of the liquid equals the vapor pressure of the solid. At this point, more solid molecules have more kinetic energy and therefore are more likely to escape into the liquid state. Intermolecular bonds are those bonds?"}, {"title": "Vapor pressure.txt", "text": "Melting point is the temperature at which the vapor pressure of the liquid equals the vapor pressure of the solid. At this point, more solid molecules have more kinetic energy and therefore are more likely to escape into the liquid state. Intermolecular bonds are those bonds? Noncovalent bonds that hold molecules together in the liquid and solid states. Stronger. Tim."}, {"title": "Vapor pressure.txt", "text": "Noncovalent bonds that hold molecules together in the liquid and solid states. Stronger. Tim. Molecular bonds will require more energy and a higher temperature to break. That means the boiling point will increase. That's why Alcohol, which Has Very Weaker Tummolcular Bonds, will Boil At A Lower Temperature."}, {"title": "Factors that Affect Solubility.txt", "text": "I've outlined four different factors. The first thing that influences a gasolubility within a given liquid is something called Henry's Law. And Henry's law is a formula that relates to solubility of the gas, to the pressure of the gas. And the high the pressure, the higher the solubility. For example, suppose we have a soda can. Now."}, {"title": "Factors that Affect Solubility.txt", "text": "And the high the pressure, the higher the solubility. For example, suppose we have a soda can. Now. How are they able to get so much gas in the liquid in a soda can? Well, that's because they decreased the volume to such a small volume found above the liquid that the increase in pressure Increase in pressure. Causes solubility of the gas to increase."}, {"title": "Factors that Affect Solubility.txt", "text": "How are they able to get so much gas in the liquid in a soda can? Well, that's because they decreased the volume to such a small volume found above the liquid that the increase in pressure Increase in pressure. Causes solubility of the gas to increase. And so a lot of the carbon dioxide found above the liquid in a soda can dissolves in the liquid. But when you open the liquid, the pressure decreases because there's an influx or an alflux of molecules out of the can. And therefore the alflux creates a lower pressure."}, {"title": "Factors that Affect Solubility.txt", "text": "And so a lot of the carbon dioxide found above the liquid in a soda can dissolves in the liquid. But when you open the liquid, the pressure decreases because there's an influx or an alflux of molecules out of the can. And therefore the alflux creates a lower pressure. A lower pressure means Solubility decreases. A lot of the molecules will leave the liquid and you will get a stale tasting soda. The second thing that influences solubility of gases is temperature."}, {"title": "Factors that Affect Solubility.txt", "text": "A lower pressure means Solubility decreases. A lot of the molecules will leave the liquid and you will get a stale tasting soda. The second thing that influences solubility of gases is temperature. The higher the temperature, the less soluble something is. Or a gas is. And that's because it has more kinetic energy and will be less likely to stay within the liquid."}, {"title": "Factors that Affect Solubility.txt", "text": "The higher the temperature, the less soluble something is. Or a gas is. And that's because it has more kinetic energy and will be less likely to stay within the liquid. Sites also influence the solubility. Larger molecules, experience larger vandal forces and tend to increase in solubility. And that's because a larger or heavier molecule will be pulled more strongly by the liquid and therefore will be more likely to dissolve within the liquid."}, {"title": "Factors that Affect Solubility.txt", "text": "Sites also influence the solubility. Larger molecules, experience larger vandal forces and tend to increase in solubility. And that's because a larger or heavier molecule will be pulled more strongly by the liquid and therefore will be more likely to dissolve within the liquid. Finally, gases that chemically react with liquids dissolve more readily in a liquid. Now for liquids and solids, it's a little different. Now."}, {"title": "Factors that Affect Solubility.txt", "text": "Finally, gases that chemically react with liquids dissolve more readily in a liquid. Now for liquids and solids, it's a little different. Now. Liquids and solids are not compressible and therefore changes in pressure has no effect on Solubility. So there is no Henry's Law for liquids and solids. However, temperature now increases Solubility."}, {"title": "Factors that Affect Solubility.txt", "text": "Liquids and solids are not compressible and therefore changes in pressure has no effect on Solubility. So there is no Henry's Law for liquids and solids. However, temperature now increases Solubility. For gases. We saw there is a decrease in Solubility when the temperatures increased. For liquids and solids."}, {"title": "Factors that Affect Solubility.txt", "text": "For gases. We saw there is a decrease in Solubility when the temperatures increased. For liquids and solids. It's the opposite. And that's because, for example, when salts dissolve in a liquid, entropy increases. Entropy becomes positive."}, {"title": "Factors that Affect Solubility.txt", "text": "It's the opposite. And that's because, for example, when salts dissolve in a liquid, entropy increases. Entropy becomes positive. And so if we look at Git's free energy at higher temperatures, this component becomes positive. At higher temperatures. This guy will be negative."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's begin. Here we have A-B-C and DME. So we have five examples. So let's begin. So, in part a, we want to draw the lewis structure for ch three. Now our first step is to figure out the amount or the number of balanced electrons in this molecule, in this ch three molecule."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's begin. So, in part a, we want to draw the lewis structure for ch three. Now our first step is to figure out the amount or the number of balanced electrons in this molecule, in this ch three molecule. So let's begin by drawing out our carbon electron configuration. So we have two electrons that go into one s orbital, two electrons that go into the two s orbital, and two more electrons that go into our two p orbital. For h, we have only one electron, and that one electron goes into our one s orbital."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's begin by drawing out our carbon electron configuration. So we have two electrons that go into one s orbital, two electrons that go into the two s orbital, and two more electrons that go into our two p orbital. For h, we have only one electron, and that one electron goes into our one s orbital. Now, to count the number of balanced electrons, we simply have to figure out what the number of electrons are for carbon that are located in the outermost shell. The outermost shell is the n equals two shell. And in the N equals two shell, we have two plus two four electrons."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "Now, to count the number of balanced electrons, we simply have to figure out what the number of electrons are for carbon that are located in the outermost shell. The outermost shell is the n equals two shell. And in the N equals two shell, we have two plus two four electrons. So that means we have four bounce electrons for carbon for h, because we have only one electron, and this one electron is an outermost one s shell, we have one balance electron for each H atom. So four plus three, because we have three h atoms, and each h atom donates one balance electrons, we have a total of seven electrons. Now, notice that this guy is neutral, so this stays the same."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So that means we have four bounce electrons for carbon for h, because we have only one electron, and this one electron is an outermost one s shell, we have one balance electron for each H atom. So four plus three, because we have three h atoms, and each h atom donates one balance electrons, we have a total of seven electrons. Now, notice that this guy is neutral, so this stays the same. If, for example, this had a negative one on top, that means we have one extra electron. So if there would be a negative one here, this would have been eight electrons. But since this is neutral, there's no charge on top."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "If, for example, this had a negative one on top, that means we have one extra electron. So if there would be a negative one here, this would have been eight electrons. But since this is neutral, there's no charge on top. This stays at seven electrons. So let's begin drawing our lewiston structure. So we counted four balanced electrons for carbon."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "This stays at seven electrons. So let's begin drawing our lewiston structure. So we counted four balanced electrons for carbon. So we're going to have three bonding electrons for carbon and one non bonding. What that basically means is that three electrons that carbon will donate will be part of a bond, while one electron, which will be found here, won't be part of that bond. Now, each h atom will donate one electron each."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So we're going to have three bonding electrons for carbon and one non bonding. What that basically means is that three electrons that carbon will donate will be part of a bond, while one electron, which will be found here, won't be part of that bond. Now, each h atom will donate one electron each. So let's draw our one electron each. So these two electrons, where one came from h and one came for c, will be shared by these two atoms, and they will create a sigma bond, or also known as a covalent bond. And we can represent that by simply drawing this dashed line here."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's draw our one electron each. So these two electrons, where one came from h and one came for c, will be shared by these two atoms, and they will create a sigma bond, or also known as a covalent bond. And we can represent that by simply drawing this dashed line here. So here's our depiction, or our low style structure of ch three. Then let's make sure that this is, in fact, the correct picture. Once again, we have seven electrons."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So here's our depiction, or our low style structure of ch three. Then let's make sure that this is, in fact, the correct picture. Once again, we have seven electrons. So 123-4567. This line represents two electrons. So this satisfies our condition."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So 123-4567. This line represents two electrons. So this satisfies our condition. And this is, in fact, the correct drawing for the lewis dot structure for this guy. So let's go to part b. In part in part b, we have a ch two atom."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "And this is, in fact, the correct drawing for the lewis dot structure for this guy. So let's go to part b. In part in part b, we have a ch two atom. So we have one less h atom as compared to part A. So once again, let's draw our electron configuration for both atoms, which is exactly the same as in part A, we have two electrons in the one s and four electrons in the balanced shell. So in the two s and the two p. Now we have two h atoms."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So we have one less h atom as compared to part A. So once again, let's draw our electron configuration for both atoms, which is exactly the same as in part A, we have two electrons in the one s and four electrons in the balanced shell. So in the two s and the two p. Now we have two h atoms. So that means we're going to have one less electron than in part A. And notice that this is a neutral atom. So we're going to have four plus two electrons."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So that means we're going to have one less electron than in part A. And notice that this is a neutral atom. So we're going to have four plus two electrons. So six electrons. So let's look at our depiction here. So once again we're going to get four electrons that come from the h atom from the C atom."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So six electrons. So let's look at our depiction here. So once again we're going to get four electrons that come from the h atom from the C atom. And two of these electrons will be bonding electrons. That is, they will bond or be shared with two atoms, these two h atoms. And now we have two non bonding electrons left over."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "And two of these electrons will be bonding electrons. That is, they will bond or be shared with two atoms, these two h atoms. And now we have two non bonding electrons left over. So let's put them in the bond up here. Now we have two electrons coming from h, so let's fill them in and let's draw our bond. So here we have two electrons coming from here and two electrons here and two electrons here."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's put them in the bond up here. Now we have two electrons coming from h, so let's fill them in and let's draw our bond. So here we have two electrons coming from here and two electrons here and two electrons here. So two two and two gives us six. So this is in fact, a correct picture. Now this is once again a sigma bond, and this is our non bonding pair of electrons."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So two two and two gives us six. So this is in fact, a correct picture. Now this is once again a sigma bond, and this is our non bonding pair of electrons. So let's go to part C.\nIn part C we have NH three or ammonia. So in this case, we have an N atom. So let's throw out our electron configuration for N.\nSo two electrons go into the one s, two electrons going to the two s, and now we have three electrons going to our two p. So altogether we have five balanced electrons because we have five electrons in the outermost shell for anna."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's go to part C.\nIn part C we have NH three or ammonia. So in this case, we have an N atom. So let's throw out our electron configuration for N.\nSo two electrons go into the one s, two electrons going to the two s, and now we have three electrons going to our two p. So altogether we have five balanced electrons because we have five electrons in the outermost shell for anna. Now we have the same story as before. For h, we have one electron that goes into the one s. And because we have three h's, that the five plus three gives us eight electrons. So let's throw out our depiction here."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "Now we have the same story as before. For h, we have one electron that goes into the one s. And because we have three h's, that the five plus three gives us eight electrons. So let's throw out our depiction here. So once again, anna will start with n will start with a central carbon central atom. So we have five electrons. So we're going to have three bonding electrons."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So once again, anna will start with n will start with a central carbon central atom. So we have five electrons. So we're going to have three bonding electrons. So let's show them here. And then we're going to have two electrons left over. So let's put them in our orbital here, and let's also add our electrons that come from our h atom."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's show them here. And then we're going to have two electrons left over. So let's put them in our orbital here, and let's also add our electrons that come from our h atom. So here we have this situation. So let's still in our line. So here we have the following lewis structure for ammonia."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So here we have this situation. So let's still in our line. So here we have the following lewis structure for ammonia. Now we have two electrons that are shared here, two electrons shared here, and two electrons shared here. So that's two plus two plus two, six plus two in the non bonding. So we have a total of eight electrons, and this is in fact, a neutral ammonia atom."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "Now we have two electrons that are shared here, two electrons shared here, and two electrons shared here. So that's two plus two plus two, six plus two in the non bonding. So we have a total of eight electrons, and this is in fact, a neutral ammonia atom. So let's go to part D and part D. We basically have a combination of part A and an N atom. So once again, from part A and from part C, we have the following electron configuration for our HC and N atom. Respectfully now, a total of twelve electrons because we're going to have three electrons that come from our H. We have five electrons coming from our N, and we have four electrons coming from our carbon atom."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's go to part D and part D. We basically have a combination of part A and an N atom. So once again, from part A and from part C, we have the following electron configuration for our HC and N atom. Respectfully now, a total of twelve electrons because we're going to have three electrons that come from our H. We have five electrons coming from our N, and we have four electrons coming from our carbon atom. So here's our depiction. So once again, let's begin with the central atom. Let's begin with our carbon."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So here's our depiction. So once again, let's begin with the central atom. Let's begin with our carbon. We're going to have four electrons coming for a carbon, and this time all of them will be part of a bond or they're going to be bonding electrons. Let's show our three electrons, one each from the H. And here we have like so let's fill in our lines. And now we're going to have five electrons coming from our N.\nSo one of these electrons will come from one of these electrons will be a bonding electron."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "We're going to have four electrons coming for a carbon, and this time all of them will be part of a bond or they're going to be bonding electrons. Let's show our three electrons, one each from the H. And here we have like so let's fill in our lines. And now we're going to have five electrons coming from our N.\nSo one of these electrons will come from one of these electrons will be a bonding electron. The other two electrons will be part of the non bonding configuration. So they're going to be in the orbital and they won't be bonding. So let's fill this guy in."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "The other two electrons will be part of the non bonding configuration. So they're going to be in the orbital and they won't be bonding. So let's fill this guy in. And here's our depiction. So as you would expect, N should have five electrons to have a neutral charge, c should have four electrons to have a neutral charge and H should have one each. And this is the same exact picture."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "And here's our depiction. So as you would expect, N should have five electrons to have a neutral charge, c should have four electrons to have a neutral charge and H should have one each. And this is the same exact picture. So that means we have a neutral charge. And this is the same thing as this. So let's make sure we have twelve electrons."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So that means we have a neutral charge. And this is the same thing as this. So let's make sure we have twelve electrons. 1234-5678, 910, eleven, one, two. So this satisfies our condition, our neutral condition. So we're done here."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "1234-5678, 910, eleven, one, two. So this satisfies our condition, our neutral condition. So we're done here. So let's go to part E. This is a more complicated example. We have the following atom or molecule. We have 123456 h atoms."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So let's go to part E. This is a more complicated example. We have the following atom or molecule. We have 123456 h atoms. That means we're going to have six bounds electrons altogether coming from the H atoms. And we have 1234. We have four carbon atoms."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "That means we're going to have six bounds electrons altogether coming from the H atoms. And we have 1234. We have four carbon atoms. So each carbon atom donates four bounds electrons four times 416 plus 622 electrons altogether. So let's begin. So once again, we're going to have four electrons coming from our carbon atom."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So each carbon atom donates four bounds electrons four times 416 plus 622 electrons altogether. So let's begin. So once again, we're going to have four electrons coming from our carbon atom. Okay? So we have one, two, and then we're going to have four more. Now remember an important point."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "Okay? So we have one, two, and then we're going to have four more. Now remember an important point. Carbon is able to double bond. So in this case, we're going to have double bonds. So let's let's first draw out all our bounce electrons for carbon."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "Carbon is able to double bond. So in this case, we're going to have double bonds. So let's let's first draw out all our bounce electrons for carbon. So we have 1234 for this carbon. We're going to have one, two, three and four for this carbon. We're going to have one, two, three and four for this carbon."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So we have 1234 for this carbon. We're going to have one, two, three and four for this carbon. We're going to have one, two, three and four for this carbon. And we're going to have well, actually, let's put this guy here and then we're going to have 1234 for that carbon and let's finish our HS. So one each. So one, one and one."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "And we're going to have well, actually, let's put this guy here and then we're going to have 1234 for that carbon and let's finish our HS. So one each. So one, one and one. So now let's fill in our lines. So we have like so like so there you go. We have one here."}, {"title": "Drawing Lewis Structures Example #1 .txt", "text": "So now let's fill in our lines. So we have like so like so there you go. We have one here. We have one here, one here. We have two here, one here and one here. So we're going to have two double bonds here."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "In this lecture, I'd like to examine a few important reactive intermediates that can be produced from a methane molecule. Now, reactive intermediates are simply molecules or compounds that are too unstable and reactive to exist for a very long time. So they exist for a very short time, and they react with other compounds and molecules to produce new compounds. So let's look at the following methane molecule. So, in this react here, we have the following ch bond that dissociates. So this bond breaks in such a way that the two electrons go directly onto our H atom."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So let's look at the following methane molecule. So, in this react here, we have the following ch bond that dissociates. So this bond breaks in such a way that the two electrons go directly onto our H atom. So we get the following two reactive intermediates. We get the methylcat ion, and we get this Hydride ion. So this Hydride molecule has two electrons in the one S state, in the one S orbital."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So we get the following two reactive intermediates. We get the methylcat ion, and we get this Hydride ion. So this Hydride molecule has two electrons in the one S state, in the one S orbital. So it has a negative charge. This methylcat ion has a positive charge on the carbon. That means it has one less electron than it should in its neutral state."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So it has a negative charge. This methylcat ion has a positive charge on the carbon. That means it has one less electron than it should in its neutral state. Now, this guy will be very active because it will tend to act as a lewis base donating that pair of electrons. Likewise, this guy will also be very active. It will tend to accept a pair of electrons into its empty orbital, which we'll see in just a second."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "Now, this guy will be very active because it will tend to act as a lewis base donating that pair of electrons. Likewise, this guy will also be very active. It will tend to accept a pair of electrons into its empty orbital, which we'll see in just a second. So let's look at the second type of reaction. Now, this methane molecule, the ch bond, also dissociates, but now it associates in a way such that the pair of electrons go onto the carbon atoms. So now we get a methyl anion and a positively charged H atom."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So let's look at the second type of reaction. Now, this methane molecule, the ch bond, also dissociates, but now it associates in a way such that the pair of electrons go onto the carbon atoms. So now we get a methyl anion and a positively charged H atom. So this guy has no electrons in the one s orbital. So that means it will be very active and will tend to act as a lewis acid. Now, this methyl and ion, since now it has a pair of electrons inside its orbital, it will tend to act as a lewis base donating those electrons."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So this guy has no electrons in the one s orbital. So that means it will be very active and will tend to act as a lewis acid. Now, this methyl and ion, since now it has a pair of electrons inside its orbital, it will tend to act as a lewis base donating those electrons. And because it has one more electron than it should in its neutral states, it has a negative one charge. So let's look at the third type of reaction with the methane molecule. In this reaction, the following ch bond associates."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "And because it has one more electron than it should in its neutral states, it has a negative one charge. So let's look at the third type of reaction with the methane molecule. In this reaction, the following ch bond associates. But it associates in a way such that one electron goes onto the carbon and one electron goes onto the H atom. So we get the following two radicals. We get the methyl radical and the H radical."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "But it associates in a way such that one electron goes onto the carbon and one electron goes onto the H atom. So we get the following two radicals. We get the methyl radical and the H radical. So each molecule has one electron in its orbital. So now let's examine the three dimensional shapes of these reactive intermediates, and let's compare and contrast them. So let's go back to the methylcanion."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So each molecule has one electron in its orbital. So now let's examine the three dimensional shapes of these reactive intermediates, and let's compare and contrast them. So let's go back to the methylcanion. So, I said earlier that this carbon will have an extra or an empty two p orbital. And because it has an empty two p orbital, it will act as a lewis acid. So let's see what that means."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So, I said earlier that this carbon will have an extra or an empty two p orbital. And because it has an empty two p orbital, it will act as a lewis acid. So let's see what that means. Notice also that it has three identical ch bonds. And that means this carbon will be SP two hybridized. So all these bonds will be SP two hybridized."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "Notice also that it has three identical ch bonds. And that means this carbon will be SP two hybridized. So all these bonds will be SP two hybridized. And that means that all these bonds will lie on the same plane, let's say the XY plane. And the angle. Since we have three angles here, the angle between each adjacent ch bond will be 120 degrees."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "And that means that all these bonds will lie on the same plane, let's say the XY plane. And the angle. Since we have three angles here, the angle between each adjacent ch bond will be 120 degrees. So this guy is 120. This angle is 120. And the angle in the back is also 120."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So this guy is 120. This angle is 120. And the angle in the back is also 120. Now, this line, the solid line, simply means the H is coming out of the board. The dashed line simply means the H is going into the board. Notice this pure two p orbital."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "Now, this line, the solid line, simply means the H is coming out of the board. The dashed line simply means the H is going into the board. Notice this pure two p orbital. This orbital is not SP two hybridized. This is a pure two p orbital. And that's exactly why this guy will act as a lewis acid."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "This orbital is not SP two hybridized. This is a pure two p orbital. And that's exactly why this guy will act as a lewis acid. So if you take this molecule or atom and combine it with this molecule, this lewis base will tend to donate this pair of electrons to this orbital forming back our methane molecule. So now let's look at the methylamion. Now, we have a similar picture, but we have a pair of electrons within this two p orbital."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So if you take this molecule or atom and combine it with this molecule, this lewis base will tend to donate this pair of electrons to this orbital forming back our methane molecule. So now let's look at the methylamion. Now, we have a similar picture, but we have a pair of electrons within this two p orbital. And that means the pair of electrons will be found within this green positive region. And now we're no longer going to have SP two hybridized orbitals. In fact, this pair of electrons will act as if it was a bond because this pair of electrons will increase the negative charge, thereby increasing the electron density."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "And that means the pair of electrons will be found within this green positive region. And now we're no longer going to have SP two hybridized orbitals. In fact, this pair of electrons will act as if it was a bond because this pair of electrons will increase the negative charge, thereby increasing the electron density. And this will make the ch bonds move downward in an umbrella like fashion. So now our bonds are going to be very close to SP three hybridized. So we're no longer at SP two."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "And this will make the ch bonds move downward in an umbrella like fashion. So now our bonds are going to be very close to SP three hybridized. So we're no longer at SP two. We're SP three in this methyl, anion molecule. And now this pair of electrons will tend to act as a lewis base, donating these two electrons to some other atom. Now let's look at the final methyl radical."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "We're SP three in this methyl, anion molecule. And now this pair of electrons will tend to act as a lewis base, donating these two electrons to some other atom. Now let's look at the final methyl radical. Now, the methyl radical is a combination of the following two drawings. Now, at one given point, it's in this state, and another given point, it's in this state. So it inverts from this form to this form."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "Now, the methyl radical is a combination of the following two drawings. Now, at one given point, it's in this state, and another given point, it's in this state. So it inverts from this form to this form. So it exists somewhere in between. And in fact, we can approximate that the three dimensional picture of our methyl radical looks something like this. So it's very similar to this methylcat ion."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "So it exists somewhere in between. And in fact, we can approximate that the three dimensional picture of our methyl radical looks something like this. So it's very similar to this methylcat ion. And for most purposes, we can approximate that this has the same exact three dimensional drawing as this methylcat ion. So we assume for the most part that all these bonds are SP two hybridized. In other words, because we only have one electron."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "And for most purposes, we can approximate that this has the same exact three dimensional drawing as this methylcat ion. So we assume for the most part that all these bonds are SP two hybridized. In other words, because we only have one electron. And now two, that one electron does not have enough power to bend this shape as much as these two electrons bend the shape here. So they will exist between this and this guy. And an approximation somewhere in between is this drawing here."}, {"title": "Methyl Cation, Methyl Anion, and Methyl Radical Intermediates .txt", "text": "And now two, that one electron does not have enough power to bend this shape as much as these two electrons bend the shape here. So they will exist between this and this guy. And an approximation somewhere in between is this drawing here. So, once again, all these three molecules are very reactive. They're called reactive income medians. And they won't persist in a stable state for a very long time."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And thus far, we've said that, according to quantum mechanics, the number of atomic orbitals that we're combining must exactly equal the number of molecular orbitals that we are forming. Once again, the number of orbitals you are combining must equal the number that you are forming. Now. So far, we have tried to combine the one s atomic orbitals of two identical H atoms. So they're both neutral. That means they have one electron and one proton each."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "So far, we have tried to combine the one s atomic orbitals of two identical H atoms. So they're both neutral. That means they have one electron and one proton each. And so far we saw that when we combine these two atomic orbitals of the one of the H atom one that's orbitals, we form something called a five bonding molecular orbital. Or simply a bonding molecular orbital. And we said that the energy of this molecular orbital, of the Bonding molecular orbital, is lower than either of the atomic orbitals from which we're forming."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And so far we saw that when we combine these two atomic orbitals of the one of the H atom one that's orbitals, we form something called a five bonding molecular orbital. Or simply a bonding molecular orbital. And we said that the energy of this molecular orbital, of the Bonding molecular orbital, is lower than either of the atomic orbitals from which we're forming. So, according to this diagram, we see that the y axis is energy. And so the lower we go, the less energy we have. The higher we go, the more energy we have."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "So, according to this diagram, we see that the y axis is energy. And so the lower we go, the less energy we have. The higher we go, the more energy we have. Now, once again, according to Quantum Mechanics, if we begin with two atomic orbitals, we have to form two molecular orbitals. So far we've only seen one. Well, the second one is shown right up here, and it's called the phi antibonding molecular orbital."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "Now, once again, according to Quantum Mechanics, if we begin with two atomic orbitals, we have to form two molecular orbitals. So far we've only seen one. Well, the second one is shown right up here, and it's called the phi antibonding molecular orbital. And we'll see why it's called antibonding in just a second. The first thing you have to notice about this antibody molecular orbital. If that is higher in energy than either of these guys, So let's move to this diagram here."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And we'll see why it's called antibonding in just a second. The first thing you have to notice about this antibody molecular orbital. If that is higher in energy than either of these guys, So let's move to this diagram here. So here we have the combination of two of these atomic orbitals to form our fine bonding molecular orbital. Now, notice the following this positive and negative sign designates the sign of the Orbitals. And Orbitals are simply wave functions."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "So here we have the combination of two of these atomic orbitals to form our fine bonding molecular orbital. Now, notice the following this positive and negative sign designates the sign of the Orbitals. And Orbitals are simply wave functions. So these guys represent the sign of the orbitals or the wave functions. They do not represent charge. Now."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "So these guys represent the sign of the orbitals or the wave functions. They do not represent charge. Now. So when we combine these two guys, what we're combining is we're combining two orbitals. Two? One as orbitals of the same sign and we form an overlapping molecular orbital called the bonding."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "So when we combine these two guys, what we're combining is we're combining two orbitals. Two? One as orbitals of the same sign and we form an overlapping molecular orbital called the bonding. Molecular orbital. So to better understand what this is, let's draw out our proton. An electron diagram."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "Molecular orbital. So to better understand what this is, let's draw out our proton. An electron diagram. So here we have the nucleus of our atom. H A. And here we have the nucleus of our atom HB."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "So here we have the nucleus of our atom. H A. And here we have the nucleus of our atom HB. Now the two electrons are found somewhere in the middle. Now, what these electrons do is they stabilize these protons. In other words, let's suppose we take these two electrons away."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "Now the two electrons are found somewhere in the middle. Now, what these electrons do is they stabilize these protons. In other words, let's suppose we take these two electrons away. According to Coulomb's law, these two positively charged nuclei would begin to repel one another and would travel in opposite directions. But what these electrons do is they stabilize these two positively charged nuclei. And because this nucleus is attracted to these two electrons, and this nucleus is attracted to these two electrons, these two protons or nuclei will be held in place, and therefore a bond or a codalin bond will be formed."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "According to Coulomb's law, these two positively charged nuclei would begin to repel one another and would travel in opposite directions. But what these electrons do is they stabilize these two positively charged nuclei. And because this nucleus is attracted to these two electrons, and this nucleus is attracted to these two electrons, these two protons or nuclei will be held in place, and therefore a bond or a codalin bond will be formed. And this is known as the bonding molecular orbital. So let's look at the Creation of the Antibinding molecular orbital. Now, what this means is that we're combining two orbitals where one of the orbitals one of the one s orbital is of a positive sign and the other atomic orbital, the other one s atomic orbital is of a negative sign."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And this is known as the bonding molecular orbital. So let's look at the Creation of the Antibinding molecular orbital. Now, what this means is that we're combining two orbitals where one of the orbitals one of the one s orbital is of a positive sign and the other atomic orbital, the other one s atomic orbital is of a negative sign. And remember, orbitals are wave functions. And that simply means that when our orbitals go from a positive sign to a negative sign, they must go through the .0. And this is known as the Nodal plane."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And remember, orbitals are wave functions. And that simply means that when our orbitals go from a positive sign to a negative sign, they must go through the .0. And this is known as the Nodal plane. And we'll see what that is in effect. So we're combining these two oppositely charged One s atomic orbitals. So We Form a positively Charged atomic orbital and a Negative charged atomic orbital."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And we'll see what that is in effect. So we're combining these two oppositely charged One s atomic orbitals. So We Form a positively Charged atomic orbital and a Negative charged atomic orbital. And this is altogether known as the molecular orbital. And notice what this region here in the middle is. This is called the nodal plane."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And this is altogether known as the molecular orbital. And notice what this region here in the middle is. This is called the nodal plane. Or simply the node. This is the .0. In other words, nodal plane is the region where the electron density is zero."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "Or simply the node. This is the .0. In other words, nodal plane is the region where the electron density is zero. What that means is that we will Never find an electron or electrons in this region here. And so let's see What that means by drawing out our Proton electron diagram. So here we have two protons."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "What that means is that we will Never find an electron or electrons in this region here. And so let's see What that means by drawing out our Proton electron diagram. So here we have two protons. Our nucleus one, our nucleus two. And electrons. Now, because The Electrons can Never Be found in this region here, that means that there is nothing stabilizing the electrostatic repulsion between these two positively charged nuclei."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "Our nucleus one, our nucleus two. And electrons. Now, because The Electrons can Never Be found in this region here, that means that there is nothing stabilizing the electrostatic repulsion between these two positively charged nuclei. And that means in this case, in the Anti bonding case, the Two nuclei will actually try to Repel One another, and Our Bond Will be broken. And that's exactly why it's called antibonding. Our bond is broken in the antibonding molecular orbital."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "And that means in this case, in the Anti bonding case, the Two nuclei will actually try to Repel One another, and Our Bond Will be broken. And that's exactly why it's called antibonding. Our bond is broken in the antibonding molecular orbital. So, once again, to Recap, electrons found In The bonding molecular orbitals will Stabilize One another because of the stabilizing formation of electrons and protons. And this will tend to hold the atoms together. Electrons In The antibinding molecular orbital cause the Dissociation of the atoms."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "So, once again, to Recap, electrons found In The bonding molecular orbitals will Stabilize One another because of the stabilizing formation of electrons and protons. And this will tend to hold the atoms together. Electrons In The antibinding molecular orbital cause the Dissociation of the atoms. Because of this Noteplane region, electrons cannot be found in the Nodal Plane region, as that means these two protons will repel and they will move away from one Another, dissociating breaking that covalent Bond. So note this a important point. Electrons can be found both in the bonding and the antibonding molecular orbitals."}, {"title": "Bonding and Antibonding Molecular Orbitals .txt", "text": "Because of this Noteplane region, electrons cannot be found in the Nodal Plane region, as that means these two protons will repel and they will move away from one Another, dissociating breaking that covalent Bond. So note this a important point. Electrons can be found both in the bonding and the antibonding molecular orbitals. However, there's a big difference between the bonding and the antibodies. When the electrons are in the bonding orbital the bonding molecular orbital. They will stabilize the bond."}, {"title": "Introduction to Lewis Structure.txt", "text": "So let's begin with examples. So let's suppose we have this fluorine atom. Now, a neutral fluorine atom has nonprofit tons and nine electrons. And the electron configuration of our fluorine atom is as follows. Two electrons will go into the one s orbital, two electrons will go into the two s orbital. Two electrons will go into the two PX, two electrons will go into the two PY, and one electron will go into the two PV."}, {"title": "Introduction to Lewis Structure.txt", "text": "And the electron configuration of our fluorine atom is as follows. Two electrons will go into the one s orbital, two electrons will go into the two s orbital. Two electrons will go into the two PX, two electrons will go into the two PY, and one electron will go into the two PV. So we have a total of 2222 and one nine electrons. So if we were to actually draw out the atomic orbitals of our fluorine atom, we would get the following picture. So within this picture, our one s orbital is found within our two s orbital."}, {"title": "Introduction to Lewis Structure.txt", "text": "So we have a total of 2222 and one nine electrons. So if we were to actually draw out the atomic orbitals of our fluorine atom, we would get the following picture. So within this picture, our one s orbital is found within our two s orbital. So I did not draw it. But this circular series here is our two s orbital. So we're placing two electrons within this two x orbital shown here."}, {"title": "Introduction to Lewis Structure.txt", "text": "So I did not draw it. But this circular series here is our two s orbital. So we're placing two electrons within this two x orbital shown here. Now we have the two p x orbital shown here. We have the two p y orbital shown here, and the two PZ orbital shown here, coming out of the board and going into the board. So this is the full depiction of the atomic orbitals of flooring."}, {"title": "Introduction to Lewis Structure.txt", "text": "Now we have the two p x orbital shown here. We have the two p y orbital shown here, and the two PZ orbital shown here, coming out of the board and going into the board. So this is the full depiction of the atomic orbitals of flooring. Now, this is very tedious and will take a long time to draw out when you're solving problems. So that's why the Lewis Structure method was developed. And what we do in the Lewis Structure method is the following."}, {"title": "Introduction to Lewis Structure.txt", "text": "Now, this is very tedious and will take a long time to draw out when you're solving problems. So that's why the Lewis Structure method was developed. And what we do in the Lewis Structure method is the following. We only worry about the balanced electrons. The balanced electrons are the electrons found in the outermost subshell, and this is this guy here. So this N equals to a principal number will include all these electrons here."}, {"title": "Introduction to Lewis Structure.txt", "text": "We only worry about the balanced electrons. The balanced electrons are the electrons found in the outermost subshell, and this is this guy here. So this N equals to a principal number will include all these electrons here. So that's two plus two plus two plus one. That's seven electrons in the balance shell in the balanced electron shell. So that means we only include these guys."}, {"title": "Introduction to Lewis Structure.txt", "text": "So that's two plus two plus two plus one. That's seven electrons in the balance shell in the balanced electron shell. So that means we only include these guys. And these one s you could think of as being inside the flooring. So they're not depicted. So we only worry about these guys."}, {"title": "Introduction to Lewis Structure.txt", "text": "And these one s you could think of as being inside the flooring. So they're not depicted. So we only worry about these guys. So two will go to two s.\nSo two here, two will go into two PX, two go here, two goes to two PY, which go here, and one goes into here. Now, notice it doesn't really matter. I could have took one away from here and placed one on the bottom."}, {"title": "Introduction to Lewis Structure.txt", "text": "So two will go to two s.\nSo two here, two will go into two PX, two go here, two goes to two PY, which go here, and one goes into here. Now, notice it doesn't really matter. I could have took one away from here and placed one on the bottom. As long as I have two two and a two and a one in any arrangement, I will have the Lewis Structure for flooring. And the same thing could be said for oxygen. So oxygen has eight protons, and that means a neutral oxygen has eight electrons."}, {"title": "Introduction to Lewis Structure.txt", "text": "As long as I have two two and a two and a one in any arrangement, I will have the Lewis Structure for flooring. And the same thing could be said for oxygen. So oxygen has eight protons, and that means a neutral oxygen has eight electrons. So let's once again draw our electron configuration for oxygen. So we have one s. So two electrons go to the one s.\nWe have two s, two electrons go to the two s. We have two PX and two PY and two PV. So two go in here, one goes in here, and one goes in the last two PZ."}, {"title": "Introduction to Lewis Structure.txt", "text": "So let's once again draw our electron configuration for oxygen. So we have one s. So two electrons go to the one s.\nWe have two s, two electrons go to the two s. We have two PX and two PY and two PV. So two go in here, one goes in here, and one goes in the last two PZ. Now, once again, we draw out the full atomic orbital picture for our oxygen. So two go into the one S, which we did not draw because it's within our two S. So two go into the two S.\nSo here's our sphere to go inside the two S, two go to the two PX. So two go into the two PX."}, {"title": "Introduction to Lewis Structure.txt", "text": "Now, once again, we draw out the full atomic orbital picture for our oxygen. So two go into the one S, which we did not draw because it's within our two S. So two go into the two S.\nSo here's our sphere to go inside the two S, two go to the two PX. So two go into the two PX. One goes into two PY. So one goes into this one, and finally, one goes into the two PZ. So once again, this took a lot of time."}, {"title": "Introduction to Lewis Structure.txt", "text": "One goes into two PY. So one goes into this one, and finally, one goes into the two PZ. So once again, this took a lot of time. There's a shortcut method now that we know the Lewis Structure method. And so we draw an oxygen atom. So the o here, we pretend that we put two electrons inside the oxygen, which means we put the two one S electrons inside the oxygen."}, {"title": "Introduction to Lewis Structure.txt", "text": "There's a shortcut method now that we know the Lewis Structure method. And so we draw an oxygen atom. So the o here, we pretend that we put two electrons inside the oxygen, which means we put the two one S electrons inside the oxygen. And now we only worry about our balanced electrons, the outermost electrons. And so we need to place two electrons that represent the two S. So you could either place them here, here, or here or here. I place them, say here."}, {"title": "Introduction to Lewis Structure.txt", "text": "And now we only worry about our balanced electrons, the outermost electrons. And so we need to place two electrons that represent the two S. So you could either place them here, here, or here or here. I place them, say here. Now we have the two PX, so I placed them here. Now I have the two P-Y-I placed one here and two PD. I placed one here."}, {"title": "Introduction to Lewis Structure.txt", "text": "Now we have the two PX, so I placed them here. Now I have the two P-Y-I placed one here and two PD. I placed one here. Now, once again, it does not matter where I place this electron pair. I could place this electron pair here, and I could place this one here, and I could switch them around in any way I want to, as long as I have one pair, one pair and two singles. So this is my depiction for oxygen."}, {"title": "Introduction to Lewis Structure.txt", "text": "Now, once again, it does not matter where I place this electron pair. I could place this electron pair here, and I could place this one here, and I could switch them around in any way I want to, as long as I have one pair, one pair and two singles. So this is my depiction for oxygen. So Lewis structure for fluorine and Lewis structure for oxygen. Now. I could also draw Lewis structures for molecules."}, {"title": "Introduction to Lewis Structure.txt", "text": "So Lewis structure for fluorine and Lewis structure for oxygen. Now. I could also draw Lewis structures for molecules. So suppose I have two fluorine atoms. So I take these two fluorine atoms, so two of them, and they interact with one another to form a diatomic fluoride. Now, what that basically means is that two of these guys will orient in a way such that these two single electrons will interact and they will form a cobaltin bond in which there is a sharing of electrons."}, {"title": "Introduction to Lewis Structure.txt", "text": "So suppose I have two fluorine atoms. So I take these two fluorine atoms, so two of them, and they interact with one another to form a diatomic fluoride. Now, what that basically means is that two of these guys will orient in a way such that these two single electrons will interact and they will form a cobaltin bond in which there is a sharing of electrons. So that means this electron will be donated to this orbital, and this electron will be donated to this orbital. So two fluoride will interact to form an electron configuration of the nearest noble gas. The nearest noble gas is neon."}, {"title": "Introduction to Lewis Structure.txt", "text": "So that means this electron will be donated to this orbital, and this electron will be donated to this orbital. So two fluoride will interact to form an electron configuration of the nearest noble gas. The nearest noble gas is neon. Neon has ten protons and ten electrons, while fluorine has or a single fluorine has nine electrons and nine protons. So since one of the fluorine will donate an electron to this guy, this will gain one electron. So this will gain ten electron, half ten electrons, and it will form electron configuration of our neon noble gas."}, {"title": "Introduction to Lewis Structure.txt", "text": "Neon has ten protons and ten electrons, while fluorine has or a single fluorine has nine electrons and nine protons. So since one of the fluorine will donate an electron to this guy, this will gain one electron. So this will gain ten electron, half ten electrons, and it will form electron configuration of our neon noble gas. And likewise, the same story for this flooring. And once again, this picture can be depicted in a shortcut way using this flooring or Lewis Structure depiction for flooring. So, once again, two electrons here, here, and here, like we drew here."}, {"title": "Introduction to Lewis Structure.txt", "text": "And likewise, the same story for this flooring. And once again, this picture can be depicted in a shortcut way using this flooring or Lewis Structure depiction for flooring. So, once again, two electrons here, here, and here, like we drew here. So this represents the one S, this the two PX, the two PY and the two PZ. So we orient them, and these electrons are shared. And so now we should actually redraw these guys, because since there's a sharing of electrons, they should draw we should draw them like this so that now each one has eight electrons as the valid electrons plus two one s electrons."}, {"title": "Introduction to Lewis Structure.txt", "text": "So this represents the one S, this the two PX, the two PY and the two PZ. So we orient them, and these electrons are shared. And so now we should actually redraw these guys, because since there's a sharing of electrons, they should draw we should draw them like this so that now each one has eight electrons as the valid electrons plus two one s electrons. So each guy has ten electrons, and that means each guy have each flooring has a has achieved electron configuration of a noble gas. Now, this is the same thing as writing this guy. Now, this is even a more shortcut method, because now we don't have to add these electrons on the outside."}, {"title": "Introduction to Lewis Structure.txt", "text": "So each guy has ten electrons, and that means each guy have each flooring has a has achieved electron configuration of a noble gas. Now, this is the same thing as writing this guy. Now, this is even a more shortcut method, because now we don't have to add these electrons on the outside. We kind of assume that they are there, but we don't draw them. What we do draw is this line here. This line simply means that there is an equal sharing of electrons."}, {"title": "Introduction to Lewis Structure.txt", "text": "We kind of assume that they are there, but we don't draw them. What we do draw is this line here. This line simply means that there is an equal sharing of electrons. And this can further be depicted by simply writing f two. And this simply means that there are two foreign atoms and they're interacting in an equivalent way. So they're sharing these two electrons."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "Heat Capacity is not the Mounted heat that Im Istoria Object remember. Heat Is the fall of Energy from a high Temperature to a Low Temperature. So what Heat Capacity actually is it's the Amount of Energy Quined to change an Object Temperature by some Amount. So this is the formula of Le Capacity. Where Capital ce is Heat capacity that equal the change in Energy or Energy Imp\u00f4t overchangein Temperature and The Units or Jel Sava jesper celles and the reason we could interchange calvince is because Changes and Temperature on the Scales are equivalent s five degree change on a cabinscaleisoa five degree change on a Celsa scale Suppose we take an object thel and we want to find the Energy Importat required to change this Object Temperature by Five degrees Celsius the way we find that is well, use this formula. Look of the particular se value for this substance we Plat get in we plan the change in Temperature In and we find the Energy input required to effect this object and Change Is objective degrees Cel siet Capacity is an extensive Property and that simply means if we Have some object the Sea shell in this hand and the 2\u1d48 Object seashell twice the Size and this one the Mathavenergy required to change the largest sech elte five the grease less will be twice as large as this one and that means if the Taquine as large for the Same Changing Temperature Will Be Large as well, so Heat Capacity or Sea changes with a change in Size of Our System."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "So this is the formula of Le Capacity. Where Capital ce is Heat capacity that equal the change in Energy or Energy Imp\u00f4t overchangein Temperature and The Units or Jel Sava jesper celles and the reason we could interchange calvince is because Changes and Temperature on the Scales are equivalent s five degree change on a cabinscaleisoa five degree change on a Celsa scale Suppose we take an object thel and we want to find the Energy Importat required to change this Object Temperature by Five degrees Celsius the way we find that is well, use this formula. Look of the particular se value for this substance we Plat get in we plan the change in Temperature In and we find the Energy input required to effect this object and Change Is objective degrees Cel siet Capacity is an extensive Property and that simply means if we Have some object the Sea shell in this hand and the 2\u1d48 Object seashell twice the Size and this one the Mathavenergy required to change the largest sech elte five the grease less will be twice as large as this one and that means if the Taquine as large for the Same Changing Temperature Will Be Large as well, so Heat Capacity or Sea changes with a change in Size of Our System. So it's An Extent of property number intense of properties of Those Properties such as Temperature that Do not Change when there is A change in Size two types of Het Capacitives that exist constant Valie Capacitives or Ce Laura tian constant Pressure het Capacitives parano. Let's talk about Constant Valetcapacities. Let's go back to our First War Terminant."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "So it's An Extent of property number intense of properties of Those Properties such as Temperature that Do not Change when there is A change in Size two types of Het Capacitives that exist constant Valie Capacitives or Ce Laura tian constant Pressure het Capacitives parano. Let's talk about Constant Valetcapacities. Let's go back to our First War Terminant. Which Basically States that Total change Energy Bar System is equal to the change in Internal Angier System plus Pv Work Done so with this Circle is Our System and all these Malts are famil tou Circle. The change in Energy is equal to the Changing all the potential Energies And Conn etic Energydata System and the work done or the work Done by these molecules on the Surrounding Molecules in Expanding or Compressing this System here. So when there is a Constant volume, lots look our equation this guy goes to zero because at Constant volume."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "Which Basically States that Total change Energy Bar System is equal to the change in Internal Angier System plus Pv Work Done so with this Circle is Our System and all these Malts are famil tou Circle. The change in Energy is equal to the Changing all the potential Energies And Conn etic Energydata System and the work done or the work Done by these molecules on the Surrounding Molecules in Expanding or Compressing this System here. So when there is a Constant volume, lots look our equation this guy goes to zero because at Constant volume. What Happens to change in Pressure. Well, it Zero so Change and Energy Is simply change in Internal Energy. So there is a Transfert Energy into a system."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "What Happens to change in Pressure. Well, it Zero so Change and Energy Is simply change in Internal Energy. So there is a Transfert Energy into a system. All the Energy is transferer In Increasing Internal Energy Bar System and Because Internal Energy Is related to Kinetic Energy and Kinetic Energy is Related to Temperature increasing Internal Energy increases America Energy and that Increases Temperature. So all the Energy In it Cas Valium Goes in Through Increasing Temperature no Pv Work is Done no expansion is done it constant pressure. However according to the ideal gasa temperature must change as well and a value must change."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "All the Energy is transferer In Increasing Internal Energy Bar System and Because Internal Energy Is related to Kinetic Energy and Kinetic Energy is Related to Temperature increasing Internal Energy increases America Energy and that Increases Temperature. So all the Energy In it Cas Valium Goes in Through Increasing Temperature no Pv Work is Done no expansion is done it constant pressure. However according to the ideal gasa temperature must change as well and a value must change. So that means that some energy goes into increasing temperature and some energy goes into increasing value. Okay, so this term is no longer zero this term exists and work is done. And that's why p values or the heat capacity at constant pressure values are usually larger than constant valium hit capacity values."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "So that means that some energy goes into increasing temperature and some energy goes into increasing value. Okay, so this term is no longer zero this term exists and work is done. And that's why p values or the heat capacity at constant pressure values are usually larger than constant valium hit capacity values. And that's because if we go back to this term here. And we look at this equation for consent valium this last part laser for this guy we simply plug in change in new for the 2\u1d48 question however we have this last term this here this anti secher will be greater. So some number over changent temperature verses a smaller number over the same changer temperature will give a smaller sea value."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "And that's because if we go back to this term here. And we look at this equation for consent valium this last part laser for this guy we simply plug in change in new for the 2\u1d48 question however we have this last term this here this anti secher will be greater. So some number over changent temperature verses a smaller number over the same changer temperature will give a smaller sea value. So this cy will be smaller than this qui srecoacapacity is an extensive property. So an will increase with increase and sides of the system. No specific capacity and mali capacitives were developed a intense of properties."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "So this cy will be smaller than this qui srecoacapacity is an extensive property. So an will increase with increase and sides of the system. No specific capacity and mali capacitives were developed a intense of properties. And that means that these two guys will save the same when the size will increase let's definie capacity specific capacity is the month of algerie to change a specific and multi mass of an object by some temperature. The formula is lower. Ces equal to put an energy over changing seperature times mass."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "And that means that these two guys will save the same when the size will increase let's definie capacity specific capacity is the month of algerie to change a specific and multi mass of an object by some temperature. The formula is lower. Ces equal to put an energy over changing seperature times mass. Now, we have this mass component in arden ome. And the units are jelspergamtimes calvin or calorifornimes celles now it's an sens of property. Why well, because is an extensive property."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "Now, we have this mass component in arden ome. And the units are jelspergamtimes calvin or calorifornimes celles now it's an sens of property. Why well, because is an extensive property. And is an extent of property. And from our lecture on extensive intent of properties, we saw that will we divide and extent of property by another extent of property, we get an in sets of property. And that's because if you increase the size of the object qu increase and but cris by simera mount and sell the cu m ratio stars the same so stars the same."}, {"title": "Heat Capacity, Specific Heat Capacity and Molar Heat Capacity.txt", "text": "And is an extent of property. And from our lecture on extensive intent of properties, we saw that will we divide and extent of property by another extent of property, we get an in sets of property. And that's because if you increase the size of the object qu increase and but cris by simera mount and sell the cu m ratio stars the same so stars the same. Or a specific et capacity stars the same. So let's define moloicapacity moloicapacity the mount of lj rir to change some mount a molles of an object by some temperature. And the formula is almost the same thing as this one except the mass is replaced with number of mills and kilogram gram is replaced by molle."}, {"title": "Partial Pressure Equilibrium Constant .txt", "text": "Now, I'm not going to go into much detail about this guy because I just want you to be aware that this guy exists. Now if reactants and products are both in the gas state, we can still use the equilibrium constant and the expression equilibrium constant. But now we have to incorporate the partial pressures of our gases. For example, suppose we have the following reaction in which gas A react with gas B to produce gas C and gas D, where it lowercase A-B-C and D are the coefficients of each respective atom that represent the moles of each atom. For example, lowercase a number of moles of A gas reacts with B number of moles of B gas produces C number of moles of C gas and B number of moles of D gas. Now our equilibrium constant for this gaseous reaction is the following our equilibrium constant KP is equal to the partial pressure of our C of our gas product C to the C power."}, {"title": "Partial Pressure Equilibrium Constant .txt", "text": "For example, suppose we have the following reaction in which gas A react with gas B to produce gas C and gas D, where it lowercase A-B-C and D are the coefficients of each respective atom that represent the moles of each atom. For example, lowercase a number of moles of A gas reacts with B number of moles of B gas produces C number of moles of C gas and B number of moles of D gas. Now our equilibrium constant for this gaseous reaction is the following our equilibrium constant KP is equal to the partial pressure of our C of our gas product C to the C power. Now that's lowercase C, an abia should represent it with the black marker. So lowercase C times the partial pressure of D gas to the D coefficient. So d exponent divided by the partial pressure of this gas A to the lowercase A power times the partial pressure of our B reactants our gas to the B lowercase B power."}, {"title": "Molality Example .txt", "text": "Molality is simply another way of measuring the concentration of a solution. The symbol for molality is lowercase letter M and the formula is moles of solute over kilograms of solvent. Now let's do an example. Using molality. The question tells us that we have 90 grams of glucose and 200 grams of H 20. And we want to find the molality of the solution."}, {"title": "Molality Example .txt", "text": "Using molality. The question tells us that we have 90 grams of glucose and 200 grams of H 20. And we want to find the molality of the solution. We're basically taking glucose, the salute and water, the solvent. We're mixing it and we want to find the concentration. In terms of molality, the first step is to find the molecular weight of glucose."}, {"title": "Molality Example .txt", "text": "We're basically taking glucose, the salute and water, the solvent. We're mixing it and we want to find the concentration. In terms of molality, the first step is to find the molecular weight of glucose. This will help us find the number of moles of glucose within our solution. The first step is to find the atomic weight of each atom and multiplied by each subscript. So the atomic weight of carbon is 12 grams/mol, the atomic weight of H is 1 gram/mol, and the atomic weight of oxygen is 16 grams/mol."}, {"title": "Molality Example .txt", "text": "This will help us find the number of moles of glucose within our solution. The first step is to find the atomic weight of each atom and multiplied by each subscript. So the atomic weight of carbon is 12 grams/mol, the atomic weight of H is 1 gram/mol, and the atomic weight of oxygen is 16 grams/mol. So we multiply six by twelve, add that to twelve times one and add that to six times 16 and we get approximately 180 grams/mol. Now, using this number, we can take our 90 grams of glucose and find the number of moles. 90 grams of glucose divided by 180 grams of glucose per mole gets us 0.5\nor one half moles of glucose."}, {"title": "Molality Example .txt", "text": "So we multiply six by twelve, add that to twelve times one and add that to six times 16 and we get approximately 180 grams/mol. Now, using this number, we can take our 90 grams of glucose and find the number of moles. 90 grams of glucose divided by 180 grams of glucose per mole gets us 0.5\nor one half moles of glucose. That's our number of moles within this solution. The final step is to use the formula for molality. Lowercase M for molality is equal to one half number of moles of salute divided by now notice we're getting 200 grams and molality ditosomality are moles per kilogram."}, {"title": "Halogens .txt", "text": "In this lecture we're going to talk about group seven, a elements, also known as group 17, according to our newer system. Now, these guides are known as the halogens, and we're going to look at five halogens. Now, fluorine, chlorine, bromine and iodine are all very common, non metallic halogens. The thick talent, which is a metal lloyd, is the least common halogen, and that's called astonine with a symbol at. Now, we're not even really talk about this guy too much because he's not very common. But you should know that aside from these talents, there's also the thick talent called Astane."}, {"title": "Halogens .txt", "text": "The thick talent, which is a metal lloyd, is the least common halogen, and that's called astonine with a symbol at. Now, we're not even really talk about this guy too much because he's not very common. But you should know that aside from these talents, there's also the thick talent called Astane. Now, halogens in general are very radioactive. And what that means is basically it reacts, or halogens react with even the most stable compounds causing damage. For example, halogens are deadly to humans or other organisms."}, {"title": "Halogens .txt", "text": "Now, halogens in general are very radioactive. And what that means is basically it reacts, or halogens react with even the most stable compounds causing damage. For example, halogens are deadly to humans or other organisms. And if we inhale some of the halogen gas, we will suffocate and die, because halogens react in our body in a very detrimental way. Now let's look specifically at fluorine. Now, fluorine exists in at room temperature in a diatomic gas form, f two form."}, {"title": "Halogens .txt", "text": "And if we inhale some of the halogen gas, we will suffocate and die, because halogens react in our body in a very detrimental way. Now let's look specifically at fluorine. Now, fluorine exists in at room temperature in a diatomic gas form, f two form. So if we inhale, for example, fluorine, the gas, that gas will cause damage in our body and we will suffocate. Now, fluorine is the most electronegative atom out there. In other words, it has the highest affinity for electrons."}, {"title": "Halogens .txt", "text": "So if we inhale, for example, fluorine, the gas, that gas will cause damage in our body and we will suffocate. Now, fluorine is the most electronegative atom out there. In other words, it has the highest affinity for electrons. It is more likely to take electrons away than any other atom. Now, fluorine also only has one oxidation state, negative one. And that means it only forms one bond."}, {"title": "Halogens .txt", "text": "It is more likely to take electrons away than any other atom. Now, fluorine also only has one oxidation state, negative one. And that means it only forms one bond. No more, only one bond. Now we'll see that chlorine, BromIn and iodine forms more than one bond. But know that fluorine, the most electronegative atom, forms only one bond, and its oxidation state is negative one."}, {"title": "Halogens .txt", "text": "No more, only one bond. Now we'll see that chlorine, BromIn and iodine forms more than one bond. But know that fluorine, the most electronegative atom, forms only one bond, and its oxidation state is negative one. For example, it forms bonds with h. It forms bonds with NA and CA, but it only forms one bond. Now let's look at chlorine. Chlorine is a diatomic gas at room temperature, just like fluorine."}, {"title": "Halogens .txt", "text": "For example, it forms bonds with h. It forms bonds with NA and CA, but it only forms one bond. Now let's look at chlorine. Chlorine is a diatomic gas at room temperature, just like fluorine. Bromine is a diatomic liquid at room temperature, and iodine is a diatomic solid at room temperature. Now, these three guys, chlorine, bromine and iodine, unlike chlorine, can have an oxidation state of up to plus seven, and that means they can all form up to seven bonds. The last important thing you should know about halogens is that halogens can react with the hion."}, {"title": "Theoretical and Percent Yield .txt", "text": "Today we're going to talk about reactions and their yield. Now, the amount of product produced when all the reactants are completely used up in other words, when our reaction runs to completion is known as the theoretical yield. Note that most reactions don't actually run to completion and that's because our reactions usually reach equilibrium before any of the reactants are depleted and that means that theoretical yield will not be possible. We'll get a number below theoretical yield and this number is known as the actual yield of our experiment. Now, to see how far a reaction proceeded, in other words, how close it got to our theoretical yield we simply find actual yield and theoretical yield. We divide two."}, {"title": "Theoretical and Percent Yield .txt", "text": "We'll get a number below theoretical yield and this number is known as the actual yield of our experiment. Now, to see how far a reaction proceeded, in other words, how close it got to our theoretical yield we simply find actual yield and theoretical yield. We divide two. So actual yield divided by theoretical yield multiplies that by 100 to get our percent because otherwise this guy is a fraction and that will give you percent yield. Now, let's look at the following reaction. So in the following reaction, 1 mol of methane reacts with two moles of oxygen to produce 1 mol of carbon dioxide and two moles of water."}, {"title": "Theoretical and Percent Yield .txt", "text": "So actual yield divided by theoretical yield multiplies that by 100 to get our percent because otherwise this guy is a fraction and that will give you percent yield. Now, let's look at the following reaction. So in the following reaction, 1 mol of methane reacts with two moles of oxygen to produce 1 mol of carbon dioxide and two moles of water. Now, let's suppose we react 1 mol of methane with two moles of oxygen, and we find that our actual yield in grams was 38 grams of carbon dioxide. Now, if we find our theoretical yield, we could then find our percent yield. How would we go about finding our theoretical yield?"}, {"title": "Theoretical and Percent Yield .txt", "text": "Now, let's suppose we react 1 mol of methane with two moles of oxygen, and we find that our actual yield in grams was 38 grams of carbon dioxide. Now, if we find our theoretical yield, we could then find our percent yield. How would we go about finding our theoretical yield? Well, we have to realize that we have 1 mol reacting in two moles. That means a theoretical yield of this guy is 1 mol. Because one to two gives you one of this guy and two of this guy."}, {"title": "Theoretical and Percent Yield .txt", "text": "Well, we have to realize that we have 1 mol reacting in two moles. That means a theoretical yield of this guy is 1 mol. Because one to two gives you one of this guy and two of this guy. And 1 mol of carbon dioxide is simply well, we take our molecular weight of our carbon dioxide which is 12 grams/mol for carbon dioxide and 32 grams/mol for two oxygens and we get 32 plus twelve gives us 44 grams/mol. This is our molecular weight of carbon dioxide. So you multiply by 1 mol and we get the theoretical yield is 44 grams of carbon dioxide."}, {"title": "Theoretical and Percent Yield .txt", "text": "And 1 mol of carbon dioxide is simply well, we take our molecular weight of our carbon dioxide which is 12 grams/mol for carbon dioxide and 32 grams/mol for two oxygens and we get 32 plus twelve gives us 44 grams/mol. This is our molecular weight of carbon dioxide. So you multiply by 1 mol and we get the theoretical yield is 44 grams of carbon dioxide. So now we follow our formula. We take our 38 grams our actual yield that we found from the experiment divide that by our 44 grams, our theoretical yield. This guy multiply that by 100% to find a percentage and we find that our percent yield is 86.4%."}, {"title": "Introduction to Alkanes .txt", "text": "Now, alkanes are a member of a family of hydrocarbons, and that simply means alkanes are composed entirely of carbons and H atoms. Now, all linear alkanes have the following molecular formula c subscript NH, subscript two N plus two, where N designates the number of carbons. For example, if N is ten, if we have an alkane composed of ten carbon atoms, that means we're going to have 22 H atoms. Two times ten plus two is 22. The simplest alkane is methane. Methane is composed of one carbon and four hydrogens."}, {"title": "Introduction to Alkanes .txt", "text": "Two times ten plus two is 22. The simplest alkane is methane. Methane is composed of one carbon and four hydrogens. Now, the molecular formula is ch four. This can be represented another way using the following diagram. So, the carbon central atom is attached to four identical H atoms in the following manner, where each bond is a covalent bond and each bond is SP three hybridized."}, {"title": "Introduction to Alkanes .txt", "text": "Now, the molecular formula is ch four. This can be represented another way using the following diagram. So, the carbon central atom is attached to four identical H atoms in the following manner, where each bond is a covalent bond and each bond is SP three hybridized. Now, what happens, for example, if we take away this H atom and along with the H atom, we take away one electron and we leave one electron on this molecule. Now, let's suppose I take two of such molecules and I interact them in such a way so that I form the following compound. So, these two electrons interact to form a covalent SP three hybridized bond, and I form a compound with two carbons."}, {"title": "Introduction to Alkanes .txt", "text": "Now, what happens, for example, if we take away this H atom and along with the H atom, we take away one electron and we leave one electron on this molecule. Now, let's suppose I take two of such molecules and I interact them in such a way so that I form the following compound. So, these two electrons interact to form a covalent SP three hybridized bond, and I form a compound with two carbons. So now I have Ethane, which has a molecular formula c, two H six. Now, I can continue doing this to produce propane, butane pentane, hexane, heptane, octane and so on. Now, butane for example, is another type of alkane that's composed of four carbon atoms."}, {"title": "Introduction to Alkanes .txt", "text": "So now I have Ethane, which has a molecular formula c, two H six. Now, I can continue doing this to produce propane, butane pentane, hexane, heptane, octane and so on. Now, butane for example, is another type of alkane that's composed of four carbon atoms. So I essentially took four of these molecules, I combined them and I got the following butane molecule. Now, another type of alkane are cycloalkanes. These are simply cyclic versions of their alkanes."}, {"title": "Introduction to Alkanes .txt", "text": "So I essentially took four of these molecules, I combined them and I got the following butane molecule. Now, another type of alkane are cycloalkanes. These are simply cyclic versions of their alkanes. And here I have two examples. For example, cyclopropane and cyclohexane. So essentially propane, pro means three a refers to the alkane, and cycle means it's in a cyclic fashion."}, {"title": "Introduction to Alkanes .txt", "text": "And here I have two examples. For example, cyclopropane and cyclohexane. So essentially propane, pro means three a refers to the alkane, and cycle means it's in a cyclic fashion. So here I have cyclopropane and cyclohexane, six carbon atoms and three carbon atoms here. Now, the formula for cycle alkanes is not the same as the formula for linear alkanes. The formula for cycloalkanes are CNH, two N. And we can see that this, in fact, is true."}, {"title": "Introduction to Alkanes .txt", "text": "So here I have cyclopropane and cyclohexane, six carbon atoms and three carbon atoms here. Now, the formula for cycle alkanes is not the same as the formula for linear alkanes. The formula for cycloalkanes are CNH, two N. And we can see that this, in fact, is true. Here we have six H atoms and we have three carbon atoms. So three carbon atoms, that means we'll have two times three six H atoms. Likewise, here we have six carbons."}, {"title": "Introduction to Alkanes .txt", "text": "Here we have six H atoms and we have three carbon atoms. So three carbon atoms, that means we'll have two times three six H atoms. Likewise, here we have six carbons. That means we will have twelve H atoms. That's exactly what we have. Two, four, 6810, twelve."}, {"title": "pH Example #2.txt", "text": "What we want to find is the hydronium concentration of our blood and the hydroxide concentration of our blood. So before we begin a luxury view exponents. Remember exponents is simply another way of writing logs. And logs is simply another way of writing exponents. We can convert from one form to the other whenever it's convenient. For example, let's look at this above statement."}, {"title": "pH Example #2.txt", "text": "And logs is simply another way of writing exponents. We can convert from one form to the other whenever it's convenient. For example, let's look at this above statement. Here we have our base is x, our exponent is e and our y is our result. So this guy equals this guy. Now we can convert from this form to the log form."}, {"title": "pH Example #2.txt", "text": "Here we have our base is x, our exponent is e and our y is our result. So this guy equals this guy. Now we can convert from this form to the log form. And the logform basically states that our log of our base x in parentheses is our result y and this equals our exponent. So why is this useful? Well let's see."}, {"title": "pH Example #2.txt", "text": "And the logform basically states that our log of our base x in parentheses is our result y and this equals our exponent. So why is this useful? Well let's see. If we know our base and we know our y, the result and we don't know our e, there's nothing we can do in this form to solve this problem. But what we can do is simply convert this guy to log form and we get log of x, which is what we know. In parentheses is our y, which is also what we know."}, {"title": "pH Example #2.txt", "text": "If we know our base and we know our y, the result and we don't know our e, there's nothing we can do in this form to solve this problem. But what we can do is simply convert this guy to log form and we get log of x, which is what we know. In parentheses is our y, which is also what we know. And we want to find our e. So now we have this entire guy. We don't count this guy. So now what we can do is simply plug this guy to the calculator, get an answer, and that's our e. Now suppose we were in the opposite situation."}, {"title": "pH Example #2.txt", "text": "And we want to find our e. So now we have this entire guy. We don't count this guy. So now what we can do is simply plug this guy to the calculator, get an answer, and that's our e. Now suppose we were in the opposite situation. Suppose we had our log x our base and we had our exponent and we didn't have our y. Well now we don't have this entire thing. So we can plug this into our calculator."}, {"title": "pH Example #2.txt", "text": "Suppose we had our log x our base and we had our exponent and we didn't have our y. Well now we don't have this entire thing. So we can plug this into our calculator. What we must do is convert from this guy to the exponent form because we have the base x, we have the y, the e to exponent. Now we have x to the e, plug that into the calculator and we get our y. And this is exactly what we do in part one."}, {"title": "pH Example #2.txt", "text": "What we must do is convert from this guy to the exponent form because we have the base x, we have the y, the e to exponent. Now we have x to the e, plug that into the calculator and we get our y. And this is exactly what we do in part one. So in part one, let's first write down our formula for PH. We know PH is equal to the negative log of our hydronium concentration. But that's the same thing as saying negative log of the hydride ion concentration."}, {"title": "pH Example #2.txt", "text": "So in part one, let's first write down our formula for PH. We know PH is equal to the negative log of our hydronium concentration. But that's the same thing as saying negative log of the hydride ion concentration. So that means this guy equals this guy. And we know from what skewed them that our PH is 7.41. So this guy equals 7.41."}, {"title": "pH Example #2.txt", "text": "So that means this guy equals this guy. And we know from what skewed them that our PH is 7.41. So this guy equals 7.41. So let's go back here for a second. So we have our exponents, we have e, we have our log, which is log ten. We have the ten, we have the x."}, {"title": "pH Example #2.txt", "text": "So let's go back here for a second. So we have our exponents, we have e, we have our log, which is log ten. We have the ten, we have the x. We don't have our inside, so we don't have our y. So we're in a situation where we can't solve in this form for the y. What we must do is we must convert from this form to the exponent form."}, {"title": "pH Example #2.txt", "text": "We don't have our inside, so we don't have our y. So we're in a situation where we can't solve in this form for the y. What we must do is we must convert from this form to the exponent form. So the equivalent way of writing logs and exponents is taking our base, base x, writing it here. Okay then taking our exponents, our exponent and writing it here but wait, we have a negative log. So this negative, we bring it over to here."}, {"title": "pH Example #2.txt", "text": "So the equivalent way of writing logs and exponents is taking our base, base x, writing it here. Okay then taking our exponents, our exponent and writing it here but wait, we have a negative log. So this negative, we bring it over to here. By simply dividing or multiplying both sides by negative one, this guy cancels becomes a positive. This guy becomes a negative. We get ten to the negative 7.41 is equal."}, {"title": "pH Example #2.txt", "text": "By simply dividing or multiplying both sides by negative one, this guy cancels becomes a positive. This guy becomes a negative. We get ten to the negative 7.41 is equal. Now we take our calculator, we simply plug it into our calculator and we find our result, namely 3.89 times ten to the negative eight molar. And this is equivalent to this guy and this guy. This is our hydrogen ion concentration as well as our hydronium ion concentration."}, {"title": "pH Example #2.txt", "text": "Now we take our calculator, we simply plug it into our calculator and we find our result, namely 3.89 times ten to the negative eight molar. And this is equivalent to this guy and this guy. This is our hydrogen ion concentration as well as our hydronium ion concentration. So we found part one. Let's move on to part two. So, the first step in part two is writing the equation for our dissociation of water."}, {"title": "pH Example #2.txt", "text": "So we found part one. Let's move on to part two. So, the first step in part two is writing the equation for our dissociation of water. So, water in the liquid states dissociates into two ions, namely H plus ion in the aqueous state and oh minus ion in the aqueous state. Notice that we have 1 mol becomes 1 mol in 1 mol. So now let's write the equilibrium constant equation."}, {"title": "pH Example #2.txt", "text": "So, water in the liquid states dissociates into two ions, namely H plus ion in the aqueous state and oh minus ion in the aqueous state. Notice that we have 1 mol becomes 1 mol in 1 mol. So now let's write the equilibrium constant equation. So our constant kw, which is what? We don't know where we could find that in a textbook or online. So this is something we actually know equals concentration of hydroxide times concentration of Hydride."}, {"title": "pH Example #2.txt", "text": "So our constant kw, which is what? We don't know where we could find that in a textbook or online. So this is something we actually know equals concentration of hydroxide times concentration of Hydride. So we know this. This is what we found here upstairs, right? We don't know this."}, {"title": "pH Example #2.txt", "text": "So we know this. This is what we found here upstairs, right? We don't know this. This is what we want to find. And we know this as well. This, if you look up online at 25 degrees Celsius, is simply 1.0 times ten to the negative 14."}, {"title": "pH Example #2.txt", "text": "This is what we want to find. And we know this as well. This, if you look up online at 25 degrees Celsius, is simply 1.0 times ten to the negative 14. And that always holds true. So you should probably remember that. So this guy equals our unknown times what?"}, {"title": "pH Example #2.txt", "text": "And that always holds true. So you should probably remember that. So this guy equals our unknown times what? These data from part one, 3.89\ntimes ten to negative eight. We bring this guy over here, we solve divide 1.0 times ten to the negative 14 divided by 3.89\ntimes ten to negative eight, we get 2.57 times ten to the negative seven. So this is our concentration of our hydroxide within our blood."}, {"title": "pH Example #2.txt", "text": "These data from part one, 3.89\ntimes ten to negative eight. We bring this guy over here, we solve divide 1.0 times ten to the negative 14 divided by 3.89\ntimes ten to negative eight, we get 2.57 times ten to the negative seven. So this is our concentration of our hydroxide within our blood. So we did part one. We did part two. Part three basically asks us to say if our blood is basic or acidic."}, {"title": "Ethane.txt", "text": "It's composed of two carbons and a total of six H atoms. A methane molecule, however, is the simplest alkane. It's composed of one carbon and four H atoms, identical H atoms attached to the carbon via SP three hybridized orbitals. So we're basically going to take two methane molecules, combine them, and form our ethylene. But before we actually combine them, we have to turn them into methyl radicals. A methyl radical is simply a methane molecule minus an h atom and an electron."}, {"title": "Ethane.txt", "text": "So we're basically going to take two methane molecules, combine them, and form our ethylene. But before we actually combine them, we have to turn them into methyl radicals. A methyl radical is simply a methane molecule minus an h atom and an electron. So to create a methyl radical from a methane molecule, we take away an h atom along with one electron. But the same thing goes for this second methane molecule. We take away an h atom along with one electron."}, {"title": "Ethane.txt", "text": "So to create a methyl radical from a methane molecule, we take away an h atom along with one electron. But the same thing goes for this second methane molecule. We take away an h atom along with one electron. So now we have a methylradical in this radical. These bonds between the carbon and the H are approximately SP two hybridyne. And this orbital is approximately a two P orbital."}, {"title": "Ethane.txt", "text": "So now we have a methylradical in this radical. These bonds between the carbon and the H are approximately SP two hybridyne. And this orbital is approximately a two P orbital. And our electron, the single electron, will be found within this two p orbital. The same thing goes for the second methyl radical. We have SP two hybridized bond between the carbon and H, and we have one electron found within this two p orbital."}, {"title": "Ethane.txt", "text": "And our electron, the single electron, will be found within this two p orbital. The same thing goes for the second methyl radical. We have SP two hybridized bond between the carbon and H, and we have one electron found within this two p orbital. Now, when these two methyl radicals begin to approach one another, the lobes, which are approximately two p, become approximately SP three hybridized. And that simply means that as they approach one another, these green sections become larger and larger, and these blue sections become smaller and smaller. And eventually, when they come close enough, there is an overlap between these two SP three hybridized orbitals, and they form what we know as a covalent bond."}, {"title": "Ethane.txt", "text": "Now, when these two methyl radicals begin to approach one another, the lobes, which are approximately two p, become approximately SP three hybridized. And that simply means that as they approach one another, these green sections become larger and larger, and these blue sections become smaller and smaller. And eventually, when they come close enough, there is an overlap between these two SP three hybridized orbitals, and they form what we know as a covalent bond. And we get our ethane molecule, in which every single bond is SP three hybridized. Once again, to build methane I'm sorry, to build this should be ethane. To build ethane, we begin with two methyl radicals."}, {"title": "Ethane.txt", "text": "And we get our ethane molecule, in which every single bond is SP three hybridized. Once again, to build methane I'm sorry, to build this should be ethane. To build ethane, we begin with two methyl radicals. They're orbitals containing the electrons react, forming an SP three sigma bond, as we saw here. Once again, individually, these methyl roticals contain SP two hybridized bonds with approximately two p orbital that contains our electrons. But when they come closer, these orbitals become SP three hybridized, which eventually form the bond, forming our ethane molecule."}, {"title": "Ethane.txt", "text": "They're orbitals containing the electrons react, forming an SP three sigma bond, as we saw here. Once again, individually, these methyl roticals contain SP two hybridized bonds with approximately two p orbital that contains our electrons. But when they come closer, these orbitals become SP three hybridized, which eventually form the bond, forming our ethane molecule. Now, let's look at our electron diagram or energy diagram for these two bonds. So here we have our SP three hybridized bond. They interact."}, {"title": "Ethane.txt", "text": "Now, let's look at our electron diagram or energy diagram for these two bonds. So here we have our SP three hybridized bond. They interact. They each have one electron each. And then they form two molecular orbitals. One is a sigma bonding molecular orbital, and the second one is a sigma antibonding molecular orbital."}, {"title": "Ethane.txt", "text": "They each have one electron each. And then they form two molecular orbitals. One is a sigma bonding molecular orbital, and the second one is a sigma antibonding molecular orbital. Now, this guy will be lower in energy. This guy will be larger. It will have larger amount of energy."}, {"title": "Ethane.txt", "text": "Now, this guy will be lower in energy. This guy will be larger. It will have larger amount of energy. And that means that our electrons will tend to go into the orbital that is lower in energy. So our electrons will tend to go into this lower bonding sigma orbital. And so this guy is lower in energy than both this and this atomic orbital."}, {"title": "Ethane.txt", "text": "And that means that our electrons will tend to go into the orbital that is lower in energy. So our electrons will tend to go into this lower bonding sigma orbital. And so this guy is lower in energy than both this and this atomic orbital. So, what is the meaning of this? Well, let's look at our reaction. So, our reaction takes place in the following way."}, {"title": "Ethane.txt", "text": "So, what is the meaning of this? Well, let's look at our reaction. So, our reaction takes place in the following way. Our reactants A, which are simply two methyl radicals, react. They have to surmel this activation barrier. They reach some transition states down here, and then they drop down and form our product, our essay molecule."}, {"title": "Ethane.txt", "text": "Our reactants A, which are simply two methyl radicals, react. They have to surmel this activation barrier. They reach some transition states down here, and then they drop down and form our product, our essay molecule. And because this final bond is lower in energy than either of these two atomic orbitals, we release energy and we release our energy. Change in enthalpy. So change in energy simply change in enthalpy."}, {"title": "Ethane.txt", "text": "And because this final bond is lower in energy than either of these two atomic orbitals, we release energy and we release our energy. Change in enthalpy. So change in energy simply change in enthalpy. So, as our reaction progresses, our energy drops by this amount, and it's actually 900 kilogows per mole. In other words, if we have 1 mol of methyl radical reaction with 1 mol of methyl radical, we form 1 mol of essay and we release energy. So we release 90 energy."}, {"title": "Heat .txt", "text": "Today we're only going to talk about heat. Now, heat is a transfer of energy from a hot object to a cold object. And this occurs naturally or spontaneously. There are three ways by which heat occurs. The first one is conduction. Conduction is energy transfer due to physical contact between the two objects or the two systems."}, {"title": "Heat .txt", "text": "There are three ways by which heat occurs. The first one is conduction. Conduction is energy transfer due to physical contact between the two objects or the two systems. Here I've outlined two situations. In the first situation, I have two blocks of the same size. The first block is at a higher temperature, Th."}, {"title": "Heat .txt", "text": "Here I've outlined two situations. In the first situation, I have two blocks of the same size. The first block is at a higher temperature, Th. The second block is at a lower temperature, TL. And nothing is connecting the two objects. No physical barrier, no physical connection occurs here."}, {"title": "Heat .txt", "text": "The second block is at a lower temperature, TL. And nothing is connecting the two objects. No physical barrier, no physical connection occurs here. In situation two, however, I have the same two blocks at the same two temperatures and a solid bridge connecting the two blocks. Okay? In this situation, conduction will occur because of this metal bridge, and energy will be transferred."}, {"title": "Heat .txt", "text": "In situation two, however, I have the same two blocks at the same two temperatures and a solid bridge connecting the two blocks. Okay? In this situation, conduction will occur because of this metal bridge, and energy will be transferred. Energy flow will occur in this direction. And in this situation, there is no physical bridge that connects the two situations. Yes, there are air molecules flying around in this space here, but there is no solid physical thing that's connecting the two systems, and therefore, conduction will not occur."}, {"title": "Heat .txt", "text": "Energy flow will occur in this direction. And in this situation, there is no physical bridge that connects the two situations. Yes, there are air molecules flying around in this space here, but there is no solid physical thing that's connecting the two systems, and therefore, conduction will not occur. A visible example of conduction is when a person takes a pencil or a pen or a marker in their hand, holds it tightly for 30 seconds, two minutes, let it go, and notices that the temperature rises in the marker or the pencil. And this occurs because there's an energy transfer or flow of energy from my body, from my hand, from my palm into this marker, thereby raising the temperature of this marker. That is conduction."}, {"title": "Heat .txt", "text": "A visible example of conduction is when a person takes a pencil or a pen or a marker in their hand, holds it tightly for 30 seconds, two minutes, let it go, and notices that the temperature rises in the marker or the pencil. And this occurs because there's an energy transfer or flow of energy from my body, from my hand, from my palm into this marker, thereby raising the temperature of this marker. That is conduction. And it only occurs when there is physical contact. So when people talk about conduction, they usually talk about energy flow or the rate of energy flow within the system. In this situation outlined here, I have a hot system, a cold system, the physical contact, the bridge that's connecting the two systems, okay?"}, {"title": "Heat .txt", "text": "And it only occurs when there is physical contact. So when people talk about conduction, they usually talk about energy flow or the rate of energy flow within the system. In this situation outlined here, I have a hot system, a cold system, the physical contact, the bridge that's connecting the two systems, okay? And energy flow will occur, will travel from here to here within this bridge. And the rate of that flow can be given by this equation here. Okay?"}, {"title": "Heat .txt", "text": "And energy flow will occur, will travel from here to here within this bridge. And the rate of that flow can be given by this equation here. Okay? So Q over T, where T is the time it takes for it to travel this distance. Q is the flow of heat or flow of energy equals K times A over L times the change in temperature. Okay?"}, {"title": "Heat .txt", "text": "So Q over T, where T is the time it takes for it to travel this distance. Q is the flow of heat or flow of energy equals K times A over L times the change in temperature. Okay? And K here is simply a constant called a thermal connectivity constant that depends on the object or the system used. And it differs from one object to the second object. It basically depends on the composition of the object."}, {"title": "Heat .txt", "text": "And K here is simply a constant called a thermal connectivity constant that depends on the object or the system used. And it differs from one object to the second object. It basically depends on the composition of the object. It also depends slightly on the temperature of the object being used. Okay, l here is the distance, the distance from this object to the second object. Therefore, the entire bridge distance l would be plugged in here okay?"}, {"title": "Heat .txt", "text": "It also depends slightly on the temperature of the object being used. Okay, l here is the distance, the distance from this object to the second object. Therefore, the entire bridge distance l would be plugged in here okay? Now, the area isn't the area of this. It isn't the area of this. It's the area of the bridge that's connecting it's the area of the face side."}, {"title": "Heat .txt", "text": "Now, the area isn't the area of this. It isn't the area of this. It's the area of the bridge that's connecting it's the area of the face side. The face side is this side. If you take this out, this is the area we want, okay? And this is the length that we're talking about here."}, {"title": "Heat .txt", "text": "The face side is this side. If you take this out, this is the area we want, okay? And this is the length that we're talking about here. Now, the difference in temperature is simply this temperature minus this temperature, when you plot the things in, you get the rate of heat flow, okay? And this will become important only when you're dealing with conduction and nothing else. This formula only applies for conduction."}, {"title": "Heat .txt", "text": "Now, the difference in temperature is simply this temperature minus this temperature, when you plot the things in, you get the rate of heat flow, okay? And this will become important only when you're dealing with conduction and nothing else. This formula only applies for conduction. The second way by which energy is transferred due to heat is called convection. Convection is a transfer of energy due to the movement of fluids. And fluids can both include gas and liquid, okay?"}, {"title": "Heat .txt", "text": "The second way by which energy is transferred due to heat is called convection. Convection is a transfer of energy due to the movement of fluids. And fluids can both include gas and liquid, okay? And convection absolutely requires the presence of molecules or atoms within the medium. There is no longer a need for a physical presence or a physical bridge connecting two objects or two systems, but there is a need for those molecules, okay? And here we have a common example used to describe or convey convection, okay?"}, {"title": "Heat .txt", "text": "And convection absolutely requires the presence of molecules or atoms within the medium. There is no longer a need for a physical presence or a physical bridge connecting two objects or two systems, but there is a need for those molecules, okay? And here we have a common example used to describe or convey convection, okay? Here we have a cup of coffee. You know that if you leave a cup of coffee, as in temperature, eventually it will cool down. That's because energy transferred from the cup of coffee to the air, okay?"}, {"title": "Heat .txt", "text": "Here we have a cup of coffee. You know that if you leave a cup of coffee, as in temperature, eventually it will cool down. That's because energy transferred from the cup of coffee to the air, okay? What actually happens on a microscopic level is that the atoms, for example, diatomic oxygen, hit the surface of the water, okay? When they hit the surface of the water, they bounce back with a higher kinetic energy, okay? So the oxygen gain kinetic energy while this lost kinetic energy, so it lost the heat."}, {"title": "Heat .txt", "text": "What actually happens on a microscopic level is that the atoms, for example, diatomic oxygen, hit the surface of the water, okay? When they hit the surface of the water, they bounce back with a higher kinetic energy, okay? So the oxygen gain kinetic energy while this lost kinetic energy, so it lost the heat. It lost energy because of conservation of energy, okay? Eventually, enough of these atoms will attack or hit the surface of the water that it will lose most of its energy, and this will cool it down, okay? And that's why convection can only occur in the presence of molecules."}, {"title": "Heat .txt", "text": "It lost energy because of conservation of energy, okay? Eventually, enough of these atoms will attack or hit the surface of the water that it will lose most of its energy, and this will cool it down, okay? And that's why convection can only occur in the presence of molecules. And conduction requires physical contact. The third and final type of energy transfer that occurs due to heat is called radiation. Radiation is the energy transfer that occurs due to the presence of electromagnetic waves, okay?"}, {"title": "Heat .txt", "text": "And conduction requires physical contact. The third and final type of energy transfer that occurs due to heat is called radiation. Radiation is the energy transfer that occurs due to the presence of electromagnetic waves, okay? These electromagnetic waves can include UV waves, radio waves, microwaves and infrared waves, and other waves as well. So all objects on Earth and in space radiate heat or radiate energy. And that's because every single object has some temperature and zero Kelvin, the absolute zero temperature, is unattainable."}, {"title": "Heat .txt", "text": "These electromagnetic waves can include UV waves, radio waves, microwaves and infrared waves, and other waves as well. So all objects on Earth and in space radiate heat or radiate energy. And that's because every single object has some temperature and zero Kelvin, the absolute zero temperature, is unattainable. So all objects vibrate or move to a certain extent, and therefore all objects at all times radiate energy. Okay? One cool thing about radiation, unlike convection or conduction, is that radiation can occur in the absence of molecules or in the absence of a physical barrier."}, {"title": "Heat .txt", "text": "So all objects vibrate or move to a certain extent, and therefore all objects at all times radiate energy. Okay? One cool thing about radiation, unlike convection or conduction, is that radiation can occur in the absence of molecules or in the absence of a physical barrier. And this is seen in space when light travels from the sun, the hot object to the Earth, the cooler object, the waves travel in a vacuum in the absence of molecules. And that's because waves, electromagnetic waves, are actually energy bundles that carry themselves from a hot object to a cool object, okay? And it is radiation that heats the Earth."}, {"title": "Heat .txt", "text": "And this is seen in space when light travels from the sun, the hot object to the Earth, the cooler object, the waves travel in a vacuum in the absence of molecules. And that's because waves, electromagnetic waves, are actually energy bundles that carry themselves from a hot object to a cool object, okay? And it is radiation that heats the Earth. When we spoke earlier about conduction, we also mentioned rate of energy transfer. Here we can also talk about a similar concept, rate of energy transfer. But the formula here, it's different."}, {"title": "Heat .txt", "text": "When we spoke earlier about conduction, we also mentioned rate of energy transfer. Here we can also talk about a similar concept, rate of energy transfer. But the formula here, it's different. The formula here is power, which is actually change in energy over time, which is a rate flow is equal to sigma, which is a constant and will be given to you admissitivity value A and T. T is the temperature of the object. A is the area of the face that's radiating the heat, and this is a value between zero and one. Now, where this value is one, that that simply means that the object is absorbing all the heat, is absorbing all the radiation, which means that it also releases all the radiation, okay?"}, {"title": "Heat .txt", "text": "The formula here is power, which is actually change in energy over time, which is a rate flow is equal to sigma, which is a constant and will be given to you admissitivity value A and T. T is the temperature of the object. A is the area of the face that's radiating the heat, and this is a value between zero and one. Now, where this value is one, that that simply means that the object is absorbing all the heat, is absorbing all the radiation, which means that it also releases all the radiation, okay? Now, a black body object is an object that has this value that equals one. Now, technically, such an object does not exist, and that's because if such an object did exist, it would not reflect any of the waves and would not reflect light waves, and so we would not be able to see it. It would be invisible to us."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "Well, electronegativity is usually defined as the ability of atoms to attract electrons. In other words, how likely is it that the protons found in the nucleus will attract our electrons in the atom? So we know as electronegativity increases. According to our definition, the atoms ability to attract electrons also increases. And generally speaking, according to our periodic trends, as we go up a group or across a period from left to right on our periodic table, the electrodegativity of the atoms increases. In other words, their ability to attract electrons increases."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "According to our definition, the atoms ability to attract electrons also increases. And generally speaking, according to our periodic trends, as we go up a group or across a period from left to right on our periodic table, the electrodegativity of the atoms increases. In other words, their ability to attract electrons increases. So, for example, let's take the second period. As we go from Lithium to Beryllium all the way to fluorine. The electronegativity of these atoms increases."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "So, for example, let's take the second period. As we go from Lithium to Beryllium all the way to fluorine. The electronegativity of these atoms increases. And in fact, fluorine and oxygen are the two most electronegative atoms found on our periodic table. So we just defined what electronegativity is. But where does it come from?"}, {"title": "Electronegativity and Polar Bonds.txt", "text": "And in fact, fluorine and oxygen are the two most electronegative atoms found on our periodic table. So we just defined what electronegativity is. But where does it come from? What determines electronegativity well, it turns out another important principle known as effective nuclear charge determines or creates electronegativity. Now effective nuclear charge is simply the charge that the balanced electrons found in the outermost shell feel due to the protons found in the nucleus. Remember, electrons and protons attract one another."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "What determines electronegativity well, it turns out another important principle known as effective nuclear charge determines or creates electronegativity. Now effective nuclear charge is simply the charge that the balanced electrons found in the outermost shell feel due to the protons found in the nucleus. Remember, electrons and protons attract one another. Now, normally, the effective nuclear charge is smaller than the actual nuclear charge found in the nucleus. And that's because some of the electrons found in the innermost shell shield the charge. They decrease the charge, creating a shielding effect."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "Now, normally, the effective nuclear charge is smaller than the actual nuclear charge found in the nucleus. And that's because some of the electrons found in the innermost shell shield the charge. They decrease the charge, creating a shielding effect. So, in other words, the more balanced electrons we have, the more our nuclear charge is. Effective nuclear charge. The more inner most electrons we have in the innermost shell, the smaller our effective nuclear charge."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "So, in other words, the more balanced electrons we have, the more our nuclear charge is. Effective nuclear charge. The more inner most electrons we have in the innermost shell, the smaller our effective nuclear charge. So once again, electronegativity arises from effective nuclear charge. The more dense electrons an atom has, the larger the effective nuclear charge is, the more electronegative the atom. And that's exactly why electronegativity increases as we go from lithium to fluorine."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "So once again, electronegativity arises from effective nuclear charge. The more dense electrons an atom has, the larger the effective nuclear charge is, the more electronegative the atom. And that's exactly why electronegativity increases as we go from lithium to fluorine. Because the number of bounced electrons found in each atom increases as we go from left to right on our periodic table. Also the ratio of bounce electrons to the innermost electrons also increases. So let's take lithium, for example, and let's compare it to fluorine."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "Because the number of bounced electrons found in each atom increases as we go from left to right on our periodic table. Also the ratio of bounce electrons to the innermost electrons also increases. So let's take lithium, for example, and let's compare it to fluorine. So Lithium has one bounced electron in the two S shell, while this fluorine has seven bounced electrons. So because this has more balanced electrons and because the ratio is seven to two versus one to two, because the ratio here is much larger, this has much more effective nuclear charge and therefore it's more electronegative. So electronegativity in polar bonds go hand in hand."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "So Lithium has one bounced electron in the two S shell, while this fluorine has seven bounced electrons. So because this has more balanced electrons and because the ratio is seven to two versus one to two, because the ratio here is much larger, this has much more effective nuclear charge and therefore it's more electronegative. So electronegativity in polar bonds go hand in hand. In fact, polar bonds or polar codalent bonds are formed because of differences in electronegativity. Once again, polar bonds are created due to electronegativity differences between the atoms composing the bond. So, as an example, let's look at HF."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "In fact, polar bonds or polar codalent bonds are formed because of differences in electronegativity. Once again, polar bonds are created due to electronegativity differences between the atoms composing the bond. So, as an example, let's look at HF. So h donates one electron and the fluorine also donates one electron. But because this fluorine atom is more electronegative, it attracts electrons more strongly. That means that electron density will be closer."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "So h donates one electron and the fluorine also donates one electron. But because this fluorine atom is more electronegative, it attracts electrons more strongly. That means that electron density will be closer. The electrons will be closer to our fluorine. And for the fluorine will develop a partial negative charge. While this h will develop a partial positive charge."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "The electrons will be closer to our fluorine. And for the fluorine will develop a partial negative charge. While this h will develop a partial positive charge. The electrons will not be shared equally. They're going to be closer to the fluorine. So if we examine the molecular orbital of this bond, we see that the one s orbital combines overlaps with the SP three orbital of the fluorine forming this hybridized or this molecular orbital."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "The electrons will not be shared equally. They're going to be closer to the fluorine. So if we examine the molecular orbital of this bond, we see that the one s orbital combines overlaps with the SP three orbital of the fluorine forming this hybridized or this molecular orbital. And the electrons found in this molecular orbital will be closer to the fluorine nucleus than the h nucleus. And once again, that's why we have a partial negative on the fluorine and a partial positive on the h.\nSo recall that Bronsted larry acids are compounds that donate an h ion. So, for example, this hydrofluoric acid is a Bronxed larry acid because it has an H that it can donate."}, {"title": "Electronegativity and Polar Bonds.txt", "text": "And the electrons found in this molecular orbital will be closer to the fluorine nucleus than the h nucleus. And once again, that's why we have a partial negative on the fluorine and a partial positive on the h.\nSo recall that Bronsted larry acids are compounds that donate an h ion. So, for example, this hydrofluoric acid is a Bronxed larry acid because it has an H that it can donate. So what determines if a compound is a good Bronsted larry acid? So, good Bronsted larry acids have very polar bonds, which means that the h atoms are held very weakly and therefore are donated very readily. So this polar bond is very polar, and that's because we have a very electronegative atom and not so electronegative atom."}, {"title": "Chalcogen\u2019s and Noble Gases .txt", "text": "In this lecture, we're going to talk about two more groups or families found on our Fiana table known as the calculus, or group 16, six A, and the noble gas group, or group 18 or eight A. Now, let's begin with our calculations. Now, there are five main calculations. Oxygen, sulfur and selenium are all nonmetals, while our metalloids are polonium and talorium. Now notice, unlike group five A and four A, which both have metals, group six A, or the calcigens, don't have any metals. Now, let's begin with oxygen."}, {"title": "Chalcogen\u2019s and Noble Gases .txt", "text": "Oxygen, sulfur and selenium are all nonmetals, while our metalloids are polonium and talorium. Now notice, unlike group five A and four A, which both have metals, group six A, or the calcigens, don't have any metals. Now, let's begin with oxygen. Now, in this lecture, we're only really going to look at oxygen and sulfur because they occur way more frequently than any of these three guys. So let's look at oxygen. Oxygen is the second most electronegative atom."}, {"title": "Chalcogen\u2019s and Noble Gases .txt", "text": "Now, in this lecture, we're only really going to look at oxygen and sulfur because they occur way more frequently than any of these three guys. So let's look at oxygen. Oxygen is the second most electronegative atom. In other words, it likes to take electrons. Now, we'll talk more about electronegativity in a future lecture. For now, it's sufficient to say that oxygen is the second most electronegative."}, {"title": "Chalcogen\u2019s and Noble Gases .txt", "text": "In other words, it likes to take electrons. Now, we'll talk more about electronegativity in a future lecture. For now, it's sufficient to say that oxygen is the second most electronegative. It's second to fluorine. And we'll talk more about fluorine when we'll talk about the seven A group elements or the halogens. Now, oxygen, like carbon, also forms strong bonds, strong double bonds."}, {"title": "Chalcogen\u2019s and Noble Gases .txt", "text": "It's second to fluorine. And we'll talk more about fluorine when we'll talk about the seven A group elements or the halogens. Now, oxygen, like carbon, also forms strong bonds, strong double bonds. And oxygen exists in nature in a diatomic or a triathlonic form, namely ozone or O two oxygen. Now, O two, or oxygen, reacts with metals, alkaline metals and earth metals and transition metals to form metal oxides. Now, whenever our alkaline metals react with our oxygen, they form peroxide, for example, li two or ma two, et cetera."}, {"title": "Chalcogen\u2019s and Noble Gases .txt", "text": "And oxygen exists in nature in a diatomic or a triathlonic form, namely ozone or O two oxygen. Now, O two, or oxygen, reacts with metals, alkaline metals and earth metals and transition metals to form metal oxides. Now, whenever our alkaline metals react with our oxygen, they form peroxide, for example, li two or ma two, et cetera. Now let's look at sulphur. Now, sulphur can form two, three, four or even six covalent bonds with other atoms or molecules, and also like oxygen, can form strong pi bonds, namely the double bond. So now let's look at our noble gases."}, {"title": "Chalcogen\u2019s and Noble Gases .txt", "text": "Now let's look at sulphur. Now, sulphur can form two, three, four or even six covalent bonds with other atoms or molecules, and also like oxygen, can form strong pi bonds, namely the double bond. So now let's look at our noble gases. Noble gases are group 18 or eight eight, and they're also known as the inner gases. Now, these guys are extremely stable and therefore, they're very unreactive. And they exist unlike every other atom or most other atoms."}, {"title": "Bomb Calorimeter Example.txt", "text": "The heat capacity of our bomb is 840 joules per Celsius. And our specific heat capacity for water is 4.18 joules per Celsius times ground. We want to find the final changes, the energy of our surroundings when 1 mol of glucose is combusted. So let's visualize the example. So here's our bomb kilometer, and we're basically taking the red oxygen molecules and the purple glucose molecules. We're mixing them, combusting them in our bomb or steel bomb."}, {"title": "Bomb Calorimeter Example.txt", "text": "So let's visualize the example. So here's our bomb kilometer, and we're basically taking the red oxygen molecules and the purple glucose molecules. We're mixing them, combusting them in our bomb or steel bomb. And we want to find the amount of energy released that goes into heating the bomb, and then the amount of energy released that goes into heating the water. So you have to find two things energy that goes into the bomb and the energy that goes into the water. Okay, first, let's look at the equation at hand."}, {"title": "Bomb Calorimeter Example.txt", "text": "And we want to find the amount of energy released that goes into heating the bomb, and then the amount of energy released that goes into heating the water. So you have to find two things energy that goes into the bomb and the energy that goes into the water. Okay, first, let's look at the equation at hand. So we have 1 mol of glucose and six mole of oxygen combusting into six moles of carbon dioxide and six moles of water. Okay? So our second step in our second step, we want to find the change in energy that goes from the system, from these two guys, from our glucose and oxygen into our bomb."}, {"title": "Bomb Calorimeter Example.txt", "text": "So we have 1 mol of glucose and six mole of oxygen combusting into six moles of carbon dioxide and six moles of water. Okay? So our second step in our second step, we want to find the change in energy that goes from the system, from these two guys, from our glucose and oxygen into our bomb. Okay? Now remember, our bomb is given a heat capacity of 840 joules per Celsius. So we simply use the formula for heat capacity or Q equals C times change in T. And we present this Q as Q one."}, {"title": "Bomb Calorimeter Example.txt", "text": "Okay? Now remember, our bomb is given a heat capacity of 840 joules per Celsius. So we simply use the formula for heat capacity or Q equals C times change in T. And we present this Q as Q one. And we find that 840 Joules per Celsius times the change in our temperature, or eleven degrees Celsius, gives us 9240 joules. So this million Joules goes from the system into the bomb, into heating the bomb. Now let's calculate that there are the change in energy that goes from the system to our water."}, {"title": "Bomb Calorimeter Example.txt", "text": "And we find that 840 Joules per Celsius times the change in our temperature, or eleven degrees Celsius, gives us 9240 joules. So this million Joules goes from the system into the bomb, into heating the bomb. Now let's calculate that there are the change in energy that goes from the system to our water. Okay? And we're given a specific heat of water, 4.8\njoules per Celsius times gram. So we have to use this formula because we're using little C. So we let this change down GB, Q Two."}, {"title": "Bomb Calorimeter Example.txt", "text": "Okay? And we're given a specific heat of water, 4.8\njoules per Celsius times gram. So we have to use this formula because we're using little C. So we let this change down GB, Q Two. We basically plug in our Master of Water, 800 grams times 4.8 joules per gram times Celsius times our difference in temperature. So eleven degrees Celsius, and we get 36,784 joules. So this main joules goes into heating the water, increasing temperature of water by eleven degrees."}, {"title": "Bomb Calorimeter Example.txt", "text": "We basically plug in our Master of Water, 800 grams times 4.8 joules per gram times Celsius times our difference in temperature. So eleven degrees Celsius, and we get 36,784 joules. So this main joules goes into heating the water, increasing temperature of water by eleven degrees. Okay, the final step is basically to add up the two energies. Because energy in both cases is released, you must add them up. So 9240 plus 36,784 joules gives you 46,024 joules."}, {"title": "Bomb Calorimeter Example.txt", "text": "Okay, the final step is basically to add up the two energies. Because energy in both cases is released, you must add them up. So 9240 plus 36,784 joules gives you 46,024 joules. But there's one last thing left to do. Notice that this amount of Joules corresponds to 3.5\ngrams of glucose, and this amount is much less than 1 mol of glucose. Okay?"}, {"title": "Bomb Calorimeter Example.txt", "text": "But there's one last thing left to do. Notice that this amount of Joules corresponds to 3.5\ngrams of glucose, and this amount is much less than 1 mol of glucose. Okay? So let's see how many moles 3.5 grams of glucose actually corresponds to. So we take our grams, 3.5 grams, divide that by our molecular formula for glucose, which is 180 grams/mol, and we get zero point 19 moles. So this number corresponds to only zero point 19 moles."}, {"title": "Bomb Calorimeter Example.txt", "text": "So let's see how many moles 3.5 grams of glucose actually corresponds to. So we take our grams, 3.5 grams, divide that by our molecular formula for glucose, which is 180 grams/mol, and we get zero point 19 moles. So this number corresponds to only zero point 19 moles. We want to find how many joules corresponds to 1 mol of glucose. Okay? So we basically take our number and divide that by our moles, and that gives us 2,422,316 joules per mole."}, {"title": "Bomb Calorimeter Example.txt", "text": "We want to find how many joules corresponds to 1 mol of glucose. Okay? So we basically take our number and divide that by our moles, and that gives us 2,422,316 joules per mole. So this number is the amount per mole of glucose. So that's how much energy is released. And because it's released, this means it's negative."}, {"title": "Empirical and Molecular Formula .txt", "text": "For example, let's look at a water molecule. The formula for water is H 20. And the ratio of atom our H molecules to our water molecules is two to one. In other words, for every O molecule, there are two H molecules. Now, this above ratio is known as the empirical formula. Now, on the contrary, there's also something called the molecular formula."}, {"title": "Empirical and Molecular Formula .txt", "text": "In other words, for every O molecule, there are two H molecules. Now, this above ratio is known as the empirical formula. Now, on the contrary, there's also something called the molecular formula. And the molecular formula gives you the exact number of atoms found in each compound. Now, sometimes your empirical formula will be the same as your molecular formula. As is in the case of water."}, {"title": "Empirical and Molecular Formula .txt", "text": "And the molecular formula gives you the exact number of atoms found in each compound. Now, sometimes your empirical formula will be the same as your molecular formula. As is in the case of water. This is both an empirical formula and the molecular formula. Other times, these two guys will differ. For example, suppose we look at the hydrocarbon ch 16."}, {"title": "Empirical and Molecular Formula .txt", "text": "This is both an empirical formula and the molecular formula. Other times, these two guys will differ. For example, suppose we look at the hydrocarbon ch 16. The molecular formula gives you the exact amount of atoms. In other words, this is our molecular formula. In this hydrocarbon, there are eight C molecules, carbon molecules and 16 H molecules."}, {"title": "Empirical and Molecular Formula .txt", "text": "The molecular formula gives you the exact amount of atoms. In other words, this is our molecular formula. In this hydrocarbon, there are eight C molecules, carbon molecules and 16 H molecules. However, the empirical formula for this molecule is C one, H two. In other words, for every two H molecules, there is one C molecule. And if we divide 16 by eight, we also get two the same way we divide two by one."}, {"title": "Empirical and Molecular Formula .txt", "text": "However, the empirical formula for this molecule is C one, H two. In other words, for every two H molecules, there is one C molecule. And if we divide 16 by eight, we also get two the same way we divide two by one. So the way you basically go from a molecular to an empirical formula is you find a common number and you divide it by that common number, making sure that we get whole numbers. So in this case, eight goes into 16 twice and eight goes into eight once. Now, why is an empirical formula useful?"}, {"title": "Empirical and Molecular Formula .txt", "text": "So the way you basically go from a molecular to an empirical formula is you find a common number and you divide it by that common number, making sure that we get whole numbers. So in this case, eight goes into 16 twice and eight goes into eight once. Now, why is an empirical formula useful? Well, an empirical formula can be used to find the percent by mass of any atom in our compound. Now, let's look at this example again. How can we find the percent by mass of hydrogen in our molecule in our compound?"}, {"title": "Empirical and Molecular Formula .txt", "text": "Well, an empirical formula can be used to find the percent by mass of any atom in our compound. Now, let's look at this example again. How can we find the percent by mass of hydrogen in our molecule in our compound? Well, what we do is we take our atomic weight of our H, which is 1 gram/mol and divide that by our molecular weight of this guy or the empirical weight of this guy. And what we get is 14 grams/mol. These guys, these units cancel, and we get one over 14, which is 0.7\n114."}, {"title": "Empirical and Molecular Formula .txt", "text": "Well, what we do is we take our atomic weight of our H, which is 1 gram/mol and divide that by our molecular weight of this guy or the empirical weight of this guy. And what we get is 14 grams/mol. These guys, these units cancel, and we get one over 14, which is 0.7\n114. Now, to get the percent, we multiply by 100 and we get 7.14% by mass of H in ch two. So in this context, our H takes up 7.14% by mass. So now, suppose I want to find my empirical formula given percent by mass."}, {"title": "Empirical and Molecular Formula .txt", "text": "Now, to get the percent, we multiply by 100 and we get 7.14% by mass of H in ch two. So in this context, our H takes up 7.14% by mass. So now, suppose I want to find my empirical formula given percent by mass. So suppose I'm given that my compound is eleven point 11% hydrogen and 88.89% oxygen. I have to follow three steps to find the empirical formula. In my first step, I make the assumption that I have 100 grams of my compound."}, {"title": "Empirical and Molecular Formula .txt", "text": "So suppose I'm given that my compound is eleven point 11% hydrogen and 88.89% oxygen. I have to follow three steps to find the empirical formula. In my first step, I make the assumption that I have 100 grams of my compound. And what I do is I convert this guy to fraction and this guy to fraction by dividing each guy by 100. And then I multiply each guy by my 100 grams of compound to find how many grams of each is in my compound. So, 100 grams of my compound multiplied by 0.111 which I got this guy divided by 100 equals eleven point 11 grams of h in my 100 grams of compound."}, {"title": "Empirical and Molecular Formula .txt", "text": "And what I do is I convert this guy to fraction and this guy to fraction by dividing each guy by 100. And then I multiply each guy by my 100 grams of compound to find how many grams of each is in my compound. So, 100 grams of my compound multiplied by 0.111 which I got this guy divided by 100 equals eleven point 11 grams of h in my 100 grams of compound. Now, likewise, I multiply 100 grams times 0.8\n889. This gives me 88.89 grams of oxygen in my 100 grams of compound. Now I have to convert my grams to moles."}, {"title": "Empirical and Molecular Formula .txt", "text": "Now, likewise, I multiply 100 grams times 0.8\n889. This gives me 88.89 grams of oxygen in my 100 grams of compound. Now I have to convert my grams to moles. In other words, I want to find how many moles of each guy of each atom is in my compound. Now, eleven point 11 grams of h divided by the molecular or atomic weight of h 1 gram/mol gives me eleven point eleven moles of h.\nLikewise, I do the same thing for oxygen. 88.89 grams of oxygen divided by the atomic weight for oxygen 16 grams/mol grams cancel and I get 5.56\nmole."}, {"title": "Empirical and Molecular Formula .txt", "text": "In other words, I want to find how many moles of each guy of each atom is in my compound. Now, eleven point 11 grams of h divided by the molecular or atomic weight of h 1 gram/mol gives me eleven point eleven moles of h.\nLikewise, I do the same thing for oxygen. 88.89 grams of oxygen divided by the atomic weight for oxygen 16 grams/mol grams cancel and I get 5.56\nmole. Now, I simply take this guy, the moles of my h, divide that by the moles of my oxygen and I get two, or approximately two. And this and this means that for every two moles of h, I have 1 mol of oxygen. And therefore, I could write my empirical formula in the following way."}, {"title": "Bomb Calorimeter .txt", "text": "And when we talk about gases, we need to remember that gases compress and expand. So what Bond kilometers do is they intentionally keep the volume constant by letting the reaction occur in a steel container, in a steel cylinder. So no expansion in the steel cylinder occurs. So let's compare box perimeters to coffee cup calorimeters. In a coffee cup, caliber, expansion or compression was not a problem. And that's because our reactants were liquids and solids."}, {"title": "Bomb Calorimeter .txt", "text": "So let's compare box perimeters to coffee cup calorimeters. In a coffee cup, caliber, expansion or compression was not a problem. And that's because our reactants were liquids and solids. Now, liquids and solids don't compress. And so when we speak about coffee cup kilometers, we talk about constant pressure. In a Bond kilometer, we talk about constant volume and pressure is allowed to change."}, {"title": "Bomb Calorimeter .txt", "text": "Now, liquids and solids don't compress. And so when we speak about coffee cup kilometers, we talk about constant pressure. In a Bond kilometer, we talk about constant volume and pressure is allowed to change. So let's look at the structure of a bomb kilometer. Now, bomb calorimeters are composed of two cylinders, a large one and a small one. The outer one is called the insulated chamber and it insulates the entire system."}, {"title": "Bomb Calorimeter .txt", "text": "So let's look at the structure of a bomb kilometer. Now, bomb calorimeters are composed of two cylinders, a large one and a small one. The outer one is called the insulated chamber and it insulates the entire system. So energy is not allowed to leave our bomb kilometer. The intermodal chamber is called a steel bomb and it's basically the location of our reaction. This is where our reaction occurs and it's made from steel so that the gas can't expand our system."}, {"title": "Bomb Calorimeter .txt", "text": "So energy is not allowed to leave our bomb kilometer. The intermodal chamber is called a steel bomb and it's basically the location of our reaction. This is where our reaction occurs and it's made from steel so that the gas can't expand our system. Now this thermometer is placed into the water and everything is sealed off. And our goal is to basically change or measure the change in temperature of our reaction, okay? The same way we do in a coffee cup calorimeter."}, {"title": "Bomb Calorimeter .txt", "text": "Now this thermometer is placed into the water and everything is sealed off. And our goal is to basically change or measure the change in temperature of our reaction, okay? The same way we do in a coffee cup calorimeter. So let's look at this section here. A sample of a known mass is added to a dish. So let's look at the combustion of glucose in the presence of oxygen."}, {"title": "Bomb Calorimeter .txt", "text": "So let's look at this section here. A sample of a known mass is added to a dish. So let's look at the combustion of glucose in the presence of oxygen. Plus some heat gives us carbon dioxide and water. So let's place a small sample of glucose into our dish found inside the steel bomb and let's fill our bomb with oxygen, right, because oxygen is required for burning to occur. Next, let's place the steel bomb into a waterfilled the cylinder and let's seal off the top."}, {"title": "Bomb Calorimeter .txt", "text": "Plus some heat gives us carbon dioxide and water. So let's place a small sample of glucose into our dish found inside the steel bomb and let's fill our bomb with oxygen, right, because oxygen is required for burning to occur. Next, let's place the steel bomb into a waterfilled the cylinder and let's seal off the top. Let's place a thermometer inside this section. Okay? Now let's ignite the sample using some electrical spark and this will allow this reaction to occur."}, {"title": "Bomb Calorimeter .txt", "text": "Let's place a thermometer inside this section. Okay? Now let's ignite the sample using some electrical spark and this will allow this reaction to occur. Now, when this reaction begins occurring, carbon dioxide is produced, so more gas produced. So this system, the pressure increases. And increase in pressure means there is an increase in temperature when the volume is held constant."}, {"title": "Bomb Calorimeter .txt", "text": "Now, when this reaction begins occurring, carbon dioxide is produced, so more gas produced. So this system, the pressure increases. And increase in pressure means there is an increase in temperature when the volume is held constant. And the increase in temperature will transfer energy from this section to the surroundings, to the water. And you could measure the change in temperature using this thermometer. So the initial and final temperature, just like you would in a coffee cup kilometer."}, {"title": "Bomb Calorimeter .txt", "text": "And the increase in temperature will transfer energy from this section to the surroundings, to the water. And you could measure the change in temperature using this thermometer. So the initial and final temperature, just like you would in a coffee cup kilometer. So finally, you can use the change in temperature and the formula to calculate our energy change. Now let's look at our first law of thermodynamics which states that change in energy is equal to heat plus work. Now, in this case, no mechanical work is done and no PV work is done because change in volume is zero."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "Both guys are at one ATM. That means constant pressure. So we have to use a coffee cup calorimeter. Now, we are also given a T initial of 23 degrees Celsius and a T final of 31 one degree Celsius. Our specific heat capacity for both guys is 4.18\njoules per gram times Celsius. We want to find our goal is to find the change in enthalpy or change in energy of our solution."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "Now, we are also given a T initial of 23 degrees Celsius and a T final of 31 one degree Celsius. Our specific heat capacity for both guys is 4.18\njoules per gram times Celsius. We want to find our goal is to find the change in enthalpy or change in energy of our solution. So we take our coffee cup calorimeter, we mix the two different types of compounds. We wait a little. We wait for the observed change in temperature to occur, and we have to calculate the change in energy that corresponds to that change in temperature."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "So we take our coffee cup calorimeter, we mix the two different types of compounds. We wait a little. We wait for the observed change in temperature to occur, and we have to calculate the change in energy that corresponds to that change in temperature. So our first step is to see exactly what's going on within our coffee cup calorimeter. So we're basically mixing 1 mol of sodium hydroxide and 1 mol of hydrochloric acid. We get dissociated ions, 1 mol of each."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "So our first step is to see exactly what's going on within our coffee cup calorimeter. So we're basically mixing 1 mol of sodium hydroxide and 1 mol of hydrochloric acid. We get dissociated ions, 1 mol of each. And then these guys, we combine to form 1 mol of salt and 1 mol of water. So our second step and third step involves finding the total grams of our mixture. So you want to find the grams of sodium hydroxide and the grams of hydrochloric acid."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "And then these guys, we combine to form 1 mol of salt and 1 mol of water. So our second step and third step involves finding the total grams of our mixture. So you want to find the grams of sodium hydroxide and the grams of hydrochloric acid. This will become important in using this equation. So, this M corresponds to the total mass. That's why we need to find the total mass."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "This will become important in using this equation. So, this M corresponds to the total mass. That's why we need to find the total mass. So first, before we find the grams of sodium hydroxide, we have to find the moles of sodium hydroxide. To find the moles, we have to take our one molar solution of sodium hydroxide. We multiply that by our volume in liters."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "So first, before we find the grams of sodium hydroxide, we have to find the moles of sodium hydroxide. To find the moles, we have to take our one molar solution of sodium hydroxide. We multiply that by our volume in liters. So, multiplied by zero point 250 liters, and we get zero point 25 moles of sodium hydroxide. Because the liters cancel, the second step is to find the grams of sodium hydroxide. To find the grams, we must first find the molecular weight of sodium hydroxide."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "So, multiplied by zero point 250 liters, and we get zero point 25 moles of sodium hydroxide. Because the liters cancel, the second step is to find the grams of sodium hydroxide. To find the grams, we must first find the molecular weight of sodium hydroxide. To find a molecular weight, we simply add up the atomic weight of each atom. So, 23 moles per gram, or grams per mole for sodium, plus 16 grams/mol for oxygen, plus 1 gram/mol for h, and we get 40 grams/mol times the number of moles. Moles cancel, and we get 10 grams of sodium hydroxide mixed into our solution."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "To find a molecular weight, we simply add up the atomic weight of each atom. So, 23 moles per gram, or grams per mole for sodium, plus 16 grams/mol for oxygen, plus 1 gram/mol for h, and we get 40 grams/mol times the number of moles. Moles cancel, and we get 10 grams of sodium hydroxide mixed into our solution. Our third step is to follow the same exact steps that we just followed just for hydrophoric acid. So find the moles. 1 mol per liter times zero point 25 liters gets you zero point 25 moles of hydrofluoric acid."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "Our third step is to follow the same exact steps that we just followed just for hydrophoric acid. So find the moles. 1 mol per liter times zero point 25 liters gets you zero point 25 moles of hydrofluoric acid. Now, the same step, we have to find a molecular weight of hydrochloric acid. Multiplies that by our moles, and we get grams. So 1 gram/mol for h plus 35.5 grams per mole for chlorine, and we get 36.5\ngrams per mole times zero point 25 moles."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "Now, the same step, we have to find a molecular weight of hydrochloric acid. Multiplies that by our moles, and we get grams. So 1 gram/mol for h plus 35.5 grams per mole for chlorine, and we get 36.5\ngrams per mole times zero point 25 moles. Moles cancel, and we get 9.13 grams of hydrochloric acid. Finally, we use our formula to find the energy change, and we get the total mass ten plus 9.13. So 19.13 grams times 4.18\njoules per celsius times gram, which we get from here."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "Moles cancel, and we get 9.13 grams of hydrochloric acid. Finally, we use our formula to find the energy change, and we get the total mass ten plus 9.13. So 19.13 grams times 4.18\njoules per celsius times gram, which we get from here. And our t final minus t initial so 31 -23 and we get 640 joules. But this corresponds to 00:25 moles of HCL and 00:25 moles of soyu hydroxide. We want to find what corresponds to 1 mol of each."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "And our t final minus t initial so 31 -23 and we get 640 joules. But this corresponds to 00:25 moles of HCL and 00:25 moles of soyu hydroxide. We want to find what corresponds to 1 mol of each. So we basically have to multiply this guy by four, or we could divide this guy by 00:25. We get the same answer. So 640 joules divided by zero point 250 mole gets you 2560 joules per mole."}, {"title": "Coffee Cup Calorimeter Example .txt", "text": "So we basically have to multiply this guy by four, or we could divide this guy by 00:25. We get the same answer. So 640 joules divided by zero point 250 mole gets you 2560 joules per mole. And that's our final answer. So 2560 Joules are released when this guy is mixed. So that's why our water inside is heated to this temperature."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "Composed of quantum numbers. So quantum numbers are simply the identification numbers for electrons in atoms. And each unique electron in any given atom has exactly four quantum numbers which make it unique. So these four quantum numbers is like the ID badge of that unique electron. So let's look at these four quantum numbers. The first quantum number is called the principal quantum number, N. So it's represented by the lowercase letter N. And this number begins with one and could be 2345."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "So these four quantum numbers is like the ID badge of that unique electron. So let's look at these four quantum numbers. The first quantum number is called the principal quantum number, N. So it's represented by the lowercase letter N. And this number begins with one and could be 2345. So N has to be a positive whole number. Now, this quantum number, the principal quantum number, designates the shell level or the energy level in which the electron is located. The Larger the n, the Larger Our size or The Larger The Size Of The Atom and The Greater The energy."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "So N has to be a positive whole number. Now, this quantum number, the principal quantum number, designates the shell level or the energy level in which the electron is located. The Larger the n, the Larger Our size or The Larger The Size Of The Atom and The Greater The energy. In other words, an electron found in n equals one is lower in energy than if an electron is found in N equals two. So the higher the n, the greater the size of the atom and the greater the energy level in which our electron is in. The second quantum number is known as the osmic quantum number and is represented by the lowercase cursive l.\nNow, this quantum number designates the subshell in which our electron is in."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "In other words, an electron found in n equals one is lower in energy than if an electron is found in N equals two. So the higher the n, the greater the size of the atom and the greater the energy level in which our electron is in. The second quantum number is known as the osmic quantum number and is represented by the lowercase cursive l.\nNow, this quantum number designates the subshell in which our electron is in. And the number of subshells. This letter N depends on the principal quantum number, and it's given by the following equation l equals n equals one. In other words, if we know that our principal quantum number in which our electron is located is n equals four, then our subshell is n minus one So four minus one L equals three."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "And the number of subshells. This letter N depends on the principal quantum number, and it's given by the following equation l equals n equals one. In other words, if we know that our principal quantum number in which our electron is located is n equals four, then our subshell is n minus one So four minus one L equals three. And so let's look at some examples of our subshells. S is a subshell, P is a subshell, D is a subshell, s is a subshell, and so on. Now, these S and P and D are the most familiar ones."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "And so let's look at some examples of our subshells. S is a subshell, P is a subshell, D is a subshell, s is a subshell, and so on. Now, these S and P and D are the most familiar ones. These guys, you should definitely know. Now, for example, s corresponds to l equals zero. When our quantum number is N equals one, we get one minus 10."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "These guys, you should definitely know. Now, for example, s corresponds to l equals zero. When our quantum number is N equals one, we get one minus 10. So if our electron is found with a principal quantum number of one, that means it must be in subsshell l equals zero, because l equals one minus one gives you zero. Likewise, p represents a subshell of l equals 1d represents a subsshell of l equals two, f represents a subshell of l equals three, and so on. Now, every subshell has a certain shape."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "So if our electron is found with a principal quantum number of one, that means it must be in subsshell l equals zero, because l equals one minus one gives you zero. Likewise, p represents a subshell of l equals 1d represents a subsshell of l equals two, f represents a subshell of l equals three, and so on. Now, every subshell has a certain shape. In other words, every subshell has an orbital that has a certain shape. And the shape of the orbitals in any subshell represents the most probable location of our electrons. So shapes are based on mathematical probabilities where our electrons are located."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "In other words, every subshell has an orbital that has a certain shape. And the shape of the orbitals in any subshell represents the most probable location of our electrons. So shapes are based on mathematical probabilities where our electrons are located. So let's look at the shape of our S. S has a spherical shape. And what that basically means what this shape means is that there's a 90% chance that our electron will be found within this sphere. In other words, if we know that our N is equal to zero, or actually M equals one, m can't be zero."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "So let's look at the shape of our S. S has a spherical shape. And what that basically means what this shape means is that there's a 90% chance that our electron will be found within this sphere. In other words, if we know that our N is equal to zero, or actually M equals one, m can't be zero. If M equals one, that means our L equals zero. So that means our subshale must be S. And that means that if this was our proton, there is a 90% chance, 90% probability that our electron is found within this skier. Likewise, let's look at the P orbital."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "If M equals one, that means our L equals zero. So that means our subshale must be S. And that means that if this was our proton, there is a 90% chance, 90% probability that our electron is found within this skier. Likewise, let's look at the P orbital. The P orbital has the following shape. And what this P orbital states is there's a 90% probability that our electron is found within this weird shape. Now, for this guy, N must be equal to two."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "The P orbital has the following shape. And what this P orbital states is there's a 90% probability that our electron is found within this weird shape. Now, for this guy, N must be equal to two. Because if M equals to two, our L will be one. And we see that P has a subshell of one. So this guy corresponds to M equals one, and actually M equals two, and L equals one."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "Because if M equals to two, our L will be one. And we see that P has a subshell of one. So this guy corresponds to M equals one, and actually M equals two, and L equals one. Now let's look at the third quantum number. The third quantum number is called the magnetic quantum number. And this guy is designated by a lowercase M with a subscript L.\nSo ML for magnetic quantum number."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "Now let's look at the third quantum number. The third quantum number is called the magnetic quantum number. And this guy is designated by a lowercase M with a subscript L.\nSo ML for magnetic quantum number. The Magnetic Quantum Number designates the exact orbital in which our electron is in. Now recall that every subshell L has a certain amount of orbitals, right? When our M equals one, they were talking about the S orbital or L equals zero subshell."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "The Magnetic Quantum Number designates the exact orbital in which our electron is in. Now recall that every subshell L has a certain amount of orbitals, right? When our M equals one, they were talking about the S orbital or L equals zero subshell. And this guy has only one orbital, namely the S orbital. If our principal quantum number, N is two, that means two minus one. Our L, our second quantum number, is one."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "And this guy has only one orbital, namely the S orbital. If our principal quantum number, N is two, that means two minus one. Our L, our second quantum number, is one. And that means there are three orbitals. Now, these guys are the p orbitals. Remember, there are three P orbitals."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "And that means there are three orbitals. Now, these guys are the p orbitals. Remember, there are three P orbitals. There's one in the X direction. So if we draw our XYZ axis, our three dimensional axis, that means our P orbital is in the X direction, our PY orbital is in the Y direction, and our PD orbital is in the Z direction. In other words, if we put all these guys together, we get three orbitals."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "There's one in the X direction. So if we draw our XYZ axis, our three dimensional axis, that means our P orbital is in the X direction, our PY orbital is in the Y direction, and our PD orbital is in the Z direction. In other words, if we put all these guys together, we get three orbitals. And these guys are all perpendicular to each other. Why? Well, because these lines, the Z and the X axis, the X and the Y axis, and the Z and the Y axis are all perpendicular to each other."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "And these guys are all perpendicular to each other. Why? Well, because these lines, the Z and the X axis, the X and the Y axis, and the Z and the Y axis are all perpendicular to each other. So all these three guys PX, PY, and PZ must be perpendicular. Now, our range for our Magnetic quantum numbers can be derived using our L. Our range begins at minus L and goes to plus L. So our PX begins at negative L.\nAnd since this guy represents L equals one, that means our PX must be L minus one or minus one. So our ML is minus one."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "So all these three guys PX, PY, and PZ must be perpendicular. Now, our range for our Magnetic quantum numbers can be derived using our L. Our range begins at minus L and goes to plus L. So our PX begins at negative L.\nAnd since this guy represents L equals one, that means our PX must be L minus one or minus one. So our ML is minus one. Now, the next number after minus one is zero. Right? We're adding increments of one."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "Now, the next number after minus one is zero. Right? We're adding increments of one. And that means our PY must have an ML of equals zero. And our final number is plus L, so plus one. So our ML for PD, the final orbital is plus one."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "And that means our PY must have an ML of equals zero. And our final number is plus L, so plus one. So our ML for PD, the final orbital is plus one. Now let's look at the final quantum number. The final quantum number is known as the electron spin quantum number. And this guy is represented by also a lowercase M but with a subscript S S for spin."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "Now let's look at the final quantum number. The final quantum number is known as the electron spin quantum number. And this guy is represented by also a lowercase M but with a subscript S S for spin. Now, every orbital can have a maximum of two electrons. And what the Pole Exclusion Principle tells us is it states that any two electrons in any given atom can never have the same four quantum numbers. Remember, just like any car in any given state can only have one license plate one unique license plate any unique electron in any given atom will only have four quantum numbers unique to that electron."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "Now, every orbital can have a maximum of two electrons. And what the Pole Exclusion Principle tells us is it states that any two electrons in any given atom can never have the same four quantum numbers. Remember, just like any car in any given state can only have one license plate one unique license plate any unique electron in any given atom will only have four quantum numbers unique to that electron. So that's what the poly exclusion principle tells us. And that means since any given orbital can have a maximum of two electrons there are two possible spin numbers or quantum spin numbers plus one two and minus one two. So I want to mention briefly two more important notes about quantum numbers."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "So that's what the poly exclusion principle tells us. And that means since any given orbital can have a maximum of two electrons there are two possible spin numbers or quantum spin numbers plus one two and minus one two. So I want to mention briefly two more important notes about quantum numbers. Now, to find a total number of orbitals in any given shell level with any given principal quantum number we simply take that principal quantum number and square it. So for N equals one there are one squared to one orbital. And that makes sense because when N equals one L equals zero and we have only the S orbital."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "Now, to find a total number of orbitals in any given shell level with any given principal quantum number we simply take that principal quantum number and square it. So for N equals one there are one squared to one orbital. And that makes sense because when N equals one L equals zero and we have only the S orbital. For M equals two there are two squared to four total orbitals. And that also makes sense that we have the SPSP one and PZ a total of four orbitals. For M equals three, there are nine orbitals."}, {"title": "Principal Azimuthal Magnetic and Spin Quantum Numbers .txt", "text": "For M equals two there are two squared to four total orbitals. And that also makes sense that we have the SPSP one and PZ a total of four orbitals. For M equals three, there are nine orbitals. For M equals four, there are 16 orbitals and so on. Now, one last thing I want to mention is if we look at our periodic table what each period represents is a new energy level, a new shell level. So period number one, all the elements have a shell level of one."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So let's recall what anisomer is. So two compounds are said to be isomers, where they have the same exact molecular formula but different structures. So in this lecture, we're going to focus on cystrend ze isomers. So let's begin by looking at the following alkane that has the following molecular formula c two, H two to XY, where XY are simply two different compounds, molecules or atoms that are not H atoms. And let's begin by trying to figure out the three dimensional structure of this alkene. So, alkines have the following structure."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So let's begin by looking at the following alkane that has the following molecular formula c two, H two to XY, where XY are simply two different compounds, molecules or atoms that are not H atoms. And let's begin by trying to figure out the three dimensional structure of this alkene. So, alkines have the following structure. So we have a double bond. The lower bond is a sigma bond. The upper bond is our Pi bond."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So we have a double bond. The lower bond is a sigma bond. The upper bond is our Pi bond. So lower bond, sigma, upper pi bond. We have the two intersections. So these are the carbon atoms shown here."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So lower bond, sigma, upper pi bond. We have the two intersections. So these are the carbon atoms shown here. And here we have the two H atoms given the red and the x atom, the green atom and the Y atom, the blue atom. Now, one important detail about the structure of alkanes is that they have planar symmetry. In other words, all these four atoms are found on the same exact plane."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "And here we have the two H atoms given the red and the x atom, the green atom and the Y atom, the blue atom. Now, one important detail about the structure of alkanes is that they have planar symmetry. In other words, all these four atoms are found on the same exact plane. So this was the XY plane. All these four atoms would be on the same plane. And these four bonds are also on the same plane."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So this was the XY plane. All these four atoms would be on the same plane. And these four bonds are also on the same plane. Now, whenever we have the two HS on the same side of the plane, or the x and Y, the heavier atoms on the same side of the plane, we have CITs or Z compounds or molecules. In other words, we define C compounds as hydrogens being on the same side. And we define Z compounds as having the heavier groups on the same side."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "Now, whenever we have the two HS on the same side of the plane, or the x and Y, the heavier atoms on the same side of the plane, we have CITs or Z compounds or molecules. In other words, we define C compounds as hydrogens being on the same side. And we define Z compounds as having the heavier groups on the same side. Remember, we define x and Y to be any compound, molecule or atom that is different than H. And that means our x and Y are both heavier than H's. So this compound is one isomer of this molecular formula and it's the sys or Z isomer. What about trans or e?"}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "Remember, we define x and Y to be any compound, molecule or atom that is different than H. And that means our x and Y are both heavier than H's. So this compound is one isomer of this molecular formula and it's the sys or Z isomer. What about trans or e? Well, let's recall another important detail about alkenes. Alkines have double bonds, and that means we can't rotate this molecule. The only way we rotate this lower bond is if we break the Pi bond."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "Well, let's recall another important detail about alkenes. Alkines have double bonds, and that means we can't rotate this molecule. The only way we rotate this lower bond is if we break the Pi bond. The only way we break the Pi bond is if we input energy. So let's say we input energy, exactly 66 energy. Then we'll break the Pi bond and our bond, our CC bond, the lower CC bond will rotate."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "The only way we break the Pi bond is if we input energy. So let's say we input energy, exactly 66 energy. Then we'll break the Pi bond and our bond, our CC bond, the lower CC bond will rotate. And then let's suppose it rotates. We take away the energy and our Pi bond reforms. We will get the following molecule."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "And then let's suppose it rotates. We take away the energy and our Pi bond reforms. We will get the following molecule. So, in this molecule, we have the HS are on the opposite sides, and the x and Y are also on the opposite sides. So we have the following picture. So we have the two HS being on different sides, and we have the Y and the X also being on different sides."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So, in this molecule, we have the HS are on the opposite sides, and the x and Y are also on the opposite sides. So we have the following picture. So we have the two HS being on different sides, and we have the Y and the X also being on different sides. So trans is defined the following way. Hydrogens are on different sides of the plane, x and x, I mean H and H different sides. And E is defined as the heavier groups are in different sides."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So trans is defined the following way. Hydrogens are on different sides of the plane, x and x, I mean H and H different sides. And E is defined as the heavier groups are in different sides. So, once again, our X and Y are in different signs. Now, these two guides are isomers to one another, and they're separated by 66 kilotons of mole. So let's look at the following few examples, and let's try to figure out which ones are E and which ones are Z."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So, once again, our X and Y are in different signs. Now, these two guides are isomers to one another, and they're separated by 66 kilotons of mole. So let's look at the following few examples, and let's try to figure out which ones are E and which ones are Z. So we're looking for the heavier groups and the lighter groups. So we have two methyl groups and two Ethyl groups. Ethyl is heavier."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So we're looking for the heavier groups and the lighter groups. So we have two methyl groups and two Ethyl groups. Ethyl is heavier. That means these guys found on the same side are the heavier groups. And so we have the heavier groups on the same side. So that means we have our Z."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "That means these guys found on the same side are the heavier groups. And so we have the heavier groups on the same side. So that means we have our Z. So this must be Z. So what about this guy here? We also have methyl and Ethyl."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So this must be Z. So what about this guy here? We also have methyl and Ethyl. But the Ethyl groups are different size, and the methyl groups are the same size. So that means the lighter are on the same sides. On different sides, the heavier are a different size."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "But the Ethyl groups are different size, and the methyl groups are the same size. So that means the lighter are on the same sides. On different sides, the heavier are a different size. So that means we must have our E.\nSo this is E. What about this guy? Well, now we have a methyl. We have an h group."}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "So that means we must have our E.\nSo this is E. What about this guy? Well, now we have a methyl. We have an h group. We have a methyl and an Ethyl. So once again, we have the two ethyls on different sides, the two heavier groups. And that means we must have our E.\nWell, what about this last one?"}, {"title": "Cis-Trans and Z-E Isomerism .txt", "text": "We have a methyl and an Ethyl. So once again, we have the two ethyls on different sides, the two heavier groups. And that means we must have our E.\nWell, what about this last one? Here we have two methyl groups attached to the same carbon. So these two groups are identical. And that means there's no way to distinguish between these two guys because they have the same exact weight."}, {"title": "Phase Change .txt", "text": "Change is a physical process by which a substance goes from one phase to another phase. Now let's look at melting, freezing, vaporization and condensation. Melting is the process by which a solid becomes a liquid. And in this process, energy input is required. So the process is endothelial. Thermic melting is endothermic."}, {"title": "Phase Change .txt", "text": "And in this process, energy input is required. So the process is endothelial. Thermic melting is endothermic. The enthalpy of fusion is simply the energy input required to melt something. And this guy is the same as this guy, so it's also positive. So in melting, the potential energy of our substance, of our solid, is increased."}, {"title": "Phase Change .txt", "text": "The enthalpy of fusion is simply the energy input required to melt something. And this guy is the same as this guy, so it's also positive. So in melting, the potential energy of our substance, of our solid, is increased. And the atrium molecular bonds connecting the solid are broken so that eventually they can become liquid molecules. So freezing is the opposite of melting. Freezing is the process by which a liquid becomes a solid."}, {"title": "Phase Change .txt", "text": "And the atrium molecular bonds connecting the solid are broken so that eventually they can become liquid molecules. So freezing is the opposite of melting. Freezing is the process by which a liquid becomes a solid. In this process, internal energy of our system, of our liquid, decreases until eventually it becomes a solid. That means energy is released the environment. Okay, so it's exothermic."}, {"title": "Phase Change .txt", "text": "In this process, internal energy of our system, of our liquid, decreases until eventually it becomes a solid. That means energy is released the environment. Okay, so it's exothermic. So the entropy of freezing is negative. It's the same magnitude as the enthalpy of fusion, but it's negative. So now let's look at vaporization."}, {"title": "Phase Change .txt", "text": "So the entropy of freezing is negative. It's the same magnitude as the enthalpy of fusion, but it's negative. So now let's look at vaporization. Vaporization is a process by which a liquid becomes a gas. And in this situation, energy input is also required, just like in melting. And that's because the potential edge of the bonds is required to increase."}, {"title": "Phase Change .txt", "text": "Vaporization is a process by which a liquid becomes a gas. And in this situation, energy input is also required, just like in melting. And that's because the potential edge of the bonds is required to increase. So energy input into our system from the environment is required. So the enthalpy of vaporization is positive. Now let's look at condensation."}, {"title": "Phase Change .txt", "text": "So energy input into our system from the environment is required. So the enthalpy of vaporization is positive. Now let's look at condensation. Condensation is the opposite of vaporization. It's the process by which a gas molecule goes back into the liquid state and condensation, just like freezing, is exothermic. When a gas becomes a liquid, energy is released of the environment because the internal energy of our system decreases."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Now, these questions and many others can be answered using a principle known as Lusciously as principle. And what this principle states is that whenever a system at equilibrium is stressed it will respond by shifting equilibrium in the direction that tends to decrease that stress. Now, let's look at three stresses temperature, pressure and concentrations. So let's begin with temperature. Suppose we have the following reaction in which 1 mol of among the gas state reacts with three moles of diatomic H two plus the gas state to produce two moles of ammonia alter the gas state and heat. So this is an exothermic reaction."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "So let's begin with temperature. Suppose we have the following reaction in which 1 mol of among the gas state reacts with three moles of diatomic H two plus the gas state to produce two moles of ammonia alter the gas state and heat. So this is an exothermic reaction. Now, whenever we're talking about lushly as principle and we see that we have heat produced or heat in the beginning, that means we can treat this heat sky as a product. So our heat is tangible, okay? So now let's increase the temperature of our system of our reaction."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Now, whenever we're talking about lushly as principle and we see that we have heat produced or heat in the beginning, that means we can treat this heat sky as a product. So our heat is tangible, okay? So now let's increase the temperature of our system of our reaction. Now, we increase temperature by adding energy and that means we're adding heat. Now, if we're adding heat, that means one of our products, namely the heat guide is increased. So the concentration is not really concentration but you could think of energy concentration of our heat increases."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Now, we increase temperature by adding energy and that means we're adding heat. Now, if we're adding heat, that means one of our products, namely the heat guide is increased. So the concentration is not really concentration but you could think of energy concentration of our heat increases. Now, if you think of this as the reaction quotient, that means our ratio will increase. So our Q will become greater than our K.\nAnd that means as K is bigger than Q, there will be a leftward shift of equilibrium. In other words, these guys will tend to react to convert back to our reactants, namely these two guys."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Now, if you think of this as the reaction quotient, that means our ratio will increase. So our Q will become greater than our K.\nAnd that means as K is bigger than Q, there will be a leftward shift of equilibrium. In other words, these guys will tend to react to convert back to our reactants, namely these two guys. So we see that whenever we're talking about an exothermic reaction in which he's produced increasing temperature of our system will shift equilibrium to the left this way. So in this case, the reverse reaction is more favored than a four reaction. So let's reverse the case."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "So we see that whenever we're talking about an exothermic reaction in which he's produced increasing temperature of our system will shift equilibrium to the left this way. So in this case, the reverse reaction is more favored than a four reaction. So let's reverse the case. Let's say we're dealing with an endothermic reaction in which heat is added to our ammonia molecules to create our products. So now these guys are products and these guys are reactants. Now, what happens if we increase temperature?"}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Let's say we're dealing with an endothermic reaction in which heat is added to our ammonia molecules to create our products. So now these guys are products and these guys are reactants. Now, what happens if we increase temperature? Well, now we have the reverse case. Now, one of our reactants, its concentration increases. And that means our Q, our quotient reaction quotient decreases, becomes smaller than K. And that means increasing temperature of an endothermic reaction will shift it to the right."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Well, now we have the reverse case. Now, one of our reactants, its concentration increases. And that means our Q, our quotient reaction quotient decreases, becomes smaller than K. And that means increasing temperature of an endothermic reaction will shift it to the right. We're going to see a leftward shift in equilibrium. So the reactants will tend to react to produce our products, our end in the gas state and three moles of H two in the gas state. Now, some exceptions to Lusciously as principle exists."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "We're going to see a leftward shift in equilibrium. So the reactants will tend to react to produce our products, our end in the gas state and three moles of H two in the gas state. Now, some exceptions to Lusciously as principle exists. Now, at very high temperatures, entropy takes over. And that's because entropy is dictated by the following reaction in which change in gigs, free energy is equal to change in entropy minus temperature times change in entropy. So, if temperature is high enough and we have a positive change in entropy, even if we have an endothermic or an exothermic reaction, it doesn't matter what reaction we have."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Now, at very high temperatures, entropy takes over. And that's because entropy is dictated by the following reaction in which change in gigs, free energy is equal to change in entropy minus temperature times change in entropy. So, if temperature is high enough and we have a positive change in entropy, even if we have an endothermic or an exothermic reaction, it doesn't matter what reaction we have. At a very high temperature with very positive entropy, we're always going to get a negative gives free energy. And so, our reaction will always tend to be this way in a rightward direction. Now, let's talk about pressure."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "At a very high temperature with very positive entropy, we're always going to get a negative gives free energy. And so, our reaction will always tend to be this way in a rightward direction. Now, let's talk about pressure. Suppose once again we have the following exostobic reaction. And now, suppose we try to increase our pressure at constant temperature, so we decrease our volume. And so, the pressure of our system, the pressure of the molecules exert on the wall of the container increases."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Suppose once again we have the following exostobic reaction. And now, suppose we try to increase our pressure at constant temperature, so we decrease our volume. And so, the pressure of our system, the pressure of the molecules exert on the wall of the container increases. So what will happen? Well, increasing, or according to Lush of Global principle, increasing pressure by decreasing volume at constant temperature, by making our system smaller, by shrinking, it will shift the equilibrium in a direction where there are less molecules present. In other words, let's see how many molecules we have on the reactant side and how many molecules we have on the product side."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "So what will happen? Well, increasing, or according to Lush of Global principle, increasing pressure by decreasing volume at constant temperature, by making our system smaller, by shrinking, it will shift the equilibrium in a direction where there are less molecules present. In other words, let's see how many molecules we have on the reactant side and how many molecules we have on the product side. Well, according to this formula, we have one and three. So, four molecules, four moles of molecules on our reactive side, and only two moles of molecules on our product side. So that means, by decreasing volume, thereby increasing pressure, our system will shift this way, rightward?"}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Well, according to this formula, we have one and three. So, four molecules, four moles of molecules on our reactive side, and only two moles of molecules on our product side. So that means, by decreasing volume, thereby increasing pressure, our system will shift this way, rightward? Because if our system goes this way, it's going to produce on average, less moles of molecules than on this side. So, our system will try to counter the increase in pressure by decreasing the pressure by going this way to decrease the number of total molecules found in our system. If we have less molecules in our system, that means less molecules are colliding with the walls of our container."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Because if our system goes this way, it's going to produce on average, less moles of molecules than on this side. So, our system will try to counter the increase in pressure by decreasing the pressure by going this way to decrease the number of total molecules found in our system. If we have less molecules in our system, that means less molecules are colliding with the walls of our container. And so, our overall total pressure is less. And finally, let's look at how concentration affects our equilibrium of our reaction according to Leslie Lee's principle. Now, let's look at the same reaction."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "And so, our overall total pressure is less. And finally, let's look at how concentration affects our equilibrium of our reaction according to Leslie Lee's principle. Now, let's look at the same reaction. 1 mol of them react to three moles of H two and gas state to produce two moles of ammonia and heat an exothermic reaction. So, how will increasing reactants change our equilibrium? Well, increasing the concentration of either or both the reactants shifts equilibrium to the right."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "1 mol of them react to three moles of H two and gas state to produce two moles of ammonia and heat an exothermic reaction. So, how will increasing reactants change our equilibrium? Well, increasing the concentration of either or both the reactants shifts equilibrium to the right. So, a right workshift in equilibrium, while increasing concentration of our products, any of the products of both the products shifts our equilibrium to the left. So, increasing this guy will shift our equilibrium to the left, producing more of our reactants. So, why is that?"}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "So, a right workshift in equilibrium, while increasing concentration of our products, any of the products of both the products shifts our equilibrium to the left. So, increasing this guy will shift our equilibrium to the left, producing more of our reactants. So, why is that? Well, let's see once again, what happens when we increase the concentrations of this guy and this guy. Well, recall what the reaction quotient states. If you increase the concentration of our reactants, then our reaction quotient will become larger than our equilibrium constant K. And if Q is larger than K, that means we're going to observe a rightward shift in equilibrium in other words, the forward reaction will proceed this way at a higher rate than the reverse reaction."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "Well, let's see once again, what happens when we increase the concentrations of this guy and this guy. Well, recall what the reaction quotient states. If you increase the concentration of our reactants, then our reaction quotient will become larger than our equilibrium constant K. And if Q is larger than K, that means we're going to observe a rightward shift in equilibrium in other words, the forward reaction will proceed this way at a higher rate than the reverse reaction. So these guys will want to convert to decrease the overall concentration of our reactants so that our Q can become our K again. In other words, Q will tend to move towards K. Likewise, if we increase the concentration of products. That means our q will become a less than K.\nAnd since q will want to move towards K, this concentration will tend to decreasing therefore providing a leftward shift in equilibrium."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "So these guys will want to convert to decrease the overall concentration of our reactants so that our Q can become our K again. In other words, Q will tend to move towards K. Likewise, if we increase the concentration of products. That means our q will become a less than K.\nAnd since q will want to move towards K, this concentration will tend to decreasing therefore providing a leftward shift in equilibrium. I want to mention one last important note what will happen if we add some other arbitrary molecule that is not present in our reaction? Such as for example, diatomic CO2 in our gas state? Well, to answer this question, let's look at our partial pressure equilibrium constant."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "I want to mention one last important note what will happen if we add some other arbitrary molecule that is not present in our reaction? Such as for example, diatomic CO2 in our gas state? Well, to answer this question, let's look at our partial pressure equilibrium constant. And the reason we're using partial pressures is because every single reactant is in the gas state. Let me just add the G and G for gas. So our equilibrium or partial pressure equilibrium constant given by KP equals the partial pressure of our product to the second power because we have a two coefficient divided by the partial pressure due to this gas molecule times the partial pressure due to this gas molecule to the third power."}, {"title": "Le Chatelier\u2019s Principle .txt", "text": "And the reason we're using partial pressures is because every single reactant is in the gas state. Let me just add the G and G for gas. So our equilibrium or partial pressure equilibrium constant given by KP equals the partial pressure of our product to the second power because we have a two coefficient divided by the partial pressure due to this gas molecule times the partial pressure due to this gas molecule to the third power. The third comes from the coefficient and the one here comes from the one coefficient. Notice that our gas molecule, this one does not appear in our equilibrium expression and that means adding this guide will have no effect on our chemical equilibrium and in fact, adding any other molecule, for example helium to our mixture will not do anything to our chemical equilibrium. And that's an important point to understand only increasing the concentrations of molecules that are actually present in our reaction in our equilibrium expression will affect our chemical equilibrium."}, {"title": "pH example #1.txt", "text": "So we basically begin with pure water, which has a PH of seven. And we added acid, a Bronxed Lyric acid, which will basically dissociate into H plus and the base. That means that concentration of our H ions within our solution will increase. So our PH should decrease. Let's see what we get. Exactly."}, {"title": "pH example #1.txt", "text": "So our PH should decrease. Let's see what we get. Exactly. So we want to use the formula for a PH. But before we use that formula, we have to find the molarity of our solution. And the way we find molarity is first we want to find the molecular weight of our solid, but then we want to find the moles of our solid."}, {"title": "pH example #1.txt", "text": "So we want to use the formula for a PH. But before we use that formula, we have to find the molarity of our solution. And the way we find molarity is first we want to find the molecular weight of our solid, but then we want to find the moles of our solid. And finally we find the kilograms of solvent. We divide moles of solid by kilograms of solvent. We get molarity, plug that into our PH formula, and we find our result."}, {"title": "pH example #1.txt", "text": "And finally we find the kilograms of solvent. We divide moles of solid by kilograms of solvent. We get molarity, plug that into our PH formula, and we find our result. So let's begin. Molecular weight of our nitric acid is simply calculated by adding up the atomic weights of each respective atom. So 1 gram/mol for H plus 14 grams/mol for N plus three times."}, {"title": "pH example #1.txt", "text": "So let's begin. Molecular weight of our nitric acid is simply calculated by adding up the atomic weights of each respective atom. So 1 gram/mol for H plus 14 grams/mol for N plus three times. Because we have a subscription of 316 grams per mole for O, gives you a molecular weight of 63 grams/mol for nitric acid. Now, to find the moles of nitric acid, we take our grams of nitric acid, divide that by our molecular weight grams cancel, moles goes on top, and we get zero point 73 moles of nitric acid. So finally, we have to find the kilograms of solvent."}, {"title": "pH example #1.txt", "text": "Because we have a subscription of 316 grams per mole for O, gives you a molecular weight of 63 grams/mol for nitric acid. Now, to find the moles of nitric acid, we take our grams of nitric acid, divide that by our molecular weight grams cancel, moles goes on top, and we get zero point 73 moles of nitric acid. So finally, we have to find the kilograms of solvent. To find the kilograms of solvent, we take our density of water and multiply it by the volume of water. So our density of water, we can look that up, and that's 1 gram per ML times the amount in volume of water. 300 MLS MLS cancel."}, {"title": "pH example #1.txt", "text": "To find the kilograms of solvent, we take our density of water and multiply it by the volume of water. So our density of water, we can look that up, and that's 1 gram per ML times the amount in volume of water. 300 MLS MLS cancel. We get 300 grams of water. Now we need kilograms. To find kilograms, we simply take this, divide it by 1000, we get 0.3 kg."}, {"title": "pH example #1.txt", "text": "We get 300 grams of water. Now we need kilograms. To find kilograms, we simply take this, divide it by 1000, we get 0.3 kg. Now, to find Molarity, we simply take our moles of solute, divide that by kilograms of solvent, and we get zero point 24 three molar. So that's our Molarity. And finally we use our PH formula which basically states PH equals negative log ten of 0.243."}, {"title": "pH example #1.txt", "text": "Now, to find Molarity, we simply take our moles of solute, divide that by kilograms of solvent, and we get zero point 24 three molar. So that's our Molarity. And finally we use our PH formula which basically states PH equals negative log ten of 0.243. Our Molarity, plug that into our calculator and we get 1.61. So clearly it went from a PH of seven, a neutral, so a very low PH 1.61. And this means it's a very acidic solution."}, {"title": "Autoionization of Water .txt", "text": "And since water is the only molecule present in our mixture, that means water must associate. And let's see exactly what happens. Remember, water can act as both a base and an acid. So what actually happens is two water molecules interact in such a way that one acts as a base and one act as an acid. So one donates H, and the other one accepts H, forming a conjugate acid and a conjugate base, namely hydronium and hydroxide. And this process of dissociation is called autoionization, where the auto basically means that two water molecules interact."}, {"title": "Autoionization of Water .txt", "text": "So what actually happens is two water molecules interact in such a way that one acts as a base and one act as an acid. So one donates H, and the other one accepts H, forming a conjugate acid and a conjugate base, namely hydronium and hydroxide. And this process of dissociation is called autoionization, where the auto basically means that two water molecules interact. And ionization means the dissociation of the two water molecules. Now, at equilibrium, reactants are favored. And that means equilibrium lies far to the left."}, {"title": "Autoionization of Water .txt", "text": "And ionization means the dissociation of the two water molecules. Now, at equilibrium, reactants are favored. And that means equilibrium lies far to the left. So that means there will be many more water molecules than ions at any given temperature. So, equilibrium construction. So let's write the equilibrium construction for our reaction above."}, {"title": "Autoionization of Water .txt", "text": "So that means there will be many more water molecules than ions at any given temperature. So, equilibrium construction. So let's write the equilibrium construction for our reaction above. So remember, liquids are not included. Only eighthway solutions are included. So our KWR constant is equal to the concentration of hydronium times the concentration of hydroxide."}, {"title": "Autoionization of Water .txt", "text": "So remember, liquids are not included. Only eighthway solutions are included. So our KWR constant is equal to the concentration of hydronium times the concentration of hydroxide. So kw in this case, when we talk about autoionization, is called the ionization constant for H 20. And this depends on temperature. And at 25 degrees Celsius, you should know that kw is 1.0 times ten to negative 14."}, {"title": "Autoionization of Water .txt", "text": "So kw in this case, when we talk about autoionization, is called the ionization constant for H 20. And this depends on temperature. And at 25 degrees Celsius, you should know that kw is 1.0 times ten to negative 14. Now, notice that our exponents here are one and one. That's because we have 1 mol and 1 mol. So, since we know our kw, and from experiments, we know that at a PH of seven, our concentration of hydronium is 1.0\ntimes ten to negative seven."}, {"title": "Autoionization of Water .txt", "text": "Now, notice that our exponents here are one and one. That's because we have 1 mol and 1 mol. So, since we know our kw, and from experiments, we know that at a PH of seven, our concentration of hydronium is 1.0\ntimes ten to negative seven. That means, since we know this guy and this guy, we can find the concentration of hydroxide. And we can do it by simply dividing through by 1.0\ntimes ten negative seven. And we get x equals 1.0 times ten to negative seven."}, {"title": "Autoionization of Water .txt", "text": "That means, since we know this guy and this guy, we can find the concentration of hydroxide. And we can do it by simply dividing through by 1.0\ntimes ten negative seven. And we get x equals 1.0 times ten to negative seven. So at a PH of seven, our concentration of hydronium equals that of hydroxide. Now, at any given temperature, the ionization constant will always remain the same. So at 25 degrees Celsius, this guy will always be 1.0\ntimes ten to negative 14."}, {"title": "Autoionization of Water .txt", "text": "So at a PH of seven, our concentration of hydronium equals that of hydroxide. Now, at any given temperature, the ionization constant will always remain the same. So at 25 degrees Celsius, this guy will always be 1.0\ntimes ten to negative 14. This will only increase with increase in temperature or decrease with decreasing temperature. This does not depend on the concentration of this guy, nor on this guy. Now, if our hydronium concentration is greater than 1.0 times ten to negative seven, that means our solution is acidic."}, {"title": "Autoionization of Water .txt", "text": "This will only increase with increase in temperature or decrease with decreasing temperature. This does not depend on the concentration of this guy, nor on this guy. Now, if our hydronium concentration is greater than 1.0 times ten to negative seven, that means our solution is acidic. If it's less than 1.0\ntimes ten to seven, it's basic. Likewise, if our hydroxide concentration is less than 1.0 times six, negative seven, it's an acidic solution. If it's more than 1.0 times six mega seven, it's a basic solution."}, {"title": "Autoionization of Water .txt", "text": "If it's less than 1.0\ntimes ten to seven, it's basic. Likewise, if our hydroxide concentration is less than 1.0 times six, negative seven, it's an acidic solution. If it's more than 1.0 times six mega seven, it's a basic solution. Now, in this last step, I want to look at the expression for kw again. And my goal is to take this expression and convert it to this expression. This expression will be convenient when we know the PH and we want to find the poh."}, {"title": "Autoionization of Water .txt", "text": "Now, in this last step, I want to look at the expression for kw again. And my goal is to take this expression and convert it to this expression. This expression will be convenient when we know the PH and we want to find the poh. And when we know that poh and we want to find the PH. So to go from this guy to this guy, let's first look at a few things. Let's review."}, {"title": "Autoionization of Water .txt", "text": "And when we know that poh and we want to find the PH. So to go from this guy to this guy, let's first look at a few things. Let's review. One of the logs of log states that log of X times Y equals log x plus log y. The second part I want to look at is the formulas for PH and poh. PH is equal to negative log of the hydronium concentration, and Poh is equal to the negative log of the hydroxide concentration."}, {"title": "Autoionization of Water .txt", "text": "One of the logs of log states that log of X times Y equals log x plus log y. The second part I want to look at is the formulas for PH and poh. PH is equal to negative log of the hydronium concentration, and Poh is equal to the negative log of the hydroxide concentration. Now, in Part C, remember, at 25 degrees Celsius, our Kw will always be 1.0\ntimes 10th to negative 14. Now, if we want to take the PH of kw, which is PKW, we simply write PKW is equal to negative log of one times ten to the negative 14. Right?"}, {"title": "Autoionization of Water .txt", "text": "Now, in Part C, remember, at 25 degrees Celsius, our Kw will always be 1.0\ntimes 10th to negative 14. Now, if we want to take the PH of kw, which is PKW, we simply write PKW is equal to negative log of one times ten to the negative 14. Right? Now, I could either plug this guy into the calculator, or I could solve it in my head. Now, since we have a negative here and a negative here, negatives cancel. So we simply have one times 10th and negative four."}, {"title": "Autoionization of Water .txt", "text": "Now, I could either plug this guy into the calculator, or I could solve it in my head. Now, since we have a negative here and a negative here, negatives cancel. So we simply have one times 10th and negative four. It's the 14th. So since our base is ten, I'll write a base ten. Since our exponent is what we don't know, we leave that blank."}, {"title": "Autoionization of Water .txt", "text": "It's the 14th. So since our base is ten, I'll write a base ten. Since our exponent is what we don't know, we leave that blank. And since this is our result, we plug this into the result. So 10th to what power gives you ten to the 14th? Well, 14."}, {"title": "Autoionization of Water .txt", "text": "And since this is our result, we plug this into the result. So 10th to what power gives you ten to the 14th? Well, 14. So x is 14. So PKW is 14. So let's go back to here."}, {"title": "Autoionization of Water .txt", "text": "So x is 14. So PKW is 14. So let's go back to here. Kw is equal to hydronium times hydroxide log of base ten of Kw is equal to log of base ten of this entire guy. Using part A, we simply distribute, and we get log of kw is equal to log of this guy plus log of this guy. Now, to use part B, I need to multiply the entire statement by negative one."}, {"title": "Autoionization of Water .txt", "text": "Kw is equal to hydronium times hydroxide log of base ten of Kw is equal to log of base ten of this entire guy. Using part A, we simply distribute, and we get log of kw is equal to log of this guy plus log of this guy. Now, to use part B, I need to multiply the entire statement by negative one. And now I can substitute. I bring this guy, the PH, into here, and this guy is simply poach goes here, and this guy is simply PKW. And now from Part C, I know that PKW is equal to 14."}, {"title": "Intensive and Extensive Properties.txt", "text": "Today we're going to talk about extensive properties, intensive properties, state functions of a system of interest. OK? Before we talk about these guys, let's define what a system is. A system is simply something that a person or a scientist or researcher wants to study. It's the object of interest, okay? And the object could be as simple as the eraser."}, {"title": "Intensive and Extensive Properties.txt", "text": "A system is simply something that a person or a scientist or researcher wants to study. It's the object of interest, okay? And the object could be as simple as the eraser. It could be this marker, it could be this whiteboard, it could be me, it could be the room, the building, this system. For example, if we chose this eraser to be the system has certain properties, it has a certain weight, it has a certain mass, it has a certain volume, certain density, certain pressure, certain height, length and width. It has certain properties that are used to quantify this system."}, {"title": "Intensive and Extensive Properties.txt", "text": "It could be this marker, it could be this whiteboard, it could be me, it could be the room, the building, this system. For example, if we chose this eraser to be the system has certain properties, it has a certain weight, it has a certain mass, it has a certain volume, certain density, certain pressure, certain height, length and width. It has certain properties that are used to quantify this system. And these properties, if you list them all in order, could be subdivided into two categories extensive properties and intensive properties. Okay? Extensive properties are those properties that depend on size and intensive properties do not depend on size."}, {"title": "Intensive and Extensive Properties.txt", "text": "And these properties, if you list them all in order, could be subdivided into two categories extensive properties and intensive properties. Okay? Extensive properties are those properties that depend on size and intensive properties do not depend on size. So let's go through the examples. Extensive properties, let's talk about mass. So this eraser has a certain mass, okay?"}, {"title": "Intensive and Extensive Properties.txt", "text": "So let's go through the examples. Extensive properties, let's talk about mass. So this eraser has a certain mass, okay? And what happens to the mass if you double the size of this eraser? Well, the mass would also double. Suppose you add or you stack a second eraser on top of this eraser, what will happen to the mass?"}, {"title": "Intensive and Extensive Properties.txt", "text": "And what happens to the mass if you double the size of this eraser? Well, the mass would also double. Suppose you add or you stack a second eraser on top of this eraser, what will happen to the mass? It will double. How about the force or the weight felt by a scale? Right?"}, {"title": "Intensive and Extensive Properties.txt", "text": "It will double. How about the force or the weight felt by a scale? Right? If you take this eraser and you weight it, you get one force. If you weight this or if you weigh two erasers, you will get a force twice the size of the original. So force is also dependent on size, volume as well."}, {"title": "Intensive and Extensive Properties.txt", "text": "If you take this eraser and you weight it, you get one force. If you weight this or if you weigh two erasers, you will get a force twice the size of the original. So force is also dependent on size, volume as well. Volume is pretty intuitive. If you increase the size of this twice the size, the volume will also increase. Same with area, same with length, width and height, right?"}, {"title": "Intensive and Extensive Properties.txt", "text": "Volume is pretty intuitive. If you increase the size of this twice the size, the volume will also increase. Same with area, same with length, width and height, right? If you take this eraser, if you stack another eraser on top of it, the height will increase. So increase in size increases height. How about moles?"}, {"title": "Intensive and Extensive Properties.txt", "text": "If you take this eraser, if you stack another eraser on top of it, the height will increase. So increase in size increases height. How about moles? Well, this is composed of certain bonds, certain atoms that are bonded together, right? Covalently well, number of moles. Suppose this is composed of a certain number of moles."}, {"title": "Intensive and Extensive Properties.txt", "text": "Well, this is composed of certain bonds, certain atoms that are bonded together, right? Covalently well, number of moles. Suppose this is composed of a certain number of moles. Suppose it's x number of moles. You take a second eraser identical to this one, stack it on top of this one, and what will you get? You will get an eraser with x number of moles."}, {"title": "Intensive and Extensive Properties.txt", "text": "Suppose it's x number of moles. You take a second eraser identical to this one, stack it on top of this one, and what will you get? You will get an eraser with x number of moles. A second eraser with x number of moles. The result is two x number of moles. So increasing this size or this system to twice its size increases the number of moles to twice number of original moles, number of moles increases."}, {"title": "Intensive and Extensive Properties.txt", "text": "A second eraser with x number of moles. The result is two x number of moles. So increasing this size or this system to twice its size increases the number of moles to twice number of original moles, number of moles increases. And remember, internal energy depends on the number of moles of the atom. So if the number of atoms increases or the number of moles increase, the internal energy also increases. Okay?"}, {"title": "Intensive and Extensive Properties.txt", "text": "And remember, internal energy depends on the number of moles of the atom. So if the number of atoms increases or the number of moles increase, the internal energy also increases. Okay? So all these guys, all these properties are properties that can be used to describe the system of interest, to quantify it depends on the size of the system. Increasing the size of this increases these properties. How about intensive properties?"}, {"title": "Intensive and Extensive Properties.txt", "text": "So all these guys, all these properties are properties that can be used to describe the system of interest, to quantify it depends on the size of the system. Increasing the size of this increases these properties. How about intensive properties? Earlier we said instant properties are those properties that do not depend on the size. They're independent. So let's see why temperature is independent property."}, {"title": "Intensive and Extensive Properties.txt", "text": "Earlier we said instant properties are those properties that do not depend on the size. They're independent. So let's see why temperature is independent property. So let's not use this as an example, but let's use something more intuitive. Let's use our human body, okay? So let's take three different individuals."}, {"title": "Intensive and Extensive Properties.txt", "text": "So let's not use this as an example, but let's use something more intuitive. Let's use our human body, okay? So let's take three different individuals. Let's take a baby, let's take myself and let's take our seven foot player, okay? Seven foot basketball player. We are all different in size, right?"}, {"title": "Intensive and Extensive Properties.txt", "text": "Let's take a baby, let's take myself and let's take our seven foot player, okay? Seven foot basketball player. We are all different in size, right? The baby is really small on average. And then the height of the basketball player is really big. So there's clearly a difference in size."}, {"title": "Intensive and Extensive Properties.txt", "text": "The baby is really small on average. And then the height of the basketball player is really big. So there's clearly a difference in size. So we're all going to have different masses, different weights, different volumes, compose a different number of spells or moles, have a different area, have a different length or height width, have different internal energy. But our temperature will stay the same. If you check using a thermometer, the baby will have approximately 37 Celsius or 98 Fahrenheit."}, {"title": "Intensive and Extensive Properties.txt", "text": "So we're all going to have different masses, different weights, different volumes, compose a different number of spells or moles, have a different area, have a different length or height width, have different internal energy. But our temperature will stay the same. If you check using a thermometer, the baby will have approximately 37 Celsius or 98 Fahrenheit. I will have that same temperature and so will that giant dusky clip. So the temperature remains the same the same way that if you return to this system, that might be less intuitive, but still remains true that the temperature of this, regardless of size, remains the same. How about pressure?"}, {"title": "Intensive and Extensive Properties.txt", "text": "I will have that same temperature and so will that giant dusky clip. So the temperature remains the same the same way that if you return to this system, that might be less intuitive, but still remains true that the temperature of this, regardless of size, remains the same. How about pressure? Well, pressure from chemistry or from physics, we know to be force over area, correct? And the force and the area we saw are both extensive properties. In fact, it's true that if we take the ratio of two extensive properties or if we divide an extensive property by second extensive property, what we get is an incentive property."}, {"title": "Intensive and Extensive Properties.txt", "text": "Well, pressure from chemistry or from physics, we know to be force over area, correct? And the force and the area we saw are both extensive properties. In fact, it's true that if we take the ratio of two extensive properties or if we divide an extensive property by second extensive property, what we get is an incentive property. That always holds true. And let's see why. Well, priced is just a ratio, right?"}, {"title": "Intensive and Extensive Properties.txt", "text": "That always holds true. And let's see why. Well, priced is just a ratio, right? So if we increase this by, say, two, this increases by two and these guys cancel out. So no matter what, we increase this by, say by three, by four, by five, by six and so on, this will increase by a similar increment. Meaning that you could cancel these guys out and you get the same ratio over and over and over."}, {"title": "Intensive and Extensive Properties.txt", "text": "So if we increase this by, say, two, this increases by two and these guys cancel out. So no matter what, we increase this by, say by three, by four, by five, by six and so on, this will increase by a similar increment. Meaning that you could cancel these guys out and you get the same ratio over and over and over. So pressure always stays the same no matter of size. Density stays the same for the same reason. Because density is mass over volume, it's also a ratio of two extensive properties."}, {"title": "Intensive and Extensive Properties.txt", "text": "So pressure always stays the same no matter of size. Density stays the same for the same reason. Because density is mass over volume, it's also a ratio of two extensive properties. And so the result is an intensive property. So let's talk about what state functions are. So state functions are also properties of a system."}, {"title": "Intensive and Extensive Properties.txt", "text": "And so the result is an intensive property. So let's talk about what state functions are. So state functions are also properties of a system. They're either excessive properties or incentive properties. And they're basically values that do not change regardless of the path taken, okay? These properties only depend on the current state of the system."}, {"title": "Intensive and Extensive Properties.txt", "text": "They're either excessive properties or incentive properties. And they're basically values that do not change regardless of the path taken, okay? These properties only depend on the current state of the system. Okay, let's take this as our example again, the eraser. And let's think about how it was made. So it was made in different factories across the world, right?"}, {"title": "Intensive and Extensive Properties.txt", "text": "Okay, let's take this as our example again, the eraser. And let's think about how it was made. So it was made in different factories across the world, right? Suppose Factory A is in China. Factory B is in California. And suppose that these factories have different methods of creating these erasers."}, {"title": "Intensive and Extensive Properties.txt", "text": "Suppose Factory A is in China. Factory B is in California. And suppose that these factories have different methods of creating these erasers. Okay? Suppose a factory A divides this in half or creates two erasers, and then it combines them in some machine, okay? And then suppose that the factory in California takes five different small pieces of the eraser and then combines them at the end, okay?"}, {"title": "Intensive and Extensive Properties.txt", "text": "Okay? Suppose a factory A divides this in half or creates two erasers, and then it combines them in some machine, okay? And then suppose that the factory in California takes five different small pieces of the eraser and then combines them at the end, okay? The end result is the same. The end result is an erasure approximately the size, this weight, this volume, the same dimensions. The pathway by which it was created differs from factory A and factory B."}, {"title": "Intensive and Extensive Properties.txt", "text": "The end result is the same. The end result is an erasure approximately the size, this weight, this volume, the same dimensions. The pathway by which it was created differs from factory A and factory B. But the end result is the same. So let's talk about mass and why. Mass is a state function or a function or a property that does not depend on the pathway taken."}, {"title": "Intensive and Extensive Properties.txt", "text": "But the end result is the same. So let's talk about mass and why. Mass is a state function or a function or a property that does not depend on the pathway taken. It only depends on the system at hand, the current system at hand. So let's take our eraser again, okay? This is our eraser."}, {"title": "Intensive and Extensive Properties.txt", "text": "It only depends on the system at hand, the current system at hand. So let's take our eraser again, okay? This is our eraser. And let's say our eraser is 500 grams, okay? So that's its mass. Suppose once again, factory A takes a 250 grams part and a 250 grams part, okay?"}, {"title": "Intensive and Extensive Properties.txt", "text": "And let's say our eraser is 500 grams, okay? So that's its mass. Suppose once again, factory A takes a 250 grams part and a 250 grams part, okay? And suppose that it uses glue. It glues this thing and it creates this eraser, okay? That's how factory A makes it."}, {"title": "Intensive and Extensive Properties.txt", "text": "And suppose that it uses glue. It glues this thing and it creates this eraser, okay? That's how factory A makes it. What is the end result? The end result is a 500 grams eraser. Now suppose factory B makes the eraser, but it makes it different."}, {"title": "Intensive and Extensive Properties.txt", "text": "What is the end result? The end result is a 500 grams eraser. Now suppose factory B makes the eraser, but it makes it different. It makes it in sections of 100 grams, okay? So it basically takes five sections of 100 grams each and glues them together and it forms our 500 grams eraser, right? The end result is also a 500 grams eraser."}, {"title": "Intensive and Extensive Properties.txt", "text": "It makes it in sections of 100 grams, okay? So it basically takes five sections of 100 grams each and glues them together and it forms our 500 grams eraser, right? The end result is also a 500 grams eraser. So the end result, the end mass of the eraser is identical. It's the same matter what path taken. No matter how you create their eraser, if you took 100 grams and put them together, and you put 250 grams and put them together, the pathways differ."}, {"title": "Base Ionization Constant .txt", "text": "In this lecture we're going to talk about ionization constants of basis known as KB. But before we talk about this guy, we really have to understand these two components. So, if you have already done so, watch the video below on Otoinisation of water and Annetzation of acids. Now, so let's begin. Let's suppose we have some hypothetical base, let's call it A. And this base in aqueous state reacts with a single water molecule in a liquid state."}, {"title": "Base Ionization Constant .txt", "text": "Now, so let's begin. Let's suppose we have some hypothetical base, let's call it A. And this base in aqueous state reacts with a single water molecule in a liquid state. So what will happen? Well, our base will act as a base trying to take that ho away from our acid, namely our water molecule. And these guys will create a conjugate acid ha and a conjugate base, our hydroxide molecule."}, {"title": "Base Ionization Constant .txt", "text": "So what will happen? Well, our base will act as a base trying to take that ho away from our acid, namely our water molecule. And these guys will create a conjugate acid ha and a conjugate base, our hydroxide molecule. Both guys are in aqueous state. The same way we wrote equilibrium expressions for acid and for water, we can also write equilibrium expressions for bases as well. Except now we replace our ka and Kw with KB or Dionization constant for our base."}, {"title": "Base Ionization Constant .txt", "text": "Both guys are in aqueous state. The same way we wrote equilibrium expressions for acid and for water, we can also write equilibrium expressions for bases as well. Except now we replace our ka and Kw with KB or Dionization constant for our base. So KB is equal to the conjugate acid or the concentration of the conjugate acid times the concentration of our hydroxide divided by our conjugate base. So, in terms of this reaction, it's this guy. The concentration of this guy times the concentration of this guy."}, {"title": "Base Ionization Constant .txt", "text": "So KB is equal to the conjugate acid or the concentration of the conjugate acid times the concentration of our hydroxide divided by our conjugate base. So, in terms of this reaction, it's this guy. The concentration of this guy times the concentration of this guy. So our products go to the denominator over our conjugate base, this guy. So our reactants go on the bottom. Our products go to the top just like they would for acids and for water."}, {"title": "Base Ionization Constant .txt", "text": "So our products go to the denominator over our conjugate base, this guy. So our reactants go on the bottom. Our products go to the top just like they would for acids and for water. Now, the same way we spoke about ka's being ratios in terms of ionization of acids, we could also talk about KBS or ionization of bases being ratios. Their ratios of amount of product formed over the amount of react is left. So that means the greater our value for KB is, the more favorable our reaction is this way."}, {"title": "Base Ionization Constant .txt", "text": "Now, the same way we spoke about ka's being ratios in terms of ionization of acids, we could also talk about KBS or ionization of bases being ratios. Their ratios of amount of product formed over the amount of react is left. So that means the greater our value for KB is, the more favorable our reaction is this way. And this means the better or stronger our base is. If this guy is greater than one, we can say that it's a strong base. If it's less than one, it's a weak base."}, {"title": "Base Ionization Constant .txt", "text": "And this means the better or stronger our base is. If this guy is greater than one, we can say that it's a strong base. If it's less than one, it's a weak base. And that's because if it's less than one, that means amount of reacting left is much greater than the amount of product formed. And that means this guy hasn't really reacted yet. And that's because our base is a poor base."}, {"title": "Base Ionization Constant .txt", "text": "And that's because if it's less than one, that means amount of reacting left is much greater than the amount of product formed. And that means this guy hasn't really reacted yet. And that's because our base is a poor base. It's not very good at what it does. It's not very good at taking that H away from that acid. So once again, if a KB is high, then that means it's a good base."}, {"title": "Base Ionization Constant .txt", "text": "It's not very good at what it does. It's not very good at taking that H away from that acid. So once again, if a KB is high, then that means it's a good base. If it's low, it's a bad base. Now, let's see what happens when we multiply ka times KB or Dionization cost of assets times Dionization constant basis. Well, let's rewrite Ka."}, {"title": "Base Ionization Constant .txt", "text": "If it's low, it's a bad base. Now, let's see what happens when we multiply ka times KB or Dionization cost of assets times Dionization constant basis. Well, let's rewrite Ka. Or let's first rewrite KB. Well, KB is simply this whole guy here. So Ha multiplied by oh of the concentration divided by concentration of eight."}, {"title": "Base Ionization Constant .txt", "text": "Or let's first rewrite KB. Well, KB is simply this whole guy here. So Ha multiplied by oh of the concentration divided by concentration of eight. Now, if we go back to the video for ionization of acids, which you'll find right here, you'll notice that ka is equal to this guy times the hydronium concentration divided by Ha or the concentration of ha. Now let's see what happens. Well, this guy and this guy cancel, right?"}, {"title": "Base Ionization Constant .txt", "text": "Now, if we go back to the video for ionization of acids, which you'll find right here, you'll notice that ka is equal to this guy times the hydronium concentration divided by Ha or the concentration of ha. Now let's see what happens. Well, this guy and this guy cancel, right? So this guy not the bad marker. This guy cancels, and this guy cancels. Also this guy cancels, and this guy cancels."}, {"title": "Base Ionization Constant .txt", "text": "So this guy not the bad marker. This guy cancels, and this guy cancels. Also this guy cancels, and this guy cancels. So we're left with hydronium concentration times the hydroxide concentration. And if you go back to the video for Autoinization of War, which you'll find right here, you'll see that this guy simply equals kw. So we see that when we multiply ka times KB, what we get is kw."}, {"title": "Base Ionization Constant .txt", "text": "So we're left with hydronium concentration times the hydroxide concentration. And if you go back to the video for Autoinization of War, which you'll find right here, you'll see that this guy simply equals kw. So we see that when we multiply ka times KB, what we get is kw. So this is an important relation. Now, what happens if we take the log of both sides? Well, if we take the log of both sides, we get PKW is equal to PKA plus PKB."}, {"title": "Base Ionization Constant .txt", "text": "So this is an important relation. Now, what happens if we take the log of both sides? Well, if we take the log of both sides, we get PKW is equal to PKA plus PKB. Remember? According to the laws of logs, when we take the log of this, this becomes addition, right? And this equals 14."}, {"title": "Base Ionization Constant .txt", "text": "Remember? According to the laws of logs, when we take the log of this, this becomes addition, right? And this equals 14. Now, if you're not sure where this came from, how I got from this point to this point, check out the video below. At the end of the video, I talk about the laws of logs. And I show you how I go from this to this guy, or something similar to this to this guy."}, {"title": "Naming of Alkanes.txt", "text": "Rule number one, find the longest straight chain of carbons. So, to demonstrate this rule, let's look at a few examples. So, let's create an alkane, and then then let's try to find the longest straight chain of carbons on that alkane. So let's suppose we have an alkane that looks like that. Okay, so we want to find the longest straight chain of carbons. So let's begin counting our carbons."}, {"title": "Naming of Alkanes.txt", "text": "So let's suppose we have an alkane that looks like that. Okay, so we want to find the longest straight chain of carbons. So let's begin counting our carbons. So one carbon, two carbon, three carbon. Now, since we want to find the longest straight chain of carbons, we have to choose in which direction we want to go to find that longer straight chain. Well, if we choose this way, we get four."}, {"title": "Naming of Alkanes.txt", "text": "So one carbon, two carbon, three carbon. Now, since we want to find the longest straight chain of carbons, we have to choose in which direction we want to go to find that longer straight chain. Well, if we choose this way, we get four. If we choose this way, we get five. So we go this way, four and five. So, this is the longest straight chain of carbons, and this is called the carbon backbone."}, {"title": "Naming of Alkanes.txt", "text": "If we choose this way, we get five. So we go this way, four and five. So, this is the longest straight chain of carbons, and this is called the carbon backbone. And we name our alkane. The last part of our name for the alkane should refer to this carbon backbone. So that means we'll have a pentane."}, {"title": "Naming of Alkanes.txt", "text": "And we name our alkane. The last part of our name for the alkane should refer to this carbon backbone. So that means we'll have a pentane. Now, we're not done with naming this molecule. We'll come back to it after rule number two. So penth means five."}, {"title": "Naming of Alkanes.txt", "text": "Now, we're not done with naming this molecule. We'll come back to it after rule number two. So penth means five. Aim means it's an alkane. So let's come up with a second alkane. So, now let's suppose we have the following."}, {"title": "Naming of Alkanes.txt", "text": "Aim means it's an alkane. So let's come up with a second alkane. So, now let's suppose we have the following. So, let's say we have one more carbon. So let's begin naming in the same manner. So, one, two, three."}, {"title": "Naming of Alkanes.txt", "text": "So, let's say we have one more carbon. So let's begin naming in the same manner. So, one, two, three. Now, we want to go this way. Why? Well, because if we go this way, we'll end at four."}, {"title": "Naming of Alkanes.txt", "text": "Now, we want to go this way. Why? Well, because if we go this way, we'll end at four. If we go this way, we'll end at 6545 and six. So, once again, our longest carbon backbone has six. So that means we write Hexane."}, {"title": "Naming of Alkanes.txt", "text": "If we go this way, we'll end at 6545 and six. So, once again, our longest carbon backbone has six. So that means we write Hexane. Hex simply means six, and A means it's an alkane. So let's get another alkane in. So let's say we have this one."}, {"title": "Naming of Alkanes.txt", "text": "Hex simply means six, and A means it's an alkane. So let's get another alkane in. So let's say we have this one. So let's say we have the following alkane. So, once again, let's begin labeling our carbons numbering our carbon. So one carbon, two, three, four."}, {"title": "Naming of Alkanes.txt", "text": "So let's say we have the following alkane. So, once again, let's begin labeling our carbons numbering our carbon. So one carbon, two, three, four. If we go up, we end at six. If we go sideways, we end at seven. So we choose sideways."}, {"title": "Naming of Alkanes.txt", "text": "If we go up, we end at six. If we go sideways, we end at seven. So we choose sideways. So, our longest carbon backbone has seven carbons, and so we name it Heptane. hept means seven. A means how can."}, {"title": "Naming of Alkanes.txt", "text": "So, our longest carbon backbone has seven carbons, and so we name it Heptane. hept means seven. A means how can. Now, once again, we're not done naming these guys. We'll come back to these guys after rule number two. So let's look at rule number two."}, {"title": "Naming of Alkanes.txt", "text": "Now, once again, we're not done naming these guys. We'll come back to these guys after rule number two. So let's look at rule number two. In substituted alkanes, the substituent is given a number based on the position on that carbon backbone, and the number value should be as low as possible. So let's go to example number one. So, our substituent is this methyl group."}, {"title": "Naming of Alkanes.txt", "text": "In substituted alkanes, the substituent is given a number based on the position on that carbon backbone, and the number value should be as low as possible. So let's go to example number one. So, our substituent is this methyl group. Now, if we start from this side, we go one, two, three. So, our methyl is on the third position of the carbon backbone. If we start from this side, it's also on the third."}, {"title": "Naming of Alkanes.txt", "text": "Now, if we start from this side, we go one, two, three. So, our methyl is on the third position of the carbon backbone. If we start from this side, it's also on the third. So this has symmetry. And so we simply write three methyl pentane. Example number two."}, {"title": "Naming of Alkanes.txt", "text": "So this has symmetry. And so we simply write three methyl pentane. Example number two. Now in this one, it's a bit more tricky because this doesn't have symmetry the same way that this one has. So if we begin on this side of our backbone, we go one, two, three. If we begin on that side, we get 1234."}, {"title": "Naming of Alkanes.txt", "text": "Now in this one, it's a bit more tricky because this doesn't have symmetry the same way that this one has. So if we begin on this side of our backbone, we go one, two, three. If we begin on that side, we get 1234. Remember, we want the lowest possible number, so we should start from this way. And that's exactly why I began labeling on this side of the chain of carbons. So since we have a methyl group, we have three methylhexane."}, {"title": "Naming of Alkanes.txt", "text": "Remember, we want the lowest possible number, so we should start from this way. And that's exactly why I began labeling on this side of the chain of carbons. So since we have a methyl group, we have three methylhexane. And finally, the last example, like this one, also has symmetry down the middle. It doesn't matter if we begin here or here, 1234-1234. So we write four, but now we have not a methyl, but an ethyl, because we have two carbons."}, {"title": "Naming of Alkanes.txt", "text": "And finally, the last example, like this one, also has symmetry down the middle. It doesn't matter if we begin here or here, 1234-1234. So we write four, but now we have not a methyl, but an ethyl, because we have two carbons. So that concludes these examples. Let's go to these two examples. So once again, according to rule number one, we want to find the longest chain of carbon."}, {"title": "Naming of Alkanes.txt", "text": "So that concludes these examples. Let's go to these two examples. So once again, according to rule number one, we want to find the longest chain of carbon. So we can either have 1234 or 1234. Now we want the lowest value for our substituent. So we begin here, 1234."}, {"title": "Naming of Alkanes.txt", "text": "So we can either have 1234 or 1234. Now we want the lowest value for our substituent. So we begin here, 1234. So 1234, and let's name them. Now, since it has four carbons, that means our backbone is butane. So we have butane for the backbone."}, {"title": "Naming of Alkanes.txt", "text": "So 1234, and let's name them. Now, since it has four carbons, that means our backbone is butane. So we have butane for the backbone. And then our second position, position two is a bromine. So we name it bromo. Bromo is short for bromine."}, {"title": "Naming of Alkanes.txt", "text": "And then our second position, position two is a bromine. So we name it bromo. Bromo is short for bromine. Now let's look at the the second one. Once again, this has symmetry. It doesn't matter if we begin here or here."}, {"title": "Naming of Alkanes.txt", "text": "Now let's look at the the second one. Once again, this has symmetry. It doesn't matter if we begin here or here. So 1234 and 1234. Either way, this is the fourth position on our carbon backbone. So let's begin 123456 and seven."}, {"title": "Naming of Alkanes.txt", "text": "So 1234 and 1234. Either way, this is the fourth position on our carbon backbone. So let's begin 123456 and seven. So it's a heptane and it's a four position substituted. So that means four chloroheptane, four chlorohessane. Okay, let's go to rule number three."}, {"title": "Naming of Alkanes.txt", "text": "So it's a heptane and it's a four position substituted. So that means four chloroheptane, four chlorohessane. Okay, let's go to rule number three. In multisubstituted alkanes, substituents are ordered alphabetically, and identical substituents receive prefix, die, try, tetra, et cetera. So let's look at the last three examples. So once again, we have to keep these two rules in mind."}, {"title": "Naming of Alkanes.txt", "text": "In multisubstituted alkanes, substituents are ordered alphabetically, and identical substituents receive prefix, die, try, tetra, et cetera. So let's look at the last three examples. So once again, we have to keep these two rules in mind. Our first goal is to find the longest straight chain of carbon, so 12345 or 123456. So we choose to go this way. And remember, since we have two substituents, we have a bromo and a methyl."}, {"title": "Naming of Alkanes.txt", "text": "Our first goal is to find the longest straight chain of carbon, so 12345 or 123456. So we choose to go this way. And remember, since we have two substituents, we have a bromo and a methyl. We want to follow an alphabetical rule. So that means since B comes before AMA, we have to start this way. So 1234, so 123456."}, {"title": "Naming of Alkanes.txt", "text": "We want to follow an alphabetical rule. So that means since B comes before AMA, we have to start this way. So 1234, so 123456. Okay, so we have a hexane. So our backbone is hexane, and we have three bromo and four methyl. I'm not going to have room, but this should be methyl."}, {"title": "Naming of Alkanes.txt", "text": "Okay, so we have a hexane. So our backbone is hexane, and we have three bromo and four methyl. I'm not going to have room, but this should be methyl. Okay, let's go to this. 11234 or 1234. Which way do we go?"}, {"title": "Naming of Alkanes.txt", "text": "Okay, let's go to this. 11234 or 1234. Which way do we go? Well, once again, we want the alphabetical rule. So since B comes before C, that means we start here. So let's start on that side."}, {"title": "Naming of Alkanes.txt", "text": "Well, once again, we want the alphabetical rule. So since B comes before C, that means we start here. So let's start on that side. We have 1234, and so we're going to have two bromo, three chloro, and then since we have four chain carbon backbone, we have butane. Okay, so now let's look at the final molecule. In this final molecule, we have two identical substituents, and that means we're going to want to use a diprefix."}, {"title": "Naming of Alkanes.txt", "text": "We have 1234, and so we're going to have two bromo, three chloro, and then since we have four chain carbon backbone, we have butane. Okay, so now let's look at the final molecule. In this final molecule, we have two identical substituents, and that means we're going to want to use a diprefix. So this is a terriblel substituent. So let's find our longest straight chain. So 1234-5678."}, {"title": "Naming of Alkanes.txt", "text": "So this is a terriblel substituent. So let's find our longest straight chain. So 1234-5678. So that means we're going to have an octane. So let's label 123-4567 and eight. So that means we're going to have an octane carbon backbone and we're going to have dye."}, {"title": "Naming of Alkanes.txt", "text": "So that means we're going to have an octane. So let's label 123-4567 and eight. So that means we're going to have an octane carbon backbone and we're going to have dye. So dye meaning two TURPE, butyl octane. And notice this is not done because we want to actually label on which carbon are our two substituents are located. So four and five."}, {"title": "Naming of Alkanes Examples .txt", "text": "Let's begin with example number one. So first we want to find the longest possible carbon backbone, so 1234. Or we can also have 1234. Either way, we have a four carbon backbone. So that means we're dealing with butane. Now, which way do we want to go starting for this carbon or this carbon?"}, {"title": "Naming of Alkanes Examples .txt", "text": "Either way, we have a four carbon backbone. So that means we're dealing with butane. Now, which way do we want to go starting for this carbon or this carbon? Well, remember, our substituents should have the lowest possible number. So that means we begin on this end. So 1234."}, {"title": "Naming of Alkanes Examples .txt", "text": "Well, remember, our substituents should have the lowest possible number. So that means we begin on this end. So 1234. And that means we're going to name it. So our substituents are both located on the second position. That means we have two, we have bromo, so dibromo."}, {"title": "Naming of Alkanes Examples .txt", "text": "And that means we're going to name it. So our substituents are both located on the second position. That means we have two, we have bromo, so dibromo. And then butane. So 22 simply means our two bromos are located on the second position. And our chain is a butane 1234."}, {"title": "Naming of Alkanes Examples .txt", "text": "And then butane. So 22 simply means our two bromos are located on the second position. And our chain is a butane 1234. So let's look at the second example. So once again, we want to find the longest possible carbon backbone. Carbon chain, 1234 or 12345."}, {"title": "Naming of Alkanes Examples .txt", "text": "So let's look at the second example. So once again, we want to find the longest possible carbon backbone. Carbon chain, 1234 or 12345. Now, since this guy is symmetrical, it doesn't matter if we start from this end or this end. So let's begin from this end. 12345"}, {"title": "Naming of Alkanes Examples .txt", "text": "Now, since this guy is symmetrical, it doesn't matter if we start from this end or this end. So let's begin from this end. 12345 And our substituents are both on the third carbon. So that means we're going to have three three dimethyl. They're both methyl compounds, and we have five."}, {"title": "Naming of Alkanes Examples .txt", "text": "And our substituents are both on the third carbon. So that means we're going to have three three dimethyl. They're both methyl compounds, and we have five. So that means we have pentane. Let's look at a third example. So once again, longest possible carbon backbone."}, {"title": "Naming of Alkanes Examples .txt", "text": "So that means we have pentane. Let's look at a third example. So once again, longest possible carbon backbone. One, two, three, or one, two, three, or one, two, three. Doesn't matter which way we go. So let's go straight across."}, {"title": "Naming of Alkanes Examples .txt", "text": "One, two, three, or one, two, three, or one, two, three. Doesn't matter which way we go. So let's go straight across. One, two, three. So three carbon backbone. That means we're dealing with a propane."}, {"title": "Naming of Alkanes Examples .txt", "text": "One, two, three. So three carbon backbone. That means we're dealing with a propane. And on the second physician, we have two methyl groups. So we have two two dimethyl propane. Let's look at the fourth example."}, {"title": "Naming of Alkanes Examples .txt", "text": "And on the second physician, we have two methyl groups. So we have two two dimethyl propane. Let's look at the fourth example. So here we have a very long carbon backbone. So we can either begin counting this way or this way or this way. So we can say 123-45-6789 or 1234-5678 910."}, {"title": "Naming of Alkanes Examples .txt", "text": "So here we have a very long carbon backbone. So we can either begin counting this way or this way or this way. So we can say 123-45-6789 or 1234-5678 910. So it looks like if we go along this carbon backbone, we're going to have the longest carbon backbone. So we can either start from this end and go all the way down, or from this end and go all the way down. Since we want our substituent to have the lowest possible number, we should probably start to this end."}, {"title": "Naming of Alkanes Examples .txt", "text": "So it looks like if we go along this carbon backbone, we're going to have the longest carbon backbone. So we can either start from this end and go all the way down, or from this end and go all the way down. Since we want our substituent to have the lowest possible number, we should probably start to this end. And so let's start labeling or numbering. 1234-5678, 910. Okay, so that means we have ten, so that's decay."}, {"title": "Naming of Alkanes Examples .txt", "text": "And so let's start labeling or numbering. 1234-5678, 910. Okay, so that means we have ten, so that's decay. And we have on the third carbon, we have so three methyl decay. All right, so let's look at the final example. So here we can start from either this position, this position, this guy, this guy, or this guy."}, {"title": "Naming of Alkanes Examples .txt", "text": "And we have on the third carbon, we have so three methyl decay. All right, so let's look at the final example. So here we can start from either this position, this position, this guy, this guy, or this guy. So if we start from this position, we have 1234-561-2345 or 123456, or we can have 123-4567. So if we start from this end or this end, it looks like we get the longest possible carbon backbone. So let's begin from this end."}, {"title": "Naming of Alkanes Examples .txt", "text": "So if we start from this position, we have 1234-561-2345 or 123456, or we can have 123-4567. So if we start from this end or this end, it looks like we get the longest possible carbon backbone. So let's begin from this end. So 123-4567. Okay, and now these guys are our substituents. So we have three substituents."}, {"title": "Naming of Alkanes Examples .txt", "text": "So 123-4567. Okay, and now these guys are our substituents. So we have three substituents. Two of them are identical, so they're methyl. And one of them is ethyl. So ethyl comes before methyl because E comes before M. So that means this guy will become will come before these two guys."}, {"title": "Naming of Alkanes Examples .txt", "text": "Two of them are identical, so they're methyl. And one of them is ethyl. So ethyl comes before methyl because E comes before M. So that means this guy will become will come before these two guys. So let's first make sure we have seven. So that means we have Heftane. And we're going to start with one, two so four."}, {"title": "Naming of Alkanes Examples .txt", "text": "So let's first make sure we have seven. So that means we have Heftane. And we're going to start with one, two so four. So we're going to have poor Ethyl. Then we're going to have two, three, so two, three, dimethyl. Four, Ethyl."}, {"title": "Naming of Alkanes Examples .txt", "text": "So we're going to have poor Ethyl. Then we're going to have two, three, so two, three, dimethyl. Four, Ethyl. Two, three, dimethyl. Heptane. So we have our seven chain carbon backbone, two identical 23 dimethyls, and we have one ethyl."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So here we have parts of our periodic table. So we have hydrogen, lithium, sodium. We have beryllium and magnesium. And this side we have have three noble gases fluorine, fluorine. And we have oxygen and sulfur. So let's look at what ionization energy of an atom is."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And this side we have have three noble gases fluorine, fluorine. And we have oxygen and sulfur. So let's look at what ionization energy of an atom is. Now, according to my definition, ionization energy is the energy required to remove an electron from an atom. So to demonstrate that, we're going to use the following Beryllium atom. So Beryllium has 12344 Protons."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Now, according to my definition, ionization energy is the energy required to remove an electron from an atom. So to demonstrate that, we're going to use the following Beryllium atom. So Beryllium has 12344 Protons. So there are four protons within the nucleus, also four neutrons. And since our neutral atom has a charge of zero, that means if we have four protons, we must have four electrons. So two electrons are placed into the one s orbital, and two electrons are placed into the two s orbital."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So there are four protons within the nucleus, also four neutrons. And since our neutral atom has a charge of zero, that means if we have four protons, we must have four electrons. So two electrons are placed into the one s orbital, and two electrons are placed into the two s orbital. So what this basically means is that in order to remove an electron from the outermost shell of my Beryllium atom, I must input energy. I must do work on my atom to take that electron away. So if I input enough energy, I will be able to pluck this outermost electron away to form the following Beryllium atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So what this basically means is that in order to remove an electron from the outermost shell of my Beryllium atom, I must input energy. I must do work on my atom to take that electron away. So if I input enough energy, I will be able to pluck this outermost electron away to form the following Beryllium atom. Now, the number of protons is the same. Before, we have four protons, and now we have four protons. So that means that we still have a Beryllium atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Now, the number of protons is the same. Before, we have four protons, and now we have four protons. So that means that we still have a Beryllium atom. But since we took that electron away, that means that we're going to have a cation. We're going to have a net or an overall positive charge, because now we have four protons each have a plus one charge. But three electrons each have a negative one charge."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "But since we took that electron away, that means that we're going to have a cation. We're going to have a net or an overall positive charge, because now we have four protons each have a plus one charge. But three electrons each have a negative one charge. So negative three plus four gives us a positive one charge. So this barrel has a plus one charge. And this is now a cat ion."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So negative three plus four gives us a positive one charge. So this barrel has a plus one charge. And this is now a cat ion. Now, my next question is where did this energy go? Well, this energy actually went into my atom system, into this atom. So that means that this cation has more energy than this neutral species, neutral Beryllium atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Now, my next question is where did this energy go? Well, this energy actually went into my atom system, into this atom. So that means that this cation has more energy than this neutral species, neutral Beryllium atom. So that means because this has a lower energy, this is more stable than this cation. So once again, ionization energy is the energy required to take away an electron from a neutral, in this case, a neutral atom. So on my periodic table, the ionization energies are labeled with the green numerical value."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So that means because this has a lower energy, this is more stable than this cation. So once again, ionization energy is the energy required to take away an electron from a neutral, in this case, a neutral atom. So on my periodic table, the ionization energies are labeled with the green numerical value. So for example, for lithium, well, actually, let's take Beryllium. Beryllium requires 9.32\nenergy to take away this electron. And this energy goes in to this atom, forming this cation."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So for example, for lithium, well, actually, let's take Beryllium. Beryllium requires 9.32\nenergy to take away this electron. And this energy goes in to this atom, forming this cation. This beryllium positively charged species. So now let's talk about electron affinity. Electron affinity is the energy that is released when an electron is added into my neutral or whatever type of atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "This beryllium positively charged species. So now let's talk about electron affinity. Electron affinity is the energy that is released when an electron is added into my neutral or whatever type of atom. So now let's work backwards. Let's start from this case. Suppose we have this same positively charged Beryllium atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So now let's work backwards. Let's start from this case. Suppose we have this same positively charged Beryllium atom. And let's suppose I take an electron and I put the electron back into my outermost shell of this positively charged cation. What I get is the following beryllium atom. This is a neutral beryllium atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And let's suppose I take an electron and I put the electron back into my outermost shell of this positively charged cation. What I get is the following beryllium atom. This is a neutral beryllium atom. Now, notice what happened here when we were going in this direction, I needed to input energy into my atom to take away that electron. And I said that this had a higher energy than this atom. So now we're working backwards."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Now, notice what happened here when we were going in this direction, I needed to input energy into my atom to take away that electron. And I said that this had a higher energy than this atom. So now we're working backwards. Now we have an atom, a beryllium atom that has a positive charge that is higher in energy than this product, than this neutral beryllium. So, in other words, when I take an electron, I put the electron inside my atom. Energy is released."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Now we have an atom, a beryllium atom that has a positive charge that is higher in energy than this product, than this neutral beryllium. So, in other words, when I take an electron, I put the electron inside my atom. Energy is released. And this energy is known as the electron affinity. Okay? And I labeled the electron affinity values with the brown color."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And this energy is known as the electron affinity. Okay? And I labeled the electron affinity values with the brown color. So, for example, for Beryllium, beryllium has approximately a zero for electron affinity. So it's very, very small. And that basically means that Beryllium does not like to gain electrons."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So, for example, for Beryllium, beryllium has approximately a zero for electron affinity. So it's very, very small. And that basically means that Beryllium does not like to gain electrons. Okay. And this concept will become important in a second when we talk about ionic bonds. So let's see the conclusion."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Okay. And this concept will become important in a second when we talk about ionic bonds. So let's see the conclusion. So atoms that have very low ionization energies. That basically means that electrons can be easily removed from those atoms. In other words, if we go to this table and we find atoms that have low ionization energies, for example, lithium lithium has 5.39 energy of ionization energy, while Fluorine, for example, has almost or has three times the value 17.42."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So atoms that have very low ionization energies. That basically means that electrons can be easily removed from those atoms. In other words, if we go to this table and we find atoms that have low ionization energies, for example, lithium lithium has 5.39 energy of ionization energy, while Fluorine, for example, has almost or has three times the value 17.42. That means that Lithium is much more prone to losing an electron than its fluorine. Same thing goes for sodium and same thing goes for chlorine. Chlorine has a higher value, higher ionization energy than sodium."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "That means that Lithium is much more prone to losing an electron than its fluorine. Same thing goes for sodium and same thing goes for chlorine. Chlorine has a higher value, higher ionization energy than sodium. And that means that sodium will be more likely to lose that electron compared to Chlorine. So let's go back to this guy. High electron affinity means that atoms will easily accept electrons."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And that means that sodium will be more likely to lose that electron compared to Chlorine. So let's go back to this guy. High electron affinity means that atoms will easily accept electrons. So if we go back to this table, we see that the electron affinity is labeled with the brown color. And notice that for flooring, electro affinity is 3.3, while the electron affinity for lithium is year zero 62. So because the electron affinity is much higher for fluorine, for chlorine, that means that these guys will be much more likely to gain an electron than will lithium or sodium."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So if we go back to this table, we see that the electron affinity is labeled with the brown color. And notice that for flooring, electro affinity is 3.3, while the electron affinity for lithium is year zero 62. So because the electron affinity is much higher for fluorine, for chlorine, that means that these guys will be much more likely to gain an electron than will lithium or sodium. So let's see what ionic bonds are. So ionic bonds. Now atoms with very low ionization energies."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So let's see what ionic bonds are. So ionic bonds. Now atoms with very low ionization energies. That basically means that very small green values. So atoms on the left side of the periodic table will form bonds with atoms that have high electron affinities. So atoms that have very high brown numbers."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "That basically means that very small green values. So atoms on the left side of the periodic table will form bonds with atoms that have high electron affinities. So atoms that have very high brown numbers. So notice that fluorine, chlorine, oxygen and sulfur all have much higher values than brown values compared to this side, where we have zero point 62 for lithium, almost zero for both magnesium Beryllium and 00:55 for sodium. In other words, ionic bonds will only form between atoms found on the left side and atoms found on the right side of our periodic table. Now notice I didn't label anything for our noble gases."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So notice that fluorine, chlorine, oxygen and sulfur all have much higher values than brown values compared to this side, where we have zero point 62 for lithium, almost zero for both magnesium Beryllium and 00:55 for sodium. In other words, ionic bonds will only form between atoms found on the left side and atoms found on the right side of our periodic table. Now notice I didn't label anything for our noble gases. And that's because noble gases, as we know, have perfect electron configuration of electrons. And that means that they won't like to gain electrons. But they won't like to lose electrons either."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And that's because noble gases, as we know, have perfect electron configuration of electrons. And that means that they won't like to gain electrons. But they won't like to lose electrons either. So let's take an example. Let's see how an ionic bond is in fact formed and how are the atoms actually held. What forces hold these two atoms together in an ionic bond?"}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So let's take an example. Let's see how an ionic bond is in fact formed and how are the atoms actually held. What forces hold these two atoms together in an ionic bond? So let's look at a neutral lithium atom. And let's look at a neutral fluorine atom. So we're taking one atom found on the left side and one atom found on the right side."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So let's look at a neutral lithium atom. And let's look at a neutral fluorine atom. So we're taking one atom found on the left side and one atom found on the right side. So we're taking lithium, which has one, two, three protons. So three protons are in the nucleus, and that means if it's a mutual atom, it must have the same number of electrons. So three electrons."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So we're taking lithium, which has one, two, three protons. So three protons are in the nucleus, and that means if it's a mutual atom, it must have the same number of electrons. So three electrons. Two goes into the one s orbital and one goes into the two s orbital. So now we're taking fluorine. Fluorine has nine protons, so it must have nine electrons for it to be a mutual atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Two goes into the one s orbital and one goes into the two s orbital. So now we're taking fluorine. Fluorine has nine protons, so it must have nine electrons for it to be a mutual atom. So that means two go into the one s orbital, two goes into the two s orbital and five go into our two p orbital. Okay, so what happens? Well, we set an ionic bond is formed between atoms that have low ionization energies and high electron athenity energy."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So that means two go into the one s orbital, two goes into the two s orbital and five go into our two p orbital. Okay, so what happens? Well, we set an ionic bond is formed between atoms that have low ionization energies and high electron athenity energy. So low ionization energies. That means that very little energy is required to take this electron away from the outermost shell, while this flooring atom has a high electron affinity, which means that it is very likely that it will gain an electron. So what happens is one of these electrons found on the outermost shell is removed, and this electron is placed on the outermost shell in the two p orbital of the fluorine atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So low ionization energies. That means that very little energy is required to take this electron away from the outermost shell, while this flooring atom has a high electron affinity, which means that it is very likely that it will gain an electron. So what happens is one of these electrons found on the outermost shell is removed, and this electron is placed on the outermost shell in the two p orbital of the fluorine atom. And now what happens is, since this still has three protons, but it now has two electrons, this develops a positive charge. While this Fluorine atom, which Had Nine protons, nine Electrons, now has nine protons, ten electrons. So it develops a negative charge."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And now what happens is, since this still has three protons, but it now has two electrons, this develops a positive charge. While this Fluorine atom, which Had Nine protons, nine Electrons, now has nine protons, ten electrons. So it develops a negative charge. And let's recall what Coulomb's Law tells us. So, from physics, we know that Coulomb's law simply states that constant K times charge one times charge two divided by the distance between the center of charges squared, gives us the force due to a charge on another charge. So here we have one charge species a cation."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And let's recall what Coulomb's Law tells us. So, from physics, we know that Coulomb's law simply states that constant K times charge one times charge two divided by the distance between the center of charges squared, gives us the force due to a charge on another charge. So here we have one charge species a cation. And here we have a second charge. Species an ion. And because these guys have now developed charges, different charges."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And here we have a second charge. Species an ion. And because these guys have now developed charges, different charges. These charges, positive or negative, will attract each other according to Coulomb's law. So we basically find the distance between their sensor charges. We plug that in here."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "These charges, positive or negative, will attract each other according to Coulomb's law. So we basically find the distance between their sensor charges. We plug that in here. We plug in the charges for both species. We multiply by the constant and we get the force that each atom feels due to the other atoms. So because there is charge now."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "We plug in the charges for both species. We multiply by the constant and we get the force that each atom feels due to the other atoms. So because there is charge now. There is a force between them. And this force holds them together. And this is known as an ionic bond."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "There is a force between them. And this force holds them together. And this is known as an ionic bond. So, once again, to recap an Ionic bond is a bond between a count ion and an anion. And what that basically means that neutral charges or neutral atoms will not be able to form ionic bonds. In order to form an ionic bond, you have to have a positive atom and a negative atom."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "So, once again, to recap an Ionic bond is a bond between a count ion and an anion. And what that basically means that neutral charges or neutral atoms will not be able to form ionic bonds. In order to form an ionic bond, you have to have a positive atom and a negative atom. And the only way ionic bonds form is if you mix atoms from this side, from the left side of the periodic table with atoms on the right side of the periodic table. So one last important thing that I'd like to mention. So let's look at the electron configuration of these two atoms."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "And the only way ionic bonds form is if you mix atoms from this side, from the left side of the periodic table with atoms on the right side of the periodic table. So one last important thing that I'd like to mention. So let's look at the electron configuration of these two atoms. Notice what happened. Remember we said that these guys, the noble gases, have perfect electron configurations. All their shells are filled."}, {"title": "Ionic Bonds, Ionization Energy, Electron Affinity.txt", "text": "Notice what happened. Remember we said that these guys, the noble gases, have perfect electron configurations. All their shells are filled. So let's look at what happened to the electro configuration of lithium and fluorine. Well, when lithium lost and lost an electron, right? When it lost an electron, it gained the electron configuration of helium, noble gas, while fluorine gained an electron and it formed the electron configuration of neon."}, {"title": "Graham\u2019s Law .txt", "text": "So earlier we spoke about the concept of the kinetic theory of ideal gases. And what this guy is is basically a bunch of assumptions made by scientists to help us explain how gases function and how they behave. They help us develop a better understanding of gas functionality. Now, one thing that we deduced from the kinetic theory was that kinetic energy, then your molecule, is directly proportional to temperature. And that means if we have a gas system, a system of gas molecules, if we increase the overall temperature of our system, then on average, every molecule will have a higher kinetic energy. And that's because kinetic energy is proportional to temperature."}, {"title": "Graham\u2019s Law .txt", "text": "Now, one thing that we deduced from the kinetic theory was that kinetic energy, then your molecule, is directly proportional to temperature. And that means if we have a gas system, a system of gas molecules, if we increase the overall temperature of our system, then on average, every molecule will have a higher kinetic energy. And that's because kinetic energy is proportional to temperature. So what happens in a system of gas molecules that is under constant temperature? Well, then, on average, the kinetic energy of any gas molecule, say gas molecule one, is the same as the kinetic energy, gas molecule number two. So let's equate these guys and let's see where that leads us."}, {"title": "Graham\u2019s Law .txt", "text": "So what happens in a system of gas molecules that is under constant temperature? Well, then, on average, the kinetic energy of any gas molecule, say gas molecule one, is the same as the kinetic energy, gas molecule number two. So let's equate these guys and let's see where that leads us. Let's suppose the kinetic energy of some gas molecule one is equal to kinetic energy of gas molecule two because they have the same temperature. Now, let's also assume that gas molecule two has a larger mass than gas molecule one. So gas molecule two is heavier."}, {"title": "Graham\u2019s Law .txt", "text": "Let's suppose the kinetic energy of some gas molecule one is equal to kinetic energy of gas molecule two because they have the same temperature. Now, let's also assume that gas molecule two has a larger mass than gas molecule one. So gas molecule two is heavier. So let's rewrite these guys in terms of our mathematical formula for kinetic energy, namely one two, MV squared. The one subscript simply represents gas molecule one, and the two subscript simply represents gas molecule two. So blue is one, red is two."}, {"title": "Graham\u2019s Law .txt", "text": "So let's rewrite these guys in terms of our mathematical formula for kinetic energy, namely one two, MV squared. The one subscript simply represents gas molecule one, and the two subscript simply represents gas molecule two. So blue is one, red is two. So one two m one, V one squared is equal to one two M two V two squared. So let's cancel out the twos by multiplying each side by two. And then let's bring our V two over and our m one over this way."}, {"title": "Graham\u2019s Law .txt", "text": "So one two m one, V one squared is equal to one two M two V two squared. So let's cancel out the twos by multiplying each side by two. And then let's bring our V two over and our m one over this way. So we want velocities on the left side and masses on the right side. We get the following formula v one squared divided by V two squared equals m two squared over m one squared. And finally, let's take the square root of both sides."}, {"title": "Graham\u2019s Law .txt", "text": "So we want velocities on the left side and masses on the right side. We get the following formula v one squared divided by V two squared equals m two squared over m one squared. And finally, let's take the square root of both sides. What we get is these twos cancel. Well, that's v one over V two. And then we have the square root of m two divided by square root of m one."}, {"title": "Graham\u2019s Law .txt", "text": "What we get is these twos cancel. Well, that's v one over V two. And then we have the square root of m two divided by square root of m one. Now, what does this tell us? Well, to gain a better understanding about what this tells us, we have to talk about a concept called effusion. But before we talk about that, notice that velocity one will be greater if m one is smaller."}, {"title": "Graham\u2019s Law .txt", "text": "Now, what does this tell us? Well, to gain a better understanding about what this tells us, we have to talk about a concept called effusion. But before we talk about that, notice that velocity one will be greater if m one is smaller. And likewise, velocity two will be greater if m two is smaller. And we'll see that more clearly in the process of infusion. So infusion is the movement of gas particles or molecules from high pressure to low pressure via a very, very tiny hole."}, {"title": "Graham\u2019s Law .txt", "text": "And likewise, velocity two will be greater if m two is smaller. And we'll see that more clearly in the process of infusion. So infusion is the movement of gas particles or molecules from high pressure to low pressure via a very, very tiny hole. So to imagine this situation, let's look at this illustration. Suppose we have a system, a Q. And we have a very tiny hole in this queue."}, {"title": "Graham\u2019s Law .txt", "text": "So to imagine this situation, let's look at this illustration. Suppose we have a system, a Q. And we have a very tiny hole in this queue. And we have lots of molecules found inside our queue. So we have two molecules. The red molecules and the blue molecules."}, {"title": "Graham\u2019s Law .txt", "text": "And we have lots of molecules found inside our queue. So we have two molecules. The red molecules and the blue molecules. Now, notice we have a high pressure on the inside and low pressure on the outside because our outside, we're assuming, is a vacuum. Now, we make a very tiny incision, a very tiny hole. And the diameter of this hole is much smaller than the distance between a two molecules."}, {"title": "Graham\u2019s Law .txt", "text": "Now, notice we have a high pressure on the inside and low pressure on the outside because our outside, we're assuming, is a vacuum. Now, we make a very tiny incision, a very tiny hole. And the diameter of this hole is much smaller than the distance between a two molecules. That's our assumption. So eventually, some of these molecules will hit the walls of the container. At any given time, some of these molecules will hit the wall at exactly the spot where this hole is, and they will exit or escape the cube."}, {"title": "Graham\u2019s Law .txt", "text": "That's our assumption. So eventually, some of these molecules will hit the walls of the container. At any given time, some of these molecules will hit the wall at exactly the spot where this hole is, and they will exit or escape the cube. What this formula tells us is that the higher your velocity is, the more likely that you will collide with the wall at this position, at this point in time, and the more likely you will escape into the vacuum. So we can rewrite these velocities as rates. So the rate of effusion for gas one divided by the rate of infusion for gas two is equal to the square root of mass one divided by the square root of mass two."}, {"title": "Graham\u2019s Law .txt", "text": "What this formula tells us is that the higher your velocity is, the more likely that you will collide with the wall at this position, at this point in time, and the more likely you will escape into the vacuum. So we can rewrite these velocities as rates. So the rate of effusion for gas one divided by the rate of infusion for gas two is equal to the square root of mass one divided by the square root of mass two. And what this equation tells us, which is by the way, grams equation or Gram's law, is that the heavier the molecule is, the smaller its speed and therefore, the smaller its rate. And our individual proportion is the following rate is directly proportional to one over square root of mass. In other words, if our M is larger, then the denominator becomes larger, and so our fracture becomes smaller."}, {"title": "Graham\u2019s Law .txt", "text": "And what this equation tells us, which is by the way, grams equation or Gram's law, is that the heavier the molecule is, the smaller its speed and therefore, the smaller its rate. And our individual proportion is the following rate is directly proportional to one over square root of mass. In other words, if our M is larger, then the denominator becomes larger, and so our fracture becomes smaller. And so this guy is smaller. Likewise, if if M is smaller, if it's less heavier, then this fraction becomes smaller, and one over a smaller number is a larger number. And that means our rate is larger."}, {"title": "Ideal Gas Law .txt", "text": "We have spoken about three gas laws that help us explain in their own individual way how gas molecules function on a macroscopic scale in the macroscopic gas system. Now, we've looked at Boyle's Law which answers questions such as why do balloons blow up or explode when you compress them by increasing pressure? We've spoken about Charles Law which answers questions such as why do cartels deflate in the winter and inflate or expand in the summer? And we've also looked at Avocadoso's Law which helps explain questions such as why do balloons become larger when you blow an air? Now, all these three laws help explain macroscopic concepts of gases. Now, now we are ready to combine all three laws into a single super law that puts all three relationships into a single formula, into a single equation."}, {"title": "Ideal Gas Law .txt", "text": "And we've also looked at Avocadoso's Law which helps explain questions such as why do balloons become larger when you blow an air? Now, all these three laws help explain macroscopic concepts of gases. Now, now we are ready to combine all three laws into a single super law that puts all three relationships into a single formula, into a single equation. And this equation is known as the ideal gas law. Now, recall that pressure is inversely proportional to volume which is exactly what this guy says. Pressure is inverse proportional to volume and directly proportional to N and T.\nThat's exactly what Charles Law in Evagondjo's Law tells us."}, {"title": "Ideal Gas Law .txt", "text": "And this equation is known as the ideal gas law. Now, recall that pressure is inversely proportional to volume which is exactly what this guy says. Pressure is inverse proportional to volume and directly proportional to N and T.\nThat's exactly what Charles Law in Evagondjo's Law tells us. Now, the only problem in this top portion is that we want to go from a proportionality sign to an equal sign. The way we adjust this guy, this top guy, so that we can equal them is by multiplying our right side or adjusting our right side by some constant R.\nThis is known as the ideal gas constant. So what problem remains?"}, {"title": "Ideal Gas Law .txt", "text": "Now, the only problem in this top portion is that we want to go from a proportionality sign to an equal sign. The way we adjust this guy, this top guy, so that we can equal them is by multiplying our right side or adjusting our right side by some constant R.\nThis is known as the ideal gas constant. So what problem remains? What is R? So, from experimental results, data shows that a temperature of zero Celsius and a pressure of one ATM and any 1 Mol of any gas will give us 22.4 liters of volume. And that means we have four unknowns."}, {"title": "Ideal Gas Law .txt", "text": "What is R? So, from experimental results, data shows that a temperature of zero Celsius and a pressure of one ATM and any 1 Mol of any gas will give us 22.4 liters of volume. And that means we have four unknowns. I mean, we have four knowns and one unknown. And that means if we plug the four knowns in, we'll get our unknown, namely our constant R.\nSo let's plug these guys in P times V divided by T times N. Simply rearrange these guys, plug our values in, and we get our gas constant R is equal to 0.826 atmospheres times literally by Kelvin times Mo. Now, this number might change if we play around with the units."}, {"title": "Ideal Gas Law .txt", "text": "I mean, we have four knowns and one unknown. And that means if we plug the four knowns in, we'll get our unknown, namely our constant R.\nSo let's plug these guys in P times V divided by T times N. Simply rearrange these guys, plug our values in, and we get our gas constant R is equal to 0.826 atmospheres times literally by Kelvin times Mo. Now, this number might change if we play around with the units. But for these units, this will be our gas constant. So the question is how does an ideal gas law help us? Well, it helps us in the following way."}, {"title": "Ideal Gas Law .txt", "text": "But for these units, this will be our gas constant. So the question is how does an ideal gas law help us? Well, it helps us in the following way. If we know three parameters, we can find and find them unknown. That's exactly what this formula tells us. If we know, for example, Nt and D, we can find P.\nOr if we know p, d and T will find N. Or if we know p, n and T will find D and so on."}, {"title": "Ideal Gas Law .txt", "text": "If we know three parameters, we can find and find them unknown. That's exactly what this formula tells us. If we know, for example, Nt and D, we can find P.\nOr if we know p, d and T will find N. Or if we know p, n and T will find D and so on. So let's do an example. Our example states that we want to calculate volume that 0.5 grams of methane occupied at 25 degrees Celsius and one atmospheric pressure. So we have one known, we have two knowns our our pressure."}, {"title": "Ideal Gas Law .txt", "text": "So let's do an example. Our example states that we want to calculate volume that 0.5 grams of methane occupied at 25 degrees Celsius and one atmospheric pressure. So we have one known, we have two knowns our our pressure. We have three knowns grams because remember, grams will translate into moles. We can find moles using grams. So our first step will be to find our third unknown."}, {"title": "Ideal Gas Law .txt", "text": "We have three knowns grams because remember, grams will translate into moles. We can find moles using grams. So our first step will be to find our third unknown. To find our number of moles. If we find that we have three knowns and one unknown, we'll suffer that unknown, namely this guy, and we'll get our answer. So to find the number of moles n, we simply take our number 0.5\ngrams of methane."}, {"title": "Ideal Gas Law .txt", "text": "To find our number of moles. If we find that we have three knowns and one unknown, we'll suffer that unknown, namely this guy, and we'll get our answer. So to find the number of moles n, we simply take our number 0.5\ngrams of methane. Divide that by our molecular weight of methane grams cancel and we get moles on top. So 0.5\ngrams divided by 16 is 132, which is 0.3125 moles. We take this moles, plug it into our formula along with not 25 degrees Celsius, but 273 plus 25."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "On the contrary, electrolytic cells use up electrical work to power unfavorable redox reactions, such as as decomposition reactions. Now, we already spoke about the decomposition of molten sodium chloride. Now we're going to talk about the decomposition of aqueous sodium chloride. And this guy is a little bit different because our solvent is water. When we talk about molten sodium chloride, we didn't have any solvent. So let's see how our solvent, namely water, affects our reaction."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "And this guy is a little bit different because our solvent is water. When we talk about molten sodium chloride, we didn't have any solvent. So let's see how our solvent, namely water, affects our reaction. So first, let's look at our electrolytic cell. So in this cell, we have a battery, a voltage cell that powers our reaction, and it powers it by sending electrons this way. So that means this electrode will collect these electrons, making it negatively charged."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "So first, let's look at our electrolytic cell. So in this cell, we have a battery, a voltage cell that powers our reaction, and it powers it by sending electrons this way. So that means this electrode will collect these electrons, making it negatively charged. At the same time, this electrode will lose electrons, making this guy positively charged, because electrons will want to flow from this guy to this guy. So let's look at our solution. Within our solution, we have a bunch of sodium ions dissolved and chloride ions dissolved."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "At the same time, this electrode will lose electrons, making this guy positively charged, because electrons will want to flow from this guy to this guy. So let's look at our solution. Within our solution, we have a bunch of sodium ions dissolved and chloride ions dissolved. The positively charged ions will be attracted to the negatively charged electrode. So a lot of the sodium ions will move to the left side next to this negatively charged electrode. Likewise, the negatively charged ions will move towards a positively charged electrode."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "The positively charged ions will be attracted to the negatively charged electrode. So a lot of the sodium ions will move to the left side next to this negatively charged electrode. Likewise, the negatively charged ions will move towards a positively charged electrode. So all the corridions will collect next to the on the right side. So anytime this positively charged sodium ion hits this negatively charged electrode, a transfer of electrons will occur. So a sodium ion will gain electrons, and that means it's reduced."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "So all the corridions will collect next to the on the right side. So anytime this positively charged sodium ion hits this negatively charged electrode, a transfer of electrons will occur. So a sodium ion will gain electrons, and that means it's reduced. And by definition, wherever reduction occurs, that's our cathode. So this side of the cell is our cathode, and that means this side of the cell is the amo, because that's where oxidation takes place. The chloride ions are oxidized and lose electrons, and these electrons jump into the electrode and travel on to this way, thus creating a circuit that moves in this direction."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "And by definition, wherever reduction occurs, that's our cathode. So this side of the cell is our cathode, and that means this side of the cell is the amo, because that's where oxidation takes place. The chloride ions are oxidized and lose electrons, and these electrons jump into the electrode and travel on to this way, thus creating a circuit that moves in this direction. Now, now that we know what our layout of our electrolytic cell is, let's talk about the oxidation and reduction reactions. Now, when we spoke about molten sodium chloride, only one oxidation reaction was possible, and only one reduction reaction was possible. But now we're in the presence of a solvent, namely water."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "Now, now that we know what our layout of our electrolytic cell is, let's talk about the oxidation and reduction reactions. Now, when we spoke about molten sodium chloride, only one oxidation reaction was possible, and only one reduction reaction was possible. But now we're in the presence of a solvent, namely water. So water is also capable of undergoing oxidation, and water likewise, is also capable of undergoing reduction. So let's look at oxidation first. So what atoms are able to oxidize?"}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "So water is also capable of undergoing oxidation, and water likewise, is also capable of undergoing reduction. So let's look at oxidation first. So what atoms are able to oxidize? Well, this atom could oxidize, and a water atom could oxidize. Now, if this atom decides to oxidize, this is our oxidation reaction. Now, if water decides to oxidize, namely the oxygen decides to oxidize, then this is our oxidation reactions."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "Well, this atom could oxidize, and a water atom could oxidize. Now, if this atom decides to oxidize, this is our oxidation reaction. Now, if water decides to oxidize, namely the oxygen decides to oxidize, then this is our oxidation reactions. How do we choose which one occurs? Well, the one that's more negative than one that's lower on the reduction HAP reaction table. That's the one that is more likely to undergo oxidation."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "How do we choose which one occurs? Well, the one that's more negative than one that's lower on the reduction HAP reaction table. That's the one that is more likely to undergo oxidation. And since this guy has a more negative cell voltage. That means this guy is more likely to undergo oxidation. And so this will be our oxidation reaction."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "And since this guy has a more negative cell voltage. That means this guy is more likely to undergo oxidation. And so this will be our oxidation reaction. Likewise, let's look at reduction. We have two possible reactions for a reduction. Now, our sodium ion could be reduced into sodium solid, or our water molecule could be reduced into this guy."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "Likewise, let's look at reduction. We have two possible reactions for a reduction. Now, our sodium ion could be reduced into sodium solid, or our water molecule could be reduced into this guy. Now let's look at which one is more likely to occur. Now, for reduction, it's the opposite. The more positive it is, the more likely it is to occur."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "Now let's look at which one is more likely to occur. Now, for reduction, it's the opposite. The more positive it is, the more likely it is to occur. So whatever is higher up on the reduction half reaction table, that guy will be more likely to undergo reduction. So since this is more positive, that means this will be our reduction reaction. So now we must add up all these reactions except this one, because this one won't occur, this one will occur, this one will occur, and this one won't occur."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "So whatever is higher up on the reduction half reaction table, that guy will be more likely to undergo reduction. So since this is more positive, that means this will be our reduction reaction. So now we must add up all these reactions except this one, because this one won't occur, this one will occur, this one will occur, and this one won't occur. And that means we're still going to have these atoms down in our solution. So we add this guy up, this guy up, and this guy up, and we get the following net reaction. So when this guy and this guy react, they will create sodium or salt within solution."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "And that means we're still going to have these atoms down in our solution. So we add this guy up, this guy up, and this guy up, and we get the following net reaction. So when this guy and this guy react, they will create sodium or salt within solution. Plus two molecules of water in a liquid state will produce diatomic gas, plus another diatomic gas, plus this two hydroxide acreage state and two sodium molecules in the acreage state. We also add up these cell voltages and we get this following cell voltage. So this is the amount of energy electrical work that needs to be put in by the battery to power this reaction."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "Plus two molecules of water in a liquid state will produce diatomic gas, plus another diatomic gas, plus this two hydroxide acreage state and two sodium molecules in the acreage state. We also add up these cell voltages and we get this following cell voltage. So this is the amount of energy electrical work that needs to be put in by the battery to power this reaction. So we see that decomposition reactions require work. Well, why? Well, that's because they're unfavorable redox reactions."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "So we see that decomposition reactions require work. Well, why? Well, that's because they're unfavorable redox reactions. So that's the difference between molten sodium chloride and aqueous sodium chloride. In molten sodium chloride, the salvage wasn't present, water wasn't present. And so there was only one oxidation and one reduction reaction in this and aqueous sodium chloride."}, {"title": "Electrolysis of Aqueous Sodium Chloride.txt", "text": "So that's the difference between molten sodium chloride and aqueous sodium chloride. In molten sodium chloride, the salvage wasn't present, water wasn't present. And so there was only one oxidation and one reduction reaction in this and aqueous sodium chloride. There are two possibilities for oxidation and reduction. And you have to look at their cell voltages to choose. For reduction, you choose the one that's more positive or less negative."}, {"title": "Phase Diagram.txt", "text": "Which simply means if I'm given a system and the system increases in size, well, our pressure and temperature will stay the same. Now phase diagrams use pressure and temperature to help us determine the phase of substance. So given some pressure and temperature, I could use the phase diagram to see what Theta substance is in. So here's our phase diagram. The y axis is pressure. The x axis is temperature."}, {"title": "Phase Diagram.txt", "text": "So here's our phase diagram. The y axis is pressure. The x axis is temperature. The phase diagram is broken down into three sections solid, liquid and gas. So to figure out what theta substance is in, we find the temperature and pressure. Suppose we're given some t and some p. So here's our T-R-P and here's our t. Now we first must find the point of intersection."}, {"title": "Phase Diagram.txt", "text": "The phase diagram is broken down into three sections solid, liquid and gas. So to figure out what theta substance is in, we find the temperature and pressure. Suppose we're given some t and some p. So here's our T-R-P and here's our t. Now we first must find the point of intersection. So we draw a line up or draw a line across? Here's our point intersection. And this point lies in a solid phase."}, {"title": "Phase Diagram.txt", "text": "So we draw a line up or draw a line across? Here's our point intersection. And this point lies in a solid phase. And that means it must be a solid at this particular temperature and this particular pressure. Now, if we keep the pressure the same and, say, increase our temperature to say, this point, well, then we have to draw the line all the way here. And then the section will be here at this pressure."}, {"title": "Phase Diagram.txt", "text": "And that means it must be a solid at this particular temperature and this particular pressure. Now, if we keep the pressure the same and, say, increase our temperature to say, this point, well, then we have to draw the line all the way here. And then the section will be here at this pressure. At this temperature, it will be in a gas phase. Now, the lines or boundaries represent or signify dynamic equilibrium between two or more phases. For example, suppose I'm on this boundary and on this line well, at this line, the rate at which the soil becomes a liquid is the same as the rate at which the liquid becomes a solid."}, {"title": "Phase Diagram.txt", "text": "At this temperature, it will be in a gas phase. Now, the lines or boundaries represent or signify dynamic equilibrium between two or more phases. For example, suppose I'm on this boundary and on this line well, at this line, the rate at which the soil becomes a liquid is the same as the rate at which the liquid becomes a solid. Okay? Now only one point exists called a triple point. The black tie."}, {"title": "Phase Diagram.txt", "text": "Okay? Now only one point exists called a triple point. The black tie. The middle. And at this point, all three phases are in dynamic equilibrium with each other. Now, another important point exists called a critical point."}, {"title": "Phase Diagram.txt", "text": "The middle. And at this point, all three phases are in dynamic equilibrium with each other. Now, another important point exists called a critical point. And a critical point corresponds to a certain pressure and temperature above which our substance has both properties of liquid and gas. Okay? So this fluid is now called supercritical fluid."}, {"title": "Phase Diagram.txt", "text": "And a critical point corresponds to a certain pressure and temperature above which our substance has both properties of liquid and gas. Okay? So this fluid is now called supercritical fluid. And this is our critical point. So our critical temperature is we draw a line. Down is right here."}, {"title": "Regioselective and Regiospecific.txt", "text": "So let's imagine for a moment that we're taking a walk on a straight path and eventually we come to a fourth in a row. Now, at the fourth, we have several options. We can either take one pathway that will lead us to one place, or we can take a different pathway that will lead us to a different place. Now, in the same way, when certain compounds react, different pathways are possible, and each pathway will lead us to a different compound. So let's examine the following example. Let's say we have the following two compounds."}, {"title": "Regioselective and Regiospecific.txt", "text": "Now, in the same way, when certain compounds react, different pathways are possible, and each pathway will lead us to a different compound. So let's examine the following example. Let's say we have the following two compounds. We have hydrochloric acid reacts with an alkene. Now, two different pathways are possible. In other words, this lewis base can take this H atom and follow pathway one, in which the H atom goes on this side of the carbon."}, {"title": "Regioselective and Regiospecific.txt", "text": "We have hydrochloric acid reacts with an alkene. Now, two different pathways are possible. In other words, this lewis base can take this H atom and follow pathway one, in which the H atom goes on this side of the carbon. So on this carbon, it attaches to the second carbon of the double bond. Or it can take pathway two, in which the lewis acid takes this H and attaches that H to the other side of the double bond to this carbon. So we can either take pathway one and form a tertiary carbocation, or we can take pathway two and form a primary carbocation."}, {"title": "Regioselective and Regiospecific.txt", "text": "So on this carbon, it attaches to the second carbon of the double bond. Or it can take pathway two, in which the lewis acid takes this H and attaches that H to the other side of the double bond to this carbon. So we can either take pathway one and form a tertiary carbocation, or we can take pathway two and form a primary carbocation. In each reaction, a different product is formed. So when our reaction takes pathway one, we form one product. When our reaction takes place and follows pathway two, we form a different product."}, {"title": "Regioselective and Regiospecific.txt", "text": "In each reaction, a different product is formed. So when our reaction takes pathway one, we form one product. When our reaction takes place and follows pathway two, we form a different product. So if a reaction takes place on an unsymmetrical compound such as this alkane, unsymmetrical simply means if we cut this in half, the left is different from the right. So if a reaction takes place on an oxymetrical compound, different pathways are possible. And this is known as regiochemistry."}, {"title": "Regioselective and Regiospecific.txt", "text": "So if a reaction takes place on an unsymmetrical compound such as this alkane, unsymmetrical simply means if we cut this in half, the left is different from the right. So if a reaction takes place on an oxymetrical compound, different pathways are possible. And this is known as regiochemistry. Regeochemistry is basically the study of the different pathways that our two reactants can take. Now, in the reaction above, pathway one is more favorable, it's more likely to occur. And that's because this intermediate carbocation, this tertiary carbocation is more stable than this primary carbocation."}, {"title": "Regioselective and Regiospecific.txt", "text": "Regeochemistry is basically the study of the different pathways that our two reactants can take. Now, in the reaction above, pathway one is more favorable, it's more likely to occur. And that's because this intermediate carbocation, this tertiary carbocation is more stable than this primary carbocation. And we'll examine that in a future lecture. So because pathway one is more favorable, it's more stable, it's more likely to take place. And generally, when a reaction can occur in several ways and one of the pathways predominates the reaction is called regioselective."}, {"title": "Regioselective and Regiospecific.txt", "text": "And we'll examine that in a future lecture. So because pathway one is more favorable, it's more stable, it's more likely to take place. And generally, when a reaction can occur in several ways and one of the pathways predominates the reaction is called regioselective. So this reaction is regioselective because pathway one predominates over pathway two. Pathway one is more likely to take place because this intermediate is more stable than this intermediate. Now, in such a reaction, one of the final products predominates, and this is known as reggiospecific reaction."}, {"title": "Clausius Clapeyron Equation .txt", "text": "In this lecture, we're going to talk about a concept called a closed cliperon relation. Now, vapor pressure is very important when we talk about this relation. So if you're not sure about vapor pressure, check out the video below. Now, the vapor pressure of a liquid increases with temperature. That means if you heat up a liquid, more molecules will evaporate, thereby increasing the vapor pressure of our liquid. Now, you could also graph vapor pressure versus temperature on an x y plane."}, {"title": "Clausius Clapeyron Equation .txt", "text": "Now, the vapor pressure of a liquid increases with temperature. That means if you heat up a liquid, more molecules will evaporate, thereby increasing the vapor pressure of our liquid. Now, you could also graph vapor pressure versus temperature on an x y plane. And what we get is an exponential curve. Now, the y axis is pressure, the X axis is temperature. What this curve tells us is that identical changes on the X axis produces different changes on the y axis."}, {"title": "Clausius Clapeyron Equation .txt", "text": "And what we get is an exponential curve. Now, the y axis is pressure, the X axis is temperature. What this curve tells us is that identical changes on the X axis produces different changes on the y axis. So, for example, if we take an identical change at higher pressures, we see that that produces a greater pressure difference than at lower temperatures. And that is clearly seen because this guy is much bigger than this guy. Now, what this relation doesn't tell us is the slope of the line."}, {"title": "Clausius Clapeyron Equation .txt", "text": "So, for example, if we take an identical change at higher pressures, we see that that produces a greater pressure difference than at lower temperatures. And that is clearly seen because this guy is much bigger than this guy. Now, what this relation doesn't tell us is the slope of the line. It would be very nice if we could somehow find the slope of the line. Unfortunately, the only way to find the slope of the line in this situation is to use calculus and find the slope of the line tangent to the curve at each point. But that's very tedious and takes a lot of work."}, {"title": "Clausius Clapeyron Equation .txt", "text": "It would be very nice if we could somehow find the slope of the line. Unfortunately, the only way to find the slope of the line in this situation is to use calculus and find the slope of the line tangent to the curve at each point. But that's very tedious and takes a lot of work. So it would be very nice if we could somehow represent vapor pressure and temperature in an easier relation, in an easier function. And mathematically, one of the most simplest functions is a linear function. And a linear function is represented by this equation y equals MX plus B, where y is our range value, x is our domain value, m is a constant slope, and B is the y intercept."}, {"title": "Clausius Clapeyron Equation .txt", "text": "So it would be very nice if we could somehow represent vapor pressure and temperature in an easier relation, in an easier function. And mathematically, one of the most simplest functions is a linear function. And a linear function is represented by this equation y equals MX plus B, where y is our range value, x is our domain value, m is a constant slope, and B is the y intercept. Now, this is exactly what the closest Clayton relation does. These were two scientists who developed mathematically and the experimental result, this relationship where this is our y value, this is our slope, this is our X value, and this is our Y intercept. This is the natural log of vapor pressure of our liquid."}, {"title": "Clausius Clapeyron Equation .txt", "text": "Now, this is exactly what the closest Clayton relation does. These were two scientists who developed mathematically and the experimental result, this relationship where this is our y value, this is our slope, this is our X value, and this is our Y intercept. This is the natural log of vapor pressure of our liquid. This, our slope, is the negative of the change in enthalpy about vaporization of the liquid over our gas constant R. Our X value is one over t or one over temperature of the liquid. And this is a constant that depends on the liquid being used. And if you graph this, you get exactly a linear line that has a negative slope shown by this negative sign here."}, {"title": "Clausius Clapeyron Equation .txt", "text": "This, our slope, is the negative of the change in enthalpy about vaporization of the liquid over our gas constant R. Our X value is one over t or one over temperature of the liquid. And this is a constant that depends on the liquid being used. And if you graph this, you get exactly a linear line that has a negative slope shown by this negative sign here. So the magnitude of this value is this value here. And in fact, this is very useful because now if we're given some vapor pressure and we're given the temperature and we know the constant, we can easily find our slope. That is, we can find the change in enthalpy of vaporization."}, {"title": "Clausius Clapeyron Equation .txt", "text": "So the magnitude of this value is this value here. And in fact, this is very useful because now if we're given some vapor pressure and we're given the temperature and we know the constant, we can easily find our slope. That is, we can find the change in enthalpy of vaporization. Likewise, if we know this guy and this guy, we could find this guy. And if we know this guy and this guy, we can find the vapor pressure. So it's very useful."}, {"title": "Clausius Clapeyron Equation .txt", "text": "Likewise, if we know this guy and this guy, we could find this guy. And if we know this guy and this guy, we can find the vapor pressure. So it's very useful. And if you want to practice using different problems, check out the link below. Now, let's talk about these graphs for a quick second. Let's compare this graph and this graph."}, {"title": "Clausius Clapeyron Equation .txt", "text": "And if you want to practice using different problems, check out the link below. Now, let's talk about these graphs for a quick second. Let's compare this graph and this graph. Well, this graph we see that this way increases in temperature. But on this side, this way is the increase in temperature because one over a larger number is a smaller number than one over a smaller number. So the larger the temperature, the closer the value will be an x axis to zero."}, {"title": "Clausius Clapeyron Equation .txt", "text": "Well, this graph we see that this way increases in temperature. But on this side, this way is the increase in temperature because one over a larger number is a smaller number than one over a smaller number. So the larger the temperature, the closer the value will be an x axis to zero. And in fact, what happens when this T becomes very large, when it tends to zero, when it tends to a very large number, when it tends to infinity, well, this becomes zero. And we see that at zero we get our Y intercept that is C.\nAnd this equation shows the same exact thing. When T becomes a very large number, this whole number tends to zero."}, {"title": "Clausius Clapeyron Equation .txt", "text": "And in fact, what happens when this T becomes very large, when it tends to zero, when it tends to a very large number, when it tends to infinity, well, this becomes zero. And we see that at zero we get our Y intercept that is C.\nAnd this equation shows the same exact thing. When T becomes a very large number, this whole number tends to zero. And so we simply get our log of our vapor pressure is equal to a constant C and we see exactly that on this graph. Another thing to realize is that the slope or the magnitude of the slope represents our change in enthalpy of vaporization. So for one liquid, this might be the slope for another liquid, say, if this was the slope for water, then alcohol would have a slope of this because alcohol has a smaller change in entry enthalpy of vaporization."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "So let's begin by first writing out the two half reactions of this reduction reaction. So let's see which guy is oxidized and which guy is reduced. Well, our iron atom goes from a neutral charge to a plus two charge, while our cavmium atom goes from a plus two charge to a neutral charge. That means this atom loses two electrons and this atom gains those same two electrons. So this is our oxidized atom and our reduced atom or our reducing agent and oxidizing agent. So let's go to step one and let's see our two half reactions."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "That means this atom loses two electrons and this atom gains those same two electrons. So this is our oxidized atom and our reduced atom or our reducing agent and oxidizing agent. So let's go to step one and let's see our two half reactions. So our oxidation half reaction is the following. Our solid iron becomes a positively charged molecule plus two electrons because it releases those two electrons while our cadmium aqueous atom gains those two electrons forming our cadmium solid. So this is our reduction reaction and oxidation reaction."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "So our oxidation half reaction is the following. Our solid iron becomes a positively charged molecule plus two electrons because it releases those two electrons while our cadmium aqueous atom gains those two electrons forming our cadmium solid. So this is our reduction reaction and oxidation reaction. So let's look at the cell diagram for this electrochemical cell. So remember, these two vertical lines represent the sole bridge and these guys simply represent separations of phases. So this and this are in different phases and these guys are in different phases also."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "So let's look at the cell diagram for this electrochemical cell. So remember, these two vertical lines represent the sole bridge and these guys simply represent separations of phases. So this and this are in different phases and these guys are in different phases also. So this is our anode and this is our cathode. So what happens is two electrons leave this atom forming our aqueous iron atom and these two electrons travel via the conductor to this guy reacting with this positively charged atom forming our solid cadmium. So let's go to step two."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "So this is our anode and this is our cathode. So what happens is two electrons leave this atom forming our aqueous iron atom and these two electrons travel via the conductor to this guy reacting with this positively charged atom forming our solid cadmium. So let's go to step two. Now, in step two and three, what we want to do or do is find a cell voltage of our electrochemical cell and then use the cell voltage to find our equilibrium constant KC. So let's go to step two. Now, this is our formula that we want to use to find the cell voltage where this is the cell voltage of the reduction reaction and the cell voltage of the oxidation reaction."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "Now, in step two and three, what we want to do or do is find a cell voltage of our electrochemical cell and then use the cell voltage to find our equilibrium constant KC. So let's go to step two. Now, this is our formula that we want to use to find the cell voltage where this is the cell voltage of the reduction reaction and the cell voltage of the oxidation reaction. Now, we basically look these guys up on our table for reduction half reactions on the statement conditions and we find that our reduction cell voltage is zero point 43 negative, while our oxidation half reaction is negative zero point 44. Well, actually our reduction going this way because only reduction half reactions are listed. So we have to look at the guy going this way."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "Now, we basically look these guys up on our table for reduction half reactions on the statement conditions and we find that our reduction cell voltage is zero point 43 negative, while our oxidation half reaction is negative zero point 44. Well, actually our reduction going this way because only reduction half reactions are listed. So we have to look at the guy going this way. So that is negative zero point 44. Now, I put this negative here in here because we want to convert this to an oxidation because in this anode oxidation that reduction occurs and that's why we have the negative sign here. So what we get is these negatives become a positive and we basically add this guy to this guy and we get zero point 37 volts."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "So that is negative zero point 44. Now, I put this negative here in here because we want to convert this to an oxidation because in this anode oxidation that reduction occurs and that's why we have the negative sign here. So what we get is these negatives become a positive and we basically add this guy to this guy and we get zero point 37 volts. This is our cell voltage of our electrochemical cell. Now, in the previous lecture we learned that there's a relationship between our cell voltage and our equilibrium constant, namely this equation here. Now, we also saw in that same lecture that we can convert this formula at 25 degrees Celsius to the following formula."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "This is our cell voltage of our electrochemical cell. Now, in the previous lecture we learned that there's a relationship between our cell voltage and our equilibrium constant, namely this equation here. Now, we also saw in that same lecture that we can convert this formula at 25 degrees Celsius to the following formula. Log k equals number of moles times our cell voltage divided by this number here. Zero point 52. Now, this number comes from the fact that both r and F are constants."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "Log k equals number of moles times our cell voltage divided by this number here. Zero point 52. Now, this number comes from the fact that both r and F are constants. And at 25 degrees Celsius, t is also constant. Now t is in Kelvin. And we also basically converted our natural log to log of base ten."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "And at 25 degrees Celsius, t is also constant. Now t is in Kelvin. And we also basically converted our natural log to log of base ten. Now, let's plug our guides in. So our E is from here, and n, we look at this equation. We see that N represents two moles of electrons."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "Now, let's plug our guides in. So our E is from here, and n, we look at this equation. We see that N represents two moles of electrons. So n is two. That's what we get. Two times zero point 37 results from this guy divided by zero point 52, and we get 1.25 equals log k.\nNow, we change this entire thing to exponents, and we get ten to the 1.5."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "So n is two. That's what we get. Two times zero point 37 results from this guy divided by zero point 52, and we get 1.25 equals log k.\nNow, we change this entire thing to exponents, and we get ten to the 1.5. You plug that into the calculator, and it's approximately 17.8. So our K is 17.8. And what does that mean?"}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "You plug that into the calculator, and it's approximately 17.8. So our K is 17.8. And what does that mean? Well, remember we said if our K is above one, that means our reaction is product favorite. It's spontaneous. So this guy, the equilibrium lies on the right."}, {"title": "Equilibrium constant of Redox reaction example .txt", "text": "Well, remember we said if our K is above one, that means our reaction is product favorite. It's spontaneous. So this guy, the equilibrium lies on the right. That means almost all of these guys are converted to our products. And this is the same thing as our e. Remember, our E says what our E gives us a positive value for cell voltage and what the a positive value for cell voltage mean? Remember, a positive value for cell voltage means our reaction is product favored, and a negative value means it's reacting favored."}, {"title": "Raoul\u2019s Law Example .txt", "text": "And we're told that the vapor pressure of pure water is 500 millimeter D.\nAnd what that basically says is the pressure exerted by these molecules, the gas molecules is is 500. What we want to do is we want to mix these two containers and then find the bigger pressure of water. So we want to mix them and find the pressure exerted by these water molecules on the walls of the container. The first step is to find the moles of water. The second step is to find the moles of ethylene glycol. And the third step is to use the formula to find vapor pressure."}, {"title": "Raoul\u2019s Law Example .txt", "text": "The first step is to find the moles of water. The second step is to find the moles of ethylene glycol. And the third step is to use the formula to find vapor pressure. Now, in the first step, before we find the moles, we first have to find the amount in grams of water. To find the amount of grams of water, we must first look up the density of water. The density of water is 1 gram per ML."}, {"title": "Raoul\u2019s Law Example .txt", "text": "Now, in the first step, before we find the moles, we first have to find the amount in grams of water. To find the amount of grams of water, we must first look up the density of water. The density of water is 1 gram per ML. So we take our volume of 100 MLS multiplied by 1 gram per ML the MLS can't sell and we get 100 grams of water in our beaker. Now, in the second step, we have to find the molecular weight of water. To find a molecular weight of water, we simply add up the atomic weights."}, {"title": "Raoul\u2019s Law Example .txt", "text": "So we take our volume of 100 MLS multiplied by 1 gram per ML the MLS can't sell and we get 100 grams of water in our beaker. Now, in the second step, we have to find the molecular weight of water. To find a molecular weight of water, we simply add up the atomic weights. So oxygen is 16 grams/mol plus two times because we have a substitute to 1 gram per mole, gives us 18 grams/mol. So the molecular weight of water is 18 grams/mol. Finally, we take this guy divided by this guy and we get 5.56\nmoles of water."}, {"title": "Raoul\u2019s Law Example .txt", "text": "So oxygen is 16 grams/mol plus two times because we have a substitute to 1 gram per mole, gives us 18 grams/mol. So the molecular weight of water is 18 grams/mol. Finally, we take this guy divided by this guy and we get 5.56\nmoles of water. That's how you find moles. The second step is the same exact step, except it's for ethylene glycol. Now we have to look up the density of ethylene glycol and the molecular formula for ethylene glycol."}, {"title": "Raoul\u2019s Law Example .txt", "text": "That's how you find moles. The second step is the same exact step, except it's for ethylene glycol. Now we have to look up the density of ethylene glycol and the molecular formula for ethylene glycol. So we follow the same exact steps, 100 ML, because we have 100 times 1.15 gives us the amount of cancer, 115 grams of ethylene glycol. So a little bit more than water. Our second step is to find the molecular weight of ethylene glycol."}, {"title": "Raoul\u2019s Law Example .txt", "text": "So we follow the same exact steps, 100 ML, because we have 100 times 1.15 gives us the amount of cancer, 115 grams of ethylene glycol. So a little bit more than water. Our second step is to find the molecular weight of ethylene glycol. We follow the same exact step, we add up the atomic weights and we get 62 grams/mol. Finally, to find the moles, we take our amount in grams divided by our molecular formula and we get 115 divided by 62 equals 1.85 moles of ethylene glycol. Finally, we use our formula to find out vapor pressure of the water molecules in this system."}, {"title": "Raoul\u2019s Law Example .txt", "text": "We follow the same exact step, we add up the atomic weights and we get 62 grams/mol. Finally, to find the moles, we take our amount in grams divided by our molecular formula and we get 115 divided by 62 equals 1.85 moles of ethylene glycol. Finally, we use our formula to find out vapor pressure of the water molecules in this system. We simply multiply the mole fraction of water times the vapor pressure of water in the initial system and that is 5.51 divided by the total number of moles. The mole fraction 5.56\nplus 1.85 gives us some fraction. That fraction is multiplied by 500 and that gives us 375.17\nmmhg and that's the bigger pressure of these water molecules."}, {"title": "Raoul\u2019s Law Example .txt", "text": "We simply multiply the mole fraction of water times the vapor pressure of water in the initial system and that is 5.51 divided by the total number of moles. The mole fraction 5.56\nplus 1.85 gives us some fraction. That fraction is multiplied by 500 and that gives us 375.17\nmmhg and that's the bigger pressure of these water molecules. Now it should be less. And that's because we have less water molecules on the surface, because now we have ethylene glycol molecules on the surface. And so less water molecules are evaporating."}, {"title": "Acidity of Alkynes .txt", "text": "And let's compare to the acidity of alkanes. So, as an example, let's use the simplest alkyne acetylene, and let's use the simplest alkane known as methane. So we basically want to determine which one of these two hydrocarbon compounds is a better Bronx Laric acid. Recall that a bronciid Laric acid is a compound that is capable of donating NH plus ion. So we want to figure out which one of these compounds is better at donating NH plus atom. So let's look at reaction one."}, {"title": "Acidity of Alkynes .txt", "text": "Recall that a bronciid Laric acid is a compound that is capable of donating NH plus ion. So we want to figure out which one of these compounds is better at donating NH plus atom. So let's look at reaction one. In reaction one, we have a methane molecule or a methane compound dissociating or donating NH plus atom and also creating a methion anion. So this method has a net charge of negative one because it has a lone pair of electrons, non bonding electrons found on the carbon. So let's look at reaction two in reaction to our acetylene, dissociates into an H plus atom, and it also creates an anion called acetylene."}, {"title": "Acidity of Alkynes .txt", "text": "In reaction one, we have a methane molecule or a methane compound dissociating or donating NH plus atom and also creating a methion anion. So this method has a net charge of negative one because it has a lone pair of electrons, non bonding electrons found on the carbon. So let's look at reaction two in reaction to our acetylene, dissociates into an H plus atom, and it also creates an anion called acetylene. Now, this acetylene also has a net charge of negative one because it also has a lone pair of non bonding electrons on this first carbon. So which one of these reactions is more likely to occur? In other words, which one of these acids is a better acid?"}, {"title": "Acidity of Alkynes .txt", "text": "Now, this acetylene also has a net charge of negative one because it also has a lone pair of non bonding electrons on this first carbon. So which one of these reactions is more likely to occur? In other words, which one of these acids is a better acid? Well, it turns out from experimental results, we know that acetylene is a better acid. In fact, acetylene is ten to 30 times better at donating an H plus atom than this methane compound. So why is that?"}, {"title": "Acidity of Alkynes .txt", "text": "Well, it turns out from experimental results, we know that acetylene is a better acid. In fact, acetylene is ten to 30 times better at donating an H plus atom than this methane compound. So why is that? Why is it that this hydrocarbon is so much better at donating an H plus atom than this simpler hydrocarbon? Well, it turns out the reason is this anion is much more stable than this anion, and therefore, this reaction is much more likely to take place. So now let's examine why this compound, this anion, is much more stable."}, {"title": "Acidity of Alkynes .txt", "text": "Why is it that this hydrocarbon is so much better at donating an H plus atom than this simpler hydrocarbon? Well, it turns out the reason is this anion is much more stable than this anion, and therefore, this reaction is much more likely to take place. So now let's examine why this compound, this anion, is much more stable. And let's look at the orbitals of these two compounds. So let's begin with our methi. So the lone pair of electrons in the methyde molecule in the methi compound are found in the SP three hybridized orbital."}, {"title": "Acidity of Alkynes .txt", "text": "And let's look at the orbitals of these two compounds. So let's begin with our methi. So the lone pair of electrons in the methyde molecule in the methi compound are found in the SP three hybridized orbital. Remember, SP three contains 25% S character and 75% P character. Now let's look at our second anion. So in this anion, our lone pair of electrons are found in an SP orbital."}, {"title": "Acidity of Alkynes .txt", "text": "Remember, SP three contains 25% S character and 75% P character. Now let's look at our second anion. So in this anion, our lone pair of electrons are found in an SP orbital. In an SP hybridized orbital, an SP orbital is an orbital that has 50% S character and 50% P character. So this has much more S character than this compound. And recall that the more S character a bond or an orbital has, the more stabilizing or the more stable the electrons in that bond will be."}, {"title": "Acidity of Alkynes .txt", "text": "In an SP hybridized orbital, an SP orbital is an orbital that has 50% S character and 50% P character. So this has much more S character than this compound. And recall that the more S character a bond or an orbital has, the more stabilizing or the more stable the electrons in that bond will be. And because this has much more S character than this, the electrons here are more stable. And therefore, this compound is more stable than this anion. And that means this reaction will be more likely to take place."}, {"title": "Acidity of Alkynes .txt", "text": "And because this has much more S character than this, the electrons here are more stable. And therefore, this compound is more stable than this anion. And that means this reaction will be more likely to take place. Recall that electrons found closer to our proton, to our nucleus are more stable. They're lower in energy. And that's exactly why when electrons are found in the s orbital, they're more stable than if they're found in the p orbital, because the S orbital is closer to the nucleus than the p orbital."}, {"title": "Acidity of Alkynes .txt", "text": "Recall that electrons found closer to our proton, to our nucleus are more stable. They're lower in energy. And that's exactly why when electrons are found in the s orbital, they're more stable than if they're found in the p orbital, because the S orbital is closer to the nucleus than the p orbital. And therefore, if we have more S character, which we have in this SP hyperbased orbital, those electrons will be more stable because they are closer to the nucleus. So once again, electrons found closer to the nucleus are more stable. Therefore, since that seed ali contains more S character 60% in the SP orbital than the methi, which contains 25% S character, acetyllide is more stable."}, {"title": "Acidity of Alkynes .txt", "text": "And therefore, if we have more S character, which we have in this SP hyperbased orbital, those electrons will be more stable because they are closer to the nucleus. So once again, electrons found closer to the nucleus are more stable. Therefore, since that seed ali contains more S character 60% in the SP orbital than the methi, which contains 25% S character, acetyllide is more stable. So this anion is more stable than this anion once again, because those electrons are found in a 50% S character versus a 25% S character orbital. So therefore, acetylalene, this guy is more likely to release an H atom or an H ion than our methane. Now, notice one important detail in this lecture."}, {"title": "Acidity of Alkynes .txt", "text": "So this anion is more stable than this anion once again, because those electrons are found in a 50% S character versus a 25% S character orbital. So therefore, acetylalene, this guy is more likely to release an H atom or an H ion than our methane. Now, notice one important detail in this lecture. We've compared the acetylene to methane. So we said that this is a relatively good acid. Relatively."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "So let's begin. Let's look at a combination reaction. Now, a combination reaction is exactly what the name applies. Two or more reactants combined by forming bonds to form a new compound with a new molecular formula. Now let's look at our hypothetical example. In our example, we have two reactants, A and B react combined to produce a new product, a new compound, namely AB."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Two or more reactants combined by forming bonds to form a new compound with a new molecular formula. Now let's look at our hypothetical example. In our example, we have two reactants, A and B react combined to produce a new product, a new compound, namely AB. Now let's look at a very common example. Whenever 1 mol of solid burns with 1 mol of diatomic oxygen in the gas state, it produces 1 mol of carbon dioxide in the gas state. Now, this is a very common combination reaction."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Now let's look at a very common example. Whenever 1 mol of solid burns with 1 mol of diatomic oxygen in the gas state, it produces 1 mol of carbon dioxide in the gas state. Now, this is a very common combination reaction. And note that this triangle on top of the arrow means that heat must be added for our reaction to occur. That's what this triangle means. It doesn't mean change."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "And note that this triangle on top of the arrow means that heat must be added for our reaction to occur. That's what this triangle means. It doesn't mean change. It means energy heat. Now let's look at a very common second type of chemical reaction known as decomposition reactions. Now, decomposition reactions are essentially the reverse of combination reactions, where in combination reactions, two reactants combine in decomposition reactions, a reactant, decomposes or dissociates into two or more new compounds."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "It means energy heat. Now let's look at a very common second type of chemical reaction known as decomposition reactions. Now, decomposition reactions are essentially the reverse of combination reactions, where in combination reactions, two reactants combine in decomposition reactions, a reactant, decomposes or dissociates into two or more new compounds. So in our hypothetical example, we have reactant AB, breaks down or decomposes into two new products, A plus B. Now let's look at a very common example. Now, the hydrolysis of water or the decomposition of water."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "So in our hypothetical example, we have reactant AB, breaks down or decomposes into two new products, A plus B. Now let's look at a very common example. Now, the hydrolysis of water or the decomposition of water. In other words, two water molecules or two moles of H two in a liquid state react to produce 1 mol of O, two in a gas state, and two moles of diatomic H, two in the gas state. Now let's look at a third type and a fourth type of chemical reactions, namely single and double displacement. Now, single displacement reactions are also known as substitution reactions."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "In other words, two water molecules or two moles of H two in a liquid state react to produce 1 mol of O, two in a gas state, and two moles of diatomic H, two in the gas state. Now let's look at a third type and a fourth type of chemical reactions, namely single and double displacement. Now, single displacement reactions are also known as substitution reactions. And double displacement reactions are also known as metaphosis reactions. So let's look at a single or substitution chemical reaction. So A plus B or AB plus C. What happens is that the bond between A and B is broken."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "And double displacement reactions are also known as metaphosis reactions. So let's look at a single or substitution chemical reaction. So A plus B or AB plus C. What happens is that the bond between A and B is broken. So B dissociates from A forming its own reactant, while A C combines with A displacing this B. So basically, C kicks out our B and forms a bond with A forming AC plus B. And this is a single displacement reaction or substitution reaction."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "So B dissociates from A forming its own reactant, while A C combines with A displacing this B. So basically, C kicks out our B and forms a bond with A forming AC plus B. And this is a single displacement reaction or substitution reaction. Now, a very common example is when two moles of hydrochloric acid in an Aqueous state react with 1 mol of magnesium solid. This produces 1 mol of Mg CL, two in the Aqueous state, and 1 mol of diatomic H two in a gas state. So this is a single display for reaction in which our Mg kicks off this H, forming our reactant, and this H combines with another H forming a diatomic gas."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Now, a very common example is when two moles of hydrochloric acid in an Aqueous state react with 1 mol of magnesium solid. This produces 1 mol of Mg CL, two in the Aqueous state, and 1 mol of diatomic H two in a gas state. So this is a single display for reaction in which our Mg kicks off this H, forming our reactant, and this H combines with another H forming a diatomic gas. Now let's look at a double displacement or a metaphor reaction. Now, we have reactance A and B react with reactants CD. What happens now is our C displaces this B kicks off this B, forming a bond with a, and at the same time, this B kicks off this C, forming a bond with D. And what we get is the following."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Now let's look at a double displacement or a metaphor reaction. Now, we have reactance A and B react with reactants CD. What happens now is our C displaces this B kicks off this B, forming a bond with a, and at the same time, this B kicks off this C, forming a bond with D. And what we get is the following. Or an example of this is the following. Now, notice what happened. This hydroxide in our first reactant kicked off this so four molecule, forming a bond with H and forming our first product, our water molecule."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Or an example of this is the following. Now, notice what happened. This hydroxide in our first reactant kicked off this so four molecule, forming a bond with H and forming our first product, our water molecule. Likewise, this K molecule bonded with the molecule that got kicked off this so four. So this so four molecule at the same time kicked off this oh or replaced this oh and bonded with H. And we got K two So four. So this is an example of a metaphosis or a double displacement reaction."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Likewise, this K molecule bonded with the molecule that got kicked off this so four. So this so four molecule at the same time kicked off this oh or replaced this oh and bonded with H. And we got K two So four. So this is an example of a metaphosis or a double displacement reaction. Now, the last type of reaction we're going to look at is called a combustion reaction. And in this reaction, hydrocarbons are burned in the presence of diatomic gas or diatomic oxygen gas to form carbon dioxide and water molecules. Now, a common example is burning of methane ch four, where our X is one and Y is four in the presence of oxygen."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Now, the last type of reaction we're going to look at is called a combustion reaction. And in this reaction, hydrocarbons are burned in the presence of diatomic gas or diatomic oxygen gas to form carbon dioxide and water molecules. Now, a common example is burning of methane ch four, where our X is one and Y is four in the presence of oxygen. And heat from the triangle symbolizes heat to form carbon dioxide and water molecules. Now, other types of reactions exist. Let's look at some of them."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "And heat from the triangle symbolizes heat to form carbon dioxide and water molecules. Now, other types of reactions exist. Let's look at some of them. Reduce reactions or oxidation reduction reactions in which the oxidation states of atoms change. Acid based reactions, hydrolysis reactions and isomerization reaction are very common examples of some other types of chemical reactions. Now, note that a certain reaction can be labeled as more than one."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "Reduce reactions or oxidation reduction reactions in which the oxidation states of atoms change. Acid based reactions, hydrolysis reactions and isomerization reaction are very common examples of some other types of chemical reactions. Now, note that a certain reaction can be labeled as more than one. In other words, let's look at the following decomposition reaction. This decomposition reaction is also an oxidation reduction reaction because our O gets oxidized while our H gets reduced. In other words, in this molecule, our O has an oxidation state of negative two."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "In other words, let's look at the following decomposition reaction. This decomposition reaction is also an oxidation reduction reaction because our O gets oxidized while our H gets reduced. In other words, in this molecule, our O has an oxidation state of negative two. And here our O has an oxidation state of zero. So if it goes from negative two to zero, it's oxidized, while this guy goes from positive one to zero. So it's reduced, it gains electrons."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "And here our O has an oxidation state of zero. So if it goes from negative two to zero, it's oxidized, while this guy goes from positive one to zero. So it's reduced, it gains electrons. So this reaction is not only a decomposition reaction, but also a reduction reaction. Likewise, let's look at this double displacement or a metaphor reaction. Now, notice this guy."}, {"title": "Combination, Decomposition, Displacement and Combustion Reactions .txt", "text": "So this reaction is not only a decomposition reaction, but also a reduction reaction. Likewise, let's look at this double displacement or a metaphor reaction. Now, notice this guy. We have a base potassium hydroxide react with this acid. That means this guy is not only a double displacement reaction or a metastasis reaction, it's also an acid based reaction. Now, let's look at the reverse of this guy."}, {"title": "Evaporation and Condensation.txt", "text": "Evaporation is the process by which liquid molecules gain enough kinetic energy to escape the liquid state and become a gas molecule. Now. Condensation is the opposite. It's the process by which gas molecules hit the liquid, lose kinetic energy, and get stuck in a liquid. Now, these two processes occur naturally at the surface of a liquid. Suppose I take a liquid and examine the surface of that liquid."}, {"title": "Evaporation and Condensation.txt", "text": "It's the process by which gas molecules hit the liquid, lose kinetic energy, and get stuck in a liquid. Now, these two processes occur naturally at the surface of a liquid. Suppose I take a liquid and examine the surface of that liquid. What I will see is a line of molecules connected together by intermolecular bonds or non covalent bonds. Now, if I somehow increase the kinetic energy of these molecules, what will happen? Well, if one of these molecules gained enough kinetic energy, that energy will be able to overcome the energy due to the bonding."}, {"title": "Evaporation and Condensation.txt", "text": "What I will see is a line of molecules connected together by intermolecular bonds or non covalent bonds. Now, if I somehow increase the kinetic energy of these molecules, what will happen? Well, if one of these molecules gained enough kinetic energy, that energy will be able to overcome the energy due to the bonding. And this will allow the molecule to escape into the space above. So that means if I somehow increase the kinetic energy of this system, evaporation will occur. And if I let them evaporate and then I open, I should hear a hissing sound."}, {"title": "Evaporation and Condensation.txt", "text": "And this will allow the molecule to escape into the space above. So that means if I somehow increase the kinetic energy of this system, evaporation will occur. And if I let them evaporate and then I open, I should hear a hissing sound. So let's try to increase the kinetic energy of this system by shaking it. If I shake this, a lot of the molecules are on the surface and within will gain enough kinetic energy, will escape into the gas phase. And if right, then I open this, I should hear the gas molecules escaping."}, {"title": "Evaporation and Condensation.txt", "text": "So let's try to increase the kinetic energy of this system by shaking it. If I shake this, a lot of the molecules are on the surface and within will gain enough kinetic energy, will escape into the gas phase. And if right, then I open this, I should hear the gas molecules escaping. So I should hear a hissing sound. So let's try to shake this. And that hissing sound was because of the escaping molecule."}, {"title": "Evaporation and Condensation.txt", "text": "So I should hear a hissing sound. So let's try to shake this. And that hissing sound was because of the escaping molecule. So I shook it. I increased kinetic energy. A lot of the molecules escaped into this space."}, {"title": "Evaporation and Condensation.txt", "text": "So I shook it. I increased kinetic energy. A lot of the molecules escaped into this space. Here. When I opened it, it evaporated. And the hissing sound was due to the evaporation."}, {"title": "Evaporation and Condensation.txt", "text": "Here. When I opened it, it evaporated. And the hissing sound was due to the evaporation. Now, another way that water can be represented is this way. H 20 in a liquid state evaporates into H two and a gas state and H so in a gas state, condenses into H 20 in a liquid state. Now in a closed container."}, {"title": "Evaporation and Condensation.txt", "text": "Now, another way that water can be represented is this way. H 20 in a liquid state evaporates into H two and a gas state and H so in a gas state, condenses into H 20 in a liquid state. Now in a closed container. Eventually, if I leave this alone, this will form Dynamic Equilibrium. Or it will achieve Dynamic Equilibrium. And what that simply means is the rate at which the molecules are evaporating is equal to the rate at which they are condensing."}, {"title": "Evaporation and Condensation.txt", "text": "Eventually, if I leave this alone, this will form Dynamic Equilibrium. Or it will achieve Dynamic Equilibrium. And what that simply means is the rate at which the molecules are evaporating is equal to the rate at which they are condensing. And so the volume at that point will remain the same. So why is it that puddles found outside evaporate completely? All the water eventually goes away."}, {"title": "Evaporation and Condensation.txt", "text": "And so the volume at that point will remain the same. So why is it that puddles found outside evaporate completely? All the water eventually goes away. Well, why is that? If eventually, the water molecules found the puddle should be in dynamic equilibrium with the gas molecules above well, the answer is that's because this system is not a closed system. It's an open system and molecules are allowed to leave."}, {"title": "Evaporation and Condensation.txt", "text": "Well, why is that? If eventually, the water molecules found the puddle should be in dynamic equilibrium with the gas molecules above well, the answer is that's because this system is not a closed system. It's an open system and molecules are allowed to leave. For example, suppose our dynamic equilibrium is reached. And so we have a bunch of water molecules above, floating above. Now, wind, for example, can blow these molecules away."}, {"title": "Evaporation and Condensation.txt", "text": "For example, suppose our dynamic equilibrium is reached. And so we have a bunch of water molecules above, floating above. Now, wind, for example, can blow these molecules away. And if it blows the molecules away, these guys will go away. And so our H 20 in the gas state will decrease. And this will shift equilibrium."}, {"title": "Evaporation and Condensation.txt", "text": "And if it blows the molecules away, these guys will go away. And so our H 20 in the gas state will decrease. And this will shift equilibrium. This way. So these guys will continue to evaporate. The wind will continue belongs in a way."}, {"title": "Evaporation and Condensation.txt", "text": "This way. So these guys will continue to evaporate. The wind will continue belongs in a way. And eventually, every single molecule will evaporate. Now, let's look at the boiling or the Boiling Point. The Boiling Point is the temperature at which the molecules in the liquid gain enough kinetic energy."}, {"title": "Evaporation and Condensation.txt", "text": "And eventually, every single molecule will evaporate. Now, let's look at the boiling or the Boiling Point. The Boiling Point is the temperature at which the molecules in the liquid gain enough kinetic energy. And now, I'm not only talking about the molecules on the surface of the water. Now I'm talking all the molecules found within the entire system gain enough kinetic energy. So when you boil something, you see bubbles forming from the inside that travel to the up."}, {"title": "Evaporation and Condensation.txt", "text": "And now, I'm not only talking about the molecules on the surface of the water. Now I'm talking all the molecules found within the entire system gain enough kinetic energy. So when you boil something, you see bubbles forming from the inside that travel to the up. It's the outside. And that's because those molecules found inside gain enough kinetic energy, they become gas molecules, and they travel up. The Boiling point is also the point at which vapor pressure created by the liquid is equal to or greater than the total pressure found above the system."}, {"title": "Evaporation and Condensation.txt", "text": "It's the outside. And that's because those molecules found inside gain enough kinetic energy, they become gas molecules, and they travel up. The Boiling point is also the point at which vapor pressure created by the liquid is equal to or greater than the total pressure found above the system. So the atmospheric pressure. So, for example, suppose this is our system and this is our vapor pressure created by the system, and this is our vapor, our pressure, our atmospheric pressure. Now, if our atmospheric pressure is greater than our vapor pressure, pressure created by a liquid, the vapor pressure will not allow these molecules to escape."}, {"title": "Colligative Properties .txt", "text": "So in liquid solutions solve particles or ions disrupt the noncovalent solvent solvent tractions, thereby creating changes in various properties of the pure solvent. For example, if a solvent is added to a pure substance, the boiling point is higher and the melting point is lower than that of the pure substance. How much these properties change depend on the concentration of the solute use. Colligative properties are those properties that depend solely on the concentration of the solid use. They depend on the number or amount of solid particles used. They don't depend on the nature or type of particle use."}, {"title": "Colligative Properties .txt", "text": "Colligative properties are those properties that depend solely on the concentration of the solid use. They depend on the number or amount of solid particles used. They don't depend on the nature or type of particle use. Now, four major colligator properties are known vapor pressure, osmotic pressure, boiling point elevation, and melting point depression. I'm going to briefly talk about the first collage of property, vapor pressure. If you want to go into more detail about vapor pressure, check out my link below."}, {"title": "Colligative Properties .txt", "text": "Now, four major colligator properties are known vapor pressure, osmotic pressure, boiling point elevation, and melting point depression. I'm going to briefly talk about the first collage of property, vapor pressure. If you want to go into more detail about vapor pressure, check out my link below. So vapor pressure is simply the pressure due to the gas molecules found in dynamic equilibrium with their liquid molecules. Now, we see that adding non volatile solute decreases vapor pressure according to Rolls Law. Now, if you want to learn more about Rolls Law, check out my link below."}, {"title": "Colligative Properties .txt", "text": "So vapor pressure is simply the pressure due to the gas molecules found in dynamic equilibrium with their liquid molecules. Now, we see that adding non volatile solute decreases vapor pressure according to Rolls Law. Now, if you want to learn more about Rolls Law, check out my link below. Now, recall that a non volatile solute is a solid that will not evaporate. So let's see two systems. We have two systems."}, {"title": "Colligative Properties .txt", "text": "Now, recall that a non volatile solute is a solid that will not evaporate. So let's see two systems. We have two systems. The first system is a system before we add our non volatile solute. The second system is after the addition of the non volatile solid. Before addition, a certain vapor pressure is created due to the molecules found in space above."}, {"title": "Colligative Properties .txt", "text": "The first system is a system before we add our non volatile solute. The second system is after the addition of the non volatile solid. Before addition, a certain vapor pressure is created due to the molecules found in space above. After addition, some of the solid molecules will be replaced by the non volatile cellulose, and so less will evaporate, and therefore the vapor pressure will be less. Now, adding volatile cells or molecules that do evaporate increases vapor pressure according to a modified version of Rolls Law. And that's simply because now we have the solid molecules evaporating and the volatile solid molecules evaporating."}, {"title": "Colligative Properties .txt", "text": "After addition, some of the solid molecules will be replaced by the non volatile cellulose, and so less will evaporate, and therefore the vapor pressure will be less. Now, adding volatile cells or molecules that do evaporate increases vapor pressure according to a modified version of Rolls Law. And that's simply because now we have the solid molecules evaporating and the volatile solid molecules evaporating. So the final pressure is the sum of the two pressures. The second collage of property is called the boiling point. The boiling point is the temperature at which the liquid molecules have enough kinetic energy to escape into the gas state."}, {"title": "Colligative Properties .txt", "text": "So the final pressure is the sum of the two pressures. The second collage of property is called the boiling point. The boiling point is the temperature at which the liquid molecules have enough kinetic energy to escape into the gas state. Now, this occurs when the vapor pressure, the liquid, equals the atmospheric pressure. So we see that the boiling point is related to that vapor pressure. Adding non volatile solution lowers vapor pressure."}, {"title": "Colligative Properties .txt", "text": "Now, this occurs when the vapor pressure, the liquid, equals the atmospheric pressure. So we see that the boiling point is related to that vapor pressure. Adding non volatile solution lowers vapor pressure. And now a higher temperature is required to raise our vapor pressure to that final atmospheric pressure. So more energy is required. In fact, we could use a formula to find the change in the boiling point."}, {"title": "Colligative Properties .txt", "text": "And now a higher temperature is required to raise our vapor pressure to that final atmospheric pressure. So more energy is required. In fact, we could use a formula to find the change in the boiling point. So change in boiling point is equal to a constant KB, which is related to the substance being boiled times. The morality so moles over a kilogram of solid times I I is called vonkov factor, and this is the number of particles a single solid molecule dissociates into. So if our solid is sodium chloride, we see that sodium chloride dissociates into exactly two particles."}, {"title": "Colligative Properties .txt", "text": "So change in boiling point is equal to a constant KB, which is related to the substance being boiled times. The morality so moles over a kilogram of solid times I I is called vonkov factor, and this is the number of particles a single solid molecule dissociates into. So if our solid is sodium chloride, we see that sodium chloride dissociates into exactly two particles. So our eye in an ideal solution will be two. In a non ideal solution, there is Something Called ion pairing or the momentary aggregation of ions in a solution. And this will cause I to slightly lower."}, {"title": "Colligative Properties .txt", "text": "So our eye in an ideal solution will be two. In a non ideal solution, there is Something Called ion pairing or the momentary aggregation of ions in a solution. And this will cause I to slightly lower. And that means, in an ideal solution, this will be two. In a non ideal solution, this will be slightly below two. The third color gate of property is called a freezing point."}, {"title": "Colligative Properties .txt", "text": "And that means, in an ideal solution, this will be two. In a non ideal solution, this will be slightly below two. The third color gate of property is called a freezing point. Or the melting point. This is the temperature at which the liquid molecules lose the kinetic energy and cannot sustain the liquid state. And they form a crystalline structure called a solid."}, {"title": "Colligative Properties .txt", "text": "Or the melting point. This is the temperature at which the liquid molecules lose the kinetic energy and cannot sustain the liquid state. And they form a crystalline structure called a solid. And a freezing or melting point is not related to vigor pressure the same way that boiling points are. By adding a solid or an impurity to a pure substance, we increase entropy and make it difficult to form a crystal structure. Therefore, more energy needs to be taken away from our system."}, {"title": "Colligative Properties .txt", "text": "And a freezing or melting point is not related to vigor pressure the same way that boiling points are. By adding a solid or an impurity to a pure substance, we increase entropy and make it difficult to form a crystal structure. Therefore, more energy needs to be taken away from our system. And this means that our freezing and melting point lowers. Now that melt by which our freezing and melting point lowers can be found by a similar formula. The change in temperature is equal to a new constant that depends on the substance being melted or frozen."}, {"title": "Colligative Properties .txt", "text": "And this means that our freezing and melting point lowers. Now that melt by which our freezing and melting point lowers can be found by a similar formula. The change in temperature is equal to a new constant that depends on the substance being melted or frozen. Times the morality of our substance. Times I the number of particles a soluble dissociates into. Now we have to be careful with freezing point depression because adding a salute to a pure substance will only decrease the freezing point to a certain extent."}, {"title": "Colligative Properties .txt", "text": "Times the morality of our substance. Times I the number of particles a soluble dissociates into. Now we have to be careful with freezing point depression because adding a salute to a pure substance will only decrease the freezing point to a certain extent. Eventually our solvent will become our impurity. Because if we add enough solute, the amount of solute will surpass the amount of solvent. And when that happens, adding an immortal solute will actually make our freezing point rise and our melting point rise."}, {"title": "Colligative Properties .txt", "text": "Eventually our solvent will become our impurity. Because if we add enough solute, the amount of solute will surpass the amount of solvent. And when that happens, adding an immortal solute will actually make our freezing point rise and our melting point rise. The fourth Caligra property is called astronomic pressure. Asthmatic pressure is a tendency of water or some other solvent to flow into an area with a higher solute concentration. Now, to demonstrate as much like pressure, let's build a system."}, {"title": "Colligative Properties .txt", "text": "The fourth Caligra property is called astronomic pressure. Asthmatic pressure is a tendency of water or some other solvent to flow into an area with a higher solute concentration. Now, to demonstrate as much like pressure, let's build a system. In this system, we have a cell that's embellished in a semipermeable membrane. Now, this membrane allows the flow of water, but it does not allow any site molecules to pass through. Now, on the outside, we have many more site molecules than on the inside."}, {"title": "Colligative Properties .txt", "text": "In this system, we have a cell that's embellished in a semipermeable membrane. Now, this membrane allows the flow of water, but it does not allow any site molecules to pass through. Now, on the outside, we have many more site molecules than on the inside. So let's see what entropy tells us about such a system. Well, entropy dictates that a system will only strive to even out. And that means in this system, the site molecules will want to move inside the cell."}, {"title": "Colligative Properties .txt", "text": "So let's see what entropy tells us about such a system. Well, entropy dictates that a system will only strive to even out. And that means in this system, the site molecules will want to move inside the cell. But remember, because this is semipermeable membrane side molecules will not be able to go inside. And that means instead, water will try to even out the system and water will flow from a lower concentration to a higher salary of concentration. So we can define as much as I pressure in a second way."}, {"title": "Colligative Properties .txt", "text": "But remember, because this is semipermeable membrane side molecules will not be able to go inside. And that means instead, water will try to even out the system and water will flow from a lower concentration to a higher salary of concentration. So we can define as much as I pressure in a second way. Astronic pressure is the pressure that needs to be applied to a membrane to stop the flow of solvent into an area of a higher solid concentration. So whenever we talk about an ideal solution that has a very low solid concentration, we can find osmotic pressure on one side of the equation or on one side of the membrane by using this formula. And this formula states that osmotic pressure is equal to Bonhopper, which is the number of particles a single molecule or a single site molecule breaks into times molarity of our solution times the gas constant or r times temperature in Kelvin."}, {"title": "Colligative Properties .txt", "text": "Astronic pressure is the pressure that needs to be applied to a membrane to stop the flow of solvent into an area of a higher solid concentration. So whenever we talk about an ideal solution that has a very low solid concentration, we can find osmotic pressure on one side of the equation or on one side of the membrane by using this formula. And this formula states that osmotic pressure is equal to Bonhopper, which is the number of particles a single molecule or a single site molecule breaks into times molarity of our solution times the gas constant or r times temperature in Kelvin. Now, now that we talk about osmotic pressure, we also talk about osmotic potential. Now, any pure substance, such as pure water, is automatically given an asthmatic potential of zero and any impure substance. So if we add a solution to water, we give it an asthmatic potential of less than deer, so it has a negative asthmatic potential."}, {"title": "Balancing Redox Reactions .txt", "text": "Going from an unbalanced redox reaction to a balanced net redox reaction is not always an easy process. Now, equations for redox reactions often involve water molecules, hydronium molecules and hydroxide molecules. And therefore, determining the number of these molecules in a balanced equation can be very tedious. Fortunately, a systematic approach exists to finding the balanced redox reactions and I've outlined seven steps that you can follow when determining or balancing redox reactions. So let's see these steps. In the first step, you basically have to recognize what atom is oxidized and what atom is reduced."}, {"title": "Balancing Redox Reactions .txt", "text": "Fortunately, a systematic approach exists to finding the balanced redox reactions and I've outlined seven steps that you can follow when determining or balancing redox reactions. So let's see these steps. In the first step, you basically have to recognize what atom is oxidized and what atom is reduced. Next, you break down the unbalanced reaction into the two half reactions an oxidation reaction and a reduction reaction. In the third step, you have to balance atoms in each reaction. So in the reduction reaction and the oxidation reaction and the atoms you're balancing are all the atoms other than oxygen and hydrogen."}, {"title": "Balancing Redox Reactions .txt", "text": "Next, you break down the unbalanced reaction into the two half reactions an oxidation reaction and a reduction reaction. In the third step, you have to balance atoms in each reaction. So in the reduction reaction and the oxidation reaction and the atoms you're balancing are all the atoms other than oxygen and hydrogen. Next, you have to balance the oxygen by adding water and hydrogen by adding H plus ions. Now, this only works for acidic solutions. In basic solutions, you balance the H atoms, the hydrogen atoms by adding oh minus."}, {"title": "Balancing Redox Reactions .txt", "text": "Next, you have to balance the oxygen by adding water and hydrogen by adding H plus ions. Now, this only works for acidic solutions. In basic solutions, you balance the H atoms, the hydrogen atoms by adding oh minus. So by adding hydroxide ions in the fifth step you balance the charge by adding electrons to one side. Remember, charge is concerned so whatever charge is lost must be gained. So that's why you need to balance the charge by adding electrons to one side of the reaction."}, {"title": "Balancing Redox Reactions .txt", "text": "So by adding hydroxide ions in the fifth step you balance the charge by adding electrons to one side. Remember, charge is concerned so whatever charge is lost must be gained. So that's why you need to balance the charge by adding electrons to one side of the reaction. Next, you multiply each half reaction by appropriate factor so that electrons gain equals electrons lost. Once again, this comes from the conservation of energy or the conservation of charge. In the final step, you basically add the two half reactions and you check to make sure that they are, in fact, balanced."}, {"title": "Mass percent example #2 .txt", "text": "The first step is to find the molecular weight of water and this can be done by adding the atomic weight of oxygen plus two times the atomic weight of h to leo, a subscript of two. So two times 1 gram per mole plus 16 grams/mol gives us 18 grams/mol of water. Now we can use this molecular mass, we can multiply that by ten moles to get the amount in grams of water in our solution. So 18 times ten gives us 180. The moles cancel, we're left with grams. So 180 grams of water in our beaker."}, {"title": "Mass percent example #2 .txt", "text": "So 18 times ten gives us 180. The moles cancel, we're left with grams. So 180 grams of water in our beaker. Now we know we want to create a 5% bimass solution of sodium chloride, so we use the formula 5% is equal to the number of sodium chloride we need to add in terms of grams divided by the total number of grams. So 180 grams of water that we already have, plus the x. The x is the number of sodium chloride in grams that we need to add multiply the whole thing by 100."}, {"title": "Mass percent example #2 .txt", "text": "Now we know we want to create a 5% bimass solution of sodium chloride, so we use the formula 5% is equal to the number of sodium chloride we need to add in terms of grams divided by the total number of grams. So 180 grams of water that we already have, plus the x. The x is the number of sodium chloride in grams that we need to add multiply the whole thing by 100. This is simply the formula for mass percentage. Next we do a little algebra. We bring the 100 over."}, {"title": "Mass percent example #2 .txt", "text": "This is simply the formula for mass percentage. Next we do a little algebra. We bring the 100 over. So we divide five by 100, get 0.5,\nthen we bring the bottom over, multiply this whole bottom by 0.5, so we get 0.05 times 180 plus x. When we multiply this out, we get nine plus 0.5 x. We equate that to x because the x was left on the other side."}, {"title": "Mass percent example #2 .txt", "text": "So we divide five by 100, get 0.5,\nthen we bring the bottom over, multiply this whole bottom by 0.5, so we get 0.05 times 180 plus x. When we multiply this out, we get nine plus 0.5 x. We equate that to x because the x was left on the other side. And now we simply solve for x and we get nine equals zero point 95 x. We divide by zero point 95 x and we get 9.47 grams of sodium chloride that we need to add to the solution of ten moles to get a 5% bimass solution of sodium chloride. The final step is to find the number of moles of sodium chloride."}, {"title": "Mass percent example #2 .txt", "text": "And now we simply solve for x and we get nine equals zero point 95 x. We divide by zero point 95 x and we get 9.47 grams of sodium chloride that we need to add to the solution of ten moles to get a 5% bimass solution of sodium chloride. The final step is to find the number of moles of sodium chloride. So we have the number of grams we need to add. But our question asks us to find the number of moles. To find that, we need to find the molecular weight of sodium chloride."}, {"title": "Mass percent example #2 .txt", "text": "So we have the number of grams we need to add. But our question asks us to find the number of moles. To find that, we need to find the molecular weight of sodium chloride. To find the molecular weight of sodium chloride, we simply add the atomic weights of sodium and chloride. So 23 grams/mol plus 35.5\ngrams/mol gives us 58.5 grams/mol. This is a molecular weight of sodium chloride."}, {"title": "Mass percent example #2 .txt", "text": "To find the molecular weight of sodium chloride, we simply add the atomic weights of sodium and chloride. So 23 grams/mol plus 35.5\ngrams/mol gives us 58.5 grams/mol. This is a molecular weight of sodium chloride. Now, to find a number of moles, we divide our molecular weight or we divide our amount that we have by our molecular weight. The grams cancel, the moles go on top. So 9.47\ndivided by 58.5 gives us 0.16 moles of sodium chloride."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So for example, let's look at HCL. HCL dissociates in water into an H plus ion and a chloride ion. And that means because HCL increases the concentration of our H plus ions found in water, by definition, HCL must be an Iranian acid. Now, Iranian bases are those substances that increase oh concentration in water. So for example, let's look at sodium Hydroxide. Sodium Hydroxide, when mixed with water, dissociates into sodium in an oh ion."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "Now, Iranian bases are those substances that increase oh concentration in water. So for example, let's look at sodium Hydroxide. Sodium Hydroxide, when mixed with water, dissociates into sodium in an oh ion. And that means it increases the concentration of our oh ions found in water. Therefore, this guide by the submission must be an Iranius base. Now, Arrangeous Aspen bases are always found in water that's by definition, whenever we talk about our radius acid bases, we talk about solutions in which our solvent is water."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "And that means it increases the concentration of our oh ions found in water. Therefore, this guide by the submission must be an Iranius base. Now, Arrangeous Aspen bases are always found in water that's by definition, whenever we talk about our radius acid bases, we talk about solutions in which our solvent is water. So let's look at this reaction again. So we know that HCL, when mixed in water, dissociates into a Hydride ion and a chloride ion. So now we have a solution that contains a bunch of H plus ions flowing around next to water."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So let's look at this reaction again. So we know that HCL, when mixed in water, dissociates into a Hydride ion and a chloride ion. So now we have a solution that contains a bunch of H plus ions flowing around next to water. So what actually happens is a base pair or a pair of electrons down on the water grabs the H ion, producing hydronium ion. So actually, the Hydride ion produced by an Iranius acid is always associated with a hydronium molecule. And that's because our solvent is always water."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So what actually happens is a base pair or a pair of electrons down on the water grabs the H ion, producing hydronium ion. So actually, the Hydride ion produced by an Iranius acid is always associated with a hydronium molecule. And that's because our solvent is always water. According to the Iranian acid based concept in Aqueous Solutions, the hydronium ions are responsible for acidic properties and the oh ions are responsible for the basic properties. Now, one problem with such a definition exists molecules such as ammonia, molecules that produce basic solutions and react with acids and yet they have no oh ions. So if we look at the structure of NH three Ammonium, we see that it contains three HS, one N and a pair of electrons."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "According to the Iranian acid based concept in Aqueous Solutions, the hydronium ions are responsible for acidic properties and the oh ions are responsible for the basic properties. Now, one problem with such a definition exists molecules such as ammonia, molecules that produce basic solutions and react with acids and yet they have no oh ions. So if we look at the structure of NH three Ammonium, we see that it contains three HS, one N and a pair of electrons. Clearly, it has no oh ions. So according to our Iranians Acid Based definition, this cannot be a base, yet it has basic properties. So this definition isn't that good."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "Clearly, it has no oh ions. So according to our Iranians Acid Based definition, this cannot be a base, yet it has basic properties. So this definition isn't that good. So they come up with a better definition, a definition that includes many more bases and many more acids. And this concept is called the Broccolari acid based concept. So Broccollory bases are those molecules that can accept a Hydride ion using a lone pair of electrons."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So they come up with a better definition, a definition that includes many more bases and many more acids. And this concept is called the Broccolari acid based concept. So Broccollory bases are those molecules that can accept a Hydride ion using a lone pair of electrons. So let's look at the reaction of Ammonia and water. So in ammonia, we have an extra pair of electrons. And in fact, this pair of electrons will take away an H ion from the water molecule and this will produce Ammonium and hydroxide."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So let's look at the reaction of Ammonia and water. So in ammonia, we have an extra pair of electrons. And in fact, this pair of electrons will take away an H ion from the water molecule and this will produce Ammonium and hydroxide. So by definition, because this guy has an extra pair of electrons, a lone pair of electrons, it acts as a bronzedidlower base. Now, a Bronzedidloric acid, are those molecules that can donate a Hydride ion. So, for example, let's look at the reaction of nitric acid and water."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So by definition, because this guy has an extra pair of electrons, a lone pair of electrons, it acts as a bronzedidlower base. Now, a Bronzedidloric acid, are those molecules that can donate a Hydride ion. So, for example, let's look at the reaction of nitric acid and water. So, nitric acid has an extra ion, extra hydrate ion. And in fact, the water has an extra pair of electrons that will take away this ion. And this will produce hydronium ion."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So, nitric acid has an extra ion, extra hydrate ion. And in fact, the water has an extra pair of electrons that will take away this ion. And this will produce hydronium ion. And this ion here, that's negatively charged. So, by the dimension nitric acid, because it could donate NH, it has an extra H. It act as a bronze of lyric acid. Now, notice something interesting."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "And this ion here, that's negatively charged. So, by the dimension nitric acid, because it could donate NH, it has an extra H. It act as a bronze of lyric acid. Now, notice something interesting. In this reaction, our water acted as an acid, a bronzed larry acid, because it donated NH, right? It gave an H to this guy producing ammonium and became hydroxide itself. So it was a base here I'm sorry, an acid here."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "In this reaction, our water acted as an acid, a bronzed larry acid, because it donated NH, right? It gave an H to this guy producing ammonium and became hydroxide itself. So it was a base here I'm sorry, an acid here. In this reaction, it took away an H.\nAnd because it took away an H, it was acting as a base. It accepted an age. So in this case, it was a base."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "In this reaction, it took away an H.\nAnd because it took away an H, it was acting as a base. It accepted an age. So in this case, it was a base. In this case, it was an acid. One interesting thing about a water molecule is that under basic conditions, it acts as an acid, and under acidic conditions, it acts as a base. Our third and final definition of an acid and a base is the most basic definition."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "In this case, it was an acid. One interesting thing about a water molecule is that under basic conditions, it acts as an acid, and under acidic conditions, it acts as a base. Our third and final definition of an acid and a base is the most basic definition. And this concept is called a lewis acidbased concept. Now, this includes all the bronzed Larry acid bases and many, many more. So let's look at a lewis acid."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "And this concept is called a lewis acidbased concept. Now, this includes all the bronzed Larry acid bases and many, many more. So let's look at a lewis acid. A lewis acid is anything that accepts a lone pair of electrons to form a new bond. And this bond is usually called a cotton covalent bond. And we'll see why in a second."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "A lewis acid is anything that accepts a lone pair of electrons to form a new bond. And this bond is usually called a cotton covalent bond. And we'll see why in a second. So we have BF three. Plus an ace that has a lone pair of electrons forms a cordon covalent bond, HBF three. And it's called Cortis covalent and not Covalent because both electrons come from a single atom."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So we have BF three. Plus an ace that has a lone pair of electrons forms a cordon covalent bond, HBF three. And it's called Cortis covalent and not Covalent because both electrons come from a single atom. Remember, in a Covalent bond, one electron comes from here, and one electron comes from here. But in this case, both electrons come from a single atom. So let's look at the structural depiction of BF three."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "Remember, in a Covalent bond, one electron comes from here, and one electron comes from here. But in this case, both electrons come from a single atom. So let's look at the structural depiction of BF three. BF three has three HS attached to the boron, and the boron has an empty orbital. And this empty orbital is what interacts with the lone pair of electrons on our H, forming our corticovalent bond. So in this case, the BF three accepted the pair of electrons because it had that anti orbital."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "BF three has three HS attached to the boron, and the boron has an empty orbital. And this empty orbital is what interacts with the lone pair of electrons on our H, forming our corticovalent bond. So in this case, the BF three accepted the pair of electrons because it had that anti orbital. So the BF three is our lewis acid. So another way to talk about a lewis acid is to say that a lewis acid is either neutral or a cation. In this case, it was neutral that has an empty orbital."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "So the BF three is our lewis acid. So another way to talk about a lewis acid is to say that a lewis acid is either neutral or a cation. In this case, it was neutral that has an empty orbital. Now let's look at lewis bases. A lewis base is anything that can donate a pair of electrons to form a new bond, a cordon covalent bond. Let's look at one of the most basic reactions."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "Now let's look at lewis bases. A lewis base is anything that can donate a pair of electrons to form a new bond, a cordon covalent bond. Let's look at one of the most basic reactions. An H atom plus a water molecule gives you a hydronium ion. So this is a cation that has an empty orbital. So this must be a lewis acid."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "An H atom plus a water molecule gives you a hydronium ion. So this is a cation that has an empty orbital. So this must be a lewis acid. Remember a cation and an empty orbital. Plus water. Well, water has what?"}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "Remember a cation and an empty orbital. Plus water. Well, water has what? It has a pair of electrons. It means it can donate a pair of electrons. So water must be our Lewis base."}, {"title": "Arrhenius, Bronsted-Lowry Example and Lecids Acids and Bases .txt", "text": "It has a pair of electrons. It means it can donate a pair of electrons. So water must be our Lewis base. And these guys interact to form hydronium ion. So in this case, in this reaction, this was our Lewis acid, and this was our Lewis base. So, once again, whenever we talk about Lewis Acid bases, we talk about a transfer of electrons."}, {"title": "Oxidation Numbers .txt", "text": "Now, this electrical charge is known as oxidation number or oxidation state of our atom. So oxidation states are just the different possible charge values that one can assign to specific atom. Now, these oxidation states help us keep track of these electrons. But in order to use these oxidation states, we must learn a few rules. Let's look at these rules. So, all atoms in their elemental state are given an oxidation state of zero."}, {"title": "Oxidation Numbers .txt", "text": "But in order to use these oxidation states, we must learn a few rules. Let's look at these rules. So, all atoms in their elemental state are given an oxidation state of zero. There are no exceptions to this rule. So atoms called fluorine always get oxidation states of negative one. There are no exceptions to this rule as well."}, {"title": "Oxidation Numbers .txt", "text": "There are no exceptions to this rule. So atoms called fluorine always get oxidation states of negative one. There are no exceptions to this rule as well. So atoms called hydrogen are given the oxidation state of plus one. But there's one exception to this rule. They're given oxidation state of negative one."}, {"title": "Oxidation Numbers .txt", "text": "So atoms called hydrogen are given the oxidation state of plus one. But there's one exception to this rule. They're given oxidation state of negative one. When combined with metals such as, for example, sodium or calcium, the atom oxygen is given to oxidation state of negative two. But there's an exception. It's given oxidation state of negative one when combined in the form H 202."}, {"title": "Oxidation Numbers .txt", "text": "When combined with metals such as, for example, sodium or calcium, the atom oxygen is given to oxidation state of negative two. But there's an exception. It's given oxidation state of negative one when combined in the form H 202. But that's because this rule is more important than this rule. So on this table, every rule that comes before it is more important. For example, this is the most important rule, the second most important rule, third most important rule, and fourth most important rule."}, {"title": "Oxidation Numbers .txt", "text": "But that's because this rule is more important than this rule. So on this table, every rule that comes before it is more important. For example, this is the most important rule, the second most important rule, third most important rule, and fourth most important rule. So when you're assigning these activation numbers, you must keep that in mind. Now, let's look at a few more guidelines that we can use when assigning oxidation states. Now, this table is meant to be a guideline."}, {"title": "Oxidation Numbers .txt", "text": "So when you're assigning these activation numbers, you must keep that in mind. Now, let's look at a few more guidelines that we can use when assigning oxidation states. Now, this table is meant to be a guideline. It's meant to help you assign oxidation states. They're not rules that will always work. And in fact, the table given before this table is more important."}, {"title": "Oxidation Numbers .txt", "text": "It's meant to help you assign oxidation states. They're not rules that will always work. And in fact, the table given before this table is more important. And so it precedes this table. But let's look at the general guidelines. So the group number one, the alkali metals are assigned an oxidation state of plus one."}, {"title": "Oxidation Numbers .txt", "text": "And so it precedes this table. But let's look at the general guidelines. So the group number one, the alkali metals are assigned an oxidation state of plus one. Group number two, the alkali earth metals are assigned oxidation state of plus two. So group number 15, including nitrogen, phosphorus, and so on, are assigned an oxidation state of negative three. Group number 16, including oxygen, sulfur, and so on, are a sign oxidation state of negative two."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "We're given the density bar of water to be 1 gram/ML and our constants for boiling and for freezing are 0.5 and 1.8. So our goal is to find the final freezing and boiling point of the solution. So in our initial condition, we have 800 our salad in our water and the boiling point of water is 100 celsius and deer Celsius. Now we want to find what the final freezing and boiling point of our solution is after we mix our solute into the mixture. So we expect our boiling point to rise and our freezing point to depress or decrease. Because when you add Solutes to pure substances, that decreases vapor pressure."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "Now we want to find what the final freezing and boiling point of our solution is after we mix our solute into the mixture. So we expect our boiling point to rise and our freezing point to depress or decrease. Because when you add Solutes to pure substances, that decreases vapor pressure. And if you want to learn more about vapor pressure and how it affects boiling points or freezing points, check out the video below. Now let's get back to our problem. So, before we can apply our two formulas for freezing and boiling points, or the change in temperature for freezing boiling points, we have to find the morality."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "And if you want to learn more about vapor pressure and how it affects boiling points or freezing points, check out the video below. Now let's get back to our problem. So, before we can apply our two formulas for freezing and boiling points, or the change in temperature for freezing boiling points, we have to find the morality. Because look, we have the I for both cases, it's simply one. We have our constants KB and KF they're given here. What we don't have is our molarity."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "Because look, we have the I for both cases, it's simply one. We have our constants KB and KF they're given here. What we don't have is our molarity. So as soon as we find our molarity, we could plug that in into our formulas and we get our results. So let's find our morality. In step one, we find the numerator of our molarity."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "So as soon as we find our molarity, we could plug that in into our formulas and we get our results. So let's find our morality. In step one, we find the numerator of our molarity. In step two, we find the denominator of our morality. So that means first we have to find the moles of our solute. But before we get to the moles of solute, we have to find the molar mass."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "In step two, we find the denominator of our morality. So that means first we have to find the moles of our solute. But before we get to the moles of solute, we have to find the molar mass. And you'll see why in a second. So let's find the molar mass of our ethylene glycol, our solute. So basically, we add up the atomic mass of each atom."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "And you'll see why in a second. So let's find the molar mass of our ethylene glycol, our solute. So basically, we add up the atomic mass of each atom. So, since we have two carbon atoms, we multiply two times 12 grams/mol for carbon. Since we have two O atoms, we multiply two times 16 grams/mol for oxygen. And since we have six H atoms, we multiply six times 1 gram/mol for h and we get all these guys up 62 grams/mol."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "So, since we have two carbon atoms, we multiply two times 12 grams/mol for carbon. Since we have two O atoms, we multiply two times 16 grams/mol for oxygen. And since we have six H atoms, we multiply six times 1 gram/mol for h and we get all these guys up 62 grams/mol. So that's our molar mass of our solute. Now, to find the moles of Solute, we take our grams of solute and divide that by molar mass and we should get 44 grams divided by 62 grams. The grams cancel moles dose on top and we get zero point 71 moles of our solute."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "So that's our molar mass of our solute. Now, to find the moles of Solute, we take our grams of solute and divide that by molar mass and we should get 44 grams divided by 62 grams. The grams cancel moles dose on top and we get zero point 71 moles of our solute. So now we have the top component of our mortality. Let's find the bottom. Remember, it's moles of solute divided by kilogram of solvent."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "So now we have the top component of our mortality. Let's find the bottom. Remember, it's moles of solute divided by kilogram of solvent. To find the kilograms of our solvent, we have to use the density of water, the fact that we have 800. Then we have to divide by a thousand because we want to go from grams to kilograms. So our density of water times milliliters of solvent divided by 1000."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "To find the kilograms of our solvent, we have to use the density of water, the fact that we have 800. Then we have to divide by a thousand because we want to go from grams to kilograms. So our density of water times milliliters of solvent divided by 1000. Once again, 1000. We divide by 1000 because we want to go from grams to kilograms. So 1 gram over ML, which is our density times 800 ML over 1000, gives us the MLS cancel grams becomes kilogram because we're dividing by 1000 and that gets us zero 8 solid."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "Once again, 1000. We divide by 1000 because we want to go from grams to kilograms. So 1 gram over ML, which is our density times 800 ML over 1000, gives us the MLS cancel grams becomes kilogram because we're dividing by 1000 and that gets us zero 8 solid. So now we have our Molarity, namely zero point 71 moles divided by 0.8\nkg, so a bit less than one. So we go to our final step. In our final step we simply write the formulas for the change in boiling point and changing freezing point when we add a salute."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "So now we have our Molarity, namely zero point 71 moles divided by 0.8\nkg, so a bit less than one. So we go to our final step. In our final step we simply write the formulas for the change in boiling point and changing freezing point when we add a salute. So let's find the change in boiling point first. Changing boiling point is equal to our constant times molarity times I. I is simply the number of particles that our solid associates into. And since this doesn't associate into anything, we simply put I equals one, so this equals zero five."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "So let's find the change in boiling point first. Changing boiling point is equal to our constant times molarity times I. I is simply the number of particles that our solid associates into. And since this doesn't associate into anything, we simply put I equals one, so this equals zero five. Our constant Celsius times kilogram over mole multiplied by zero 71 moles over zero eight our mole ounce the kilograms cancel, the moles cancel, we're left with 44 degrees Celsius. So our temperature, our boiling point increases by 00:44 degree Celsius degree and that basically bumps our boiling point up to 100.44\ndegrees Celsius. Now our change in freezing point is equal to our constant KF times molarity times i."}, {"title": "Boiling Point Elevation and Freezing Point Depression.txt", "text": "Our constant Celsius times kilogram over mole multiplied by zero 71 moles over zero eight our mole ounce the kilograms cancel, the moles cancel, we're left with 44 degrees Celsius. So our temperature, our boiling point increases by 00:44 degree Celsius degree and that basically bumps our boiling point up to 100.44\ndegrees Celsius. Now our change in freezing point is equal to our constant KF times molarity times i. Once again I is one. Our molarity stays the same, our constant changes. Now we plug in 1.8, we get 1.8\ntimes our molarity and we get approximately 1.6 degrees Celsius."}, {"title": "Nernst Equation Part I .txt", "text": "So far we have only spoken about electrochemical cells in which the reactants and products are both at standard state conditions. This means when I showed you the relationship between the concentration of reactants and products and the electrochemical cells cell voltage I did this on the standard state conditions. However, most reactions actions do not occur under standard state conditions. This means we must revisit the topic of the relationship between the concentrations and the cell voltage and adjusted for nonstandard state conditions. So, to begin, let's first talk about standard state conditions. So, standard state conditions simply means a concentration of one molar and a pressure of 1 bar and the temperature could be anything."}, {"title": "Nernst Equation Part I .txt", "text": "This means we must revisit the topic of the relationship between the concentrations and the cell voltage and adjusted for nonstandard state conditions. So, to begin, let's first talk about standard state conditions. So, standard state conditions simply means a concentration of one molar and a pressure of 1 bar and the temperature could be anything. Normally the temperature is 25 degrees Celsius or 298 degrees Kelvin. But it could really be anything. So, in a previous lecture we saw that the relationship between equilibrium constant K on the standard conditions is the following log of base ten of k is equal to number of moles of electrons and times our cell voltage on the same state conditions divided by 0.0592."}, {"title": "Nernst Equation Part I .txt", "text": "Normally the temperature is 25 degrees Celsius or 298 degrees Kelvin. But it could really be anything. So, in a previous lecture we saw that the relationship between equilibrium constant K on the standard conditions is the following log of base ten of k is equal to number of moles of electrons and times our cell voltage on the same state conditions divided by 0.0592. Now, this guy is at equilibrium is when our reaction is at equilibrium. And if our reaction has reactants A and b and C and D which are all in an Aqueous state, then our equilibrium constant expression is the following if k equals the ratio of the concentration of products divided by the ratio of the concentration of reactants. Now, to begin our process of adjusting for the new relationship between the concentration and the cell potential, let's look at gives free energy."}, {"title": "Nernst Equation Part I .txt", "text": "Now, this guy is at equilibrium is when our reaction is at equilibrium. And if our reaction has reactants A and b and C and D which are all in an Aqueous state, then our equilibrium constant expression is the following if k equals the ratio of the concentration of products divided by the ratio of the concentration of reactants. Now, to begin our process of adjusting for the new relationship between the concentration and the cell potential, let's look at gives free energy. Now, under nonstandard state conditions we can make an adjustment to gifts free energy formula. Remember, before under standard state conditions our formula was the following. It was simply gifts free energy or change in gifts free energy under standard state conditions is equal to negative of this guy."}, {"title": "Nernst Equation Part I .txt", "text": "Now, under nonstandard state conditions we can make an adjustment to gifts free energy formula. Remember, before under standard state conditions our formula was the following. It was simply gifts free energy or change in gifts free energy under standard state conditions is equal to negative of this guy. But now we are not under Stan and state conditions. And we must make that adjustment. So, the adjustment is the following."}, {"title": "Nernst Equation Part I .txt", "text": "But now we are not under Stan and state conditions. And we must make that adjustment. So, the adjustment is the following. Now, this guy is now our term used to adjust for non equilibrium and nonstandard state conditions. Now, this is still our free energy at equilibrium for state and state conditions. And this is the actual free energy at some point in our reaction."}, {"title": "Nernst Equation Part I .txt", "text": "Now, this guy is now our term used to adjust for non equilibrium and nonstandard state conditions. Now, this is still our free energy at equilibrium for state and state conditions. And this is the actual free energy at some point in our reaction. Now, let's look at this reaction. Suppose we had some reaction that wasn't at Stan and state condition and that wasn't an equilibrium. So our reactant A was an acre state, reactant B was an acre state, our product C was also an acrylic state."}, {"title": "Nernst Equation Part I .txt", "text": "Now, let's look at this reaction. Suppose we had some reaction that wasn't at Stan and state condition and that wasn't an equilibrium. So our reactant A was an acre state, reactant B was an acre state, our product C was also an acrylic state. And our product D was also an acreage state. Now, this Q has the same meaning as this K. It's also a constant. And we can use or develop an expression for this reaction for this Q in the same way that we did here."}, {"title": "Nernst Equation Part I .txt", "text": "And our product D was also an acreage state. Now, this Q has the same meaning as this K. It's also a constant. And we can use or develop an expression for this reaction for this Q in the same way that we did here. And here it is. So, q is equal to the product of the concentrations of the products divided by the product of the concentrations of reactants. Now, what happens when these guys begin reacting forming C and D, while the concentration of C and D Begins to Increase."}, {"title": "Nernst Equation Part I .txt", "text": "And here it is. So, q is equal to the product of the concentrations of the products divided by the product of the concentrations of reactants. Now, what happens when these guys begin reacting forming C and D, while the concentration of C and D Begins to Increase. While the concentration of A and B begins to decrease. And so our Top begins to increase and bottom Begins to Increase. Our Q becomes larger."}, {"title": "Nernst Equation Part I .txt", "text": "While the concentration of A and B begins to decrease. And so our Top begins to increase and bottom Begins to Increase. Our Q becomes larger. But notice that according to Lush at Leer Principle, if our concentration of products begin to Increase, what will tend to happen? Well, if we have more of this guy equilibrium will slowly begin to shift to the Left. This is according to Le Chaplier principle."}, {"title": "Nernst Equation Part I .txt", "text": "But notice that according to Lush at Leer Principle, if our concentration of products begin to Increase, what will tend to happen? Well, if we have more of this guy equilibrium will slowly begin to shift to the Left. This is according to Le Chaplier principle. Now what this equation does is it agrees with Chiplier Principle. With this idea that if we have more products, our equilibrium will shift this way. Now let's see how exactly it agrees."}, {"title": "Nernst Equation Part I .txt", "text": "Now what this equation does is it agrees with Chiplier Principle. With this idea that if we have more products, our equilibrium will shift this way. Now let's see how exactly it agrees. Well, let's look. This is our change in Gifs free energy at equilibrium at Seven state conditions. And whenever it's Negative, that means our reaction is product Favored, right?"}, {"title": "Nernst Equation Part I .txt", "text": "Well, let's look. This is our change in Gifs free energy at equilibrium at Seven state conditions. And whenever it's Negative, that means our reaction is product Favored, right? It's spontaneous at those conditions. So if this is A negative term and this is A positive term. Then together they will be less negative and more positive."}, {"title": "Nernst Equation Part I .txt", "text": "It's spontaneous at those conditions. So if this is A negative term and this is A positive term. Then together they will be less negative and more positive. So this term adjusts for lusha clear's principle. Because look, if our Q Begins to increase when more products are formed, that means Q is say bigger than One. And whenever we take the natural log of any number that's bigger than One, we get A positive number."}, {"title": "Nernst Equation Part I .txt", "text": "So this term adjusts for lusha clear's principle. Because look, if our Q Begins to increase when more products are formed, that means Q is say bigger than One. And whenever we take the natural log of any number that's bigger than One, we get A positive number. So this whole thing is A positive number. Because R is positive to constant and T is positive to temperature. So this guy will be Positive, right?"}, {"title": "Nernst Equation Part I .txt", "text": "So this whole thing is A positive number. Because R is positive to constant and T is positive to temperature. So this guy will be Positive, right? That means our equilibrium, some negative number plus A positive number will make A more positive number. So let's principle says that when we when we form more products, our reaction will shift to the Left. That's exactly what this guy says as well."}, {"title": "Nernst Equation Part I .txt", "text": "That means our equilibrium, some negative number plus A positive number will make A more positive number. So let's principle says that when we when we form more products, our reaction will shift to the Left. That's exactly what this guy says as well. If this is A negative number and you add A positive number to it, we're going to get A more positive result. More positive, free energy. And what does gifts free energy tell us when it's positive?"}, {"title": "Nernst Equation Part I .txt", "text": "If this is A negative number and you add A positive number to it, we're going to get A more positive result. More positive, free energy. And what does gifts free energy tell us when it's positive? It tells us our reaction is react in favor. So this is an equation for gifts free energy under non thin state conditions. So now we can use this formula."}, {"title": "Nernst Equation Part I .txt", "text": "It tells us our reaction is react in favor. So this is an equation for gifts free energy under non thin state conditions. So now we can use this formula. And in Part C we can plug things in. Let's look. So in another lecture we saw that changing gifts the energy on the same and state Conditions is equal to negative number of Moles and of electrons times our Faraday's constant times cell Voltage."}, {"title": "Nernst Equation Part I .txt", "text": "And in Part C we can plug things in. Let's look. So in another lecture we saw that changing gifts the energy on the same and state Conditions is equal to negative number of Moles and of electrons times our Faraday's constant times cell Voltage. Well, we can write the same equation for non Simmon state Conditions. Except now we're not using Dot. Because now we're not at SIM and state conditions."}, {"title": "Nernst Equation Part I .txt", "text": "Well, we can write the same equation for non Simmon state Conditions. Except now we're not using Dot. Because now we're not at SIM and state conditions. So Now let's plug these guys in into this formula. And this is what we get. Well, we plug this guy in into this guy here and this guy into this guy here."}, {"title": "Nernst Equation Example .txt", "text": "So zinc solid reacts with nickel in the April state to produce zinc in the April state and nickel solid. We are also given our cell potential or cell voltage of our electrochemical cell under stanley conditions of 1 bar pressure and one molar concentration. And this is zero point 51 volts. We're also given the concentrations of nickel of this guy and the concentration of the zinc in the acreage state of this guy. So notice that two and three represent nonstandard conditions. And in fact, our goal is to find the cell potential under these nonstandard state conditions."}, {"title": "Nernst Equation Example .txt", "text": "We're also given the concentrations of nickel of this guy and the concentration of the zinc in the acreage state of this guy. So notice that two and three represent nonstandard conditions. And in fact, our goal is to find the cell potential under these nonstandard state conditions. So whenever we hear the word nonstandard and cell potential, we have to think nurse equation. That's exactly what we do in step two. But before we can use this equation, we have to find our Q."}, {"title": "Nernst Equation Example .txt", "text": "So whenever we hear the word nonstandard and cell potential, we have to think nurse equation. That's exactly what we do in step two. But before we can use this equation, we have to find our Q. So in step one, that's what we do. Remember, Q is a ratio of the concentration of products over the concentration of reactants. And our expression is similar to K equilibrium constant, except this condition represents a situation that is not an equilibrium."}, {"title": "Nernst Equation Example .txt", "text": "So in step one, that's what we do. Remember, Q is a ratio of the concentration of products over the concentration of reactants. And our expression is similar to K equilibrium constant, except this condition represents a situation that is not an equilibrium. So we basically take our concentration of this guy and divide it by the concentration of reactants of this guy. Now, we don't include this guy or this guy why? Or because they're in the solid states."}, {"title": "Nernst Equation Example .txt", "text": "So we basically take our concentration of this guy and divide it by the concentration of reactants of this guy. Now, we don't include this guy or this guy why? Or because they're in the solid states. Solid atoms or liquid atoms are not included in our expression. Only gas molecules and aqueous atoms are included in our expression. And notice that we're giving these guys, right?"}, {"title": "Nernst Equation Example .txt", "text": "Solid atoms or liquid atoms are not included in our expression. Only gas molecules and aqueous atoms are included in our expression. And notice that we're giving these guys, right? This guy is 0.5\nmolar, while this guy is five molar. So we divide 0.5 divided by five. That gives us one over 100, which is 0.1 the M cancel."}, {"title": "Nernst Equation Example .txt", "text": "This guy is 0.5\nmolar, while this guy is five molar. So we divide 0.5 divided by five. That gives us one over 100, which is 0.1 the M cancel. So this guy is unitless. And recall what Q tells us about our reaction. If Q is less than one, that means we have a lot of this guy and none of this guy."}, {"title": "Nernst Equation Example .txt", "text": "So this guy is unitless. And recall what Q tells us about our reaction. If Q is less than one, that means we have a lot of this guy and none of this guy. In this situation, for every one of this guy, for every one of the in the acre state, we have 100 of nickels in the Aqueous state. And that means we have much more reactants than products. And according to Leslie Clay's principle, this reaction will be very product favorite."}, {"title": "Nernst Equation Example .txt", "text": "In this situation, for every one of this guy, for every one of the in the acre state, we have 100 of nickels in the Aqueous state. And that means we have much more reactants than products. And according to Leslie Clay's principle, this reaction will be very product favorite. It will want to form the products. So that's what Q tells us. Let's see what we can deduce from number two and number two."}, {"title": "Nernst Equation Example .txt", "text": "It will want to form the products. So that's what Q tells us. Let's see what we can deduce from number two and number two. We're going to find the cell potential. Now, e is equal to e this guy minus r times t divided by N times F times Ln natural log of Q. Now, under a pressure or under a temperature of 25 degrees Celsius, we can rewrite this equation into this format."}, {"title": "Nernst Equation Example .txt", "text": "We're going to find the cell potential. Now, e is equal to e this guy minus r times t divided by N times F times Ln natural log of Q. Now, under a pressure or under a temperature of 25 degrees Celsius, we can rewrite this equation into this format. Now this guy we know, the Q we know. So what is the N? Well, the N is the number of moles of electrons released by this equation."}, {"title": "Nernst Equation Example .txt", "text": "Now this guy we know, the Q we know. So what is the N? Well, the N is the number of moles of electrons released by this equation. So notice that our zinc solid is the guy that gets oxidized. It releases two moles of electrons, right? So that means this guy accepts two moles."}, {"title": "Nernst Equation Example .txt", "text": "So notice that our zinc solid is the guy that gets oxidized. It releases two moles of electrons, right? So that means this guy accepts two moles. It's reduced into zinc solid. So our N is two. So let's plug all our values into our formula and we get zero point 51 volts minus this number times log of 0.1\ndivided by our moles two."}, {"title": "Nernst Equation Example .txt", "text": "It's reduced into zinc solid. So our N is two. So let's plug all our values into our formula and we get zero point 51 volts minus this number times log of 0.1\ndivided by our moles two. Now what we get is a number zero point 56 nine two. That's greater than our cell potential at equilibrium. What that means is that this condition of this concentrations of nonstandard conditions represent a situation in which the reaction is more product favored than at equilibrium."}, {"title": "Nernst Equation Example .txt", "text": "Now what we get is a number zero point 56 nine two. That's greater than our cell potential at equilibrium. What that means is that this condition of this concentrations of nonstandard conditions represent a situation in which the reaction is more product favored than at equilibrium. And that makes sense because this guy represents a condition at the beginning. So when we first add our Z and our nickel, what will happen? Well, we won't have a lot of products form, we're going to have a lot of reactants."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "So, in this lecture, we're going to look at a relationship between the equilibrium constant of a reduction reaction found in the electrochemical cell and that electrochemical cells cell voltage or its electromotive force. Now, before we look at the relationship, let's remember what equilibrium constant is. Suppose we have the following expression or reaction in which we have two reactants and two products. Now our A reactant is in the gas state, our B reactor is in the liquid state, our C product is in the Aqueous state and our D product is in the Aqueous state. So let's write the equilibrium constant expression for this reaction. And let's assume that equilibrium has been reached."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Now our A reactant is in the gas state, our B reactor is in the liquid state, our C product is in the Aqueous state and our D product is in the Aqueous state. So let's write the equilibrium constant expression for this reaction. And let's assume that equilibrium has been reached. So our equilibrium constant under standard conditions is equal to the concentration of product C times the concentration of product D divided by the concentration of product A. Now remember, whenever we're writing our equilibrium constant expressions, only aqueous molecules and gas molecules count in our equilibrium expression. We don't include liquid molecules or solid molecules."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "So our equilibrium constant under standard conditions is equal to the concentration of product C times the concentration of product D divided by the concentration of product A. Now remember, whenever we're writing our equilibrium constant expressions, only aqueous molecules and gas molecules count in our equilibrium expression. We don't include liquid molecules or solid molecules. And that's exactly why we don't include the B molecule. Because it's in a liquid state. So we only get this expression no b is included."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And that's exactly why we don't include the B molecule. Because it's in a liquid state. So we only get this expression no b is included. Now notice that this k is simply a ratio between the product concentration and the reactant concentration. So we say that if our K is greater than one, that means our reaction is favorable. If our K is less than one, our reaction is unfavorable."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Now notice that this k is simply a ratio between the product concentration and the reactant concentration. So we say that if our K is greater than one, that means our reaction is favorable. If our K is less than one, our reaction is unfavorable. In other words, if it's greater than one, that means our equilibrium lies to the right. Because these guys completely react to form our products. In other words, if K is much larger than one, we have very little of our concentration of A and a lot of concentration of products, namely C and D.\nLikewise, if K is much less than one, that means we have lots of this guy left over at equilibrium and very little affirmation of our products."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "In other words, if it's greater than one, that means our equilibrium lies to the right. Because these guys completely react to form our products. In other words, if K is much larger than one, we have very little of our concentration of A and a lot of concentration of products, namely C and D.\nLikewise, if K is much less than one, that means we have lots of this guy left over at equilibrium and very little affirmation of our products. And that means equilibrium lies to the left, which also means we have very little of this guy and a lot of this guy. Now, let's look at B. So we just learned that the change in Gibbs free energy is equal to negative."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And that means equilibrium lies to the left, which also means we have very little of this guy and a lot of this guy. Now, let's look at B. So we just learned that the change in Gibbs free energy is equal to negative. N the number of moles of electrons times staradate constant times the cell voltage. Now, from before, we know that the change in Gates free Energy can Also Be Expressed as negative r times T.\nWell, R is the Gas constant at T's temperature in Kelvin times our Natural log of our K, our equilibrium constant expression on The Standard Conditions. Now, what we can do is set these guys equal."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "N the number of moles of electrons times staradate constant times the cell voltage. Now, from before, we know that the change in Gates free Energy can Also Be Expressed as negative r times T.\nWell, R is the Gas constant at T's temperature in Kelvin times our Natural log of our K, our equilibrium constant expression on The Standard Conditions. Now, what we can do is set these guys equal. Why? Well, because this and this are the same expressions, because they both equal the same thing. Namely change in Gibbs free energy."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Why? Well, because this and this are the same expressions, because they both equal the same thing. Namely change in Gibbs free energy. So let's set these guys equal. So we get negative MF times E equals negative RT, notch and log of K.\nSo we can basically rearrange this guy and solve for our cell voltage. And what we get is the negatives cancel and we bring the M and the F to this side."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "So let's set these guys equal. So we get negative MF times E equals negative RT, notch and log of K.\nSo we can basically rearrange this guy and solve for our cell voltage. And what we get is the negatives cancel and we bring the M and the F to this side. And we get our cell voltage equal to gas constant times temperature tail and divided by moles of electrons times paradise constant times natural log of our equilibrium constant. So this expression basically relates our cells voltage and the concentration of our reactants and products or the ratio of the concentration. Now, notice in this equation we have R is our constant and F is also constant."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And we get our cell voltage equal to gas constant times temperature tail and divided by moles of electrons times paradise constant times natural log of our equilibrium constant. So this expression basically relates our cells voltage and the concentration of our reactants and products or the ratio of the concentration. Now, notice in this equation we have R is our constant and F is also constant. And if we assume constant temperature of, say, 25 degrees Celsius, we can simplify this expression into the following expression. So, let's rewrite this expression. Cell voltage E is equal to a constant times a constant outcase because T is 298 in Kelvin."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And if we assume constant temperature of, say, 25 degrees Celsius, we can simplify this expression into the following expression. So, let's rewrite this expression. Cell voltage E is equal to a constant times a constant outcase because T is 298 in Kelvin. Since we assume that temperature is constant and temperatures of 25 degrees Celsius, that's our assumption. That means in Kelvin, 25 plus a 73 is 298. So a constant times our temperature that's constant divided by another constant called Faraday's constant, we get something called 0.257 a number, right?"}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Since we assume that temperature is constant and temperatures of 25 degrees Celsius, that's our assumption. That means in Kelvin, 25 plus a 73 is 298. So a constant times our temperature that's constant divided by another constant called Faraday's constant, we get something called 0.257 a number, right? So this expression under these conditions gives us the following simplified expression. And this expression is nice because here we have three unknowns. This is our unknown, this is our unknown, and this is our unknown."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "So this expression under these conditions gives us the following simplified expression. And this expression is nice because here we have three unknowns. This is our unknown, this is our unknown, and this is our unknown. So if we, for example, know our cell voltage and we know our number of moles, we can rearrange this equation and solve for K in the following way. Natural log of K is equal to N times cell voltage divided by zero point 25 seven. And suppose now we know our K and we know our end."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "So if we, for example, know our cell voltage and we know our number of moles, we can rearrange this equation and solve for K in the following way. Natural log of K is equal to N times cell voltage divided by zero point 25 seven. And suppose now we know our K and we know our end. That means we can use our equilibrium constant and our end to find our cell voltage. So these two equations become very useful. And these two equations build a relationship between the concentration of reactants and products and the cell's voltage."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "That means we can use our equilibrium constant and our end to find our cell voltage. So these two equations become very useful. And these two equations build a relationship between the concentration of reactants and products and the cell's voltage. Now, notice one interesting thing. Whenever we take natural log of a number that's bigger than one, this expression becomes positive. And that means if this expression is positive, then this is positive."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Now, notice one interesting thing. Whenever we take natural log of a number that's bigger than one, this expression becomes positive. And that means if this expression is positive, then this is positive. And recall that if our cell voltage is positive, that means we have a product favorite reaction. It's spontaneous. And that makes sense because earlier we said if we have a K greater than one, that means we have a favorable reaction."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And recall that if our cell voltage is positive, that means we have a product favorite reaction. It's spontaneous. And that makes sense because earlier we said if we have a K greater than one, that means we have a favorable reaction. Our reaction favors this going this way. Likewise, if this K is less than one, if it's a zero nine, then this guy, our natural log of a number that's less than one becomes negative. So our e becomes negative."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Our reaction favors this going this way. Likewise, if this K is less than one, if it's a zero nine, then this guy, our natural log of a number that's less than one becomes negative. So our e becomes negative. And that means if E is negative, our reaction is not product favorite, it's reactant favorite. And that means it's not favorable. And just like our K is less than one means that our reaction is unfavorable."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And that means if E is negative, our reaction is not product favorite, it's reactant favorite. And that means it's not favorable. And just like our K is less than one means that our reaction is unfavorable. The following thing I want to talk to you about is the following. I want to show you how we can convert the formula we just found to something simpler. Now, notice we're dealing with natural logs."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "The following thing I want to talk to you about is the following. I want to show you how we can convert the formula we just found to something simpler. Now, notice we're dealing with natural logs. That means if we were to convert this log to exponents, we would have to deal with bases of E.\nNow, bases Of E Aren't Very Easy To Calculate. You need a calculator to calculate basis of E. For example, you don't know what E to the one, E to the two. Each of the three."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "That means if we were to convert this log to exponents, we would have to deal with bases of E.\nNow, bases Of E Aren't Very Easy To Calculate. You need a calculator to calculate basis of E. For example, you don't know what E to the one, E to the two. Each of the three. Well, maybe e to the one you do but e to the two e to the three to the four you don't really know what that is with that using calculator. But bases like Base Ten, that's easy to calculate, right? Ten to the 110 to the 210 to the three."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Well, maybe e to the one you do but e to the two e to the three to the four you don't really know what that is with that using calculator. But bases like Base Ten, that's easy to calculate, right? Ten to the 110 to the 210 to the three. That's easy to calculate. You don't need a calculator. And therefore, our goal is to convert this natural log into base ten log?"}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "That's easy to calculate. You don't need a calculator. And therefore, our goal is to convert this natural log into base ten log? And the way we do it is we have to remember from algebra what the base conversion formula is for logs. In other words, this is a formula that basically tells us we can convert any log of any base of any inside to the base of ten. So the way we do it is log of base X of Y is equal to log of base ten of y divided by log of base ten of X."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And the way we do it is we have to remember from algebra what the base conversion formula is for logs. In other words, this is a formula that basically tells us we can convert any log of any base of any inside to the base of ten. So the way we do it is log of base X of Y is equal to log of base ten of y divided by log of base ten of X. Now, in our case, we have a natural log of some Y. And that's equivalent to saying log of base e of Y equals. That means we have to divide log of base ten of Y divided by log of base ten of E. So this is exactly what we follow in this process here."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Now, in our case, we have a natural log of some Y. And that's equivalent to saying log of base e of Y equals. That means we have to divide log of base ten of Y divided by log of base ten of E. So this is exactly what we follow in this process here. This step natural log of K is equal to log of base. E of K is equal to same exact process. This guy over this guy."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "This step natural log of K is equal to log of base. E of K is equal to same exact process. This guy over this guy. Same thing that we did here is equal to now, this is something we know. We know what e is. E to the one is some number."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Same thing that we did here is equal to now, this is something we know. We know what e is. E to the one is some number. So we basically plug this guy into the calculator and we find that it's zero point 43 four. So now we have a lot of base. Ten of K is equal to this whole guy."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "So we basically plug this guy into the calculator and we find that it's zero point 43 four. So now we have a lot of base. Ten of K is equal to this whole guy. And this guy is over this guy, right? So what we do next is we bring this guy over to this side. And that's what we do in step e. And we get late of base."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "And this guy is over this guy, right? So what we do next is we bring this guy over to this side. And that's what we do in step e. And we get late of base. Ten of K equals this guy is brought over to here. And what we get is zero point 43 four divided by zero point 25 seven times N times e. Now we plug this guy to the calculator, and we get 16.89\nmoles of electrons. Times or times moles electrons times cell voltage."}, {"title": "Cell voltage and Equilibrium constant .txt", "text": "Ten of K equals this guy is brought over to here. And what we get is zero point 43 four divided by zero point 25 seven times N times e. Now we plug this guy to the calculator, and we get 16.89\nmoles of electrons. Times or times moles electrons times cell voltage. Now, in textbooks you'll find this formula. But this guy and this guy are the same. Now, the way you go from this formula to this formula is you simply take this number and you reach the negative one power and you get this expression."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Now, examples include melting, freezing, evaporation and condensation. Now note that other examples exist. Now, let's look at evaporation of water. Whenever a water molecule in the liquid state gains enough kinetic energy, it escapes the bonds of the liquid molecules and becomes a gas molecule. And that means our compound goes from the liquid state to the gas state. But note that our molecular formula remains H 20."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Whenever a water molecule in the liquid state gains enough kinetic energy, it escapes the bonds of the liquid molecules and becomes a gas molecule. And that means our compound goes from the liquid state to the gas state. But note that our molecular formula remains H 20. And that means evaporation is a physical reaction. Now, on the contrary, whenever a compound goes from one molecular formula to a different compound with a different molecular formula, this type of reaction is known as a chemical reaction. Now, examples include combustion of hydrocarbons in which a hydrocarbon burns in the presence of oxygen to produce carbon dioxide and water."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "And that means evaporation is a physical reaction. Now, on the contrary, whenever a compound goes from one molecular formula to a different compound with a different molecular formula, this type of reaction is known as a chemical reaction. Now, examples include combustion of hydrocarbons in which a hydrocarbon burns in the presence of oxygen to produce carbon dioxide and water. Other examples include oxidation reduction reactions, combination and addition reactions. Now, let's look at the following chemical reaction in which the following hydrocarbon C, two S, H, six reacts in the presence of oxygen to form carbon dioxide and a water molecule. Now, remember what the conservation of energy and the conservation of mass tells us."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Other examples include oxidation reduction reactions, combination and addition reactions. Now, let's look at the following chemical reaction in which the following hydrocarbon C, two S, H, six reacts in the presence of oxygen to form carbon dioxide and a water molecule. Now, remember what the conservation of energy and the conservation of mass tells us. Mass cannot be destroyed or created. It always exists. And that means whatever we put into our equation, we have to take out."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Mass cannot be destroyed or created. It always exists. And that means whatever we put into our equation, we have to take out. In other words, the amount of atoms of each respective atom the carbon, the H and the O. Whatever we put in, we must get out. Now, notice in this equation we put in two C atoms and get one C. We put in six H atoms and only get two H atoms back."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "In other words, the amount of atoms of each respective atom the carbon, the H and the O. Whatever we put in, we must get out. Now, notice in this equation we put in two C atoms and get one C. We put in six H atoms and only get two H atoms back. And somehow we put in only two O atoms and we get three O atoms back. This type of an equation where our numbers are unbalanced, our coefficients and atoms are imbalanced, is known as an unbalanced equation. To balance this equation, we basically have to multiply each atom by some number so that the number on this side of each atom equals the number on this side."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "And somehow we put in only two O atoms and we get three O atoms back. This type of an equation where our numbers are unbalanced, our coefficients and atoms are imbalanced, is known as an unbalanced equation. To balance this equation, we basically have to multiply each atom by some number so that the number on this side of each atom equals the number on this side. So let's balance this out. Now, let's begin with carbon. Now, on this side, we have two carbons, while on this side, we only have one carbon."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "So let's balance this out. Now, let's begin with carbon. Now, on this side, we have two carbons, while on this side, we only have one carbon. To balance this out, we multiply this side, this CO2, by two. If we multiply this by two, we get two carbons on this side and two carbons on this side. So now, whatever amount of carbons we put in, namely two, we get two back."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "To balance this out, we multiply this side, this CO2, by two. If we multiply this by two, we get two carbons on this side and two carbons on this side. So now, whatever amount of carbons we put in, namely two, we get two back. So that makes sense. Next, let's balance out the H atoms. Now, we put in six H atoms and only get two back."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "So that makes sense. Next, let's balance out the H atoms. Now, we put in six H atoms and only get two back. So let's multiply this water molecule by a coefficient of three. So if we multiply it by three, we get six H atoms and six H atoms. So we're putting in two C atoms and six H atoms and we're getting back two C atoms and six H atoms."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "So let's multiply this water molecule by a coefficient of three. So if we multiply it by three, we get six H atoms and six H atoms. So we're putting in two C atoms and six H atoms and we're getting back two C atoms and six H atoms. Finally, let's balance the oxygen out. So notice we have two times two oxygens. So four oxygen here, and three times one oxygen, three oxygens here."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Finally, let's balance the oxygen out. So notice we have two times two oxygens. So four oxygen here, and three times one oxygen, three oxygens here. So we have a total of seven oxygens here. That means we have to multiply this guy by seven over two, because seven over two times two, the twos cancel and we are left with seven oxygen molecules here. So, in other words, we put in two carbons, six HS, and seven OS, and we get back two carbons, six H's, and three times one, two times four, seven O's."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "So we have a total of seven oxygens here. That means we have to multiply this guy by seven over two, because seven over two times two, the twos cancel and we are left with seven oxygen molecules here. So, in other words, we put in two carbons, six HS, and seven OS, and we get back two carbons, six H's, and three times one, two times four, seven O's. So this type of reaction is called a balanced equation. Now, notice that what these coefficients represent is moles or molecules or atoms. But what they can't never represent is mass."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "So this type of reaction is called a balanced equation. Now, notice that what these coefficients represent is moles or molecules or atoms. But what they can't never represent is mass. So these numbers, these coefficients, in this case, one, one, one, in this case, one seven over two, two and three will never represent mass. So never kilograms, never grams, never pounds, never any type of mass. Only moles atoms or molecules."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "So these numbers, these coefficients, in this case, one, one, one, in this case, one seven over two, two and three will never represent mass. So never kilograms, never grams, never pounds, never any type of mass. Only moles atoms or molecules. In other words, 1 mol of this hydrocarbon reacts with seven over two moles of diatomic oxygen to produce two moles of carbon dioxide and three moles of water. We can never say 1 gram of this car react with seven over 2 grams. That makes no sense."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "In other words, 1 mol of this hydrocarbon reacts with seven over two moles of diatomic oxygen to produce two moles of carbon dioxide and three moles of water. We can never say 1 gram of this car react with seven over 2 grams. That makes no sense. These coefficients represent moles, molecules or atom, never any mass. So whenever a reaction is set to run to completion, what that basically means is that one of the reactants is completely used up. It's completely depleted."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "These coefficients represent moles, molecules or atom, never any mass. So whenever a reaction is set to run to completion, what that basically means is that one of the reactants is completely used up. It's completely depleted. Note, however, most reactions do not run to completion. And that's because before one of the reactants is used up, equilibrium is established. Now, suppose the following reaction x plus Y react to form our product XY."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Note, however, most reactions do not run to completion. And that's because before one of the reactants is used up, equilibrium is established. Now, suppose the following reaction x plus Y react to form our product XY. And suppose this reaction achieves equilibrium before one of the reactants is used up. What that means is that the forward reaction rate is equal to the reverse reaction rate. So the rate at which our reactants react to produce our product is the same as the rate at which the product associates into our reactants."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "And suppose this reaction achieves equilibrium before one of the reactants is used up. What that means is that the forward reaction rate is equal to the reverse reaction rate. So the rate at which our reactants react to produce our product is the same as the rate at which the product associates into our reactants. In other words, the concentration of these guys and this guy remains the same even though the reactions are occurring. And that's because they're occurring at the same rates. Now, suppose we have the following combustion reaction in which methane combusts in the presence of oxygen to produce carbon dioxide and water."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "In other words, the concentration of these guys and this guy remains the same even though the reactions are occurring. And that's because they're occurring at the same rates. Now, suppose we have the following combustion reaction in which methane combusts in the presence of oxygen to produce carbon dioxide and water. Suppose this reaction is allowed to go to completion. Now, what happens if two moles of methane of this guy react with six moles of water? So what happens?"}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Suppose this reaction is allowed to go to completion. Now, what happens if two moles of methane of this guy react with six moles of water? So what happens? Well, let's see. Let's first notice that the ratio of the number of moles of methane, the number of moles of oxygen is one to two. So that means whenever 1 mol reacts with two moles of this guy, they produce carbon dioxide, 1 mol and two moles of water."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "Well, let's see. Let's first notice that the ratio of the number of moles of methane, the number of moles of oxygen is one to two. So that means whenever 1 mol reacts with two moles of this guy, they produce carbon dioxide, 1 mol and two moles of water. So now we have two moles of methane reacting with oxygen. That means, since not one, but two moles of methane are reacting, that means not two, but four moles of oxygen will react to produce our products. That's because two times two is four."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "So now we have two moles of methane reacting with oxygen. That means, since not one, but two moles of methane are reacting, that means not two, but four moles of oxygen will react to produce our products. That's because two times two is four. That means we're going to produce two moles of carbon dioxide and four moles of oxygen. Now, it's exactly what I said here. Once our reaction runs to completion, two moles of methane and four moles of oxygen are used up to produce two moles of carbon dioxide and four moles of H 20."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "That means we're going to produce two moles of carbon dioxide and four moles of oxygen. Now, it's exactly what I said here. Once our reaction runs to completion, two moles of methane and four moles of oxygen are used up to produce two moles of carbon dioxide and four moles of H 20. But notice we begin with six moles of oxygen. That means if only four moles of oxygen were used up, we have two moles of oxygen left over. And that means one of the reactants is completely used up."}, {"title": "Balancing Chemical Reactions and Limiting Reagents .txt", "text": "But notice we begin with six moles of oxygen. That means if only four moles of oxygen were used up, we have two moles of oxygen left over. And that means one of the reactants is completely used up. It's depleted while the other reactants is still left. So some of this guy is left over. That means this guy is our limiting reagent."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "I will show you that the movement of ions across the cell membrane in a neuron cell is identical to the movement of ions in a concentration cell. And in fact, a neuron cell is a more complex version of a concentration cell. So let's begin. So let's this is our cell membrane in a neuron cell. Here's our outside, the cell and the inside of the cell. Now, assuming our cell reaches a rusting electrical potential, we're going to have many more of these sodium ions on the outside than on the inside."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So let's this is our cell membrane in a neuron cell. Here's our outside, the cell and the inside of the cell. Now, assuming our cell reaches a rusting electrical potential, we're going to have many more of these sodium ions on the outside than on the inside. Now, what will entropy tell us about this situation? Well, entropy will tell us that our system is not even and we will want to move towards a more even system. In other words, some of these guys will want to move inside the cell so that we have the same number of ions on the inside as outside."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "Now, what will entropy tell us about this situation? Well, entropy will tell us that our system is not even and we will want to move towards a more even system. In other words, some of these guys will want to move inside the cell so that we have the same number of ions on the inside as outside. So at this point, we say there's a high concentration gradient or high chemical gradient on the outside. And the concentration gradient simply comes from the number of molecules. The more molecules they are, the higher the concentration gradient."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So at this point, we say there's a high concentration gradient or high chemical gradient on the outside. And the concentration gradient simply comes from the number of molecules. The more molecules they are, the higher the concentration gradient. And things tend to move from a high concentration gradient to a low concentration gradient. So ions will want to move from the outside to the inside due to a chemical or concentration gradient. Now notice what happens as our positively charged ions begin to move to the inside."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And things tend to move from a high concentration gradient to a low concentration gradient. So ions will want to move from the outside to the inside due to a chemical or concentration gradient. Now notice what happens as our positively charged ions begin to move to the inside. Well, as this one moves, then this one moves, and the third one moves, and the fourth one moves. What happens? Well, there's a build up of positive charge on this side."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "Well, as this one moves, then this one moves, and the third one moves, and the fourth one moves. What happens? Well, there's a build up of positive charge on this side. As these guys go across the membrane, there's a build up of positive charge. And this slows down the movement because positive charge and positive charge repel. And so eventually, these guys will stop moving this way."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "As these guys go across the membrane, there's a build up of positive charge. And this slows down the movement because positive charge and positive charge repel. And so eventually, these guys will stop moving this way. And that means now we can talk about another gradient called the electrical gradient. Now, the electrical gradient is due to charge and not number. So the more charge there is here, the more our electrical gradient will play a role in dictating in which direction our molecule moves."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And that means now we can talk about another gradient called the electrical gradient. Now, the electrical gradient is due to charge and not number. So the more charge there is here, the more our electrical gradient will play a role in dictating in which direction our molecule moves. And in fact, when the chemical gradient or the concentration gradient equals the electrical gradient, the rates that these molecules are moving inside is going to equal to the rates at which they're moving to the outside. And that's exactly what a rusting potential is. It's in the equilibrium state."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And in fact, when the chemical gradient or the concentration gradient equals the electrical gradient, the rates that these molecules are moving inside is going to equal to the rates at which they're moving to the outside. And that's exactly what a rusting potential is. It's in the equilibrium state. So now let's look at our electrochemical cells or a concentration cell. In the first half cell. This is our anode oxidation takes place."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So now let's look at our electrochemical cells or a concentration cell. In the first half cell. This is our anode oxidation takes place. And now cathode reduction takes place. So notice that these guys so we have the two in an electrodes. We have the conductor and the sole bridge."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And now cathode reduction takes place. So notice that these guys so we have the two in an electrodes. We have the conductor and the sole bridge. These guys can be thought of as being the membrane. Okay? Now what happens in this cell?"}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "These guys can be thought of as being the membrane. Okay? Now what happens in this cell? Well, if in this cell oxidation takes place, then our sodium side molecules oh, by the way, these guys are made from sodium solid. So our sodium solid molecules are oxidized, releasing our NA molecules into our solution and releasing electrons into this conductor. These electrons then transfer all the way to this electrode, react with a positively charged ion, which is taken off from the solution into our electrode."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "Well, if in this cell oxidation takes place, then our sodium side molecules oh, by the way, these guys are made from sodium solid. So our sodium solid molecules are oxidized, releasing our NA molecules into our solution and releasing electrons into this conductor. These electrons then transfer all the way to this electrode, react with a positively charged ion, which is taken off from the solution into our electrode. They react to produce sodium solid. So really, as our reaction progresses, our concentration of sodium ions increases in beaker one and decreases in beaker two. So, in a way, we can think of our sodium ions traveling from this speaker beaker two to beaker one, from the cathode to the anode."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "They react to produce sodium solid. So really, as our reaction progresses, our concentration of sodium ions increases in beaker one and decreases in beaker two. So, in a way, we can think of our sodium ions traveling from this speaker beaker two to beaker one, from the cathode to the anode. And that is exactly why we consider this speaker one to be the inside and this speaker two to be the outside, right? Because initially on the inside, we have less sodium ions than on the outside. Here, we have more."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And that is exactly why we consider this speaker one to be the inside and this speaker two to be the outside, right? Because initially on the inside, we have less sodium ions than on the outside. Here, we have more. And when we have our right potential, this is the amount of sodium ions on the inside, and this is the amount of sodium ions on the outside. Now notice what happens. Our sodium ions tend to move from the outside to the inside, and the same thing is dictated by this concentration system."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And when we have our right potential, this is the amount of sodium ions on the inside, and this is the amount of sodium ions on the outside. Now notice what happens. Our sodium ions tend to move from the outside to the inside, and the same thing is dictated by this concentration system. Also, electrons tend to move from the abode to the cathode. Likewise, electrons move from a high electrical gradient to a low electrical gradient, right? What happens as the ions move from this guy to this guy?"}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "Also, electrons tend to move from the abode to the cathode. Likewise, electrons move from a high electrical gradient to a low electrical gradient, right? What happens as the ions move from this guy to this guy? Well, there is a build up of electrical gradient on this side. It increases, and then electrons begin to transfer from this side to this side, right? And that's exactly what happens in this system, too."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "Well, there is a build up of electrical gradient on this side. It increases, and then electrons begin to transfer from this side to this side, right? And that's exactly what happens in this system, too. The solar lines move this way. The solar move this way, and electrons move the other way, and electrons move the other way. So a neuron cell is simply a complex version of a concentration cell."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "The solar lines move this way. The solar move this way, and electrons move the other way, and electrons move the other way. So a neuron cell is simply a complex version of a concentration cell. And the ions and electrons move in the same way in a cell membrane of a neuron cell as they do in a concentration cell. And now we can use this concentration cell set up with the oxidation reduction reactions to calculate our cell voltage or rustic potential due to these sodium ions. And let's do exactly that."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And the ions and electrons move in the same way in a cell membrane of a neuron cell as they do in a concentration cell. And now we can use this concentration cell set up with the oxidation reduction reactions to calculate our cell voltage or rustic potential due to these sodium ions. And let's do exactly that. So here's our oxidation reaction and our reduction reaction. We add these guys up to find a net reaction. Now, these guys cancel out."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So here's our oxidation reaction and our reduction reaction. We add these guys up to find a net reaction. Now, these guys cancel out. The electrons cancel out, and we are left with the sodium ions in the cathode and the sodium ions in the anode. And so according to our net reduction, there's a transfer of these guys from the cathode to the anode. And that's exactly what we see in this system and in this system."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "The electrons cancel out, and we are left with the sodium ions in the cathode and the sodium ions in the anode. And so according to our net reduction, there's a transfer of these guys from the cathode to the anode. And that's exactly what we see in this system and in this system. So now what we do is we simply realize that our Q for this system is the concentration of this guy over the concentration of this guy. So let's go back to our nurse equation. Notice that our e for standard conditions cancels out because we have the same type of equation, and so they have the same magnitude of different signs."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So now what we do is we simply realize that our Q for this system is the concentration of this guy over the concentration of this guy. So let's go back to our nurse equation. Notice that our e for standard conditions cancels out because we have the same type of equation, and so they have the same magnitude of different signs. So when you look the value up on a table you add them up and they cancel out. That's why that e goes away. And now we have E equals negative RTln of Q over N F. Now, F is a constant."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So when you look the value up on a table you add them up and they cancel out. That's why that e goes away. And now we have E equals negative RTln of Q over N F. Now, F is a constant. R is a constant. Since this is a neuron cell, it's by temperature. So 37 degrees Celsius."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "R is a constant. Since this is a neuron cell, it's by temperature. So 37 degrees Celsius. 37 plus 273 is 310. Now, let's see. R is a constant."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "37 plus 273 is 310. Now, let's see. R is a constant. F is a constant. N is the number of electrons. In our case, we have 1 mol of electron on this side and on this side."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "F is a constant. N is the number of electrons. In our case, we have 1 mol of electron on this side and on this side. So n is one. Now, Q, we said, is the concentration in the anode over the concentration in the cathode? So 0.018 molar over zero point 150 molar."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So n is one. Now, Q, we said, is the concentration in the anode over the concentration in the cathode? So 0.018 molar over zero point 150 molar. The amp cancel and we get this guy here. Now we plug this into the calculator. We get a negative number and there's a negative on the outside."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "The amp cancel and we get this guy here. Now we plug this into the calculator. We get a negative number and there's a negative on the outside. So the negatives cancel. And this is our final answer. Now, once again, I want to really talk about what this guy means, because this guy is important."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "So the negatives cancel. And this is our final answer. Now, once again, I want to really talk about what this guy means, because this guy is important. This is what is achieved at Equilibrium, right? This means that when equilibrium is achieved, the chemical gradient or the concentration gradient equals the electrical gradient. So their rates are equal."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "This is what is achieved at Equilibrium, right? This means that when equilibrium is achieved, the chemical gradient or the concentration gradient equals the electrical gradient. So their rates are equal. And this is exactly what it is. So our concentration gradient is zero point 57 volts. But our electrical gradient is negative."}, {"title": "Electrochemical Gradient of Cell Membrane.txt", "text": "And this is exactly what it is. So our concentration gradient is zero point 57 volts. But our electrical gradient is negative. Zero point 57. Remember? Because they're reversed."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Now, in this lecture, I'd like to continue our discussion on Lewis dot structures. Now, before, we only spoke about neutral species or neutral molecules. Now we're going to draw Lewis dot structures for charged species or charged molecules. So let's begin with A. In a, we have an oh or a hydroxide molecule that has a negative one charge on the oxygen. Now, that negative one charge on the oxygen simply means that oxygen has one more electron than it does in its neutral state."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So let's begin with A. In a, we have an oh or a hydroxide molecule that has a negative one charge on the oxygen. Now, that negative one charge on the oxygen simply means that oxygen has one more electron than it does in its neutral state. In other words, in its neutral state, oxygen has eight electrons and eight protons. Now, a charge negative one oxygen molecule has eight protons, but now it has nine electrons. So let's draw our electron configuration for oxygen."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "In other words, in its neutral state, oxygen has eight electrons and eight protons. Now, a charge negative one oxygen molecule has eight protons, but now it has nine electrons. So let's draw our electron configuration for oxygen. So two electrons go into the one s, two electrons go into the two s, and five electrons go into the two p. Now, H stays the same because H is neutral. And so we place one electron into our one s orbital. So let's begin by first counting the total amount of balanced electrons that we have."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So two electrons go into the one s, two electrons go into the two s, and five electrons go into the two p. Now, H stays the same because H is neutral. And so we place one electron into our one s orbital. So let's begin by first counting the total amount of balanced electrons that we have. Remember, balanced electrons are those electrons found in the outermost energy shell. For oxygen, that happens to be the N equals two shell. So two plus five equals seven."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Remember, balanced electrons are those electrons found in the outermost energy shell. For oxygen, that happens to be the N equals two shell. So two plus five equals seven. So seven balance electrons for o, and we have one electron in the balanced electron of one s for H. So we have all together eight electrons, eight balanced electrons that we will have to place around our atoms. So let's begin by placing our oxygen and H adjacent to one another. So we have eight electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So seven balance electrons for o, and we have one electron in the balanced electron of one s for H. So we have all together eight electrons, eight balanced electrons that we will have to place around our atoms. So let's begin by placing our oxygen and H adjacent to one another. So we have eight electrons. Let's begin by drawing a sigma oracovalent bond between oxygen and nitrogen. So once we draw this line, that basically means that one electron is being donated by H and one electron is being donated by o. Now, this means that all the orbitals, meaning this one s orbital, is completely filled for H. Because before, when H is by itself, it has one electron in its one s shell."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Let's begin by drawing a sigma oracovalent bond between oxygen and nitrogen. So once we draw this line, that basically means that one electron is being donated by H and one electron is being donated by o. Now, this means that all the orbitals, meaning this one s orbital, is completely filled for H. Because before, when H is by itself, it has one electron in its one s shell. But now this electron is being shared. So we have two electrons in the one s shell. And that means all electrons or all the orbitals of the age are completely filled."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "But now this electron is being shared. So we have two electrons in the one s shell. And that means all electrons or all the orbitals of the age are completely filled. So we can't place any more electrons around our age. How about oxygen? Well, oxygen has more orbitals, right?"}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So we can't place any more electrons around our age. How about oxygen? Well, oxygen has more orbitals, right? When this orbital is still it has three more orbitals. So that means we can place the remaining six valve electrons into or around our oxygen. So we place a pair here."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "When this orbital is still it has three more orbitals. So that means we can place the remaining six valve electrons into or around our oxygen. So we place a pair here. A place we place a pair here, and we place a pair here. Now, notice in its mutual state, oxygen has six electrons. But because we have one plus one plus one plus one plus one plus one plus one, this gives us six or seven electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "A place we place a pair here, and we place a pair here. Now, notice in its mutual state, oxygen has six electrons. But because we have one plus one plus one plus one plus one plus one plus one, this gives us six or seven electrons. That means we're going to have a negative one charge on the oxygen. So this concludes our Lewis dot structure. Just to make sure, let's make sure we have the right amount of electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "That means we're going to have a negative one charge on the oxygen. So this concludes our Lewis dot structure. Just to make sure, let's make sure we have the right amount of electrons. So our electron counts, we have two electrons in the bonding or covalent bond, and six non bonding electrons surrounding our oxygen. So two plus six is eight. We begin with eight balance electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So our electron counts, we have two electrons in the bonding or covalent bond, and six non bonding electrons surrounding our oxygen. So two plus six is eight. We begin with eight balance electrons. We end with eight balance electrons. Oxygen has a negative one charge. So this concludes part a."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "We end with eight balance electrons. Oxygen has a negative one charge. So this concludes part a. Let's move to part b. In part B, we have this BH four molecule. Now b is boron."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Let's move to part b. In part B, we have this BH four molecule. Now b is boron. Normally, boron has five protons and five electrons in its neutral state. But because this has a negative one charge, it has one more electron. So that means it has a total of six electrons and five protons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Normally, boron has five protons and five electrons in its neutral state. But because this has a negative one charge, it has one more electron. So that means it has a total of six electrons and five protons. So two going to the one s, two going to the two s, and two going to the two p. H still has that one electron in the one s because it's neutral. Now, notice, however, now we have four h atoms. And that means when we're counting our balanced electrons, we have not one electron from h, but four electrons from h. So four times one."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So two going to the one s, two going to the two s, and two going to the two p. H still has that one electron in the one s because it's neutral. Now, notice, however, now we have four h atoms. And that means when we're counting our balanced electrons, we have not one electron from h, but four electrons from h. So four times one. So we have four electrons coming from h, and two plus two four electrons coming from our boron. So altogether, we have eight balanced electrons. So once again, we place our b in the middle, our central atom, and we place our h atoms around our b atom."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So we have four electrons coming from h, and two plus two four electrons coming from our boron. So altogether, we have eight balanced electrons. So once again, we place our b in the middle, our central atom, and we place our h atoms around our b atom. So we begin by first creating sigma or covalent bonds. So we connect our b's and HS, and now we have four covalent bonds. So let's count how many electrons we have."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So we begin by first creating sigma or covalent bonds. So we connect our b's and HS, and now we have four covalent bonds. So let's count how many electrons we have. So we begin with eight. And now we have 1234-5678. So we have completely used up all our balance electrons, and that means that this is our lewis structure."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So we begin with eight. And now we have 1234-5678. So we have completely used up all our balance electrons, and that means that this is our lewis structure. Now, normally, boron could form three bonds, so it has three electrons. But in this case, it has 12341 more electron than in its neutral state. And that means that it has a negative one charge."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Now, normally, boron could form three bonds, so it has three electrons. But in this case, it has 12341 more electron than in its neutral state. And that means that it has a negative one charge. So let's do our electron count. We have eight bonding electrons, right? 2468 and zero non bonding."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So let's do our electron count. We have eight bonding electrons, right? 2468 and zero non bonding. So we have a net of eight electrons. And that concludes our picture for b. Let's jump to part C. In part C, we have this ammonium atom, or NH, four positive."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So we have a net of eight electrons. And that concludes our picture for b. Let's jump to part C. In part C, we have this ammonium atom, or NH, four positive. That means nitrogen has one less electron than it does in its neutral state. So let's draw our electron configuration. For n.\nOne or two electrons go into the two s, two electrons go to the two or two electrons going to the one s, two electrons go to the two s, and two electrons go into the two p now, normally, in a neutral atom, we would have three electrons going to the two P. But because it has a plus one, that means it has one more proton in an electron."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "That means nitrogen has one less electron than it does in its neutral state. So let's draw our electron configuration. For n.\nOne or two electrons go into the two s, two electrons go to the two or two electrons going to the one s, two electrons go to the two s, and two electrons go into the two p now, normally, in a neutral atom, we would have three electrons going to the two P. But because it has a plus one, that means it has one more proton in an electron. So it has only two electrons in the two p.\nH, once again, is neutral, so it has one electron in the one s orbital. Once again, we count our balanced electrons. We have two and two."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So it has only two electrons in the two p.\nH, once again, is neutral, so it has one electron in the one s orbital. Once again, we count our balanced electrons. We have two and two. So four electrons coming from n, and one times four. So eight balanced electrons altogether. So once again, the same story."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So four electrons coming from n, and one times four. So eight balanced electrons altogether. So once again, the same story. We draw out our N, that central atom in the middle. We draw our four HS around and we connect our HS and NS. So notice we have two, four, six and eight."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "We draw out our N, that central atom in the middle. We draw our four HS around and we connect our HS and NS. So notice we have two, four, six and eight. So we have a total of eight balanced electrons. So we have used up our balanced electrons. And this must be the electron configuration for NH four."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So we have a total of eight balanced electrons. So we have used up our balanced electrons. And this must be the electron configuration for NH four. Notice that once again, n is used to having five electrons. Here. It has four electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Notice that once again, n is used to having five electrons. Here. It has four electrons. 1234 So it has a plus one charge on the N, and these HS are neutral. So a net charge of plus one."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "1234 So it has a plus one charge on the N, and these HS are neutral. So a net charge of plus one. Once again our electron count. Eight electrons coming from the bonding orbitals. Or the bonding electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "Once again our electron count. Eight electrons coming from the bonding orbitals. Or the bonding electrons. And we have zero non bonding electrons, just like we had in this picture here. So let's go to the last one. Part D.\nIn Part D, we have NH Two with a minus one."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "And we have zero non bonding electrons, just like we had in this picture here. So let's go to the last one. Part D.\nIn Part D, we have NH Two with a minus one. So right here we had a plus one and here we have a minus one on the end. That means it will have one more electron than a dot in its neutral state. So instead of having three electrons in its two p it's going to have four electrons in its two p orbital."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So right here we had a plus one and here we have a minus one on the end. That means it will have one more electron than a dot in its neutral state. So instead of having three electrons in its two p it's going to have four electrons in its two p orbital. So two going to the one S, two going to the two S, and four going to the two P. H is once again neutral. It has one electron in the one s. So we have two h atoms. So two times one, we have two electrons balance electrons coming from h and two plus four six electrons coming from n.\nSo altogether, once again we have eight balance electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So two going to the one S, two going to the two S, and four going to the two P. H is once again neutral. It has one electron in the one s. So we have two h atoms. So two times one, we have two electrons balance electrons coming from h and two plus four six electrons coming from n.\nSo altogether, once again we have eight balance electrons. So once again we begin our drawing by writing or drawing n in the middle of the central atom and h is adjacent to it. So we connect our atoms, and now we are left with four balance electrons. Because we begin with eight balance electrons."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "So once again we begin our drawing by writing or drawing n in the middle of the central atom and h is adjacent to it. So we connect our atoms, and now we are left with four balance electrons. Because we begin with eight balance electrons. We use up 12348. Minus four is four. And because these orbitals are filled, we are left with filling."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "We use up 12348. Minus four is four. And because these orbitals are filled, we are left with filling. The N orbitals. So we basically put two here. We place two here and we conclude our drawing."}, {"title": "Drawing Lewis Structures Example #2.txt", "text": "The N orbitals. So we basically put two here. We place two here and we conclude our drawing. Because now we have 1234-5678 balance electrons. We have four bonding and four non bonding electrons. And because nitrogen is normally used to having five electrons, and in this case it has 123-4456, this end will have a negative one charge."}, {"title": "Root Mean Square Velocity .txt", "text": "Now let's look at two sets of values. In my first set I have 21012, in my second set I have negative two, two negative 1012. I want to take the average of, or find the average of this set. And this set, well, I add up all my values, divide them by five and get six over five. And the same thing for this guy, negative one, negative one, negative two minus negative one plus zero, plus one plus two divided by five. Well, these guys cancel and I get zero."}, {"title": "Root Mean Square Velocity .txt", "text": "And this set, well, I add up all my values, divide them by five and get six over five. And the same thing for this guy, negative one, negative one, negative two minus negative one plus zero, plus one plus two divided by five. Well, these guys cancel and I get zero. Now look, in this set all my numbers have the same magnitude as these guys. But in this set, these first two numbers have the same magnitude as these first two numbers, but different signs. And that's why my average is zero."}, {"title": "Root Mean Square Velocity .txt", "text": "Now look, in this set all my numbers have the same magnitude as these guys. But in this set, these first two numbers have the same magnitude as these first two numbers, but different signs. And that's why my average is zero. So in some cases, this type of average won't make sense. Now, from a fitness perspective, let's look at moving cars. Suppose a car is traveling in this direction at 60."}, {"title": "Root Mean Square Velocity .txt", "text": "So in some cases, this type of average won't make sense. Now, from a fitness perspective, let's look at moving cars. Suppose a car is traveling in this direction at 60. Suppose in this direction is a positive direction. Now suppose another car is also traveling 60 mph, but in the other direction, so it's negative 60, where 60 is our magnitude and direction is our negative sign. So if someone asks you what is the average of the first two cars, from a physics perspective you will say 60 mph."}, {"title": "Root Mean Square Velocity .txt", "text": "Suppose in this direction is a positive direction. Now suppose another car is also traveling 60 mph, but in the other direction, so it's negative 60, where 60 is our magnitude and direction is our negative sign. So if someone asks you what is the average of the first two cars, from a physics perspective you will say 60 mph. Because if one car is going 60 and the other car is going 60, then the average must be 60. Well, if you use the formula to find the average from a mathematics point of view, you will see that it's 60 plus -60, gives you zero divided by two, which is zero. So sometimes from a mathematics point of view, taking the average in the same way that you did here doesn't make sense."}, {"title": "Root Mean Square Velocity .txt", "text": "Because if one car is going 60 and the other car is going 60, then the average must be 60. Well, if you use the formula to find the average from a mathematics point of view, you will see that it's 60 plus -60, gives you zero divided by two, which is zero. So sometimes from a mathematics point of view, taking the average in the same way that you did here doesn't make sense. Because in the real world, if you have two cars traveling at some speed and you take their average, how can the average be zero? So that's where the root mean square formula comes from. What this formula does is it takes away these negatives and gives you the value of this type of average without the negatives."}, {"title": "Root Mean Square Velocity .txt", "text": "Because in the real world, if you have two cars traveling at some speed and you take their average, how can the average be zero? So that's where the root mean square formula comes from. What this formula does is it takes away these negatives and gives you the value of this type of average without the negatives. So let's look at the formula. VRS is equal to, you take the squares of every single velocity or every single value and then you divide that by n. So almost the same thing as you did here, except you square every value. The reason you square every value is because a square will take away that negative sign and then you divide by n and you take the square root of that."}, {"title": "Root Mean Square Velocity .txt", "text": "So let's look at the formula. VRS is equal to, you take the squares of every single velocity or every single value and then you divide that by n. So almost the same thing as you did here, except you square every value. The reason you square every value is because a square will take away that negative sign and then you divide by n and you take the square root of that. So this will always give you a positive value. So let's take this example and let's use these two values to find our average. So BRMs is equal to 60 mph squared plus negative 60 mph squared."}, {"title": "Root Mean Square Velocity .txt", "text": "So this will always give you a positive value. So let's take this example and let's use these two values to find our average. So BRMs is equal to 60 mph squared plus negative 60 mph squared. This gives you 3600 plus negative canceled. So 3600 gives you 7200 divided by two and square root, that gives you 60 mph. So you see, once again, that the purpose of the route means square is to take away those negative signs and just give you the magnitude."}, {"title": "Root Mean Square Velocity .txt", "text": "This gives you 3600 plus negative canceled. So 3600 gives you 7200 divided by two and square root, that gives you 60 mph. So you see, once again, that the purpose of the route means square is to take away those negative signs and just give you the magnitude. And this becomes useful when you're talking about velocities. Because if a molecule is traveling with one velocity this way and another molecule is traveling with the same velocity but the direction the other way, you want to take those averages and you want those averages to give you a positive value, not zero. That's exactly why you use root mean square velocity."}, {"title": "Root Mean Square Velocity .txt", "text": "And this becomes useful when you're talking about velocities. Because if a molecule is traveling with one velocity this way and another molecule is traveling with the same velocity but the direction the other way, you want to take those averages and you want those averages to give you a positive value, not zero. That's exactly why you use root mean square velocity. So you see that this guy gives you 60 just like it would from a logical physics perspective. From a pure mathematical perspective, it gives you zero if you use the formula. Now, one last thing I want to mention is that this definition of velocity is not actually correct."}, {"title": "Root Mean Square Velocity .txt", "text": "So you see that this guy gives you 60 just like it would from a logical physics perspective. From a pure mathematical perspective, it gives you zero if you use the formula. Now, one last thing I want to mention is that this definition of velocity is not actually correct. Because remember, velocity is a vector. That means it has both magnitude and direction. While speed is a scaling, it only has magnitude."}, {"title": "Root Mean Square Velocity .txt", "text": "Because remember, velocity is a vector. That means it has both magnitude and direction. While speed is a scaling, it only has magnitude. Now, this velocity only has magnitude. It doesn't have direction. Because when we take the square root, we can't get a negative, so we always get positive."}, {"title": "First Law of Thermodynamics .txt", "text": "So today we're going to talk about the first law of thermodynamics, the second law of thermodynamics and the heat engine. So the first law of thermodynamics is an extension from the law of conservation of energy which states that energy cannot be created, it cannot be destroyed, it must be transformed from one form to another. Now, when we talk about the first law of thermodynamics, we basically, basically talk about closed systems. Now, aside from closed systems, they're open systems as well as isolated systems. Within a closed system, matter or mass is not allowed to exchange. The mass remains constant within the system."}, {"title": "First Law of Thermodynamics .txt", "text": "Now, aside from closed systems, they're open systems as well as isolated systems. Within a closed system, matter or mass is not allowed to exchange. The mass remains constant within the system. What is allowed to exchange, however, is energy. So energy flows into the system or it can flow out of the system. In an isolated system, mass and energy is not allowed to flow anywhere."}, {"title": "First Law of Thermodynamics .txt", "text": "What is allowed to exchange, however, is energy. So energy flows into the system or it can flow out of the system. In an isolated system, mass and energy is not allowed to flow anywhere. So everything remains constant. In an open system, both matter and energy is allowed to exchange. Okay?"}, {"title": "First Law of Thermodynamics .txt", "text": "So everything remains constant. In an open system, both matter and energy is allowed to exchange. Okay? Now when we talk about the transfer of energy, remember that there are only two types of transfer of energy work and heat. Now heat can be subdivided into three categories convection, conduction and radiation. When we talk about chemical work, we talk about this type of work pressure times change in volume."}, {"title": "First Law of Thermodynamics .txt", "text": "Now when we talk about the transfer of energy, remember that there are only two types of transfer of energy work and heat. Now heat can be subdivided into three categories convection, conduction and radiation. When we talk about chemical work, we talk about this type of work pressure times change in volume. And when pressure remains constant, we use this equation. When pressure isn't constant, we use calculus and integrate from the initial to the final. Now this law can be summarized in this equation."}, {"title": "First Law of Thermodynamics .txt", "text": "And when pressure remains constant, we use this equation. When pressure isn't constant, we use calculus and integrate from the initial to the final. Now this law can be summarized in this equation. Okay? This law basically translates into this equation. And what this equation states is pretty simple."}, {"title": "First Law of Thermodynamics .txt", "text": "Okay? This law basically translates into this equation. And what this equation states is pretty simple. All it states is that the energy transfer or energy flow into a system is equal to the heat flow into the system plus the work done on the system. What it basically says is that a transfer of energy amounts to two types of transfers. Transfers due to heat or transfers due to work."}, {"title": "First Law of Thermodynamics .txt", "text": "All it states is that the energy transfer or energy flow into a system is equal to the heat flow into the system plus the work done on the system. What it basically says is that a transfer of energy amounts to two types of transfers. Transfers due to heat or transfers due to work. Okay, no other transfer of energy exists. So let's see what heat engines are. Heat engines are basically systems or mechanisms that convert one form of energy into a second form of energy, namely heat into work."}, {"title": "First Law of Thermodynamics .txt", "text": "Okay, no other transfer of energy exists. So let's see what heat engines are. Heat engines are basically systems or mechanisms that convert one form of energy into a second form of energy, namely heat into work. And this occurs under constant temperature. So let's see what the layout of a heat engine is. A heat engine is composed of a long cylindrical tube that contains molecules inside this area and that contains a movable piston controlled by an outside force, maybe your hand that's moving it up or down."}, {"title": "First Law of Thermodynamics .txt", "text": "And this occurs under constant temperature. So let's see what the layout of a heat engine is. A heat engine is composed of a long cylindrical tube that contains molecules inside this area and that contains a movable piston controlled by an outside force, maybe your hand that's moving it up or down. It also contains a hot body connected to the bottom. This hot body is important in conduction because remember, conduction requires for physical contact between two systems. And conduction allows heat transfer or energy transfer from a hot object to a cold object."}, {"title": "First Law of Thermodynamics .txt", "text": "It also contains a hot body connected to the bottom. This hot body is important in conduction because remember, conduction requires for physical contact between two systems. And conduction allows heat transfer or energy transfer from a hot object to a cold object. So let's see what the result is of constant pressure. From this formula we see that kinetic energy is related to KB, which is a constant and number of particles and T temperature. Now, the number of particles remains constant because this is a closed system."}, {"title": "First Law of Thermodynamics .txt", "text": "So let's see what the result is of constant pressure. From this formula we see that kinetic energy is related to KB, which is a constant and number of particles and T temperature. Now, the number of particles remains constant because this is a closed system. Remember, a closed system is a system in which the mass or matter or the number of particles remains constant. So n remains constant. These guys remain constant, and temperature remains constant."}, {"title": "First Law of Thermodynamics .txt", "text": "Remember, a closed system is a system in which the mass or matter or the number of particles remains constant. So n remains constant. These guys remain constant, and temperature remains constant. So that means kinetic energy also must remain constant. Okay, so what's the result? Remember, normally, when there's a transfer of energy, the energy is transferred into increasing the kinetic energy."}, {"title": "First Law of Thermodynamics .txt", "text": "So that means kinetic energy also must remain constant. Okay, so what's the result? Remember, normally, when there's a transfer of energy, the energy is transferred into increasing the kinetic energy. But in this situation, there is no increase in kinetic energy. So the question remains, where does this energy transfer go into if it doesn't go into the kinetic energy? Okay, the answer lies in this equation."}, {"title": "First Law of Thermodynamics .txt", "text": "But in this situation, there is no increase in kinetic energy. So the question remains, where does this energy transfer go into if it doesn't go into the kinetic energy? Okay, the answer lies in this equation. We see that by the ideal Gas Law, PV, or pressure times volume equals number of moles times the constant times the temperature in Kelvin. Okay? So this guy remains constant because the temperature remains constant."}, {"title": "First Law of Thermodynamics .txt", "text": "We see that by the ideal Gas Law, PV, or pressure times volume equals number of moles times the constant times the temperature in Kelvin. Okay? So this guy remains constant because the temperature remains constant. And these guys are constants. This is constant because this is a closed system. So this must mean that the energy transfer must go into increasing the volume or expanding it."}, {"title": "First Law of Thermodynamics .txt", "text": "And these guys are constants. This is constant because this is a closed system. So this must mean that the energy transfer must go into increasing the volume or expanding it. And the expansion creates a larger D or a larger volume. And because this guy is constant, this must be constant. So an increase in D must mean a decrease in P. A decrease in P will decrease."}, {"title": "First Law of Thermodynamics .txt", "text": "And the expansion creates a larger D or a larger volume. And because this guy is constant, this must be constant. So an increase in D must mean a decrease in P. A decrease in P will decrease. Obviously, this P, and look, p, or pressure, is equal to force comes area. Area remains constant, because if you take the cross sectional area of this cylinder, it doesn't change as the piston moves up or down. So if this guy is constant and this guy is decreasing, the pressure is decreasing, then the force must also decrease."}, {"title": "First Law of Thermodynamics .txt", "text": "Obviously, this P, and look, p, or pressure, is equal to force comes area. Area remains constant, because if you take the cross sectional area of this cylinder, it doesn't change as the piston moves up or down. So if this guy is constant and this guy is decreasing, the pressure is decreasing, then the force must also decrease. So we see that in a heat engine, when the piston is moving, the force is changing, and so is pressure, and so is volume, but temperature remains the same. Okay, so what this basically means, or what this basically implies, is that heat must be converted to work, and all the heat must be converted to work. But that's actually not true, and only about 10% to 20% normally is converted to work."}, {"title": "First Law of Thermodynamics .txt", "text": "So we see that in a heat engine, when the piston is moving, the force is changing, and so is pressure, and so is volume, but temperature remains the same. Okay, so what this basically means, or what this basically implies, is that heat must be converted to work, and all the heat must be converted to work. But that's actually not true, and only about 10% to 20% normally is converted to work. Okay, and let's see why. Well, let's go back here. When the energy is transferred into this system, this system, the volume increases because the piston starts moving this way, and it continues moving this way until when?"}, {"title": "First Law of Thermodynamics .txt", "text": "Okay, and let's see why. Well, let's go back here. When the energy is transferred into this system, this system, the volume increases because the piston starts moving this way, and it continues moving this way until when? Until it hits this limit here, because the cylinder eventually ends when it reaches this place. You have this picture here, okay? And now what happens?"}, {"title": "First Law of Thermodynamics .txt", "text": "Until it hits this limit here, because the cylinder eventually ends when it reaches this place. You have this picture here, okay? And now what happens? Now we need to somehow move this piston back to its original location so the process could repeat. So this is a cyclic process, right? We want energy to go into here to move the piston this way."}, {"title": "First Law of Thermodynamics .txt", "text": "Now we need to somehow move this piston back to its original location so the process could repeat. So this is a cyclic process, right? We want energy to go into here to move the piston this way. Then we want to move the piston back this way, and this to continues, okay? Indefinitely. But let's see what happens here."}, {"title": "First Law of Thermodynamics .txt", "text": "Then we want to move the piston back this way, and this to continues, okay? Indefinitely. But let's see what happens here. When we have a force and we start pushing with the force this way, what happens to the kinetic energy or the temperature of the particles within this system? Well, when you move this way, pressure increases, volume decreases, and the particles get closer together. They start banging against each other more violently, more frequently."}, {"title": "First Law of Thermodynamics .txt", "text": "When we have a force and we start pushing with the force this way, what happens to the kinetic energy or the temperature of the particles within this system? Well, when you move this way, pressure increases, volume decreases, and the particles get closer together. They start banging against each other more violently, more frequently. This, in turn, increases temperature, which in turn increases kinetic energy, but we saw that we want to have constant temperature. Okay, so what does this mean? While this increase in temperature means that there is an increase in pressure and there's an increase in the force, so not the force that we require or the net force that we require to move from this point to this point is greater than the force required to move from this point to this point."}, {"title": "First Law of Thermodynamics .txt", "text": "This, in turn, increases temperature, which in turn increases kinetic energy, but we saw that we want to have constant temperature. Okay, so what does this mean? While this increase in temperature means that there is an increase in pressure and there's an increase in the force, so not the force that we require or the net force that we require to move from this point to this point is greater than the force required to move from this point to this point. So the work that's required to move depiction from this position to this position is less than the work required to move from this position to this position. And that's not something we want. We want to be efficient, and that's not very efficient."}, {"title": "First Law of Thermodynamics .txt", "text": "So the work that's required to move depiction from this position to this position is less than the work required to move from this position to this position. And that's not something we want. We want to be efficient, and that's not very efficient. So how do we fix this system? Well, we fix the system by adding another object to the heat engine, something called a cold body. We saw that we had a hot body for the specific purpose of conduction."}, {"title": "First Law of Thermodynamics .txt", "text": "So how do we fix this system? Well, we fix the system by adding another object to the heat engine, something called a cold body. We saw that we had a hot body for the specific purpose of conduction. Well, now we have a cold body that also is specific for conduction. In other words, when this piston moves this way, there's an increase in temperature. And the increase in temperature causes energy to transfer via conduction from this place to this place."}, {"title": "First Law of Thermodynamics .txt", "text": "Well, now we have a cold body that also is specific for conduction. In other words, when this piston moves this way, there's an increase in temperature. And the increase in temperature causes energy to transfer via conduction from this place to this place. And this makes the temperature stay constant. Okay, so the final outcome is that a heat engine needs to look like this. A heat engine needs to have a hot body, a cold body, a cylindrical tube, as well as a piston that's controlled by some outside force."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "A positive cell voltage indicates that our reaction within our electrochemical cell is product favorite, so it's spontaneous. Likewise, a negative cell voltage indicates a reactant favorite reaction. Recall that Gibbs free energy builds a relationship between entropy and entropy, and it also dictates whether a reaction is product favorite or reactant favorite. So notice that both cell voltage of an electrochemical cell and GIBS free energy dictate whether or not a reaction is product favorite. So therefore, we can imagine that there's some type of relationship between cell voltage and gives free energy. Now, when gives free energy is positive, that means our reaction is reacting favorite."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So notice that both cell voltage of an electrochemical cell and GIBS free energy dictate whether or not a reaction is product favorite. So therefore, we can imagine that there's some type of relationship between cell voltage and gives free energy. Now, when gives free energy is positive, that means our reaction is reacting favorite. It's not spontaneous. But when it's negative, the reaction is product favorite. It is spontaneous."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "It's not spontaneous. But when it's negative, the reaction is product favorite. It is spontaneous. Now, before we actually build a relationship and show how the cell voltage relates to GIBS free energy, let's examine something called electrical work. Now, we already spoke about electrical work when we spoke about electromotive force, but let's revisit this topic because it becomes important. So voltaic cells or electrochemical cells convert chemical energy into electrical potential energy that can be used to do work, for example, para light bulb or para motor."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Now, before we actually build a relationship and show how the cell voltage relates to GIBS free energy, let's examine something called electrical work. Now, we already spoke about electrical work when we spoke about electromotive force, but let's revisit this topic because it becomes important. So voltaic cells or electrochemical cells convert chemical energy into electrical potential energy that can be used to do work, for example, para light bulb or para motor. Now, this electrical potential energy is also known as electrical work. And we can express electrical work mathematically by equaling this guy to charge of an object or a system times our cell voltage. So if we're talking about cells and electrochemical cells, our charge refers to all the possible charge in our battery in our chemical cell or electric chemical cells."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Now, this electrical potential energy is also known as electrical work. And we can express electrical work mathematically by equaling this guy to charge of an object or a system times our cell voltage. So if we're talking about cells and electrochemical cells, our charge refers to all the possible charge in our battery in our chemical cell or electric chemical cells. Now, recall that a single electron has a charge of 1.622\ntimes ten to negative 19 Coulombs. This is per electron per one electron. But obviously, a single battery, a single electrochemical cell, has many electrons within itself."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Now, recall that a single electron has a charge of 1.622\ntimes ten to negative 19 Coulombs. This is per electron per one electron. But obviously, a single battery, a single electrochemical cell, has many electrons within itself. That means we're dealing with a lot of these guys. Now, let's find how much charge is in 1 mol of electrons. Right?"}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "That means we're dealing with a lot of these guys. Now, let's find how much charge is in 1 mol of electrons. Right? The way we do it is we use avogadjo's number. Remember, avogadjo's number refers to the number of atoms or electrons found in 1 mol of anything. Now, we're talking about 1 mol of electrons."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "The way we do it is we use avogadjo's number. Remember, avogadjo's number refers to the number of atoms or electrons found in 1 mol of anything. Now, we're talking about 1 mol of electrons. That means we take this number of electrons and we multiply by our charge per electron. We see that averagadro's number of electrons per mole times 1.622 times ten to the negative 19 kilos per electron, which we got from here. The electrons cancel, and we get our units to be Coulomb per mole."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "That means we take this number of electrons and we multiply by our charge per electron. We see that averagadro's number of electrons per mole times 1.622 times ten to the negative 19 kilos per electron, which we got from here. The electrons cancel, and we get our units to be Coulomb per mole. So the charge of a single mole of electron electrons is 96,484 Coulombs per mole of electron. So when we have 6.22 times ten to the 23 electrons in our cell, that means our cell has a charge of 96,484 klombs. So we can represent this equation in another way."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So the charge of a single mole of electron electrons is 96,484 Coulombs per mole of electron. So when we have 6.22 times ten to the 23 electrons in our cell, that means our cell has a charge of 96,484 klombs. So we can represent this equation in another way. So, this equation electrical work is equal to n, the number of moles times f. Now, f is Faraday's constant. This is known as this number is known as Faraday's constant. And it talks about a charge of one coulomb per mole of electrons."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So, this equation electrical work is equal to n, the number of moles times f. Now, f is Faraday's constant. This is known as this number is known as Faraday's constant. And it talks about a charge of one coulomb per mole of electrons. So, once again, f is Faradays constant times our cell voltage. So this n times f is simply charge. It's the same thing."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So, once again, f is Faradays constant times our cell voltage. So this n times f is simply charge. It's the same thing. Now, for example, if we have 1 mol in our cell, that means we multiply 1 mol times f.\nSo 1 mol times this guy gives us the most cancel, and we simply get our charge. So n times f is our charge. Now, this equation becomes important when we build a relationship between our cell voltage and gear spree energy."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Now, for example, if we have 1 mol in our cell, that means we multiply 1 mol times f.\nSo 1 mol times this guy gives us the most cancel, and we simply get our charge. So n times f is our charge. Now, this equation becomes important when we build a relationship between our cell voltage and gear spree energy. So let's see what the relationship is. So, let's finally examine and see what the relationship is between cell voltage and changing gear free energy. Well, we see that changing gear's free energy is equal to negative of electrical work done."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So let's see what the relationship is. So, let's finally examine and see what the relationship is between cell voltage and changing gear free energy. Well, we see that changing gear's free energy is equal to negative of electrical work done. Well, why is that? Well, this guy is simply the amount of free energy available to do useful work, while with this guy, the electrical work is the amount of energy transformed from chemical energy to electrical energy in the form of moving electrons. And this is also energy used to do useful work."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Well, why is that? Well, this guy is simply the amount of free energy available to do useful work, while with this guy, the electrical work is the amount of energy transformed from chemical energy to electrical energy in the form of moving electrons. And this is also energy used to do useful work. So that's why the magnitude of these guys is the same, but the sign is different. So, why is it that we have a negative sign in front of the electrical work? Well, let's see why."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So that's why the magnitude of these guys is the same, but the sign is different. So, why is it that we have a negative sign in front of the electrical work? Well, let's see why. The negative sign accounts for the fact that for a product favorite reaction, changing gibsi energy is negative while cell voltage is positive. So that's why we need to add that negative sign in. So, from before, we saw that electrical work is equal to number of moles times starbase constant times our cell voltage."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "The negative sign accounts for the fact that for a product favorite reaction, changing gibsi energy is negative while cell voltage is positive. So that's why we need to add that negative sign in. So, from before, we saw that electrical work is equal to number of moles times starbase constant times our cell voltage. So change and gives the energy is equal to negative m times f times cell voltage. So that's our reaction. That's our equation."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So change and gives the energy is equal to negative m times f times cell voltage. So that's our reaction. That's our equation. Now, suppose we have an electrochemical cell able to take cell that's composed of the following redox reaction. So it has an oxidation reaction and a reduction reaction. Our zinc solid is oxidized into zinc ion, while our copper ion is reduced to copper solid."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Now, suppose we have an electrochemical cell able to take cell that's composed of the following redox reaction. So it has an oxidation reaction and a reduction reaction. Our zinc solid is oxidized into zinc ion, while our copper ion is reduced to copper solid. So, this guy releases two electrons, while this guy takes those two electrons, gains two electrons forming our copper solid. Now, these electrons in the end, end up crossing out. Now, our cell voltage is 1.1 volt."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So, this guy releases two electrons, while this guy takes those two electrons, gains two electrons forming our copper solid. Now, these electrons in the end, end up crossing out. Now, our cell voltage is 1.1 volt. So let's use this formula to approximate how much energy this battery or this electrochemical cell can produce with the charge that it has. So, the change in gates free energy is equal to negative of the number of moles of electrons. Well, in this case, I specifically left these guys in to show you how many moles of electrons we have."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So let's use this formula to approximate how much energy this battery or this electrochemical cell can produce with the charge that it has. So, the change in gates free energy is equal to negative of the number of moles of electrons. Well, in this case, I specifically left these guys in to show you how many moles of electrons we have. Well, zinc loses two moles of electrons, and this guy gains two moles of electrons. That means we're dealing with two moles of electrons times our Faradays constant 96,500 approximately Coulombs per mole of electron times our cell voltage or electromotive force 1.1 volt. We see that these guys cancel out and our units are left Coulombs times voltage, which is simply joules."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Well, zinc loses two moles of electrons, and this guy gains two moles of electrons. That means we're dealing with two moles of electrons times our Faradays constant 96,500 approximately Coulombs per mole of electron times our cell voltage or electromotive force 1.1 volt. We see that these guys cancel out and our units are left Coulombs times voltage, which is simply joules. So in the end we get 212,300 joules of work. It's produced is released in the form of moving electrons or moving charge from the anode to the capital from this battery. And this amount of energy can be used to do useful work."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "So in the end we get 212,300 joules of work. It's produced is released in the form of moving electrons or moving charge from the anode to the capital from this battery. And this amount of energy can be used to do useful work. For example, run a motor or power a light bulb, for example. So this becomes very useful and we'll do many more examples using this in the future. So I already spoke about this briefly in another lecture, but I want to talk about it once more."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "For example, run a motor or power a light bulb, for example. So this becomes very useful and we'll do many more examples using this in the future. So I already spoke about this briefly in another lecture, but I want to talk about it once more. Now, the reason that a D battery has more energy than AAA battery is not because a difference of cell voltage. In fact, the cell voltage for this guy and this guy are the same. It's 1.5\nvolts."}, {"title": "Cell voltage and Gibbs free energy .txt", "text": "Now, the reason that a D battery has more energy than AAA battery is not because a difference of cell voltage. In fact, the cell voltage for this guy and this guy are the same. It's 1.5\nvolts. You could check the label of the batteries. The reason this is more expensive is because it can do more work. But how can it do more work?"}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So let's begin by looking at the following diagram. So, we're basically taking two identical H atoms. We're combining them to form a Diatomic H two Model molecule. So, here are our two identical H atoms. So, this is a one s Atomic orbital of the first H atom and the second Atomic orbital of the second H atom. So we have an electron in each atomic orbital."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So, here are our two identical H atoms. So, this is a one s Atomic orbital of the first H atom and the second Atomic orbital of the second H atom. So we have an electron in each atomic orbital. So, if we combine two Atomic orbitals according to quantum mechanics, we're going to form two molecular orbitals. One will be the phi B, which is the Bonding molecular orbital. And the second will be the phi A, which is the Anti Bonding molecular orbital."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So, if we combine two Atomic orbitals according to quantum mechanics, we're going to form two molecular orbitals. One will be the phi B, which is the Bonding molecular orbital. And the second will be the phi A, which is the Anti Bonding molecular orbital. Now, the bonding molecular orbital is lower in energy. It's more stabilizing. And, in fact, this is the bond responsible for forming our covalent bond."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "Now, the bonding molecular orbital is lower in energy. It's more stabilizing. And, in fact, this is the bond responsible for forming our covalent bond. So our electrons will be found in the Lower Energy bonding molecular orbital. So phi b. Now, phi A, or Antiboming Molecular Orbital, is responsible for breaking the bond."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So our electrons will be found in the Lower Energy bonding molecular orbital. So phi b. Now, phi A, or Antiboming Molecular Orbital, is responsible for breaking the bond. If the electrons are found within this bond, that means those electrons will play a role in Destabilizing our molecule in breaking that Covalent bond. Now, so electrons will be found in this bonding molecular orbital. And notice what happens."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "If the electrons are found within this bond, that means those electrons will play a role in Destabilizing our molecule in breaking that Covalent bond. Now, so electrons will be found in this bonding molecular orbital. And notice what happens. Because these two Atomic orbitals are higher in energy than this molecular orbital, energy will Be lost. So, there is some change in energy that occurs when these two atomic orbitals form this molecular orbital or this molecular Diatomic molecule. Now, this energy is Released into the environment."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "Because these two Atomic orbitals are higher in energy than this molecular orbital, energy will Be lost. So, there is some change in energy that occurs when these two atomic orbitals form this molecular orbital or this molecular Diatomic molecule. Now, this energy is Released into the environment. In other words, when my two H atoms form to create a bond, energy is released. And, in fact, any time we form bonds, energy will always Be Released into the environment. So let's look at this in a more simplified fashion."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "In other words, when my two H atoms form to create a bond, energy is released. And, in fact, any time we form bonds, energy will always Be Released into the environment. So let's look at this in a more simplified fashion. So, here we have two H molecules. At some level, they react. They surmount this Activation barrier, and they form our Diatomic H two molecule, which is lower in energy than the initial."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So, here we have two H molecules. At some level, they react. They surmount this Activation barrier, and they form our Diatomic H two molecule, which is lower in energy than the initial. So our products are lower than Our reactants. And this change in energy is the same as the change in energy that we saw here. Another name for that is change in enthalpy."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So our products are lower than Our reactants. And this change in energy is the same as the change in energy that we saw here. Another name for that is change in enthalpy. So change in H. And from 1 mol, where whenever 1 mol of H reacts with Another Mole of H to form 1 Mol of H 2104, energy will Be Released Into The environment. And this is known as an exothermic reaction. In other words, exothermic reaction is A reaction in which the energy of products is lower than the energy of reactants, and the energy is released into the environment."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So change in H. And from 1 mol, where whenever 1 mol of H reacts with Another Mole of H to form 1 Mol of H 2104, energy will Be Released Into The environment. And this is known as an exothermic reaction. In other words, exothermic reaction is A reaction in which the energy of products is lower than the energy of reactants, and the energy is released into the environment. Now, let's look at the same Exact diagram. But now we're going to work backwards. So we basically want to begin with a diatomic H two molecule, and we want to somehow get to this H two."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "Now, let's look at the same Exact diagram. But now we're going to work backwards. So we basically want to begin with a diatomic H two molecule, and we want to somehow get to this H two. So we essentially want to break this covalent bond and form two separate H molecules or h atoms. So that means since we go from a lower energy to a higher energy, we have to input energy to go from this guy to this guy. So we input energy to break the bond to form two individual H molecules."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So we essentially want to break this covalent bond and form two separate H molecules or h atoms. So that means since we go from a lower energy to a higher energy, we have to input energy to go from this guy to this guy. So we input energy to break the bond to form two individual H molecules. And this is known as an endothermic reaction. So once again, endothermic reaction is a reaction in which the energy of product is higher than the end of reactants and energy is used up or inputted into the system for that reaction to take place. So once again, anytime we have an endothermic reaction going this way, going the forward direction, we have an endothermic reaction going in the backward or reverse direction."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "And this is known as an endothermic reaction. So once again, endothermic reaction is a reaction in which the energy of product is higher than the end of reactants and energy is used up or inputted into the system for that reaction to take place. So once again, anytime we have an endothermic reaction going this way, going the forward direction, we have an endothermic reaction going in the backward or reverse direction. So once again, to sum this information up, anytime we form a bond, energy is released. Anytime we want to break a bond, energy needs to be inputted into our system, into the molecule. Now, what is bond association energy?"}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "So once again, to sum this information up, anytime we form a bond, energy is released. Anytime we want to break a bond, energy needs to be inputted into our system, into the molecule. Now, what is bond association energy? Well, bond association energy is the amount of energy required to break a bond. In other words, if we want to break this bond, we need to input some amount of energy. And this is known as the bond association energy."}, {"title": "Exothermic Reactions, Endothermic Reactions and BDE.txt", "text": "Well, bond association energy is the amount of energy required to break a bond. In other words, if we want to break this bond, we need to input some amount of energy. And this is known as the bond association energy. So for example, let's compare two molecules or two bonds. That Ch bond and the CF bond. Now, Ch bond has a bond association of 103 kg/mol."}, {"title": "Homo-Lumo Examples .txt", "text": "So in this reaction, we're going to identify the homo and luma orbitals of a few reactions. Now, before we look at that, let's recall what homo and luma orbitals are. Now, recall that a lewis acidbased reaction is simply a reaction between the highest occupied molecular orbital known as the homo of one kind compound, and the lowest unoccupied molecular orbital of a second compound known as the lumo. So we basically have a pair of electrons, grams, another molecule, or compound. So let's look at the following reaction. So here we have an amine."}, {"title": "Homo-Lumo Examples .txt", "text": "So we basically have a pair of electrons, grams, another molecule, or compound. So let's look at the following reaction. So here we have an amine. We have ammonia that reacts with our H plus ion. So our lewis acid and lewis base. So here we have our molecular orbital that is occupied."}, {"title": "Homo-Lumo Examples .txt", "text": "We have ammonia that reacts with our H plus ion. So our lewis acid and lewis base. So here we have our molecular orbital that is occupied. It has a pair of electrons which grab this unoccupied orbital, one s orbital. So this SP three hybridized molecular orbital is our homo. It's the highest occupied molecular orbital."}, {"title": "Homo-Lumo Examples .txt", "text": "It has a pair of electrons which grab this unoccupied orbital, one s orbital. So this SP three hybridized molecular orbital is our homo. It's the highest occupied molecular orbital. While on this molecule, or actually atom, our lowest unoccupied molecular orbital is the one S orbital. Because we're missing an electron, we have a cat ion. So we have an MP, one S orbital."}, {"title": "Homo-Lumo Examples .txt", "text": "While on this molecule, or actually atom, our lowest unoccupied molecular orbital is the one S orbital. Because we're missing an electron, we have a cat ion. So we have an MP, one S orbital. So this is our lumo, or lowest unoccupied molecular orbital. So when these two orbitals interact, so we have the SP three interacting with the one S, and we form the following two orbitals. We form the bonding molecular orbital and the anti bonding molecular orbital."}, {"title": "Homo-Lumo Examples .txt", "text": "So this is our lumo, or lowest unoccupied molecular orbital. So when these two orbitals interact, so we have the SP three interacting with the one S, and we form the following two orbitals. We form the bonding molecular orbital and the anti bonding molecular orbital. Now, since this orbital is lower in energy, that means our electron pair coming from this highest occupied molecular orbital goes directly into this bonding molecular orbital. So let's look at a few more examples. Now, let's suppose we have example a."}, {"title": "Homo-Lumo Examples .txt", "text": "Now, since this orbital is lower in energy, that means our electron pair coming from this highest occupied molecular orbital goes directly into this bonding molecular orbital. So let's look at a few more examples. Now, let's suppose we have example a. In this reaction, we have a hydroxide with a negative sign interacting with a sodium ion with a positive sign, and we form sodium hydroxide. So which one has the homo and which one has the lumo? So clearly, this is our base."}, {"title": "Homo-Lumo Examples .txt", "text": "In this reaction, we have a hydroxide with a negative sign interacting with a sodium ion with a positive sign, and we form sodium hydroxide. So which one has the homo and which one has the lumo? So clearly, this is our base. Why? Well, because it has a pair of electrons that it can donate that it can use to grab another molecule, in this case, atom. So that means this pair of electrons, the orbital that is found in, must be the homo."}, {"title": "Homo-Lumo Examples .txt", "text": "Why? Well, because it has a pair of electrons that it can donate that it can use to grab another molecule, in this case, atom. So that means this pair of electrons, the orbital that is found in, must be the homo. So the highest occupied molecular orbital is the field non bonding SP three hybridized orbital molecular orbital. So what's the lumo in our case? Well, if this is the homo, this must be the lumo."}, {"title": "Homo-Lumo Examples .txt", "text": "So the highest occupied molecular orbital is the field non bonding SP three hybridized orbital molecular orbital. So what's the lumo in our case? Well, if this is the homo, this must be the lumo. So the lowest unoccupied molecular orbital is the empty three s orbital of our sodium atom. Let's move on to example two. Reaction two."}, {"title": "Homo-Lumo Examples .txt", "text": "So the lowest unoccupied molecular orbital is the empty three s orbital of our sodium atom. Let's move on to example two. Reaction two. Here we have a hydronium acid reacting with our hydroxide base, forming two H, two O molecules. So which one is the homo and which one is the lumo? Now, this one is a bit tricky, and we'll see why in a second."}, {"title": "Homo-Lumo Examples .txt", "text": "Here we have a hydronium acid reacting with our hydroxide base, forming two H, two O molecules. So which one is the homo and which one is the lumo? Now, this one is a bit tricky, and we'll see why in a second. So let's begin with homo. So the homo has the highest or is the highest occupied molecular orbital. And that means our hydroxide has a lone pair of electrons, so it must have the homo."}, {"title": "Homo-Lumo Examples .txt", "text": "So let's begin with homo. So the homo has the highest or is the highest occupied molecular orbital. And that means our hydroxide has a lone pair of electrons, so it must have the homo. So our filled non bonding orbital of the oh is our homo and it's SP three hybridized. So it's the same exact homo that we saw in example A. Now, what's the lumo?"}, {"title": "Homo-Lumo Examples .txt", "text": "So our filled non bonding orbital of the oh is our homo and it's SP three hybridized. So it's the same exact homo that we saw in example A. Now, what's the lumo? Well, if we examine our hydronium ion, we see that every single orbital is taken. Every single bonding orbital is taken. We have the three bonds."}, {"title": "Homo-Lumo Examples .txt", "text": "Well, if we examine our hydronium ion, we see that every single orbital is taken. Every single bonding orbital is taken. We have the three bonds. And we have a pair of electrons which aren't drawn. Here, let me fill them in. We have a pair of electrons on this oxygen."}, {"title": "Homo-Lumo Examples .txt", "text": "And we have a pair of electrons which aren't drawn. Here, let me fill them in. We have a pair of electrons on this oxygen. So all four types of orbitals are filled. So that means if we don't have our bonding orbitals our Sigma bonding, we must use our Sigma antibodies. So the lowest unoccupied molecular orbital is the anti unoccupied Sigma antibonding orbital of one of the ho bonds."}, {"title": "Homo-Lumo Examples .txt", "text": "So all four types of orbitals are filled. So that means if we don't have our bonding orbitals our Sigma bonding, we must use our Sigma antibodies. So the lowest unoccupied molecular orbital is the anti unoccupied Sigma antibonding orbital of one of the ho bonds. Now, let's go to example C. An example C.\nWe have an alkane reacting with our hydrobromic acid to form the following carbocation with a positive charge. And our bromine an ion. So which one is the homo?"}, {"title": "Homo-Lumo Examples .txt", "text": "Now, let's go to example C. An example C.\nWe have an alkane reacting with our hydrobromic acid to form the following carbocation with a positive charge. And our bromine an ion. So which one is the homo? Which one is the lumo? So which one of these is doing the donating? Which one is using its electron pair?"}, {"title": "Homo-Lumo Examples .txt", "text": "Which one is the lumo? So which one of these is doing the donating? Which one is using its electron pair? So clearly it's the alkene. This pair of electrons in the Pi bond is used to attract or take this H atom from our bromine. So that means our highest occupied molecular orbital is the pi bond."}, {"title": "Homo-Lumo Examples .txt", "text": "So clearly it's the alkene. This pair of electrons in the Pi bond is used to attract or take this H atom from our bromine. So that means our highest occupied molecular orbital is the pi bond. The pi bonding of the carbon carbon double bond. So our homo is the pi bond. What about our lumo?"}, {"title": "Homo-Lumo Examples .txt", "text": "The pi bonding of the carbon carbon double bond. So our homo is the pi bond. What about our lumo? So once again, our lumo must be on this molecule. But notice that the Sigma bonding orbital is taking. So that means we must go to the next unoccupied orbital."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "So here we have our two H atoms, the one S orbital and the one S orbital. We combine the two atomic orbitals to form two molecular orbitals, one molecular orbital. The bonding molecular orbital is lower in energy and is stabilizing, while the higher in energy and the destabilizing molecular orbital is called the antibonding molecular orbital. Now, what we haven't done so far is use our electrons. Remember, an electron is found in this atomic orbital, and one electron is found in this atomic orbital. Because we have two identical H atoms, and each are neutral."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Now, what we haven't done so far is use our electrons. Remember, an electron is found in this atomic orbital, and one electron is found in this atomic orbital. Because we have two identical H atoms, and each are neutral. So each have one neutron, one proton, and one electron. So here we have a positive one half spin. Here we have a negative one half spin."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "So each have one neutron, one proton, and one electron. So here we have a positive one half spin. Here we have a negative one half spin. When these two guys combine, where would these two electrons want to go? In the higher energy or the lower energy? Remember, nature likes stabilizing states."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "When these two guys combine, where would these two electrons want to go? In the higher energy or the lower energy? Remember, nature likes stabilizing states. They like low energy. So that means these two electrons will combine and will go into this bonding molecular orbital. Now, according to the poly exclusion principle, two things must happen."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "They like low energy. So that means these two electrons will combine and will go into this bonding molecular orbital. Now, according to the poly exclusion principle, two things must happen. A maximum of two electrons should be found in this orbital, and these guys should have positive one half and negative one half. So opposite spins. And that's exactly what we have here."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "A maximum of two electrons should be found in this orbital, and these guys should have positive one half and negative one half. So opposite spins. And that's exactly what we have here. So these electrons will not want to go into this orbital because this is higher in energy, and it causes the bond to destabilize or break apart. So, once again, this was combining two atomic one s orbitals to form two molecular orbitals. Now, let's combine one S orbital and a two p orbital."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "So these electrons will not want to go into this orbital because this is higher in energy, and it causes the bond to destabilize or break apart. So, once again, this was combining two atomic one s orbitals to form two molecular orbitals. Now, let's combine one S orbital and a two p orbital. Remember, a two p orbital has this eight shape, right? Okay, so let's look at A. And let's look at B."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Remember, a two p orbital has this eight shape, right? Okay, so let's look at A. And let's look at B. Here I have two ways two potential ways in which our two atomic orbitals can interact in space. In part A, we have an orthogonal interaction. In part B, we have a non orthogonal interaction."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Here I have two ways two potential ways in which our two atomic orbitals can interact in space. In part A, we have an orthogonal interaction. In part B, we have a non orthogonal interaction. Orthogonal simply means perpendicular. So let's see which one of these is the correct interaction. In other words, which one of these will form molecular bonds, and which one of these will not form molecular bonds."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Orthogonal simply means perpendicular. So let's see which one of these is the correct interaction. In other words, which one of these will form molecular bonds, and which one of these will not form molecular bonds. So let's begin with A. Now, remember, this plus and minus does not mean charge. It means, for example, the plus means that we're combining the one s positive orbital and the two p positive orbital."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "So let's begin with A. Now, remember, this plus and minus does not mean charge. It means, for example, the plus means that we're combining the one s positive orbital and the two p positive orbital. Now, when we combine the two positive orbitals, we get the following figure. When we combine the positive one S orbital and the negative one two p orbital, we get the following interaction. To get the negative two p, we simply switch it or flip it."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Now, when we combine the two positive orbitals, we get the following figure. When we combine the positive one S orbital and the negative one two p orbital, we get the following interaction. To get the negative two p, we simply switch it or flip it. So here we have our two interactions. So let's look at this guy. So, our positive will want to interact in a bonding way with the positive two p. So positive one s wants to interact with the positive sign of the two P orbital."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "So here we have our two interactions. So let's look at this guy. So, our positive will want to interact in a bonding way with the positive two p. So positive one s wants to interact with the positive sign of the two P orbital. And likewise at the same time when these guys are interacting in a bonding way, these guys are interacting in an anti bonding way. And that's because we have a positive and a negative. Remember, positive and positive orbitals create bonding interactions."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "And likewise at the same time when these guys are interacting in a bonding way, these guys are interacting in an anti bonding way. And that's because we have a positive and a negative. Remember, positive and positive orbitals create bonding interactions. Positive and negative create antibodies. So here we have a bonding and an antiboding. Let's go to this one."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Positive and negative create antibodies. So here we have a bonding and an antiboding. Let's go to this one. Here we have the same exact thing. Even though we flipped our two P orbital, we still have a bonding orbital between the positive one S and the positive two P and we have a negative interaction or an antibiotic interaction because we have a positive one S and a negative two P. So what happens when we have bonding and antiboding? Well, the bonding will exactly cancel out the antibonding."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Here we have the same exact thing. Even though we flipped our two P orbital, we still have a bonding orbital between the positive one S and the positive two P and we have a negative interaction or an antibiotic interaction because we have a positive one S and a negative two P. So what happens when we have bonding and antiboding? Well, the bonding will exactly cancel out the antibonding. And that means we will have a net interaction of zero. So orthogonal approach of orbitals. Or this guy does not allow for bonding because the bonding interactions are canceled out by the antibonding interaction."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "And that means we will have a net interaction of zero. So orthogonal approach of orbitals. Or this guy does not allow for bonding because the bonding interactions are canceled out by the antibonding interaction. So there will be no interaction when our two atomic orbitals, the one S and the two P approach in this orthogonal fashion. So now let's look at part B. Now we have the following non orthogonal interaction."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "So there will be no interaction when our two atomic orbitals, the one S and the two P approach in this orthogonal fashion. So now let's look at part B. Now we have the following non orthogonal interaction. So once again, let's show our pictures out. So we have the positive one S interact with the positive two P. So we keep the two orientations and we have the following picture. So this is one molecular orbital."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "So once again, let's show our pictures out. So we have the positive one S interact with the positive two P. So we keep the two orientations and we have the following picture. So this is one molecular orbital. And now let's try to do the negative. So we have the one S positive and the two P negative. So we flip the two P and we have the following depiction."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "And now let's try to do the negative. So we have the one S positive and the two P negative. So we flip the two P and we have the following depiction. Now we have a note or nodal plane symbolized by this black dash here. So we have the positive oneness interacts in an anti bonding fashion with the negative two P and we create this nodal plane which is once again simply our region where the electron density is zero. In other words, electrons will never will have a zero probability to be found in this region here."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "Now we have a note or nodal plane symbolized by this black dash here. So we have the positive oneness interacts in an anti bonding fashion with the negative two P and we create this nodal plane which is once again simply our region where the electron density is zero. In other words, electrons will never will have a zero probability to be found in this region here. And notice that we don't have the same situation as we have here. In other words, we simply have bonding and then we have antibonding. So that means that this will be the correct interaction."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "And notice that we don't have the same situation as we have here. In other words, we simply have bonding and then we have antibonding. So that means that this will be the correct interaction. This is how our two or how our one S and our two P orbitals will interact to form our two molecular orbitals. So we put in two atomic orbitals and we get out two molecular orbitals. And this is our picture here."}, {"title": "Orthogonal Molecular Orbitals .txt", "text": "This is how our two or how our one S and our two P orbitals will interact to form our two molecular orbitals. So we put in two atomic orbitals and we get out two molecular orbitals. And this is our picture here. So this is our energy diagram. So this is our one S orbital. And remember, the two P orbital is slightly higher on the energy level."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "So suppose we begin with a neutral atom in which the number of electrons equals number of protons. Now, what that basically means is that if we add up our charges due to our electrons, with the charges due to our protons, we're going to get a net charge or an overall charge of zero. Now, that's exactly what a neutral atom is. It's an atom. Atom in which our charge is zero. Well, now, suppose we take our atom and our neutral atom gains or loses electrons."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "It's an atom. Atom in which our charge is zero. Well, now, suppose we take our atom and our neutral atom gains or loses electrons. Well, now we have a case in which the number of electrons is no longer equal to the number of protons. And so we're going to get a charge species called an ion. Now, whenever we're dealing with metals, be it transition metals, alkaline metals or alkaline earth metals, these guys tend to lose electrons because they can hold onto an electrons very tightly."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "Well, now we have a case in which the number of electrons is no longer equal to the number of protons. And so we're going to get a charge species called an ion. Now, whenever we're dealing with metals, be it transition metals, alkaline metals or alkaline earth metals, these guys tend to lose electrons because they can hold onto an electrons very tightly. They're not very electronegative. And what that basically means is that they will form ions with positive charges. And these guys are called Cations."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "They're not very electronegative. And what that basically means is that they will form ions with positive charges. And these guys are called Cations. Now. On the contrary. Nonmetal such as the halogens or oxygen or sulfur."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "Now. On the contrary. Nonmetal such as the halogens or oxygen or sulfur. These guys have a very high affinity for electrons. And that means they will tend to take away electrons from other less electronegative atoms. And that means they will form ions in which there is a negative charge."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "These guys have a very high affinity for electrons. And that means they will tend to take away electrons from other less electronegative atoms. And that means they will form ions in which there is a negative charge. And these guys are called nions. Now, let's look at the following illustration. Suppose we have a neutral atom x and this guy takes away the electron from some other atom forming an anion."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "And these guys are called nions. Now, let's look at the following illustration. Suppose we have a neutral atom x and this guy takes away the electron from some other atom forming an anion. Well, this guy will simply have a negative charge represented in the following way now, most nonmetals follow this pathway. Now, suppose we have the reverse. Suppose we have an atom x that loses an electrons or loses an electron."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "Well, this guy will simply have a negative charge represented in the following way now, most nonmetals follow this pathway. Now, suppose we have the reverse. Suppose we have an atom x that loses an electrons or loses an electron. Now, this guy will form a cation or an ion with a positive charge. Now, most metals follow this pathway. Now, one note about transition metals."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "Now, this guy will form a cation or an ion with a positive charge. Now, most metals follow this pathway. Now, one note about transition metals. Transition metals. Whenever they lose electrons, they first lose electrons from the s orbital and therefore the d orbital. And that's because the S orbital is at a higher state than the D orbital."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "Transition metals. Whenever they lose electrons, they first lose electrons from the s orbital and therefore the d orbital. And that's because the S orbital is at a higher state than the D orbital. And electrons lose or electrons leave the higher levels first before the lower levels. Now we're going to talk more about orbitals. The s orbitals and d orbitals in a future lecture."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "And electrons lose or electrons leave the higher levels first before the lower levels. Now we're going to talk more about orbitals. The s orbitals and d orbitals in a future lecture. So stick around. Now, what happens to the size of our atom when it loses or gains an electron? Well, let's suppose we have a neutral atom that has two protons and two electrons."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "So stick around. Now, what happens to the size of our atom when it loses or gains an electron? Well, let's suppose we have a neutral atom that has two protons and two electrons. And let's look at our illustration. So one electron on the atomo shell, one electron on the innermost shell, and two plus or two protons down there a nucleus. So its charge is zero."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "And let's look at our illustration. So one electron on the atomo shell, one electron on the innermost shell, and two plus or two protons down there a nucleus. So its charge is zero. Now, suppose it loses an electron. Let's say it loses our atomos electron. Well, that means it's going to shrink in size."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "Now, suppose it loses an electron. Let's say it loses our atomos electron. Well, that means it's going to shrink in size. And this is why. Remember, most atoms or atoms are generally composed of empty space. So when this guy disappears, all this empty space goes with it."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "And this is why. Remember, most atoms or atoms are generally composed of empty space. So when this guy disappears, all this empty space goes with it. And so it becomes much smaller. So now the plus two charge will be greater than our negative one charge. And that means this electron will be pulled even closer to the nucleus, decreasing in size even further."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "And so it becomes much smaller. So now the plus two charge will be greater than our negative one charge. And that means this electron will be pulled even closer to the nucleus, decreasing in size even further. So when we talk about cations, or when neutral atoms become can ions, there is a loss of electron, and this shrinks the element to a smaller size. Now, on the contrary, let's look at anions. So what happens to anion?"}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "So when we talk about cations, or when neutral atoms become can ions, there is a loss of electron, and this shrinks the element to a smaller size. Now, on the contrary, let's look at anions. So what happens to anion? Suppose now we have a neutral atom with a plus one charge and a minus one charge, forming a neutral charge. And now suppose it gains an electron from some other atom, probably metal. Now what will happen is it will gain a new outer shell, forming this outer shell and forming this empty space in between."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "Suppose now we have a neutral atom with a plus one charge and a minus one charge, forming a neutral charge. And now suppose it gains an electron from some other atom, probably metal. Now what will happen is it will gain a new outer shell, forming this outer shell and forming this empty space in between. And that means anions. When they gain electrons, this causes our atom or element, to grow in size. So cations, or the formation of cations, decrease the size of our atom, while the formation of anions increases the size of our atom."}, {"title": "Cations, Anions and Isoelectronic Atoms .txt", "text": "And that means anions. When they gain electrons, this causes our atom or element, to grow in size. So cations, or the formation of cations, decrease the size of our atom, while the formation of anions increases the size of our atom. Now let's look at one last thing. Now isoelectronic atoms are those atoms that have the same number of electrons, but different number of protons. Suppose we have an atom with ten electrons and ten protons, and an atom with ten electrons and eleven protons."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "Now as of now we've only really spoken about oneway elementary actions in which reactants become products in a single forward step. Now we're going to look at complex reactions and we're going to find the rate laws for complex reactions. Now remember, in complex reactions reactants convert to products in more than one step. So the mechanism of our conversion will require more than one step. Now let's look at the following complex reaction. This is the overall reaction in which five molecules, two of these guys, one of this guy and two of these guys react to produce four moles of this guy and 1 mol of this guy."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "So the mechanism of our conversion will require more than one step. Now let's look at the following complex reaction. This is the overall reaction in which five molecules, two of these guys, one of this guy and two of these guys react to produce four moles of this guy and 1 mol of this guy. So let's break this overall net reaction into its individual steps. So step one is the following 1 mol of hydrogen peroxide reacts with 1 mol of this guy in a slow step producing 1 mol of hydroxide and 1 mol of hoi. Now this slow step will be important in determining the rate law and we'll see in a second why."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "So let's break this overall net reaction into its individual steps. So step one is the following 1 mol of hydrogen peroxide reacts with 1 mol of this guy in a slow step producing 1 mol of hydroxide and 1 mol of hoi. Now this slow step will be important in determining the rate law and we'll see in a second why. Now the second step is the following. This intermediate or this intermediate reacts with 1 mol reactants to produce again one molar hydroxide and one of our products. Now notice that this is one of our products."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "Now the second step is the following. This intermediate or this intermediate reacts with 1 mol reactants to produce again one molar hydroxide and one of our products. Now notice that this is one of our products. So step two is responsible for producing this product, namely this guy. And now these two guys, two moles of hydroxide finally react with two moles of hydronium to produce our four moles of water, the second product. So the second step is responsible for producing one of the products while the third step is responsible for producing the second type of product."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "So step two is responsible for producing this product, namely this guy. And now these two guys, two moles of hydroxide finally react with two moles of hydronium to produce our four moles of water, the second product. So the second step is responsible for producing one of the products while the third step is responsible for producing the second type of product. Now let's go back to this step. Now, relative to the rates of these two steps, this is a very, very slow step and in fact these steps can be assumed to be instantaneous very quick. So what's the significance of this slow step?"}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "Now let's go back to this step. Now, relative to the rates of these two steps, this is a very, very slow step and in fact these steps can be assumed to be instantaneous very quick. So what's the significance of this slow step? This low step is called a rate determining step. And this step limits the rate at which products are produced. And this equation or this reaction can be used to find the rate law."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "This low step is called a rate determining step. And this step limits the rate at which products are produced. And this equation or this reaction can be used to find the rate law. Now remember, the rate law can only really be found experimentally via results. But this is a second way with which you can find the rate law. But you still need to actually find the rate law using the results."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "Now remember, the rate law can only really be found experimentally via results. But this is a second way with which you can find the rate law. But you still need to actually find the rate law using the results. And then you can check the two and see if they coincide. And usually for the most part they will coincide. So this is a second way that you could find rate laws in complex reactions by using the rate law for the slow step, the rate determining step."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "And then you can check the two and see if they coincide. And usually for the most part they will coincide. So this is a second way that you could find rate laws in complex reactions by using the rate law for the slow step, the rate determining step. So since this is a bimolecular elementary reaction we can use the coefficients as the exponents. In other words, our rate of reaction is equal to KR rate constant times the concentration of hydrogen peroxide times the concentration of iodine. And each of the exponents is one because we have 1 mol of this guy and 1 mol of this guy react to produce these two intermediates."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "So since this is a bimolecular elementary reaction we can use the coefficients as the exponents. In other words, our rate of reaction is equal to KR rate constant times the concentration of hydrogen peroxide times the concentration of iodine. And each of the exponents is one because we have 1 mol of this guy and 1 mol of this guy react to produce these two intermediates. And this is indeed a balanced equation. So once again, the reason we're allowed to find a weight loss so quickly and without experimental results is because step one is an elementary bimolecular reaction with an overall reaction order of two. In other words, this has one coefficient and a second coefficient."}, {"title": "Rate Determining Step and Rate Law Part 1.txt", "text": "And this is indeed a balanced equation. So once again, the reason we're allowed to find a weight loss so quickly and without experimental results is because step one is an elementary bimolecular reaction with an overall reaction order of two. In other words, this has one coefficient and a second coefficient. So one plus one, two. That's why. Now, once again, once you find this guy, you have to check to make sure this is in fact the rate law."}, {"title": "Octet Rule.txt", "text": "So in this lecture, we're going to talk about the octave rule. But before we talk about the octet rule, let's recall what the electron configuration of an atom atom is. So let's look at helium, neon, and argon. So the electron configuration is simply the layout of the electrons found within that atom. So let's look at helium. Now, helium has two protons found in the nucleus and two electrons."}, {"title": "Octet Rule.txt", "text": "So the electron configuration is simply the layout of the electrons found within that atom. So let's look at helium. Now, helium has two protons found in the nucleus and two electrons. These two electrons are going to be found in the one s orbital. So the one s orbital can have a maximum of two electrons. And because helium has two electrons, these two electrons will be found in the one s orbital."}, {"title": "Octet Rule.txt", "text": "These two electrons are going to be found in the one s orbital. So the one s orbital can have a maximum of two electrons. And because helium has two electrons, these two electrons will be found in the one s orbital. So let's go to neon. Now, neon has ten protons down in the nucleus and ten electrons down in the orbital surrounding our nucleus. So our electron configuration for neon will be one f, two."}, {"title": "Octet Rule.txt", "text": "So let's go to neon. Now, neon has ten protons down in the nucleus and ten electrons down in the orbital surrounding our nucleus. So our electron configuration for neon will be one f, two. So two electrons will be in the one s orbital, two electrons will be in a two s orbital, and six electrons will be in the two p orbital. Remember, there are actually three p orbitals. There's PX, PY, and PZ."}, {"title": "Octet Rule.txt", "text": "So two electrons will be in the one s orbital, two electrons will be in a two s orbital, and six electrons will be in the two p orbital. Remember, there are actually three p orbitals. There's PX, PY, and PZ. And each orbital can hold a maximum of two electrons. So that means we're going to have a total of six electrons in our p orbitals. So, once again, our orbitals within neon are completely filled, just like they were in helium."}, {"title": "Octet Rule.txt", "text": "And each orbital can hold a maximum of two electrons. So that means we're going to have a total of six electrons in our p orbitals. So, once again, our orbitals within neon are completely filled, just like they were in helium. So let's look at argon. Argon has 18 protons found in the nucleus, 18 electrons found in the surrounding orbital. So we're going to have two electrons in the one s orbital, two electrons in the two s orbital, six electrons in the three p orbitals."}, {"title": "Octet Rule.txt", "text": "So let's look at argon. Argon has 18 protons found in the nucleus, 18 electrons found in the surrounding orbital. So we're going to have two electrons in the one s orbital, two electrons in the two s orbital, six electrons in the three p orbitals. So a total of six in the p.\nNow we're going to have two electrons in the three s orbital, and three p will have six electrons. So, once again, every orbital within our gun is completely filled, just like it was in neon and helium. And in fact, these three atoms are noble gases."}, {"title": "Octet Rule.txt", "text": "So a total of six in the p.\nNow we're going to have two electrons in the three s orbital, and three p will have six electrons. So, once again, every orbital within our gun is completely filled, just like it was in neon and helium. And in fact, these three atoms are noble gases. Noble gases are atoms that have electron configurations that are completely filled. So they have a perfect electron configuration. All the orbitals that can possibly be filled are filled."}, {"title": "Octet Rule.txt", "text": "Noble gases are atoms that have electron configurations that are completely filled. So they have a perfect electron configuration. All the orbitals that can possibly be filled are filled. And that's exactly why noble gases are very stable. They have very stable electron configurations. And in fact, any atom that is not a noble gas will try to attain."}, {"title": "Octet Rule.txt", "text": "And that's exactly why noble gases are very stable. They have very stable electron configurations. And in fact, any atom that is not a noble gas will try to attain. So it will try to either lose or gain electrons such that its electron configuration matches that of one of the noble gases. And this is what we call the octave rule. So the octave rule is simply the process of filling electron shells or electron orbitals in a way such that an electron configuration that matches one of the noble gases is reached."}, {"title": "Octet Rule.txt", "text": "So it will try to either lose or gain electrons such that its electron configuration matches that of one of the noble gases. And this is what we call the octave rule. So the octave rule is simply the process of filling electron shells or electron orbitals in a way such that an electron configuration that matches one of the noble gases is reached. So let's do an example. So, here we have a carbon atom, and we have a fluorine atom. So this carbon atom has six protons and six electrons."}, {"title": "Octet Rule.txt", "text": "So let's do an example. So, here we have a carbon atom, and we have a fluorine atom. So this carbon atom has six protons and six electrons. So it has six protons within the nucleus along with the six neutrons. And we have six electrons surrounding our nucleus. So two electrons will be in the one s orbital, two electrons will be in the two s orbital, and only two electrons will be in the two p orbital."}, {"title": "Octet Rule.txt", "text": "So it has six protons within the nucleus along with the six neutrons. And we have six electrons surrounding our nucleus. So two electrons will be in the one s orbital, two electrons will be in the two s orbital, and only two electrons will be in the two p orbital. So that means that there are four more electrons that can fit into our p orbitals because the p orbital can have a maximum of six electrons. So if we go up here, that means that we want, or carbon wants to gain four more electrons so that its electron configuration matches that of neon. So we basically want to attain this electron configuration."}, {"title": "Octet Rule.txt", "text": "So that means that there are four more electrons that can fit into our p orbitals because the p orbital can have a maximum of six electrons. So if we go up here, that means that we want, or carbon wants to gain four more electrons so that its electron configuration matches that of neon. So we basically want to attain this electron configuration. Carbon wants to attain this electron configuration. So that means that it is very likely that it will gain four electrons to form our perfect electron configuration. So if this carbon had four electrons floating around this atom, that means it would gain those electrons, and these electrons would steal the remaining two p orbitals."}, {"title": "Octet Rule.txt", "text": "Carbon wants to attain this electron configuration. So that means that it is very likely that it will gain four electrons to form our perfect electron configuration. So if this carbon had four electrons floating around this atom, that means it would gain those electrons, and these electrons would steal the remaining two p orbitals. And that means once it gains the four electrons, it will have the one f two, two f two, and two p six the perfect configuration that matches neon. So notice that a neutral atom of carbon has a charge of zero. That's because it has six protons and six electrons."}, {"title": "Octet Rule.txt", "text": "And that means once it gains the four electrons, it will have the one f two, two f two, and two p six the perfect configuration that matches neon. So notice that a neutral atom of carbon has a charge of zero. That's because it has six protons and six electrons. Now, we have six protons. By the way, this six, this subscript means six protons. It's the atomic number which corresponds to the number of protons."}, {"title": "Octet Rule.txt", "text": "Now, we have six protons. By the way, this six, this subscript means six protons. It's the atomic number which corresponds to the number of protons. So we have six protons, but now we have four plus six electrons. So we have a total of ten electrons, just like neon does. And that means our charge will be negative ten plus six."}, {"title": "Octet Rule.txt", "text": "So we have six protons, but now we have four plus six electrons. So we have a total of ten electrons, just like neon does. And that means our charge will be negative ten plus six. So our charge will be four. So this is an anion. So now let's look at Florida."}, {"title": "Octet Rule.txt", "text": "So our charge will be four. So this is an anion. So now let's look at Florida. Now, fluorine has nine protons and nine electrons. So once again, our nucleus will have nine protons, and our mutual fluorine atom will have nine electrons floating around the nucleus. So once again, let's apply our octave rule and let's see how many electrons this fluorine atom, this neutral fluorine atom needs to gain to obtain a noble gas configuration."}, {"title": "Octet Rule.txt", "text": "Now, fluorine has nine protons and nine electrons. So once again, our nucleus will have nine protons, and our mutual fluorine atom will have nine electrons floating around the nucleus. So once again, let's apply our octave rule and let's see how many electrons this fluorine atom, this neutral fluorine atom needs to gain to obtain a noble gas configuration. So since nine is very close to ten, so this fluorine has nine electrons, which is only one away, one less than neon, that means fluorine will tend to gain an electron to form our neon noble gas configuration. So, once again, if we have an electron floating around this fluorine nucleus, it will tend to take that electron and place it into one of the two p orbitals so that once it places it there, it has a noble gas configuration. So, once again, this neutral fluorine atom gains one electron, puts it into the two p orbital, and it becomes an electron configuration that matches that of neon."}, {"title": "Octet Rule.txt", "text": "So since nine is very close to ten, so this fluorine has nine electrons, which is only one away, one less than neon, that means fluorine will tend to gain an electron to form our neon noble gas configuration. So, once again, if we have an electron floating around this fluorine nucleus, it will tend to take that electron and place it into one of the two p orbitals so that once it places it there, it has a noble gas configuration. So, once again, this neutral fluorine atom gains one electron, puts it into the two p orbital, and it becomes an electron configuration that matches that of neon. So it has two electrons in the one s. It has two electrons in the two s and six electrons in the two p. Now, this guy is very happy because he's very stable, and that's because noble gases have very stable electron configuration. And that's basically what the Octave Rule is. It's basically a procedure, a process that you follow that will give you a noble gas electron configuration."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "For example, I could take this water bottle, and I could measure the amount of water inside in terms of volume. I could measure the pressure. I could also measure the temperature. What I can say is, oh, this water bottle has x amount of enthalpy, okay? And that's because enthalpy is defined using a formula. So when somebody asks you what enthalpy is, you tell them it's a formula."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "What I can say is, oh, this water bottle has x amount of enthalpy, okay? And that's because enthalpy is defined using a formula. So when somebody asks you what enthalpy is, you tell them it's a formula. And the formula is this the change in enthalpy is equal to change in U plus P times change in B. So let's see what this guy is. This guy is just a total internal energy of the system, okay?"}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And the formula is this the change in enthalpy is equal to change in U plus P times change in B. So let's see what this guy is. This guy is just a total internal energy of the system, okay? And if you look at this system here, which is a circle or a sphere, it's basically the collective energy of all the different molecules found within the system. So you sum their kinetic energies and you sum their potential energies, and that gives you, this guy here, the total infernal energy of the system. What this is is it's the work done to displace environment in creating the current state of the system, okay?"}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And if you look at this system here, which is a circle or a sphere, it's basically the collective energy of all the different molecules found within the system. So you sum their kinetic energies and you sum their potential energies, and that gives you, this guy here, the total infernal energy of the system. What this is is it's the work done to displace environment in creating the current state of the system, okay? And what that basically means is in order to get from nothing to this volume and pressure, these molecules have to do work on the environment and displacing them in forming this structure, this sphere or the circle. So what enthalpy is, it's the total energy of the system plus the work that the system has to do on the environment in creating that system. And that's what enthalpy is."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And what that basically means is in order to get from nothing to this volume and pressure, these molecules have to do work on the environment and displacing them in forming this structure, this sphere or the circle. So what enthalpy is, it's the total energy of the system plus the work that the system has to do on the environment in creating that system. And that's what enthalpy is. Now, enthalpy is a state function, which means that enthalpy is independent of the pathway and the reactions that lead to the system. In other words, enthalpy only depends on the current state of the system, okay? Now, enthalpy is also an external property, which also means that entropy depends on the amount of the system."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "Now, enthalpy is a state function, which means that enthalpy is independent of the pathway and the reactions that lead to the system. In other words, enthalpy only depends on the current state of the system, okay? Now, enthalpy is also an external property, which also means that entropy depends on the amount of the system. So the more of the system that we have, the largest system, the more enthalpy we have. And that's because enthalpy is related to internal energy, and internal energy is related to number of moles, number of particles. So the more particles we have, the more enthalpy we have."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "So the more of the system that we have, the largest system, the more enthalpy we have. And that's because enthalpy is related to internal energy, and internal energy is related to number of moles, number of particles. So the more particles we have, the more enthalpy we have. And that's why it's an external property. So because enthalpy is a man made concept, we can't talk about absolute values of enthalpy. They simply do not exist."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And that's why it's an external property. So because enthalpy is a man made concept, we can't talk about absolute values of enthalpy. They simply do not exist. What we do talk about are changes in enthalpy. And in order for changes in enthalpy to exist, we must assign zero values to Cherton atoms with the churchin's elements, okay? And in fact, under a standard state of 1 bar or 750 tor and a certain temperature such as 25 degrees Celsius, all elements such as diatomic, hydrogen diatomic, oxygen, carbon diatonic, fluorine and so on are assigned a value of zero entropy, zero joules per mole."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "What we do talk about are changes in enthalpy. And in order for changes in enthalpy to exist, we must assign zero values to Cherton atoms with the churchin's elements, okay? And in fact, under a standard state of 1 bar or 750 tor and a certain temperature such as 25 degrees Celsius, all elements such as diatomic, hydrogen diatomic, oxygen, carbon diatonic, fluorine and so on are assigned a value of zero entropy, zero joules per mole. Okay? So when a reaction or a compound is created from its constituents or from its raw elements, there's a change in entropy. And this change in entropy is known as the standard entropy of formation."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "Okay? So when a reaction or a compound is created from its constituents or from its raw elements, there's a change in entropy. And this change in entropy is known as the standard entropy of formation. And it's given this value where the small zero here called not is basically represents standard state or a 1 bar. Okay? And this f is formation."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And it's given this value where the small zero here called not is basically represents standard state or a 1 bar. Okay? And this f is formation. So change in enthalpy, okay? So standard enthalpy of formation and it basically means it's the change in enthalpy for a reaction that creates 1 mol of compound from its constituent elements. For example, let's take the creation or formation of water from its constituents, from hydrogen and from oxygen, okay?"}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "So change in enthalpy, okay? So standard enthalpy of formation and it basically means it's the change in enthalpy for a reaction that creates 1 mol of compound from its constituent elements. For example, let's take the creation or formation of water from its constituents, from hydrogen and from oxygen, okay? These guys combine to form water and there is a decrease in enthalpy. And that means that the internal energy of the system decreases. There are less molecules within the system and so there is less Enthropy within the system."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "These guys combine to form water and there is a decrease in enthalpy. And that means that the internal energy of the system decreases. There are less molecules within the system and so there is less Enthropy within the system. Now, different types of reactions exist. Some reactions are in a gas phase when other reactions are in a liquid and solid phases. Now, when you're dealing with reactions in a gas phase, you have to realize that gaseous systems are compressible."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "Now, different types of reactions exist. Some reactions are in a gas phase when other reactions are in a liquid and solid phases. Now, when you're dealing with reactions in a gas phase, you have to realize that gaseous systems are compressible. You can expand them and compress them. And so pressure changes because volume changes, so pressure isn't constant. And if pressure changes, if volume changes, then PV work is done."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "You can expand them and compress them. And so pressure changes because volume changes, so pressure isn't constant. And if pressure changes, if volume changes, then PV work is done. In other words, the system does work to expand by doing work in the surroundings, to move the surroundings away so that the system could expand. Same thing with compression. Therefore, when you're dealing with reactions of gases, this equation holds the change in enthalpy is equal to a change in internal energy plus p times change in volume."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "In other words, the system does work to expand by doing work in the surroundings, to move the surroundings away so that the system could expand. Same thing with compression. Therefore, when you're dealing with reactions of gases, this equation holds the change in enthalpy is equal to a change in internal energy plus p times change in volume. But when you're dealing with reactions in liquid and solid phases, you have to realize that these type of systems are not compressible. If you can't compress them or expand them, that means no PD work is done. Okay?"}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "But when you're dealing with reactions in liquid and solid phases, you have to realize that these type of systems are not compressible. If you can't compress them or expand them, that means no PD work is done. Okay? And that means this guy can go to zero. And now we could approximate the change in enthalpy to just simply change in internal energy, which is heat or the exchange of energy. Okay?"}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And that means this guy can go to zero. And now we could approximate the change in enthalpy to just simply change in internal energy, which is heat or the exchange of energy. Okay? So under lab conditions, normally this condition holds. We could approximate this, we could approximate change in Enthalpy as just simply being the change in internal energy. So what is the heat of reaction?"}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "So under lab conditions, normally this condition holds. We could approximate this, we could approximate change in Enthalpy as just simply being the change in internal energy. So what is the heat of reaction? The heat of reaction is simply the change in internal energy or change in enthalpy of the reaction. And it's found using this formula. For example, let's take this equation here we have reactants and we have products."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "The heat of reaction is simply the change in internal energy or change in enthalpy of the reaction. And it's found using this formula. For example, let's take this equation here we have reactants and we have products. To find the heat of reaction, we simply find a change in atrial energy or change in enthalpy of the product. We find the change in entropy or, I'm sorry, enthalpy of the reactants and we subtract this from this and that will give us the heat of reaction. So recall it, enthalpy is a state function and what that basically means that the pathway taken to get to the final product or to get to the final system does not affect the final change in enthalpy."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "To find the heat of reaction, we simply find a change in atrial energy or change in enthalpy of the product. We find the change in entropy or, I'm sorry, enthalpy of the reactants and we subtract this from this and that will give us the heat of reaction. So recall it, enthalpy is a state function and what that basically means that the pathway taken to get to the final product or to get to the final system does not affect the final change in enthalpy. And this jumps directly into something called Hess's Law. Now, Hessa's Law basically states that the steps taken to get from the reactant to the products does not affect the final change in enthalpy, okay? And Hess's Law can be applied to reactions."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And this jumps directly into something called Hess's Law. Now, Hessa's Law basically states that the steps taken to get from the reactant to the products does not affect the final change in enthalpy, okay? And Hess's Law can be applied to reactions. Now, here we have an imaginary reaction. That's a two step reaction. And each step has its corresponding change in enthalpy."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "Now, here we have an imaginary reaction. That's a two step reaction. And each step has its corresponding change in enthalpy. Now, what Hess Law basically states is that in order to find the final change in enthalpy, you simply add each corresponding enthalpy. Together we get the final result. So ten kilojoules per mole plus 20 kilojoules per mole gives you 30 kilojoules per mole."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "Now, what Hess Law basically states is that in order to find the final change in enthalpy, you simply add each corresponding enthalpy. Together we get the final result. So ten kilojoules per mole plus 20 kilojoules per mole gives you 30 kilojoules per mole. And this is their final answer. Now, yield this side, you basically the same exact thing. You use basic algebra."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "And this is their final answer. Now, yield this side, you basically the same exact thing. You use basic algebra. You add these guys up. You add these guys up, except instead of using equal sign like you would use in algebra, you use arrows. Okay?"}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "You add these guys up. You add these guys up, except instead of using equal sign like you would use in algebra, you use arrows. Okay? So A plus B plus B plus C gives you a plus two b plus C arrows. C plus D. Now, one last thing left. You have to notice the molecules that appear on both sides of the equation."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "So A plus B plus B plus C gives you a plus two b plus C arrows. C plus D. Now, one last thing left. You have to notice the molecules that appear on both sides of the equation. If they appear on both sides of the equation, you could simply cross them out, okay? So A appears only on this side, you leave it alone. Two B appears only on this side, you leave it alone."}, {"title": "Enthalpy, Heat of Reaction and Hess Law .txt", "text": "If they appear on both sides of the equation, you could simply cross them out, okay? So A appears only on this side, you leave it alone. Two B appears only on this side, you leave it alone. C appears on this side, and on this side, you cross them out. And D appears only on this side. So you leave it alone."}, {"title": "pH indicators .txt", "text": "In this lecture we're going to talk a bit more about titrations and equivalence points. Now, earlier in another video, we saw that we can actually calculate our equivalence point using a formula. Now, if you want to learn more about that, check out the link above. Now, actually calculating your equivalence point isn't very useful when you're conducting your experiment. When you're titrating, you want to actually visualize the equivalence point. And to visualize it, you use two things."}, {"title": "pH indicators .txt", "text": "Now, actually calculating your equivalence point isn't very useful when you're conducting your experiment. When you're titrating, you want to actually visualize the equivalence point. And to visualize it, you use two things. The first thing you use is a PH meter. The PH meter will automatically tell you the PH of your solution. For example, suppose this is your PH meter and it's attached to your solution at any given time, you can look at it and it will tell you the PH of your buffer solution."}, {"title": "pH indicators .txt", "text": "The first thing you use is a PH meter. The PH meter will automatically tell you the PH of your solution. For example, suppose this is your PH meter and it's attached to your solution at any given time, you can look at it and it will tell you the PH of your buffer solution. Now, the second thing you need to use is called an acid based indicator. Acid based indicators are compounds that change color in solution when they convert to their conjugate acid based form. Now, what you do is you take your indicator, you take a very small amount of it and you place it into your solution before you begin saturation."}, {"title": "pH indicators .txt", "text": "Now, the second thing you need to use is called an acid based indicator. Acid based indicators are compounds that change color in solution when they convert to their conjugate acid based form. Now, what you do is you take your indicator, you take a very small amount of it and you place it into your solution before you begin saturation. And what the indicator does, it reacts with the H plus ions in the following way. When there is a lot of H plus ions associated in our solution, aka, when it's acidic, that means equilibrium will shift this way. Now, when there is very little H plus ions or our concentration of H plus ion is very low, equilibrium will shift this way."}, {"title": "pH indicators .txt", "text": "And what the indicator does, it reacts with the H plus ions in the following way. When there is a lot of H plus ions associated in our solution, aka, when it's acidic, that means equilibrium will shift this way. Now, when there is very little H plus ions or our concentration of H plus ion is very low, equilibrium will shift this way. Meaning we're going to have more of the conjugate base of our conjugate acid. Note that this represents one color and this represents a different color. So in solution, when this guy predominates, our solution will be of this color."}, {"title": "pH indicators .txt", "text": "Meaning we're going to have more of the conjugate base of our conjugate acid. Note that this represents one color and this represents a different color. So in solution, when this guy predominates, our solution will be of this color. And when this guy is now a solution and this guy predominates, the solution B will be of a different color. So once again, in acidic solutions, this guy dominates. And in basic solutions, this guy dominates."}, {"title": "pH indicators .txt", "text": "And when this guy is now a solution and this guy predominates, the solution B will be of a different color. So once again, in acidic solutions, this guy dominates. And in basic solutions, this guy dominates. So, we can look at the ratio of the concentration of this guy to this guy. And when the ratio is ten or above, that means we're going to be of this guy, of this color, so of the active color, because there's much more of this than this in our solution. Now, the opposite is true as well."}, {"title": "pH indicators .txt", "text": "So, we can look at the ratio of the concentration of this guy to this guy. And when the ratio is ten or above, that means we're going to be of this guy, of this color, so of the active color, because there's much more of this than this in our solution. Now, the opposite is true as well. When this guy will dominate, when there is more of this guy than this guy. Now, ratio is 0.1 or less. That means the base color will dominate."}, {"title": "pH indicators .txt", "text": "When this guy will dominate, when there is more of this guy than this guy. Now, ratio is 0.1 or less. That means the base color will dominate. So this guy will dominate. Now, note that the end point is the point at which you observe a change in color. And the end point is not the same thing as the equivalence point."}, {"title": "titration.txt", "text": "Now, whatever we're adding to our unknown sample, it's called a titren, and a titrate be their base or an acid. So let's illustrate Titration using a picture and then a graph. So suppose we have some known volume of acid, say, I don't know, hydrochloric acid in our flask, and we have a PH reader attached to the inside of our flask. So at any given time, we could measure our PH. Now suppose we have some known concentration of base and we begin adding this base drop by drop. Now, before we add the base, our PH is say, I don't know, two."}, {"title": "titration.txt", "text": "So at any given time, we could measure our PH. Now suppose we have some known concentration of base and we begin adding this base drop by drop. Now, before we add the base, our PH is say, I don't know, two. Okay, that's our PH. Now, as I begin slowly adding drop by drop, what will happen to our PH? Well, it will increase."}, {"title": "titration.txt", "text": "Okay, that's our PH. Now, as I begin slowly adding drop by drop, what will happen to our PH? Well, it will increase. Right, but how will it increase? Well, let's look at the graph of PH versus volume added, where PH is the Y axis, and volume added of our base in this case is the x axis. So according to this graph, we see that the relationship is based on a sigmoidal curve."}, {"title": "titration.txt", "text": "Right, but how will it increase? Well, let's look at the graph of PH versus volume added, where PH is the Y axis, and volume added of our base in this case is the x axis. So according to this graph, we see that the relationship is based on a sigmoidal curve. And what that means is that initially, when you first add some known amount of concentration of base, the PH change will be very little. The PH won't change by that much. And in fact, from this much to this much volume added, the PH only rises by maybe 0.5."}, {"title": "titration.txt", "text": "And what that means is that initially, when you first add some known amount of concentration of base, the PH change will be very little. The PH won't change by that much. And in fact, from this much to this much volume added, the PH only rises by maybe 0.5. But eventually we come to a point where any more volume added will increase the PH dramatically. And we will come to a point where the PH will increase by ten units. So why is that?"}, {"title": "titration.txt", "text": "But eventually we come to a point where any more volume added will increase the PH dramatically. And we will come to a point where the PH will increase by ten units. So why is that? Well, let's see. Now let's look at the reaction of sodium hydroxide with hydrochloric acid. Initially, we only have HCL in our mixture and HCL dissociates into H plus ion and the chloride ion."}, {"title": "titration.txt", "text": "Well, let's see. Now let's look at the reaction of sodium hydroxide with hydrochloric acid. Initially, we only have HCL in our mixture and HCL dissociates into H plus ion and the chloride ion. Now, the H plus ion is what's responsible for creating the acidic solution for lowering our PH to a PH of two. So what will happen when we first dissociate or add NaOH to our mixture? Well, initially we only have a small amount of NaOH, and NaOH will dissociate into Ma plus and oh minus."}, {"title": "titration.txt", "text": "Now, the H plus ion is what's responsible for creating the acidic solution for lowering our PH to a PH of two. So what will happen when we first dissociate or add NaOH to our mixture? Well, initially we only have a small amount of NaOH, and NaOH will dissociate into Ma plus and oh minus. But since we only have a small amount of NaOH, that means we only have a small amount of these guys. And these guys are the ones that associate with HMCL to form water and NaCl. So initially, only a small percentage of these guys will reassociate into this form."}, {"title": "titration.txt", "text": "But since we only have a small amount of NaOH, that means we only have a small amount of these guys. And these guys are the ones that associate with HMCL to form water and NaCl. So initially, only a small percentage of these guys will reassociate into this form. And so we will see a decrease in the H plus ion, but a very small decrease. And that means we're only going to see a very small change or increase in PH initially. What will happen when the ratio of moles is one to one?"}, {"title": "titration.txt", "text": "And so we will see a decrease in the H plus ion, but a very small decrease. And that means we're only going to see a very small change or increase in PH initially. What will happen when the ratio of moles is one to one? So suppose we have some ratio of HCL, some unknown in our solution and suppose we add that same amount of moles, of NaOH. Well, that means we're going to have a ratio one to one to one to one. And all these guys are going to reassociate to form water and NaCl."}, {"title": "titration.txt", "text": "So suppose we have some ratio of HCL, some unknown in our solution and suppose we add that same amount of moles, of NaOH. Well, that means we're going to have a ratio one to one to one to one. And all these guys are going to reassociate to form water and NaCl. And what will happen then? Well, then our PH will become seven, right? And in fact, at this point, when the ratio of moles added is the same as the ratio of this guy, this point is called the equivalence point."}, {"title": "titration.txt", "text": "And what will happen then? Well, then our PH will become seven, right? And in fact, at this point, when the ratio of moles added is the same as the ratio of this guy, this point is called the equivalence point. And for NaOH and HCL, for a strong acid and strong base, this is a PH of seven. So we see that this point is the point at which we add enough Maoh that our ratio begins to equal out. Our ratio becomes closer and closer to one and one."}, {"title": "titration.txt", "text": "And for NaOH and HCL, for a strong acid and strong base, this is a PH of seven. So we see that this point is the point at which we add enough Maoh that our ratio begins to equal out. Our ratio becomes closer and closer to one and one. And as we add more, as we get closer to a ratio of one to one, we get closer to our PH of seven. And that's why we see this large increase in PH. Now, if we begin to add more of our sodium hydroxide, our solution becomes basic."}, {"title": "Introduction to PV work.txt", "text": "Recall what the conservation of energy tells us. According to the conservation of energy, energy cannot be destroyed and it cannot be created. Energy always exists, and energy of the Universe is constant. It does not change. But what energy can do is it can be transformed from one form to another. For example, from kinetic energy to another."}, {"title": "Introduction to PV work.txt", "text": "It does not change. But what energy can do is it can be transformed from one form to another. For example, from kinetic energy to another. Another type of energy. Now there are two types of energy transfers. There's heat and work."}, {"title": "Introduction to PV work.txt", "text": "Another type of energy. Now there are two types of energy transfers. There's heat and work. Heat is a natural Transfer of energy from an object with a Higher temperature to an object with a Lower temperature. And if you want to learn more about heat, check out the lecture on heat in the chemistry section. Here."}, {"title": "Introduction to PV work.txt", "text": "Heat is a natural Transfer of energy from an object with a Higher temperature to an object with a Lower temperature. And if you want to learn more about heat, check out the lecture on heat in the chemistry section. Here. We're only going to talk about work. Now work is the transfer of energy due to forces. It could be individual forces acting on objects or the net forces acting on the entire object."}, {"title": "Introduction to PV work.txt", "text": "We're only going to talk about work. Now work is the transfer of energy due to forces. It could be individual forces acting on objects or the net forces acting on the entire object. Now, work is a scalar, and that means it only has magnitude, no direction. And the units of work is like the units of energy. It's joules."}, {"title": "Introduction to PV work.txt", "text": "Now, work is a scalar, and that means it only has magnitude, no direction. And the units of work is like the units of energy. It's joules. And we could also use electron volts when we're talking about very small microscopic objects. Now. Work does have negative and a positive."}, {"title": "Introduction to PV work.txt", "text": "And we could also use electron volts when we're talking about very small microscopic objects. Now. Work does have negative and a positive. Negative work simply means our object loses energy, while positive work means our object gains energy. Now, whenever a nonfictional force is applied, work can be found using the following equation work is equal to the force applied on our object, times the distance our object moves. And since work is, in fact a scalar, a vector times a vector gives us a dot product."}, {"title": "Introduction to PV work.txt", "text": "Negative work simply means our object loses energy, while positive work means our object gains energy. Now, whenever a nonfictional force is applied, work can be found using the following equation work is equal to the force applied on our object, times the distance our object moves. And since work is, in fact a scalar, a vector times a vector gives us a dot product. So that means we take the cosine of the angle. If you want to learn more about dot product and cross product, check out the lecture on that. So, basically, if we have some object, say, this block, and we apply a force with an angle theta to the horizontal of that object, and our object moves a distance d we can find the amount of work done on the object."}, {"title": "Introduction to PV work.txt", "text": "So that means we take the cosine of the angle. If you want to learn more about dot product and cross product, check out the lecture on that. So, basically, if we have some object, say, this block, and we apply a force with an angle theta to the horizontal of that object, and our object moves a distance d we can find the amount of work done on the object. The amount of energy that object gain by using the following formula force times, distance travel times, cosine of the angle between them. Now, if we do not neglect friction, if we actually include friction, for example, when this box is traveling along this horizontal surface, there is friction in a real world situation. So if we do not neglect friction, we find our overall work by following the following formula the work done is equal to change in our potential energy of the object plus change in the kinetic energy of the object plus the change in the internal energy of our object due to friction."}, {"title": "Introduction to PV work.txt", "text": "The amount of energy that object gain by using the following formula force times, distance travel times, cosine of the angle between them. Now, if we do not neglect friction, if we actually include friction, for example, when this box is traveling along this horizontal surface, there is friction in a real world situation. So if we do not neglect friction, we find our overall work by following the following formula the work done is equal to change in our potential energy of the object plus change in the kinetic energy of the object plus the change in the internal energy of our object due to friction. Now, for the most part, whenever friction acts on an object, it changes the object's internal energy. And internal energy is simply the energy of the individual molecules found in that object. If you sum up all the different types of energies on those individual molecules, you will get the internal energy of that system."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So in this lecture we're going to look at three examples of redox reactions. And we're going to assign oxidation states to each atom in our molecule. So let's begin with the first one. So we begin with copper in its elemental state. And that means copper gets a charge of zero according to our rules. Let's look at nitric acid."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So we begin with copper in its elemental state. And that means copper gets a charge of zero according to our rules. Let's look at nitric acid. Now, nitric acid, it contains an hmm and an O. That means we first assign to H, then we assign to O, then we assign to the N. Now our entire molecule must be neutral, so we must take that into consideration when assigning. So let's give our H a plus one according to our table."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Now, nitric acid, it contains an hmm and an O. That means we first assign to H, then we assign to O, then we assign to the N. Now our entire molecule must be neutral, so we must take that into consideration when assigning. So let's give our H a plus one according to our table. And let's give an O a minus two according to our table. So our N should be a number, such that when you add that number to one and when you subtract negative six, we get here. Why negative six?"}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "And let's give an O a minus two according to our table. So our N should be a number, such that when you add that number to one and when you subtract negative six, we get here. Why negative six? Well, because there are three oxygen molecules. So three times negative two gives us a total of negative six. So negative six plus one gives us a five."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Well, because there are three oxygen molecules. So three times negative two gives us a total of negative six. So negative six plus one gives us a five. So our N should be five. Let's check. One plus five."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So our N should be five. Let's check. One plus five. Six minus six gives us a zero. So that works. Let's go to this side."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Six minus six gives us a zero. So that works. Let's go to this side. So here we have copper nitrate. So from this part we know that N and three is a negative one, right, because the entire thing to balance a neutral charge, this must be a plus one. So this guy is minus one."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So here we have copper nitrate. So from this part we know that N and three is a negative one, right, because the entire thing to balance a neutral charge, this must be a plus one. So this guy is minus one. So our three is minus one. So no three is minus one times two such molecules, this whole thing means it's negative two. So our copper must be plus two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So our three is minus one. So no three is minus one times two such molecules, this whole thing means it's negative two. So our copper must be plus two. Our copper is plus two. Now that means our oxygen is negative two, and our N must be plus five. So this whole thing must be minus two, including this two here."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Our copper is plus two. Now that means our oxygen is negative two, and our N must be plus five. So this whole thing must be minus two, including this two here. So let's go to nitric oxide. So our O, we first assign our oxidation state to O, and our O gets a minus two. So that means our N must be plus two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So let's go to nitric oxide. So our O, we first assign our oxidation state to O, and our O gets a minus two. So that means our N must be plus two. Why is it plus two? Well, because this molecule is neutral and that means charge of zero. So to balance this negative two, this N must be a plus two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Why is it plus two? Well, because this molecule is neutral and that means charge of zero. So to balance this negative two, this N must be a plus two. And finally, let's look at water the oxygen. Well, we first assign to our H. So that gets a plus one. Plus one times two gives us a plus two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "And finally, let's look at water the oxygen. Well, we first assign to our H. So that gets a plus one. Plus one times two gives us a plus two. So this guy must be a negative two to balance out the mutual charge. So now let's look at what gets oxidized and what gets reduced. So our copper goes from a zero charge to a plus two charge."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So this guy must be a negative two to balance out the mutual charge. So now let's look at what gets oxidized and what gets reduced. So our copper goes from a zero charge to a plus two charge. And that means copper is oxidized because electrons are removed and these electrons transfer to some other atom. So let's look at what atom is reduced. So our N goes from a plus five to a plus two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "And that means copper is oxidized because electrons are removed and these electrons transfer to some other atom. So let's look at what atom is reduced. So our N goes from a plus five to a plus two. That means all the electrons move from copper to N, creating that plus two charge. So our N is what's reduced. So our oxidizing agent is this guy nitric acid, the N specifically."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "That means all the electrons move from copper to N, creating that plus two charge. So our N is what's reduced. So our oxidizing agent is this guy nitric acid, the N specifically. And our reducing agent is copper. So let's write that. So this guy is oxidized and this guy is reduced."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "And our reducing agent is copper. So let's write that. So this guy is oxidized and this guy is reduced. And once again, our reducing agent, our oxidizing agent. So let's move on to example two. Now, in example two, we begin with the copper sulfide."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "And once again, our reducing agent, our oxidizing agent. So let's move on to example two. Now, in example two, we begin with the copper sulfide. So an oxygen in its elemental state. So oxygen automatically gets a charge of zero because it's in its elemental state. And let's look at these guys."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So an oxygen in its elemental state. So oxygen automatically gets a charge of zero because it's in its elemental state. And let's look at these guys. First, we assign our charge to our sulfur. Now, sulfur is in the same group as oxygen. That means it gets a negative two charge."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "First, we assign our charge to our sulfur. Now, sulfur is in the same group as oxygen. That means it gets a negative two charge. So our sulfur is negative two. Now, to balance this guy off, because this entire molecule is neutral charge of zero, our copper must be a plus one because we have two copper molecules. So plus one, let's check two times one, two minus 20 works."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So our sulfur is negative two. Now, to balance this guy off, because this entire molecule is neutral charge of zero, our copper must be a plus one because we have two copper molecules. So plus one, let's check two times one, two minus 20 works. So we're done with our reactant side. Let's go to product side. So now copper is in our elemental state."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So we're done with our reactant side. Let's go to product side. So now copper is in our elemental state. So our copper must be neutral zero. And let's look at so, oxygen gets assigned first. So oxygen gets a negative two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So our copper must be neutral zero. And let's look at so, oxygen gets assigned first. So oxygen gets a negative two. A negative two times two is negative four. So what must s be? Well, if this whole thing is neutral, this guy must be plus four."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "A negative two times two is negative four. So what must s be? Well, if this whole thing is neutral, this guy must be plus four. Why? Well, because plus four minus four negative two times two gives us zero. So let's see what's reduced and what's oxidized."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Why? Well, because plus four minus four negative two times two gives us zero. So let's see what's reduced and what's oxidized. Our oxygen goes from zero to negative two. And that means this guy is reduced. So let's write this."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Our oxygen goes from zero to negative two. And that means this guy is reduced. So let's write this. So it's reduced. Our oxygen is reduced. So what gets oxidized?"}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So it's reduced. Our oxygen is reduced. So what gets oxidized? If something is reduced, something must be oxidized. So our copper goes from a plus one to zero. And that means, actually, copper also gets reduced."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "If something is reduced, something must be oxidized. So our copper goes from a plus one to zero. And that means, actually, copper also gets reduced. And so copper is actually also reduced. So let's write reduced for our copper. Now, sulfur goes from negative two to positive four."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "And so copper is actually also reduced. So let's write reduced for our copper. Now, sulfur goes from negative two to positive four. That means our sulfur is the atom that gets oxidized. So let's write that. So this guy gets oxidized."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "That means our sulfur is the atom that gets oxidized. So let's write that. So this guy gets oxidized. So two atoms get reduced, one atom gets oxidized. So this guy, that means this guy's our reducing agent. While our oxygen must be the oxidizing agent because it obtains those electrons."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So two atoms get reduced, one atom gets oxidized. So this guy, that means this guy's our reducing agent. While our oxygen must be the oxidizing agent because it obtains those electrons. It takes those electrons away from the sulfur atom. So let's look at our final example. So zinc sulfide plus oxygen gives us this guy."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "It takes those electrons away from the sulfur atom. So let's look at our final example. So zinc sulfide plus oxygen gives us this guy. So our oxygen once again is in its elemental state. So our oxygen gets a charge of zero, a neutral charge. Let's look at these guys."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So our oxygen once again is in its elemental state. So our oxygen gets a charge of zero, a neutral charge. Let's look at these guys. So, we assign our charge to sulfur first. So sulfur gets a charge of negative two because it's in the same group as oxygen. So this guy gets a negative two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So, we assign our charge to sulfur first. So sulfur gets a charge of negative two because it's in the same group as oxygen. So this guy gets a negative two. And our zinc must have a charge of plus two. So plus two minus two gives us neutral. Makes sense."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "And our zinc must have a charge of plus two. So plus two minus two gives us neutral. Makes sense. Let's go to this guy. So first we assign our charge to oxygen. Oxygen gets a charge of negative two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Let's go to this guy. So first we assign our charge to oxygen. Oxygen gets a charge of negative two. So this guy gets a charge of negative two. Now our zinc stays at positive two. So zinc remains at positive two."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So this guy gets a charge of negative two. Now our zinc stays at positive two. So zinc remains at positive two. So in order for this whole molecule to be to have a charge of zero, that means all these guys must add up to zero. So negative two times eight times four gives us negative eight. Plus two gives us negative six."}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "So in order for this whole molecule to be to have a charge of zero, that means all these guys must add up to zero. So negative two times eight times four gives us negative eight. Plus two gives us negative six. Sulfur must have a positive six charge. So what's oxidized? What's reduced?"}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Sulfur must have a positive six charge. So what's oxidized? What's reduced? Well, our oxygen goes from zero to negative two. So that means oxygen gets reduced. Okay?"}, {"title": "Oxidation-Reduction Reactions Example .txt", "text": "Well, our oxygen goes from zero to negative two. So that means oxygen gets reduced. Okay? Now, our sulfur goes from negative two to plus six. And that means our sulfur gets oxidized. So our sulfur gets oxidized."}, {"title": "Solutions and Solvation.txt", "text": "In this lecture, I will talk to you about solutions. So, what are solutions? Solutions are simply mixtures of two more compounds found in the same state. They're also called homogeneous solutions. Homogeneous simply means in the same state. Now, since there are three states possible, three types of solutions exist solid solutions, liquid solutions and gas solutions."}, {"title": "Solutions and Solvation.txt", "text": "They're also called homogeneous solutions. Homogeneous simply means in the same state. Now, since there are three states possible, three types of solutions exist solid solutions, liquid solutions and gas solutions. An example of a solid solution is brass. Brass is a metal composed of zinc and copper. An example of a liquid solution is sodium chloride or salt, found in water."}, {"title": "Solutions and Solvation.txt", "text": "An example of a solid solution is brass. Brass is a metal composed of zinc and copper. An example of a liquid solution is sodium chloride or salt, found in water. An example of a gas solution is air. Air is composed of nitrogen and oxygen. Now, whenever we talk about solutions, it's important to differentiate between the solvent and a solute."}, {"title": "Solutions and Solvation.txt", "text": "An example of a gas solution is air. Air is composed of nitrogen and oxygen. Now, whenever we talk about solutions, it's important to differentiate between the solvent and a solute. The solvent is a compound of which there is more. And the solute is a compound of which there is less. So let's go back to our examples in brass, brass is composed of 55% copper and 45% zinc."}, {"title": "Solutions and Solvation.txt", "text": "The solvent is a compound of which there is more. And the solute is a compound of which there is less. So let's go back to our examples in brass, brass is composed of 55% copper and 45% zinc. So there's more copper, which means copper is a solvent and zinc is a solute. In this example, there's more water than sodium, chloride or salt. So the solvent is water and the solute is the salt."}, {"title": "Solutions and Solvation.txt", "text": "So there's more copper, which means copper is a solvent and zinc is a solute. In this example, there's more water than sodium, chloride or salt. So the solvent is water and the solute is the salt. Now let's go to our gas example. In our gas solution, there's 79% nitrogen. So this guy is a solvent and 21% oxygen."}, {"title": "Solutions and Solvation.txt", "text": "Now let's go to our gas example. In our gas solution, there's 79% nitrogen. So this guy is a solvent and 21% oxygen. So oxygen is a solve. Now let's look at ideal dilute solutions. So what are ideal dilute solutions?"}, {"title": "Solutions and Solvation.txt", "text": "So oxygen is a solve. Now let's look at ideal dilute solutions. So what are ideal dilute solutions? These solutions are simply solutions in which every single sole molecule is separated by a solvent molecule. So there is no interaction between neighboring soluble molecules. Let's look at an example."}, {"title": "Solutions and Solvation.txt", "text": "These solutions are simply solutions in which every single sole molecule is separated by a solvent molecule. So there is no interaction between neighboring soluble molecules. Let's look at an example. So this is an example of an ideal dilute solution. In this example, the sodium and a chloride are separated by water molecules so they can't interact. This is an example of a nonideal dilute solution."}, {"title": "Solutions and Solvation.txt", "text": "So this is an example of an ideal dilute solution. In this example, the sodium and a chloride are separated by water molecules so they can't interact. This is an example of a nonideal dilute solution. In this example, the sodium and the chloride are able to interact with each other, so therefore, it's non. Ideal dissolving is a process by which solvent molecules break apart solvent molecules from one another. Now, in the solvent, there's water."}, {"title": "Solutions and Solvation.txt", "text": "In this example, the sodium and the chloride are able to interact with each other, so therefore, it's non. Ideal dissolving is a process by which solvent molecules break apart solvent molecules from one another. Now, in the solvent, there's water. This is called hydration. Now remember, light dissolves like so. Polar molecules dissolve polar molecules and nonpolar molecules will dissolve other nonpolar molecules."}, {"title": "Solutions and Solvation.txt", "text": "This is called hydration. Now remember, light dissolves like so. Polar molecules dissolve polar molecules and nonpolar molecules will dissolve other nonpolar molecules. Nonpolar will not be able to dissolve polar, and that's because they won't be able to overcome the large dipole moments of the polar molecules. And these dipole moments come from large differences in electronegativity. Now, salvation is the process of breaking down ionic compounds, bipolar compounds such as H 20."}, {"title": "Solutions and Solvation.txt", "text": "Nonpolar will not be able to dissolve polar, and that's because they won't be able to overcome the large dipole moments of the polar molecules. And these dipole moments come from large differences in electronegativity. Now, salvation is the process of breaking down ionic compounds, bipolar compounds such as H 20. Now, the results are aqueous solutions. Within an aqueous solution, the ions are able to move freely, so that electrons are also able to move freely. And because, because of this, aqueous solutions conduct electricity very well."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "So let's look at the following complex reaction in which two no molecules react with one BR two molecule to produce two moles of Nobr molecules. Now from experimental results we find that our rate law is k times concentration of no squared times the concentration of BR two squared. Now this guy comes from experiment and our goal in this lecture will be to find the rate law using the second step, the slow step and compare that rate law to our experimental rate law and see if they coincide. So first let's examine that situation at hand. Notice our first step is now the fast step and that means 1 mol of N o will react with 1 mol of BR two to produce 1 mol of no BR two. And this step will be really quick meaning that by the time it gets here it's going to go back."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "So first let's examine that situation at hand. Notice our first step is now the fast step and that means 1 mol of N o will react with 1 mol of BR two to produce 1 mol of no BR two. And this step will be really quick meaning that by the time it gets here it's going to go back. And in fact we're going to assume that this step, this reaction first reaction is at equilibrium before this reaction even begins. And we're going to assume that the concentration of this guy is very, very small. Now let's look at the second step."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "And in fact we're going to assume that this step, this reaction first reaction is at equilibrium before this reaction even begins. And we're going to assume that the concentration of this guy is very, very small. Now let's look at the second step. In the second step, notice that the reactant, this guy is the intermediate, this guy. So the product of step one is the reaction of step two and that means step two will be dependent on step one and that's because the intermediate is part of the reactive part of the slow step process. So the slow step is still the rate determining step and we're still going to use this step to find our rate law."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "In the second step, notice that the reactant, this guy is the intermediate, this guy. So the product of step one is the reaction of step two and that means step two will be dependent on step one and that's because the intermediate is part of the reactive part of the slow step process. So the slow step is still the rate determining step and we're still going to use this step to find our rate law. But now we're going to have to take this guide into consideration from the first step. So let's write the rate determining or let's write the rate law for our first step. So remember, we're dealing with an elementary bimolecular reaction, the first step as well as a second step."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "But now we're going to have to take this guide into consideration from the first step. So let's write the rate determining or let's write the rate law for our first step. So remember, we're dealing with an elementary bimolecular reaction, the first step as well as a second step. So we can write our rate laws in the following manner. The rate of the first step is equal to k one. Our constant going this way times the concentration of this guy to the first power because we have a coefficient of one times the concentration of BR also to the first power because we have a coefficient of 1 mol."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "So we can write our rate laws in the following manner. The rate of the first step is equal to k one. Our constant going this way times the concentration of this guy to the first power because we have a coefficient of one times the concentration of BR also to the first power because we have a coefficient of 1 mol. Now this equals to the reverse reaction because we're assuming our equilibrium or the first reaction reached equilibrium. And so this rate equals this rate, the reverse rate. So this guy equals the constant minus one going this way times the concentration of this guy."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "Now this equals to the reverse reaction because we're assuming our equilibrium or the first reaction reached equilibrium. And so this rate equals this rate, the reverse rate. So this guy equals the constant minus one going this way times the concentration of this guy. Now when we go backwards this is already active. So these guys are equal. All right, now we're done."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "Now when we go backwards this is already active. So these guys are equal. All right, now we're done. Let's write the rate law for our second step to slow step. So rate is equal to k two or some constant for going this way times the concentration of Nobr two times the concentration of no. Now, notice that this guy appears here and here."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "Let's write the rate law for our second step to slow step. So rate is equal to k two or some constant for going this way times the concentration of Nobr two times the concentration of no. Now, notice that this guy appears here and here. Therefore, we can represent this guy in terms of all these guys. So let's bring the k negative one over and we get the following. Nobr or the concentration of Nobr equals k one divided by k minus one, right times no or concentration of no times the concentration of BR two."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "Therefore, we can represent this guy in terms of all these guys. So let's bring the k negative one over and we get the following. Nobr or the concentration of Nobr equals k one divided by k minus one, right times no or concentration of no times the concentration of BR two. And now we want to take this whole guy and plug it into our rate law for our step number two. For our slow step, we plug this guy in and we get the following our rate for the second step, the slow step is equal to k two. This guy times k one."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "And now we want to take this whole guy and plug it into our rate law for our step number two. For our slow step, we plug this guy in and we get the following our rate for the second step, the slow step is equal to k two. This guy times k one. This guy divided by k minus one. This guy times two of these because one comes from the second step and one comes from the first step times this guy. The concentration of BR two from the first step gives you this."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "This guy divided by k minus one. This guy times two of these because one comes from the second step and one comes from the first step times this guy. The concentration of BR two from the first step gives you this. Now we can combine these to give exponent of two. And we can let this guy be some other constants, say KC. And so our rate becomes KC times the concentration of NL squared times the concentration of BR."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "Now we can combine these to give exponent of two. And we can let this guy be some other constants, say KC. And so our rate becomes KC times the concentration of NL squared times the concentration of BR. And this is exactly what we get from experiments some constant k which we can say they're equal times the concentration of this guy squared times the concentration of BR two. This is exactly what we get from experiments. And so our rate laws do coincide."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "And this is exactly what we get from experiments some constant k which we can say they're equal times the concentration of this guy squared times the concentration of BR two. This is exactly what we get from experiments. And so our rate laws do coincide. Now, once again, whenever our rate law is not the first step, it's the second or third step. We have to take into consideration the intermediate guys because the intermediate guys will determine the concentration of reactants for the slow step. And that's why they need to be taken into consideration."}, {"title": "Rate Determining Step and Rate Law Part 2.txt", "text": "Now, once again, whenever our rate law is not the first step, it's the second or third step. We have to take into consideration the intermediate guys because the intermediate guys will determine the concentration of reactants for the slow step. And that's why they need to be taken into consideration. So what we basically did is first we found the rate law for our first reaction, for our fast reaction for going the forward and going backwards. And then we found the rate law for the second reaction going one way because it's the slow step. And then we basically represent the intermediate concentration as everything else."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "So let ka be 1.8\ntimes ten to negative five. You can look that up in a textbook. So we get 1.8\ntimes ten to negative five gives you X times X divided by 0.02 minus X, that gives you X squared divided by 0.2 minus X. Now we bring this guy over on this side, and we get 1.8 times ten to negative five multiplied by 0.2 minus X gives you X squared. Now distribute, and we get 1.8 times ten to negative five multiplied by 0.2 minus this guy multiplied by X gives you 1.8\ntimes ten to negative five. X equals X squared."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Now we bring this guy over on this side, and we get 1.8 times ten to negative five multiplied by 0.2 minus X gives you X squared. Now distribute, and we get 1.8 times ten to negative five multiplied by 0.2 minus this guy multiplied by X gives you 1.8\ntimes ten to negative five. X equals X squared. Now we bring everything to one side, and we equate it to zero. So we want a positive X squared, because it's easier to work with. So we bring these guys over on this side."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Now we bring everything to one side, and we equate it to zero. So we want a positive X squared, because it's easier to work with. So we bring these guys over on this side. So X squared plus 1.8\ntimes ten to the negative five. Because when we bring this guy over, this becomes a positive minus, because this guy was initially a plus. So when we bring it over, it will be a -3.6\ntimes ten to the negative seven equals zero."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "So X squared plus 1.8\ntimes ten to the negative five. Because when we bring this guy over, this becomes a positive minus, because this guy was initially a plus. So when we bring it over, it will be a -3.6\ntimes ten to the negative seven equals zero. Now we use a quadratic formula which states negative B squared plus minus square root of B squared minus four AC divided by two A. So in our case, our A is one. Our b is this guy?"}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Now we use a quadratic formula which states negative B squared plus minus square root of B squared minus four AC divided by two A. So in our case, our A is one. Our b is this guy? 1.8 times tens, negative five. And our C is negative 3.6\ntimes ten to the negative seven. So let's plug everything in."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "1.8 times tens, negative five. And our C is negative 3.6\ntimes ten to the negative seven. So let's plug everything in. We get negative 1.8\ntimes 10th to negative five squared plus minus a square root of 1.8 times ten to negative five squared minus four times one times A times C times this guy. So divided by two. So now we simplify these guys, and we get negative 3.24 times ten to negative ten plus minus 0012 divided by two."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "We get negative 1.8\ntimes 10th to negative five squared plus minus a square root of 1.8 times ten to negative five squared minus four times one times A times C times this guy. So divided by two. So now we simplify these guys, and we get negative 3.24 times ten to negative ten plus minus 0012 divided by two. Now we get two answers. One is negative. One is positive."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Now we get two answers. One is negative. One is positive. Remember, we're dealing with concentrations. That means concentrations can't be negative. So we reject this negative one, and we accept this guy, which is 6.7 times ten to negative seven."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Remember, we're dealing with concentrations. That means concentrations can't be negative. So we reject this negative one, and we accept this guy, which is 6.7 times ten to negative seven. So our initial concentration of our acetic acid was 0.02. Our final concentration is 0.02 minus this guy gives you 0.0194. So they're very close, only a little bit dissociated."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "So our initial concentration of our acetic acid was 0.02. Our final concentration is 0.02 minus this guy gives you 0.0194. So they're very close, only a little bit dissociated. And that makes sense because this acetic acid is a very weak acid. Its ka is very small. So that was the first way, first method of finding our concentration."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "And that makes sense because this acetic acid is a very weak acid. Its ka is very small. So that was the first way, first method of finding our concentration. What's the second way? Well, the second way is a much better way, but it is an approximation. So if you are allowed to approximate, you should definitely use this way, because it's shorter."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "What's the second way? Well, the second way is a much better way, but it is an approximation. So if you are allowed to approximate, you should definitely use this way, because it's shorter. Now, let me show you what happens. So let's go back to our first step. So 1.8\ntimes ten to negative five equals X squared divided by 0.2 minus X."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Now, let me show you what happens. So let's go back to our first step. So 1.8\ntimes ten to negative five equals X squared divided by 0.2 minus X. Now, remember, our initial concentration is 0.2. And our final concentration is very small. It's very small compared to our initial."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Now, remember, our initial concentration is 0.2. And our final concentration is very small. It's very small compared to our initial. That's because we have a very weak acid, not a lot of the acid will dissociate. And so our final concentration, our X, will be much smaller than 0.2, our initial concentration. So we can approximate this guy to be equal to X squared over 0.2\nbecause our concentration will change by that much because we see, look, 0.94 and 0.2 and only changed by 0.6."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "That's because we have a very weak acid, not a lot of the acid will dissociate. And so our final concentration, our X, will be much smaller than 0.2, our initial concentration. So we can approximate this guy to be equal to X squared over 0.2\nbecause our concentration will change by that much because we see, look, 0.94 and 0.2 and only changed by 0.6. That's a very small amount. So now we have this very simple situation in which we simply bring over the 0.2, multiply it by 1.8\ntimes ten to negative five equals X squared. Take the radical, and we get 1.8 times ten to negative five times zero."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "That's a very small amount. So now we have this very simple situation in which we simply bring over the 0.2, multiply it by 1.8\ntimes ten to negative five equals X squared. Take the radical, and we get 1.8 times ten to negative five times zero. Two radical gives you two answers a positive and a negative. We reject the negative because concentrations can't be negative and we take the positive guy. Now look how close this guy is to this guy."}, {"title": "Determining pH of Strong and Weak Acids and Bases .txt", "text": "Two radical gives you two answers a positive and a negative. We reject the negative because concentrations can't be negative and we take the positive guy. Now look how close this guy is to this guy. There's a difference of only zero. That's a very small difference between our approximation, our shorter method, and our exact value, the longer method. So if you're allowed to approximate, you should definitely use this method."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "Ethane is a two carbon alkane. So it's composed of two carbons and six H atoms. Now, let's begin by examining the three dimensional picture of our ethane molecule. So ethane looks something like this, where these two black intersections are our carbons. So carbon one and carbon two given by these two carbons on the board. Now, these rest spheres are our H atoms."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "So ethane looks something like this, where these two black intersections are our carbons. So carbon one and carbon two given by these two carbons on the board. Now, these rest spheres are our H atoms. So we have six altogether. Now, these black solid wedges are our Sigma bonds, the Covalent bonds coming out of the board given by these two bonds here. These dashed wedges are given by these two bonds, sigma bonds in the back."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "So we have six altogether. Now, these black solid wedges are our Sigma bonds, the Covalent bonds coming out of the board given by these two bonds here. These dashed wedges are given by these two bonds, sigma bonds in the back. They're going into the board. These two sigma bonds are on the plane of the page, on the plane of the board. So if this was the XY axis, that means these two sigma bonds would be on the plane, on the XY plane."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "They're going into the board. These two sigma bonds are on the plane of the page, on the plane of the board. So if this was the XY axis, that means these two sigma bonds would be on the plane, on the XY plane. Now, one thing you should know about coding single sigma bonds is that these bonds are able to rotate in space. They could rotate 360 degrees around. So since all these guides are sigma bonds, all of these bonds are able to rotate."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "Now, one thing you should know about coding single sigma bonds is that these bonds are able to rotate in space. They could rotate 360 degrees around. So since all these guides are sigma bonds, all of these bonds are able to rotate. And in fact, what confirmations are there are three dimensional structures related to one another by sigma bond rotations. For example, here I have one confirmation. If I rotate this bond some given amount of degrees, I will have a second type of confirmation."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "And in fact, what confirmations are there are three dimensional structures related to one another by sigma bond rotations. For example, here I have one confirmation. If I rotate this bond some given amount of degrees, I will have a second type of confirmation. So these two molecules are related to one another by the fact that they're rotated some amount of degrees about this carbon carbon bond. And these guys are also known as confirmers. Now, there are many different types of conference that exist, right?"}, {"title": "Structural Conformations and Newman Projections .txt", "text": "So these two molecules are related to one another by the fact that they're rotated some amount of degrees about this carbon carbon bond. And these guys are also known as confirmers. Now, there are many different types of conference that exist, right? There's one conference, a second conference, a third conference, a fourth conference, and so on. Two important conference that you should know are the Eclipse conference and the staggered confirmer. Now, Eclipsed simply means that these two ch bonds are eclipsed."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "There's one conference, a second conference, a third conference, a fourth conference, and so on. Two important conference that you should know are the Eclipse conference and the staggered confirmer. Now, Eclipsed simply means that these two ch bonds are eclipsed. They're on the same plane. So if we look this way on our molecule, so here is our molecule and we're looking this way, we're going to see that this carbon bond, this carbon H bond exactly aligns with this carbon H bond. And likewise this carbon H bond aligns with this and this carbon H bond aligns with, aligns with this one."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "They're on the same plane. So if we look this way on our molecule, so here is our molecule and we're looking this way, we're going to see that this carbon bond, this carbon H bond exactly aligns with this carbon H bond. And likewise this carbon H bond aligns with this and this carbon H bond aligns with, aligns with this one. Now this is eclipse. What happens if we take the sigma bond and we rotate the sigma bond 180 degrees in this fashion? So we rotated 180 degrees."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "Now this is eclipse. What happens if we take the sigma bond and we rotate the sigma bond 180 degrees in this fashion? So we rotated 180 degrees. Well, we get the following staggered confirmation. Staggered simply means that this sigma bond, carbon, carbon sigma bond rotate 180 degrees. And so these angles are now a 60 degree angle to one another."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "Well, we get the following staggered confirmation. Staggered simply means that this sigma bond, carbon, carbon sigma bond rotate 180 degrees. And so these angles are now a 60 degree angle to one another. So before we had zero degrees between each, between each carbon H bond. But now we have a measure of 60 degrees between this bond and this bond. Now this is a more stable confirmation and we'll see why in a second."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "So before we had zero degrees between each, between each carbon H bond. But now we have a measure of 60 degrees between this bond and this bond. Now this is a more stable confirmation and we'll see why in a second. So now let's talk about Newman projections. Now, Newman projections are simply a way to visualize these three dimensional structures on a two dimensional plane, like this whiteboard or a sheet of paper. So once again, let's take our eclipse three dimensional structure and let's look at the structure from this way down."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "So now let's talk about Newman projections. Now, Newman projections are simply a way to visualize these three dimensional structures on a two dimensional plane, like this whiteboard or a sheet of paper. So once again, let's take our eclipse three dimensional structure and let's look at the structure from this way down. So when you look at it this way, in an eclipsed fashion, in an eclipse confirmation, all these ch bonds are aligned exactly across one another. So if we look this way down, all we'll see is this carbon bond and these three ch bonds. Because this ch bond, for example, will exactly cancel out the one in the back."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "So when you look at it this way, in an eclipsed fashion, in an eclipse confirmation, all these ch bonds are aligned exactly across one another. So if we look this way down, all we'll see is this carbon bond and these three ch bonds. Because this ch bond, for example, will exactly cancel out the one in the back. And the same thing goes for these other two. So that means looking this way, this is exactly what we see. Well, this is not a very good presentation because we can see the carbon, the back, we can't see the three ch bonds in the back."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "And the same thing goes for these other two. So that means looking this way, this is exactly what we see. Well, this is not a very good presentation because we can see the carbon, the back, we can't see the three ch bonds in the back. They do exist, but we can't really see them. And that's exactly where newer projections come in. It simply gives us a better way of visualizing this three dimensional structure."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "They do exist, but we can't really see them. And that's exactly where newer projections come in. It simply gives us a better way of visualizing this three dimensional structure. And it's given by the following depiction. So we simply take three ch bonds and we connect them like so, where each bond here is 120 degrees. Now, this intersection in the middle represents our first carbon atom."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "And it's given by the following depiction. So we simply take three ch bonds and we connect them like so, where each bond here is 120 degrees. Now, this intersection in the middle represents our first carbon atom. Now, to visualize the back carbon atom, we simply draw this blue circle, which symbolizes this blue carbon here. And then, because these ch bonds are right across on the same plane of these ch bonds, we simply shift them slightly this way so that we can visualize it. Now, you should know these ch bonds in the back are actually aligned exactly with this ch bond."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "Now, to visualize the back carbon atom, we simply draw this blue circle, which symbolizes this blue carbon here. And then, because these ch bonds are right across on the same plane of these ch bonds, we simply shift them slightly this way so that we can visualize it. Now, you should know these ch bonds in the back are actually aligned exactly with this ch bond. And the same goes for these two. But in order for us to visualize it, we shift the angle. We cheat a little bit."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "And the same goes for these two. But in order for us to visualize it, we shift the angle. We cheat a little bit. We shift the angle so that we can actually see them. So our carbon one, our carbon two and the ch bonds in the back, like we have here now. Or for the standard confirmation, it gets a little bit easier."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "We shift the angle so that we can actually see them. So our carbon one, our carbon two and the ch bonds in the back, like we have here now. Or for the standard confirmation, it gets a little bit easier. Because if we look at the staggered confirmation, which looks like this, what we see is this picture here. So we have the carbon atom and it's attached to three ch bonds. So three HS."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "Because if we look at the staggered confirmation, which looks like this, what we see is this picture here. So we have the carbon atom and it's attached to three ch bonds. So three HS. And then we have that bad carbon that's also attached to these three ch bonds. So in this picture, we can't really see the carbon atom. So that's exactly why we want the human projection."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "And then we have that bad carbon that's also attached to these three ch bonds. So in this picture, we can't really see the carbon atom. So that's exactly why we want the human projection. In the human projection, we can visualize this back carbon blue atom that's given here, as well as this first carbon atom. Green one here, given by this intersection. So the angle between any two bonds here is 60 degrees, as we said earlier for the staggered."}, {"title": "Structural Conformations and Newman Projections .txt", "text": "In the human projection, we can visualize this back carbon blue atom that's given here, as well as this first carbon atom. Green one here, given by this intersection. So the angle between any two bonds here is 60 degrees, as we said earlier for the staggered. So the angle here between any two ch bonds is zero. The angle here is 60 degrees. And the angle between any two of these bonds, for example, this green ch bond."}, {"title": "Isomers of Heptane.txt", "text": "So, isomers are compounds that have the same molecular formula but different structures. So whenever we try to find isomers of alkanes, it usually helps to have a systematic approach, and that's exactly what we use in this example. Here, we have a set of four steps that we're going to follow to find our isomers. And, in fact, we can always use these same steps whenever we're trying to find isomers of alkanes. So let's begin with step one. Now, step one happens to be the easiest one, and you'll see why in just a second."}, {"title": "Isomers of Heptane.txt", "text": "And, in fact, we can always use these same steps whenever we're trying to find isomers of alkanes. So let's begin with step one. Now, step one happens to be the easiest one, and you'll see why in just a second. Begin with the longest carbon chain. So what's the longest carbon chain of our heptane molecule? Well, it's simply heptane itself."}, {"title": "Isomers of Heptane.txt", "text": "Begin with the longest carbon chain. So what's the longest carbon chain of our heptane molecule? Well, it's simply heptane itself. So in step one, our first isomer is heptane itself. So here we have our heptane, and this is our first isomer. Let's go to step two."}, {"title": "Isomers of Heptane.txt", "text": "So in step one, our first isomer is heptane itself. So here we have our heptane, and this is our first isomer. Let's go to step two. So, in step two, shorten the chain by removing exactly one carbon from the end of our chain. So we remove this carbon, and then we place that carbon onto a different position. So let's remove it."}, {"title": "Isomers of Heptane.txt", "text": "So, in step two, shorten the chain by removing exactly one carbon from the end of our chain. So we remove this carbon, and then we place that carbon onto a different position. So let's remove it. And now let's take that methyl group and place it onto the first position or the second position here. Now, if we place it onto this position, we get back our heptane. The goal is to develop as many structural different compounds as possible."}, {"title": "Isomers of Heptane.txt", "text": "And now let's take that methyl group and place it onto the first position or the second position here. Now, if we place it onto this position, we get back our heptane. The goal is to develop as many structural different compounds as possible. So here's one more isomer. Notice that these two guys have the same exact molecular formula, but they have different structural forms. So now let's take this methyl and place it onto this carbon, and we will get another isomer."}, {"title": "Isomers of Heptane.txt", "text": "So here's one more isomer. Notice that these two guys have the same exact molecular formula, but they have different structural forms. So now let's take this methyl and place it onto this carbon, and we will get another isomer. So we place this methyl group here, and we get a third isomer. And notice if we draw one more guy, and then we place our methyl onto this position. This guy is exactly the same as this guy."}, {"title": "Isomers of Heptane.txt", "text": "So we place this methyl group here, and we get a third isomer. And notice if we draw one more guy, and then we place our methyl onto this position. This guy is exactly the same as this guy. And so this is not an isomer, because these two guys have the same molecular formula and the same structural forms. So this is not an isomer. So let's remove this guy."}, {"title": "Isomers of Heptane.txt", "text": "And so this is not an isomer, because these two guys have the same molecular formula and the same structural forms. So this is not an isomer. So let's remove this guy. So, it looks like we're done with step two. Let's go to step three. So, in step three, it says shorten the chain by removing two carbons."}, {"title": "Isomers of Heptane.txt", "text": "So, it looks like we're done with step two. Let's go to step three. So, in step three, it says shorten the chain by removing two carbons. Not one, but two. So now we're removing this carbon as well as this carbon. So we're developing a pentane molecule, and now we have two carbon atoms at our disposal, methyl groups at our disposal."}, {"title": "Isomers of Heptane.txt", "text": "Not one, but two. So now we're removing this carbon as well as this carbon. So we're developing a pentane molecule, and now we have two carbon atoms at our disposal, methyl groups at our disposal. So let's first point the two methyl groups here. That's one isomer. Let's show our pentane again."}, {"title": "Isomers of Heptane.txt", "text": "So let's first point the two methyl groups here. That's one isomer. Let's show our pentane again. And now take one methyl, place it here. The second methyl. Place it here."}, {"title": "Isomers of Heptane.txt", "text": "And now take one methyl, place it here. The second methyl. Place it here. That's a second isomer. Let's show one more pentane. And let's leave this methyl here and place this methyl here."}, {"title": "Isomers of Heptane.txt", "text": "That's a second isomer. Let's show one more pentane. And let's leave this methyl here and place this methyl here. That's yet another isomer. So let's see if we can get more isomers. Let's actually take both of these guys and place them here."}, {"title": "Isomers of Heptane.txt", "text": "That's yet another isomer. So let's see if we can get more isomers. Let's actually take both of these guys and place them here. And let's take one more. And place it like so all these guys have different structures, but identical molecular formulas, so they're all isomers. So let's go to step four, because we're essentially done here, if we rearrange them in some other way, we're going to get back on the same molecule as before."}, {"title": "Isomers of Heptane.txt", "text": "And let's take one more. And place it like so all these guys have different structures, but identical molecular formulas, so they're all isomers. So let's go to step four, because we're essentially done here, if we rearrange them in some other way, we're going to get back on the same molecule as before. So let's go to step four. In step four, I have dot, dot, dot, because we're simply removing not one or two, but three carbons from the end. So we're moving one, two, three."}, {"title": "Isomers of Heptane.txt", "text": "So let's go to step four. In step four, I have dot, dot, dot, because we're simply removing not one or two, but three carbons from the end. So we're moving one, two, three. So here we have one. So we have one, two, three. And now we have three carbons at our disposal, three methyl groups."}, {"title": "Isomers of Heptane.txt", "text": "So here we have one. So we have one, two, three. And now we have three carbons at our disposal, three methyl groups. So let's place one here, a second one here, and let's place a third one here. And this concludes our isomers. We should have a total of nine, so 123-45-6789."}, {"title": "Electrochemical cell .txt", "text": "Now, in this lecture, we're going to talk about something called electrochemical cells, also known as Galvanic cells. Now, recall that redox reactions are chemical reactions in which electrons flow from one atom to another. From physics, we know that moving charge such as electrons can be used to do useful work. Therefore, the flow of electrons found in redox reactions can somehow be transformed to do useful work. So an electrochemical cell is simply a way to capture this useful work produced by the movement of electrons from one atom to another. Now, we're going to talk about special types of electrochemical cells called voltaic cells, also known as batteries."}, {"title": "Electrochemical cell .txt", "text": "Therefore, the flow of electrons found in redox reactions can somehow be transformed to do useful work. So an electrochemical cell is simply a way to capture this useful work produced by the movement of electrons from one atom to another. Now, we're going to talk about special types of electrochemical cells called voltaic cells, also known as batteries. Now, these types of cells contain oxidizing reducing agent pairs connected by a conductor such as a wire. And this conductor allows electrons to flow from one atom to a second atom. Now, let's look at a layout of a voltaic cell."}, {"title": "Electrochemical cell .txt", "text": "Now, these types of cells contain oxidizing reducing agent pairs connected by a conductor such as a wire. And this conductor allows electrons to flow from one atom to a second atom. Now, let's look at a layout of a voltaic cell. Voltaic cells broken down into two half cells. So one half cell in a second half cell. In the first half cell, one half reaction takes place called oxidation."}, {"title": "Electrochemical cell .txt", "text": "Voltaic cells broken down into two half cells. So one half cell in a second half cell. In the first half cell, one half reaction takes place called oxidation. And the second half reaction takes place in a second half cell called reduction. Now, this wire connects the two cells. This wire is our conductor allowing electrons to flow."}, {"title": "Electrochemical cell .txt", "text": "And the second half reaction takes place in a second half cell called reduction. Now, this wire connects the two cells. This wire is our conductor allowing electrons to flow. This is a light bulb that lights up when there is a flow of electrons. And this salt bridge becomes important in allowing these electrons to continually flow. Now, let's look at this picture in more detail and let's see exactly what voltaic cells are and how they function."}, {"title": "Electrochemical cell .txt", "text": "This is a light bulb that lights up when there is a flow of electrons. And this salt bridge becomes important in allowing these electrons to continually flow. Now, let's look at this picture in more detail and let's see exactly what voltaic cells are and how they function. So let's examine the following reduction reaction. So, zinc solid reacts with aqueous copper to form a crease zinc and solid copper. Notice that our zinc solid is oxidized."}, {"title": "Electrochemical cell .txt", "text": "So let's examine the following reduction reaction. So, zinc solid reacts with aqueous copper to form a crease zinc and solid copper. Notice that our zinc solid is oxidized. It loses two electrons to form a plus two ana while those two same electrons are taken up by our copper molecule in the Aqueous state. And so this guy is reduced from a plus two to a neutral atom. Now, oxidation of zinc occurs in half cell number one."}, {"title": "Electrochemical cell .txt", "text": "It loses two electrons to form a plus two ana while those two same electrons are taken up by our copper molecule in the Aqueous state. And so this guy is reduced from a plus two to a neutral atom. Now, oxidation of zinc occurs in half cell number one. And zinc solid becomes zinc in the Aqueous state with a plus two charge, and it releases two electrons while in half cell number two reduction occurs. An aqueous copper takes up two electrons to form solid copper. So let's examine these reactions as they occur within our electrochemical cell, our voltaic cell."}, {"title": "Electrochemical cell .txt", "text": "And zinc solid becomes zinc in the Aqueous state with a plus two charge, and it releases two electrons while in half cell number two reduction occurs. An aqueous copper takes up two electrons to form solid copper. So let's examine these reactions as they occur within our electrochemical cell, our voltaic cell. So in eco number one and half cell number one, this red bar corresponds to our zinc solid. So zinc solid releases two electrons and it also releases our zinc ion. Now, this zinc ion is released into our solution from our metal bar."}, {"title": "Electrochemical cell .txt", "text": "So in eco number one and half cell number one, this red bar corresponds to our zinc solid. So zinc solid releases two electrons and it also releases our zinc ion. Now, this zinc ion is released into our solution from our metal bar. So the solution that has water as solvent increases in its concentration of zinc ion and at the same time, it increases the positive charge found within our solution in beaker one in half cell number one. Now, these electrons travel through the conductor and across and into the other side. Now, as it travels from this side to this side, this light bulb lights up."}, {"title": "Electrochemical cell .txt", "text": "So the solution that has water as solvent increases in its concentration of zinc ion and at the same time, it increases the positive charge found within our solution in beaker one in half cell number one. Now, these electrons travel through the conductor and across and into the other side. Now, as it travels from this side to this side, this light bulb lights up. And therefore, this light bulb allows us to visualize the movement of these electrons. As soon as it lights up, we know that electrons are traveling from this side to this side. Now let's look at hop cell number two."}, {"title": "Electrochemical cell .txt", "text": "And therefore, this light bulb allows us to visualize the movement of these electrons. As soon as it lights up, we know that electrons are traveling from this side to this side. Now let's look at hop cell number two. So this metal bar corresponds to our copper solid. And what happens is these two electrons combine with this copper ions down within the aqueous solution to form our copper solid. So in this solution, the copper ions move from in the solution to inside this metal bar."}, {"title": "Electrochemical cell .txt", "text": "So this metal bar corresponds to our copper solid. And what happens is these two electrons combine with this copper ions down within the aqueous solution to form our copper solid. So in this solution, the copper ions move from in the solution to inside this metal bar. So our concentration of copper ions found in solution decreases. And that means our plus charge found in this solution also decreases. So now let's look at a few terms that we need to know."}, {"title": "Electrochemical cell .txt", "text": "So our concentration of copper ions found in solution decreases. And that means our plus charge found in this solution also decreases. So now let's look at a few terms that we need to know. Electrodes are metals that conduct electrical current into out of the solution. So in this case, this is our electrode and this is our electrode. So our zinc solid and copper solid are our electrodes because they're metals that allow electrons to flow in or out."}, {"title": "Electrochemical cell .txt", "text": "Electrodes are metals that conduct electrical current into out of the solution. So in this case, this is our electrode and this is our electrode. So our zinc solid and copper solid are our electrodes because they're metals that allow electrons to flow in or out. So the anode is defined to be the half cell where oxidation takes place. So the anode is this guy. It includes beaker one, the aqueous solution, as well as the electrode."}, {"title": "Electrochemical cell .txt", "text": "So the anode is defined to be the half cell where oxidation takes place. So the anode is this guy. It includes beaker one, the aqueous solution, as well as the electrode. Beaker one, the capital, is defined to be the half cell where reduction takes place. So this is our cathode. It includes the aqueous solution, the beaker, as well as the electrode down and beaker two."}, {"title": "Electrochemical cell .txt", "text": "Beaker one, the capital, is defined to be the half cell where reduction takes place. So this is our cathode. It includes the aqueous solution, the beaker, as well as the electrode down and beaker two. So electrons travel from our anode to our cathode. Now let's look at the salt bridge. We still have discussed what this guy here is."}, {"title": "Electrochemical cell .txt", "text": "So electrons travel from our anode to our cathode. Now let's look at the salt bridge. We still have discussed what this guy here is. This is our salt bridge. Now, our salt bridge is composed of a solution of salt. For example, k two, so four."}, {"title": "Electrochemical cell .txt", "text": "This is our salt bridge. Now, our salt bridge is composed of a solution of salt. For example, k two, so four. So what's the purpose? What's the function of our salt bridge? Well, let's look at this picture here."}, {"title": "Electrochemical cell .txt", "text": "So what's the purpose? What's the function of our salt bridge? Well, let's look at this picture here. Eventually, when enough electrons travel to this location, what will happen to our positive charge in this speaker and our positive charge in this speaker? Well, we're going to have a build up, a positive charge in this speaker because this metal bar releases ions, right while in this speaker, these ions found within our solution are taken up by this metal bar, meaning there's a decrease in positive charge found on this side. So eventually, if we don't have anything connecting them like a salt bridge, the electrons will stop flowing."}, {"title": "Electrochemical cell .txt", "text": "Eventually, when enough electrons travel to this location, what will happen to our positive charge in this speaker and our positive charge in this speaker? Well, we're going to have a build up, a positive charge in this speaker because this metal bar releases ions, right while in this speaker, these ions found within our solution are taken up by this metal bar, meaning there's a decrease in positive charge found on this side. So eventually, if we don't have anything connecting them like a salt bridge, the electrons will stop flowing. So in order for the electrons to continue to flow, the circuit, this circuit must be closed. And it's closed using this sold bridge. And what happens is this salt dissociates into k plus."}, {"title": "Electrochemical cell .txt", "text": "So in order for the electrons to continue to flow, the circuit, this circuit must be closed. And it's closed using this sold bridge. And what happens is this salt dissociates into k plus. And so for minus ions, and the positively charged ions begin to flow into the second half cell into the cathode. And this increases the positive charge found in this cathode in this half cell two. Now, the same happens with the so minus four."}, {"title": "The Periodic Table .txt", "text": "That means we need a very good way of organizing all these elements. And the periodic table does just that. What it does is it organizes our elements or atoms into columns and rows. Now the columns are known as groups or families, while the rows are called periods. So let's zoom in on our periodic table. So this is a general representation of our periodic table."}, {"title": "The Periodic Table .txt", "text": "Now the columns are known as groups or families, while the rows are called periods. So let's zoom in on our periodic table. So this is a general representation of our periodic table. I did not include the names of our atoms and I also did not include other elements usually found in two rows on the bottom. Now that's simply for simplification purposes. If you'd like to see the actual table, Google it or check out a chemistry textbook."}, {"title": "The Periodic Table .txt", "text": "I did not include the names of our atoms and I also did not include other elements usually found in two rows on the bottom. Now that's simply for simplification purposes. If you'd like to see the actual table, Google it or check out a chemistry textbook. Now these guys, these columns are known as groups. So group one, group two, group three, group four, all the way up to group 18, while these rows are known as periods. So period one, period two, period 3456, all the way up to period seven."}, {"title": "The Periodic Table .txt", "text": "Now these guys, these columns are known as groups. So group one, group two, group three, group four, all the way up to group 18, while these rows are known as periods. So period one, period two, period 3456, all the way up to period seven. Now these guys, or this table is divided into three main divisions known as metals, nonmetals and metalloids. Now the white squares or the white elements are known as metals. And they're found from left all the way up to this section here."}, {"title": "The Periodic Table .txt", "text": "Now these guys, or this table is divided into three main divisions known as metals, nonmetals and metalloids. Now the white squares or the white elements are known as metals. And they're found from left all the way up to this section here. Now the orange guys are known as metalloids, while the red guys are known as nonmetals. Now group 18 is called the Noble gas group, while group 17, the group right next to group 18, are known as halogens. Now these guys here from this group to group number three are called transition metals."}, {"title": "The Periodic Table .txt", "text": "Now the orange guys are known as metalloids, while the red guys are known as nonmetals. Now group 18 is called the Noble gas group, while group 17, the group right next to group 18, are known as halogens. Now these guys here from this group to group number three are called transition metals. Group number one are known as alkaline metals, and group number two are known as alkaline earth metals. Now we're going to go into more detail in our next lecture about what the alkaline, alkaline, earth, noble halogens and transition metals are. In this lecture, we're only going to look at the three divisions that exist, namely metals, metalloids and nonmetals."}, {"title": "The Periodic Table .txt", "text": "Group number one are known as alkaline metals, and group number two are known as alkaline earth metals. Now we're going to go into more detail in our next lecture about what the alkaline, alkaline, earth, noble halogens and transition metals are. In this lecture, we're only going to look at the three divisions that exist, namely metals, metalloids and nonmetals. So let's zoom out. Now. Let's examine our divisions."}, {"title": "The Periodic Table .txt", "text": "So let's zoom out. Now. Let's examine our divisions. Let's look at the metals. Metals are large atoms that tend to lose electrons with great ease, forming ions in which the oxidation state is positive. Now within a metal, electrons move with great ease from one point to another."}, {"title": "The Periodic Table .txt", "text": "Let's look at the metals. Metals are large atoms that tend to lose electrons with great ease, forming ions in which the oxidation state is positive. Now within a metal, electrons move with great ease from one point to another. And that means our metals are able to conduct electricity very well. So metals generally have high connectivity rates. Now metals are also malleable, which means you can hammer them into very thin strips."}, {"title": "The Periodic Table .txt", "text": "And that means our metals are able to conduct electricity very well. So metals generally have high connectivity rates. Now metals are also malleable, which means you can hammer them into very thin strips. Examples include wires. Now metals are also or have high ductility rates. In other words, they're stretchy or stretchable."}, {"title": "The Periodic Table .txt", "text": "Examples include wires. Now metals are also or have high ductility rates. In other words, they're stretchy or stretchable. Now metals, whenever they form compounds with oxygen, they form or bond non Covalently. They create ionic oxides. The one exception is Beryllium."}, {"title": "The Periodic Table .txt", "text": "Now metals, whenever they form compounds with oxygen, they form or bond non Covalently. They create ionic oxides. The one exception is Beryllium. Beryllium bonds with oxygen Covalently. And that's the only exception known. Now let's look at the second type of division, namely the nonmetals."}, {"title": "The Periodic Table .txt", "text": "Beryllium bonds with oxygen Covalently. And that's the only exception known. Now let's look at the second type of division, namely the nonmetals. Now nonmetals, which are found on the right side of the table. The red guys have very diverse chemical characteristics. And these guys form negative ions."}, {"title": "The Periodic Table .txt", "text": "Now nonmetals, which are found on the right side of the table. The red guys have very diverse chemical characteristics. And these guys form negative ions. They don't lose electrons very easily. In fact, they gain electrons more easily than metals and that's why they form negative oxidation states. Now, these guys, when they combine with oxygen, they form covalent oxides."}, {"title": "The Periodic Table .txt", "text": "They don't lose electrons very easily. In fact, they gain electrons more easily than metals and that's why they form negative oxidation states. Now, these guys, when they combine with oxygen, they form covalent oxides. Examples include carbon dioxide or carbon monoxide. Now, let's look at the third type division called the metalloids. metalloids are interesting in that they have characteristics of both metals and nonmetals."}, {"title": "Avogadro's Law .txt", "text": "And we saw that these laws both helped us explain exactly how gas molecules interact and function on the macroscopic level. Now we're going to look at a third law called Avocadoso's Law. And we're going to see how this law also helps us gain more intuition about the interactions of gas molecules on a macroscopic level. So let's look at the conditions under which this law holds. So this law will only work when our pressure and temperature are both held constant. And what this law tells us is that volume is directly proportional to the number of moles."}, {"title": "Avogadro's Law .txt", "text": "So let's look at the conditions under which this law holds. So this law will only work when our pressure and temperature are both held constant. And what this law tells us is that volume is directly proportional to the number of moles. And what that means. If these guys are held constant, the only way we can increase our volume of our system is to add more gas or to add more moles of gas. Now, if we multiply this side by some constant, this we set equal and our N comes to this side."}, {"title": "Avogadro's Law .txt", "text": "And what that means. If these guys are held constant, the only way we can increase our volume of our system is to add more gas or to add more moles of gas. Now, if we multiply this side by some constant, this we set equal and our N comes to this side. We get the following relation. Volume over N, our number of moles equals a constant. Now this constant, which we will see next when we learn about the ideal gas law, depends on temperature and pressure."}, {"title": "Avogadro's Law .txt", "text": "We get the following relation. Volume over N, our number of moles equals a constant. Now this constant, which we will see next when we learn about the ideal gas law, depends on temperature and pressure. In other words, if our pressure and temperature are the same, then our constant will always be the same. But if we increase or decrease our temperature or pressure, our constant will also change. So that brings up an interesting relation."}, {"title": "Avogadro's Law .txt", "text": "In other words, if our pressure and temperature are the same, then our constant will always be the same. But if we increase or decrease our temperature or pressure, our constant will also change. So that brings up an interesting relation. That basically means as long as our pressure and temperature are the same, this will always be true for any volume or for any number of moles. This will always be our constant. And we'll see why this is important in Part D. Let's look at C for a second."}, {"title": "Avogadro's Law .txt", "text": "That basically means as long as our pressure and temperature are the same, this will always be true for any volume or for any number of moles. This will always be our constant. And we'll see why this is important in Part D. Let's look at C for a second. It's important to mention that experimental results show that a zero degree Celsius or 273 Kelvin and 1 ATM and pressure 1 mol of any gas, any gas whatsoever, will always correspond to 22.4\nliters. And that's because according to our Kinetic Molecular Theory, volume of gas is zero. It's assumed to be zero."}, {"title": "Avogadro's Law .txt", "text": "It's important to mention that experimental results show that a zero degree Celsius or 273 Kelvin and 1 ATM and pressure 1 mol of any gas, any gas whatsoever, will always correspond to 22.4\nliters. And that's because according to our Kinetic Molecular Theory, volume of gas is zero. It's assumed to be zero. And so it doesn't matter what type of gas you use, if it's large or small, it will have a volume of 22.4 liters. Now, this is according to experimental results. Once again, something we observe experimentally, we turn into a theory."}, {"title": "Avogadro's Law .txt", "text": "And so it doesn't matter what type of gas you use, if it's large or small, it will have a volume of 22.4 liters. Now, this is according to experimental results. Once again, something we observe experimentally, we turn into a theory. And that's where our assumptions from our kinetic theory came. Let's look at Part D.\nSo earlier I said that for any given temperature and pressure, as long as they are held constant at that same temperature and pressure, our constant will always be the same. So suppose that's our case."}, {"title": "Avogadro's Law .txt", "text": "And that's where our assumptions from our kinetic theory came. Let's look at Part D.\nSo earlier I said that for any given temperature and pressure, as long as they are held constant at that same temperature and pressure, our constant will always be the same. So suppose that's our case. Suppose I have a system under which I have constant temperature and pressure. Now, suppose I have a balloon at some volume one and some volume two. And suppose I have three molecules or three moles on my gas inside my balloon."}, {"title": "Avogadro's Law .txt", "text": "Suppose I have a system under which I have constant temperature and pressure. Now, suppose I have a balloon at some volume one and some volume two. And suppose I have three molecules or three moles on my gas inside my balloon. What if I increase the number of moles to six moles? What will happen to my volume? Well, suppose I take a balloon and I put a liter of water into my balloon, it's going to fill up to a certain volume."}, {"title": "Avogadro's Law .txt", "text": "What if I increase the number of moles to six moles? What will happen to my volume? Well, suppose I take a balloon and I put a liter of water into my balloon, it's going to fill up to a certain volume. Suppose I put one more liter of water into my balloon. Well, it's going to take up twice as much volume because we're assuming, of course, that temperature and pressure is the same. So our Law of Evangels Law becomes the following this law holds for two sets of different conditions under which temperature and pressure is held the same."}, {"title": "Avogadro's Law .txt", "text": "Suppose I put one more liter of water into my balloon. Well, it's going to take up twice as much volume because we're assuming, of course, that temperature and pressure is the same. So our Law of Evangels Law becomes the following this law holds for two sets of different conditions under which temperature and pressure is held the same. So for one condition, for one volume in one mode, this guy equals the second condition, v two over n two. The same thing. The same results we saw in Charles Law and also in Boyle's Law."}, {"title": "Avogadro's Law .txt", "text": "So for one condition, for one volume in one mode, this guy equals the second condition, v two over n two. The same thing. The same results we saw in Charles Law and also in Boyle's Law. Except in Boyle's Law it was p times V equals p two times V two. Now this guy is equal to the constant, because remember, no matter what volume or number of mold we're talking about, as long as this is true, our constant will be the same. So both this guy and this guy equals the same constant."}, {"title": "Avogadro's Law .txt", "text": "Except in Boyle's Law it was p times V equals p two times V two. Now this guy is equal to the constant, because remember, no matter what volume or number of mold we're talking about, as long as this is true, our constant will be the same. So both this guy and this guy equals the same constant. Now, this formula can be applied in many different examples. Let's see one, an easy one in Part E. Suppose I'm given that for three moles, my volume is 22 liters. Now suppose my second volume in my second condition is 44 liters."}, {"title": "Avogadro's Law .txt", "text": "Now, this formula can be applied in many different examples. Let's see one, an easy one in Part E. Suppose I'm given that for three moles, my volume is 22 liters. Now suppose my second volume in my second condition is 44 liters. What is my mole? What? I basically plug in my values."}, {"title": "Avogadro's Law .txt", "text": "What is my mole? What? I basically plug in my values. 22 over three equals 44 over n two. I solve for n and I find six. That's exactly how I use Avogadro's Law to solve problems."}, {"title": "Avogadro's Law .txt", "text": "22 over three equals 44 over n two. I solve for n and I find six. That's exactly how I use Avogadro's Law to solve problems. So now let's explain Avogadro's Law, a macro scale concept using a microscale concept or the kinetic theory or the Kinetic molecular theory. Now, Kinetic theory explains of Agadro's Law in the following way. Now if temperature and pressure is to remain constant, an increase in the number of moles will increase volume."}, {"title": "Avogadro's Law .txt", "text": "So now let's explain Avogadro's Law, a macro scale concept using a microscale concept or the kinetic theory or the Kinetic molecular theory. Now, Kinetic theory explains of Agadro's Law in the following way. Now if temperature and pressure is to remain constant, an increase in the number of moles will increase volume. In other words, if our kinetic energy or average kinetic energy of our molecules is to remain the same, and the pressure or the force per unit area exerted by or on the walls of my container is to remain the same, that means when we increase the number of moles, we also increase the number of molecules hitting the walls. And that means the only way to keep these two guys constant is if we increase the volume. That's exactly how our kinetic theory explains Abogro's Law."}, {"title": "Avogadro's Law .txt", "text": "In other words, if our kinetic energy or average kinetic energy of our molecules is to remain the same, and the pressure or the force per unit area exerted by or on the walls of my container is to remain the same, that means when we increase the number of moles, we also increase the number of molecules hitting the walls. And that means the only way to keep these two guys constant is if we increase the volume. That's exactly how our kinetic theory explains Abogro's Law. Now, once again, to recap, kinetic theory explains microscopic concepts. It explains how two individual gas molecules interact. The fact that their volume is so small that it's assumed to be zero."}, {"title": "Avogadro's Law .txt", "text": "Now, once again, to recap, kinetic theory explains microscopic concepts. It explains how two individual gas molecules interact. The fact that their volume is so small that it's assumed to be zero. The fact that individual molecules travel at very high speeds, about 1000 these laws avoidros Charles and Boyle's Law all explain macroscopic concepts, things that you could see and feel and hear. For example, a balloon popping when you're putting pressure on it, or a balloon inflating when you're putting in more molds. Things like that are explained by these three laws."}, {"title": "Sp3 Hybridization.txt", "text": "Well, this is defined as simply the combination of four atomic orbitals in a given atom to produce four hybridized orbitals which then can interact with other atomic orbitals of other atoms to produce codalent bonds. So to show this, let's examine the following methane molecule. So, our goal will be to produce this methane molecule composed of one carbon and four H atoms. So the carbon has the following electron configuration. It has a total of six electrons. Two electrons go to the one S, two electrons go to the two S and two electrons still to two P. Now, each H atom has one electron each."}, {"title": "Sp3 Hybridization.txt", "text": "So the carbon has the following electron configuration. It has a total of six electrons. Two electrons go to the one S, two electrons go to the two S and two electrons still to two P. Now, each H atom has one electron each. And that means that electron goes into the one x orbital. So we have one balance electron per H atom and four balance electrons two plus two four for the carbon atom. Now, before the carbon can combine with the H atoms to form our methane, hybridization must take place."}, {"title": "Sp3 Hybridization.txt", "text": "And that means that electron goes into the one x orbital. So we have one balance electron per H atom and four balance electrons two plus two four for the carbon atom. Now, before the carbon can combine with the H atoms to form our methane, hybridization must take place. Remember, hybridization occurs because it increases the volume of the lobe interacting with the other atomic orbitals. And this increase in overlap will increase the strength of the bond. So hybridization takes place so that there is a better overlap between atomic orbitals and this stabilizes the bond."}, {"title": "Sp3 Hybridization.txt", "text": "Remember, hybridization occurs because it increases the volume of the lobe interacting with the other atomic orbitals. And this increase in overlap will increase the strength of the bond. So hybridization takes place so that there is a better overlap between atomic orbitals and this stabilizes the bond. So before hybridization took place, we had the following picture of our carbon atom. So, the carbon atom has four bounce electrons. Two bounce electrons are in the two S orbitals shown here."}, {"title": "Sp3 Hybridization.txt", "text": "So before hybridization took place, we had the following picture of our carbon atom. So, the carbon atom has four bounce electrons. Two bounce electrons are in the two S orbitals shown here. One bounce electron is in the two PX, one balanced electron is in the two PY and no electrons are in the two PZ. So how does hybridization take place? Well, first we must ask the following question how many hybrid orbitals should carbon develop so that it can create the methane molecule?"}, {"title": "Sp3 Hybridization.txt", "text": "One bounce electron is in the two PX, one balanced electron is in the two PY and no electrons are in the two PZ. So how does hybridization take place? Well, first we must ask the following question how many hybrid orbitals should carbon develop so that it can create the methane molecule? The answer lies in this picture. How many bonds are created between the carbon and the H? Well, since we have one carbon and four HS, there are four bonds."}, {"title": "Sp3 Hybridization.txt", "text": "The answer lies in this picture. How many bonds are created between the carbon and the H? Well, since we have one carbon and four HS, there are four bonds. So that means we need four hybrid orbitals. So that means we have to use the two x and all the three P, x's, y's and Z's to form our four hybrid orbitals. In other words, for hybridization to take place, the two S must combine the two PX that must combine the two PY and the two PZ."}, {"title": "Sp3 Hybridization.txt", "text": "So that means we need four hybrid orbitals. So that means we have to use the two x and all the three P, x's, y's and Z's to form our four hybrid orbitals. In other words, for hybridization to take place, the two S must combine the two PX that must combine the two PY and the two PZ. If we combine all these four atomic orbitals, we will get four hybrid orbitals that are all identical and look like this. So we get four SP, three hybridized orbitals in which we have 25% S character and 75% P character. So this guy undergoes hybridization, we get the following depiction."}, {"title": "Sp3 Hybridization.txt", "text": "If we combine all these four atomic orbitals, we will get four hybrid orbitals that are all identical and look like this. So we get four SP, three hybridized orbitals in which we have 25% S character and 75% P character. So this guy undergoes hybridization, we get the following depiction. So now our carbon atom no longer has that individual two S and these individual two PX two PY, two PZ. Instead, we have four identical XP, three hybridized orbitals. And so and since we have four balanced electrons, each bounced electron goes into each of the four identical SP three hybridized orbitals."}, {"title": "Sp3 Hybridization.txt", "text": "So now our carbon atom no longer has that individual two S and these individual two PX two PY, two PZ. Instead, we have four identical XP, three hybridized orbitals. And so and since we have four balanced electrons, each bounced electron goes into each of the four identical SP three hybridized orbitals. So one goes into here, one goes into here, one in here and one in here. Now, the carbon, which has undergone hybridization, is ready to interact with four other one s orbitals. So here we take the four one s orbitals."}, {"title": "Sp3 Hybridization.txt", "text": "So one goes into here, one goes into here, one in here and one in here. Now, the carbon, which has undergone hybridization, is ready to interact with four other one s orbitals. So here we take the four one s orbitals. We place each on to the positive green lobe, and we get our methane molecule. So a simpler way of looking at it is via this black diagram. So here we have our carbon nucleus."}, {"title": "Sp3 Hybridization.txt", "text": "We place each on to the positive green lobe, and we get our methane molecule. So a simpler way of looking at it is via this black diagram. So here we have our carbon nucleus. We have these SP three lobes, which each have an electron. They interact with a one s orbital. So H, with one electron, they bind or bonds."}, {"title": "Sp3 Hybridization.txt", "text": "We have these SP three lobes, which each have an electron. They interact with a one s orbital. So H, with one electron, they bind or bonds. And we create the following picture. Notice that these guys are identical. Now, for our methane molecule."}, {"title": "Sp3 Hybridization.txt", "text": "And we create the following picture. Notice that these guys are identical. Now, for our methane molecule. Experimentally, we know that the bond between a two ch and C H is 109 degrees. And this takes the form of a tetrahedron. So now let's look at the energy diagram."}, {"title": "Sp3 Hybridization.txt", "text": "Experimentally, we know that the bond between a two ch and C H is 109 degrees. And this takes the form of a tetrahedron. So now let's look at the energy diagram. So, let's say we want to combine one of these one SS with the SP three hypothetical orbital. So that one s will be slightly lower in energy than the SP three. The SP three will be slightly higher."}, {"title": "Sp3 Hybridization.txt", "text": "So, let's say we want to combine one of these one SS with the SP three hypothetical orbital. So that one s will be slightly lower in energy than the SP three. The SP three will be slightly higher. They will combine to form a bonding and an antibanding orbital or a molecular orbital. So here we have the bonding, and the electrons will go into this orbital. And here we have the antibinding."}, {"title": "Sp3 Hybridization.txt", "text": "They will combine to form a bonding and an antibanding orbital or a molecular orbital. So here we have the bonding, and the electrons will go into this orbital. And here we have the antibinding. Electrons will not want to go into this orbital. They will stay in this bonding orbital. And so this is exactly what happens in this picture."}, {"title": "Nernst Equation Part II .txt", "text": "So these guys cancel, this guy becomes negative. And now we have the denominator on this term here. This equation is called a nurse equation and it can be used, used to find a cell voltage under non standard state conditions whereas this guy is our cell voltage under nonsense state conditions. So we basically plug this guy in and all these guys in and we get our cell voltage for nonstand state conditions. Now, this guy can be simplified even further because notice we have a constant R, a constant F. And if we're given some constant temperature, say 25 degrees Celsius, the most common temperature, room temperature, we can plug these guys in and simplify this whole guy to simply this guy here. Look, we rewrite this formula, we get this."}, {"title": "Nernst Equation Part II .txt", "text": "So we basically plug this guy in and all these guys in and we get our cell voltage for nonstand state conditions. Now, this guy can be simplified even further because notice we have a constant R, a constant F. And if we're given some constant temperature, say 25 degrees Celsius, the most common temperature, room temperature, we can plug these guys in and simplify this whole guy to simply this guy here. Look, we rewrite this formula, we get this. We plug in our constant, our gas constant, our temperature in Kelvin and our Faraday's constant. We plug this into the calculator and we get this number to be zero point 25 seven. So this is a simplified version of this guy."}, {"title": "Nernst Equation Part II .txt", "text": "We plug in our constant, our gas constant, our temperature in Kelvin and our Faraday's constant. We plug this into the calculator and we get this number to be zero point 25 seven. So this is a simplified version of this guy. But we're not done. Notice we have natural logs. We never want to deal with natural logs."}, {"title": "Nernst Equation Part II .txt", "text": "But we're not done. Notice we have natural logs. We never want to deal with natural logs. We always want to convert natural logs into easier logs, say log base ten. So that means we have to use this mathematical formula that relates bases of logs. And what we basically have to realize is that natural log of Q is the same thing as this guy divided by this guy."}, {"title": "Nernst Equation Part II .txt", "text": "We always want to convert natural logs into easier logs, say log base ten. So that means we have to use this mathematical formula that relates bases of logs. And what we basically have to realize is that natural log of Q is the same thing as this guy divided by this guy. It's just a formula. You can look that up online. So we take this guy, we plug it into this Lnq and that means we are left with log base ten Q on the top."}, {"title": "Nernst Equation Part II .txt", "text": "It's just a formula. You can look that up online. So we take this guy, we plug it into this Lnq and that means we are left with log base ten Q on the top. And on the bottom we have this log of base ten of E. We plug this into the calculator and then we divide zero point 25 by this number to get our 0.0\n59 two. So once again, we take this guy, we plug it into this LMQ, we use a calculator to find this number and our final nurse equation at 25 degrees Celsius is this following equation. Now, if this was a different temperature, I'd go back to my equation in part E and I'd plug in a different temperature here and solve it the same way and get a new value here."}, {"title": "Nernst Equation Part II .txt", "text": "And on the bottom we have this log of base ten of E. We plug this into the calculator and then we divide zero point 25 by this number to get our 0.0\n59 two. So once again, we take this guy, we plug it into this LMQ, we use a calculator to find this number and our final nurse equation at 25 degrees Celsius is this following equation. Now, if this was a different temperature, I'd go back to my equation in part E and I'd plug in a different temperature here and solve it the same way and get a new value here. Now, last thing I want to mention is notice that if Q is equal to one, that means our log of one is zero. So what we guess is our cell voltage is simply cell voltage under state and state conditions. And that means this holds true as well."}, {"title": "Nernst Equation Part II .txt", "text": "Now, last thing I want to mention is notice that if Q is equal to one, that means our log of one is zero. So what we guess is our cell voltage is simply cell voltage under state and state conditions. And that means this holds true as well. Well, why is this the case? Well, if our Q is one, that means this Q is one. So our ratio of concentration of products or reactions is one."}, {"title": "Nernst Equation Part II .txt", "text": "Well, why is this the case? Well, if our Q is one, that means this Q is one. So our ratio of concentration of products or reactions is one. That means this guy, this guy and this guy. This guy has molarity of one. But that simply stands the condition, right?"}, {"title": "Nernst Equation Part II .txt", "text": "That means this guy, this guy and this guy. This guy has molarity of one. But that simply stands the condition, right? It's assumed that molarity is one under those conditions. That's exactly what this states. Now, also notice what this formula states."}, {"title": "Introduction to Gas State .txt", "text": "So gas molecules have generally different properties than that of liquid molecules or solid molecules. Now, in this lecture, I'm going to give a brief introduction to the gas state. So, gas molecules travel at very, very high velocities at approximately 1000 mph or 480 meters/second. Now, that means if you allow single gas molecule to travel from New York to California without being interrupted, it would take it about 3 hours to get there versus a car that would take days and days. And so, because of this high speed, they feel very little force from other gas molecules and that means very little intermolecular bonding. Remember, the reason water is held together or any other solid is due to intermolecular bonding between the molecules."}, {"title": "Introduction to Gas State .txt", "text": "Now, that means if you allow single gas molecule to travel from New York to California without being interrupted, it would take it about 3 hours to get there versus a car that would take days and days. And so, because of this high speed, they feel very little force from other gas molecules and that means very little intermolecular bonding. Remember, the reason water is held together or any other solid is due to intermolecular bonding between the molecules. Now, in gas molecules we don't have that because the molecules move at high speeds when they pass each other, they don't really feel too much force, so they don't bond. Let's look at the second thing. So gases are compressible and that's because they take up much less volume than the volume of the container they are in."}, {"title": "Introduction to Gas State .txt", "text": "Now, in gas molecules we don't have that because the molecules move at high speeds when they pass each other, they don't really feel too much force, so they don't bond. Let's look at the second thing. So gases are compressible and that's because they take up much less volume than the volume of the container they are in. For example, let's look at this big ball, right? So inside this ball we have air. And if you were to push this guy in, I could easily to some point push this ball in."}, {"title": "Introduction to Gas State .txt", "text": "For example, let's look at this big ball, right? So inside this ball we have air. And if you were to push this guy in, I could easily to some point push this ball in. Now, if this, if the inside was replaced with, say, salad or liquid, I wouldn't be able to push it without changing the volume. Notice how when I'm pushing this I'm not really affecting the volume too much, right? And that's why I'm able to push it, I'm compressing it."}, {"title": "Introduction to Gas State .txt", "text": "Now, if this, if the inside was replaced with, say, salad or liquid, I wouldn't be able to push it without changing the volume. Notice how when I'm pushing this I'm not really affecting the volume too much, right? And that's why I'm able to push it, I'm compressing it. And the reason for that is this following idea. Now, the volume of the inside of the ball is much greater than the volume that is taken up by these molecules. In other words, any two molecules at any given time are very far apart."}, {"title": "Introduction to Gas State .txt", "text": "And the reason for that is this following idea. Now, the volume of the inside of the ball is much greater than the volume that is taken up by these molecules. In other words, any two molecules at any given time are very far apart. So when I compress this ball, these molecules have lots of room to get closer, right? So this model gets closer, this one gets closer, this one gets closer and eventually they all get closer so I could push them in. Now, if this guy was replaced with liquid or solid, liquid and solid is much more dense and that means I would not be able to push it together because all the molecules are close together."}, {"title": "Introduction to Gas State .txt", "text": "So when I compress this ball, these molecules have lots of room to get closer, right? So this model gets closer, this one gets closer, this one gets closer and eventually they all get closer so I could push them in. Now, if this guy was replaced with liquid or solid, liquid and solid is much more dense and that means I would not be able to push it together because all the molecules are close together. And that leads straight to the third point. Gas molecules exert a force on whatever they hit. And this force can be calculated in terms of pressure."}, {"title": "Introduction to Gas State .txt", "text": "And that leads straight to the third point. Gas molecules exert a force on whatever they hit. And this force can be calculated in terms of pressure. So let's go back to this example, right? So when I take my ball and I push this ball, I can only push it to a certain extent at some point. I can't push it any further unless I exert a stronger force."}, {"title": "Introduction to Gas State .txt", "text": "So let's go back to this example, right? So when I take my ball and I push this ball, I can only push it to a certain extent at some point. I can't push it any further unless I exert a stronger force. Well, why is that? Well, the reason for that is because all these molecules are moving at tremendous velocities and every time they hit the container, hit the walls, they exert a force or a pressure. Pressure is simply force per some given area."}, {"title": "Introduction to Gas State .txt", "text": "Well, why is that? Well, the reason for that is because all these molecules are moving at tremendous velocities and every time they hit the container, hit the walls, they exert a force or a pressure. Pressure is simply force per some given area. And that means the added effect of all these molecules hitting the wall will be tremendous. And that's exactly why I can't push this ball any further than, say this, right? So whenever we talk about force of molecules, force of gas molecules, we really want to talk about the pressure."}, {"title": "Introduction to Gas State .txt", "text": "And that means the added effect of all these molecules hitting the wall will be tremendous. And that's exactly why I can't push this ball any further than, say this, right? So whenever we talk about force of molecules, force of gas molecules, we really want to talk about the pressure. And so when we talk about things like vapor pressure of a gas, vapor pressure is exactly this. It's the force that these molecules exert on the walls of the container at equilibrium that's vapor pressure. So let's jump into number or part D.\nSo gas molecules expand into container."}, {"title": "Introduction to Gas State .txt", "text": "And so when we talk about things like vapor pressure of a gas, vapor pressure is exactly this. It's the force that these molecules exert on the walls of the container at equilibrium that's vapor pressure. So let's jump into number or part D.\nSo gas molecules expand into container. And that's because whenever I have this container and does this container out of air, if I open this up, what will happen to gas molecules? Well, remember, gas molecules are moving at high speeds. So if I open something up, they will escape."}, {"title": "Introduction to Gas State .txt", "text": "And that's because whenever I have this container and does this container out of air, if I open this up, what will happen to gas molecules? Well, remember, gas molecules are moving at high speeds. So if I open something up, they will escape. And also, because they're moving at high speeds, they don't feel intermocular attractions, so there's nothing stopping them from exiting into the container, so they can completely expand the container. And that means, on a similar note, different gases will mix completely because there is no inter molecular interactions between any two gas molecules. Now, compare this to liquids and solids, right?"}, {"title": "Introduction to Gas State .txt", "text": "And also, because they're moving at high speeds, they don't feel intermocular attractions, so there's nothing stopping them from exiting into the container, so they can completely expand the container. And that means, on a similar note, different gases will mix completely because there is no inter molecular interactions between any two gas molecules. Now, compare this to liquids and solids, right? If I take the same container of liquid and I spill it into the room, what will happen? Well, they won't escape. They will just create a small puddle and they will stay there, right?"}, {"title": "Introduction to Gas State .txt", "text": "If I take the same container of liquid and I spill it into the room, what will happen? Well, they won't escape. They will just create a small puddle and they will stay there, right? That's because they do experience intermolecular attractions and they don't have high velocities or high translational speeds. And that means these intermolecular attractions and low velocities will keep the molecules together. Now, of course, some of these guys will have enough kinetic energy to escape this environment, but that's a different idea."}, {"title": "Introduction to Gas State .txt", "text": "That's because they do experience intermolecular attractions and they don't have high velocities or high translational speeds. And that means these intermolecular attractions and low velocities will keep the molecules together. Now, of course, some of these guys will have enough kinetic energy to escape this environment, but that's a different idea. That's called evaporation. And when you evaporate from a liquid to a gas, the gas gains that high velocity escapes. And now it no longer experiences any intermolecular attractions."}, {"title": "Introduction to Gas State .txt", "text": "That's called evaporation. And when you evaporate from a liquid to a gas, the gas gains that high velocity escapes. And now it no longer experiences any intermolecular attractions. And the final thing we want to look at is the following. Now, we normally describe gas as using a few things temperature. And later we'll see how temperature and kinetic energy speed of our molecules are related."}, {"title": "Osmotic Pressure Example .txt", "text": "So before we begin, I want to illustrate exactly, exactly what's meant by asthmaic pressure. So let's look at this system. Suppose we have two sides, one side and a second side. And the two sides are blocked off by a membrane. This membrane allows one to pass through, but does not allow our solid molecules to pass through. Now this side simply contains 20 solvent."}, {"title": "Osmotic Pressure Example .txt", "text": "And the two sides are blocked off by a membrane. This membrane allows one to pass through, but does not allow our solid molecules to pass through. Now this side simply contains 20 solvent. This side contains a solution of 20 ML that contains 3 grams of our polymer. So the red dots are our polymer dots. Now what will happen?"}, {"title": "Osmotic Pressure Example .txt", "text": "This side contains a solution of 20 ML that contains 3 grams of our polymer. So the red dots are our polymer dots. Now what will happen? Well, since the concentration of our solute is greater in this side, then water will tend to move from this side to this side, right? So osmotic pressure basically says well, I need to apply this much pressure on this side of the membrane to stop the movement of that water, to stop osmosis from occurring. So that's what is meant by asthmatic pressure."}, {"title": "Osmotic Pressure Example .txt", "text": "Well, since the concentration of our solute is greater in this side, then water will tend to move from this side to this side, right? So osmotic pressure basically says well, I need to apply this much pressure on this side of the membrane to stop the movement of that water, to stop osmosis from occurring. So that's what is meant by asthmatic pressure. So now we basically want to use all the given info and use our formula for asthmatic pressure and find our molar mass of polymer. Now, first step is let's write down our formula and let's see which ones we have and which ones we don't. So I we have I just simply won."}, {"title": "Osmotic Pressure Example .txt", "text": "So now we basically want to use all the given info and use our formula for asthmatic pressure and find our molar mass of polymer. Now, first step is let's write down our formula and let's see which ones we have and which ones we don't. So I we have I just simply won. And that's because our polymer doesn't associate into anything. So I is one. Our key in Kelvin is 25 plus 273 gives us 298 Kelvin."}, {"title": "Osmotic Pressure Example .txt", "text": "And that's because our polymer doesn't associate into anything. So I is one. Our key in Kelvin is 25 plus 273 gives us 298 Kelvin. Our R is a constant. So it's zero point 81 litres times ATM over moles times Kelvin. And our molarity, well, we don't have our molarity, but we do have our asthma pressure."}, {"title": "Osmotic Pressure Example .txt", "text": "Our R is a constant. So it's zero point 81 litres times ATM over moles times Kelvin. And our molarity, well, we don't have our molarity, but we do have our asthma pressure. So this is what we want to find. And we want to find this guy and use that to find our grams per mole or molar mass. Because remember, molar mass is the amount of mass per 1 mol."}, {"title": "Osmotic Pressure Example .txt", "text": "So this is what we want to find. And we want to find this guy and use that to find our grams per mole or molar mass. Because remember, molar mass is the amount of mass per 1 mol. So let's plug these guys in into this equation and we got this by rearranging this, by bringing all these guys on this side. So we got that. So we plug in our atmosphere pressure zero point 13 ATMs over our constant times, our temperature times I or one."}, {"title": "Osmotic Pressure Example .txt", "text": "So let's plug these guys in into this equation and we got this by rearranging this, by bringing all these guys on this side. So we got that. So we plug in our atmosphere pressure zero point 13 ATMs over our constant times, our temperature times I or one. So ATMs cancel, the Kelvins cancel and we get 5.31 times ten to negative four moles per liter. So this is our molarity. Now let's find the number of moles of polymer."}, {"title": "Osmotic Pressure Example .txt", "text": "So ATMs cancel, the Kelvins cancel and we get 5.31 times ten to negative four moles per liter. So this is our molarity. Now let's find the number of moles of polymer. If we find the number of moles of polymer, we could take our amount in grams, divide that by our moles and we get mol and mass or mass per mole. So 5.31 times ten to negative four molar which we got from here, times the amount of solution we have, which is 20 ML or 0.2 liters because we have to convert milliliters to liters. We divide this by thousand and we get to zero two the liters cancel and we get 1.6 times ten to negative four moles of polymer."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "Let's remember what a non ideal fluid is. A non ideal fluid is a fluid in which the volume of molecules is not neglected and intermolecular forces connecting the molecules exist. And these forces will play a role in decreasing the vapor repression of the fluid. And we'll see that in a second. Now, before we talk about that, let's talk about the heat of solution. Now, we have already spoken about heat of solution in another video, so let's briefly discuss it."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "And we'll see that in a second. Now, before we talk about that, let's talk about the heat of solution. Now, we have already spoken about heat of solution in another video, so let's briefly discuss it. So the heat of solution is the sum of energy required to break the intermolecular forces of compound one. Plus the sum of the energy is required to break the intermolecular forces of compound two plus the energy release when you form the new solution. And that gives you the enthalpy of solution or heat of solution."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "So the heat of solution is the sum of energy required to break the intermolecular forces of compound one. Plus the sum of the energy is required to break the intermolecular forces of compound two plus the energy release when you form the new solution. And that gives you the enthalpy of solution or heat of solution. Now, when this guy is negative, when he's exothermic the bonds form are stronger than the bonds broken, so they're more stable. Now, when it's endothermic, when it's positive, the bonds form are weaker and less stable than the bonds that were broken. Now let's compare ideal and non ideal situations."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "Now, when this guy is negative, when he's exothermic the bonds form are stronger than the bonds broken, so they're more stable. Now, when it's endothermic, when it's positive, the bonds form are weaker and less stable than the bonds that were broken. Now let's compare ideal and non ideal situations. Now, when we talk about ideal situations, we can graph Roth's law. And the Y axis is vapor pressure and the x axis is mole fraction. Now, we have two compounds here, compound A and compound B."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "Now, when we talk about ideal situations, we can graph Roth's law. And the Y axis is vapor pressure and the x axis is mole fraction. Now, we have two compounds here, compound A and compound B. Now for compound A when we go from here to here along the x axis the mole fraction increases. For compound B, when we go from here to here on the x axis the mole fraction decreases. The brown line represents the pressure."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "Now for compound A when we go from here to here along the x axis the mole fraction increases. For compound B, when we go from here to here on the x axis the mole fraction decreases. The brown line represents the pressure. It's the slope for compound A and the green line represents the pressure of the slope for compound B. Now, as we go as we increase the mole concentration in compound one or compound A, the pressure of it increases. And as we go this way, the mole fraction of this guy compound B decreases."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "It's the slope for compound A and the green line represents the pressure of the slope for compound B. Now, as we go as we increase the mole concentration in compound one or compound A, the pressure of it increases. And as we go this way, the mole fraction of this guy compound B decreases. So as we go this way, the vapor pressure of compound B decreases. Now, at any given time, we could find a final vapor pressure and we find the final vapor pressure by summing the two pressures. So for example, say at this point the total pressure is this guy plus this guy gives you this plus this."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "So as we go this way, the vapor pressure of compound B decreases. Now, at any given time, we could find a final vapor pressure and we find the final vapor pressure by summing the two pressures. So for example, say at this point the total pressure is this guy plus this guy gives you this plus this. So it's somewhere over here and that's what the red line is. Now this is for an ideal fluid. Now, in an ideal fluid the surface molecules are not connected by any intermolecular forces because remember, in an ideal fluid we're neglecting those intermolecular forces."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "So it's somewhere over here and that's what the red line is. Now this is for an ideal fluid. Now, in an ideal fluid the surface molecules are not connected by any intermolecular forces because remember, in an ideal fluid we're neglecting those intermolecular forces. So nothing holds the molecules together so they could freely escape into the environment if they have the kinetic energy. However, in a non ideal fluid there are intermolecular forces and these forces hold the molecules together. They inhibit the molecules from escaping into the environment."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "So nothing holds the molecules together so they could freely escape into the environment if they have the kinetic energy. However, in a non ideal fluid there are intermolecular forces and these forces hold the molecules together. They inhibit the molecules from escaping into the environment. So at any given time, less molecules will be able to escape in a non ideal fluid than in an ideal fluid. And because there are less molecules present in the states above the pressure will be less. Therefore, the stronger the intermolecular forces, the less likely that they will evaporate."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "So at any given time, less molecules will be able to escape in a non ideal fluid than in an ideal fluid. And because there are less molecules present in the states above the pressure will be less. Therefore, the stronger the intermolecular forces, the less likely that they will evaporate. Likewise, the weaker the intermolecular forces holding the molecules together, the more likely that they will evaporate. Now, let's look at some graphs for non ideal fluids. Now in this graph we're going to compare two situations."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "Likewise, the weaker the intermolecular forces holding the molecules together, the more likely that they will evaporate. Now, let's look at some graphs for non ideal fluids. Now in this graph we're going to compare two situations. We're going to compare positive or endothermic cube solution and negative or exothermicial solution. We're also going to compare non ideal situations we're presented to by dashed lines and ideal situations they're presented by solid lines. Now, in this graph, the y axis is vapor pressure and the x axis is the mole fraction."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "We're going to compare positive or endothermic cube solution and negative or exothermicial solution. We're also going to compare non ideal situations we're presented to by dashed lines and ideal situations they're presented by solid lines. Now, in this graph, the y axis is vapor pressure and the x axis is the mole fraction. We're dealing with two compounds, compounds A, compound B. Compound B is the green line. Compound A is the brown line."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "We're dealing with two compounds, compounds A, compound B. Compound B is the green line. Compound A is the brown line. Now, in an endothermic reaction the bonds form are weaker than the bonds broken. And that means the bonds formed are going to be less likely to hold the molecules on the surface. That means more molecules will evaporate."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "Now, in an endothermic reaction the bonds form are weaker than the bonds broken. And that means the bonds formed are going to be less likely to hold the molecules on the surface. That means more molecules will evaporate. And if more molecules evaporate, more molecules will be present in the space above, more gas molecules. And if more gas molecules are present in the space above, the vapor pressure is higher. Therefore, compared to ideal situations for each case in a non ideal situation that's endothermic the pressure for each will be higher and that's why they curve upward."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "And if more molecules evaporate, more molecules will be present in the space above, more gas molecules. And if more gas molecules are present in the space above, the vapor pressure is higher. Therefore, compared to ideal situations for each case in a non ideal situation that's endothermic the pressure for each will be higher and that's why they curve upward. Now, in an exothermic, the opposite holds. In an exothermic reaction, the bonds formed are stronger than the bonds broken. Therefore, the final solution will hold its molecules very tightly and will not let them go."}, {"title": "Raoul\u2019s Law for Non-Ideal Fluids .txt", "text": "Now, in an exothermic, the opposite holds. In an exothermic reaction, the bonds formed are stronger than the bonds broken. Therefore, the final solution will hold its molecules very tightly and will not let them go. They will not be able to escape, will be less likely to go into the gas state. And that's why less gas molecules will be present in the space above and the deeper pressure will be lower. And that's why in each situation they're curved downward."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "Firstly, whenever we talk about gives free energy, we talk about closed systems or reactions occurring under closed systems. So what is a closed system? A closed system simply means that no mass is exchanged, only energy is exchanged. And since mass is matter, and matter are molecules, that means no molecules go into the system and no molecules lead the system. So the number of moles remains constant. Secondly, reactions are under constant temperature and pressure."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "And since mass is matter, and matter are molecules, that means no molecules go into the system and no molecules lead the system. So the number of moles remains constant. Secondly, reactions are under constant temperature and pressure. If this wasn't true, the equation would break down. It would not work. So according to the ideal gas law, when number of moles is held constant, temperature is held constant, and pressure is held constant, then volume must remain constant."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "If this wasn't true, the equation would break down. It would not work. So according to the ideal gas law, when number of moles is held constant, temperature is held constant, and pressure is held constant, then volume must remain constant. If volume is constant, then change in volume is zero, because the final and the initial are the same. So the TV work done is zero. And that means that we could approximate change in enthalpy."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "If volume is constant, then change in volume is zero, because the final and the initial are the same. So the TV work done is zero. And that means that we could approximate change in enthalpy. And simply change in enthalpy of the system is equal to change in internal energy or simply change in energy. The PV term disappears because it's zero. Now thirdly, reactions are reversible, which means that if this wasn't true, the equation would break down as well."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "And simply change in enthalpy of the system is equal to change in internal energy or simply change in energy. The PV term disappears because it's zero. Now thirdly, reactions are reversible, which means that if this wasn't true, the equation would break down as well. Now, since reactions are reversible and enthalpy is a state function, hess's law tells us the change in enthalpy of the four reaction is equal to the negative change in enthalpy of the reverse reaction. So if we look at this reversible reaction here, a plus B is equal to C plus D, where the change in enthalpy is negative ten. And the reverse reaction, according to Hess's Law, is C plus D equal A plus B, and this is negative of this."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "Now, since reactions are reversible and enthalpy is a state function, hess's law tells us the change in enthalpy of the four reaction is equal to the negative change in enthalpy of the reverse reaction. So if we look at this reversible reaction here, a plus B is equal to C plus D, where the change in enthalpy is negative ten. And the reverse reaction, according to Hess's Law, is C plus D equal A plus B, and this is negative of this. So it's positive ten kilojoules per mole. The last thing we should mention is the mathematical formula for entropy, which basically states that change in entropy is the heat or change in energy over time, over temperature, and temperature is in Kelvin. Step one."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "So it's positive ten kilojoules per mole. The last thing we should mention is the mathematical formula for entropy, which basically states that change in entropy is the heat or change in energy over time, over temperature, and temperature is in Kelvin. Step one. In step one, I apply the formula for a change in entropy. That is, change in entropy surroundings is equal to change in energy over temperature, as seen here. Step two remember when we talk about free energy, or gifts free energy, we talk about a closed system, constant pressure and constant temperature."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "In step one, I apply the formula for a change in entropy. That is, change in entropy surroundings is equal to change in energy over temperature, as seen here. Step two remember when we talk about free energy, or gifts free energy, we talk about a closed system, constant pressure and constant temperature. And what this basically means, according to the ideal gas law, is that volume is constant. So change in volume is zero. So the PV work done is zero."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "And what this basically means, according to the ideal gas law, is that volume is constant. So change in volume is zero. So the PV work done is zero. So when we look at the equation for change in enthalpy, the PV work term disappears. And that means that our equation becomes change in enthalpy. The surroundings is equal to change in energy or heat."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "So when we look at the equation for change in enthalpy, the PV work term disappears. And that means that our equation becomes change in enthalpy. The surroundings is equal to change in energy or heat. So this becomes this. Now, step three remember when we talk about Gibbs free energy, we talk about a reversible reaction. And because enthalpy is a state function, according to Hess's law, hess's law states that the entropy change of the forward reaction is equal to the negative entropy change of the reverse reaction."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "So this becomes this. Now, step three remember when we talk about Gibbs free energy, we talk about a reversible reaction. And because enthalpy is a state function, according to Hess's law, hess's law states that the entropy change of the forward reaction is equal to the negative entropy change of the reverse reaction. So the amount of energy that leaves the system is the same amount of energy, except negative that into the system. Okay? So the magnitude remains the same, but the signs change."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "So the amount of energy that leaves the system is the same amount of energy, except negative that into the system. Okay? So the magnitude remains the same, but the signs change. Therefore, the change in entropy of the surroundings is equal to negative change in entropy of the system. Step four. Remember, change in entropy of the universe is equal to change in entropy the surroundings plus change in entropy of the system."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "Therefore, the change in entropy of the surroundings is equal to negative change in entropy of the system. Step four. Remember, change in entropy of the universe is equal to change in entropy the surroundings plus change in entropy of the system. Now, this guy equals this guy. Our next step will be to see that this guy can be replaced by this guy, okay? And that's exactly what we do in step five."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "Now, this guy equals this guy. Our next step will be to see that this guy can be replaced by this guy, okay? And that's exactly what we do in step five. We simply take this and plug it into here. And that's what we get. Here."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "We simply take this and plug it into here. And that's what we get. Here. The next step we multiply out by negative t.\nAnd that's because we want to get rid of this t, okay? So let's do that. Negative t times this gives you this negative t times this gives you that negative t times a negative whatever gives us a positive."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "The next step we multiply out by negative t.\nAnd that's because we want to get rid of this t, okay? So let's do that. Negative t times this gives you this negative t times this gives you that negative t times a negative whatever gives us a positive. So plus this whole term, step seven, is basically the TS cancel out. So we get this whole thing minus these t's. Next step, what I do is I switch these guys to make it look more like the actual equation."}, {"title": "Gibbs Free Energy Derivation .txt", "text": "So plus this whole term, step seven, is basically the TS cancel out. So we get this whole thing minus these t's. Next step, what I do is I switch these guys to make it look more like the actual equation. And now I equate this to changing Gibbs energy. And now I get our final equation change. In Gibbs, the energy is equal to change in entropy of the system minus temperature times change in entropy of the system."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "In this lecture we're going to look at and compare two more important groups found in our periodic table. We're going to look at group 14 or four A and group 15 and five A elements. Now let's begin with the group 14 or four A elements. Now this group consists of five elements or at least five elements. And the important elements are are listed below. We have carbon, which is a nonmetal."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "Now this group consists of five elements or at least five elements. And the important elements are are listed below. We have carbon, which is a nonmetal. We have silicon, which is a metalloid. We have germanium, which is also a metalloid. And we have two metals, tin and lead."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "We have silicon, which is a metalloid. We have germanium, which is also a metalloid. And we have two metals, tin and lead. Now notice that in our group we have at least one of each. We have at least one nonmetal, at least one metalloid and at least one metal. Now every single element within this group can form four Covalent bonds with other nonmetals."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "Now notice that in our group we have at least one of each. We have at least one nonmetal, at least one metalloid and at least one metal. Now every single element within this group can form four Covalent bonds with other nonmetals. And every atom except the sea atom, except our carbon can form two more additional bonds with lowest bases. Now remember what a lewis bases. It's simply an atom or a molecule that has an extra lone pair of electrons that it can donate to some other lewis acid."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "And every atom except the sea atom, except our carbon can form two more additional bonds with lowest bases. Now remember what a lewis bases. It's simply an atom or a molecule that has an extra lone pair of electrons that it can donate to some other lewis acid. Now let's look at D. Now, only carbon is capable of forming strong pi bonds. In other words, it can form a strong double bond or a strong triple bond. No other atom found in our group 14 or four A has that capability."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "Now let's look at D. Now, only carbon is capable of forming strong pi bonds. In other words, it can form a strong double bond or a strong triple bond. No other atom found in our group 14 or four A has that capability. Now let's jump to group 15. In group 15 or five A, we have also at least five atoms. These are the important atoms listed."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "Now let's jump to group 15. In group 15 or five A, we have also at least five atoms. These are the important atoms listed. So we have nitrogen and phosphorus, which is a nonmetal or both nonmetals. We have arsenic and antimony, which are both metalloids, and bismuth, which is a metal. Now once again, just like group 14 or four A, we have at least one of each in our group 15 or five day."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "So we have nitrogen and phosphorus, which is a nonmetal or both nonmetals. We have arsenic and antimony, which are both metalloids, and bismuth, which is a metal. Now once again, just like group 14 or four A, we have at least one of each in our group 15 or five day. Now, unlike this group which forms four Covalent bonds with other nonmetals, group 15 or five day forms three Covalent bonds and in addition, every single atom, except that nitrogen can form two more bonds using their higher D orbitals. Now we're going to talk more about D orbitals and P orbitals and S orbitals in another lecture. Let's look at C. Now nitrogen, this atom can form pi bonds just like carbon can form pi bonds in group 14."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "Now, unlike this group which forms four Covalent bonds with other nonmetals, group 15 or five day forms three Covalent bonds and in addition, every single atom, except that nitrogen can form two more bonds using their higher D orbitals. Now we're going to talk more about D orbitals and P orbitals and S orbitals in another lecture. Let's look at C. Now nitrogen, this atom can form pi bonds just like carbon can form pi bonds in group 14. And these pi bonds can be triple bonds or double bonds. And they're relatively strong. But unlike this group, we have one more atom here, our phosphorus, that can form a double bond, but it's a weak double bond."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "And these pi bonds can be triple bonds or double bonds. And they're relatively strong. But unlike this group, we have one more atom here, our phosphorus, that can form a double bond, but it's a weak double bond. But regardless, it's still a double bond, a pi bond. Now let's look at one more thing. Let's look at our N. Now normally when we have nitrogen in a molecule or compound that nitrogen has a lone pair of electrons."}, {"title": "Carbon and Nitrogen Groups .txt", "text": "But regardless, it's still a double bond, a pi bond. Now let's look at one more thing. Let's look at our N. Now normally when we have nitrogen in a molecule or compound that nitrogen has a lone pair of electrons. For example, in ammonia. And ammonia can take one more H because it has a lone pair of electrons. In other words, our nitrogen has the capability of forming not three Covalent bonds, but four Covalent bonds."}, {"title": "Henry\u2019s Law Example .txt", "text": "Now, as an example, let's take a section of our capillary found within our lung. And let's assume this section contains exactly one liter of blood or one liter of the red molecules. Now, a capillary is simply a very, very small vessel that carries blood. And within a capillary, oxygen and nitrogen can be exchanged with the environment. And we're using lungs because gases are exchanged within our lungs. So we want to find a number of blue molecules and a number of green molecules in terms of grounds found within one liter of this blood."}, {"title": "Henry\u2019s Law Example .txt", "text": "And within a capillary, oxygen and nitrogen can be exchanged with the environment. And we're using lungs because gases are exchanged within our lungs. So we want to find a number of blue molecules and a number of green molecules in terms of grounds found within one liter of this blood. So the first step is to realize that air is composed approximately 79% nitrogen and approximately 21% oxygen. Now, this translates to zero 79 mole fraction nitrogen and 0.2 1 mol fraction of oxygen. So the first step is to use rolled slow."}, {"title": "Henry\u2019s Law Example .txt", "text": "So the first step is to realize that air is composed approximately 79% nitrogen and approximately 21% oxygen. Now, this translates to zero 79 mole fraction nitrogen and 0.2 1 mol fraction of oxygen. So the first step is to use rolled slow. But before we use a roll slow, we must assume that there is dynamic equilibriums between the gas molecules in the space above and the gas molecules dissolved within our blood. So let's make that assumption. Now, we can find the partial pressures of each gas by simply using the formula."}, {"title": "Henry\u2019s Law Example .txt", "text": "But before we use a roll slow, we must assume that there is dynamic equilibriums between the gas molecules in the space above and the gas molecules dissolved within our blood. So let's make that assumption. Now, we can find the partial pressures of each gas by simply using the formula. Now, before we use the formula, let's let's realize why we're using 760 mmhg. So the total pressure above our system is the air that we're breathing in is the pressure of the air that we're breathing in. And the air we're breathing in is at atmospheric pressure."}, {"title": "Henry\u2019s Law Example .txt", "text": "Now, before we use the formula, let's let's realize why we're using 760 mmhg. So the total pressure above our system is the air that we're breathing in is the pressure of the air that we're breathing in. And the air we're breathing in is at atmospheric pressure. And atmospheric pressure is one ATM or 760 mercury. So defined partial pressure of oxygen, we simply multiply mole fraction by our total pressure and we get 159.6 mmhg for oxygen and 600.4\nmmhg for nitrogen. So second step is to find the Molarity."}, {"title": "Henry\u2019s Law Example .txt", "text": "And atmospheric pressure is one ATM or 760 mercury. So defined partial pressure of oxygen, we simply multiply mole fraction by our total pressure and we get 159.6 mmhg for oxygen and 600.4\nmmhg for nitrogen. So second step is to find the Molarity. We can use these guys, the partial pressures to find the Molarity. Now, Henry's loss states. So we can find the Molarity by simply taking our constant and multiplying it by the partial pressure of the gas."}, {"title": "Henry\u2019s Law Example .txt", "text": "We can use these guys, the partial pressures to find the Molarity. Now, Henry's loss states. So we can find the Molarity by simply taking our constant and multiplying it by the partial pressure of the gas. Therefore, we get 1.66\ntimes ten to negative six times 1.59.6 and we get 2.64 times ten to negative four molar for two or moles per liter. Now, we do the same thing, except we change the partial pressure for nitrogen and the concept of nitrogen and we get 5.55 times ten to negative four molar or moles per liter. The third step is to find the moles of oxygen and the moles of nitrogen dissolved within our blood."}, {"title": "Henry\u2019s Law Example .txt", "text": "Therefore, we get 1.66\ntimes ten to negative six times 1.59.6 and we get 2.64 times ten to negative four molar for two or moles per liter. Now, we do the same thing, except we change the partial pressure for nitrogen and the concept of nitrogen and we get 5.55 times ten to negative four molar or moles per liter. The third step is to find the moles of oxygen and the moles of nitrogen dissolved within our blood. To find the moles of oxygen, we simply take our Molar concentration, multiply that by one liter of total solution and we get 0.000,264 moles of O two and 0.0055 moles for M two. Finally, we can use the moles, multiplying that by the molecular weight of each respective compound. That will give us the number in grams of each compound resolved within our system."}, {"title": "Henry\u2019s Law Example .txt", "text": "To find the moles of oxygen, we simply take our Molar concentration, multiply that by one liter of total solution and we get 0.000,264 moles of O two and 0.0055 moles for M two. Finally, we can use the moles, multiplying that by the molecular weight of each respective compound. That will give us the number in grams of each compound resolved within our system. So molecular weight of oxygen, which is 32 grams/mol multiplied by the moles, which is 0.00,134\nmoles, gets us oh, I'm sorry. I made a mistake here. This should be 0.00,264\nmoles."}, {"title": "Henry\u2019s Law Example .txt", "text": "So molecular weight of oxygen, which is 32 grams/mol multiplied by the moles, which is 0.00,134\nmoles, gets us oh, I'm sorry. I made a mistake here. This should be 0.00,264\nmoles. So it's 32 centrist is right. 32 times .0,006,264. And that is correct."}, {"title": "Mole Fraction Example .txt", "text": "Molality is simply another way of measuring the concentration of a solution. The symbol for molality is lowercase letter M and the formula is moles of solute over kilograms of solvent. Now let's do an example. Using molality. The question tells us that we have 90 grams of glucose and 200 grams of H 20. And we want to find the molality of the solution."}, {"title": "Mole Fraction Example .txt", "text": "Using molality. The question tells us that we have 90 grams of glucose and 200 grams of H 20. And we want to find the molality of the solution. We're basically taking glucose, the salute and water, the solvent. We're mixing it and we want to find the concentration. In terms of molality, the first step is to find the molecular weight of glucose."}, {"title": "Mole Fraction Example .txt", "text": "We're basically taking glucose, the salute and water, the solvent. We're mixing it and we want to find the concentration. In terms of molality, the first step is to find the molecular weight of glucose. This will help us find the number of moles of glucose within our solution. The first step is to find the atomic weight of each atom and multiplied by each subscript. So the atomic weight of carbon is 12 grams/mol, the atomic weight of H is 1 gram/mol, and the atomic weight of oxygen is 16 grams/mol."}, {"title": "Mole Fraction Example .txt", "text": "This will help us find the number of moles of glucose within our solution. The first step is to find the atomic weight of each atom and multiplied by each subscript. So the atomic weight of carbon is 12 grams/mol, the atomic weight of H is 1 gram/mol, and the atomic weight of oxygen is 16 grams/mol. So we multiply six by twelve, add that to twelve times one and add that to six times 16 and we get approximately 180 grams/mol. Now, using this number, we can take our 90 grams of glucose and find the number of moles. 90 grams of glucose divided by 180 grams of glucose per mole gets us 0.5\nor one half moles of glucose."}, {"title": "Mole Fraction Example .txt", "text": "So we multiply six by twelve, add that to twelve times one and add that to six times 16 and we get approximately 180 grams/mol. Now, using this number, we can take our 90 grams of glucose and find the number of moles. 90 grams of glucose divided by 180 grams of glucose per mole gets us 0.5\nor one half moles of glucose. That's our number of moles within this solution. The final step is to use the formula for molality. Lowercase M for molality is equal to one half number of moles of salute divided by now notice we're getting 200 grams and molality ditosomality are moles per kilogram."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "So in this lecture we're going to look at something called the Hamilton Hasselbach equation. Now this equation has two uses. First, we can use it to find the PH of the buffet system. And second, we can use it to find the ratio of conjugate base to conjugate acid or the ratio of conjugate acid to conjugate base of our buffet system. Now let's see where this equation comes from. Let's derive it."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "And second, we can use it to find the ratio of conjugate base to conjugate acid or the ratio of conjugate acid to conjugate base of our buffet system. Now let's see where this equation comes from. Let's derive it. But first, let's look at the reaction of an acid in an aqueous state with the water molecule in the liquid state. Well, what we get is a conjugate base in the acreage state and a conjugate acid in the aqueous states. So let's write the equilibrium constant expression for this reaction."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "But first, let's look at the reaction of an acid in an aqueous state with the water molecule in the liquid state. Well, what we get is a conjugate base in the acreage state and a conjugate acid in the aqueous states. So let's write the equilibrium constant expression for this reaction. So we get the acid ionization constant for this acid is equal to the concentration of hydronium times the concentration of the conjugate base. These two guys divided by the concentration of the conjugate acid for this guy. Now, if you are confused where this expression comes from, check out the link below."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "So we get the acid ionization constant for this acid is equal to the concentration of hydronium times the concentration of the conjugate base. These two guys divided by the concentration of the conjugate acid for this guy. Now, if you are confused where this expression comes from, check out the link below. Now remember, our goal is to find an expression that we can use to solve for the PH. So which one of these guys determines PH? Well, it's this guy."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "Now remember, our goal is to find an expression that we can use to solve for the PH. So which one of these guys determines PH? Well, it's this guy. Remember, PH is equal to negative log of the hydronium concentration. So our goal will be to isolate this guy first. So let's isolate him."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "Remember, PH is equal to negative log of the hydronium concentration. So our goal will be to isolate this guy first. So let's isolate him. We get ka times the concentration of conjugate acid. This guy divided by the concentration of conjugate base for this guy equals the hydronium concentration. Now our next step is to convert these guys into logs because our PH involves logs."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "We get ka times the concentration of conjugate acid. This guy divided by the concentration of conjugate base for this guy equals the hydronium concentration. Now our next step is to convert these guys into logs because our PH involves logs. So we take the logs of both sides and we get log of base ten of the hydronium concentration of this guy equals log of base ten of this whole guy. Now our next goal is to use the laws of logs to convert this guy from this more compact state to a less compact state. Now, what we get is the following log of the hydronium concentration."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "So we take the logs of both sides and we get log of base ten of the hydronium concentration of this guy equals log of base ten of this whole guy. Now our next goal is to use the laws of logs to convert this guy from this more compact state to a less compact state. Now, what we get is the following log of the hydronium concentration. This guy is equal to notice inside this log we're multiplying ka times this whole thing, this ratio. So another way of expressing this is to use the logs of logs and we get log of ka plus log of this guy. So the concentration of the conjugate axis divided by the concentration of conjugate base."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "This guy is equal to notice inside this log we're multiplying ka times this whole thing, this ratio. So another way of expressing this is to use the logs of logs and we get log of ka plus log of this guy. So the concentration of the conjugate axis divided by the concentration of conjugate base. Now our next step is to basically rewrite this guy in the form of the PH formula. We can't say this guy is equal to PH because remember, PH is negative log of this guy. So let's take the negative of the whole expression."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "Now our next step is to basically rewrite this guy in the form of the PH formula. We can't say this guy is equal to PH because remember, PH is negative log of this guy. So let's take the negative of the whole expression. This way we can get our PH. So we multiply two by negative one and we get negative this whole thing. So negative this equals negative this guy minus this guy."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "This way we can get our PH. So we multiply two by negative one and we get negative this whole thing. So negative this equals negative this guy minus this guy. So next step is to convert this guy into PH. And we get PH is equal to negative of log ka. Now notice what happens here whenever we have a negative on the outside of our log."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "So next step is to convert this guy into PH. And we get PH is equal to negative of log ka. Now notice what happens here whenever we have a negative on the outside of our log. That simply means we're taking this guy and raising it to the negative one. This is equivalent to saying plus log of this guy to the negative one. But this guy to the negative one is simply reciprocating these guys."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "That simply means we're taking this guy and raising it to the negative one. This is equivalent to saying plus log of this guy to the negative one. But this guy to the negative one is simply reciprocating these guys. So negative log of this guy is the same as saying positive log of the reciprocal of this guy. So the concentration of conjugate base divided by the concentration of conjugate acid. So Now, A final formula states we find that PH is equal to."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "So negative log of this guy is the same as saying positive log of the reciprocal of this guy. So the concentration of conjugate base divided by the concentration of conjugate acid. So Now, A final formula states we find that PH is equal to. Remember, negative log of Ka is simply PKA. And we get PH is equal to PKA plus log of the ratio of conjugate base over conjugate acid. And this is our Henderson half of block formula."}, {"title": "Henderson-Hasselbach Equation .txt", "text": "Remember, negative log of Ka is simply PKA. And we get PH is equal to PKA plus log of the ratio of conjugate base over conjugate acid. And this is our Henderson half of block formula. Now, if you want to do an example using this formula, check out the link below. But remember, this formula can Be used to find Two things. So, if You Know the PKA and you know our ratio, you can find the PH."}, {"title": "Lewis Acids and Bases .txt", "text": "So let's begin by defining what a lowest acid and lowest base is. Lewis acids are molecules containing an empty, unfilled atomic orbital, and lowest bases are molecules containing a pair of electrons, bonds, or a filled orbital. So let's look at a Lewis acid. So, this is one example of a Lewis acid. It's a methylcation. This methylcation has three SP two hybridized covalent bonds between the C and the HS."}, {"title": "Lewis Acids and Bases .txt", "text": "So, this is one example of a Lewis acid. It's a methylcation. This methylcation has three SP two hybridized covalent bonds between the C and the HS. And it has an empty pur two P orbital given here. So it has a positive charge. So here's an example of a Lewis base."}, {"title": "Lewis Acids and Bases .txt", "text": "And it has an empty pur two P orbital given here. So it has a positive charge. So here's an example of a Lewis base. Once again, a Lewis base is a molecule containing a pair of electrons, a pair of non bonding electrons in an orbital. So hydrant is an example of a Lewis base because it has two electrons, a pair of non bonding electrons within the one S orbital. Now, when the Lewis acid react with the Lewis base, that basically means their atomic orbitals interact."}, {"title": "Lewis Acids and Bases .txt", "text": "Once again, a Lewis base is a molecule containing a pair of electrons, a pair of non bonding electrons in an orbital. So hydrant is an example of a Lewis base because it has two electrons, a pair of non bonding electrons within the one S orbital. Now, when the Lewis acid react with the Lewis base, that basically means their atomic orbitals interact. They overlap, producing a bond. So in this case, when this Lewis acid interacts with this Lewis base, this one S interacts with this two p.\nThese two electrons are donated to this two P orbital. Now, as these come closer, this load becomes smaller, this becomes larger so that the overlap is better."}, {"title": "Lewis Acids and Bases .txt", "text": "They overlap, producing a bond. So in this case, when this Lewis acid interacts with this Lewis base, this one S interacts with this two p.\nThese two electrons are donated to this two P orbital. Now, as these come closer, this load becomes smaller, this becomes larger so that the overlap is better. And we form SP three hybridized bonds. So 1234 SP three hybridized bonds. And now this carbon and this H share a pair of electrons."}, {"title": "Lewis Acids and Bases .txt", "text": "And we form SP three hybridized bonds. So 1234 SP three hybridized bonds. And now this carbon and this H share a pair of electrons. So let's look at the energy diagram of our interaction of our two lowest acids and bases. So, once again, our one S is lowering energy than the two p. That's because this is closer to our nucleus than this orbital is. And so this guy will be found lower on the energy level."}, {"title": "Lewis Acids and Bases .txt", "text": "So let's look at the energy diagram of our interaction of our two lowest acids and bases. So, once again, our one S is lowering energy than the two p. That's because this is closer to our nucleus than this orbital is. And so this guy will be found lower on the energy level. So the y axis in the energy, this will be lower than our two P orbital. So the two electrons will come directly from the one S from the Hydride. And when they interact, they will form a bonding molecular orbital and an antibounding molecular orbital."}, {"title": "Lewis Acids and Bases .txt", "text": "So the y axis in the energy, this will be lower than our two P orbital. So the two electrons will come directly from the one S from the Hydride. And when they interact, they will form a bonding molecular orbital and an antibounding molecular orbital. The two electrons will go into the bonding molecular orbital, forming our SP 30 hybridized molecular orbitals. So, though, now let's define what a broncidlaric acid and a broncillary base is. A bronzedidlori acid is a molecule that donates an H ion, while a broncillary base is a molecule that accepts an H ion."}, {"title": "Lewis Acids and Bases .txt", "text": "The two electrons will go into the bonding molecular orbital, forming our SP 30 hybridized molecular orbitals. So, though, now let's define what a broncidlaric acid and a broncillary base is. A bronzedidlori acid is a molecule that donates an H ion, while a broncillary base is a molecule that accepts an H ion. So acid strength of a broncillary acid increases with increasing S character. So why is that? Well, to examine that, let's recall one simple concept."}, {"title": "Lewis Acids and Bases .txt", "text": "So acid strength of a broncillary acid increases with increasing S character. So why is that? Well, to examine that, let's recall one simple concept. So, here we have a protons. We have a nucleus and protons inside that nucleus. And this is our one S orbital, our two S orbital and the two P orbital."}, {"title": "Lewis Acids and Bases .txt", "text": "So, here we have a protons. We have a nucleus and protons inside that nucleus. And this is our one S orbital, our two S orbital and the two P orbital. So recall that as our electron gets closer and closer to our protons in the nucleus, the energy level of the entire atom decreases, so it becomes more stable. So the closer our electron is to our nucleus, the more stable that atom is. And let's look what happens when an acid, a brothed lauric acid, reacts."}, {"title": "Lewis Acids and Bases .txt", "text": "So recall that as our electron gets closer and closer to our protons in the nucleus, the energy level of the entire atom decreases, so it becomes more stable. So the closer our electron is to our nucleus, the more stable that atom is. And let's look what happens when an acid, a brothed lauric acid, reacts. So, when a brothed Laric acid, reacts, it creates a conjugate base. So a brothelary base, and it creates an hion. Now, if this has a lot of S character, that means the electron pair here will be found in that S character."}, {"title": "Lewis Acids and Bases .txt", "text": "So, when a brothed Laric acid, reacts, it creates a conjugate base. So a brothelary base, and it creates an hion. Now, if this has a lot of S character, that means the electron pair here will be found in that S character. So, the more S character we have, the closer our electrons are to our nucleus and the more stable our conjugate bases. And if we have a stable conjugate base, that means our acid will be more likely to form this conjugate base, and therefore, our acid will be more likely to dissociate. And that means our asset will be a stronger acid."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "Now, the system is used for either one of two things. It's either used in determining the absolute configuration of your enantomers to either R or S absolute configuration, or it's used to help you rank groups attached to double bonds. So in this lecture, we're going to focus primarily on this usage here. So whenever you're using this system, four important rules must be followed that will help you to get or find the highest priority groups. So let's look at rule number one. So, atom with higher atomic number receives higher priority."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So whenever you're using this system, four important rules must be followed that will help you to get or find the highest priority groups. So let's look at rule number one. So, atom with higher atomic number receives higher priority. So let's see what that means via this example. So here we have a carbon carbon double bond. So let's examine this carbon carbon number one."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So let's see what that means via this example. So here we have a carbon carbon double bond. So let's examine this carbon carbon number one. So carbon number one is attached to two groups. It's either attached to the H or it's attached to the carbon here. So which atom has a higher atomic number?"}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So carbon number one is attached to two groups. It's either attached to the H or it's attached to the carbon here. So which atom has a higher atomic number? Well, clearly the carbon has a higher atomic number, and that means this group attached this carbon has a higher priority than this group here. Likewise, let's examine the second carbon. The second carbon of the double bond is also attached to a carbon, and it's also attached to an H.\nWhich one of these two groups has a higher atomic number?"}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "Well, clearly the carbon has a higher atomic number, and that means this group attached this carbon has a higher priority than this group here. Likewise, let's examine the second carbon. The second carbon of the double bond is also attached to a carbon, and it's also attached to an H.\nWhich one of these two groups has a higher atomic number? Well, clearly, the carbon has a higher atomic number, and so the carbon wins. And let's label it with an asterisk. So notice that in this compound, our groups with the higher priorities are on the same side of the double bond."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "Well, clearly, the carbon has a higher atomic number, and so the carbon wins. And let's label it with an asterisk. So notice that in this compound, our groups with the higher priorities are on the same side of the double bond. And that means this must be a Z isomer. So let's look at rule number two for isotopes. The isotope with a higher atomic weight wins."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "And that means this must be a Z isomer. So let's look at rule number two for isotopes. The isotope with a higher atomic weight wins. Remember, two compounds or two atoms are isotopes if they have the same number of protons and electrons, but different number of neutrons. So they differ in atomic weight. So let's look at the following example."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "Remember, two compounds or two atoms are isotopes if they have the same number of protons and electrons, but different number of neutrons. So they differ in atomic weight. So let's look at the following example. Let's suppose we have a carbon carbon double bond. So let's examine this carbon here. This carbon is attached to an H group and also to a D group."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "Let's suppose we have a carbon carbon double bond. So let's examine this carbon here. This carbon is attached to an H group and also to a D group. D is simply deteriorate. It's the isotope of H. So since D has a higher atomic weight, d must have a higher priority than H.\nSo this group has a higher priority than this group. Likewise, for the second carbon in a double bond, we have the following two groups."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "D is simply deteriorate. It's the isotope of H. So since D has a higher atomic weight, d must have a higher priority than H.\nSo this group has a higher priority than this group. Likewise, for the second carbon in a double bond, we have the following two groups. Once again, we have the H and we have the D.\nSo our D, the deteriorium, has a higher atomic weight, so it has a higher priority. And now we have an allochene where our two higher priority groups are on opposite sides of the double bond. And that means this must be an Eisomer."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "Once again, we have the H and we have the D.\nSo our D, the deteriorium, has a higher atomic weight, so it has a higher priority. And now we have an allochene where our two higher priority groups are on opposite sides of the double bond. And that means this must be an Eisomer. So let's look at rule number three. If we have the same atom, if we have a tie between our atoms, we move to the next atom. So let's see exactly what that means."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So let's look at rule number three. If we have the same atom, if we have a tie between our atoms, we move to the next atom. So let's see exactly what that means. Let's look at the following example. We have a carbon carbon double bond. So let's begin with this carbon, the first carbon in the double bond."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "Let's look at the following example. We have a carbon carbon double bond. So let's begin with this carbon, the first carbon in the double bond. So this carbon is attached to a carbon of this side and a carbon on that side. So far, we have a tie. We can't determine the atomic number, so we move on to the next atom."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So this carbon is attached to a carbon of this side and a carbon on that side. So far, we have a tie. We can't determine the atomic number, so we move on to the next atom. So there is no next atom here. But here we have a following carbon atom. So that means this side wins."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So there is no next atom here. But here we have a following carbon atom. So that means this side wins. It has a higher atomic number and it also has a higher atomic weight. So this side, this group, has a higher priority than the lower group. Likewise, for this carbon, we have the same exact situation."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "It has a higher atomic number and it also has a higher atomic weight. So this side, this group, has a higher priority than the lower group. Likewise, for this carbon, we have the same exact situation. So, once again, as the first example, we have the Z Isomer, because our two higher priority groups are the same size. Finally, let's look at rule number four. A double bond to a carbon is considered as two single bonds."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So, once again, as the first example, we have the Z Isomer, because our two higher priority groups are the same size. Finally, let's look at rule number four. A double bond to a carbon is considered as two single bonds. So let's see exactly what that means. Let's suppose we have our double bond here. So a carbon and carbon double bond."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So let's see exactly what that means. Let's suppose we have our double bond here. So a carbon and carbon double bond. We want to examine the groups attached to our first carbon. So here we have a carbon in a carbon. So so far, no one wins."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "We want to examine the groups attached to our first carbon. So here we have a carbon in a carbon. So so far, no one wins. And a carbon and a carbon. So no one wins. But we have a double bond here."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "And a carbon and a carbon. So no one wins. But we have a double bond here. And this double bond, according to rule number four, is actually, or actually looks something like this. So every double bond is considered as having two single bonds. So we erase this double bond and we replace one carbon and a second carbon."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "And this double bond, according to rule number four, is actually, or actually looks something like this. So every double bond is considered as having two single bonds. So we erase this double bond and we replace one carbon and a second carbon. So two more single covalent bonds. So now this side, this group has a higher atomic weight. And so this group wins, and this group must be the higher priority group."}, {"title": "Cahn-Ingold-Prelog Priority System .txt", "text": "So two more single covalent bonds. So now this side, this group has a higher atomic weight. And so this group wins, and this group must be the higher priority group. Now, on this side of the carbon, we have two identical methyl groups. So we have carbon and carbon. So no one wins here."}, {"title": "Stability of Alkenes.txt", "text": "So let's suppose we have a certain alky, let's say hexine, for example, and let's write out all different types of isomers of hexine. Now, if we compare the stability of one isomer of hexine to another isomer of exe, we'll see a difference in stability. In other words, some isomers are more stable than other isomers. So in general, why is that? Why is it that some isomers of alkans are more stable than other isomers of that same alkane? So we're going to address that question in this lecture."}, {"title": "Stability of Alkenes.txt", "text": "So in general, why is that? Why is it that some isomers of alkans are more stable than other isomers of that same alkane? So we're going to address that question in this lecture. So let's begin by defining change in Enthalpy affirmation. So loosely speaking, enthalpy affirmation is the energy difference between the final product and its constituent elements. In other words, if this value is negative, that means our final products are stable or more stable than the constituent elements."}, {"title": "Stability of Alkenes.txt", "text": "So let's begin by defining change in Enthalpy affirmation. So loosely speaking, enthalpy affirmation is the energy difference between the final product and its constituent elements. In other words, if this value is negative, that means our final products are stable or more stable than the constituent elements. And in fact, the more negative this value is, the more stable our product is. So let's list a few isomers of hexine. So here's list of five different isomers of hexine and each corresponding change in enthalpy of formation."}, {"title": "Stability of Alkenes.txt", "text": "And in fact, the more negative this value is, the more stable our product is. So let's list a few isomers of hexine. So here's list of five different isomers of hexine and each corresponding change in enthalpy of formation. So for example, for this isomer of hexane, we have negative ten kilo cals per mole of change in enthalpy of formation. And we see that as we go from one to five, our values become more and more negative. And five is the most stable isomer and one is the least stabilizomer."}, {"title": "Stability of Alkenes.txt", "text": "So for example, for this isomer of hexane, we have negative ten kilo cals per mole of change in enthalpy of formation. And we see that as we go from one to five, our values become more and more negative. And five is the most stable isomer and one is the least stabilizomer. So stability increases as we go down from one to five. So why is that? Why is it that number one is less stable than number five?"}, {"title": "Stability of Alkenes.txt", "text": "So stability increases as we go down from one to five. So why is that? Why is it that number one is less stable than number five? So to examine this, let's recall an important detail. Remember, when electrons are found in the s orbital, those electrons are more stable than if the electrons were found in the p orbital. And generally speaking, s character is more stable than p character because of that same concept."}, {"title": "Stability of Alkenes.txt", "text": "So to examine this, let's recall an important detail. Remember, when electrons are found in the s orbital, those electrons are more stable than if the electrons were found in the p orbital. And generally speaking, s character is more stable than p character because of that same concept. So let's examine the different bonds, let's compare one and five and examine different bonds that exist within that molecule. And let's see if we can find where the difference in stability comes from. So let's look at one."}, {"title": "Stability of Alkenes.txt", "text": "So let's examine the different bonds, let's compare one and five and examine different bonds that exist within that molecule. And let's see if we can find where the difference in stability comes from. So let's look at one. So here we have the first isomer. So notice we have a double bond. And within this double bond, the Sigma bond contains SP two, SP two character."}, {"title": "Stability of Alkenes.txt", "text": "So here we have the first isomer. So notice we have a double bond. And within this double bond, the Sigma bond contains SP two, SP two character. Remember, SP two simply means there's 33% S character and 66% P character. SP three means there's only 25% S character and 75% P character. So SP two bonds are more stable than SP three bonds because SP two bonds contain more S character."}, {"title": "Stability of Alkenes.txt", "text": "Remember, SP two simply means there's 33% S character and 66% P character. SP three means there's only 25% S character and 75% P character. So SP two bonds are more stable than SP three bonds because SP two bonds contain more S character. So this Sigma bond within the double bond contains SP two SP two character. This bond, this covalent bond has SP two SP three character. And each of these three covalent bonds has SP three SP three character each."}, {"title": "Stability of Alkenes.txt", "text": "So this Sigma bond within the double bond contains SP two SP two character. This bond, this covalent bond has SP two SP three character. And each of these three covalent bonds has SP three SP three character each. So we have three SP three SP three bonds. We have one SP two SP two bond, and we have one SP two SP three bonds. So now let's compare the bonds within compound pi within itemer number five."}, {"title": "Stability of Alkenes.txt", "text": "So we have three SP three SP three bonds. We have one SP two SP two bond, and we have one SP two SP three bonds. So now let's compare the bonds within compound pi within itemer number five. So once again, we have that same exact SP two SP two bond like we have here and we have 1234 SP three SP two bonds. So notice we have more stable bonds within this compound than this compound because SP two is more stable than SP three. Here, we only have one SP two SP three bond and the rest are SP three SP threes, while here, all four bonds are SP three SP two bonds."}, {"title": "Stability of Alkenes.txt", "text": "So once again, we have that same exact SP two SP two bond like we have here and we have 1234 SP three SP two bonds. So notice we have more stable bonds within this compound than this compound because SP two is more stable than SP three. Here, we only have one SP two SP three bond and the rest are SP three SP threes, while here, all four bonds are SP three SP two bonds. So in other words, electrons are more stable in the s orbitals. Thus, the more S character in a bond, the more stable that bond. And alkines are more stable when they have bonds with a lot of S character."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Today we're going to talk about the concept of chemical equilibrium or dynamic equilibrium. Now, let's look at the following hypothetical reversible first order elementary reaction in which x converts to Y in a single step. Elementary simply means that we can use the coefficients of X and Y to produce the rate law. So let's begin at time equals zero. The concentration of our reactants is very high and the concentration of our product is very low. It's actually zero."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "So let's begin at time equals zero. The concentration of our reactants is very high and the concentration of our product is very low. It's actually zero. So at times zero before our reaction even occurred. We don't have any of our products yet the products haven't formed, so our Y is zero. Now, our X, the concentration of X of our reactants is at its maximum because none of this guy has converted to Y yet."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "So at times zero before our reaction even occurred. We don't have any of our products yet the products haven't formed, so our Y is zero. Now, our X, the concentration of X of our reactants is at its maximum because none of this guy has converted to Y yet. However, as the reaction begins to proceed, the concentration of X begins to decrease, while the concentration of Y begins to increase because some of this X is converting or becoming Y. Eventually a point is reached in which the concentration of the reactants of X and the concentration of the products Y does not change. And this is known as chemical equilibrium."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "However, as the reaction begins to proceed, the concentration of X begins to decrease, while the concentration of Y begins to increase because some of this X is converting or becoming Y. Eventually a point is reached in which the concentration of the reactants of X and the concentration of the products Y does not change. And this is known as chemical equilibrium. This is the point of greatest entropy. In other words, our entropy of our system is at its highest at equilibrium. Now, let's look at the rates of the reactions, the four reaction and the reverse reaction at each point."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "This is the point of greatest entropy. In other words, our entropy of our system is at its highest at equilibrium. Now, let's look at the rates of the reactions, the four reaction and the reverse reaction at each point. Let's go back here. Now, at this point, what's the rate of our forward reaction of X to Y? Well, our rate is determined by our rate constant as well as the concentration of X."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Let's go back here. Now, at this point, what's the rate of our forward reaction of X to Y? Well, our rate is determined by our rate constant as well as the concentration of X. And because our constant K stays the same at the same temperature going this way, that means our rate is strictly dependent on our concentration. And because we said our concentration is at its maximum at the beginning at time equals zero, that means our rate going this way forward of X converting to Y is at its highest. What about the reverse rate of going from Y to X?"}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And because our constant K stays the same at the same temperature going this way, that means our rate is strictly dependent on our concentration. And because we said our concentration is at its maximum at the beginning at time equals zero, that means our rate going this way forward of X converting to Y is at its highest. What about the reverse rate of going from Y to X? Well, at the beginning we said we don't have any of the Y, our Y or concentration of our product is zero. And that means since this guy is zero for the reverse reaction, that means our rate going backwards is zero. And that makes sense because if none of the Y has formed yet, that means none of the Y can convert back to X."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Well, at the beginning we said we don't have any of the Y, our Y or concentration of our product is zero. And that means since this guy is zero for the reverse reaction, that means our rate going backwards is zero. And that makes sense because if none of the Y has formed yet, that means none of the Y can convert back to X. So let's go to this point somewhere in between our initial and our final. Well, somewhere here, our rate begins to decrease going this way. In other words, because our concentration of X begins to slowly diminish, this guy begins to get smaller."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "So let's go to this point somewhere in between our initial and our final. Well, somewhere here, our rate begins to decrease going this way. In other words, because our concentration of X begins to slowly diminish, this guy begins to get smaller. And so, since this guy remains the same throughout the experiment going this way, that means our rate also begins to diminish going from X to Y. How about going from Y to X? Well, going from Y to X, our concentration of that product begins to increase."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And so, since this guy remains the same throughout the experiment going this way, that means our rate also begins to diminish going from X to Y. How about going from Y to X? Well, going from Y to X, our concentration of that product begins to increase. And that means as the reaction is going this way, we're getting more of y. That means our Y, our rate law for going from the products to reactants begins to increase the rate of reaction, rate of reverse reaction. That's because the concentration of Y begins to increase."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And that means as the reaction is going this way, we're getting more of y. That means our Y, our rate law for going from the products to reactants begins to increase the rate of reaction, rate of reverse reaction. That's because the concentration of Y begins to increase. Eventually, the concentration of this guy and this guy at equilibrium will be exactly constant. They won't be the same, although they could be the same, but they will be constant. The concentration of x will not change and the concentration of Y will also not change."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Eventually, the concentration of this guy and this guy at equilibrium will be exactly constant. They won't be the same, although they could be the same, but they will be constant. The concentration of x will not change and the concentration of Y will also not change. And that means our rate of the reverse reaction this way and the rate of the four reaction known this way will be the same. Now, once again, it's important to understand that the concentration of x and Y in equilibrium will not be the same. They could be the same, but it's not necessary for them to be the same."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And that means our rate of the reverse reaction this way and the rate of the four reaction known this way will be the same. Now, once again, it's important to understand that the concentration of x and Y in equilibrium will not be the same. They could be the same, but it's not necessary for them to be the same. The reason that their rates are equal is because the K constant going this way and the K constant for the reverse reaction are different values. And so the rates are equal even though the concentrations might not be equal because the K's balance the rates out. Now, once again, it's important to understand that entropy at this point is at its highest because entropy defines the most probable state and equilibrium is the most probable state."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "The reason that their rates are equal is because the K constant going this way and the K constant for the reverse reaction are different values. And so the rates are equal even though the concentrations might not be equal because the K's balance the rates out. Now, once again, it's important to understand that entropy at this point is at its highest because entropy defines the most probable state and equilibrium is the most probable state. Now, it's also important to understand that at this point it's dynamic equilibrium. In other words, the reactions are still occurring. X is still being converted to Y and y is still being converted to x."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Now, it's also important to understand that at this point it's dynamic equilibrium. In other words, the reactions are still occurring. X is still being converted to Y and y is still being converted to x. It's just because the two rates are equal, the concentrations remain the same. So even though it seems like the reactions have stopped occurring, x is not going to Y and Y is not going to x. That's not true."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "It's just because the two rates are equal, the concentrations remain the same. So even though it seems like the reactions have stopped occurring, x is not going to Y and Y is not going to x. That's not true. Reactions forward and backward are still occurring. It's just they're occurring at the same rate. And that's exactly why the concentrations are remaining the same."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Reactions forward and backward are still occurring. It's just they're occurring at the same rate. And that's exactly why the concentrations are remaining the same. Now suppose, for example, if I added more x, if I add more x, I would change my concentration of x. And that means for a few seconds or for some time, my concentration of x would change and this would shift equilibrium this way, causing x to be converted to Y at a faster rate than y converted to x. And that's called leisurely air principle and we'll talk about that in another lecture."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Now suppose, for example, if I added more x, if I add more x, I would change my concentration of x. And that means for a few seconds or for some time, my concentration of x would change and this would shift equilibrium this way, causing x to be converted to Y at a faster rate than y converted to x. And that's called leisurely air principle and we'll talk about that in another lecture. So now let's suppose we have the following elementary reaction in which our reactants x and y convert to our product z and W. Now, A-B-C and D are the coefficients that represent the moles of each respective atom. Now let's suppose our reaction is a dynamic equilibrium. And what that basically means once again is that the rates at which x and Y are converting to z and W is the same as the rate at which z and w is converting to x and Y."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "So now let's suppose we have the following elementary reaction in which our reactants x and y convert to our product z and W. Now, A-B-C and D are the coefficients that represent the moles of each respective atom. Now let's suppose our reaction is a dynamic equilibrium. And what that basically means once again is that the rates at which x and Y are converting to z and W is the same as the rate at which z and w is converting to x and Y. Now, once again, we're making an assumption that this is elementary reaction. So that means we can write the rate law for each reaction going this way and going that way by simply using the coefficient ADC and d.\nSo let's do exactly that. So the rate of our four reaction is equal to the constant for going this way."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Now, once again, we're making an assumption that this is elementary reaction. So that means we can write the rate law for each reaction going this way and going that way by simply using the coefficient ADC and d.\nSo let's do exactly that. So the rate of our four reaction is equal to the constant for going this way. K one times the concentration of A to the A power times the concentration of y to the b power, the coefficients. And this equals the rate of the reverse reaction, the backwards reaction z and w converting back to x and y. And this equals k minus one, which is a different constant than this constant."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "K one times the concentration of A to the A power times the concentration of y to the b power, the coefficients. And this equals the rate of the reverse reaction, the backwards reaction z and w converting back to x and y. And this equals k minus one, which is a different constant than this constant. Remember, the two constants going this way and this way are different times the concentration of z to the c coefficient or exponent times the concentration of w to the d power. Now this guy is equal to this guy because we're a dynamic equilibrium. And the only reason we are able to write the rate laws like this using the coefficients is because this is an elementary reaction."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Remember, the two constants going this way and this way are different times the concentration of z to the c coefficient or exponent times the concentration of w to the d power. Now this guy is equal to this guy because we're a dynamic equilibrium. And the only reason we are able to write the rate laws like this using the coefficients is because this is an elementary reaction. So now let's bring all the constant to one side and everything else the concentrations to this side. So we get the following. K one divided by k minus one equals this guy divided by this guy."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "So now let's bring all the constant to one side and everything else the concentrations to this side. So we get the following. K one divided by k minus one equals this guy divided by this guy. Now notice that these two guys are constants. They're the same or they don't change at the same temperature. And that means we can represent this guy as another constant, namely K. Now this K is known as the equilibrium constant."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Now notice that these two guys are constants. They're the same or they don't change at the same temperature. And that means we can represent this guy as another constant, namely K. Now this K is known as the equilibrium constant. And the relationship between our K, the equilibrium constant. And the chemical equation above is known as the law of mass action. So what's the meaning of K?"}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And the relationship between our K, the equilibrium constant. And the chemical equation above is known as the law of mass action. So what's the meaning of K? Well, K is simply the ratio of the concentration of products and the concentration of reactants at equilibrium when the Navy equilibrium has been established. And what K does is it tells us how far the reaction proceeded at equilibrium. In other words, we can have three situations."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Well, K is simply the ratio of the concentration of products and the concentration of reactants at equilibrium when the Navy equilibrium has been established. And what K does is it tells us how far the reaction proceeded at equilibrium. In other words, we can have three situations. K can either be greater than one and if K is greater than one, that means at equilibrium we have more products than reactants. And that means our reaction is a product favored, it's spontaneous going this way. Now, if K equals one, that means at equilibrium our concentration of products is the same as the concentration of reactants."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "K can either be greater than one and if K is greater than one, that means at equilibrium we have more products than reactants. And that means our reaction is a product favored, it's spontaneous going this way. Now, if K equals one, that means at equilibrium our concentration of products is the same as the concentration of reactants. Now, if K is less than one, that means this denominator is larger than our enumerator. And that means we have more concentration of reactants of these guys at equilibrium than of our products than these guys. And that means our reaction is not product favorite, it's not spontaneous, in fact it's reactant favorite."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Now, if K is less than one, that means this denominator is larger than our enumerator. And that means we have more concentration of reactants of these guys at equilibrium than of our products than these guys. And that means our reaction is not product favorite, it's not spontaneous, in fact it's reactant favorite. This reaction is spontaneous, but this reaction isn't if our K is less than one. And that's the meaning of K. Now, a few more important things that I want to mention about equilibrium constants. Now an equilibrium constant is unitless."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "This reaction is spontaneous, but this reaction isn't if our K is less than one. And that's the meaning of K. Now, a few more important things that I want to mention about equilibrium constants. Now an equilibrium constant is unitless. And that's because we're dividing concentration by concentration. So our units at the end will cancel out. Now, our equilibrium constant depends strictly on temperature."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And that's because we're dividing concentration by concentration. So our units at the end will cancel out. Now, our equilibrium constant depends strictly on temperature. And that's because our constant is actually a rate constant divided by a rate constant. So it's the ratio of the rate constant going this way to the rate constant going in the reverse direction. And because these guys are dependent only on temperature, these guys also depend upon temperature."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And that's because our constant is actually a rate constant divided by a rate constant. So it's the ratio of the rate constant going this way to the rate constant going in the reverse direction. And because these guys are dependent only on temperature, these guys also depend upon temperature. It does not depend on the concentration. Now note there is a big difference between equilibrium constant and chemical equilibrium. Although the two things are related, they're two different separate ideas."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "It does not depend on the concentration. Now note there is a big difference between equilibrium constant and chemical equilibrium. Although the two things are related, they're two different separate ideas. Once again, equilibrium constant is a ratio of products to reactants, and it depends on temperature. While chemical equilibrium refers to a condition, a system. And if we add, for example, more reactants to our system now our chemical equilibrium is shifted to the right, more reactants will be produced."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "Once again, equilibrium constant is a ratio of products to reactants, and it depends on temperature. While chemical equilibrium refers to a condition, a system. And if we add, for example, more reactants to our system now our chemical equilibrium is shifted to the right, more reactants will be produced. And that's because of Washacliere's principle. We'll discuss that in a bit. But remember to have this distinction between equilibrium constant and chemical equilibrium."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "And that's because of Washacliere's principle. We'll discuss that in a bit. But remember to have this distinction between equilibrium constant and chemical equilibrium. There are different things. Last thing I want to mention is about this expression, this chemical equilibrium expression. Now, notice we included every single reactant product."}, {"title": "Chemical Equilibrium and Equilibirium Constant .txt", "text": "There are different things. Last thing I want to mention is about this expression, this chemical equilibrium expression. Now, notice we included every single reactant product. And that's because we assume that X-Y-Z and W were either in the aqueous state or the gas state. Now only aqueous or gas molecules are included or expressed in our final expression. Solid molecules and liquid molecules are not included in our expression."}, {"title": "Neuron Cells Part II .txt", "text": "So let's say our outside is 0.003 molar, and on the inside is 0.135\nmolar. So how would we find the cell voltage due to the potassium ions? Will we use the Nurse equation? What this equation says is our cell voltage at any given concentration is equal to our standard cell voltage. But this guy is zero. We just said that the cell voltage of this reaction and this reaction are equal but opposite."}, {"title": "Neuron Cells Part II .txt", "text": "What this equation says is our cell voltage at any given concentration is equal to our standard cell voltage. But this guy is zero. We just said that the cell voltage of this reaction and this reaction are equal but opposite. So when you add them, this guy goes to zero. That means our cell voltage is just simply this whole guy. Where gas constant T is our temperature, n is the Mozzo electrons, epic Faradays constant."}, {"title": "Neuron Cells Part II .txt", "text": "So when you add them, this guy goes to zero. That means our cell voltage is just simply this whole guy. Where gas constant T is our temperature, n is the Mozzo electrons, epic Faradays constant. And Q is our expression. Now, let's look at Q first. What is Q?"}, {"title": "Neuron Cells Part II .txt", "text": "And Q is our expression. Now, let's look at Q first. What is Q? Well, Q is the concentration of products divided by the concentration of reactants, right? And our products is this guy, it's 0.3 molar, our 0.3 molar. Sorry."}, {"title": "Neuron Cells Part II .txt", "text": "Well, Q is the concentration of products divided by the concentration of reactants, right? And our products is this guy, it's 0.3 molar, our 0.3 molar. Sorry. And this guy is zero point 13 five molar. So our Q is 0.3\nover zero point 13 five. The M cancel out."}, {"title": "Neuron Cells Part II .txt", "text": "And this guy is zero point 13 five molar. So our Q is 0.3\nover zero point 13 five. The M cancel out. Now, our T is our temperature of our body. It's not 25 degrees Celsius, it's 37 degrees Celsius. So we have 37 to 273, and we get 310."}, {"title": "Neuron Cells Part II .txt", "text": "Now, our T is our temperature of our body. It's not 25 degrees Celsius, it's 37 degrees Celsius. So we have 37 to 273, and we get 310. So it's 310 right here. This is our gas constant. It's just a constant 8.3, 114."}, {"title": "Neuron Cells Part II .txt", "text": "So it's 310 right here. This is our gas constant. It's just a constant 8.3, 114. Our fahrenheit is constant. So what is N is the moles of electrons produced per potassium or a mole of potassium. So notice that our mole here is one."}, {"title": "Neuron Cells Part II .txt", "text": "Our fahrenheit is constant. So what is N is the moles of electrons produced per potassium or a mole of potassium. So notice that our mole here is one. It's a ratio of one to one. That means we have 1 mol of electron. So number one goes for N.\nWe plug these guys into the calculator."}, {"title": "Neuron Cells Part II .txt", "text": "It's a ratio of one to one. That means we have 1 mol of electron. So number one goes for N.\nWe plug these guys into the calculator. Notice that natural log of a number smaller than one gives you a negative number. So the negative is becoming positive. And this is our final cell voltage, zero point 102 volts."}, {"title": "Neuron Cells Part II .txt", "text": "Notice that natural log of a number smaller than one gives you a negative number. So the negative is becoming positive. And this is our final cell voltage, zero point 102 volts. So then we do the same exact thing for calcium, for sodium, and for chloride. Add all the guys up and we should get our final resting electrical potential to sell. Now, I want to talk more about the meaning of this number."}, {"title": "Neuron Cells Part II .txt", "text": "So then we do the same exact thing for calcium, for sodium, and for chloride. Add all the guys up and we should get our final resting electrical potential to sell. Now, I want to talk more about the meaning of this number. What is meant by this number? Remember, we have the electrochemical gradient of our cell. And this is the gradient due to the concentration of ions and due to charge."}, {"title": "Neuron Cells Part II .txt", "text": "What is meant by this number? Remember, we have the electrochemical gradient of our cell. And this is the gradient due to the concentration of ions and due to charge. So it's the chemical gradient and electrical gradient or voltage gradient. And these guys are opposite of each other. In other words, notice that our potassium ion, there is a larger concentration on the inside than outside."}, {"title": "Neuron Cells Part II .txt", "text": "So it's the chemical gradient and electrical gradient or voltage gradient. And these guys are opposite of each other. In other words, notice that our potassium ion, there is a larger concentration on the inside than outside. And that means these guys will tend to move down their chemical gradients, right? Because there are more of these guys on the outside. So equilibrium will want to establish and these guys will want to move to the outside down their chemical gradient."}, {"title": "Neuron Cells Part II .txt", "text": "And that means these guys will tend to move down their chemical gradients, right? Because there are more of these guys on the outside. So equilibrium will want to establish and these guys will want to move to the outside down their chemical gradient. Now, electrical gradient is the opposite of that. Because electrons travel this way, electrons will want to travel to the place where there is more positive charge. That means it's opposite."}, {"title": "Neuron Cells Part II .txt", "text": "Now, electrical gradient is the opposite of that. Because electrons travel this way, electrons will want to travel to the place where there is more positive charge. That means it's opposite. Now, what this number means is that when our electrical gradient and our chemical gradient equal to this number, when they're both this number, that means equilibrium will be established between the potassium ions and the same number of potassium ions will be going in as they will be coming out, right? So the rates will equal, and that's what this number means. So actually, our electrical potential should be negative of this because they're opposite of each other."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "So we already spoke about a process called effusion which is the movement of gas molecules from a high pressure to a low pressure via a very small hole. And we said that we can find the rates at which gas molecules effuse via something called Grams law which is given right down here. And Grams law states that rate rate of gas molecule one over rate of gas molecule two is equal to the square root of mass of two divided by the square root of mass of one. And what this says is that the lighter the molecule, the faster its rate or the higher its rate. Now we're going to talk about something called diffusion. Diffusion is the movement or the flux of one type of gas molecule into another gas or an empty space."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "And what this says is that the lighter the molecule, the faster its rate or the higher its rate. Now we're going to talk about something called diffusion. Diffusion is the movement or the flux of one type of gas molecule into another gas or an empty space. And now we can use grammar's law to approximate the rates of diffusion. Now, the reason that we approximate is because of something called mean free path. Now, the average or the mean free path of a gas molecule is the distance it travels between any two collisions."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "And now we can use grammar's law to approximate the rates of diffusion. Now, the reason that we approximate is because of something called mean free path. Now, the average or the mean free path of a gas molecule is the distance it travels between any two collisions. So let's look at this system here. In this system we have a square and we have a bunch of red molecules. We have one green molecule."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "So let's look at this system here. In this system we have a square and we have a bunch of red molecules. We have one green molecule. Suppose this green molecule wants to go from this corner to this corner. Now, the best way would be to go directly across from this point to this point. But notice that we have a bunch of red molecules in the way."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "Suppose this green molecule wants to go from this corner to this corner. Now, the best way would be to go directly across from this point to this point. But notice that we have a bunch of red molecules in the way. So this guy will have to push its way to the other side. And by pushing, I mean colliding. So it's going to travel some distance A, then distance B, then CDEF, and finally G.\nSo each time it collides and it bounces off."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "So this guy will have to push its way to the other side. And by pushing, I mean colliding. So it's going to travel some distance A, then distance B, then CDEF, and finally G.\nSo each time it collides and it bounces off. Now, to find the mean free path or the average distance between a two collisions, I simply add up all the distances up and divide by the number of collisions. And that's my mean free path. Now, because of this, we can use Grams law to approximate diffusion and we'll see why in a second."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "Now, to find the mean free path or the average distance between a two collisions, I simply add up all the distances up and divide by the number of collisions. And that's my mean free path. Now, because of this, we can use Grams law to approximate diffusion and we'll see why in a second. So let's look at an illustration of diffusion. Suppose I have a cylindrical tube and I have two claps. Suppose I take this clap, I soak it into ammonia NH three and plug it up."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "So let's look at an illustration of diffusion. Suppose I have a cylindrical tube and I have two claps. Suppose I take this clap, I soak it into ammonia NH three and plug it up. Suppose I take this cloth, I soak it up in hydrochloric acid and then plug it up as well. So it's plugged up on both sides and I have air molecules in the middle. So orange guys are air molecules, green guys armonia molecules and red guys are hydrochloric molecules."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "Suppose I take this cloth, I soak it up in hydrochloric acid and then plug it up as well. So it's plugged up on both sides and I have air molecules in the middle. So orange guys are air molecules, green guys armonia molecules and red guys are hydrochloric molecules. So what will happen? Well, some of these guys will evaporate and some of these guys will evaporate and they will begin moving. But they won't move directly from this point to this point."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "So what will happen? Well, some of these guys will evaporate and some of these guys will evaporate and they will begin moving. But they won't move directly from this point to this point. They will move via a crooked path because of something called the mean free path because there are air molecules present. And these air molecules will create collisions. And this green molecule, for example, will first collide with this guy, then move here, close the wall, then collide with this guy, and so on until it gets to some point here."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "They will move via a crooked path because of something called the mean free path because there are air molecules present. And these air molecules will create collisions. And this green molecule, for example, will first collide with this guy, then move here, close the wall, then collide with this guy, and so on until it gets to some point here. Now, when the green guy reaches the red guy, something will happen. Well, the reaction is as follows. The green guy reacts with the red guy, or ammonia reacts with hydrochloric acid."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "Now, when the green guy reaches the red guy, something will happen. Well, the reaction is as follows. The green guy reacts with the red guy, or ammonia reacts with hydrochloric acid. In a gas station to form a precipitate, it forms a solid. So when these guys meet, at whatever point they will meet, they will form a wall or a solid wall. The precipitate."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "In a gas station to form a precipitate, it forms a solid. So when these guys meet, at whatever point they will meet, they will form a wall or a solid wall. The precipitate. And my question is, at which point will the wall lie? Will it be in the middle? Will it be on this side?"}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "And my question is, at which point will the wall lie? Will it be in the middle? Will it be on this side? Or closer to hydrochloric acid? So, we can use Grounds law to approximate the position of this wall of solid. And the way we do it is the following."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "Or closer to hydrochloric acid? So, we can use Grounds law to approximate the position of this wall of solid. And the way we do it is the following. So rate one over rate two equals square root of mass of two divided by the square root of mass of one equal. So what's the molecular weight or mass of my ammonia? Well, it's 14 plus three gives us 17 on the bottom."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "So rate one over rate two equals square root of mass of two divided by the square root of mass of one equal. So what's the molecular weight or mass of my ammonia? Well, it's 14 plus three gives us 17 on the bottom. What about this guy? Well, one plus 35.5 gives us 36.5. So 36.5 on top."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "What about this guy? Well, one plus 35.5 gives us 36.5. So 36.5 on top. We plug this into our calculator, and we get 1.5. So rate of molecule one is 1.5 times larger or faster than rate of two, because two is heavier. So it's not going to travel with the same velocity."}, {"title": "Diffusion of Gas and Graham\u2019s Law .txt", "text": "We plug this into our calculator, and we get 1.5. So rate of molecule one is 1.5 times larger or faster than rate of two, because two is heavier. So it's not going to travel with the same velocity. Velocity will be smaller. Remember, we're assuming constant temperature. So kinetic energies or average kinetic energies are equal."}, {"title": "Naming of Alkenes.txt", "text": "So let's look at a few important rules that we have to use whenever we're naming alkanes. Rule number one, find the longest possible carbon chain containing all the double bonds. Rule number two, the lowest possible number value is given to the double bonds. Rule number three, if molecules contain more than one double bond, we give it a specific name. For example two double bonds, we name it a dying three double bonds, we name it a triangle. And rule number four, ring compounds containing double bonds are called cycloalkings."}, {"title": "Naming of Alkenes.txt", "text": "Rule number three, if molecules contain more than one double bond, we give it a specific name. For example two double bonds, we name it a dying three double bonds, we name it a triangle. And rule number four, ring compounds containing double bonds are called cycloalkings. So here we have six examples. So let's look at example A, in which we're going to name our alkane. So our first and second step tells us that we have to find the longest possible carbon backbone and we have to assign our double bond the lowest possible number value."}, {"title": "Naming of Alkenes.txt", "text": "So here we have six examples. So let's look at example A, in which we're going to name our alkane. So our first and second step tells us that we have to find the longest possible carbon backbone and we have to assign our double bond the lowest possible number value. So that means we have to begin on this end. So carbon number one, carbon number two, carbon number three, carbon number four. Notice in this case we have a four carbon backbone and our double bond has a one."}, {"title": "Naming of Alkenes.txt", "text": "So that means we have to begin on this end. So carbon number one, carbon number two, carbon number three, carbon number four. Notice in this case we have a four carbon backbone and our double bond has a one. It gets assigned a number one because our double bond begins on carbon one. If we begin number eight, our backbone, from this end, our carbon double bond will get a three. And since we want the lowest possible number value according to step two or rule two, this is how we label it."}, {"title": "Naming of Alkenes.txt", "text": "It gets assigned a number one because our double bond begins on carbon one. If we begin number eight, our backbone, from this end, our carbon double bond will get a three. And since we want the lowest possible number value according to step two or rule two, this is how we label it. So we name our alkene simply one. Butane so the in part simply means our double bond is found on the first position and we have a four carbon backbone. So butte means we have a four carbon backbone."}, {"title": "Naming of Alkenes.txt", "text": "So we name our alkene simply one. Butane so the in part simply means our double bond is found on the first position and we have a four carbon backbone. So butte means we have a four carbon backbone. So let's go to example B. In example B, we have a symmetrical molecule, a symmetrical compound. And that simply means that it doesn't matter if we begin on this end or this end, we get the same alkyne, the same alkyne name."}, {"title": "Naming of Alkenes.txt", "text": "So let's go to example B. In example B, we have a symmetrical molecule, a symmetrical compound. And that simply means that it doesn't matter if we begin on this end or this end, we get the same alkyne, the same alkyne name. So let's begin counting our carbons. Carbon one, carbon two, carbon three, and carbon four. Now now we have two double bonds."}, {"title": "Naming of Alkenes.txt", "text": "So let's begin counting our carbons. Carbon one, carbon two, carbon three, and carbon four. Now now we have two double bonds. So according to rule number three, we have to name this compound a dyene. So our name becomes one three. Butene so the dying part means we have two double bonds, one on the first carbon and the second one on the third carbon."}, {"title": "Naming of Alkenes.txt", "text": "So according to rule number three, we have to name this compound a dyene. So our name becomes one three. Butene so the dying part means we have two double bonds, one on the first carbon and the second one on the third carbon. Buta simply means we have a four carbon backbone. So let's go to example C.\nSo in example C, we have the following compound. So let's begin numbering."}, {"title": "Naming of Alkenes.txt", "text": "Buta simply means we have a four carbon backbone. So let's go to example C.\nSo in example C, we have the following compound. So let's begin numbering. So remember, we want to find the longest possible carbon backbone that contains all the double bonds. So that means we either begin on this end and end on this end, or we begin on this end and go to this end. Since we want to find the lowest possible number values, we begin on this end."}, {"title": "Naming of Alkenes.txt", "text": "So remember, we want to find the longest possible carbon backbone that contains all the double bonds. So that means we either begin on this end and end on this end, or we begin on this end and go to this end. Since we want to find the lowest possible number values, we begin on this end. So 123456 and seven. So we have a seven carbon backbone. Our first double bond begins on the first carbon."}, {"title": "Naming of Alkenes.txt", "text": "So 123456 and seven. So we have a seven carbon backbone. Our first double bond begins on the first carbon. The second double bond begins on the third carbon. And also on the third carbon, we have this Ethyl group. So this group comes first, so three Ethyl."}, {"title": "Naming of Alkenes.txt", "text": "The second double bond begins on the third carbon. And also on the third carbon, we have this Ethyl group. So this group comes first, so three Ethyl. And then we have one comma, three one comma, three Hecta, because we have a seven carbon backbone, dying because we have two double bonds. So once again, two double bonds, one on the first carbon, second one on the third carbon, we have an Ethyl group on the third carbon, and we have a seven carbon backbone hepta. So three Ethyl, one three Hepta, dying for this compound."}, {"title": "Naming of Alkenes.txt", "text": "And then we have one comma, three one comma, three Hecta, because we have a seven carbon backbone, dying because we have two double bonds. So once again, two double bonds, one on the first carbon, second one on the third carbon, we have an Ethyl group on the third carbon, and we have a seven carbon backbone hepta. So three Ethyl, one three Hepta, dying for this compound. So compound D.\nSo now we have a ring, a compound, a ring structure. So let's begin labeling on this guy, on this carbon. So 123456."}, {"title": "Naming of Alkenes.txt", "text": "So compound D.\nSo now we have a ring, a compound, a ring structure. So let's begin labeling on this guy, on this carbon. So 123456. So once again, I begin labeling or numbering on my double bond because I want my double bond to have the lowest possible number value. So this guy is known as one cyclone because it's a cyclic compound, and we have six, so hexene. So one cyclohexine, or simply cyclohexine."}, {"title": "Naming of Alkenes.txt", "text": "So once again, I begin labeling or numbering on my double bond because I want my double bond to have the lowest possible number value. So this guy is known as one cyclone because it's a cyclic compound, and we have six, so hexene. So one cyclohexine, or simply cyclohexine. So let's go to part E, or compound E. So once again, I want to label my compound in a way such that my double bond gets the lowest possible number value, and this methyl group also gets the lowest possible number value. So I begin on this side. So one carbon, second carbon, third carbon, fourth carbon, fifth carbon, 6th carbon."}, {"title": "Naming of Alkenes.txt", "text": "So let's go to part E, or compound E. So once again, I want to label my compound in a way such that my double bond gets the lowest possible number value, and this methyl group also gets the lowest possible number value. So I begin on this side. So one carbon, second carbon, third carbon, fourth carbon, fifth carbon, 6th carbon. So once again, I have a 6th carbon ring. And so I named this guy. So, notice that my methyl group is found on the third carbon."}, {"title": "Naming of Alkenes.txt", "text": "So once again, I have a 6th carbon ring. And so I named this guy. So, notice that my methyl group is found on the third carbon. So I name it three methyl, one cycloxine. So three methyl simply means our methyl group is in the third carbon, and one cyclohexine means that my double bond is found on the first carbon, and we have a ring. So finally, we get to compound E. So in compound E, we have a cyclic six carbon backbone, and we have three double bonds."}, {"title": "Naming of Alkenes.txt", "text": "So I name it three methyl, one cycloxine. So three methyl simply means our methyl group is in the third carbon, and one cyclohexine means that my double bond is found on the first carbon, and we have a ring. So finally, we get to compound E. So in compound E, we have a cyclic six carbon backbone, and we have three double bonds. So let's begin labeling or numbering. And actually doesn't matter where we begin numbering or labeling because this is a symmetrical compound. So 123456."}, {"title": "Naming of Alkenes.txt", "text": "So let's begin labeling or numbering. And actually doesn't matter where we begin numbering or labeling because this is a symmetrical compound. So 123456. So we have one three five so one three and five cyclotex. One three five presents our double bonds. We have three double bonds, so we have a triangle, and we have a six member cyclic ring."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "Today we're going to compare and contrast two important groups or families found on our periodic table. We're going to look at alkali metals and alkaline earth metals. So let's begin with these guys. So the alkaline metals are found on group one or group one A on our periodic table. So that means these guys are metals or part of that a metal division. So that applies that they are soft, malleable and ductile."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "So the alkaline metals are found on group one or group one A on our periodic table. So that means these guys are metals or part of that a metal division. So that applies that they are soft, malleable and ductile. Duct till simply means they're stretchy or stretchable malleable simply means we can hammer them into thin sheets of metal and soft. Well, it simply means they're soft. These guys also display lust alike properties, which simply means they are shiny."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "Duct till simply means they're stretchy or stretchable malleable simply means we can hammer them into thin sheets of metal and soft. Well, it simply means they're soft. These guys also display lust alike properties, which simply means they are shiny. Now, metals are shiny, so that makes sense. And these guys, because they're metals, they conduct electricity. In other words, electrons are capable of moving from one point to another in alkali metals very easily."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "Now, metals are shiny, so that makes sense. And these guys, because they're metals, they conduct electricity. In other words, electrons are capable of moving from one point to another in alkali metals very easily. And that means because moving charge creates electricity, these guys create or conduct electricity very well. Now, they also form ions with a positive oxidation state. In other words, they're capable of losing electrons very easily, and therefore, they usually form plus one oxidation states."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "And that means because moving charge creates electricity, these guys create or conduct electricity very well. Now, they also form ions with a positive oxidation state. In other words, they're capable of losing electrons very easily, and therefore, they usually form plus one oxidation states. So a positive oxidation state. Now, these guys are highly reactive when you mix them with nonmetals and they form ionic compounds. For example, if we react these guys with an H, they will form something called metal Hydrides nah, lih, et cetera, in which the NA and the li both have a positive one oxidation state."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "So a positive oxidation state. Now, these guys are highly reactive when you mix them with nonmetals and they form ionic compounds. For example, if we react these guys with an H, they will form something called metal Hydrides nah, lih, et cetera, in which the NA and the li both have a positive one oxidation state. Now, if you mix the metals with water, they will react exothermically to produce a metal hydroxide and H two gas. Let's put this guy's in parentheses gas. So, for example, if you make two moles of sodium in a solid state with two moles of water, you will get two moles of metal hydroxide and 1 mol of H two or diatomic gas."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "Now, if you mix the metals with water, they will react exothermically to produce a metal hydroxide and H two gas. Let's put this guy's in parentheses gas. So, for example, if you make two moles of sodium in a solid state with two moles of water, you will get two moles of metal hydroxide and 1 mol of H two or diatomic gas. Now, let's look at the second type of group right next to our alkaline metals, known as alkaline earth metals. Now, these guys are obviously in group two or group two A on our periodic table. And these guys, just like the alkaline metals, are also part of the metal division."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "Now, let's look at the second type of group right next to our alkaline metals, known as alkaline earth metals. Now, these guys are obviously in group two or group two A on our periodic table. And these guys, just like the alkaline metals, are also part of the metal division. So they're metals. But unlike these guys, which are soft, these guys are harder and more dense. That means their molecules in a solid state are closer together per unit volume."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "So they're metals. But unlike these guys, which are soft, these guys are harder and more dense. That means their molecules in a solid state are closer together per unit volume. Now, since they're metals, they're also malleable duct till and they conduct electricity well. Now, these guys, unlike these guys, form plus two oxidation state. In other words, these guys lose not one electron but two electrons when they react with our nonmetals."}, {"title": "Alkali and Alkaline Earth Metals .txt", "text": "Now, since they're metals, they're also malleable duct till and they conduct electricity well. Now, these guys, unlike these guys, form plus two oxidation state. In other words, these guys lose not one electron but two electrons when they react with our nonmetals. So that means they're more likely to lose electrons. Now, our alkaline earth metals are less reactive than the alkali metals, but still react with the nonmetals to form ionic compounds. For example, calcium reacts with hydrogen to form calcium Hydride CAH two."}, {"title": "Rate Law .txt", "text": "And we said that rate law is a mathematical representation between the relationship of the concentration of reactants and our reaction rates. Now, we also said that rate law can only be determined using experimental results. And that's exactly right. Now, in this lecture, we're going to look at the following form reaction and try to determine our rate law using some experimental data. So let's begin. 1 mol of methyl acetate react with 1 mol of hydroxide to produce 1 mol of acetate ion and 1 mol of methanol."}, {"title": "Rate Law .txt", "text": "Now, in this lecture, we're going to look at the following form reaction and try to determine our rate law using some experimental data. So let's begin. 1 mol of methyl acetate react with 1 mol of hydroxide to produce 1 mol of acetate ion and 1 mol of methanol. Now, let's conduct the following three experiments in which we measure in each experiment the concentration of methylacetate and hydroxide and we find the initial rate. Now, our goal is to see how our initial rate changes when we change our concentration of reactants. Now, the first experiment will serve as a control."}, {"title": "Rate Law .txt", "text": "Now, let's conduct the following three experiments in which we measure in each experiment the concentration of methylacetate and hydroxide and we find the initial rate. Now, our goal is to see how our initial rate changes when we change our concentration of reactants. Now, the first experiment will serve as a control. We're going to basically compare our second and third experiment to our first experiment and see how our initial rate changes. So in the first experiment, we see that we have 0.5\nmolar of initial methyl acetate and 0.5 molar of our initial hydroxide. Now, when these two concentrations are 0.5\neach, our initial rate is 0.2."}, {"title": "Rate Law .txt", "text": "We're going to basically compare our second and third experiment to our first experiment and see how our initial rate changes. So in the first experiment, we see that we have 0.5\nmolar of initial methyl acetate and 0.5 molar of our initial hydroxide. Now, when these two concentrations are 0.5\neach, our initial rate is 0.2. Next, our goal is to change one of these guys and see how our initial rate is influenced. So let's keep our initial hydroxide concentration the same and only change our initial concentration of methyl acetate. So let's double it."}, {"title": "Rate Law .txt", "text": "Next, our goal is to change one of these guys and see how our initial rate is influenced. So let's keep our initial hydroxide concentration the same and only change our initial concentration of methyl acetate. So let's double it. So we double it to 0.1 molar, and this guy stays at 0.5\nmolar, and we see that our initial rate also doubles to 0.4. Now, that means that because we double this and our initial rate doubles, these guys are directly proportional. In other words, if you double this guy, you must double the initial rate."}, {"title": "Rate Law .txt", "text": "So we double it to 0.1 molar, and this guy stays at 0.5\nmolar, and we see that our initial rate also doubles to 0.4. Now, that means that because we double this and our initial rate doubles, these guys are directly proportional. In other words, if you double this guy, you must double the initial rate. So let's conduct the same exact experiment. But now we keep our initial concentration of methyl acetate the same, and we double our concentration of our initial hydroxide. So let's stay at 0.1\nmolar and go from 0.5 molar of our hydroxide to 0.1 molar."}, {"title": "Rate Law .txt", "text": "So let's conduct the same exact experiment. But now we keep our initial concentration of methyl acetate the same, and we double our concentration of our initial hydroxide. So let's stay at 0.1\nmolar and go from 0.5 molar of our hydroxide to 0.1 molar. We see that when we double our initial concentration of hydroxide, our initial rate also doubles. That means that our hydroxide is also proportional to our rate. So now with this result, we can find our rate law."}, {"title": "Rate Law .txt", "text": "We see that when we double our initial concentration of hydroxide, our initial rate also doubles. That means that our hydroxide is also proportional to our rate. So now with this result, we can find our rate law. So, rate law is the following equation the rate of my reaction in the forward direction is equal to the rate constant of the forward reaction times the concentration of methylacetate times the concentration of hydroxide. Now, notice my exponents are each one. That means there is a direct relationship between our rate of reaction and our concentration."}, {"title": "Rate Law .txt", "text": "So, rate law is the following equation the rate of my reaction in the forward direction is equal to the rate constant of the forward reaction times the concentration of methylacetate times the concentration of hydroxide. Now, notice my exponents are each one. That means there is a direct relationship between our rate of reaction and our concentration. In other words, if we double our concentration of methyl acetate while keeping this guy the same, we double our rate of reaction. Likewise, if we double this guy while keeping our methyl acetate the same, we also double our reaction rate. But if we double each guy, if this guy is multiplied by two and this guy is multiplied by two, that means this guy is quadrupled."}, {"title": "Rate Law .txt", "text": "In other words, if we double our concentration of methyl acetate while keeping this guy the same, we double our rate of reaction. Likewise, if we double this guy while keeping our methyl acetate the same, we also double our reaction rate. But if we double each guy, if this guy is multiplied by two and this guy is multiplied by two, that means this guy is quadrupled. He's multiplied by four. So now I have this. I have this and I know my rate of reaction, but I don't know my rate constant."}, {"title": "Rate Law .txt", "text": "He's multiplied by four. So now I have this. I have this and I know my rate of reaction, but I don't know my rate constant. Now, the rate constant can also be found using our experimental results. The way we do it is we choose any experiment we like and we plug in the data points and we find our KF. So let's use the first experiment."}, {"title": "Rate Law .txt", "text": "Now, the rate constant can also be found using our experimental results. The way we do it is we choose any experiment we like and we plug in the data points and we find our KF. So let's use the first experiment. Let's plug in 0.5 for this guy, 0.5 for this guy, and 0.2\nfor our rate of reaction, the forward direction. So we plug these guys in and we solve for KF and we get 0.002 divided by zero. Five 0.5 times 0.05 gives us 0.08."}, {"title": "Rate Law .txt", "text": "Let's plug in 0.5 for this guy, 0.5 for this guy, and 0.2\nfor our rate of reaction, the forward direction. So we plug these guys in and we solve for KF and we get 0.002 divided by zero. Five 0.5 times 0.05 gives us 0.08. So that means our KF, our rate constant for this reaction going this way is 0.8. So now we plug this into our reaction and we find the final rate law to be the rate of our four reaction is equal to 0.8 times the concentration of our methyl acetate to the first power. Times the concentration of our hydroxide to the first power."}, {"title": "Rate Law .txt", "text": "So that means our KF, our rate constant for this reaction going this way is 0.8. So now we plug this into our reaction and we find the final rate law to be the rate of our four reaction is equal to 0.8 times the concentration of our methyl acetate to the first power. Times the concentration of our hydroxide to the first power. Now, let's think about it. So our initial concentration increases and that increase increases our initial rates. Why is that?"}, {"title": "Rate Law .txt", "text": "Now, let's think about it. So our initial concentration increases and that increase increases our initial rates. Why is that? Well, think about it. The only way that these guys react is if they collide. And if you increase the concentration of either guy, we have more collisions happening."}, {"title": "Rate Law .txt", "text": "Well, think about it. The only way that these guys react is if they collide. And if you increase the concentration of either guy, we have more collisions happening. And if there are more collisions happening, that means our rate should increase. In other words, these guys will convert quicker to our products from the reactants. Now, likewise, if you decrease either concentration, our rate should decrease because we have less collisions occurring."}, {"title": "Molarity Example .txt", "text": "Molarity is represented by the capital letter M, and it has the unit moles of solid over volume of solution. And this includes the volume of both the solvent and the solid. Now let's do an example using molar. The question tells us that we have 0.6 liters of a three molar solution. We need to find the number of liters that we need to add to go from three molar to a one molar solution. So we're diluting."}, {"title": "Molarity Example .txt", "text": "The question tells us that we have 0.6 liters of a three molar solution. We need to find the number of liters that we need to add to go from three molar to a one molar solution. So we're diluting. Whenever we dilute, we want to keep the number of solid the same. We want to increase the number of solids. So this is our current situation."}, {"title": "Molarity Example .txt", "text": "Whenever we dilute, we want to keep the number of solid the same. We want to increase the number of solids. So this is our current situation. We have a three molar solution where the red dots are the solid, blue dots are the solid. We want to go from a three molar to a one molar solution. So we need to ask ourselves, how many more blue dots do we need to add to go from a three molar to a one molar?"}, {"title": "Molarity Example .txt", "text": "We have a three molar solution where the red dots are the solid, blue dots are the solid. We want to go from a three molar to a one molar solution. So we need to ask ourselves, how many more blue dots do we need to add to go from a three molar to a one molar? So since the number of red dots stays the same, we need to find the constant. The constant here is the number of moles of solute. To find the number of moles of solute, we take 0.6 liters and we multiply it by three molar."}, {"title": "Molarity Example .txt", "text": "So since the number of red dots stays the same, we need to find the constant. The constant here is the number of moles of solute. To find the number of moles of solute, we take 0.6 liters and we multiply it by three molar. 0.6\nliters times three moles of solute over liters. The L's cross out 0.6 times three. We get 1.8 moles of solution."}, {"title": "Molarity Example .txt", "text": "0.6\nliters times three moles of solute over liters. The L's cross out 0.6 times three. We get 1.8 moles of solution. Now, we found the number of red dots or the moles of red dots. Now, our goal is a one molar solution. So we set up an equation."}, {"title": "Molarity Example .txt", "text": "Now, we found the number of red dots or the moles of red dots. Now, our goal is a one molar solution. So we set up an equation. One molar is equal to the thing that stays constant, one moles of solution. Over what the amount? We already have 0.6 liters plus the amount we need to add the mount of blue dots that we need to add to the system to get one molar solution."}, {"title": "Sp2 Hybridization.txt", "text": "So we essentially took one S orbital, we took one P orbital, we combine them, and we formed two different sphypedized orbitals. Now we're going to look at SP two hybridized orbitals. So let's suppose we want to construct a BH three molecule. Now, in order to construct this molecule, we need three H atoms and one boron atom. Now, boron has five protons, so has five electrons. Two go into the one S, two go into the two S, and one goes into the two piece."}, {"title": "Sp2 Hybridization.txt", "text": "Now, in order to construct this molecule, we need three H atoms and one boron atom. Now, boron has five protons, so has five electrons. Two go into the one S, two go into the two S, and one goes into the two piece. So we have three balanced electrons. Now, the H atom each has one electron. So the one electron goes into the one S orbital."}, {"title": "Sp2 Hybridization.txt", "text": "So we have three balanced electrons. Now, the H atom each has one electron. So the one electron goes into the one S orbital. Now, before these atoms can combine to form, our BH three molecule hybridization of boron must take place. The question is, how many hybrid orbitals should boron form before this molecule can be created? The answer lies in this molecule itself."}, {"title": "Sp2 Hybridization.txt", "text": "Now, before these atoms can combine to form, our BH three molecule hybridization of boron must take place. The question is, how many hybrid orbitals should boron form before this molecule can be created? The answer lies in this molecule itself. How many times does boron bond two H? Well, since there's one boron and three H atoms, that means there are three different orbitals. So that means we must develop three hybrid orbitals."}, {"title": "Sp2 Hybridization.txt", "text": "How many times does boron bond two H? Well, since there's one boron and three H atoms, that means there are three different orbitals. So that means we must develop three hybrid orbitals. So that means we can no longer use SP hybridization, because SP hybridization produces only two hybrid orbitals. And we need three, as we see in this case here. So that means we're not combining S and P, but we're combining three orbitals."}, {"title": "Sp2 Hybridization.txt", "text": "So that means we can no longer use SP hybridization, because SP hybridization produces only two hybrid orbitals. And we need three, as we see in this case here. So that means we're not combining S and P, but we're combining three orbitals. And these three orbitals are the two S orbital, the two PX orbital, and the two PY orbital. So we combine these atomic orbitals of the boron atom, and we form three identical SP, two orbitals or hybrid orbitals. And these guys look like this hybrid orbital here."}, {"title": "Sp2 Hybridization.txt", "text": "And these three orbitals are the two S orbital, the two PX orbital, and the two PY orbital. So we combine these atomic orbitals of the boron atom, and we form three identical SP, two orbitals or hybrid orbitals. And these guys look like this hybrid orbital here. The only difference is they all lie in different directions. So they point in different directions. So that means because we're combining two P orbitals and one two S orbital, we're going to have 66% P character and 33% S character."}, {"title": "Sp2 Hybridization.txt", "text": "The only difference is they all lie in different directions. So they point in different directions. So that means because we're combining two P orbitals and one two S orbital, we're going to have 66% P character and 33% S character. So now that we formed the three different hybrid orbitals, we are ready for these orbitals to interact with the one S orbitals of the H.\nSo we take three one S orbitals, we combine them with three SP two orbitals, and we form the following picture. So here we have our boron atom. The nucleus is in the middle here."}, {"title": "Sp2 Hybridization.txt", "text": "So now that we formed the three different hybrid orbitals, we are ready for these orbitals to interact with the one S orbitals of the H.\nSo we take three one S orbitals, we combine them with three SP two orbitals, and we form the following picture. So here we have our boron atom. The nucleus is in the middle here. It's not shown. We have one SP hybridized orbital pointing into the page into the board, and it's bonding to one of the S's. One is coming out of the board, and that is bonding to a second one S orbital."}, {"title": "Sp2 Hybridization.txt", "text": "It's not shown. We have one SP hybridized orbital pointing into the page into the board, and it's bonding to one of the S's. One is coming out of the board, and that is bonding to a second one S orbital. And a third is lying on the board, and it's bonding with a third one S orbital. And here we have our BH three molecule that also looks like this. So once again, the boron nucleus, one bond is coming out of the page."}, {"title": "Sp2 Hybridization.txt", "text": "And a third is lying on the board, and it's bonding with a third one S orbital. And here we have our BH three molecule that also looks like this. So once again, the boron nucleus, one bond is coming out of the page. The second bond is going into the page, and the third is going this way. Now, if we grab the XYZ axis, we see that this molecule lies on the same plane. All the bonds lie on the same plane."}, {"title": "Sp2 Hybridization.txt", "text": "The second bond is going into the page, and the third is going this way. Now, if we grab the XYZ axis, we see that this molecule lies on the same plane. All the bonds lie on the same plane. So that means since we have three angles, each angle must be 120 degrees. So, once again, let's review. So what is an SP two hybridized orbital?"}, {"title": "Sp2 Hybridization.txt", "text": "So that means since we have three angles, each angle must be 120 degrees. So, once again, let's review. So what is an SP two hybridized orbital? Well, an SP two hybridized orbital is simply the combination of three different atomic orbitals found within an atom. In this case, it was the Boron atom. When these three different atomic orbitals combined, they form identical hybrid orbitals."}, {"title": "Polyprotic Acids .txt", "text": "Now, monoprotic acid is an acid that can donate a single H plus ion. Let's look at a few examples. Hydrochloric acid, acetic acid, nitric acid, hydrobromic acid, hydrochloric acid, chloro acid, and perchloric acid are all examples of monopolic acids because they can all donate a single H plus ion. Now let's look at the ionization of these acids in water. Let's choose the hypothetical monopolic acid. So let's choose Ha to be our acid."}, {"title": "Polyprotic Acids .txt", "text": "Now let's look at the ionization of these acids in water. Let's choose the hypothetical monopolic acid. So let's choose Ha to be our acid. And Ha will react in water to produce hydronium and a conjugate base. Now, we've spoken about something called Ka, or the acid ionization concept. If you don't know what K is, check out the link above."}, {"title": "Polyprotic Acids .txt", "text": "And Ha will react in water to produce hydronium and a conjugate base. Now, we've spoken about something called Ka, or the acid ionization concept. If you don't know what K is, check out the link above. But Ka is basically the product of the concentrations of the product. So the concentration of this guy times the concentration of this guy divided by the concentration of the reaction of this guy. And we said that if our Ka value is high, if it's bigger than one, that means our acid is a strong acid."}, {"title": "Polyprotic Acids .txt", "text": "But Ka is basically the product of the concentrations of the product. So the concentration of this guy times the concentration of this guy divided by the concentration of the reaction of this guy. And we said that if our Ka value is high, if it's bigger than one, that means our acid is a strong acid. And if our Ka is low, if it's less than one, we're dealing with a weak acid. Now, by strong acid, I simply need an acid that's willing to dissociate, that's willing to go from this form to this form. So therefore, if our Ka is high, that means the concentration of these guys is high and the concentration of this guy is low."}, {"title": "Polyprotic Acids .txt", "text": "And if our Ka is low, if it's less than one, we're dealing with a weak acid. Now, by strong acid, I simply need an acid that's willing to dissociate, that's willing to go from this form to this form. So therefore, if our Ka is high, that means the concentration of these guys is high and the concentration of this guy is low. And that makes sense. Now let's look at polyphonic acid. Polyportic acids are those acids that can donate more than one H plus ion."}, {"title": "Polyprotic Acids .txt", "text": "And that makes sense. Now let's look at polyphonic acid. Polyportic acids are those acids that can donate more than one H plus ion. So let's look at a few examples. Sulfuric acid, hydronium ion, phosphoric acid and carbonic acid are all examples of polyphonic acids. And that's because they all have more than two or more than one H plus ion that they can donate."}, {"title": "Polyprotic Acids .txt", "text": "So let's look at a few examples. Sulfuric acid, hydronium ion, phosphoric acid and carbonic acid are all examples of polyphonic acids. And that's because they all have more than two or more than one H plus ion that they can donate. This guy has three H plus ions that he can donate. So in the same way that we spoke about ionization reactions of monopolytic acids, we can also talk about ionization reactions of polyphotic acids. So let's look at a hypothetical example of a polyphotic acid, h two A."}, {"title": "Polyprotic Acids .txt", "text": "This guy has three H plus ions that he can donate. So in the same way that we spoke about ionization reactions of monopolytic acids, we can also talk about ionization reactions of polyphotic acids. So let's look at a hypothetical example of a polyphotic acid, h two A. Now, for H two A, we're not going to have one reaction. We're going to have two reactions. And that's because we have two potential HS that can be lost to our environment."}, {"title": "Polyprotic Acids .txt", "text": "Now, for H two A, we're not going to have one reaction. We're going to have two reactions. And that's because we have two potential HS that can be lost to our environment. So H two A will react in water to produce hydronium and our conjugate base. Now let's look at this conjugate base. This conjugate base can either go two ways."}, {"title": "Polyprotic Acids .txt", "text": "So H two A will react in water to produce hydronium and our conjugate base. Now let's look at this conjugate base. This conjugate base can either go two ways. It can either act as a conjugate base, taking this H back and creating back our acid, or it can act as an acid itself and it can react with water to produce a second hydronium ion and the final conjugate base. Now, the same way we spoke about Ka or acidization constants for monoprotic acids, we can also talk about Ka's about polyproteic acids. Except now we're not going to have one Ka."}, {"title": "Polyprotic Acids .txt", "text": "It can either act as a conjugate base, taking this H back and creating back our acid, or it can act as an acid itself and it can react with water to produce a second hydronium ion and the final conjugate base. Now, the same way we spoke about Ka or acidization constants for monoprotic acids, we can also talk about Ka's about polyproteic acids. Except now we're not going to have one Ka. We're going to have more than one Ka. Each Ka represents a single reaction. So, for this particular polyphonic acids, we have two reactions."}, {"title": "Polyprotic Acids .txt", "text": "We're going to have more than one Ka. Each Ka represents a single reaction. So, for this particular polyphonic acids, we have two reactions. So we have two ka's. And normally, the first ka will have a higher value than the second ka. And that's because of the following fact."}, {"title": "Polyprotic Acids .txt", "text": "So we have two ka's. And normally, the first ka will have a higher value than the second ka. And that's because of the following fact. Now, when we go from this acid to this base, we get a negative one charge. When we go from this acid to this base, we get a negative two charge. Which one of these is less stable?"}, {"title": "Polyprotic Acids .txt", "text": "Now, when we go from this acid to this base, we get a negative one charge. When we go from this acid to this base, we get a negative two charge. Which one of these is less stable? Well, this guy. And that's because the more charge we have, the less stability. And so this guy will not want to exist by itself."}, {"title": "Polyprotic Acids .txt", "text": "Well, this guy. And that's because the more charge we have, the less stability. And so this guy will not want to exist by itself. This guy will want to exist in this form. And so our reaction for this guy is favored this way. And that's why our ka is much smaller for this reaction than this reaction."}, {"title": "Polyprotic Acids .txt", "text": "This guy will want to exist in this form. And so our reaction for this guy is favored this way. And that's why our ka is much smaller for this reaction than this reaction. And in fact, this reaction is 10,000 times more likely than this reaction. So, in the same way that we spoke about titration curves of monopolic acids, we can also talk about titration curves of polyphonic acids. So here's our titration curve."}, {"title": "Polyprotic Acids .txt", "text": "And in fact, this reaction is 10,000 times more likely than this reaction. So, in the same way that we spoke about titration curves of monopolic acids, we can also talk about titration curves of polyphonic acids. So here's our titration curve. While the Y axis is PH and the X axis is volume based at it. Now, this curve is for our hypothetical phypotic acid, h two A. Now, this guy will have not one equivalence point, but two equivalence points."}, {"title": "Polyprotic Acids .txt", "text": "While the Y axis is PH and the X axis is volume based at it. Now, this curve is for our hypothetical phypotic acid, h two A. Now, this guy will have not one equivalence point, but two equivalence points. And let's look at what each represents. Let's look at the first one. What the first one represents is it's the point at which all of this guy has been neutralized to this guy."}, {"title": "Polyprotic Acids .txt", "text": "And let's look at what each represents. Let's look at the first one. What the first one represents is it's the point at which all of this guy has been neutralized to this guy. So there's no more of this guy, and 100% of our solution is in this form at this point. Now, what the second point means is it's the point at which all of this guy has been neutralized into our final conjugate base. And at this point, as a second equivalence point, all of our solution is in this form."}, {"title": "Polyprotic Acids .txt", "text": "So there's no more of this guy, and 100% of our solution is in this form at this point. Now, what the second point means is it's the point at which all of this guy has been neutralized into our final conjugate base. And at this point, as a second equivalence point, all of our solution is in this form. So, at this point, we notice we have a very high PH. And a high PH means we're in a basic environment. And basic means we don't have a lot of H plus ions."}, {"title": "Polyprotic Acids .txt", "text": "So, at this point, we notice we have a very high PH. And a high PH means we're in a basic environment. And basic means we don't have a lot of H plus ions. So why is it that at this point, this guy really wants to associate into this guy? Well, let's examine why. Well, in our high PH or a basic environment, we don't have a lot of H plus, so this guy decreases."}, {"title": "Polyprotic Acids .txt", "text": "So why is it that at this point, this guy really wants to associate into this guy? Well, let's examine why. Well, in our high PH or a basic environment, we don't have a lot of H plus, so this guy decreases. So to compensate that, according to Alicia clear principle, this guy will dissociate. So our equilibrium will be favored this way. And that's exactly why at a high PH, our reaction goes this way."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now, we begin with two reactants, x and Y and they convert into W and Z. What elementary simply means is that they convert from reactants to products in a single step. Elementary reactions are some of the most basic reactions out there. Now notice that A-B-C and D are coefficients of each respective molecule and they represent the number of molecules or moles. Now, rates of reactions tell you how quickly reactants become products, so how quickly x and y become W and Z. In other words, the rate can be found by the change in concentration of reactants over some given time."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now notice that A-B-C and D are coefficients of each respective molecule and they represent the number of molecules or moles. Now, rates of reactions tell you how quickly reactants become products, so how quickly x and y become W and Z. In other words, the rate can be found by the change in concentration of reactants over some given time. Now, notice that we go from a positive amount or some amount to a small amount. So over time, our x and y will disappear. So therefore, the rate of these guys are negative because remember, we're subtracting our initial from final."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now, notice that we go from a positive amount or some amount to a small amount. So over time, our x and y will disappear. So therefore, the rate of these guys are negative because remember, we're subtracting our initial from final. An initial concentration of x is larger than final concentration of x. The final will be less. We'll have less x and y at the end because some of these guys will convert to W and Z."}, {"title": "Reaction Rates and Rate Law .txt", "text": "An initial concentration of x is larger than final concentration of x. The final will be less. We'll have less x and y at the end because some of these guys will convert to W and Z. So once again, the rate or average rate of x is negative change in concentration of x over time. Now, time is also multiplied by A. Now, for the case that A-B-C and D are one, we simply take away the A, because one times T is T. So the same thing can be said for w for Y."}, {"title": "Reaction Rates and Rate Law .txt", "text": "So once again, the rate or average rate of x is negative change in concentration of x over time. Now, time is also multiplied by A. Now, for the case that A-B-C and D are one, we simply take away the A, because one times T is T. So the same thing can be said for w for Y. The negative change in concentration of y over BT gives you the rate of disappearance of Y. Now, the rate of appearance of W and Z can be given by these formulas where now we have the positive because these guys appear and so these guys will both be positives. Now, let's look at the factors affecting our reaction rates."}, {"title": "Reaction Rates and Rate Law .txt", "text": "The negative change in concentration of y over BT gives you the rate of disappearance of Y. Now, the rate of appearance of W and Z can be given by these formulas where now we have the positive because these guys appear and so these guys will both be positives. Now, let's look at the factors affecting our reaction rates. So, we already said that temperature affects our rates of reaction because it affects our rate constant. Now, if the rate constant increases, then our rate of reaction increases. And we'll see why at the end of this lecture."}, {"title": "Reaction Rates and Rate Law .txt", "text": "So, we already said that temperature affects our rates of reaction because it affects our rate constant. Now, if the rate constant increases, then our rate of reaction increases. And we'll see why at the end of this lecture. But notice that temperature increasing. Temperature affects rate constant by affecting the kinetic energy of molecules. On average, molecules will have higher kinetic energy and a higher temperature."}, {"title": "Reaction Rates and Rate Law .txt", "text": "But notice that temperature increasing. Temperature affects rate constant by affecting the kinetic energy of molecules. On average, molecules will have higher kinetic energy and a higher temperature. And that means they will be more likely to overcome the activation energy barrier and become our products. Now, concentrations we see using these formulas affect or increase our rates. In other words, if we have higher concentration, if this guy becomes higher, this guy becomes higher, if this guy becomes higher and this guy becomes higher, our rates will increase."}, {"title": "Reaction Rates and Rate Law .txt", "text": "And that means they will be more likely to overcome the activation energy barrier and become our products. Now, concentrations we see using these formulas affect or increase our rates. In other words, if we have higher concentration, if this guy becomes higher, this guy becomes higher, if this guy becomes higher and this guy becomes higher, our rates will increase. Now, pressure also has an effect on rates and we'll see how in another video. Now we spoke about elementary reactions in which reactants become products in a single step. Now, there are also multi step reactions and those include many different steps, intermediate steps."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now, pressure also has an effect on rates and we'll see how in another video. Now we spoke about elementary reactions in which reactants become products in a single step. Now, there are also multi step reactions and those include many different steps, intermediate steps. However, we can use this guide or this formula to approximate our rates of multistep reactions as long as the concentrations of intermediates are held relatively low. Now that means suppose we have the following reaction a plus B react to form intermediate AB and then that intermediate AB reacts with C, some other guy to form ABC. Now this guy is out intermediate."}, {"title": "Reaction Rates and Rate Law .txt", "text": "However, we can use this guide or this formula to approximate our rates of multistep reactions as long as the concentrations of intermediates are held relatively low. Now that means suppose we have the following reaction a plus B react to form intermediate AB and then that intermediate AB reacts with C, some other guy to form ABC. Now this guy is out intermediate. Now, as long as the concentration of this guy is kept relatively low, we can approximate using these formulas here. So likewise, the rate of change or disappearance of A and B can be given by negative change in concentration of B divided by T. Now that only works if that intermediate concentration is kept low. Now let's talk about the rate law."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now, as long as the concentration of this guy is kept relatively low, we can approximate using these formulas here. So likewise, the rate of change or disappearance of A and B can be given by negative change in concentration of B divided by T. Now that only works if that intermediate concentration is kept low. Now let's talk about the rate law. The rate law is a mathematical representation that summarizes relationship between reactants and reaction rate. And it also builds a relationship between our rate constant and our rate of reaction. Now notice in this reaction we have a forward reaction and a reverse reaction."}, {"title": "Reaction Rates and Rate Law .txt", "text": "The rate law is a mathematical representation that summarizes relationship between reactants and reaction rate. And it also builds a relationship between our rate constant and our rate of reaction. Now notice in this reaction we have a forward reaction and a reverse reaction. Now, we're only going to worry about the forward reaction, but know that if there is a forward reaction, there could also be a reverse reaction. Now let's look at the rate law. The rate law is this entire equation or relation."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now, we're only going to worry about the forward reaction, but know that if there is a forward reaction, there could also be a reverse reaction. Now let's look at the rate law. The rate law is this entire equation or relation. So rate of the forward reaction of this reaction is given by our rate constant KF for the forward reaction times the concentration of X to the A power. In this case this A times the concentration of Y to the B power B. Now, the reason we chose A and B is because this is an elementary reaction."}, {"title": "Reaction Rates and Rate Law .txt", "text": "So rate of the forward reaction of this reaction is given by our rate constant KF for the forward reaction times the concentration of X to the A power. In this case this A times the concentration of Y to the B power B. Now, the reason we chose A and B is because this is an elementary reaction. Now, if this was not an elementary reaction, we would not be able to use this A and B the way we did here. Now for now, we're only going to worry about elementary reactions. So this is our rate law."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now, if this was not an elementary reaction, we would not be able to use this A and B the way we did here. Now for now, we're only going to worry about elementary reactions. So this is our rate law. Now this KF, our rate constant depends strictly on temperature and types of reactants. It does not depend on the concentrations of reactants nor products. Now, we saw from another video that KF is known as the Iranians equation and it depends strictly on temperature."}, {"title": "Reaction Rates and Rate Law .txt", "text": "Now this KF, our rate constant depends strictly on temperature and types of reactants. It does not depend on the concentrations of reactants nor products. Now, we saw from another video that KF is known as the Iranians equation and it depends strictly on temperature. So at the same temperature we're going to use the same reactant for some given reaction. Now, if we change our reactants, we're also going to have to change our KF because our rate constant also depends on the types of reactants used. Now, one last thing I want to mention is that this rate law, this equation is determined strictly using experimental results."}, {"title": "Reaction Rates and Rate Law .txt", "text": "So at the same temperature we're going to use the same reactant for some given reaction. Now, if we change our reactants, we're also going to have to change our KF because our rate constant also depends on the types of reactants used. Now, one last thing I want to mention is that this rate law, this equation is determined strictly using experimental results. That's the only way we can determine this relation. We can't determine this relation using the formula above."}, {"title": "Reaction Rates and Rate Law .txt", "text": "That's the only way we can determine this relation. We can't determine this relation using the formula above."}]