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Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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? |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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? |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Boyle’s Law .txt | 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. |
Structure of Atoms .txt | 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. |
Structure of Atoms .txt | 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. |
Structure of Atoms .txt | 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. |
Structure of Atoms .txt | 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
times 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. |
Structure of Atoms .txt | The charge, however, of a proton, an electron has the same magnitude 1.6
times 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. |
Structure of Atoms .txt | 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. |
Structure of Atoms .txt | 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. |
Structure of Atoms .txt | 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. |
Structure of Atoms .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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? |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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? |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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. |
Dalton’s Law on Partial Pressure .txt | 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.
And 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. |
Dalton’s Law on Partial Pressure .txt | 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.
And 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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? |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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? |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Oxidation Numbers Example .txt | 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. |
Cell diagram .txt | 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. |
Cell diagram .txt | 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. |
Cell diagram .txt | 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. |
Cell diagram .txt | 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. |
Cell diagram .txt | 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. |
Cell diagram .txt | 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. |
Cell diagram .txt | 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. |
Cell diagram .txt | 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. |
Half equivalence point .txt | 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. |
Half equivalence point .txt | 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. |
Half equivalence point .txt | 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. |
Half equivalence point .txt | 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. |
Half equivalence point .txt | 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. |
Half equivalence point .txt | 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. |
Half equivalence point .txt | 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. |
Half equivalence point .txt | 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. |