title
stringlengths 8
78
| text
stringlengths 36
1.56k
|
|---|---|
The pH scale .txt | 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
and 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. |
The pH scale .txt | 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
and 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. |
The pH scale .txt | 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. |
The pH scale .txt | 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. |
The pH scale .txt | 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. |
The pH scale .txt | 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. |
The pH scale .txt | 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. |
The pH scale .txt | 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. |
The pH scale .txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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? |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Triple bond of Acetylene.txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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
and I get poh is equal to 10.5. |
Acid Ionization Constant Example .txt | 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
and I get poh is equal to 10.5. And now I simply apply the formula for poh. So, Poh is equal to 10.5,
which is equal to negative log of the concentration of oh. Now, I convert this log to exponents. |
Acid Ionization Constant Example .txt | And now I simply apply the formula for poh. So, Poh is equal to 10.5,
which 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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? |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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
molar of our acid. We want to find the equilibrium concentration. So in equilibrium, our PH is 3.5. |
Acid Ionization Constant Example .txt | So initially, we have zero of this, zero of this and 0.3
molar 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. |
Acid Ionization Constant Example .txt | 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
to my given information is equal to negative log of the concentration of hydrogen. |
Acid Ionization Constant Example .txt | 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
to 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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
because a little bit of it is dissociated into this guy. But how much less is exactly? |
Acid Ionization Constant Example .txt | 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
because 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. |
Acid Ionization Constant Example .txt | 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. |
Acid Ionization Constant Example .txt | 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
times ten to the negative four times 3.16 times ten to negative four divided by 0.2968 gives me 3.36
times ten to negative six. So that's our Ka. |
Acid Ionization Constant Example .txt | 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
times ten to the negative four times 3.16 times ten to negative four divided by 0.2968 gives me 3.36
times 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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.
So 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. |
Degree of Unsaturation.txt | So let's go back to our rules. According to rule number three, remove nitrogens along with one H.
So 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Degree of Unsaturation.txt | 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. |
Work.txt | 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? |
Work.txt | 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.
And 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. |
Work.txt | 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.
And 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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? |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |
Work.txt | 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. |