 The first observation is fairly obvious. And that is, dead people don't move. So don't pretend you're dead any more than you need to, or you will be. This is a paper that came out that looked at 17,000 13 Canadians aged not 18 to 90 years and followed them for 12 years and asked them how much they sat down. And then there was an independent assessment, too, of how much they sat down. Some people, the working construction, for example, they aren't sitting down very much, except maybe when they get home, watch TV. Other people are sitting down a lot. For example, if you're a truck driver, you might be sitting down all day. And here is the conclusion. It's a little bit hard to read. These data demonstrate a dose response association between sitting time and mortality from all causes and cardiovascular disease independent of leisure time physical activity. That's important because that means the guys who are there on the treadmill and then sit down the rest of the day still done. It means it's more important to be active all the time than it is to make a great display and then collapse. And they conclude, in addition to the promotion of moderate to vigorous physical activity and a healthy weight, physicians should have a scourge sitting for extended periods. Well, you're going to have to probably sit down for two hours on Tuesday. But the rest of the day, you should be ambling around all the time. And the second article, which this is very recent, June 1, immune cells gobble up healthy but idle brain cells. Here's the first line. Use it or lose it. A class of immune cell demolishes idle searches and connections in the brain, even a healthy one. Understanding more about the process could help prevent the onset of degenerative brain diseases. I've seen firsthand that when people retire, they often die or decline surprisingly quickly thereafter. And when people get older, everything slows down. You can't run this fast. You can't think this fast. You can't do anything this fast, as you used to. And it could be that this immune system in older people mistakes the diminished activity in parts of their brain to mean time to get rid of this, and then overreacts, and then starts demolishing their neurons one after another. There are other mechanisms which destroy your brain as well. But this could be a new one. Until now, microplea have been dismissed as simple immune cells that do little more than protect brain cells from damage and tidy up when we after that with disease. The idea that they can clean up brain debris has been well-established in studies of brain disease as best students of Boston Children's Hospital. But now, even without damage, we found them to respond to subtle changes in synastic function. And here's what they did. They took mice. They covered one eye on the mouse, so that eye is not nearly as active. And then they looked at their brains, and they could stain the brains with dyes, and then look at them under a microscope and see which cells are getting targeted. And here's what they found. Though with the help of dyes to distinguish the signals from the left and right eyes, they saw in postmortems that microplea had preferentially proven the connection for synapses from circuits serving the underactive eye. Synapses were marked out for destruction through labeling with an immune chemical called C3. You can read this article in Neuron if you're interested. We think C3 is an eating signal, says Stevens. And here are the green cells marked for destruction. Because they aren't being used. Green for go, which means go and disappear. No danger of that today, because we're going to do some difficult problems. So we're going to be firing a lot of action potentials. Stay attention. Problem 64. Imagine this scary thing appears on an exam out of nowhere. Oh my god, oh my god. What am I going to do? What am I going to do? It's not the Vander Waals equation. It's some funky equation. Does anybody have any idea how many equations of state there are for gas? Alphabetical list of everybody's name, Ken Robinson, Guillain Ritchie, Redlich Cuang, on and on and on. They all have basically the same kinds of features. Some are more accurate than others. Some, instead of having two things, have three things. That makes them more accurate and more tedious to work with. Here's the question, first of all. What do you think the roles of A and B are? And here, Bm is the molar volume, B over n, the number of moles. But here's the first question. What do you think the roles of A and B are? What are the units of A and B? Or do they be easier? And then, suppose that B is 0.1 and A is 0.05. What molar volume, in other words, what's B sub n? Does the gas occupy at five atmospheres and 25 degrees Celsius? I didn't put the units on here, but they're going to be in leader atmospheres, blah, blah, blah. But we're going to have to figure out what they actually are. The units are not discordant units like PSI or something funny. And I'm going to do this two ways. I blew the cobwebs off this calculator. And courtesy of Kevin, who's sitting right over there, include me in on a great way to solve these. I want to go through the calculator method to solve these. Today, you've got a graphing calculator. And so you can pull that out. And if you don't, I'm going to go through the guessing method one more time, because that will be just as good. The calculator method will turn out to be very useful if you want very high accuracy. But I'm never going to demand very high accuracy. I'm not going to say, please give me this to four significant digits or something, so you're guessing correctly. So either method will work. But why not let the calculator do the work if it will do it? OK, so let's go through this. First of all, what are the roles of A and B? Well, the term B sub M is the molar volume. And we're subtracting a positive number from it. And therefore, B must have something to do with repulsive forces. And that makes sense, because at least they seem to have standardized the notation for equations of state with two parameters. Usually, B has to do with repulsive forces or excluded volume. Something to do with how small the thing can get before repulsive forces say, hey, forget it. You can press all you want. I'm already a liquid or a solid. And I'm not going to change any further. The exponential term is new. It's on the right-hand side of the equation. And it's always less than 1. But if we divide both sides by it or multiply both sides by E to the plus 8 over RTVM, then we see that it increases the pressure. It increases the pressure because the real pressure, P, is too small. And the reason why the real pressure, P, is too small is because the gas is tending to shrink. In other words, it's always attracting. So it's not hitting the sides as hard as it could. But therefore, A has something to do with attractive forces. In the Van der Waals equation, what we did is we added a term to P. We added that term A n squared over v squared. That's one way to do it. Here, we're multiplying it by something that's bigger than 1. That's a slightly different way to do it. It may have some advantages. It may have some disadvantages. We just have to see. This equation is theoretically interesting to a number of people. And there are people publishing papers in 2001, 2003, and so forth about this equation. So they're still actually looking at it. It's a topic of research in some circles. So B is repulsive or excluded volume. And A is attractive forces. Well, the molar volume has units of liters per mole. And because we're subtracting B, B has to have the same units. You can't subtract something that has different units. B must have the units of liters per mole. For A, it's a little bit more subtle. But we realize that no term in any exponential function can have any units. So A has to have the same units, since it's on the top, as RT times Vm. And I'll let you work out what the units on A are over the weekend might come in useful. Now, for the last part, remember that we want to have an equation if we're guessing. And it turns out for the calculator method that doesn't have both sides changing. And when I did previously the moles problem in the Van der Waals equation, I divided by 1 over n, because that just seemed like the most obvious thing for me to do. But it may not be the best thing to do. And now I've decided that probably the best thing to do is get everything on the left side and make the right-hand side 0. Because if your calculator says a negative number, you know you're too small. And if your calculator says a positive number, you know you're too big. And on an exam, that may be faster than comparing with 26.3472. You just try to get it to 0. And it turns out your calculator, if you're trying to get it to 0, will do that for you. So here's what we do. We take our equation which was written in this form. This is just how it was presented in a reference. And we multiply both sides by the molar volume minus b. So we bring that over here. And then we multiply both sides by e to the plus a over RTVM. And then we still end up with the RT over here. We could start guessing with this to get v, for example. Or we could go one step further and subtract RT from both sides and make both sides 0. And when you're rearranging equations, the correct way to think about it is that whatever you do on one side of the equal sign, you got to do the same exact thing on the other side. I don't know how they taught you to do it in high school. But sometimes it seems like it's a little bit muddy or iffy. The clear way to think about it is to just guarantee that you're doing the same thing to both sides. If x is equal to y, then 2 times x is equal to 2 times y. And log x is equal to log y and so forth. And 1 over x is equal to 1 over y. Just make sure that you're doing the same thing to both sides. Now with this equation, it turns out, let's see what this one is. This is not so fancy. Ti 83 plus. I don't think that's a super duper fancy calculator, but I confess I don't know. OK, so oops, excuse me. But once we have it in the form on the third line below, once we have it in this form, your calculator will solve it if you know how. However, once I get the solution for whatever I'm looking for here, I would put it back in one time to check the answer, because I don't like a little black box telling me what grade I'm going to get in a course. I like to verify myself what the solution is and make sure when I put it in, it balances out about right. So let's have a look. So you first go to the, well, first of all, I should be complete. You first turn your calculator on. If you don't, then the rest of the instructions don't work. And then you go up to this top thing where it says y equals. And you hit that. And you're going to use the first. This one here has five of them. Maybe there's more, but it's showing five, y1, y2, y3, y4. And for the first one, I enter the equation in with numbers. We'll do it in a second. I'll show you the numbers. So you have to type in your equation. You don't have to simplify it. It can be as messy as you like. But get the parentheses in the right place or it'll be wrong. Missing parentheses or a syntax error when you do the calculator way will be fatal. And then for y2, you just type in y2 equals zero for the second equation. And then for the rest of them, you make sure it's showing nothing at all, this blank. On this one here, it's highlighting the equal sign for the ones that it's actually going to graph for me. So y1 and y2 are highlighted. And it's going to show those two. And then you have to select a region to plot. We can do that in all these problems because we have an estimate for v from the ideal gas. And we can plot from smaller than v to bigger than v. If you let the calculator choose, it may make a crazy choice on its own, like from zero to some number. And then when v is zero, the equation may blow up or become nonsense. And so you may want to control what it does. And then if you want to zoom, you can select the region to plot in the zoom menu, or you can let the calculator choose it by just entering zero in the menu. And your equation has to be, the variable you're solving for has to be x, not v, or anything else. And it has to be of the form some function of x equals to zero, not some other number. But that's not a big deal, especially if you practice. Then you hit graph and your two functions come up. Your y1 function and the function y2 equals zero, which is sometimes hard to see because it might be right on the axis. And then the calculator will figure out for you where the two graphs intersect. And that's your solution. You press second, and then you go on the trace. You press second trace. And then once you get in that menu, you press five on this calculator, where it says intersect. Or in your calculator, you press whatever it takes. And then you hit enter a couple of times I found. And it may ask you a few questions, like I can't recall all of them. But one of them was important. One of them was initial gas or something like that. And I thought, perfect, because I can put in the ideal gas equation for my initial gas. And that'll pretty much guarantee that the calculator doesn't get lost, doesn't go looking way, way off somewhere else. Give me a crazy number. OK, and then it waits a while, because what it's doing is guessing, and it's not as fast as a computer, but waits a while, don't panic. And then it says, boom, number. And the third time I tried it, the number was correct. I made two mistakes. So practice, if you're going to use this on Tuesday. You're welcome to try it. So let's try it with our gas function. Here's our numerical equation. If you want, you can go ahead and enter this in. Instead of VM, use X. So mine here has 5 times quantity x minus 0.1 times e to the quantity 0.05 over 24.46544 times x. I just went ahead and multiplied r times t. And of course, I would never be in such a hurry that I would use 25 Celsius, because that would be a disaster. And then I put in this here, minus 24.46, et cetera. That's my function. And for y2, I put 0. And then on this calculator, which had some funny variables set for x, I had to go to window and tell it to plot it from x equals 1 to 10. Because it was plotting from some other problem in the ancient past, it was plotting negative values of x. And volume is positive. I just picked x equals 1 to 10. And I didn't give it a guess the first time I tried it. And bingo, it came out and it said x is 4.991. And that's exactly right. That's right on. That was very interesting that it did that so quickly. You can try this. These slides and all the other ones will be up there the second I get back to my office today. Try it and practice if you want to use this method. I would be in favor of this. I would have told you about this right at the beginning had I been using this calculator. But I tend not to use calculators much. I either use a computer or I use this thing. And that's, in fact, correct. The molar volume's about five liters. And that seems to be a very good method. And it'll work with any equation that you can enter into your calculator, which is any equation that I'll ever give you on Tuesday. Doesn't matter what it is, right? And as long as you have a good guess to start with from the ideal gas equation, you're pretty much in like flint. Now, if you don't have such a fancy calculator as that one, like I said, I don't think it's that fancy. But if your calculator will not solve it automatically, you can still get it with guessing. Let's just try the same equation here. 5 times molar volume minus 0.1 times the exponential function of this spinach minus the same thing equals to 0. And let's figure out our ideal gas estimate. The molar volume for the ideal gas is just RT over P because n is 1. And I put in R times T in Kelvin, five atmospheres. And my ideal gas estimate was 4.89309 liters. Of course, many of those digits are not important. Usually, I only need two. And let's just try guessing. So now we're going to guess and try to get it to be 0. Well, I guess the ideal gas, and I get minus about a half for the function. I want 0, so I'm too small. And I could see right away by looking at the five atmospheres times V. That if I change V by 0.1, that it's going to give me the right answer. But let's suppose we didn't know that. Let's guess a little bit bigger. Let's guess 5.1. Now I see that it's about equally on the other side a little bit bigger. Target value is 0. It's too big. Well, I'm going to guess halfway in between because this is minus a half and this is plus a half. I'm going to guess 5. And bingo, it's very close, but it's a little bit too big. And therefore, I'm going to drop it down just a smidge, 4.99. Now it's a little too small. Now I can usually stop here on an exam because I know to two digits the answer is 5.0. And if I only want two digits, I'm done. Because the true answer is between 4.99, which is too small, and 5.0, which is too big. So I know that to two digits it rounds to 5.0, and I'm done. But it's not really too much longer, depending how, what kind of facility you have with typing equations into your little calculator. You'll have to decide which method is better. And then I added just a little bit more. And boy, it's very, very close. That's certainly OK. My comment here is that if you need a lot of digits, then the calculator method is superior because it'll give you a lot of digits and it's worth doing it. But if you don't need a lot of digits, it depends how fast you can type the equation into the y1 equals whatever. How quickly you'll be able to get it. So try that and see how it works. Now when I quit last lecture, I was talking about Ka and Kb. And there are lots of subscripts on K because there are lots of different kinds of equilibrium constants, one for any kind of chemical reaction. And certain kinds of reactions turn up a lot. If you're working in a biochemistry lab, you're going to be working with acids and bases a lot, making up buffers. And you're going to want to know Ka, or as you'll see in a more advanced course, you're going to want to know Pka that you'll learn about in Chem 1c. And Ka always refers to an acid dissociating. That's what it means. And it's always this kind of equilibrium constant, concentration of H plus, concentration of A minus, divided by the concentration of the reactant acid at equilibrium. Kb always refers to base. And for example, we showed this last time, if we take some base and we add it to water, then we get the conjugate acid and hydroxide. And Kb is always the concentration of this times the concentration of hydroxide, divided by the concentration of this. And you notice that it does not, does not, does not, does not contain the concentration of water. The reason why it doesn't contain the concentration of water is a little bit of a subtle point. But just keep in mind that if you've got a pure liquid or water as the solvent, you don't include it in the equilibrium constant. And you likewise do not include solids when things are dissolving. So we don't include concentrations of water when it's the solvent, nor do we include pure solids or pure liquids in the equilibrium constant. The final K that's interesting to know about is Ksp, which is called the solubility product. For a salt-like sodium chloride dissolving in water to give sodium plus ion and chloride minus ion, Ksp is just this concentration of M plus times the concentration of X minus. And I do not divide it by the concentration of the solid, whatever that was. Part of the clue is when you look at something and you say, well, what is the concentration of the solid? You stop and you say, wait, that doesn't make any sense. How am I going to know what that is? What does it mean? And then that's the clue to leave it out. Don't put it in. If it's Mx2, then it's the concentration of M times the concentration of X squared because there's a 2 here. And if it's Mx3, it's the concentration of M times the concentration of X cubed because there's a 3 here. That's our rule for the reaction quotient. There's nothing new about that. Ksp is listed for all kinds of things. And Ksp is important to know because if we can make something insoluble, we can make it come out of the water. Therefore, if we're doing something like we're trying to remediate some pollution in water, if we can get the bad guy to precipitate by putting in some other stuff, then we may be able to then filter out the solid and let the clean water go on. And there's a lot of interest in doing stuff like that because water is being recycled more and more and more and more because we want to use it many times before it actually gets to the ocean because fresh water is very precious. If we had a better plumbing system, we would not have drinking quality water go into the toilet because first we cleaned up the water like crazy so you could drink it if you wanted. And then what happens to it? Well, it's pretty much non-drinkable fairly quickly. We could actually have two pipes coming into every house and have gray water in the toilet and drinking water only coming out of the tap. And that would be much more efficient, but we would have had to have thought about that back then when we wired up every house with plumbing and nobody did. It didn't seem important at the time, and so it's not done. Anyway, let's figure some solubility here and see how it works. Solubility just tells us how many moles per liter of compound will dissolve in a solvent. So for example, some kind of salt dissolving in water. And Ksp is going to tell us what the solubility is because once we know Ksp, we just solve an equation and we can figure out. So here's an example of this. The solubility product for calcium hydroxide I found is 5.02 times 10 to the minus 6. What is the solubility of calcium hydroxide in water? What that means is how many moles per liter of calcium hydroxide can you throw in water before you're going to have some gunk at the bottom? No more will dissolve. Let's see how to do it. So we have an Mx2 equation. Well, anything that's an equilibrium constant, if you see K on the exam, you don't start doing anything at all until you write the chemical equation. You write the chemical equation down and then everything is crystal clear. Yes? Sure. But if you don't write down the chemical equation, it seems obvious to a chemist. But it's not obvious if you're a chemist in training. You may start doing things, calculating things, and you go around in a circle like a cat chasing its own tail because you never wrote down the proper chemical reaction and balanced it and see what's what. Whenever you see K or delta G or anything, you say, what's the chemical reaction that's occurring? What is this referring to? And for solubility, it's always this stuff dissolving. I've left out the water. The water is kind of riding along in this aqueous. But obviously, we need some water on this side to dissolve it. But we don't include water in the equilibrium constant for Ksp anyhow, so it doesn't matter. And we don't include the pure solid either. It's just these guys that matter. Ksp, this number, 5 times 10 to the minus 6 is equal to the concentration of calcium times the concentration of hydroxide squared. And now we have to say, OK, we don't know what either of these are, but we know the stoichiometry. Every time one of these guys dissolves, we get one of these and we get two of these. So we know exactly what we're getting here. We let x moles per liter dissolve. We don't know what x is. We get x moles per liter of calcium 2 plus in there. And we get 2x moles per liter of hydroxide. Therefore, in terms of our variable x that we're going to have to solve for, our solubility product equation here, our algebraic equation, becomes the following. 5.02 times 10 to the minus 6 equals x. And boy, do I love the parentheses here, because I have been fiddled so many times by not squaring the 2. And if you don't square the 2, you're a factor of 2 wrong. And I think you know by now what happens on my tests if you're a factor of 2 wrong. You get x times 4x squared, which is 4x cubed. And this equation is easy to solve because there's nothing to really isolate. We just divide by 4 and take the cube root. That's what I've done here. x is 1 fourth times 5.02 times 10 to the minus 6 raised to the 1 third power, or the cube root. And I got out x is equal to 1.08 times 10 to the minus 2 molars. That's how much calcium hydroxide will dissolve in water, about a hundredth of a mole per liter. It's not very soluble. The concentration of hydroxide is twice the concentration of calcium. Therefore, if an exam problem says, instead of the solubility, if it says, what is the concentration of hydroxide, you go through all this. And at the end, you say, wait, the concentration of hydroxide is 2x. The concentration of calcium is x. And so you have to multiply this by 2. And you've got 2.16 times 10 to the minus 2 molar for the concentration of hydroxide. And I've just made a little summary here of all the common cases. For a regular 1 to 1 salt, Ksp is x squared. You want to figure the solubility, you just take the square root of Ksp. That's pretty quick. For a salt like calcium hydroxide of the formula Mx2, Ksp is 4 times x cubed because we had to square the 2. For a salt like aluminum chloride, Mx3, Ksp is 27, which comes from 3 cubed times x to the 4. And then the one I always hated when I was a student, which is the one we always got, was some salt like this, aluminum carbonate, diabolical. We have 2x squared here and 3x quantity cubed here. We have 27 times 4. We end up with x to the 5. And you can just count the number of ions. And that's the power of x. There's three guys here. There's four. There's five. That's the power of x. But this number is important. And this one, in particular, is very error prone because you have to square the 2 and you have to cube the 3. And if you don't, you get it wrong. But whatever it is, just write out the reaction, see how many moles of each ion is made, and then raise it in parentheses to the appropriate multiple, to the appropriate power. And then solve it. And they're very easy to solve because there's nothing to do except take the fifth root or the fourth root or something like that. OK, let's have a bit of a review on something else. Tantalum. We're going back a bit here. Tantalum has a body scent. Boy, this is so rich, isn't it? Every word means something. Tantalum, metal, has a body-centered cubic lattice. Wow, that now means a lot to me and should to you as well. And has a density of 16.65 grams per cubic centimeter. What is the atomic radius of tantalum in the metal? Of course, you have a periodic table because you have to know the molar mass of tantalum. Otherwise, all bets are off. But let's assume we have a periodic table and I had one. So I'll give you. Here's what we're thinking here. Body-centered cubic has how many atoms of tantalum in the little cube? Well, I have to remember that or know it. There's one in the center. There's 8 times 1 eighth on the corners. That's what body-centered cubic means. So there's 2 atoms of tantalum in the little cube. And if I know how many grams, literally, the 2 atoms of tantalum have, then I can figure out the size of the little cube because it has to come out to the same density as the overall stuff since the overall stuff is literally made out of these little cubes. And then I have to remember, gee, how do I figure out the atomic radius? Always take a path where they touch. In body-centered, they touch from one corner of the cube through the big guy in the center to the other corner. And there's 1 radius, 2, 3, 4, radii. And then the side of the cube. If it's d, this is the square root of 3 times d, which you can get by making right triangles. So when I see this, I say, gee, I could do that one. I may go on to another problem. I may say, well, I've already done that one. Of course, you do have to write it down on an exam. But sometimes you can just put a mental placeholder and go on to some other problem because I can get that one in my sleep. OK, here we've written it out. There's one atom in the center. There's eight on the corners. And that means I have to realize what BCC means, body-centered cubic. So there's two. And the molar mass of tantalum, I need to know. There's 180.9479 grams per 6.0221415 times 10 to the 23 atoms. And therefore, I can figure out the mass and grams of a single atom of tantalum. It turns out to be 3 times 10 to the minus 22 grams. Very, very light. Using this, the total mass in the unit cell, which is twice this, is about 6 times 10 to the minus 22 grams per unit cell. And now I have to figure out how big the unit cell should be so that this much mass per volume in cubic centimeters comes out to 16.65 grams per cubic centimeter. Obviously, since this is very small, the size of the unit cell is going to be very small. And that makes sense because it's atomic size. From the density, we can determine the edge length of the unit cell because we can figure out the volume. And then we take the cube root of the volume. And then the atomic radius is related by going across the body center of the cube. 4r is the square root of 3d. And I will let you work this out or not, depending on your predilection. And then check your answer. If you do work it out, look up the atomic radius of tantalum and see if it makes sense. OK, let's do another one. We're going back again. The enthalpy of vaporization, delta H vap, I should have said the molar enthalpy, but the units give it away. The molar enthalpy of vaporization of diethyl ether is 27.24 kilojoules per mole. And the normal boiling point, which is where the vapor pressure is 1 atmosphere, is 34.6 degrees C. And here's the question. Is diethyl ether likely to have extensive structure in the liquid state? Is it making long chains or elaborate patterns or hooking up to itself? Or is it random? Well, first of all, we have to make a connection between what we're being asked and what we know. Anything that has to do with randomness or structure, we immediately think of entropy. What this problem is asking about is the entropy of vaporization consistent with random order in the liquid and the gas we assume to be random. And we did that, but it's been a while, so let's go back. The condition for equilibrium between the two phases is always that delta G should be 0. And I have a relationship between delta G and delta H minus T delta S. They're equal. So delta H minus T delta S should be 0. Therefore, at the normal boiling point, delta S should be delta H upon T. Let me put in the numbers. Of course, I don't put in 34.6 degrees Celsius, because by now I automatically convert everything to Kelvin as fast as possible. Delta S of vaporization is 27.24 times 10 to the 3 joules per mole divided by 273.15 plus 34.6. And that comes out to 88.5 joules per mole per Kelvin. And a number like this should ring a bell, because we had this rough rule that if you went from a completely disordered liquid to a completely disordered gas, that that would create the same amount of chaos independent of what the stuff was. And that's called Troutin's rule. It says that the entropy of vaporization from a random liquid to a random gas would be about 88 or so. And since this is close to that, I would say diethyl ether has no structure in the liquid state. And when I look at the molecular structure of diethyl ether, I see that it can't hydrogen bond like water or methanol or some other molecules. And so that confirms it. If this number delta S of vaporization were something like 100, then I would say, aha, it has structure in the liquid. Because when I convert it to a gas, I'm creating more chaos than if it were a random liquid. Therefore, it had to be structured as a liquid. And I first had to break up the structure. And then get it to go to the gas. This gives us a useful trick, in fact. And so I'll show you what this trick is. You can estimate the boiling point of almost anything without knowing too much. Because delta H has been measured, you can look it up. If we've already got the enthalpy of vaporization and we think that the liquid will follow Troutin's rule, then we can estimate the normal boiling point by just taking whatever the enthalpy of vaporization is and dividing it by something like 88. Books vary on the number. Some books insist it's 87.5. Other books say it's 88.5. But if it's about 88 or so, it's in the ballpark. Don't worry about the exact number. Because this is not going to be that accurate anyhow. And although you can do this with vaporization and although sublimation follows exactly the same equation, the Clausius-Clapeyron equation, as a liquid, therefore, if I look at the vapor pressure above solid CO2, it'll be log P is minus delta H of sublimation over R times 1 over T plus a constant. Same equation. But there is no Troutin's rule for sublimation because solids are certainly not random condensed phases. Solids usually have structure. Because when things form a solid, they often form a repeating lattice depending how the tinker toys fit together. And oftentimes, they exclude other things that don't fit in. That's why water, when it's frozen, is pretty much pure unless you flash-freeze it. Because if you freeze it slowly, even if it's salt water, when the water forms the ice crystals, it bungs the salt out. Keep pushing it out. Get out of here. Get out of here. You're screwing up my lattice. And as it forms, it just pushes it aside like that. And that's a good way to purify things, in fact, is to crystallize them. And in organic chemistry lab next year, you'll probably do some crystallization to purify things. It's a good way to do it. OK, let's try it. Here's a problem. Estimate the boiling point of hydrogen sulfide, H2S, and argon, which is obviously a gas at room temperature, knowing that their enthalpy of vaporization values are listed as 18.6 kilojoules per mole for H2S and 6.4 kilojoules per mole for argon. Well, OK, we need two pieces of information to get it. The first is, if the two phases are present, which they are at the normal boiling point, that's why it's called a boiling point. At that exact temperature, delta G is 0. That's why they can both be there. It's a wash. You could either be the liquid or you could be the gas. And the second is, well, we'll just assume that they follow Troutin's rule. And here I've used 87.5. Therefore, at the unique temperature where delta G is 0, the boiling point, delta H is equal to 18.6 times 10 to the 3. Get the same units. I assume what delta S is by essentially saying that the liquid is totally random and the gas is random. And I get 213 Kelvin, which is minus 60 degrees C, roughly. I'm not going to worry about the 0.15 because I have no illusion that this is so accurate that I'm going to bother putting extra digits like that into this kind of problem. And for argon, doing the same thing but with a different value here of delta H, 6.4 times 10 to the 3, I get 73 Kelvin, which is about minus 200 Celsius. And I looked up the values and they aren't bad. This is just a fluke that came out that close, minus 60.2. And this one is off by less than 5% error. So it can be useful because you can guess sometimes the boiling point of things without knowing. OK, here's another one to remember. Remember this one? You might say, wait a second. That was supposed to be on midterm 2, right? There was all this groaning. The reason why it wasn't on midterm 2 is because I don't have a color printer. And if they're all gray, it's not a very fair question because you can't tell who's who. And I don't like to just make things different sizes. But I have all weekend. And so I can replace this with some pattern that you can see. OK? Of course, it may not be this exact one, barium titanate, but let's go through it again anyhow. We have a titanium 4 plus ion right in the center of this cubic lattice. We have the oxygens on each face, so they're cut in half. And we have the bariums on the corners, so they're cut in an eighth. If there were one right here, it would be cut into a fourth. You don't actually have to remember. You just have to say, look, nothing can be outside the cube. If it's outside the cube, when I put the cubes together, they aren't going to fit. So just saw off this and that and this, and you'll just see there's a little 1 eighth inside the cube. That's all I can keep. I can't have anything poking out of my unit cell because I'm going to produce the solid by pasting them together. You don't actually have to memorize which one's which. So here is what we said before. 8 times 1 eighth for the barium is 1. The titanium's in the center, so that's 1 because it's totally inside. And the six oxygens are on the six faces. They're cut in half. 6 times 1 half is 3. And the charges also help us because 2, 4, and minus 6 means it's salt is neutral, and it better be. And therefore, the formula is barium titanate. You wouldn't necessarily know from the structure that it's barium titanate, but from the name, we would put whichever one's first as the first one in the molecular formula. But I wouldn't care if you wrote ti ba o3. That's fine at this stage. Here's my observation. There are 68 problems. That's a lot, but there's more than 68 hours. An hour of problems is not that big a deal. Go over them. Make sure that if that same stupid problem appears that was just up here in Technicolor, that it's just nothing but net. Second, turn them around and solve them the other way. Solve for all the independent variables. Don't just do it one way. And then turn it sideways and solve it as well. And when you solve it, pay attention to units. Pay attention to significant figures, and pay attention to accuracy. And as far as I know, our exam is on Tuesday at 10.30 in this room. It'll be fine with me if you come a bit early, but if you come a bit late, we actually do have to get out by 12.30, so you'll be short on time. So come early. Get organized. Same seats as your last assignment.