 Today is small day, right, so there's three hot dogs today. They might have some other stuff besides hot dogs, I'm not sure, but definitely hot dogs. And that's going to be in front of HSC from 11 to 1. So if you want free lunch, you can go grab it. The midterm gain grades are posted, I guess. I forgot to release them, but I'll do that right after class. The high was a 97.4. The average is a 70. That's a high average. Higher than what I expected, actually, so that's pretty good. Unfortunately, I didn't have time to finish grading exam 2, so I couldn't put it in there, and the grades were due last night, actually. So what I'll probably be doing, I'll probably finish grading exam 2 by Monday. If you desire, you can come to my office on an individual basis, and I'll recalculate your midterm grade with exam 2 included. So if you can, to do that. How did I calculate the grade? So I calculated it, say, it tells you essentially how to calculate the final grade. Of course, the final grade has the final in it, and you guys haven't taken the final yet, so I couldn't put that 20% in there. So what I've done is I've taken that 20% and added 5% to each of the other four things. So I've taken that and taken that 20% away and added 5% to each one of these. So that's exactly how I calculated your grade. Of course, this gives exam 1 a lot of weight because I wasn't able to finish exam 2. So 40% of your grade is exam 1, so if you did poorly on exam 1, then your grade is going to be relatively lower than what you would expect. Of course, the midterm grades aren't like official or anything like that. It's just to see where your progress is right now. So don't just take it as just like a marker of where you are. At this point in time, recall, you know, we haven't dropped any tests. We haven't dropped any quizzes. We haven't dropped any HAL assignments. We haven't curved any of that stuff except for the exam. I mean, the midterm grades, it's really not all that meaningful, if you will. It just kind of is a road marker as to where you approximately are in the class. Okay, so if you've got something lower than what you would have expected or what you were looking for, don't panic. I don't think it's a time to panic. People that have, you know, low C's on the midterm get relatively high A's, you know, by the end of the class. So I mean, it is possible. So I don't think that there, I think there are very few people in the class currently that really need to be panicking about anything and most of those aren't here. So most of the people here are probably fine. So that being said, don't, you know, if you've got a higher grade than what you were expecting or something like that, don't put your chemistry to the back burner and, you know, neglect it because just like you can get a better grade, you can also get a much worse grade, okay? So whatever it is, like I said, I just neglected to release them before class. So they're turned in, but if somebody will remind me after class, which I'm sure a lot of you will do, we'll just release them. Any questions on that? Any questions? Okay. You can find that grading policy, of course, if you go to the home page and then go into the course administration folder. It's right there, just with all the other courses. Okay, so I believe we ended on talking about concentration. Remember, concentration is the amount of solute that is dissolved in a given amount of solution. So the amount of solute, this could be like moles, moles of solute or grams of solute. Oh, and I have some blank copies of the exams up here, exam A and exam B if you want to take one home. So this could be moles of solute, grams of solute. It depends really what concentration unit the problem is asking for, whether you're going to be using grams, moles, or the different units. But it's always going to be amount of solute and given amount of solvent or solution. Again, the concentration of the solution has an effect both physically and chemically. And I believe we talked about that at the end of last class. So recall the chemical equation tells us the number of moles, so this is the mole ratio, right? So if this one's relatively simple, sodium chloride, the number of moles of sodium chloride will equal the number of moles of sodium and the number of moles of chlorine. But let's look at something a little more, takes a little more work to figure out. So magnesium chloride, of course, magnesium has a two plus oxidation state and chlorine has a minus one. So you're going to have a, you know, more elaborate formula than with sodium chloride. So when we dissolve this into water, what do we get? We get the magnesium two plus ion. So recall, if I said, well, we have 0.12 moles of magnesium chloride, how many moles of chloride ions do we have? This is going to give us the ratio of moles to moles. So we're looking for chloride relative to magnesium chloride. So for every one mole, magnesium chloride is going to equal two moles. If I cancel out the moles of magnesium chloride, we've got to put moles of magnesium chloride of CO minus ion. So recall that you can do that with the chemical equation. So what you'll find is not only is it going to be moles to moles, but when we look at this next concentration here, this is the first concentration that we're going to learn, molarity, this capital M is the symbol for molarity. So molarity equals the number of moles of solute divided by the liters of solution. So let's say we, let's keep thinking about this solution here and say we had the volume of the solution, the number of moles of magnesium chloride we had in that two liters was 0.2 moles. So let's figure out the molarity here, the number of moles over the volume and liter. So the number of moles was, and all of these numbers are just going out of my head. Okay, so don't be worried where I'm getting my capital M molarity. So we say this is the, we could say also this also equals what? 6.0 times 10 to the negative 2. This would probably be the way you'd see it written on a bottle. And we'd say molar. So recall this chemical equation up here. It all is going to come back to this chemical equation. Okay, so if we say, if we have a solution that's 6.0 times 10 to the negative second molar of magnesium chloride and I ask, well, what would be the concentration of chloride ions? You can use this chemical equation just like we did here. Okay, so let's do that. Let's make sure that we can do that. Can I raise this part? Everybody's got that? So we said our molarity or our concentration, but we wanted to know when you put brackets around something, that means the concentration and molarity. It's just a symbolism that we use in chemistry. So we wanted to know the concentration of the Cl minus ion. Take what we know, the concentration of the magnesium chloride is exactly the same procedure. Okay, but instead of using this conversion factor, one mol of magnesium chloride to two mol of chloride ions, we use one molar of magnesium chloride equals two molar of chloride. So again, just keep referencing back to the chemical equations. It'll tell you everything you need. So here up there, two molar of Cl minus chloride cancels out. And we take this, multiply by that. So again, it's this simple factor of one to two that you're using. Okay, and that's coming directly from the chemical equation. So don't forget what you learned, what was it in chapter three or two or four or whatever it was, when you learned about the chemical equation and being able to convert from looking at one particle in the chemical equation relative to another particle in the chemical equation. And if you didn't grasp that back then, make sure you go back to it and learn it because you'll be using this quite a bit. So anyways, here's some sample problems for you to do. You could take the number of grams. Of course, if you wanted to calculate the molarity, you would need the number of moles. So you would need to take grams, multiply it, or divide it by the molar mass, of course, and notice here this is also in milliliters. So you're going to have to convert that to liters because this is liters of solution. So this would be a kind of a harder molarity problem. This one would be more straightforward. I'd like you to try those on your own. Okay, so let's talk about diluting solutions, diluting concentrated solutions. So concentrated solutions, that means there's a lot of solute in the solvent recall. Concentrated solutions. Dilute solutions means there's just a little bit of solute in the solvent. So what you could imagine is if you had a concentrated solution and you added more solvent to that solution, you could dilute that solution, right? So that's what people do with like, I guess, espresso or something like that, if you can imagine. One shot of espresso is like a concentrated thing and then they dump it in a bunch of hot water or something, if you want. I guess it's called an Americano, right? Or whatever. I don't drink coffee, so I don't know, but that's a good example to think of to take something very concentrated and diluting it. So how do we figure out what the new molarity is if we dilute something? So that would be something that would be very advantageous to know, especially if you're working with solutions in the chemistry lab, okay? So let's learn how to do that. And in fact, up here is the slide that talks about it, so let's refer to the slide real quick. So there's an equation, a simple equation. So it's very similar to Boyle's law, if you recall that from the gas law chapter, gas law chapter, chapter five. You can say the initial molarity times the initial volume equals the final molarity times the final volume. And let's just make up a problem. Let's say we've got that magnesium chloride solution again, but let's say, well, what was the final concentration of it? So we had molarity final equal, what, one point? Oh, that was for chloride, but so 6.0 times 10 to the negative 2 molar. Let's just say that was the final molarity. Okay, let's pretend it was a more concentrated solution and we diluted it to that final molarity. Let's figure out what the initial concentration was. And let's say our final volume, what did we say? It was 2 liters, 2.00 liters. And we'll say our initial volume. So if we had this information, we could figure out what the initial molarity was. So if I ask this question, I might say, we had diluted solution of magnesium chloride, the concentration being 6.0 times 10 to the negative 2 molar. We had 2 liters of this solution. This solution was a solution that was initially 20 milliliters in volume. What was the initial molarity or something like that? So let's figure that out. Yeah, so remember, you use this equation. Mi Vi equals Mf Vf. Okay, we want to isolate this Mi variable. So we divide both sides. So we've got to get rid of this thing, Vi. So we're going to divide both sides by Vi. Cancels out. Divide something by a tilt. It cancels out. So our new equation is... And then all we do is plug and chug. Okay? Well, no, actually. We've got to convert milliliters to liters. Yeah, sorry, I made this problem a little more difficult. But yeah, good job. So remember, how many milliliters are in a liter? A thousand. A thousand, yeah, a thousand. So we put 1,000 mills on the bottom, cancel out that, and one liter there. So 1, 2, 3. So Vi is going to be... So the negative 2 molar, Vf, divided by Vi, 0.0200 liters. Notice, we've got liters over liters here. Leaders will cancel, giving us what? Malarity is a concentration unit. So that's cool. That will give us what we want. So now let's just do this problem. Initially, this was a 6.0 molar solution of magnesium chloride. Pretty concentrated solution. Six moles per liter. Remember, a mole of magnesium chloride. How much is a mole of magnesium chloride? How much mass is a mole of magnesium chloride? I wouldn't really like that one. Yeah, look up on that table, right? 35 times 2, that's 70, plus 24, that's 90, about 95 grams, right? You can imagine 95 grams. Can anybody imagine that by now? Looking at, or noticing what you've been weighing on the balances at, yeah, it's a lot, right? But this is 6 times 95 grams, okay? So almost 600 grams, or what, 580 grams, right? And 20 mils. It's a lot, you know? Here's an easier problem. Why don't you guys try it on your own? Notice here, I haven't switched from the antiquated way that I used to do it, M1v1 to M2v2. One and I are the same thing, too, and that's what it's like. So I believe we talked about saturated, super saturated solutions last time. Just remember, we talked about how to super saturate a solution. We can put a bunch of solute in there and then heat it up, and then when the temperature is higher, the more solute will be able to be dissolved. So an unsaturated solution contains less than the maximum amount of solute. Saturated solution contains the exact maximum amount, and a super saturated solution contains more than can be dissolved at the current temperature. How does that happen? Well, it's an unstable solution that will eventually crash out the solutes. So remember, dynamic equilibrium, if excess solute is added to solvent, some dissolves. Let's see, I should have... Maybe I'll do a demonstration of this next time. Demonstration of dissolving salt into water, but it really does make the point of this. So you can imagine when you're dissolving salt into water, at first when you dissolve a little bit, that goes into solution right away. So what we say is the rate of dissolution is large. So it's going very fast. It's dissolving very fast. But later, you can dump too much salt in there. You keep dumping, dumping, dumping until no more salt can be dissolved into that solution. And then what happens? Just solid salt sits at the bottom of your container. What really is happening, though, is that solid salt isn't, what we say, static. It's not just staying still. The ions in that salt that's at the bottom. Some of those ions, in fact, an equal amount of those ions are going from the solid to the ionic phase, just as much of the ions are going back into the solid phase. So we call that a dynamic equilibrium. So just kind of a picture drawing of what I just said. You can imagine our pot of boiling water or something, or maybe not boiling yet. That's why there's salt at the bottom of it. And so pretend that's a lot of salt. After a certain point, you're going to have put too much salt into this solution. So there's going to be some solid at the bottom. We have ions that have been dissolved. These are also salt. I am then these ones. So what's happening is these ions, they don't always stay in solution, and these ones don't always stay in the solid phase. What you find is that these go out into solution and these go back into the solid phase. And an equal rate. So the rates are equal. So the rate of one is equal to the negative rate of the other. So they're doing opposite things. This is known. This happens a lot in chemistry when you've got phases, different phases, interacting with each other. This is called the dynamic opposite things. So do you think about me running just as fast as somebody else is running backwards? You want to think of it kind of like that. Like they're going the opposite direction. That's me, so they'd be negative with relative to me. Okay, we talked about solid solutes. Solid solutes increase their solubility when the temperature is raised of the solution. The opposite is true of gaseous solutes. Gaseous solutes, their solubility decreases when the temperature is increased. Just like our example last time of Coke sitting out all night. You crack open a Coke and drink one sip of it and fall asleep, sit out all night, and what happens, right? The next morning you go to drink some of it and it's all flat. So that's because, of course, before I was in the refrigerator and being cold, so all that carbon dioxide was dissolved into that solution. As you let it sit out overnight, it increased the temperature, not to mention you had the can open, so it was all being able to go out. But it increased the temperature, thus decreasing the solubility of that carbon dioxide, which we all know as a gas, and let it all escape. Okay, therefore your thing is flat. So let's talk about a new concentration unit. Very similar to molarity. It's called molarity. Instead of moles for a leader, they even have a very similar symbol. Like, remember, molarity is big and molality is little. So molality is the moles of solutes. So just like molarity, moles of solute over kilograms of solvent. Okay, so if you want to compare that, so for molarity we're talking about the solution, molality we're talking about specifically the solvent. Okay, so you've got to watch that. So let's see if we can do this one together. Calculate the molality of a KCL solution containing 52.3 grams of KCL and 301 kilograms of water. So the moles of solute here, do we know that from the problem? Does the problem tell us the moles of solute? Yeah, you've got to take the grams, change it to moles, right? So we don't know that yet, but we do know the mass of the solute equals 52. How can we go from here to here? What else do we know about the solute? No, no, what else do we know about the solute? How do we go from there to get this answer? The molar mass, right? Yeah, the molar mass. We know that. How do we know that? Because the periodic table tells us, right? Okay, so let's calculate that molar mass. 35.45, I'll tell you. And like was mentioned earlier, it gives us the mass of the solvent, which is what else we need, right? And it's 3.01 kilograms. Okay, and conveniently enough, it gave it to us in kilograms so we don't have to convert something else to kilograms. So molality, now just plug in your numbers and solve the problem. Two moles divided by... So the cool thing is, is all that stuff you learned in the other chapters, it just keeps coming back and back and back. So you've already learned everything you need to know. So if you already learned it, then you're like free-sailing. You just need to learn these new equations, okay? And just plug them in. Plug these numbers in, okay? So this is why chemistry is kind of comprehensive because it just keeps going back in circles, going back in circles, going back in circles. I'm getting a little bit harder every time, okay? I'm looking at analyzing a new thing every time. So, of course, those aren't the only concentrations that you will see. Those concentrations actually are more chemistry-oriented concentrations, okay? So for those of you who are looking to work in, you know, the various health bills or whatever probably won't see very much molar concentration and molal concentration, but probably see more of percent weight volume concentration. In fact, if you look, go look at your medicines at home. A lot of medicines that are in what we would call liquid form are just actually an aqueous solution. And it will say, you know, this is a 5.15% solution of X or whatever, okay? So you can go and look at them. I'll say that. Like, go look at your peptobismol or something. It's like, you know, bismuth chloride, some solution of bismuth chloride. I don't know. Just all of them. It starts really becoming, I don't know, you can actually start seeing chemistry within everyday society by using these concentrations, and hopefully by now you're able to start noticing more and more about this glass in your everyday life. And so you always want to keep thinking about, too, that chemistry really is everything. So if you ever thought that chemistry never affected your life, it really affects everything about your life. In fact, even you is chemistry, you know? You are like some solution, honestly, you know? We're all, I don't know, 70% water solutions. But anyways, let's look at percent weight volume. Let's see, do we have? Yeah. So remember it's going to be grams of solute over milliliters of solution. So again, this is very similar to the concentration units that we learned before, molarity. Instead of moles, it's grams. And instead of liters, it's milliliters, okay? And then you multiply by 100. So let's see. Let's try this one here. Calculate the number of grams of glucose and 750 milliliters of a 15% solution. So what did we say? Percent weight volume equals, what was this, grams of solute, right? So what do we know? Let's look at this bottom one here. We know the volume of the solution, right? And it mills conveniently. So 750. What else do we know? Do we know the grams? No. We need to find the mass. In fact, that's what we're looking for.