Lecture 24. Review Final Part I
Potma, Eric
UCI Chem 1P is a preparation go General Chemistry that covers: units of measurement, dimensional analysis, significant figures; elementary concepts of volume, mass, force, pressure, energy, density, temperature, heat, work; fundamentals of atomic and molecular structure; the mole concept, stoichiometry; properties of the states of matter; gas laws; solutions concentrations. Course may be offered online. Slides:
 Dr. Okay, here's friends. Today we are going to start with you on the material. And for the next week, we are special people finally. And that means, of course, you are going to show all the educated representatives how great the person you are. Time to get those names, right? But let me give you some information about our exam that may be helpful. So first, this class is evaluated. Every class that does the research is evaluated through this system. I think you have a couple of invitations to fill out the questionnaire. It's rather long, I mean some of them. But it's actually very useful and these evaluations are taken quite seriously. In fact, every time my performance is evaluated by my department chair, we have all these stats, all the comments that are made by students. And this course, and it doesn't look good, it's kind of a smacky. So it's a serious thought, it's not something that should be taken lightly. So fill out your impression, if you like it or don't like it, you can just fill out what you thought and your experience was. Okay, in addition, next week Laura, one of the TAs, has kind of generously offered to do a review session. Next, we, of course, are now official classes, so this is an extra review session in this auditorium, thanks for this place. It's at seven o'clock on December 10th, so she will be at your service to ask you questions, just like you film at work, to do it again. Unfortunately, on Monday I'm unable to be here, I have a very important engagement in Rochka Line and Tooth University, I can't never do it myself, that's for Laura, just stepping in and she's having a whole class event. Alright, so December 14th, it's on a Friday, it starts at 1.30. Like I said, the exam will be two hours in total, so that's much longer than the mid-furnace, and that means there's a little bit more breathing space for the protein question, and you can withhold your answers probably several times, and make sure you, you know, you get to do as best possible performance on this exam. So, while I said the more information, again, we have allocated seating, so keep an eye on the website, the class website, all the information will be there, when you post it. The practice mid-term is already been posted by the way, so you can already start looking at that. Yes? The practice exam has been posted. I just said that, and you can just go there and look, and you get an idea of how long the exam will be, what kind of questions you may expect. As I said, it's going to be slightly different, but this is a very good impression of the level of difficulty, the length of the exam. Things to bring, as usual, an index card, your calculator, cell phones, we don't like so much, so leave them in your bag, turn them off. Can we get two of these cards or two of these? Oh, very good question, people. How many index cards can you use? The answer is uno. Okay. And that is the same as before. So, I'm not, this is not because the information is here, we have two cards, one card. Okay. So, you can use that wisely. Some people write very beautifully with, you know, fine liners, they do a great job. While people just print it out on a special printer, they also do a great job. That's fine, okay? You don't have multiple liners, it's not open. That's not a lie. Okay. This is the list of topics. There's nothing new about this, I'm just going to do it for formalities sake. Okay. Everything we have talked about is going to be on final time, it's truly a cumulative kind of a Xen. There are basically a facility in Hyde Force in its entirety. And these are the things that we talked about. Basically, if you like the book as a main source, it runs from chapter one to chapter four. Okay. That includes the following topics. Dealing with numbers. Okay. SIGFIX, units, prefixes and so forth. Dealing with numbers, writing numbers in scientific notation is part of this class. So, we should do that on the exam. Even if it's not explicitly stated, we should do that because it's part of this exam. Unit conversions are ultimately important, converting one unit into another unit. Seven meters into meters, but also density into mass, or, for instance, inches into kilometers. Those kind of concerns we have had many times over. And be sure to be able to do that on the exam. Adam is an isotopes. We talked about the basic architecture of an atom, isotopes, ions and so forth. And we should know something about the orbital and where the electrons are. Remember, how many electrons a certain atom has. How many protons, how many neutrons, because the basic architecture of an atom is something that's part of this exam. As is, naming these atoms, making molecules from them, and naming their ions probably. Knowing the charges of the ions, and knowing the difference between a polyethylene and anion, and an elemental anion, for instance. Then, we started to do some calculations with molecules, atoms and ions. Namely, we started to determine the molar mass of these guys, and mass percentages of elements in compounds. These calculations are part of the final. Then, we looked at chemical reactions. So we have molecules that are actually engaging into a chemical interaction, and there are reagents and products. And we should be able to balance such a reaction, and come up with the right stoichiometry. Why is the stoichiometry so important? Because we have to be able to determine mole ratios between things. Mole ratios can be used to convert one type of compound into another type of compound. It turns out, how many of you have? But we need to be able to do the mole ratios properly, to store chemical reactions, and give you the mole ratios. Molarity, and again, stoichiometry and solution. We have done that, we look at many compounds, and it dissolves in water. And then, react in a solution. Molearity is an expression of concentration, and along with it, we should be able to calculate, for instance, how many ions of a particular kind are in a solution. If I throw a luminous chloride into a part of water, you should be able to calculate the number of chlorine ions in it. If you dilute it, put a normal linear in there, you should be able to calculate the number of chlorine ions again. For calculating the concentration, those are very standard calculations that we've done. It's part of the problem. And then finally, three types of reactions that we've wrestled with, precipitation reactions, acid-base reactions, and redox reactions, as examples of very important chemical reactions. There are more types of chemical reactions, but these are three main ones. We should be familiar with, dealing with, the stoichiometry, being comfortable with the naming of compounds, but setting off for a successful 10-1-a series after this one. Okay. Now, I'm going to say a few more words in details about exam-taking, and then I'm going to say a few more words about redox, just to wrap that up. Here's a kind of a snapshot of an exam. Two years ago, there's a question about monium and terms of molarity. We're not going to look exactly at how this is answered. I'm going to tell you that, show you work, because we are creating your work. Okay. In this particular question, you're asked to calculate the volume of a solution, and there's some details to show you. But you have to do the first thing, always have to do with the term of how many moles of the stock do I have. Always, you want to know how many moles you have. If you want to know the number of moles, you can easily calculate the degree of grams, or the molarity, so that's the first step. You have to determine the number of moles. Check. That check means, if you're on the right way, that means you get points. Okay. Then, you have the number of moles, you have to do something correct with it. Okay. You have to convert out, in this case, to volume, the molarity is given, is a ratio. Then, the definition of molarity is, number of moles or volume. That means, if the volume is unknown, and the amount you are given, you calculate the volume, that's exactly what you have to do here. And that's the second check. If you know how to do that, you get points. In other words, you get points where the seps, for instance, 5 and 5, total of 10 points. The final answer, actually, is not even the most important. Okay. So, even if it's completely wrong, you still get points where doing this correctly, the methodology is very methodology. Okay. So, a lot of time for us to do that, because we think it's very important. Okay. So, we go through all the calculations, all you guys, three monotheism exams, we go through all the questions and see if there's something correct in there. The better, the more correct we will get. If you want to help us, this will help you to get more points, be clear about what you calculate and what steps you get. If you can write it down like this, what brings in this example, where I say exactly what I'm trying to do in each step, I'm going to calculate the number of moles, near and so forth, conversion to gram, I'm going to just explicitly write out what I did, then the grade, you can see, oh, look, this is very nice. One, two, three, four steps, beautifully laid out, maybe a little mistake here, but at least the person knows what he or she is doing. This greatly helps, it's great, and helps you get more points. Because if this is all messy and jumbled up, we have a very hard time giving you partial credit. So, check, check, check, we can give you more points. That's the bottom line. It's completely fine to do this first on the user blank page, first write out the way to think the answer should be, and then reproduce the answer neatly and nicely with the correct steps in the blank that we're actually working on. Just one tip, you don't have to do this, I'm just telling you, it's an option that makes life a lot easier for us. Okie dokie, keep it in mind, I'm going to wrap up the topic of redox very briefly by the following, you know, we know what redox reactions are by now, redox reactions are reactions where there's an oxidation step and a reduction step. Something is losing electrons, something is gaining electrons. Here's an example, in this particular case, I can split this into half reactions, namely, the iron is losing electrons, so that's an oxidation step. And the copper here is gaining electrons, so that is the reduction step. So I can write this, you never know what the phrase is in the following, right, forget about this sulfate, it's a spectator anion, it doesn't do anything, it doesn't rotate the anion, what so ever, everything is achely, so I can just get rid of it, and just write this. It's very easy to see in this case that, look, this guy is changing to 2+, this 2 plus copper is changing it to solar copper, which has an oxidation state zero. In other words, this is the reduction step, copper, and that is the oxidation step, both each state plays at the same time. So you can nicely see in this way of writing that the electrons lost by the iron have been eaten up by copper. Effectively, the only thing that happened here is that two electrons lost by iron have gone to copper. So there's this whole review this week, okay, Friday, as usual, we'll have a talk, it's about the reaction in all the conditions that we have considered, namely just normal reaction, reaction in solution, two flavors, acidic and basic conditions. So here are a couple of settling questions that have come up through an office hours, and I'd like to just quickly go over, hopefully it will clarify a few things. So I'm just going to see if we can get this thing up. That's a challenge. Small screen up, down. How do we get this thing to work? Okay. So let's look at the first one, because something must stay slightly differently than we have talked about in the last. And so, you should know that. So the first one is O2 and then 4 copper and then copper oxide. So you see, first of all, what settling doesn't do so nicely, which I really don't like, is there's no physical states, okay? That's not nice. Don't do the same thing on the exam. We have to write physical states, because actually, this is something completely insurmountable in settling, but we really don't know. So, in this case, what they want to do in the first equation is to write half an answer. And the problem is, copper oxide, this thing actually is a solid, and we have said, it implies that you should keep it as a solid. But settling doesn't stick to regularity. So what they say, fine, consider this as an ionic compound, and in fact, I have two ions, namely this thing is oxygen, oxide, always 2 minus. And this copper, because this is 2 minus, the 2 of them, must be copper plus. In other words, this is copper 1, oxidation state 1, this is oxide. Standard oxide is always oxidation state 2 minus. Okay, so now I can write a half reaction, because we really have to do it in this case, oxygen, O2 turns into O2 minus. And copper, which was oxygenation in zero, turns into copper plus. And they want you to balance these two gas, relative to one another. So we can see which one is the oxidation which is the reduction. This, in fact, is the reduction step, because this is zero, this is 2 minus, so it becomes more negative. It has taken up electrons, and this is the reduction step, this is the oxidation step. Indeed, this one is losing electron. So some electron has to appear here, and some electron has to appear there. So to that, I have two oxygen here, one here, so I need to put a 2 there, to balance the oxygen. So there's no need to put water there, there's already oxygen there. It's already oxygen. You just have to balance out the oxygen, then I can do it, so I have four charges there. In total, 4 minus, that means I have to balance that out. So in this case, I have to add four electrons to that side. Four minus on each side. This is the reduction step. The tie is the reduction step, it should appear on this side. Copper, the elements are already balanced, the charges are not. I have to add an electron here. So that's it. The only thing I have to do now is to make sure that the balance is balanced to one another, except that it wants to do that. You can do that by putting a 4 there, a 4 there, and a 4 there. Four electrons in each state, and this is the final result. That's the second one, funny one. Funny, funny, funny one, and not something you can expect on the midterm, I saw it on the final. So it's very much an exception, but I just want to show you so that you don't get stuck on the whole thing. The question is, if you balance this equation, okay, plus copper again, yeah, copper again, 2 plus, and it generates S4O6 to minus plus CU plus. Okay, so what I'm going to do, I'm going to split this again in two steps. I'm going to write down copper to plus, and then I'm going to copper plus, oh, this one is taken up an electron. So an electron plus this gives me that. This is the reduction step. That means the other guy is the oxidation step. Let's just put it down. Now this is indeed an oxidation step, but there's something very funny going on with this particular compound. If you try to find the oxidation state of sulfur, you are in for a big surprise. The oxidation state of the sulfur is 2.5. That doesn't make any sense. It doesn't make any sense because it's a very particular compound where there's four different solvers actually in different roles in different oxidation states. So this goes way beyond what we should know at this point, and it's also not relevant for our discussion. So I'm going to tell you what you should focus on with these kind of questions. You balance the elements. You can do that quite easily, but according to today. So now I have four solvers, six oxygens, that's the same as here, but in doing so I haven't really balanced the charges because 2 times minus 2 is minus 4. I have minus 2 here, so I need to add 2 more there. So now it's balanced. This is the oxidation reaction. It is indeed an oxidation ad reaction because I'm generating an electron plus. To balance the balance from one another, I have to multiply this whole thing by 2. 2, 2, 2. And now I can add them up. And the final solution, I can put it in here. 2, 2, 2, and that's it. That's the answer for that. Now to the final one, 2. Why not? I'd like to do this also because I'm going to show you that even though these pressures may seem laborious, if you know your stop, you can just simply write, as if you write a letter to your mom. Okay? And in that kind of case, you can solve this problem. It is straightforward adaptation of the rules of your dealt. So here this is again for magnetase, NaO, that is only going to be the RCI. So I'm going to write down for magnetase, it's changing into 2 plus. Quickly look at the oxidation states here. This is 2 plus evidently. This whole thing has to be minus 4 times minus 2 is minus 8. It's nothing to the 7. Okay? If you don't know how to do that, teach yourself how to do it. We have done it a lot of times, it's essential you are able to determine the oxidation states of this state. Namely, oxidation states that all these guys added should add up to minus 1. This is minus 2, the store of them is minus 8 plus 7 equal to minus 1. This must be plus 7. It should take you a lot of time. Later on, the algebra should take you about 10 or 15 seconds to figure it out. You don't lose time on that. So I'm going to balance this equation. What do I have to do? It happens in the city of Riga, let's say. O4, I have to add 4 waters. But I have added also 8. H must, if I do that. So I put it on the other side. Okay? And now, I have balanced all the elements, but I've not balanced the charges. I can tell you what the charges should be. I don't even have to charge the little side. I can tell you that since this is 2 plus and this is 7 plus, it needs 5 electrons. Right? It has to be 5 electrons on this side. Because it's got the 7 plus and the 2 plus. That's a check, but I can also do it like this. The whole total charge on this side is 2 plus. The total charge on this side is 8 plus. Okay? Minus 1 is 7 plus. And I need to get to 2, so I need to subtract 5. That's exactly these 5. So either way, it doesn't matter when I get there, but do this consistently. Correct. Okay, now the second one. So this is the reduction because it eats electrons. This guy moves into N of 3 minus. What's going on here? Let's look at the oscillations. This takes very quick. The whole thing is minus 3 oscillations, minus 6, with numbers plus 5. What is going on here? This thing here is the nitride, non-A-on, okay? And H-noubles X. So this whole thing here is oscillations, say, as charge minus 1. Oxidation state minus 2 times 2 is minus 4, plus 3 is my ionic energy. This is plus 3. Oxidation state of the nitride is changing here by 2. By 2. Okay? So I know 2-in-nitrons have to material here. I can already tell you that without looking at anything else. So let's see if that actually works out. First of all, there's two oxygens here. Three there, so I can add water here to balance the oxygen. The dual-cell we can use two extra hydrogen. There's already one here. There's three in total. There's plus 3 and plus. Let's look at the charges. Did I add up correctly? No charge on this side. Okay? No charge on this side. At least it has to be no charge on this side. So I have a charge minus 1, 2 electrons, minus 3, 3 plus, correct. Right? So the charge is balance of element. Now I have two reactions. This thing has 5, this thing has 2, so this falls in reaction, multiplied by 2, this whole reaction, multiplied by 5. And then you can add them up. That's it. So you can do that for yourself. And then cross out the H pluses and cross out the water. Okay, so that's just a little detour. But you see, you can just write this at waters, at H plus, balance with electrons, and this can be done relatively quick. Okay? And doing this several times, like doing the homework, is really helpful in getting comfortable with these kind of questions. Typically people do pretty well on redox on the final. So I'm pretty confident you guys will step it up. Okay. Let's rewind and go back at the very beginning. When we talk about dealing with numbers, dealing with units, one of these things is writing numbers correctly. Significance figures is the way of writing numbers correctly. And we should do that. We should just do that. So if you do that, make sure you determine the multiplication division and the division of the subtraction. They have different rules. Occupation division says you have to give the number that you are calculating the same number as the least precise number in the calculation. But that rule is different if you add and subtract, because in that case the rule holds that your answer should have the same number of decimal places as the least precise measurement in the calculation for the number with the least amount of decimal places dictates how much you have to be answered. So let me quickly give you an example of how that plays out in practice. Let's look at the following. Can't play the math of, let me put it easier here, 0.643. Well if you do that, if you go into the figure of table and you look at the molar mass of the prism and you find 24.305. This number comes from the figure of table and has five different figures. This is not the limiting factor in the calculation. If that's five, this one has three. So the final answer is that this is a multiplication. It should have also three significant figures. Okay, I think we can all do that and most of you have done that correctly on these maps. Now there's a situation where you may go and learn what to do and that happens when you first have a calculation that involves both multiplication and an addition of some sort. For instance, calculate the total mass of this much magnesium and this much chlorine. Okay, so what you have to do is, well I have an answer here already, 15.6. I do the same thing for chlorine. I take the number of moles, I multiply that by the molar mass of chlorine that is 35.453. It gives me this number here in grams. Okay, so that is the answer but this should be given in three significant figures. Because 1 to 3 is the least precise measurement in this particular question. And so the answer here is 1.61. I have two numbers, these two I have to add up. So what I do here is, at first, finish the number that I'm getting to and I have to add up. And I'm rounding this because this multiplication step is over now. And I have to make sure that when the number rules out of the multiplication it has to be the proper number that means something. It means something if it has an x-ray number, of certain figures. So what I'm doing, I'm taking this as the input in my addition. So I'm going to add this number to it. These are two results that I added up. As you can see, this is what your calculator says but you have to realize that this number here has one decimal place and hence you should also have this number in one decimal place. That is the rule for subtraction. So the answer is 70.3. It has three sig-picks, not because there are three sig-picks here but because the addition says that the final number needs to have one decimal place. So first complete the multiplication and division. Then add them up and use the rules for each operation. Okay, now let's do an example of the kind of calculation of the F-dial. Let's determine the mass percentage of each element in this particular formula, this particular molecule. So let's do that. Here's the chemical formula of this compound. From this formula I can calculate the molar mass. 14 times the molar mass of carbon, 12 times the molar mass of hydrogen and 3 times the molar mass of oxygen, which gives me the molar mass. What I do then is 14 times the molar mass of carbon divided by the total that I just calculated, times 100, and you get a percentage. What is the number of sig-picks in this answer? Well, this is the division in the multiplication, hence the least amount of sig-picks is determined in the answer and in this case that is this number right here. Okay, this one has four. This number comes from the table. If this number would have five, then this would be in five. You would have three places after a decimal point. Okay, three. But because this is in four, your answer is going to be in four sig-picks total. Hydrogen, let's say for instance that the table you get the information from has three sig-picks for this number, the molar mass of hydrogen. Then your answer should also be in three. Okay, if this was four, the answer was in four. And you get 5.307. There's one more digit in that case. Right? And the second oxygen in this case, there's four, that means the answer should be in four. But maybe this is an example that's not well-chosen. If this is four, then you get three places after the decimal point. Right? Three digits. Which is two here, two there. So it doesn't mean these calculations are independent of how many numbers. This can have three digits after a decimal point. This one can have two, this one can have two at the same time. So you don't have to kind of line them up for some kind of symmetry reason. That's not what they say. Sig-picks. Look at each calculation set. Yeah. Is 5.307 two sig-picks? No, this zero is after a decimal point. This is significant. Yeah. Conversions evolve. You have to be able to construct or write unit factors for converting one unit into something else. And one of the most, you know, I would say, involved calculations are calculations of volume. Because you have to multiply 90 to the power three. Okay? So for instance, here you have a box. Here's a calculator of volume. The box is given in inches, as you can see. But I have to give my answer in liters. Everything okay? So liters is decimier cubed. Right? So the definition of one liter equals one decimier cubed. So I have to go from this, an inch, into something that is method. And so you want to go from inches to centimeters and centimeters to decimiers and then decimier cubed. So what we can do this in a variety of ways, this is one way to do it. These are the two definitions that you need. The relation between inches and centimeters and the relation between liter and decimier cubed. So, for instance, I can convert each length separately. That's a possibility. This is 1.5 inch. I can convert it to centimeters by multiplying by this unit factor. One inch, 2.45 centimeters. This one goes on top, this one at the bottom. Inches is crossing out. I get my answer in centimeters. Everything between parentheses here now is in centimeters. And I can do it for each of the lengths that I multiply. For instance, width, length, and depth. I do it for each one. I do each of these types. I can do a conversion on separately. That's completely fine. I could also have done it in one step, of course. I could have multiplied these numbers and then have this conversion step raised to the power of three. It's exactly the same calculation. Because, effectively, that's what you do here. One conversion, two conversion, three, all multiplication signs. You basically raise this conversion to the power of three. That's exactly the same operation. Wherever you prefer a sign, these things are equivalent. The answer is 446 centimeters cubed. Why is it cubed? Well, we have one, two, three centimeters multiplied. It becomes centimeters cubed. The next thing I have to do is relate this to decimeters. I convert that by using the conversion between centimeters and decimeters. Ten centimeters to one decimeter. One decimeter equals ten centimeters. The centimeter goes at the bottom. This one goes on top because I go from centimeters to decimeters. And, raised to the power of three, so the unit factor also goes to the power of three. That means you can convert this number now to decimeter cubed. And, one decimeter cubed equals a liter. So, you can just multiply a decimeter cubed by liter. And, this is your final answer. I'm going to put that in two stick pegs. Because this one has two stick pegs, this one too. Okay? So, be familiar with this kind of calculation. If you are only exam and you see this volume conversion, and you have no idea what to do, then you haven't prepared a solution. Okay? Will we probably do something like this on final, if you want to see if you can do this? This is how it works. This is the one way you should solve the problem. Now, throughout this class we have used different formulas. The main formulas, formulas from which you can derive everything else are equal. Density equals mass over volume. Model mass equals mass, total mass, over number of moles. Okay? That's how you calculate the model of mass. Or, you can also say, you can circle this through, right? You can basically put this mole by a key, number of moles equals total mass if I go to the model mass. That's the same. And the definition of molarity, which is the number of moles, in the number of moles of the solute, in the leaders of the solution. Almost all calculations may use either this, this or this, in one way or another, four. Okay? So, typically, something is given, two rivals are given, you calculate a third. Be it mole, be it leader or molarity, it doesn't matter. These relations can be used and rotated through to calculate any of the missing variables. Right? This is really all the math that is doing it. It's either a division or multiplication. That's it. And your relations will tell you what it is. So, don't flip things. Again, there were lots of situations on the last lecture where people simply guessed that it was a multiplication or division. If you do that, you have 50% chance of doing it. Okay? Two ways to avoid it is to probably think about these relations and you're able to tell yourself, or you can tell yourself what you have to do. If that is enough, look at the units. The units should cross out properly. Okay? If they do not, you're probably going to say. So, look at that, that your approach definitely gives you. You know what to do, of course, no issue, but if you're stuck, you feel like, I know you're good at multiplying and dividing. You have to use logic. Never ever guess, because the chances are pretty high if you make the wrong guess. Okay? This is not how we do science. The science of mechanics, if you are in a hospital and everybody says, here this person has this many molecules of his drug and you have to guess when you have to multiply, and you divide, you are in big trouble. If you guess the wrong, patient may die. Okay? You don't want that on your resume. You'll never find a job again. Okay? So, these things are things you can avoid if I just stay in calm and there is all these things on logic that are not blurry lines. They are not that. The blurry lines are only in your head. Get rid of them. Okay? An example of a quick calculation. Could be anything. I have 1.2 liters of calcium sulfate solution and a certain concentration. Now, the question could be, for instance, calculate how many moles you have or how many grams of this stuff do you have? How many grams would dissolve in this solution? How do you do that? Well, I want to know how many moles. I know, liters times molarity equals moles. Boom, boom, boom. Okay, this is something I've done so many times that I've trusted this, okay. Letters times molarity equals moles. If you don't believe that, suddenly. Okay? So, maybe I say, oh, hold on a second. I don't believe that. And you can always say, look, molarity equals moles over volume. That's the unit of molarity. Molarity is moles over volume, m over l. L is here, l is at the bottom, l crosses out. Your unit is going to be moles. That's called dimensional analysis. You can use that unit trick to figure out if you're doing the right thing. Always check out if you have time. If you know the number of moles, you can do a simple multiplication with a molar mass to find the amount of grams of calcium sulfate that would dissolve in this particular solution to get to this concentration. The number of moles times the molar mass equals your answer in grams. If you don't forget about this, look at the units. Mol here, the unit of molar mass is the amount of mole. Mole crosses out and gets to the amount of grams. All right? Another one. 1.2 liter solution is a density. It's not given in molarity, but in density. Is that a reason to mean you're panicking? Because I'm used to working with molarity and suddenly the concentration is given in density. Kilograms over liters. No? It's not a reason to depend on what's around. I can do exactly the same calculation because I can calculate how many kilograms I have in solution. Another relation of density, liter and mass. So that means that if you multiply volume with density, you should get an answer in kilograms. The unit of density is kilograms per liter. So if you multiply this with liter, L will cross out and your answer is going to be kilograms. So here I have a measure of mass. Thank you very much. And this is a measure of mass. Can be related to how many moles of that stuff are in it, for instance. Okay? All right, so this is again molarity. It was an important quantity and we should be able to do quick calculations with it. This is a typical question. This is a question that was on the midterm and the last midterm. Okay? I would say probably 50% of people did this incorrectly. And you shouldn't do that again. Okay? So this is a very standard and very practical situation. You have a certain volume of solution, 1.2 liters, and you have to add a certain amount of sodium phosphate. Actually, sodium phosphate is the main ingredient of a very important buffer system that people use in the biology lab all the time. Okay? It's also a very important buffer in medical sciences. So this is not a ridiculous question. Five grams of that stuff dissolved in 1.5 liter. Your advisor wants to know what is the molarity of that? Okay? The molarity of sodium ions in 1.5. Okay, so you have to realize that, look, sodium phosphate is this. Phosphate is 3 minus. You need 3 Na plus to create a neutral compound. So Na3PO4, that is sodium phosphate. If you dissolve that in water, what you get is 3 sodium ions, 1 phosphate, and enough. In other words, for each one mole of this, you generate 3 moles of sodium. So, 5 grams divided by the molar mass of sodium phosphate, okay, gives you the number of moles of sodium phosphate. This is the amount of sodium phosphate in moles that has been dissolved in 1.2 liters. But it does not tell you what the concentration of sodium ions are. Yes. So, you can take this and divide it by liters, that gives you the molarity of sodium phosphate. But you have 3 moles of sodium ions for each one mole of the compound. So you have to multiply that by 3. That 3 should not come as a total mystery to you. Okay? This is not a mystery. There's no other way this can be. Because there's 3 sodium ions for one mole of the compound. You have to multiply that by 3. You get 3 times as much sodium as you have this, as you have sodium phosphate. So, the answer is in this case 7.2 times 7 to the minus 2. Give your answer to the correct number of symphics, preferably in scientific notation, and with a unit. Please, add a unit so we know what you are talking about. Molts per liter is basically a unit for molarity. So you have to put a big end there. Okay. 2 more minutes. Naming compounds is something we have to be able to do. Here's one. What is that? A rillium sulfide. Yep. By the way, rillium is 2 plus. Sulfide is always 2 minus. Good to know. What is this? Phosphorous tribromide. Tribromide, because this is a cobalt compound, a molecular compound. This is not a magnetic compound. You have to tell the reader exactly how many anion's you have. In this case. How about this? Ammonium hydroxide. Yes, ammonium hydroxide. This is the ammonium cation. If the H plus splits up, you get NH3, which is a molecule, that's the ammonium molecule. The ammonium molecule we have seen many times. Ammonium is a molecule that you add a proton to it, becomes ammonium. N02, what is that? Nitrogen dioxide. Nitrogen dioxide is not nitrite. Why not? Because there is no minus there. Polyatomic anions always have their charges indicated. Always. Some people ask me, if I write down the charge of the anions, the answer is, yes! Is that clear? Yes! Always add the charge, the charge of the ion. If you don't do that, you just write down something else. For instance, if you forget the, if this was nitrite, you forget the charge, it becomes nitrogen dioxide. This is completely different. What is this? Nitrogen, nitride, sodium nitride, because this is an anion that has three negative charges, right? Let's do one more. What's that? Aluminum oxide. Yes, is aluminum oxide, do I have to say, do I have to stipulate oxidation state of aluminum? No, you don't. You don't. Because aluminum is a type one metal, not a type two. It only has one oxidation state, which is 3 plus ammonium oxide. All right, see you on Friday for the second session of the review.
 Dr.  Okay, here's friends.  Today we are going to start with you on the material.  And for the next week, we are special people finally.  And that means, of course, you are going to show all the educated representatives how great the person you are.  Time to get those names, right?  But let me give you some information about our exam that may be helpful.  So first, this class is evaluated.  Every class that does the research is evaluated through this system.  I think you have a couple of invitations to fill out the questionnaire.  It's rather long, I mean some of them.  But it's actually very useful and these evaluations are taken quite seriously.  In fact, every time my performance is evaluated by my department chair,  we have all these stats, all the comments that are made by students.  And this course, and it doesn't look good, it's kind of a smacky.  So it's a serious thought, it's not something that should be taken lightly.  So fill out your impression, if you like it or don't like it,  you can just fill out what you thought and your experience was.  Okay, in addition, next week Laura, one of the TAs,  has kind of generously offered to do a review session.  Next, we, of course, are now official classes,  so this is an extra review session in this auditorium, thanks for this place.  It's at seven o'clock on December 10th, so she will be at your service  to ask you questions, just like you film at work, to do it again.  Unfortunately, on Monday I'm unable to be here, I have a very important engagement  in Rochka Line and Tooth University, I can't never do it myself,  that's for Laura, just stepping in and she's having a whole class event.  Alright, so December 14th, it's on a Friday, it starts at 1.30.  Like I said, the exam will be two hours in total, so that's much longer than the mid-furnace,  and that means there's a little bit more breathing space for the protein question,  and you can withhold your answers probably several times,  and make sure you, you know, you get to do as best possible performance on this exam.  So, while I said the more information, again, we have allocated seating,  so keep an eye on the website, the class website, all the information will be there,  when you post it. The practice mid-term is already been posted by the way,  so you can already start looking at that. Yes?  The practice exam has been posted.  I just said that, and you can just go there and look,  and you get an idea of how long the exam will be, what kind of questions you may expect.  As I said, it's going to be slightly different, but this is a very good impression  of the level of difficulty, the length of the exam.  Things to bring, as usual, an index card, your calculator, cell phones,  we don't like so much, so leave them in your bag, turn them off.  Can we get two of these cards or two of these?  Oh, very good question, people.  How many index cards can you use? The answer is uno.  Okay. And that is the same as before.  So, I'm not, this is not because the information is here, we have two cards, one card.  Okay. So, you can use that wisely.  Some people write very beautifully with, you know, fine liners, they do a great job.  While people just print it out on a special printer, they also do a great job.  That's fine, okay? You don't have multiple liners, it's not open.  That's not a lie.  Okay. This is the list of topics.  There's nothing new about this, I'm just going to do it for formalities sake.  Okay. Everything we have talked about is going to be on final time,  it's truly a cumulative kind of a Xen.  There are basically a facility in Hyde Force in its entirety.  And these are the things that we talked about.  Basically, if you like the book as a main source, it runs from chapter one to chapter four.  Okay. That includes the following topics.  Dealing with numbers. Okay.  SIGFIX, units, prefixes and so forth.  Dealing with numbers, writing numbers in scientific notation is part of this class.  So, we should do that on the exam.  Even if it's not explicitly stated, we should do that because it's part of this exam.  Unit conversions are ultimately important, converting one unit into another unit.  Seven meters into meters, but also density into mass,  or, for instance, inches into kilometers.  Those kind of concerns we have had many times over.  And be sure to be able to do that on the exam.  Adam is an isotopes. We talked about the basic architecture of an atom, isotopes, ions and so forth.  And we should know something about the orbital and where the electrons are.  Remember, how many electrons a certain atom has.  How many protons, how many neutrons, because the basic architecture of an atom is something that's part of this exam.  As is, naming these atoms, making molecules from them, and naming their ions probably.  Knowing the charges of the ions, and knowing the difference between a polyethylene and anion, and an elemental anion, for instance.  Then, we started to do some calculations with molecules, atoms and ions.  Namely, we started to determine the molar mass of these guys, and mass percentages of elements in compounds.  These calculations are part of the final.  Then, we looked at chemical reactions.  So we have molecules that are actually engaging into a chemical interaction, and there are reagents and products.  And we should be able to balance such a reaction, and come up with the right stoichiometry.  Why is the stoichiometry so important?  Because we have to be able to determine mole ratios between things.  Mole ratios can be used to convert one type of compound into another type of compound.  It turns out, how many of you have?  But we need to be able to do the mole ratios properly, to store chemical reactions, and give you the mole ratios.  Molarity, and again, stoichiometry and solution.  We have done that, we look at many compounds, and it dissolves in water.  And then, react in a solution.  Molearity is an expression of concentration, and along with it, we should be able to calculate, for instance, how many ions of a particular kind are in a solution.  If I throw a luminous chloride into a part of water, you should be able to calculate the number of chlorine ions in it.  If you dilute it, put a normal linear in there, you should be able to calculate the number of chlorine ions again.  For calculating the concentration, those are very standard calculations that we've done.  It's part of the problem.  And then finally, three types of reactions that we've wrestled with, precipitation reactions, acid-base reactions, and redox reactions,  as examples of very important chemical reactions. There are more types of chemical reactions, but these are three main ones.  We should be familiar with, dealing with, the stoichiometry, being comfortable with the naming of compounds,  but setting off for a successful 10-1-a series after this one.  Okay.  Now, I'm going to say a few more words in details about exam-taking, and then I'm going to say a few more words about redox,  just to wrap that up.  Here's a kind of a snapshot of an exam.  Two years ago, there's a question about monium and terms of molarity.  We're not going to look exactly at how this is answered.  I'm going to tell you that, show you work, because we are creating your work.  Okay.  In this particular question, you're asked to calculate the volume of a solution, and there's some details to show you.  But you have to do the first thing, always have to do with the term of how many moles of the stock do I have.  Always, you want to know how many moles you have.  If you want to know the number of moles, you can easily calculate the degree of grams, or the molarity, so that's the first step.  You have to determine the number of moles. Check.  That check means, if you're on the right way, that means you get points.  Okay.  Then, you have the number of moles, you have to do something correct with it.  Okay. You have to convert out, in this case, to volume, the molarity is given, is a ratio.  Then, the definition of molarity is, number of moles or volume.  That means, if the volume is unknown, and the amount you are given, you calculate the volume, that's exactly what you have to do here.  And that's the second check.  If you know how to do that, you get points.  In other words, you get points where the seps, for instance, 5 and 5, total of 10 points.  The final answer, actually, is not even the most important.  Okay.  So, even if it's completely wrong, you still get points where doing this correctly, the methodology is very methodology.  Okay. So, a lot of time for us to do that, because we think it's very important.  Okay. So, we go through all the calculations, all you guys, three monotheism exams, we go through all the questions and see if there's something correct in there.  The better, the more correct we will get. If you want to help us, this will help you to get more points, be clear about what you calculate and what steps you get.  If you can write it down like this, what brings in this example, where I say exactly what I'm trying to do in each step, I'm going to calculate the number of moles,  near and so forth, conversion to gram, I'm going to just explicitly write out what I did, then the grade, you can see, oh, look, this is very nice.  One, two, three, four steps, beautifully laid out, maybe a little mistake here, but at least the person knows what he or she is doing.  This greatly helps, it's great, and helps you get more points.  Because if this is all messy and jumbled up, we have a very hard time giving you partial credit.  So, check, check, check, we can give you more points. That's the bottom line.  It's completely fine to do this first on the user blank page, first write out the way to think the answer should be, and then reproduce the answer neatly and nicely with the correct steps in the blank that we're actually working on.  Just one tip, you don't have to do this, I'm just telling you, it's an option that makes life a lot easier for us.  Okie dokie, keep it in mind, I'm going to wrap up the topic of redox very briefly by the following, you know, we know what redox reactions are by now, redox reactions are reactions where there's an oxidation step and a reduction step.  Something is losing electrons, something is gaining electrons.  Here's an example, in this particular case, I can split this into half reactions, namely, the iron is losing electrons, so that's an oxidation step.  And the copper here is gaining electrons, so that is the reduction step.  So I can write this, you never know what the phrase is in the following, right, forget about this sulfate, it's a spectator anion, it doesn't do anything, it doesn't rotate the anion, what so ever, everything is achely, so I can just get rid of it, and just write this.  It's very easy to see in this case that, look, this guy is changing to 2+, this 2 plus copper is changing it to solar copper, which has an oxidation state zero.  In other words, this is the reduction step, copper, and that is the oxidation step, both each state plays at the same time.  So you can nicely see in this way of writing that the electrons lost by the iron have been eaten up by copper.  Effectively, the only thing that happened here is that two electrons lost by iron have gone to copper.  So there's this whole review this week, okay, Friday, as usual, we'll have a talk, it's about the reaction in all the conditions that we have considered, namely just normal reaction, reaction in solution, two flavors, acidic and basic conditions.  So here are a couple of settling questions that have come up through an office hours, and I'd like to just quickly go over, hopefully it will clarify a few things.  So I'm just going to see if we can get this thing up.  That's a challenge.  Small screen up, down.  How do we get this thing to work?  Okay.  So let's look at the first one, because something must stay slightly differently than we have talked about in the last.  And so, you should know that.  So the first one is O2 and then 4 copper and then copper oxide.  So you see, first of all, what settling doesn't do so nicely, which I really don't like, is there's no physical states, okay?  That's not nice.  Don't do the same thing on the exam.  We have to write physical states, because actually, this is something completely insurmountable in settling, but we really don't know.  So, in this case, what they want to do in the first equation is to write half an answer.  And the problem is, copper oxide, this thing actually is a solid, and we have said, it implies that you should keep it as a solid.  But settling doesn't stick to regularity.  So what they say, fine, consider this as an ionic compound, and in fact, I have two ions, namely this thing is oxygen, oxide, always 2 minus.  And this copper, because this is 2 minus, the 2 of them, must be copper plus.  In other words, this is copper 1, oxidation state 1, this is oxide.  Standard oxide is always oxidation state 2 minus.  Okay, so now I can write a half reaction, because we really have to do it in this case, oxygen, O2 turns into O2 minus.  And copper, which was oxygenation in zero, turns into copper plus.  And they want you to balance these two gas, relative to one another.  So we can see which one is the oxidation which is the reduction.  This, in fact, is the reduction step, because this is zero, this is 2 minus, so it becomes more negative.  It has taken up electrons, and this is the reduction step, this is the oxidation step.  Indeed, this one is losing electron.  So some electron has to appear here, and some electron has to appear there.  So to that, I have two oxygen here, one here, so I need to put a 2 there, to balance the oxygen.  So there's no need to put water there, there's already oxygen there.  It's already oxygen.  You just have to balance out the oxygen, then I can do it, so I have four charges there.  In total, 4 minus, that means I have to balance that out.  So in this case, I have to add four electrons to that side.  Four minus on each side.  This is the reduction step.  The tie is the reduction step, it should appear on this side.  Copper, the elements are already balanced, the charges are not.  I have to add an electron here.  So that's it.  The only thing I have to do now is to make sure that the balance is balanced to one another, except that it wants to do that.  You can do that by putting a 4 there, a 4 there, and a 4 there.  Four electrons in each state, and this is the final result.  That's the second one, funny one.  Funny, funny, funny one, and not something you can expect on the midterm, I saw it on the final.  So it's very much an exception, but I just want to show you so that you don't get stuck on the whole thing.  The question is, if you balance this equation, okay, plus copper again, yeah, copper again, 2 plus,  and it generates S4O6 to minus plus CU plus.  Okay, so what I'm going to do, I'm going to split this again in two steps.  I'm going to write down copper to plus, and then I'm going to copper plus, oh, this one is taken up an electron.  So an electron plus this gives me that.  This is the reduction step.  That means the other guy is the oxidation step.  Let's just put it down.  Now this is indeed an oxidation step, but there's something very funny going on with this particular compound.  If you try to find the oxidation state of sulfur, you are in for a big surprise.  The oxidation state of the sulfur is 2.5.  That doesn't make any sense.  It doesn't make any sense because it's a very particular compound where there's four different solvers actually in different roles in different oxidation states.  So this goes way beyond what we should know at this point, and it's also not relevant for our discussion.  So I'm going to tell you what you should focus on with these kind of questions.  You balance the elements.  You can do that quite easily, but according to today.  So now I have four solvers, six oxygens, that's the same as here, but in doing so I haven't really balanced the charges because 2 times minus 2 is minus 4.  I have minus 2 here, so I need to add 2 more there.  So now it's balanced.  This is the oxidation reaction.  It is indeed an oxidation ad reaction because I'm generating an electron plus.  To balance the balance from one another, I have to multiply this whole thing by 2.  2, 2, 2.  And now I can add them up.  And the final solution, I can put it in here.  2, 2, 2, and that's it.  That's the answer for that.  Now to the final one, 2.  Why not?  I'd like to do this also because I'm going to show you that even though these pressures may seem laborious, if you know your stop, you can just simply write, as if you write a letter to your mom.  Okay?  And in that kind of case, you can solve this problem.  It is straightforward adaptation of the rules of your dealt.  So here this is again for magnetase, NaO, that is only going to be the RCI.  So I'm going to write down for magnetase, it's changing into 2 plus.  Quickly look at the oxidation states here.  This is 2 plus evidently.  This whole thing has to be minus 4 times minus 2 is minus 8.  It's nothing to the 7.  Okay? If you don't know how to do that, teach yourself how to do it.  We have done it a lot of times, it's essential you are able to determine the oxidation states of this state.  Namely, oxidation states that all these guys added should add up to minus 1.  This is minus 2, the store of them is minus 8 plus 7 equal to minus 1.  This must be plus 7.  It should take you a lot of time.  Later on, the algebra should take you about 10 or 15 seconds to figure it out.  You don't lose time on that.  So I'm going to balance this equation.  What do I have to do?  It happens in the city of Riga, let's say.  O4, I have to add 4 waters.  But I have added also 8.  H must, if I do that.  So I put it on the other side.  Okay?  And now, I have balanced all the elements, but I've not balanced the charges.  I can tell you what the charges should be.  I don't even have to charge the little side.  I can tell you that since this is 2 plus and this is 7 plus, it needs 5 electrons.  Right?  It has to be 5 electrons on this side.  Because it's got the 7 plus and the 2 plus.  That's a check, but I can also do it like this.  The whole total charge on this side is 2 plus.  The total charge on this side is 8 plus.  Okay?  Minus 1 is 7 plus.  And I need to get to 2, so I need to subtract 5.  That's exactly these 5.  So either way, it doesn't matter when I get there, but do this consistently.  Correct.  Okay, now the second one.  So this is the reduction because it eats electrons.  This guy moves into N of 3 minus.  What's going on here?  Let's look at the oscillations.  This takes very quick.  The whole thing is minus 3 oscillations, minus 6, with numbers plus 5.  What is going on here?  This thing here is the nitride, non-A-on, okay?  And H-noubles X.  So this whole thing here is oscillations, say, as charge minus 1.  Oxidation state minus 2 times 2 is minus 4, plus 3 is my ionic energy.  This is plus 3.  Oxidation state of the nitride is changing here by 2.  By 2.  Okay?  So I know 2-in-nitrons have to material here.  I can already tell you that without looking at anything else.  So let's see if that actually works out.  First of all, there's two oxygens here.  Three there, so I can add water here to balance the oxygen.  The dual-cell we can use two extra hydrogen.  There's already one here.  There's three in total.  There's plus 3 and plus.  Let's look at the charges.  Did I add up correctly?  No charge on this side.  Okay? No charge on this side.  At least it has to be no charge on this side.  So I have a charge minus 1, 2 electrons, minus 3, 3 plus, correct.  Right?  So the charge is balance of element.  Now I have two reactions.  This thing has 5, this thing has 2, so this falls in reaction,  multiplied by 2, this whole reaction, multiplied by 5.  And then you can add them up.  That's it.  So you can do that for yourself.  And then cross out the H pluses and cross out the water.  Okay, so that's just a little detour.  But you see, you can just write this at waters, at H plus,  balance with electrons, and this can be done relatively quick.  Okay?  And doing this several times, like doing the homework,  is really helpful in getting comfortable with these kind of questions.  Typically people do pretty well on redox on the final.  So I'm pretty confident you guys will step it up.  Okay.  Let's rewind and go back at the very beginning.  When we talk about dealing with numbers, dealing with units,  one of these things is writing numbers correctly.  Significance figures is the way of writing numbers correctly.  And we should do that.  We should just do that.  So if you do that, make sure you determine  the multiplication division and the division of the subtraction.  They have different rules.  Occupation division says you have to give the number that you are calculating  the same number as the least precise number in the calculation.  But that rule is different if you add and subtract,  because in that case the rule holds that your answer should have the same number  of decimal places as the least precise measurement in the calculation  for the number with the least amount of decimal places dictates  how much you have to be answered.  So let me quickly give you an example of how that plays out in practice.  Let's look at the following.  Can't play the math of, let me put it easier here, 0.643.  Well if you do that, if you go into the figure of table  and you look at the molar mass of the prism  and you find 24.305.  This number comes from the figure of table  and has five different figures.  This is not the limiting factor in the calculation.  If that's five, this one has three.  So the final answer is that this is a multiplication.  It should have also three significant figures.  Okay, I think we can all do that  and most of you have done that correctly on these maps.  Now there's a situation where you may go and learn what to do  and that happens when you first have a calculation that involves both multiplication  and an addition of some sort.  For instance, calculate the total mass of this much magnesium  and this much chlorine.  Okay, so what you have to do is,  well I have an answer here already, 15.6.  I do the same thing for chlorine.  I take the number of moles, I multiply that by the molar mass of chlorine  that is 35.453.  It gives me this number here in grams.  Okay, so that is the answer but this should be given in three significant figures.  Because 1 to 3 is the least precise measurement in this particular question.  And so the answer here is 1.61.  I have two numbers, these two I have to add up.  So what I do here is, at first,  finish the number that I'm getting to and I have to add up.  And I'm rounding this because this multiplication step is over now.  And I have to make sure that when the number rules out of the multiplication  it has to be the proper number that means something.  It means something if it has an x-ray number,  of certain figures.  So what I'm doing, I'm taking this as the input in my addition.  So I'm going to add this number to it.  These are two results that I added up.  As you can see, this is what your calculator says  but you have to realize that this number here has one decimal place  and hence you should also have this number in one decimal place.  That is the rule for subtraction.  So the answer is 70.3.  It has three sig-picks, not because there are three sig-picks here  but because the addition says that the final number needs to have one decimal place.  So first complete the multiplication and division.  Then add them up and use the rules for each operation.  Okay, now let's do an example of the kind of calculation of the F-dial.  Let's determine the mass percentage of each element in this particular formula,  this particular molecule.  So let's do that.  Here's the chemical formula of this compound.  From this formula I can calculate the molar mass.  14 times the molar mass of carbon, 12 times the molar mass of hydrogen  and 3 times the molar mass of oxygen,  which gives me the molar mass.  What I do then is 14 times the molar mass of carbon  divided by the total that I just calculated, times 100,  and you get a percentage.  What is the number of sig-picks in this answer?  Well, this is the division in the multiplication,  hence the least amount of sig-picks is determined in the answer  and in this case that is this number right here.  Okay, this one has four.  This number comes from the table.  If this number would have five, then this would be in five.  You would have three places after a decimal point.  Okay, three.  But because this is in four, your answer is going to be in four sig-picks total.  Hydrogen, let's say for instance that the table you get the information from  has three sig-picks for this number, the molar mass of hydrogen.  Then your answer should also be in three.  Okay, if this was four, the answer was in four.  And you get 5.307.  There's one more digit in that case.  Right?  And the second oxygen in this case, there's four,  that means the answer should be in four.  But maybe this is an example that's not well-chosen.  If this is four, then you get three places after the decimal point.  Right? Three digits.  Which is two here, two there.  So it doesn't mean these calculations are independent of how many numbers.  This can have three digits after a decimal point.  This one can have two, this one can have two at the same time.  So you don't have to kind of line them up for some kind of symmetry reason.  That's not what they say.  Sig-picks.  Look at each calculation set.  Yeah.  Is 5.307 two sig-picks?  No, this zero is after a decimal point.  This is significant.  Yeah.  Conversions evolve.  You have to be able to construct or write unit factors for converting one unit into something else.  And one of the most, you know, I would say, involved calculations are calculations of volume.  Because you have to multiply 90 to the power three.  Okay?  So for instance, here you have a box.  Here's a calculator of volume.  The box is given in inches, as you can see.  But I have to give my answer in liters.  Everything okay?  So liters is decimier cubed.  Right?  So the definition of one liter equals one decimier cubed.  So I have to go from this, an inch, into something that is method.  And so you want to go from inches to centimeters and centimeters to decimiers and then decimier cubed.  So what we can do this in a variety of ways, this is one way to do it.  These are the two definitions that you need.  The relation between inches and centimeters and the relation between liter and decimier cubed.  So, for instance, I can convert each length separately.  That's a possibility.  This is 1.5 inch.  I can convert it to centimeters by multiplying by this unit factor.  One inch, 2.45 centimeters.  This one goes on top, this one at the bottom.  Inches is crossing out.  I get my answer in centimeters.  Everything between parentheses here now is in centimeters.  And I can do it for each of the lengths that I multiply.  For instance, width, length, and depth.  I do it for each one.  I do each of these types.  I can do a conversion on separately.  That's completely fine.  I could also have done it in one step, of course.  I could have multiplied these numbers and then have this conversion step raised to the power of three.  It's exactly the same calculation.  Because, effectively, that's what you do here.  One conversion, two conversion, three, all multiplication signs.  You basically raise this conversion to the power of three.  That's exactly the same operation.  Wherever you prefer a sign, these things are equivalent.  The answer is 446 centimeters cubed.  Why is it cubed?  Well, we have one, two, three centimeters multiplied.  It becomes centimeters cubed.  The next thing I have to do is relate this to decimeters.  I convert that by using the conversion between centimeters and decimeters.  Ten centimeters to one decimeter.  One decimeter equals ten centimeters.  The centimeter goes at the bottom.  This one goes on top because I go from centimeters to decimeters.  And, raised to the power of three, so the unit factor also goes to the power of three.  That means you can convert this number now to decimeter cubed.  And, one decimeter cubed equals a liter.  So, you can just multiply a decimeter cubed by liter.  And, this is your final answer.  I'm going to put that in two stick pegs.  Because this one has two stick pegs, this one too.  Okay?  So, be familiar with this kind of calculation.  If you are only exam and you see this volume conversion,  and you have no idea what to do,  then you haven't prepared a solution.  Okay?  Will we probably do something like this on final,  if you want to see if you can do this?  This is how it works.  This is the one way you should solve the problem.  Now, throughout this class we have used different formulas.  The main formulas,  formulas from which you can derive everything else are equal.  Density equals mass over volume.  Model mass equals mass, total mass, over number of moles.  Okay?  That's how you calculate the model of mass.  Or, you can also say, you can circle this through, right?  You can basically put this mole by a key,  number of moles equals total mass if I go to the model mass.  That's the same.  And the definition of molarity, which is the number of moles,  in the number of moles of the solute,  in the leaders of the solution.  Almost all calculations may use either this, this or this,  in one way or another, four.  Okay?  So, typically, something is given, two rivals are given,  you calculate a third.  Be it mole, be it leader or molarity, it doesn't matter.  These relations can be used and rotated through  to calculate any of the missing variables.  Right?  This is really all the math that is doing it.  It's either a division or multiplication.  That's it.  And your relations will tell you what it is.  So, don't flip things.  Again, there were lots of situations on the last lecture  where people simply guessed that it was a multiplication or division.  If you do that, you have 50% chance of doing it.  Okay?  Two ways to avoid it is to probably think about these relations  and you're able to tell yourself,  or you can tell yourself what you have to do.  If that is enough, look at the units.  The units should cross out properly.  Okay?  If they do not, you're probably going to say.  So, look at that, that your approach definitely gives you.  You know what to do, of course, no issue,  but if you're stuck, you feel like,  I know you're good at multiplying and dividing.  You have to use logic.  Never ever guess, because the chances are pretty high  if you make the wrong guess.  Okay?  This is not how we do science.  The science of mechanics, if you are in a hospital  and everybody says,  here this person has this many molecules of his drug  and you have to guess when you have to multiply,  and you divide, you are in big trouble.  If you guess the wrong, patient may die.  Okay?  You don't want that on your resume.  You'll never find a job again.  Okay?  So, these things are things you can avoid  if I just stay in calm  and there is all these things on logic  that are not blurry lines.  They are not that.  The blurry lines are only in your head.  Get rid of them.  Okay?  An example of a quick calculation.  Could be anything.  I have 1.2 liters of calcium sulfate solution  and a certain concentration.  Now, the question could be, for instance,  calculate how many moles you have  or how many grams of this stuff do you have?  How many grams would dissolve in this solution?  How do you do that?  Well, I want to know how many moles.  I know, liters times molarity equals moles.  Boom, boom, boom.  Okay, this is something I've done so many times  that I've trusted this, okay.  Letters times molarity equals moles.  If you don't believe that, suddenly.  Okay?  So, maybe I say, oh, hold on a second.  I don't believe that.  And you can always say, look, molarity equals moles over volume.  That's the unit of molarity.  Molarity is moles over volume, m over l.  L is here, l is at the bottom, l crosses out.  Your unit is going to be moles.  That's called dimensional analysis.  You can use that unit trick to figure out  if you're doing the right thing.  Always check out if you have time.  If you know the number of moles,  you can do a simple multiplication with a molar mass  to find the amount of grams of calcium sulfate  that would dissolve in this particular solution  to get to this concentration.  The number of moles times the molar mass  equals your answer in grams.  If you don't forget about this,  look at the units.  Mol here, the unit of molar mass is the amount of mole.  Mole crosses out and gets to the amount of grams.  All right?  Another one.  1.2 liter solution is a density.  It's not given in molarity, but in density.  Is that a reason to mean you're panicking?  Because I'm used to working with molarity  and suddenly the concentration is given in density.  Kilograms over liters.  No?  It's not a reason to depend on what's around.  I can do exactly the same calculation  because I can calculate how many kilograms I have in solution.  Another relation of density,  liter and mass.  So that means that if you multiply volume with density,  you should get an answer in kilograms.  The unit of density is kilograms per liter.  So if you multiply this with liter,  L will cross out and your answer is going to be kilograms.  So here I have a measure of mass.  Thank you very much.  And this is a measure of mass.  Can be related to how many moles of that stuff are in it, for instance.  Okay?  All right, so this is again molarity.  It was an important quantity  and we should be able to do quick calculations with it.  This is a typical question.  This is a question that was on the midterm and the last midterm.  Okay?  I would say probably 50% of people did this incorrectly.  And you shouldn't do that again.  Okay?  So this is a very standard and very practical situation.  You have a certain volume of solution, 1.2 liters,  and you have to add a certain amount of sodium phosphate.  Actually, sodium phosphate is the main ingredient  of a very important buffer system that people use in the biology lab all the time.  Okay?  It's also a very important buffer in medical sciences.  So this is not a ridiculous question.  Five grams of that stuff dissolved in 1.5 liter.  Your advisor wants to know what is the molarity of that?  Okay?  The molarity of sodium ions in 1.5.  Okay, so you have to realize that, look, sodium phosphate is this.  Phosphate is 3 minus.  You need 3 Na plus to create a neutral compound.  So Na3PO4, that is sodium phosphate.  If you dissolve that in water, what you get is 3 sodium ions,  1 phosphate, and enough.  In other words, for each one mole of this,  you generate 3 moles of sodium.  So, 5 grams divided by the molar mass of sodium phosphate,  okay, gives you the number of moles of sodium phosphate.  This is the amount of sodium phosphate in moles  that has been dissolved in 1.2 liters.  But it does not tell you what the concentration of sodium ions are.  Yes.  So, you can take this and divide it by liters,  that gives you the molarity of sodium phosphate.  But you have 3 moles of sodium ions for each one mole of the compound.  So you have to multiply that by 3.  That 3 should not come as a total mystery to you.  Okay?  This is not a mystery.  There's no other way this can be.  Because there's 3 sodium ions for one mole of the compound.  You have to multiply that by 3.  You get 3 times as much sodium as you have this,  as you have sodium phosphate.  So, the answer is in this case 7.2 times 7 to the minus 2.  Give your answer to the correct number of symphics,  preferably in scientific notation,  and with a unit.  Please, add a unit so we know what you are talking about.  Molts per liter is basically a unit for molarity.  So you have to put a big end there.  Okay.  2 more minutes.  Naming compounds is something we have to be able to do.  Here's one.  What is that?  A rillium sulfide.  Yep.  By the way, rillium is 2 plus.  Sulfide is always 2 minus.  Good to know.  What is this?  Phosphorous tribromide.  Tribromide, because this is a cobalt compound,  a molecular compound.  This is not a magnetic compound.  You have to tell the reader exactly how many anion's you have.  In this case.  How about this?  Ammonium hydroxide.  Yes, ammonium hydroxide.  This is the ammonium cation.  If the H plus splits up, you get NH3,  which is a molecule, that's the ammonium molecule.  The ammonium molecule we have seen many times.  Ammonium is a molecule that you add a proton to it,  becomes ammonium.  N02, what is that?  Nitrogen dioxide.  Nitrogen dioxide is not nitrite.  Why not?  Because there is no minus there.  Polyatomic anions always have their charges indicated.  Always.  Some people ask me,  if I write down the charge of the anions,  the answer is, yes!  Is that clear?  Yes!  Always add the charge, the charge of the ion.  If you don't do that,  you just write down something else.  For instance, if you forget the,  if this was nitrite,  you forget the charge,  it becomes nitrogen dioxide.  This is completely different.  What is this?  Nitrogen, nitride,  sodium nitride,  because this is an anion  that has three negative charges, right?  Let's do one more.  What's that?  Aluminum oxide.  Yes, is aluminum oxide,  do I have to say,  do I have to stipulate oxidation state of aluminum?  No, you don't.  You don't.  Because aluminum is a type one metal,  not a type two.  It only has one oxidation state,  which is 3 plus ammonium oxide.  All right, see you on Friday  for the second session of the review.
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Dr. Okay, here's friends. Today we are going to start with you on the material. And for the next week, we are special people finally. And that means, of course, you are going to show all the educated representatives how great the person you are. Time to get those names, right? But let me give you some information about our exam that may be helpful. So first, this class is evaluated. Every class that does the research is evaluated through this system. I think you have a couple of invitations to fill out the questionnaire. It's rather long, I mean some of them. But it's actually very useful and these evaluations are taken quite seriously. In fact, every time my performance is evaluated by my department chair, we have all these stats, all the comments that are made by students. And this course, and it doesn't look good, it's kind of a smacky. So it's a serious thought, it's not something that should be taken lightly. So fill out your impression, if you like it or don't like it, you can just fill out what you thought and your experience was. Okay, in addition, next week Laura, one of the TAs, has kind of generously offered to do a review session. Next, we, of course, are now official classes, so this is an extra review session in this auditorium, thanks for this place. It's at seven o'clock on December 10th, so she will be at your service to ask you questions, just like you film at work, to do it again. Unfortunately, on Monday I'm unable to be here, I have a very important engagement in Rochka Line and Tooth University, I can't never do it myself, that's for Laura, just stepping in and she's having a whole class event. Alright, so December 14th, it's on a Friday, it starts at 1.30. Like I said, the exam will be two hours in total, so that's much longer than the mid-furnace, and that means there's a little bit more breathing space for the protein question, and you can withhold your answers probably several times, and make sure you, you know, you get to do as best possible performance on this exam. So, while I said the more information, again, we have allocated seating, so keep an eye on the website, the class website, all the information will be there, when you post it. The practice mid-term is already been posted by the way, so you can already start looking at that. Yes? The practice exam has been posted. I just said that, and you can just go there and look, and you get an idea of how long the exam will be, what kind of questions you may expect. As I said, it's going to be slightly different, but this is a very good impression of the level of difficulty, the length of the exam. Things to bring, as usual, an index card, your calculator, cell phones, we don't like so much, so leave them in your bag, turn them off. Can we get two of these cards or two of these? Oh, very good question, people. How many index cards can you use? The answer is uno. Okay. And that is the same as before. So, I'm not, this is not because the information is here, we have two cards, one card. Okay. So, you can use that wisely. Some people write very beautifully with, you know, fine liners, they do a great job. While people just print it out on a special printer, they also do a great job. That's fine, okay? You don't have multiple liners, it's not open. That's not a lie. Okay. This is the list of topics. There's nothing new about this, I'm just going to do it for formalities sake. Okay. Everything we have talked about is going to be on final time, it's truly a cumulative kind of a Xen. There are basically a facility in Hyde Force in its entirety. And these are the things that we talked about. Basically, if you like the book as a main source, it runs from chapter one to chapter four. Okay. That includes the following topics. Dealing with numbers. Okay. SIGFIX, units, prefixes and so forth. Dealing with numbers, writing numbers in scientific notation is part of this class. So, we should do that on the exam. Even if it's not explicitly stated, we should do that because it's part of this exam. Unit conversions are ultimately important, converting one unit into another unit. Seven meters into meters, but also density into mass, or, for instance, inches into kilometers. Those kind of concerns we have had many times over. And be sure to be able to do that on the exam. Adam is an isotopes. We talked about the basic architecture of an atom, isotopes, ions and so forth. And we should know something about the orbital and where the electrons are. Remember, how many electrons a certain atom has. How many protons, how many neutrons, because the basic architecture of an atom is something that's part of this exam. As is, naming these atoms, making molecules from them, and naming their ions probably. Knowing the charges of the ions, and knowing the difference between a polyethylene and anion, and an elemental anion, for instance. Then, we started to do some calculations with molecules, atoms and ions. Namely, we started to determine the molar mass of these guys, and mass percentages of elements in compounds. These calculations are part of the final. Then, we looked at chemical reactions. So we have molecules that are actually engaging into a chemical interaction, and there are reagents and products. And we should be able to balance such a reaction, and come up with the right stoichiometry. Why is the stoichiometry so important? Because we have to be able to determine mole ratios between things. Mole ratios can be used to convert one type of compound into another type of compound. It turns out, how many of you have? But we need to be able to do the mole ratios properly, to store chemical reactions, and give you the mole ratios. Molarity, and again, stoichiometry and solution. We have done that, we look at many compounds, and it dissolves in water. And then, react in a solution. Molearity is an expression of concentration, and along with it, we should be able to calculate, for instance, how many ions of a particular kind are in a solution. If I throw a luminous chloride into a part of water, you should be able to calculate the number of chlorine ions in it. If you dilute it, put a normal linear in there, you should be able to calculate the number of chlorine ions again. For calculating the concentration, those are very standard calculations that we've done. It's part of the problem. And then finally, three types of reactions that we've wrestled with, precipitation reactions, acid-base reactions, and redox reactions, as examples of very important chemical reactions. There are more types of chemical reactions, but these are three main ones. We should be familiar with, dealing with, the stoichiometry, being comfortable with the naming of compounds, but setting off for a successful 10-1-a series after this one. Okay. Now, I'm going to say a few more words in details about exam-taking, and then I'm going to say a few more words about redox, just to wrap that up. Here's a kind of a snapshot of an exam. Two years ago, there's a question about monium and terms of molarity. We're not going to look exactly at how this is answered. I'm going to tell you that, show you work, because we are creating your work. Okay. In this particular question, you're asked to calculate the volume of a solution, and there's some details to show you. But you have to do the first thing, always have to do with the term of how many moles of the stock do I have. Always, you want to know how many moles you have. If you want to know the number of moles, you can easily calculate the degree of grams, or the molarity, so that's the first step. You have to determine the number of moles. Check. That check means, if you're on the right way, that means you get points. Okay. Then, you have the number of moles, you have to do something correct with it. Okay. You have to convert out, in this case, to volume, the molarity is given, is a ratio. Then, the definition of molarity is, number of moles or volume. That means, if the volume is unknown, and the amount you are given, you calculate the volume, that's exactly what you have to do here. And that's the second check. If you know how to do that, you get points. In other words, you get points where the seps, for instance, 5 and 5, total of 10 points. The final answer, actually, is not even the most important. Okay. So, even if it's completely wrong, you still get points where doing this correctly, the methodology is very methodology. Okay. So, a lot of time for us to do that, because we think it's very important. Okay. So, we go through all the calculations, all you guys, three monotheism exams, we go through all the questions and see if there's something correct in there. The better, the more correct we will get. If you want to help us, this will help you to get more points, be clear about what you calculate and what steps you get. If you can write it down like this, what brings in this example, where I say exactly what I'm trying to do in each step, I'm going to calculate the number of moles, near and so forth, conversion to gram, I'm going to just explicitly write out what I did, then the grade, you can see, oh, look, this is very nice. One, two, three, four steps, beautifully laid out, maybe a little mistake here, but at least the person knows what he or she is doing. This greatly helps, it's great, and helps you get more points. Because if this is all messy and jumbled up, we have a very hard time giving you partial credit. So, check, check, check, we can give you more points. That's the bottom line. It's completely fine to do this first on the user blank page, first write out the way to think the answer should be, and then reproduce the answer neatly and nicely with the correct steps in the blank that we're actually working on. Just one tip, you don't have to do this, I'm just telling you, it's an option that makes life a lot easier for us. Okie dokie, keep it in mind, I'm going to wrap up the topic of redox very briefly by the following, you know, we know what redox reactions are by now, redox reactions are reactions where there's an oxidation step and a reduction step. Something is losing electrons, something is gaining electrons. Here's an example, in this particular case, I can split this into half reactions, namely, the iron is losing electrons, so that's an oxidation step. And the copper here is gaining electrons, so that is the reduction step. So I can write this, you never know what the phrase is in the following, right, forget about this sulfate, it's a spectator anion, it doesn't do anything, it doesn't rotate the anion, what so ever, everything is achely, so I can just get rid of it, and just write this. It's very easy to see in this case that, look, this guy is changing to 2+, this 2 plus copper is changing it to solar copper, which has an oxidation state zero. In other words, this is the reduction step, copper, and that is the oxidation step, both each state plays at the same time. So you can nicely see in this way of writing that the electrons lost by the iron have been eaten up by copper. Effectively, the only thing that happened here is that two electrons lost by iron have gone to copper. So there's this whole review this week, okay, Friday, as usual, we'll have a talk, it's about the reaction in all the conditions that we have considered, namely just normal reaction, reaction in solution, two flavors, acidic and basic conditions. So here are a couple of settling questions that have come up through an office hours, and I'd like to just quickly go over, hopefully it will clarify a few things. So I'm just going to see if we can get this thing up. That's a challenge. Small screen up, down. How do we get this thing to work? Okay. So let's look at the first one, because something must stay slightly differently than we have talked about in the last. And so, you should know that. So the first one is O2 and then 4 copper and then copper oxide. So you see, first of all, what settling doesn't do so nicely, which I really don't like, is there's no physical states, okay? That's not nice. Don't do the same thing on the exam. We have to write physical states, because actually, this is something completely insurmountable in settling, but we really don't know. So, in this case, what they want to do in the first equation is to write half an answer. And the problem is, copper oxide, this thing actually is a solid, and we have said, it implies that you should keep it as a solid. But settling doesn't stick to regularity. So what they say, fine, consider this as an ionic compound, and in fact, I have two ions, namely this thing is oxygen, oxide, always 2 minus. And this copper, because this is 2 minus, the 2 of them, must be copper plus. In other words, this is copper 1, oxidation state 1, this is oxide. Standard oxide is always oxidation state 2 minus. Okay, so now I can write a half reaction, because we really have to do it in this case, oxygen, O2 turns into O2 minus. And copper, which was oxygenation in zero, turns into copper plus. And they want you to balance these two gas, relative to one another. So we can see which one is the oxidation which is the reduction. This, in fact, is the reduction step, because this is zero, this is 2 minus, so it becomes more negative. It has taken up electrons, and this is the reduction step, this is the oxidation step. Indeed, this one is losing electron. So some electron has to appear here, and some electron has to appear there. So to that, I have two oxygen here, one here, so I need to put a 2 there, to balance the oxygen. So there's no need to put water there, there's already oxygen there. It's already oxygen. You just have to balance out the oxygen, then I can do it, so I have four charges there. In total, 4 minus, that means I have to balance that out. So in this case, I have to add four electrons to that side. Four minus on each side. This is the reduction step. The tie is the reduction step, it should appear on this side. Copper, the elements are already balanced, the charges are not. I have to add an electron here. So that's it. The only thing I have to do now is to make sure that the balance is balanced to one another, except that it wants to do that. You can do that by putting a 4 there, a 4 there, and a 4 there. Four electrons in each state, and this is the final result. That's the second one, funny one. Funny, funny, funny one, and not something you can expect on the midterm, I saw it on the final. So it's very much an exception, but I just want to show you so that you don't get stuck on the whole thing. The question is, if you balance this equation, okay, plus copper again, yeah, copper again, 2 plus, and it generates S4O6 to minus plus CU plus. Okay, so what I'm going to do, I'm going to split this again in two steps. I'm going to write down copper to plus, and then I'm going to copper plus, oh, this one is taken up an electron. So an electron plus this gives me that. This is the reduction step. That means the other guy is the oxidation step. Let's just put it down. Now this is indeed an oxidation step, but there's something very funny going on with this particular compound. If you try to find the oxidation state of sulfur, you are in for a big surprise. The oxidation state of the sulfur is 2.5. That doesn't make any sense. It doesn't make any sense because it's a very particular compound where there's four different solvers actually in different roles in different oxidation states. So this goes way beyond what we should know at this point, and it's also not relevant for our discussion. So I'm going to tell you what you should focus on with these kind of questions. You balance the elements. You can do that quite easily, but according to today. So now I have four solvers, six oxygens, that's the same as here, but in doing so I haven't really balanced the charges because 2 times minus 2 is minus 4. I have minus 2 here, so I need to add 2 more there. So now it's balanced. This is the oxidation reaction. It is indeed an oxidation ad reaction because I'm generating an electron plus. To balance the balance from one another, I have to multiply this whole thing by 2. 2, 2, 2. And now I can add them up. And the final solution, I can put it in here. 2, 2, 2, and that's it. That's the answer for that. Now to the final one, 2. Why not? I'd like to do this also because I'm going to show you that even though these pressures may seem laborious, if you know your stop, you can just simply write, as if you write a letter to your mom. Okay? And in that kind of case, you can solve this problem. It is straightforward adaptation of the rules of your dealt. So here this is again for magnetase, NaO, that is only going to be the RCI. So I'm going to write down for magnetase, it's changing into 2 plus. Quickly look at the oxidation states here. This is 2 plus evidently. This whole thing has to be minus 4 times minus 2 is minus 8. It's nothing to the 7. Okay? If you don't know how to do that, teach yourself how to do it. We have done it a lot of times, it's essential you are able to determine the oxidation states of this state. Namely, oxidation states that all these guys added should add up to minus 1. This is minus 2, the store of them is minus 8 plus 7 equal to minus 1. This must be plus 7. It should take you a lot of time. Later on, the algebra should take you about 10 or 15 seconds to figure it out. You don't lose time on that. So I'm going to balance this equation. What do I have to do? It happens in the city of Riga, let's say. O4, I have to add 4 waters. But I have added also 8. H must, if I do that. So I put it on the other side. Okay? And now, I have balanced all the elements, but I've not balanced the charges. I can tell you what the charges should be. I don't even have to charge the little side. I can tell you that since this is 2 plus and this is 7 plus, it needs 5 electrons. Right? It has to be 5 electrons on this side. Because it's got the 7 plus and the 2 plus. That's a check, but I can also do it like this. The whole total charge on this side is 2 plus. The total charge on this side is 8 plus. Okay? Minus 1 is 7 plus. And I need to get to 2, so I need to subtract 5. That's exactly these 5. So either way, it doesn't matter when I get there, but do this consistently. Correct. Okay, now the second one. So this is the reduction because it eats electrons. This guy moves into N of 3 minus. What's going on here? Let's look at the oscillations. This takes very quick. The whole thing is minus 3 oscillations, minus 6, with numbers plus 5. What is going on here? This thing here is the nitride, non-A-on, okay? And H-noubles X. So this whole thing here is oscillations, say, as charge minus 1. Oxidation state minus 2 times 2 is minus 4, plus 3 is my ionic energy. This is plus 3. Oxidation state of the nitride is changing here by 2. By 2. Okay? So I know 2-in-nitrons have to material here. I can already tell you that without looking at anything else. So let's see if that actually works out. First of all, there's two oxygens here. Three there, so I can add water here to balance the oxygen. The dual-cell we can use two extra hydrogen. There's already one here. There's three in total. There's plus 3 and plus. Let's look at the charges. Did I add up correctly? No charge on this side. Okay? No charge on this side. At least it has to be no charge on this side. So I have a charge minus 1, 2 electrons, minus 3, 3 plus, correct. Right? So the charge is balance of element. Now I have two reactions. This thing has 5, this thing has 2, so this falls in reaction, multiplied by 2, this whole reaction, multiplied by 5. And then you can add them up. That's it. So you can do that for yourself. And then cross out the H pluses and cross out the water. Okay, so that's just a little detour. But you see, you can just write this at waters, at H plus, balance with electrons, and this can be done relatively quick. Okay? And doing this several times, like doing the homework, is really helpful in getting comfortable with these kind of questions. Typically people do pretty well on redox on the final. So I'm pretty confident you guys will step it up. Okay. Let's rewind and go back at the very beginning. When we talk about dealing with numbers, dealing with units, one of these things is writing numbers correctly. Significance figures is the way of writing numbers correctly. And we should do that. We should just do that. So if you do that, make sure you determine the multiplication division and the division of the subtraction. They have different rules. Occupation division says you have to give the number that you are calculating the same number as the least precise number in the calculation. But that rule is different if you add and subtract, because in that case the rule holds that your answer should have the same number of decimal places as the least precise measurement in the calculation for the number with the least amount of decimal places dictates how much you have to be answered. So let me quickly give you an example of how that plays out in practice. Let's look at the following. Can't play the math of, let me put it easier here, 0.643. Well if you do that, if you go into the figure of table and you look at the molar mass of the prism and you find 24.305. This number comes from the figure of table and has five different figures. This is not the limiting factor in the calculation. If that's five, this one has three. So the final answer is that this is a multiplication. It should have also three significant figures. Okay, I think we can all do that and most of you have done that correctly on these maps. Now there's a situation where you may go and learn what to do and that happens when you first have a calculation that involves both multiplication and an addition of some sort. For instance, calculate the total mass of this much magnesium and this much chlorine. Okay, so what you have to do is, well I have an answer here already, 15.6. I do the same thing for chlorine. I take the number of moles, I multiply that by the molar mass of chlorine that is 35.453. It gives me this number here in grams. Okay, so that is the answer but this should be given in three significant figures. Because 1 to 3 is the least precise measurement in this particular question. And so the answer here is 1.61. I have two numbers, these two I have to add up. So what I do here is, at first, finish the number that I'm getting to and I have to add up. And I'm rounding this because this multiplication step is over now. And I have to make sure that when the number rules out of the multiplication it has to be the proper number that means something. It means something if it has an x-ray number, of certain figures. So what I'm doing, I'm taking this as the input in my addition. So I'm going to add this number to it. These are two results that I added up. As you can see, this is what your calculator says but you have to realize that this number here has one decimal place and hence you should also have this number in one decimal place. That is the rule for subtraction. So the answer is 70.3. It has three sig-picks, not because there are three sig-picks here but because the addition says that the final number needs to have one decimal place. So first complete the multiplication and division. Then add them up and use the rules for each operation. Okay, now let's do an example of the kind of calculation of the F-dial. Let's determine the mass percentage of each element in this particular formula, this particular molecule. So let's do that. Here's the chemical formula of this compound. From this formula I can calculate the molar mass. 14 times the molar mass of carbon, 12 times the molar mass of hydrogen and 3 times the molar mass of oxygen, which gives me the molar mass. What I do then is 14 times the molar mass of carbon divided by the total that I just calculated, times 100, and you get a percentage. What is the number of sig-picks in this answer? Well, this is the division in the multiplication, hence the least amount of sig-picks is determined in the answer and in this case that is this number right here. Okay, this one has four. This number comes from the table. If this number would have five, then this would be in five. You would have three places after a decimal point. Okay, three. But because this is in four, your answer is going to be in four sig-picks total. Hydrogen, let's say for instance that the table you get the information from has three sig-picks for this number, the molar mass of hydrogen. Then your answer should also be in three. Okay, if this was four, the answer was in four. And you get 5.307. There's one more digit in that case. Right? And the second oxygen in this case, there's four, that means the answer should be in four. But maybe this is an example that's not well-chosen. If this is four, then you get three places after the decimal point. Right? Three digits. Which is two here, two there. So it doesn't mean these calculations are independent of how many numbers. This can have three digits after a decimal point. This one can have two, this one can have two at the same time. So you don't have to kind of line them up for some kind of symmetry reason. That's not what they say. Sig-picks. Look at each calculation set. Yeah. Is 5.307 two sig-picks? No, this zero is after a decimal point. This is significant. Yeah. Conversions evolve. You have to be able to construct or write unit factors for converting one unit into something else. And one of the most, you know, I would say, involved calculations are calculations of volume. Because you have to multiply 90 to the power three. Okay? So for instance, here you have a box. Here's a calculator of volume. The box is given in inches, as you can see. But I have to give my answer in liters. Everything okay? So liters is decimier cubed. Right? So the definition of one liter equals one decimier cubed. So I have to go from this, an inch, into something that is method. And so you want to go from inches to centimeters and centimeters to decimiers and then decimier cubed. So what we can do this in a variety of ways, this is one way to do it. These are the two definitions that you need. The relation between inches and centimeters and the relation between liter and decimier cubed. So, for instance, I can convert each length separately. That's a possibility. This is 1.5 inch. I can convert it to centimeters by multiplying by this unit factor. One inch, 2.45 centimeters. This one goes on top, this one at the bottom. Inches is crossing out. I get my answer in centimeters. Everything between parentheses here now is in centimeters. And I can do it for each of the lengths that I multiply. For instance, width, length, and depth. I do it for each one. I do each of these types. I can do a conversion on separately. That's completely fine. I could also have done it in one step, of course. I could have multiplied these numbers and then have this conversion step raised to the power of three. It's exactly the same calculation. Because, effectively, that's what you do here. One conversion, two conversion, three, all multiplication signs. You basically raise this conversion to the power of three. That's exactly the same operation. Wherever you prefer a sign, these things are equivalent. The answer is 446 centimeters cubed. Why is it cubed? Well, we have one, two, three centimeters multiplied. It becomes centimeters cubed. The next thing I have to do is relate this to decimeters. I convert that by using the conversion between centimeters and decimeters. Ten centimeters to one decimeter. One decimeter equals ten centimeters. The centimeter goes at the bottom. This one goes on top because I go from centimeters to decimeters. And, raised to the power of three, so the unit factor also goes to the power of three. That means you can convert this number now to decimeter cubed. And, one decimeter cubed equals a liter. So, you can just multiply a decimeter cubed by liter. And, this is your final answer. I'm going to put that in two stick pegs. Because this one has two stick pegs, this one too. Okay? So, be familiar with this kind of calculation. If you are only exam and you see this volume conversion, and you have no idea what to do, then you haven't prepared a solution. Okay? Will we probably do something like this on final, if you want to see if you can do this? This is how it works. This is the one way you should solve the problem. Now, throughout this class we have used different formulas. The main formulas, formulas from which you can derive everything else are equal. Density equals mass over volume. Model mass equals mass, total mass, over number of moles. Okay? That's how you calculate the model of mass. Or, you can also say, you can circle this through, right? You can basically put this mole by a key, number of moles equals total mass if I go to the model mass. That's the same. And the definition of molarity, which is the number of moles, in the number of moles of the solute, in the leaders of the solution. Almost all calculations may use either this, this or this, in one way or another, four. Okay? So, typically, something is given, two rivals are given, you calculate a third. Be it mole, be it leader or molarity, it doesn't matter. These relations can be used and rotated through to calculate any of the missing variables. Right? This is really all the math that is doing it. It's either a division or multiplication. That's it. And your relations will tell you what it is. So, don't flip things. Again, there were lots of situations on the last lecture where people simply guessed that it was a multiplication or division. If you do that, you have 50% chance of doing it. Okay? Two ways to avoid it is to probably think about these relations and you're able to tell yourself, or you can tell yourself what you have to do. If that is enough, look at the units. The units should cross out properly. Okay? If they do not, you're probably going to say. So, look at that, that your approach definitely gives you. You know what to do, of course, no issue, but if you're stuck, you feel like, I know you're good at multiplying and dividing. You have to use logic. Never ever guess, because the chances are pretty high if you make the wrong guess. Okay? This is not how we do science. The science of mechanics, if you are in a hospital and everybody says, here this person has this many molecules of his drug and you have to guess when you have to multiply, and you divide, you are in big trouble. If you guess the wrong, patient may die. Okay? You don't want that on your resume. You'll never find a job again. Okay? So, these things are things you can avoid if I just stay in calm and there is all these things on logic that are not blurry lines. They are not that. The blurry lines are only in your head. Get rid of them. Okay? An example of a quick calculation. Could be anything. I have 1.2 liters of calcium sulfate solution and a certain concentration. Now, the question could be, for instance, calculate how many moles you have or how many grams of this stuff do you have? How many grams would dissolve in this solution? How do you do that? Well, I want to know how many moles. I know, liters times molarity equals moles. Boom, boom, boom. Okay, this is something I've done so many times that I've trusted this, okay. Letters times molarity equals moles. If you don't believe that, suddenly. Okay? So, maybe I say, oh, hold on a second. I don't believe that. And you can always say, look, molarity equals moles over volume. That's the unit of molarity. Molarity is moles over volume, m over l. L is here, l is at the bottom, l crosses out. Your unit is going to be moles. That's called dimensional analysis. You can use that unit trick to figure out if you're doing the right thing. Always check out if you have time. If you know the number of moles, you can do a simple multiplication with a molar mass to find the amount of grams of calcium sulfate that would dissolve in this particular solution to get to this concentration. The number of moles times the molar mass equals your answer in grams. If you don't forget about this, look at the units. Mol here, the unit of molar mass is the amount of mole. Mole crosses out and gets to the amount of grams. All right? Another one. 1.2 liter solution is a density. It's not given in molarity, but in density. Is that a reason to mean you're panicking? Because I'm used to working with molarity and suddenly the concentration is given in density. Kilograms over liters. No? It's not a reason to depend on what's around. I can do exactly the same calculation because I can calculate how many kilograms I have in solution. Another relation of density, liter and mass. So that means that if you multiply volume with density, you should get an answer in kilograms. The unit of density is kilograms per liter. So if you multiply this with liter, L will cross out and your answer is going to be kilograms. So here I have a measure of mass. Thank you very much. And this is a measure of mass. Can be related to how many moles of that stuff are in it, for instance. Okay? All right, so this is again molarity. It was an important quantity and we should be able to do quick calculations with it. This is a typical question. This is a question that was on the midterm and the last midterm. Okay? I would say probably 50% of people did this incorrectly. And you shouldn't do that again. Okay? So this is a very standard and very practical situation. You have a certain volume of solution, 1.2 liters, and you have to add a certain amount of sodium phosphate. Actually, sodium phosphate is the main ingredient of a very important buffer system that people use in the biology lab all the time. Okay? It's also a very important buffer in medical sciences. So this is not a ridiculous question. Five grams of that stuff dissolved in 1.5 liter. Your advisor wants to know what is the molarity of that? Okay? The molarity of sodium ions in 1.5. Okay, so you have to realize that, look, sodium phosphate is this. Phosphate is 3 minus. You need 3 Na plus to create a neutral compound. So Na3PO4, that is sodium phosphate. If you dissolve that in water, what you get is 3 sodium ions, 1 phosphate, and enough. In other words, for each one mole of this, you generate 3 moles of sodium. So, 5 grams divided by the molar mass of sodium phosphate, okay, gives you the number of moles of sodium phosphate. This is the amount of sodium phosphate in moles that has been dissolved in 1.2 liters. But it does not tell you what the concentration of sodium ions are. Yes. So, you can take this and divide it by liters, that gives you the molarity of sodium phosphate. But you have 3 moles of sodium ions for each one mole of the compound. So you have to multiply that by 3. That 3 should not come as a total mystery to you. Okay? This is not a mystery. There's no other way this can be. Because there's 3 sodium ions for one mole of the compound. You have to multiply that by 3. You get 3 times as much sodium as you have this, as you have sodium phosphate. So, the answer is in this case 7.2 times 7 to the minus 2. Give your answer to the correct number of symphics, preferably in scientific notation, and with a unit. Please, add a unit so we know what you are talking about. Molts per liter is basically a unit for molarity. So you have to put a big end there. Okay. 2 more minutes. Naming compounds is something we have to be able to do. Here's one. What is that? A rillium sulfide. Yep. By the way, rillium is 2 plus. Sulfide is always 2 minus. Good to know. What is this? Phosphorous tribromide. Tribromide, because this is a cobalt compound, a molecular compound. This is not a magnetic compound. You have to tell the reader exactly how many anion's you have. In this case. How about this? Ammonium hydroxide. Yes, ammonium hydroxide. This is the ammonium cation. If the H plus splits up, you get NH3, which is a molecule, that's the ammonium molecule. The ammonium molecule we have seen many times. Ammonium is a molecule that you add a proton to it, becomes ammonium. N02, what is that? Nitrogen dioxide. Nitrogen dioxide is not nitrite. Why not? Because there is no minus there. Polyatomic anions always have their charges indicated. Always. Some people ask me, if I write down the charge of the anions, the answer is, yes! Is that clear? Yes! Always add the charge, the charge of the ion. If you don't do that, you just write down something else. For instance, if you forget the, if this was nitrite, you forget the charge, it becomes nitrogen dioxide. This is completely different. What is this? Nitrogen, nitride, sodium nitride, because this is an anion that has three negative charges, right? Let's do one more. What's that? Aluminum oxide. Yes, is aluminum oxide, do I have to say, do I have to stipulate oxidation state of aluminum? No, you don't. You don't. Because aluminum is a type one metal, not a type two. It only has one oxidation state, which is 3 plus ammonium oxide. All right, see you on Friday for the second session of the review.
reactions, and give you the mole ratios. Molarity, and again, stoichiometry and solution. the oxidation states of this state. Namely, multiplication division and the division of the where you may go and learn what to do and that given in three significant figures. Because 1 to calculations are calculations of volume. Because dimensional analysis. You can use that unit trick And this is a measure of mass. Can be related to