Let me move to the middle. >> All right. >> Good afternoon. >> Just to reminder that one week from today is exam number three. Hopefully everybody's had a chance to take a look at the practice exam. But if not, I recommend doing so as soon as possible. Today, I hope to come as close as I can to finishing all of the relevant material for exam number three. So we'll work on that. And if it gets done today, great. And if not, we will finish up on Monday and then save the rest of the time on Monday for whatever questions you might have last time. Just to give you an idea of where we are as we walk through the textbook. Chapter 19, section 19.3 is titled standard reduction potentials. Big table on page 87 of standard reduction potentials. The point is these aren't numbers we did not expect you to know off the top of your head. We will provide you with a table of standard reduction potentials. But what you should realize is defined that what's called the standard cell potential for a galvanic cell, we just look up the oxidation reduction half reactions and subtract reduction minus oxidation. Take the two numbers as they are off the table and subtract them. Section 19.4 is titled spontaneity of redox reactions under standard state conditions. >> That's something we were alluding to right at the end of last class we had shown you this slide. >> The point is recently we've seen three ways of trying to determine whether or not a reaction is spontaneous. If by that we mean whether it goes to completion. Once it started. One is a large value of the equilibrium constant, large being defined as greater than one and other is a negative value of delta G. The free energy change. Then, as we described last time, the standard cell potential for any functioning battery must be positive. So a positive value of E 0 cell also means a spontaneous reaction. These three things are all basically measuring the same thing, and therefore they must be related to each other. And over here we show the equations by which they are related to each other. We saw this one earlier, delta G, the free energy change, is equal to r negative R T, where T is temperature and R is 8.314 gas constant with jewels as units times the natural logarithm of the equilibrium constant. Comparing these two things, delta g equals negative n F E cell, where n is the number of moles of electrons being transferred, E L as the standard cell potential, which we get by subtracting those two numbers from the table. And f is something called Faraday's constant. And let me just say a little bit about Faraday's constant before I go over the rest of that slide. Faraday's constant named after a British scientist by the name of Michael Faraday, who did some of the early investigations into this and many other areas of chemistry. It has a value of 96,500 coulombs per mole of electrons. Where this number comes from is two other numbers that you saw in Chemistry, 1011 of which should, I hope look pretty familiar. The other of which maybe not so much. I hope you recognize this number right? What is this Avogadro's number? The number of things in a mole. Maybe remember what this number signifies. If not, that's OK. We don't make a big deal out of this number. But the point is, when we talk about the charge on an electron as being negative one. Well, okay, that's a shorthand notation. Minus1 actually represents this number. The actual value of the charge on an electron in Coulombs is negative 1.6 times ten to the negative 19. So that's the number of coulombs there are in one electron, there are 6.02 times ten to the 23rd electrons in a mole. So basically, if we multiply these two numbers together, we get this number. But the advantage to Faraday's constant, as we'll see shortly, is that it allows us to convert back and forth between electrical units, such as Coulombs, and chemical units such as moles, specifically moles of electrons. Anyway, that's what Faraday's constant. Now the rest of this slide, if you want to knock yourself out copying all this stuff down. Okay, that's up to you. But I will point out that most of this information is in your textbook. For starters, they give you the value of Faraday's constant on page 891. And then a lot of what's on page 892 is just the derivation of what I'm about to show you. So here's what I'm just going to say about this slide before we move on. If Delta G equals negative RT natural log k EQ, and delta t equals negative n F E cell. Then it must be the case that negative RT natural log EQ is equal to negative NFO. Now dividing both sides by negative n f to get E0 cell by itself gives us this equation. But here's the thing. R, T, f all have particular value is the value of R is 8.314 joules per mole per Kelvin. The 0 in E 0 cell means we're evaluating this under standard conditions of pressure and temperature. Standard temperature is 25 degrees Celsius, which is 298 kelvin. And we just mentioned that Faraday's constant has a value of 96,500 coulombs per mole of electrons. The other thing that's done here is to take the natural logarithm and convert it to the LOG, logarithm. And the conversion factor there is a factor of 2.303. I point this out because most of these things are just numbers that we can multiply together and get one number out of them. And so if we multiply and divide all these numbers, here's the number that we get. The only thing we haven't factored in here is n, because the number of moles of electrons berries from reaction to reaction. So E cell can be said to be equal to 0.0592 volts times the LOG, log of the equilibrium constant divided by n. And if you're wondering where that 0.0592 comes from, that's just the collection of constants that appears down here at the very bottom of the slide. I will ask you to do for the moment is remember that 0.0592, and we will say more about its significance in just a few moments. Pronoun, does everybody have one inverse relation they want from this slide? And this is really just a derivation. And I'm not going to be asking you to derive this expression on the exam, but it is important for various reasons. >> Now, just as an example, >> Of how these concepts are related to each other. Last time we showed you a very straightforward schematic diagram of a galvanic or voltaic cell involving the Zinc and Copper half reactions as seen here. I mean, you set this up under standard conditions. You can measure a voltage of 1.10 volts of the volt meter. 1.10 volts doesn't really sound like all that much, but it's actually a significant amount of energy being transferred. As you can see by using this equation, delta G equals negative if E 0 cell, where n is the number of moles of electrons being transferred. >> F is Faraday's constant we just showed you a few moments ago. >> And easy, rho cell is the voltage that shows up on the voltmeter. So suppose we have a zinc copper cell looking very much like what we just showed you that produces positive 1.10 volts. >> Ok? >> If you have a battery that's producing 1.10 volts, what's delta G for the reaction that's taking place? And remember, delta G is a measure of the amount of useful work that you can get from that particular chemical reaction. Well, this isn't that hard to figure out. The only part that's even remotely tricky about this is what's the value of n? >> So what I'm gonna do is I'll put this back up here for a moment. >> So you can see the path equations down here. In each of those half equations, how many moles of electrons are being transferred? Can look down here. You can look down here. >> Doesn't make any difference. >> Anybody That's the coefficient on the E in each of those half equations. >> Two, that's the answer. In other words, each mole of zinc is losing two moles of electrons during the oxidation process. >> Or to make one mole of copper, the copper ions have to gain two moles of electrons. Either way, the answer is to all we're looking for, for the value of n is the number of moles electrons being lost or gained. And the point is, if you have a balanced cell equation, the number of moles of electrons lost must be equal to the number of moles of electrons gain. >> As is the case here, losing two moles of electrons are deeming two moles of electrons. >> So the point is for the zinc copper cell, the value of n happens to be two. Once you figure that out, the rest is pretty straight forward. You're just plugging in numbers. So negative NF E cell and is two for two moles of electrons. F is Faraday's constant, 96,500 coulombs per mole of electrons. An easier sell is positive 1.10 volts. But earlier We defined a volt in terms of joules and coolant. Specifically, one volt is the same thing as one joule per one coulomb. And the significance of that is that if you have coulombs is the denominator and columns in the numerator here, they cancel out and moles of electrons divided by moles of electrons cancels out. >> So the answer comes out in joules. >> If you multiply all this stuff out, delta G for this reaction is negative 2.13 times ten to the fifth Joules or divide by a 1000 to express it in kilojoules, negative 213 kilo joules. And the point is, it's not a huge number, but it sounds a little bit more impressive that way than it does when you just say 1.10 volts. What most people don't appreciate is that a full involves a fair amount of energy being transferred. >> So are there any questions about how to use this equation? >> How to figure out the value of in anything like that. >> Alright, Section 19.5 in the textbook. >> And just by way of comparison, section 19.4, The title is spontaneity of redox reactions under standard state conditions. Section 19.5, spontaneity of redox reactions under conditions other than the standard state. I'm going to drag in here something called the Nernst equation. Now you may recall, let me just go back for a moment that picture of the voltaic cell, you may recall that when we showed you this picture earlier and we were defining what we meant by the standard conditions. One of the things he said was that the concentrations of the ions in the solutions had to be one more for this to be under standard conditions. And the reason that's true is because if the concentrations are something other than one molar, that's going to have an impact on the voltage produced by the battery. As we're about to see. That's one thing that's meant by when things happen under other than standard conditions. And that's why we drag in this thing called the Nernst equation. This is named after a German scientist by the name of Walter Nernst, who 1D, I believe 1920 Nobel Prize in chemistry for this work. And here's what it comes down to. You can calculate E 0 cell for any battery just by taking the two equations off the table and subtracting the two numbers. But E 0 means assuming they're operating under standard conditions. What if we're not? If we're not, then we don't have the 0 up there anymore. >> It's just E cell. >> E cell is the voltage actually produced by the battery. And that's equal to E 0 cell minus, here's that 0 59 to number again that we saw about two slides ago, which is a collection of a whole bunch of constants divided by n, the number of moles electrons being transferred time the LOG, logarithm of Q. And we know what Q is from back talking about equilibria sometime last month, hopefully you haven't forgotten how to write equilibrium expressions. And the point is Q is simply the equilibrium expression, assuming that the reaction is behaving as an equilibrium concentrations of products in the numerator, concentrations of reactants in the denominator with everything raised to the appropriate exponent that matches, but it's coefficient in the balanced equation. So suppose we have a zinc copper cell. Only difference is instead of one molar, both things. Suppose it's one molar zinc ions, but only one times ten to the minus four molar copper ions. Now we know that the zinc copper cell under standard conditions produces 1.10 volts. These are not the standard conditions. Bolts are going to be produced by this battery. Well, no surprise, we're going to use the Nernst equation. And we saw in the previous example that for this particular equation, the value of n happens to be two because there are two moles of electrons being transferred. Two moles of electrons lost by the zinc, two moles of electrons gained by the copper. The rest is just entering the numbers. But the trickiest part of this is generally figuring out what we mean by log q. What I would recommend that you do to try to solve something like this is write out a balanced cell equation and include things like what's a solid, what's a liquid, What's aqueous, et cetera. You need to see, I gave it away. So let me just grab a piece of paper and show you what I mean by that. All right, let's consider this. Now the point is when you see this cell notation like this, you can pretty much read this from left to right. Zinc metal is becoming zinc ions. Copper ions are becoming copper metal. Somebody draw an arrow here. >> The reactants are the zinc metal, zinc solid, and the copper ions, copper two plus aqueous solution. >> The products are zinc ions, zinc two-plus in aqueous solution, and copper, metal, solid copper. Now, let's treat this for the moment. As an equilibrium. I hope it's obvious that it's a heterogeneous equilibrium. That being the case, how would you write the equilibrium expression for this equilibrium? Zinc two plus is on the top or the bottom. >> Yep. >> Okay. Over copper two plus. That's absolutely right. And the point is we ignore this. We ignore this because whatever we're writing an equilibrium expression for a heterogeneous equilibrium, we ignore the pure solids, pure liquids. So those go away. And you're absolutely right. It's concentration of zinc ions over concentration of copper ions. And then all we have to do is plug in the numbers. The zinc ion concentration is one molar, the copper ion concentration is one times ten to the negative formal, or divide those up, we get 1 times ten to the fourth. And the rest is pretty much just arithmetic. So here's the arithmetic. Now that we know that Q is 1 times ten to the negative four, we could just insert that in the log part of this. Everything else we know, including the value of N, which is to the logarithm of one times ten to the fourth is simply four. Multiply and divide all of this, you get 0.118. Subtract that from 1.1 to 0 volts. And taking the two sig figs were allowed. Here, we find that our battery operating under non standard conditions is producing a little bit smaller voltage than it would if it were operating under standard conditions. Standard conditions, 1.1-zero bolts here, 0.98 volts. So that's a battery that's on its way to discharging but still producing some energy due the concepts make sense. Does the arithmetic makes sense? This is probably about as complicated as the math gets for exam number three, but your liberty to disagree after you look at the practice exam. So yep. >> Questions, we'll talk about them on Monday. >> Everybody habit they need several people still writing. Okay? >> Alright. >> As I've been saying all along the way, most people interact with galvanic cells is when they use batteries for anything. That's all batteries are voltaic or galvanic cells. There are oh, let me point out one other thing before we move further on that, I will encourage you to read this part about concentration cells on page 897 in your textbook. Ideally, before you go to lab, I realize some of you have already been there pathologies, but the bottom line is, if you're wondering what a concentration cell is, what it's all about. There's some information on pages eight, ninety seven ninety eight, especially on 898, which talks about biological concentration sales that you might find interesting. >> Now, section 19.6, batteries, the diagrams that appear in Section 19.6 are much better than my hand-drawn versions. >> So I'll refer you to the artwork on page nine, L1, for example. Here's a schematic diagram of a car battery. You seen so-called dry cell batteries that looked like this? Many other better pictures than what I draw in the textbook. So I'll refer you to that. But just to give you a rough idea, and I'm sure you already know some of this. What kind of interesting reactions take place inside batteries? >> Okay? >> Your car has a battery in it. It's referred to as a lead storage battery because all of the half reactions involved there involve led in some oxidation state or another. Here's led at 0 going to lead plus two inlet sulfate. Here's lead plus four in lead oxide, going to lead plus two and led solely. The problem is that all of those things are solids. We've talked about how we transfer electrons. When we have aqueous solutions involved. How do we transfer electrons? But everything's a solid. Well, the answer is by having the solid pieces dip down into some electrolyte. I don't know. How many of you have ever picked up the hood of your car and actually looked at the battery, the thunder there these days. And you look at a car battery, you'll see big warning signs on it. >> That's a danger acid inside. >> Do not mess with this thing unless you are a trained mechanic or something like that. There's a reason for that. The electrolyte that is used is concentrated sulfuric acid, which is very corrosive and can in fact cause severe burns if it gets on your skin. So unless you are a trained mechanic, obey those warnings and don't open up the car battery if it needs work being done on it and take it to somebody who knows what they're doing. Calculators run on batteries. The so-called new CAD battery that fuels a calculator is called that because the 2.5 reactions involved are based on nickel compounds. And cadmium combats a battery that might be used for something like a Flash line is called a dry cell for a reason. You know basically what a battery looks like. It's a cylinder that has this little bump on top, maybe a little divot on the bottom. >> Okay. >> The bump comes from this carbon graphite rod that goes up through the middle of the dry cell battery. That's actually the cathode for the reaction that takes place. Although you don't see carbon anywhere in the equation here. The zinc anode is actually the jack cutting on the outside of the battery. The manganese half reaction that occurs here is happening in the form of the relatively thick manganese oxide paste that makes up most of the contents of the battery. Now, since this is a relatively thick paste and there's a problem, the point is that after a while, positive charge builds up near the zinc, negative charge builds up near the carbon. And there's no real way to redistribute those charges. So after a while, this battery dies. Okay, no big deal. You go out, buy more batteries and you recycle the old ones. Although here's a cute little trick. You can try some dive if you're on a budget these days, who isn't? When your battery dies, if you want to squeeze a little bit more life out of it, pretend this is a battery shaken up a little bit, you just shake it up a little bit, it redistributes the ions a little bit. You might squeeze a little bit more life out of the battery. But the point is that kind of battery can't be recharged like a car battery can because car battery has a liquid electrolyte in it, which means if you apply an external voltage to that battery, you can reverse the chemical reaction, make the electrons flow in the opposite direction. That doesn't work with something like this that mostly has solid contents. So you can maybe mill another 10% worth of life out of the battery by doing the shaking thing. But realistically, just go buy more batteries and recycle the old ones. Section 19.7 and your textbook is titled electrolysis. >> Very early in our discussion of electrochemistry, I mentioned that there were two kinds of electro chemical cells. >> We've been talking exclusively about galvanic cells at this point. Now we're going to say a few words about electrolytic cells. The difference, as you may recall, is an MA, galvanic cell. We take advantage of a redox reaction, set things up to force the transfer of electrons to flow through a wire, generate a current of electricity that way. In an electrolytic cell, it's exactly the opposite. We use some external current of electricity to make an oxidation reduction reaction happen. That wouldn't happen under normal circumstances. Here's my example. If you take sodium, metal, and chlorine gas and allow them to come in contact with each other, a violent reaction will ensue, and the product of that violent reaction will be. >> So it's a good thing that reaction is not reversible. >> I mean, it is under the right circumstances, but we don't normally think of that as an equilibrium. And that should make sense because if you reach for the salt shaker at the dinner table with the idea that you're going to sprinkle well solved on your french fries or something. It would really be helpful if what fell out of that salt shaker was little white crystals of salt as opposed to a greenish yellow gas, a silvery gray metal. And perhaps the explosion that would take place when you come into contact with each other. So salt is pretty stable stuff. Having said that, you can make this reaction go backwards by running a current of electricity through it. And that's what we mean by electrolysis. Let me show you a picture. Here is a schematic diagram of what's called a downs sell. And the point is what we're doing in this cell is electrolyzer, sodium chloride. Basically we're taking salt and running occurred of electricity through it to make this reaction go backwards and produce sodium and chlorine. Okay, how's this work? Well, first of all, notice that we have a big tub of molten sodium chloride here. You can't just take a big pile of sodium chloride, stick a couple of electrodes in it and run this reaction at room temperature. Because again, if we're talking about nothing but solids here, there's no way for the electrons to flow. There's no way for the ions to migrate. There's no way for any of the movement that dates to take place. They place, which is why you have to heat it up until it melts. Salt melts at about 800 degrees Celsius, so takes a fair amount of heat to get this into a liquid state. But once it's there, if you have an anode over here and a cathode over here, and you're using some external voltage source supply energy, taking electrons away from here and moving electrons over to here. >> Then two things happen. >> As electrons move over here to the cathode, the cathode becomes negative, which is why positive sodium ions migrate toward it and are, they are reduced to form sodium metal, which at the high temperatures of the molten sodium chloride bath, happens to be a liquid. Meantime, measure removing electrons from the anode. The anode is becoming positive, which causes chloride ions, which are negative, to migrate toward it and give up their electrons as well. In the process this half reaction takes place, and what's formed is chlorine gas. So basically we can make this reaction run backwards by supplying a turtle electricity. So what we mean by electrolysis, electro lysis. The lysis suffix means clean, cleaving, or tearing apart. So electro lysis involves using electricity. Convert an ionic compound into the elements from which it is made. By making reactions like the one you see at the top of the screen run backwards. So electro lysis in this case means we're using electricity to tear apart salt and make sodium and chlorine. Now that's the basic concept. And if you understand the basic concept than the rest is simply the details like how much of an electric current does this take? How long do you have to let that current, electricity run, things like that. And that's actually summarized by this equation down here at the bottom of this slide, which I call the quit equation, first of all, because it sort of looks like the word quit. And second, this is the point in the course where I'm almost quitting talking about math. The remaining six weeks or so of this course will involve very little number-crunching. >> Maybe not absolutely none, but you don't see that many equal signs after this. >> A lot of what follows is going to be very descriptive in the nature of the chemistry that we talk about. So for those of you who have not been big fans of all the math we've had to do in this course so far. One more exam and then you can almost put away the calculators. >> Anyway, what this equation means, q, as we saw before, is the charge in coulombs. >> I represents the current of electricity measured in Ampere's. And T is the amount of time that we allow that current to flow in seconds. Let's take a look at how this actually works. Suppose what you want to do. Let's make some copper. To do it, you're going to take a solution that contains copper plus two ions and run a current of electricity through it. Specifically, you're going to use one ampere of electricity and run that current for five minutes. And the question is, how many grams of copper can you make this way? Suggestions for how we might start solving this problem? >> Okay? >> The idea behind doing that, of course, is that we're going to use the q equals IT equation. And T in that equation is measured in seconds. >> Okay? >> How many seconds are there in five minutes? You could ask your calculator, of course, but you really should be able to do that one of your head. 300, yep. >> 60 seconds in a minute times five minutes. >> T here is 300 seconds. >> Okay. >> Well, once we did that, the Q equals IT part is pretty easy because I happened to be one ampere. So one times 300 is 300. The total charge transferred is 300 coulombs. >> Okay? >> Now what, what we're eventually trying to figure out is how many grams of copper we can make by doing this. Say again, moles. >> I was wondering if anybody has figured out what kind of problem this is. >> And if not, that's okay. >> I'm just going to dredge up that bad word from Kim one-on-one. >> Again, stoichiometry. >> Remember that if there's moles involved, it's a stoichiometry problem. Step one of any stoichiometry problem is convert something into moles. Now what we did here was he took the information about the electricity and figured out coulombs. But earlier today we were talking about Faraday's constant as a conversion factor between electrical units like Coulombs and chemical units like moles, especially moles of electrons. So here we have 300 Coulombs. Faraday's constant says that 96,500 coulombs is the same thing as one mole of electrons. >> Divide this out, and we find that using this much current for this much time transfers 3.11 times ten to the negative third moles of electrons. >> Ok, that step one of our stoichiometry problem. Now we have something in moles. What's step two of a stoichiometry problem? Say again, well, we're not ready for the molar mass just yet, because we have moles of electrons. What we're trying to find is grams of copper, which are kind of on the right track. We will need to use the molecular weight or the molar mass. Actually in this case, going to be the atomic weight of copper eventually. But we need to get moles of copper first from the information that you can see on the screen, how many moles of electrons does it take to make one mole of copper from the ions? >> Yes, to the key point here is the plus two charge and the copper. >> In any stoichiometry problem, step two involves referring to the balanced equation and use the coefficients in the balanced equation that convert moles of one thing into moles of something else. In this case, we have a balanced equation. We're using copper plus two ions and converting them into copper with no charge copper metal. To do that, we need two electrons for each cup, right? So the mole ratio between electrons and copper atoms is two to one. >> So we set up a conversion factor. >> Basically we divide by 2.551 times ten to the negative third moles of copper is what we get, and we divide by two. And then the last step of that is to in fact go to the periodic table, look up the atomic weight of copper, 63.54 grams per mole. Multiply through and to decreasing things were allowed. We find that we can make not quite a tenth of a gram of copper. This way, 0.0988 grams of copper. That's not much copper, but let's face it, a current of one ampere running for five minutes, that's not very much current and not very much time. If you want more copper, either use a bigger current or let it run for longer or both. >> Do the calculations make sense? >> Okay, couple people still writing. >> I have one more slide to show you and that will be it for the day. Everybody, yeah, but they need OK, why should we care about electrolysis? That depends on who you are, but I suspect everybody here has in some way, shape, or form, interacted with the products of electrolysis we just showed you in the so-called downs itself. Or you can produce elemental sodium from molten sodium chloride by running a current, electricity through it. In general, electrolysis is a good way to produce metals that don't occur as metals, but rather as their corresponding ionic compounds. In nature, sodium is very abundant in sodium chloride in seawater. But we need a down cell type of setup, like we showed you before to make metallic sodium. That way, seawater is also rich in magnesium. You can get some magnesium salts from sea water. You can run a current through what's called a Dow process cell and generate metallic magnesium that way. We'll talk later on in the course about aluminum. But suffice it to say this for right now, aluminum was once thought of as a precious metal right up there with things like silver and gold, until a young man by the name of Charles Martin Hall figured out a relatively inexpensive way to liberate aluminum from its minerals Basically by using electrolysis. And nowadays we think nothing of aluminum. There's aluminum all over the place, aluminum cans, aluminum foil, et cetera, et cetera. Three quarters of the surface of the Earth is covered with sea water, very abundant natural resource. And we've seen that salt water does conduct electricity, right? Salt as any electrolyte. We did the little demonstration in here with the light bulb and all that. >> Okay? >> You can run a current electricity through seawater if you do, you get three things, all of which turn out to be useful for reasons that we'll talk about later in the course. Hydrogen gas, chlorine gas. If you have water in there and you don't get metallic sodium, metallic sodium reacts with water, but you do get an aqueous solution of sodium hydroxide, commonly known as ly, which has its own purposes question. Whereas the chlorine coming from, oh, okay, this is not a balanced equation, but the point is the elemental form of chlorine is Cl2. So if we wanted to make a balanced equation out of it and put a two in front of the NACL. And we'd go from there. And then finally, here's one that's on a more everyday level. There's a process called electro plating that can be used to put a thin coating of a metal on some other surface. And thereby perhaps, depending on which metal you choose, protect that surface from rusting and corrosion, for example. Actually, I'll just show you all three examples right here. My schematic diagram here is sort of like the process we showed you before with the molten sodium chloride. The only difference in this case is the cathode is a fork or a knife or a spoon, whatever you would like. But the point is, everybody's used silverware before yesterday's trick question. What silverware mostly made up. It's a trick question because the answer is really not silver. It again, now that aluminum, although you can find a aluminum forks and things like that out there. >> Okay. >> Now you're just guessing the actual answer is stainless steel. If you've ever tried to pick up a fork or a knife with a magnet, you can do that because it's largely iron. >> The problem is iron rust, that would suck. >> So how do we fix that problem? We take our stainless steel silverware and make it silverware by dipping it into a solution that contains silver ions and running occurred of electricity through it. Net effect, we get a coding of silver on the silverware. Silver doesn't rust or corrode anywhere near as fast as iron does, therefore stay shiny or for longer and survived the trip through the dishwasher and things like that. You can also make gold-plated jewelry, this link. Or you can put a Chrome finish on the real work of your automobile by putting the real work into a solution that contains chromium ions that are running a current through that. That's why it's called a Chrome finish. >> It's made of shiny chromium. >> Makes sense. Ok, draw a line through your notes at this point. This is the end that what you are responsible for, for your exam. And one week from today, if there are questions between now and then, you can always email me and we will see you in class on Monday to specifically address any questions you might have. Enjoy the lovely weather today. Okay, I assume your circle the numbers of the questions you want, but that's alright. >> Yep. Well,

chem102-010-20170405-122001.mp4

From Dana Chatellier October 11, 2018

74 plays 0 comments You unliked the media.

Comments

Add a comment