When we write down, Good afternoon and for the benefit of anybody watching these videos at some future time, Today is Wednesday, February 26, which means that we are now nine days away from the first exam in this class. >> I'm hoping by the end of this week to be finished with chapter 14, which is the last chapter you'll be responsible for for your first exam. Hopefully everybody's had a chance to look at the practice exam. If not, now's a good time. But before I continue on with chapter 14, any questions about anything we've talked about so far? Woody? >> Okay. >> Alright. A brief memory refresher from last time in attempting to address the fundamental question, how long is this reaction going to take? We turned to what I call the concentration versus time equations. Your textbook calls them the integrated rate laws. But suffice it to say that they are equations like this one, which include t for time as a variable in there someplace. Because when you're asking how long is the reaction going to take, that is, how much time is it going to take? T is going to be the answer. Now one thing that's helpful is if you know what the order of the reaction is. And as we said previously, the order of the reaction is simply the sum of the exponents in the rate equation for the reaction. For example, this would be the rate equation for a first-order reaction. We worked through this problem, which appears in your lecture notes, have found that for this particular first-order reaction, The question was, how long would it take for 99% of the reactant to be consumed? And the answer turned out to be less than about eight minutes. By contrast, we looked at another reaction which happens to be a second-order reaction. Here's a typical second-order rate equation. Here is the second order rate constant. And the question was, after one hour, what would be the concentration of the one reactant given that its initial concentration was point 1-0-0 molar. And we work through the details using the second-order concentration versus time equation and found out that it was about 11% of what we started with. So the main point is the first-order reaction was 99% done in about eight minutes. This reactions only about 90% done after now, hopefully it makes sense to that means the first order reaction is going faster than the second-order reaction. I want to look at these reactions again today, but in a slightly different context. Especially useful if all you want is a rough idea as opposed to a rigorous idea of how long the reaction is going to take. And this concept is mentioned for the first time in your textbook on page 623. But you may have heard of this concept before. >> It has to do with the concept of the half-life of a reaction. >> Let me show you an illustration that I hope will help us to make sense of this concept. We were looking at a curve that looks something like this wouldn't be for, for a different reaction. The point is we're following the rate of disappearance of one of the reactants. Now in this particular example, the react that happens to be a gas. So instead of measuring concentration, they measured pressure in TOR, but it's very similar in concept. The point is we start up here at an initial pressure of a 150 Torr. >> And then the curve goes like this. >> And again, by drawing the tangent line to that curve, we can get a sense of how relatively fast the reaction is going at any given time. In the early stages, where there's a steep slope to that curve, means the reaction's going relatively rapidly. But as time goes on, the curve becomes less steep and slope, that means the reaction is slowing down because the concentration of the reactants is not what it was at the beginning of the reaction. But here's what we mean by the concept of half-life. And this is the symbol for half-life. >> It's Tea Time with a subscript 1.5. >> You can see that symbol being used on this graph. The point is I want to emphasize that this is simply a symbol. It does not mean to multiply t by 1.5 or divide t by 1.5 or anything like that. >> It's just a symbol that's used to represent the concept of half-life. >> And what we can see in this particular example is starting at a pressure of a 150 torr for the one reactant. >> After about, let's say 13 thousand seconds here, we're down to half of that, which is 75 tool. >> And then we wait another 13 thousand seconds, announcement, a total of 26 thousand seconds, we're down to half of that. Half of 75 is 37.5 and so on. In other words, for this particular reaction, everytime you wait about 13 thousand more seconds, you're down to half of whatever it was you had 13 thousand seconds ago. >> That's the definition of half-life. >> Half-life of any reaction is the amount of time it takes for the concentration of the reactive to decrease too. >> Half of its original value. >> And you can actually use the concentration versus time equations that we were looking at last time to figure out the half-life for reaction. Because by definition, the final concentration of reactant a is simply 1.5 or whatever the initial concentration was. So if you take either the first-order or second-order concentration versus time equations and simply substitute initial concentration over two for the final concentration. It turns out that that simplifies the calculations considerably. And the half-life is a good way to at least estimate how long the reaction is going to take. Here's what I mean. Here is the first-order concentration versus time equation k, the rate constant, times t, The amount of time that has elapsed equals the natural logarithm of the initial concentration of reactant a divided by the final concentration of reactive it. But if we substitute a initial over two for a final than this simply reduces down to the natural logarithm of the number two. And the natural logarithm of the number two turns out to be 0.693. So if you don't like working with an equation that's this complicated, you can work with a somewhat simpler equation. The half-life for a first order reaction is just 0.693 divided by the rate constant. And I'll give you an idea of how long it takes to get down to half of whatever it was you started with? Or do you want to go more than that? 2.5 lives, it get you down to one-quarter. What you started with. >> Three half-lives that get you down to 1 eighth of what you started with and so on. >> But here's the thing to realize about a first-order half-life. 0.693 is just a number. The rate constant k for any particular reaction is just a number. The half-life for any first-order reaction is constant throughout the lifetime of the reaction. That is not the case for a second-order reaction. Here's the second-order concentration versus time equation. K times t equals one over the final concentration minus one over the initial concentration. But again, if we substitute a initial over two for a final and do a little bit of algebra. Then the half-life equation for a second-order reaction looks like this. T1 half. The half-life is equal to the reciprocal of the rate constant k multiplied by the initial concentration. However, one difference between a first-order reaction and a second-order reaction. The second order half-life depends on the concentration, and the concentration changes as a function of time. Therefore, the half-life for a second-order reaction is not constant throughout the life of the reaction. As the concentration of the reactant decreases, that's going to increase the half-life as time goes on. Not so for a first-order reaction, for which the half-life remains constant throughout the life of the reaction. In the case we were just looking at a few moments ago. It seems like the half-life is remaining constant 13 thousand seconds, we're down to half of what we started with. 13 thousand more seconds, we're down to half of that. >> And so on. >> That being the case, I suspect that the reaction symbolized by that graph is probably a first-order reaction because its half-life remains constant throughout. >> Before we look at any specific examples, any questions about anything that's on this slide? Yes. >> Here we will talk more about nuclear chemistry later on this semester. But just to answer your question, most of the time when people think of half-life that are thinking about nuclear reactions. And one of the reasons that half-life is a popular concept when you're talking about nuclear reactions. It turns out that all nuclear reactions are all radioactive decay reactions, which is kind of what we're talking about here, do happen to follow first order reaction kinetics. So the reason the half-life is a useful concept there is that the half-life for any nuclear decay process is constant throughout the life of the reaction. Are there any other questions about anything that's on this slide? Alright, let's go back to the two reactions we were looking at last time and think about what they're half lives might be. By the way, for those of you following along in the lecture notes, the whole concept of half-life and the problems we're about to look at are summarized on page 26. Ok, so here's the reaction we were looking at before is an example of a first-order reaction rate constant, 0.010 seconds to the minus one power. >> What's the half-life of this reaction? >> Well, it's very easy to figure that out. If you know that the half-life equation for a first order reaction, it's just 0.693 divided by the rate constant. To finish off this problem, what do we need to do? >> You get the 0.693. >> Basically, it's derived from the concentration versus time equation. It just happens to be the natural logarithm of a number to it. See where two might come into play in the context of Half-Life. But what I'm saying is, given this information How would we solved as Brahmi? >> Yep, exactly. >> K is 0.010 seconds to the minus one. But K right there, 0.693 divided by 0.010 seconds to the minus one. Seconds to the minus one in the denominator becomes seconds to the plus one in the numerator. So the answer comes out at seconds, only two sig figs in this number. So when we divide it out, we only get to see things in the answer. The half-life is 69 seconds. That means after just over a minute, this reaction is half done. >> Make sense? >> Okay, here's our second-order reaction. >> Again, notice the similarity between this chemical equation and this chemical equation. There's no way that just by looking at those two, you will be able to figure out that this what is first-order or second-order. But that's why we work through the problems previously. And you can actually tell that from the units of the rate constants. This is typical units for a first-order rate constant. This is typical units for a second order rate constant. But the main point is given this value for the rate constant and this value for the initial concentration of CH3 VR. What is the half-life of this reaction? Given that the second order half-life equation looks like this, how would you solve this problem? >> Yeah, very, very similar to last time. >> The only difference is we have two numbers that we have to plug into the denominator. What is the rate constant? And the other is the initial concentration. >> So if we do that, notice that moles per liter times moles per liter to the minus one cancels out. >> And seconds to the minus one, the denominator becomes seconds in the numerator. So once again, the answer comes out at seconds. This time we're allowed three sig figs, because we have three sig figs and built the rate constant and the concentration. So it comes out 467 seconds, again, roughly eight minutes. So looking at these two results for the half-life, which reaction is faster? The first-order reaction or the second-order reaction >> Yes. >> First-order, shorter half-life, faster reaction. It takes this one only 69 seconds to be half done. It takes this 1467 seconds closer to eight minutes to be half done. Smaller half-life, faster reaction. Now this is the same conclusion we came to last time, but I'm just saying that you could look at it from the standpoint of half-life, which is somewhat easier equation to work with. And then just say, oh, okay, this has a shorter half-life of this does so this one's going faster questions we'll start with you now just for these particular examples. Obviously the actual values here depend on what the rate constants happened to be. This rate constant works out for this particular reaction, but there are others that may have different rate constants. And the same is true here. And this one has the added variable, whatever the concentration is. So it's not always the case that first-order reactions are faster, but second-order reactions, it is always the case that you can calculate the half-life for a first-order reaction using this equation. For a second-order reaction using this equation. And they just plug in the numbers and see how it works out. >> Grab question, that was same thing. >> Okay. >> Any other questions about the concept of half-life? >> Okay, section 14.4 talks about the dependence of reaction rate on temperature. >> Up until now, we've been talking about how concentration plays a role. Now it's time to say something about where temperature comes into play. And before I leave that page, they start by talking about something called Collision Theory. Let me go back for just a moment to my silly analogy involving the baby elephants. Ok, our goal is to make baby else once we have male elephants and one gauge female elephants in another cage. But it's not going to work until we open the cage doors and let the elephants mix to get suddenly open the cage doors. Then the elephant start roaming around and occasionally bumping into each other every once in awhile, a collision between two elephants may result in the formation, eventually it'll baby element. Not all collisions between elephants result in the formation baby elephants. We will ignore for the moment the obvious cases where two male elephants bumped into each other to female, Oh, it's bump into each other. >> No, those aren't going to work. But even in cases where male elephants in the female elephants bump into each other. >> There are two things that have to be true for a collision between elephants to make baby elephants. Let me illustrate one of these using this diagram, which is an illustration of the following reaction Now in this diagram on the slide, this over just a little bit, the green balls represent chlorine atoms, blue nitrogen, red oxygen. This is what a molecule of a compound called nitrosyl chloride and OCL looks like. And the goal is to try to collided with a chlorine atom to make a molecule of Cl2 and a molecule of NO, this one is labeled as an effective collision because if the collision that actually works. But here's another case where the molecules collide and it doesn't work, they just bounce off each other. >> Why? >> Well, sooner or later if you're going to make Cl2 to get the two chlorine atoms to bump into each other. During this collision, the two chlorine atoms don't come anywhere near each other. The analogous thing with the elephants is even if you have a male elephant and a female elephant and they bump into each other head on. Or you're going to get is elephants with headaches, not baby elephants. Same thing here. >> This doesn't work, but this does. >> So one thing that must be true is if collision between molecules, just like the collision between elephant must have the proper orientation or else it's not going to work and only a certain percentage of collisions have the right orientation. The other thing that must be true is that the collision between elephants must be sufficiently energetic if the office just aren't in the mood and they sort of go, Okay, that's nice. >> But there'll baby elephants coming out of that. >> But to be honest, they're excited, then maybe it'll work. >> I told you this was extraordinarily done, but there's a method to the madness. >> As a general rule, the rate of a reaction goes up by about a factor of two or three every time you increase the temperature by ten degrees Celsius. >> Okay, fine. >> But when you look at a rate equation for a chemical reaction, that's not obvious because when you look at the equations that we've seen so far, you don't see a temperature term in there anywhere. So where does temperature come into play? Iterate equations. For example, here's a generic rate equation. Rate equals the rate constant k times the concentration of a raised to some power, little a times the concentration of B raised to some power a little bit, but I don't see a temperature term in there. >> Well, here's where we have to confess to having lied to you just a little bit. >> Rate constants are constant as long as you maintain the reaction at a particular temperature. >> But rate constants change when you change the temperature. >> It turns out that the rate constant k is related to a number of other factors summarized in this equation. Which is called the Arrhenius equation. Don't know if the name Arrhenius was mentioned in your camp one-on-one class or not. We will mention him again later on in this class. Suffice to say that Svante Arrhenius was a Swedish scientist to at his finger in a lot of different aspects of chemistry over time. And it was actually one of the first ones to win a Nobel Prize for chemistry for convincing people that there were such things as ions, which a 100 years ago struck people as a weird idea. But we've kind of tough to accept that there are such things as ions these days anyway, this equation called the Arrhenius equation, shows you how the rate constant k depends on a number of factors. One of the being capital T, which is the Kelvin temperature of the reaction. Some other factors that are involved in the Arrhenius equation. We've seen E before. E is the base number for natural logarithms, 2.718, et cetera, et cetera, are, is actually the same r that comes from the ideal gas law, PV equals nRT. However, it is possible to express are not in units involving pressure units and volume units like atmospheres for pressure leaders, for value, but in energy units such as joules. And so the, our ear is typically expressed in this context using energy units. The value of r in that case is 8.314 joules per mole per Kelvin. This capital a over here involves other phenomena, one of them being the orientation of the collision between molecules. As he said before, not every collision between molecules as the correct orientation for the reaction to proceed. And this E with a subscript a here has to do with something called the energy of activation, which we'll say more about in just a moment. But suffice it to say that it does have to do with how vigorous or energetic the collision between molecules is. Come to the realization that we should be meeting in a different room. Because if a meeting in something shaped more like an auditorium, those of you in the back row would be craning your neck to see what's going on. I'd like the people in the front row, you'll have a clear view what's going up. Having said that, there are empty seats in the front row, if you really want to see what's going on, you can use them. >> Would anyone like more time with this slide cleric? >> Let's talk some more about this thing called the energy of activation. >> A moment ago, in the context of talking about the baby elephants at all, that we say that a collision between molecules or elephants must be sufficiently energetic. >> In order for, in the case of molecules boss to start forming bonds, to start breaking things like that. The energy of activation for a chemical reaction is simply the minimum amount of energy that a collision between molecules must have in order for the reaction to take place. There is a diagram that appears on page 629 in your textbook. And I'm going to show you a similar diagram here, but I'm going to wait until people are finished with the top part of this slide because we don't want you to see the whole thing here. Alright, we'll come back to that in just a few minutes. >> But for right now, let's look at this diagram. >> Now what this is trying to describe is energy. And specifically they're talking about potential energy on the y-axis. And when I say reaction pathway on the x axis, what they mean is if the reactants are over here, the products are over here. How much progress have you made and getting from the reactants to the products? In other words, you can really think about the X axis here is essentially being time. How much time has elapsed in going from reactants to products? The energy of activation is this right here. Now let me simulate what's going on using my fists to represent molecules. Suppose we're a relatively low temperature, but that means is the molecules are moving around kind of slowly like this. And maybe once in a while they'll bump into each other. Now at the moment they bump into each other, whatever kinetic energy they had before the collision becomes potential energy because at the moment of the collision that they're not moving relative to each other anymore. So all that kinetic energy becomes potential energy. Potential energy is what's being measured on this axis here. >> The analogy here is, think of this as a hill and you have a rock this side of the hill. >> And what you're trying to do is roll the rock up the hill and get it over here to the other side. However, if it's a slow moving bunch of atoms and the collision is not very energetic. >> That's sort of like rolling the rock halfway up the hill and then letting it go. >> What's going to happen? >> Suppose I'm a little guy standing right here, only this rock up the hill and then I let go. >> What happens? >> Hope it's obvious that it's not gonna be pleasant for the guy standing there because Iraq's going to roll right over Amun-Ra white back down to where you started from and you've accomplished nothing. And that's the point that collision between molecules was not sufficiently energetic to make it all the way up to the top of the energy of activation But suppose you are at a higher temperature now, the molecules move it around a little bit faster like this. And I don't know, it's a while back. They whack into each other and it's a bit more energetic collision. So now there's a lot more kinetic energy that translates to potential energy. >> That's the equivalent of rolling the rock all the way up the hill and then giving it one more shove, in which case it goes down to one of the products. That's the point. >> The energy hill here represents the minimum amount of energy that a collision between molecules must have to get down over here to the other side. >> If the collision doesn't have that much energy, nothing's going to happen. >> Now, a moment ago we were looking at the Arrhenius Equation, which includes the energy of activation as part of the equation. Here's an alternative version of Arrhenius equation, which you can get by taking the natural logarithm of both sides of the Arrhenius equation as it appeared in the previous slide. I will spare you the details of why we would do such a thing except to say this. >> Who remembers from high school math classes the basic equation of a straight line and halfway up, yes, Y equals MX plus B. >> The reason for taking the natural logarithm of both sides and rewriting the Arrhenius equation this way is that when you do this equation actually has the form of the equation of a straight line. >> Y equals MX plus B, as long as y is the natural logarithm of the rate constant k, and x is the reciprocal of the Kelvin temperature. >> Now I mentioned earlier, if you have a kinetics lab to do this week, depending on whether you have lab Thursday or Friday, that's either tomorrow or at the end of the week, you're going to be doing two basic things during that kinetics lamp. One is a method of initial rates study. You're going to be measuring the rate of the reaction under different sets of concentration conditions. You're going to change one concentration at a time or each reacted and try to figure out the exponents in the rate equation. Try to figure out the rate constants and stuff like that. But the other thing you're going to do is use the temperature dependence of the rate constant to figure out the energy of activation for a reaction. >> And here's how it works. The idea is to do the exact same reaction using the exact same concentrations at several different temperatures. And construct a graph that looks something like this with the natural logarithm of the rate constant on the y axis and one over the Kelvin temperature on the x-axis. >> You draw a straight line through your data points and measure the slope of that line. >> But if you look up here, okay, Y equals MX plus B. >> In that formulation of a straight line, m is the slope, b is the y-intercept. But we don't really care about B. We just care about the slope. Because when we're trying to find is the energy of activation and the slope of that line m is equal to the negative of the energy of activation divided by r. So once you measure the slope of that line, it's easy to find the energy of activation, just multiply by r and change the sign. By the way, when you construct such a graph, it should look something like this. That is to say, the slope of the line should be negative. >> Not like say, this pink line right here, which would be a positive slope. >> It should be like what I've drawn here on the slide, which has a negative slope because energy of activation is always going to be a positive number. So if your energy of activation comes out to be a negative number, it probably just means you forgot to do the negative side part here. >> Seen as folks are done with this slide, we will walk you through an example of how to do something like this, which will hopefully give you a better idea of what's going on in the lab for what it's worth. >> One of the illustrations on the front cover of the Kim one or two lecture notes shows you what your graph should look like. But let's look at an example which appears on page 28 of the lecture notes. And hopefully you'll get a better idea of how to do this. Alright, so in the example from age 28 were cheating. >> That is to say we only have two data points. >> Anytime you only have two data points, you can obviously draw a straight line between those two data points. Ideally, you should add like three or four data points and then draw the best straight line that you can between those data points. >> But in this case, we only have two data points. >> Whoever did this reaction measured its rate constant at two different temperatures, 25 degree Celsius, 50.1 degrees Celsius. And got these two values for the two rate constants in what appears to be a first-order reaction, because those are typical units for a first-order rate constant. So the question is, what is the energy of activation for this reaction? Alright, and the previous slide bear in mind what we wanna do. We want to make a graph of the natural logarithm of the rate constant on the y-axis versus one over the Kelvin temperature on the x axis. So what's something we need to do with this data? Before we can make our graph, yes, temperature has to be in Kelvin. That straightforward, we just add 273 to each of these numbers. >> Okay? >> And then what if necessary, look back at the previous slide, your notes. >> Once we have the temperature in Kelvin and what do we have to do to the data as it's presented? I'll give you a hint. Make a graph that looks like that. So what do we have to do to the data? >> Yes, I'm sorry, but they put it in. >> Oh, yes. Well, before we construct a graph, we have to do something else to the number that are up here. In other words, we're not going to use the numbers add up here, not even out yet. Alright, I mean, I'd be making sense. So I will simply show you the point is what we're plotting is the natural logarithm of the rate constant k versus one over the Kelvin temperature. So the point is first we add 273 to each of the Celsius temperatures to get Kelvin temperatures. Then we can use the LAN key on the calculator to convert each of these numbers into its natural logarithm. And if we do that, we get this information. Now here's one thing that's going to be important, especially when you do this in the laboratory. When you take the reciprocal of the Kelvin temperature, make sure you keep all of the sig figs you're allowed We have these temperatures to three sig figs that major allowed three sig figs. And the answers at these zeros in front are not significant. So 0.00336 is the reciprocal of 298.003100 is the reciprocal of 323. In a moment, you'll see why that's important. But the point is we're trying to make a graph of natural log of k on the y axis versus one over T on the x axis. What we really care about is not so much making the graph, but getting the slope of the line. >> Now if you have two data points, the slope of the line, he said a moment ago, slope intercept form. Okay. >> That's sort of kind of the right track. >> Yes. Okay. >> In other words, the definition of the slope is change in y divided by the change in x or y, y2 minus y1 over x2 minus x1. >> Okay? >> In other words, subtract these two numbers. That's going to be the numerator, subtract these two numbers, that's going to be the denominator. >> In short, set it up like this. Now, before we show you what the answer is going to be, let me just point something out up here. >> When I said be sure to keep all the sig figs you're allowed. Suppose you had written this number is just 0.003. And suppose you have written this number as 0.003. When you subtract them, what are you going to get? >> 0. >> And that means your denominator is going to be 0, which means the value is your fraction is going to be infinity, which doesn't make any sense. That's why you need to make sure you keep as many sig figs you're allowed, because when you subtract the numbers as they are, the numerator becomes negative 3, the denominator becomes negative zeros that are not negative 0 to six kelvin to the minus one. Well, you divide those two numbers and keep the two sig figs, you're allowed the slope of the line turns out to be negative 1.2 times ten to the fourth with units of Kelvin. >> Again, kelvin to the minus one in the denominator becomes Kelvins in anywhere. And now our goal was to find the energy of activation. >> But the point of constructing this graph at finding the slope, the slope is equal to the negative of the energy of activation divided by r, where r as the value of 8.314 joules per mole per kelvin. So to finish this problem off, all you have to do is multiply through by negative r, that cancels that negative sign. >> So the answer comes out to be a positive number, which it must >> Because all energies of activation or positive numbers. And then when you multiply this number by 8.314 joules per mole per Kelvin, kelvins divided by Kelvins cancels out. >> Nothing else does. >> So the units that you get here are joules per mole. And you could leave it pretty much as it, it'll show up in your calculator. 9.6 times ten to the fourth joules per mole would be a perfectly acceptable answer. Here. Some people like to divide by 1000 converted into kilojoules per mole, in which case, okay, 96 kilojoules per mole. >> Either one of those would be an acceptable answer to this question. So in lab, you're going to be running a series of experiments. >> Sometimes you're going to be changing concentrations, trying to figure out what the rate equation look like. Sometimes you're gonna be changing temperatures trying to figure out what the energy of activation looks like. Hopefully, based on things we've talked about here in class, got a better idea of how to do that. >> I will stop here for today. >> Good luck in lab. We'll see you on Friday. For those of you who have the Friday lab section, we'll see you right after lunch. >> Dollar recorded lectures going into your own personal EDA capture. Well, what if I want to make your brain not care about how well they let out the napkins. Computer calculator, we can really call rapeseed side way where we can do that. Alright? Alright, this is week 9.69 divided by 0.01. The answer comparable 60, right? I say, okay, they thought could work now because that's quite all right. >> So this is one divided by e to the i. >> So we'll put in one divided by 0 to one for whatever that meant, that whenever we have to divide that by one so that multiply divide rapidly about haven, like when? 1165. >> Thank you. Only mighty. >> Okay. Here's what else? You got the 1.02, wonderful. >> At 0.1 we get y. >> Let's put that in the memory. >> So with that memory, okay. >> Yeah, that would be one divided by what's in memory, right. >> Okay. >> Divided by the number of memory. >> But no, what do I tried to put it that might not believe it. Well, if you don't like working with number up here, Absolutely not. >> No, I don't have to do with the units of the actual numerical value should come out to be seconds recalculating it. >> Okay? >> I don't have to work with the horse. >> Everything will be made. Appointment comes not from a time in the class. It step-by-step. Alright? But why not have the practice problems? I don't know which investment weren't beyond. There's so many for me at this point. >> Okay. >> At this point, look at the breakfast then. >> Yes. >> Okay. >> So you have the answer. >> Just focus on fracked. >> Well, I'd say if you give me a question or you could do the ones at the margin on the answer key in the textbook we recommend. >> Or you can come and talk to me about any of those. >> We'll go with it. >> Carnegie work? >> Yeah. Okay. >> Well, don't be afraid to ask for help immediately.
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From Dana Chatellier February 26, 2020
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