 So let's do this bomb calorimeter problem. It says the burning of 1.01 grams of C12 H22011 in a bomb calorimeter causes the temperature of the water to rise from 24.92 degrees Celsius to 28.33 degrees Celsius. The calorimeter contains 980 dot grams of water. Heat capacity of the bomb calorimeter is 785 joules per degree Celsius. What is the heat of combustion of this sucrose in kilojoules per mole? So notice they're doing it in kilojoules per mole here. So that's important. So anyways, let's write down what we got here. So we've got the mass of our substance, mass of C12 H22011 is 101. And we know the initial temperature, Ti, is 24.92 degrees C. Tf is 28.33 degrees C. So we can figure out the change in T. Something I'll need to calculate. So change in T is Tf minus Ti. So it also gives us the mass of water, heat capacity of the calorimeter. So this is C cap, 85 joules per degree Celsius. And eventually it's going to want us to tell us what the heat of combustion is for sucrose in kilojoules per mole. So the one thing I hope you notice already is that this is in joules. The other thing is we need to know the heat capacity of water, which is also in joules. So let's go ahead and do this. So this is the equation you use. Q total equals the Q of the water plus the Q of the calorimeter. No, no, this is from... Similar, yeah. So what do we do? Remember Q equals mc delta T. So in this case, m for water is 98.0, I'm sorry, that's a bit, 980 grams. The heat capacity, 4.184 joules per 1 gram degree C. And the change in temperature, 3.41 degrees C. I like to put brackets around all of that stuff, because we're going to have to do another one. So again, remember I showed you, joules per degrees C. So we don't really have to worry about the mass of the same thing, or the mass of the calorimeter. So in this case, it's going to be the change in temperature is 3.41 degrees C. And the specific heat of the calorimeter is 785 joules per 1 degree C. Look at that. Let's cancel our units. Grams, grams, degrees C, degrees C, degrees C, degrees C. Hopefully you see we've got joules and joules. Joules added to joules is joules. That's a good energy unit. So 3.41 times 4.184 times 3 sig figs for this one. Well, let's just, we'll just do it the 3 sig figs. 1.67 times 10 to the 1, 2, 3, 4 kilojoules. So that's the number of kilojoules, I mean joules, sorry, sorry, sorry, sorry. The number of joules for 1.01 grams. Okay, that's not necessarily a mole. So let's go about and figure out how many moles of this stuff we got. How do we do that? So we've got to convert grams to moles. How do we do that? Molar. Molar mass, right? So the molar mass of C12H22O11 is going to be, unless somebody else fits it. Yeah, it should be something. So how do we figure out the number of moles we got? So number of moles, C12H22O11 equals 1.0 moles. Okay, so this was joules and it said kilojoules per mole, right? So let's take this, 1,000 joules, 1 kilojoule. So we've got 16.7 kilojoules for that many moles. Is everybody okay with that? That many moles gave us 16.7 kilojoules of energy. So if we wanted to know how many kilojoules per mole, we would take that, divide it by that, and that would give us the kilojoules per mole, right? So all we got to do is take that, 2.95, that's 10 to the negative 3 moles, 560 kilojoules per mole. So this will work for 660 kilojoules per mole. Okay? Is everybody okay with that? So remember, the reason we could do this is because we know how many moles we put in. Okay, so if we take that number of kilojoules, divide it by that, that's going to give us per 1 mole. Is everybody okay with that? So this is going to be the last thing you need to know. Questions? Questions?