Okay. So to start with two given information here is that when one mole of nitro Glazer and decomposes, we end up with a collection of gases that at one atmosphere and 1950 degrees Celsius take up 1323 leaders. And in the first part of the question is, if we have 0.4 moles of nitro glycerin, how many moles of gas are we going to create eso to start off? Let's figure out how many moles of gas we got from one mole of nitro glycerin. So one atmosphere 13 23 leaders 1950 Celsius. Let's go ahead and use PV equals NRT are ideal gas law we're solving for end So P V over r t equals end. And so what that tells us is our pressure is one atmosphere are volume is 1323 leaders are our value is 13230.8 to 1. And of course, as always, we need to convert into Kelvin S O. That would be 22 23. Uh, kill them. So let's go ahead and do that math out. So 13. 23 divided by 230.8 to 1 divide by to 2 to 3. That gives a 7.25 moles of gas created for every one mole of nitro glycerine. So if we take our 7.25 and multiply it by 0.4 we get 0.29 moles of hot gas being created. So that is part a of the question. Uh, part B is telling us that we're going to Ah, I would do this in a 500 milliliters flask on. Cool it to negative 10 degrees Celsius. Ah, and then the pressure inside the flask is 623 after product, A solidifies. So, uh, what we're gonna do here is we're going to look at product a solidifying some. First of all, it is how many old moles of a were present. And what is its likely identity? Well, we know that whenever we have hydrocarbons, combusting or exploding, usually we get co two and H 20 as products. And so h 20 is a very likely product that would solidify by the time we got to negative 10 degrees Celsius. But what we do need to figure out then is how many moles of gas we have left. We're going to assume that the ice, if it is ice, has no volume. It's pretty common to assume solids have no volume on and we're going to at a new page. We're going to use P V over r T equals and to figure out how many moles of gas we have left from our original 0.2 nights. So the pressure now is 623 millimeters of mercury. The volume is 0.5 leaders. The our value since now we're in millimeters of mercury is going to be 62.4 and the temperature is negative. 10 Celsius, which is 263 killed in so are in value now is zero point zero one nine. So originally it was 0.29 Now it's 0.19 which tells us that we lost zero point 01 moles of H 20 Uh uh make sure that's what the question is. Actually asked how many moles of a were present? Ah yes. So we have 0.1 moles of a were present. Uh, part C is asking us eso when when these air passed through, when what's left is passed through a tube of powdered ally to oh, gas be reacts to form Ally to CEO. Three. The remaining gas to see India collected another 500 milliliters flask, a found of a pressure of 260 millimeters of mercury at 25 degrees Celsius. How many moles of B were present and what is its likely identity? OK, well, uh, let's start with the identity cause that's pretty easy If we have alli to oh, we have to add something to it to get Ally to CEO three. Then clearly what we are adding to it is CEO, too. So for adding CO two, that's that's the identity of be there. And the question is, ah, when. So the question is how many moles of of that co two did we have. So we now have a 500 milliliters flask, 260 mil millimeters of mercury and 25 degrees Celsius. We can use the same equation using our new pressure of 260 millimeters of mercury. Another 500 millimeter flask are. Our value is still is still 62.4 and our temperature is now 25 degrees Celsius, which is 298 Kelvin. And so what we have left then to 60 times 600.5. Uh, divide 62.4. Divide to 98 gives us 0.7 is what's left. Which means that what we lost is 0.3 Moles of co two. Okay, so that's part C. Finally, we have part D on and then I guess we have party afterward. But let's move on to part D, Sephardi says, When gases cnd were passed through a hot tube of powdered copper gas, see reacted to form copper two oxide, the remaining gas D was collected in 1/3 500 militar flask found to have a mass of 5000.168 grams and a pressure of 223 millimeters of mercury, a 25 degrees Celsius. How many moles of CND were present and what are their likely identities? Okay, so moving on to a new page s So we know that when we added copper to something, we ended up with copper two oxide. So obviously this gas had to be 02 That's what gas see is and what we are going to remember here is that the remaining gas. So the nitrogen gas that's left in a 500 military flask Ah, a lot of pressure of 223 millimeters of mercury, it 25 Celsius. So going back and using our same equation. 223 millimeters millimeters of mercury millimeters of mercury s times are 0.5 leaders divide by our 62.4 and by our 298 gives us to 23 times 0.5 divide 62.4 divide to 98 point 006 moles of whatever D is, which means that since we had 0.7 before, we have point 00 one moles of oh too. And 0.6 moles of D that 0.6 moles of d we also know takes up a mass of 0.168 grams 0.168 grams. Divide by 0.6 Moles gives us a Moeller mass of 28 grams per mole and based on the ah, compounds that we have, the elements that we have in our original that tells us that are 28 grams from all is probably in two. So then the last thing we have to do is we have to add it all together. We have to get our, uh, our full balanced equation here. So we know that we end up with 0.6 moles. Pretty much everything as we've been doing this has has come out very close to the to the thousands, starting with the 0.4 moles of nitro glycerin. So if we just multiply everything through by 1000 we should be able to get our equation. So we have our four moles of the Nitro glycerin, which is C three h five in 309 And what do we get out of it? We get ah, first the water vapor and how much water vapour did we get? We got 0.10 moles, so multiply that by 1000 and we have 10 h 20 and then be waas points Here is there are three moles of CO two So that's three co two, see, was our one. Oh, too. I'm not gonna write the one. I'm just gonna right 02 And then finally six into and nice and convenient that we finished right with that end to right there. And now, let's double check. Always good to double check. Make sure that we have proper balancing here. So our carbon is not balanced properly. Why not? Did we end up with the wrong amount of carbon there? Yes, I see what I did. Okay, I subtracted this 0.0 Zahra seven from the amount of water instead of from the amount of Ah, the amount that was left the 0.19 So I should have got 0.0. Let me go ahead and fix that right here on this part. Here. It's not 0.3 It is actually 0.12 moles of carbon dioxide. And so what that's going to do here is it's gonna do same thing instead of three carbon dioxides. Of course, it is going to be 12 carbon dioxides eso Now we can see that we've got 12 carbons and we've got 12 carbons right here. We've got 20 hydrogen ins, and here they all are 20. Hydrogen is right there. We've got 12 nitrogen ins and here's 12. Nitrogen is right here and then we've got 36 oxygen's and out of those 36 Oxygen's there's 10 of them here, plus another 24 that's 34 plus another two here. That's 36. So this balances out. That is the correct equation for the explosion of nitro glycerine.