Mhm. Yeah. Okay, So this is a pretty large question, basically, it states few propane right, will combust. So the first question is too right down the balanced equation for the complete combustion of the property death, death. So, let's start with that. So, normally, right, we have the propane gas as a gas face and the combustion will be involving oxygen, right? Which is also gas face. And for the complete combustion, the product the usually carbon dioxide as gas face and water as liquid face. Now, we can try to balance the equivalent you relieve for combustion reaction. The easiest way to balance direction is to look at the fuel itself, right? The combusted reaction itself first. So let's look at the field, right? The propane. So in the propane molecule there's three carbons. Right? So for the product, we will say, right, there would be three carbon dioxide And in the propane there are eight hydrogen. So for the water as a product, right? Because each water has two hydrogen, so there should be four waters as product. And then we can try to calculate how much or how many oxygen right in the product, which is for carbon dioxide. There are three times two. And for the hydrogen for the water, right? There are four times one, so 10 In total oxygen in vote interaction. Right? So for the reactant there should be five oxygen molecules because each oxygen molecules has true oxygen. And this is a balanced equipment. Now we can look at the second Ukrainian question which is calculated the modeling of air at 25°C.. And one atmosphere that is needed to completely Combust 25 g of propane. Yeah, yeah. So this question actually related to the story symmetry of the chemical reactions. So it says it will combust 25g per pain based on that. We can actually calculate the more of the propane right, consumed in the combustion, which is the mass of the propane over the more mass of the propane And the masses 25g. It's given in the question and the more mass we can do the calculation right? There are three carbons. So it's three times 12 pause a hydrogen. So it's eight times one. It's grams or more. And the calculated results is around Yeah. Mhm. Let me say it's actually around. Yeah. A point fi 68 more. Right. And according to the equation, right? Every time you consume more and more simple pain, you will consume five more oxygen. So based on that, we can get relations right for the reaction the most of the oxygen in direction, we equal five times the most of the propane in the reaction, Which is five times .568 More. Right? And it's equal to two points. It's for one more. And based on that, we can now calculate the mass of the oxygen that needed for the reaction, which equals to the more mass of the oxygen times the moors. That was oxygen. The Morris of the map more more massive. The oxygen is certainly true. Right, grabs for more. The morse of the oxygen is 2.8 for one more. And this is actually equals two 90 0.9 g. So the volume of the oxygen. Then we'll be right. The mass of the oxygen over the density of the oxygen gas. So, plugging the number, The mass of the oxygen is 90.98g. And the density of the auction is also given the says right one liter. So the oxygen gas contains 0.275 um grams of the oxygen. So it's actually 0.275 g per liter. And this would be equal to Mhm around 330.3 Features. And this is the answer. So now let's look at the question. See So the questions say, is um questions that gives the combustion gives the uh entropy of the combustion of the propane and also keep the formation energy of the um the water and carbon dioxide. And it asks you to calculate the heat of formation with the propane. So, let's write down all the information we have. Right? So, first, it gives right the combustion entropy entropy. With the combustion it was the propane. It's equal to -2219.2 killed your for more. And we can first we need to write down its corresponding reactions right? Which we actually already ah Roldan it's here. Right? So that's just write it down again. Right? We have propane combust with oxygen and forms carbon dioxide and water. And the second piece of information is the formation energy of the water. It's equal to Miners. point a key to joe for more. And we can write down into corresponding reactions right? Which? Sorry? Which is hydrogen As a gas face react or his oxygen. And we're forms the water. And the 3rd pieces of the information is the formation energy. I was a carbon dioxide equal to -393.5 kHz Joe for more. And the corresponding reactions, Okay, is this right? The combustion of the carbon into carbon dioxide? And these are three pieces information we get and what we want to calculate, right is the formation energy of the propane. This is what we want to tackle it and saying we need to write down its corresponding um reactions. So since it's a formation energy, right? Or entropy of the formation, So the product should be the propane and reactant should be the pure substance which is carbon right? And hydrogen. And we need to make it balanced three carbons, right? And eight hydrogen. So it's four ha jin molecules. So through the observation. So you queer in four, right? The camcorder queer and four actually equals two. You can say right, He goes to Chemical Aquarium four equals 2. There are four hydrogen. You say there are one hydrogen so equals two. four times the chemical equation too. Plus three times You can call your question. three miners a chemical aquarian war and same right for the and therapy. Then the S. O. P. I. Was uh Propane will be equal to four times good entropy information entropy of the hydrogen plus three times the entropy of formation of the carbon dioxide. The miners. The entropy of the combustion. That was the answer. Uh the propane. And we can plug in the numbers here, which is four times minus 285 point a kilo job more. Right. Plus three times miners. 390 3.5. So here I will not write down the unit but usually it's necessary. And miners Right? This combustion energy which is -2200 and 19 point two. And you will find the calculated results will be around Um miners 100 and 4.5 killer joe for more. And this is the answer. So now we can move to the question t Assuming that all the heat released in burning 25g of propane is transferred to four kg of water, calculate the increased the temperature of the water. So this is a a confirmatory problem. Right? So, first, what we need to do is to calculate right the heat transferred through combustion of the propane, which will be equal to right the morse the propane used times its combustion entropy of the propane. So it says by 25 g of the propane will be combusted and the morse of the propane will be used is actually calculated in the question be which is here? Right .568. So we can write down here The more so the propane is .568 and times Yeah, the entropy of the combustion of propane which is already given right. It's give me the question which is um miners 2,219 2 And this equals two -1.26 times 10-3 killed. You remember it's killed um material. And it's a color geometry problem. So we're no right. The heat change for the solution will be minus sign he change of the reaction. So this will be the same. And then we can also write down the expression of the key change of the solution, which is right uh specific heat of the water times the mass of the water and times the temperature change. And the specific heat of the water, which is 4.184 clams as no grams. But joe's right, joe program particularly. And the mass here he says, used like four kg of water. So it's 4000 grams of the water. Right? And time scale that he which we don't know. It's what we want calculate right? And this We equal to this right? We equal to 1.26 times 10-6 Joe. Now it becomes joe, right? And we can use the calculator to do the calculation will get. Temperature change is equal to 75 .4° C. And this is an answer of the question.