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Previous PageNext PageQuestion 31 (3.5 points) Propane gas, CaHg, a hydrocarbon, is used as a fuel for many barbecuesHow many grams of carbon are in 245.5 grams of ...

Question

Previous PageNext PageQuestion 31 (3.5 points) Propane gas, CaHg, a hydrocarbon, is used as a fuel for many barbecuesHow many grams of carbon are in 245.5 grams of propane?Answer with a number only; no units:Your Answer:AnswemQuestien 32 (3.5 poinas) lproname #a5 C Hywa hvdroc Urbolis used as a fuel for many barbeculeryPage

Previous Page Next Page Question 31 (3.5 points) Propane gas, CaHg, a hydrocarbon, is used as a fuel for many barbecues How many grams of carbon are in 245.5 grams of propane? Answer with a number only; no units: Your Answer: Answem Questien 32 (3.5 poinas) lproname #a5 C Hywa hvdroc Urbolis used as a fuel for many barbeculery Page



Answers

Propane, $\mathrm{C}_{3} \mathrm{H}_{8}(\mathrm{~g})$, is used as the fuel for some modern internal combustion engines. Methane, $\mathrm{CH}_{4}(\mathrm{~g})$, has been proposed by the movie industry as the post-Apocalypse fuel when gasoline and propane are supposedly no longer available. Assume that the fuel and oxygen are completely converted into $\mathrm{CO}_{2}(\mathrm{~g})$ and $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$. For each of these fuels determine the heat released per gram of fuel burned. Compare your answers to the answers for Exercise $40 .$

Hi everyone. So in this question they asked them it's propping but cycle following equations in three agent plus by or two used ratio two plus forest or how many grams support are needed to react fully with $3.45 profit. Be how many grams of CO two can form uh If we don't find $177 province, see how many grams of water are pushed by the commotion of 4.86. Small up a propane. So no. Oh God. A port you know grand support book. Mhm. I That is equal to 3.40 for you more. Love C. three H 8 into for you more. Love a conclusion upon one mould. Mhm C three H checked indu 32 g of opposition divided by one More. Love oxygen 90 sequel to 552. Yeah grandma or pretty then be part grandma. C. 02. That is equal to 0.177 more Love C. three H 8 into three more labs. Ceo to Dubai Dubai one mould C. Three. Okay. They checked into 44.01 grandma's you want to upon one more obscure do. And that is equal to 23 point four g of Seo Do. And then oh see part gram of water. That is equal to 4.86 small of C. three. attract into formal of water. Yeah. Mhm. What does Upon one more love C. three. A. Check into 18.01 g of water Upon one mole of water. Yeah. Mhm. It is equally too 3 50 g of Work. Mhm.

Hello. Today we're doing harm 61 heart. 10 week. The first part were asked to bounce the tomorrow chemical equation. For cocaine and for cocaine is a combustion balanced equation. So what we're gonna do is dance the equation. So what you're gonna do is we're five here. Yeah three over here and four. Okay over here Thank you. This is the balanced chemical equation of conduction of propane. Okay next we're gonna do quite beat cooperate doctor E. Productive if we are given the delta HF. So products minds reacted. So if we're left with negative one one. Mhm. Zero 0.5. Mhm. Mhm minus no one. One for three. Okay. Two yeah plus 104 Mhm point one. Yeah. Oh and if you calculate this yeah you're gonna get I mean no one here. Yeah. What you're gonna get? Yeah. Oh Trixie. Well you're gonna get negative. Okay you Yeah two one nine point mhm. One current. You? Mhm. Yeah. Uh huh mm. Mr Quint answer. Okay that's it for this video.

Here, we're taking a look at the equation that represents the combustion of liquid propane. So we have C3H8 as a liquid at 502 as a gas. That's an equilibrium with three C 02 At full H 20. Where delta H not is equal to the sum of the moles of the entropy of formation of the product. Subtract the some of the moles multiplied by the entropy of formation of the reactors. This value becomes negative 2 to 19.9 kg jewels. And so next we need to calculate a few different parameters. So delta N. G. The difference of GCS moles, it's equal to three. Subtract five. We're left with minus two, Temperature is 298 Calvin. We know the ideal gas constant 8.3 one for George, McKelvin pummel. The entropy change. We just calculated it here. So now we're in good shape to plug all of our values into the following equation of delta E. Is equal to delta H minus delta and G R T. We plug in all of our values, we get negative 2-1 4.9 kg jewels. Mhm

Okay, Question 84. We have to find the amount of carbon the outset produced, then 18.9 liters. All my opinion C three h h is burned in Austrian. The witness do you take. That's why I go do given Tracy or two. Bless Paul H. Two, now 18.9 liters has a density off 0.6 to 1. So the masses multiplied. Based little 16 to 1 unions. The more amounts off Japanese 44 on that of Carbonell cities. Also 44. So one beautiful grants of propane with that rate they into 44 g off common. Guilty united you so 44 g preparing hundreds, three into 40 Program carbon. So 1 g program in rates three in the body board apart. What people That is 3 g off. Probably so. 1 18.9 in two boys, 6 to 1, 6 to 1. You know, Graham off propane billion rate pinto. Three into 44.44. That is equal to three. Yeah, I don't want to be 35 point 21 reason


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