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For 39,62gof the compound how many moles of C in it?0 17.198114600 2292 &0 4298 €None of these...

Question

For 39,62gof the compound how many moles of C in it?0 17.198114600 2292 &0 4298 €None of these

For 39,62gof the compound how many moles of C in it? 0 17.198 11460 0 2292 & 0 4298 € None of these



Answers

How many moles of compound are in (a) $46.1 \mathrm{~g} \mathrm{NH}_{4} \mathrm{Cl}$ ? (b) $22.8 \mathrm{~g} \mathrm{CH}_{4} ?$ (c) $9.63 \mathrm{~g} \mathrm{CaCO}_{3} ?$

So we were running through some basic calculations between moles molar mass and weights. So firstly we have the Maciver to that eight g. And so the molar mass is 32 g per mole. And so we can multiply the massive 02 with the conversion factor one mole of 02 divided by 32 g of CO two. To calculate the number of moles. three take 8g, multiplied by one mole, Divided by 32 g. We get 9.250 moles. Next, what we have is Ammonia NH three. So we've got 10.7 g and the molar mass is 17.03. So we take 10.7 g multiplied by one mole Divided by 17.03 g to obtain 9.6- eight moles. So this is for 02 and for NH three respectively. Yeah.

Answer for the question to the formula of the given compound. See, it will is 22 0 11. So one model of the compound contained compound contained contains 12 myself carbon, 22 more self delusion and 11 moles of oxygen, Yeah.

Hello. In each of these samples, we need to calculate the moles of carbon that is present. So let's go ahead and get started in letter. A. We have 2.5 moles, okay of nothing. Ch four. Okay, As we look at that formula for methane, we see that in every mole of ch four, there's only one carbon. So for every mole of ch four, here's one mole of carbon. Therefore, the moles of carbon are going to be the same as the number of moles of methane or 2.5. Oops. All right, so 2.5 moles of carbon in part B. We're starting with 0.115 moles of C two h six. In this formula, we can see that there are two carbon two moles of carbon for every mole of the compound. So in every mole of C two h six, there are two moles of carbon. So taking 0.115 multiplying by two gives us 0.230 moles of carbon people. I kept three significant figures because that is what we were given in the problem. Alright. In letter C, we have 5.67 moles of C four h 10. This compound has four moles of carbon in every mole of the compound. Yeah, okay, so I need to take my initial number of moles and multiply by four. Rounding to three significant figures gives me 22.7 moles of carbon. And finally, in letter d, we have octane. Check it and we have 25.1 moles off the octane C eight h 18. This has eight moles of carbon. According to the formula, we have little subscript of eight there in every mole of C eight h 18, multiplying 25.1 times eight and rounding it to three significant figures gives me 201 moles of carbon. Mhm. All right. And thank you for watching.

In this problem, I'm writing the reactions yesterday, poultry CBR In presence of CS three. See, okay here three get positive sign. We'll give CS three all to see the world bone CH two in presence of NBS. It will give CST see CH two B. Table one, Ch 2 and in presence of whole al khali KMN 04. It will give C. S. Story, see C. H two B and yes to the CH two who H. Oh to come bombs? He has one carrier carbon atom. Therefore it has to sterilize Hamas. So the answer for this problem will be equal to.


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