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Consider pcis Pcly - (9) (4) + C(zu4 ) Intally 0.60 mole ot PCls S Plaud In 1.0 L Flosp librwum lre Is 0.16 mole & ecls netlae WVCt @(e AT equi " Jtc conc...

Question

Consider pcis Pcly - (9) (4) + C(zu4 ) Intally 0.60 mole ot PCls S Plaud In 1.0 L Flosp librwum lre Is 0.16 mole & ecls netlae WVCt @(e AT equi " Jtc concentvuhon} OF Pcls and clz te eqjlibri= lbylum lonstunt kc tor *nlf uchon 2-Wkat 4 I: equi Ljil Ap_ Clz (S added @ X efu lbnum mixture 3-k 0..20 mole Df or O ecls IneeaX 0( deereale Cor GAth

Consider pcis Pcly - (9) (4) + C(zu4 ) Intally 0.60 mole ot PCls S Plaud In 1.0 L Flosp librwum lre Is 0.16 mole & ecls netlae WVCt @(e AT equi " Jtc concentvuhon} OF Pcls and clz te eqjlibri= lbylum lonstunt kc tor *nlf uchon 2-Wkat 4 I: equi Ljil Ap_ Clz (S added @ X efu lbnum mixture 3-k 0..20 mole Df or O ecls IneeaX 0( deereale Cor GAth



Answers

From the previous problem $$ C_{V}=\frac{v_{1} \frac{R}{\gamma_{1}-1}+v_{2} \frac{R}{\gamma_{2}-1}}{v_{1}+v_{2}}=15 \cdot 2 \mathrm{~J} / \mathrm{mole} . \mathrm{K} $$ and $C_{p}=\frac{v_{1} \frac{\gamma_{1} R}{\gamma_{1}-1}+v_{2} \frac{\gamma_{2} R}{\gamma_{2}-1}}{v_{1}+v_{2}}=23 \cdot 85 \mathrm{~J} / \mathrm{mole} . \mathrm{K}$ Now molar mass of the mixture $(M)=\frac{\text { Total mass }}{\text { Total number of moles }}=\frac{20+7}{\frac{1}{2}+\frac{1}{4}}=36$ Hence $c_{V}=\frac{C_{V}}{M}=0.42 \mathrm{~J} / \mathrm{g} \cdot \mathrm{K}$ and $c_{p}=\frac{C_{p}}{M}=0.66 \mathrm{~J} / \mathrm{g} \cdot \mathrm{K}$

Mixture of this compound was prepared in that solution. The first one is given there to see you next three on five. As for for we are and the second one is Ceo and it's free All five We are correct. The number of moles of I and J R both have terrible yeah, In that two liter of solution. So modernity will be 0.01 and zero point by volume. So there is the correct option.

In this problem three Bse every two three Bs here, too. Plus two and italy your food will react to give be it three Vo four. All too last six and the season. So the limiting value are I'm writing here. So just look at it carefully limiting the agent, limiting at the agent it and it too bill for 0.2 moon. We need to gibbs .1 More. Love be it today, be your food called to minus so option age, correct answer here.

Let's determine the empirical formula for each of the following a. We have 1.245 grams of nickel And 5381 grounds of iodine. Where did you beat each of these two moles? Nickel has a mouse of 58.7 g per mole Equal to .0212 malls and iodine. Yeah. has a mass of 126 .9 g per mole. Yeah. Okay. 2.42404, 2, 4 more. Were divided by the smallest. Just .0212 to be one. Yeah. and two. So the empirical formula here works out to be and I too. For A for B We have 2.677 g of barium. Yeah. And 3.115 g of roaming. From where both of these two moles. Barium has a mass of 137.3 g. Put more. Yeah, Forgive me .0195 moles. Yeah. Roaming has a mass of 79.9 g per mole. Yeah. This would give me a zero 390 rolls divide by the least. Yeah. One Mr Give Me two. Therefore, empirical formula would be a B R two and for C. Yeah, We have 2.128 g of beryllium. Yeah. 7.557g of sulfur And 15.107 g of oxygen. Yeah. Ready to these two moles, Brilliant smaller, massive nine g per mole. Yeah, We .236 moles. Yeah. Yeah. So for Last 30- one grounds for more. Yeah. Mhm .235 laws and oxygen. 16 g from all. Looks at 2.944 Walls divide by the smallest which is .236. We got 1 1 And four. Therefore the empirical foreman here would be ESO four

We were trying to figure out how many moles of the compound have We got the first figure out the molar mass of the compound and then convert grams that are no more. So you think periodic table, You just look up the molar mass for each of the I also in the compound and then multiple. I have many, um, as you have and then add it all together. That would be Imola message the compound and then we that convert from very simple. So let me show you what I need. Let's take a water H 20 as an example were first gonna figure out the molar mass of water over his state. See what the molar masses for hydrogen and most like that I choose. Since we have two hydrogen atoms, right, Molar mass for hydrogen is 1.1 grants formal and we have two hydrogen atoms, so that gets more blood. Then you add the mola master oxygen, which is 16. And when you do this way, we get 18.2 grants from old. And since we know that we have four points by rams of where you just, uh, possible slide out with the grants normal, which is the molar Mass. So for one mole of we have 18.2 grand, these grants would just cancel each other out, and he looks those with moles as the unit. So when you basically divide or 0.5 by 18 02 you get 0.25 0.25 more released you up. Let's do a couple more examples. Let's take the example of very and hydroxide remember, savior of the molar mass reach element, and that's going to give us ultimately the molar mass for the entire compound would, you know, by looking at periodic table. So we have one. Very, um, I on And the molar mass from Mariam is 1 37.33 That phrase Add that to the number of oxygen's that we have in the molar mass for oxygen. So we know that we have two oxygen here and the mullahs master, that is 16 grams. We're also going to add that to the hydrogen atomic master molar mass. We have two engines, and the master bed is one point. What? And multiply all this and Advil together we're gonna get 171.35 raised from all of the whole compounds. That's for Berrien outside. Now, Uh, let's see what we know, which is that there's 471.6 grams very hydroxide. We're gonna cross Multiply it by so we know that for every mole for everyone mold, we have 171.35 grams. So now by dividing 471.69 are sorry point. Sit by that by 171 point I can't seem to get those 2.75 Mol bearing hijack side. Let's do 1/3 1 Let's do iron. I think this iron too bossy. Take the molar mass for each element Ridge most like that by how many as we have. So for iron we have three iron mull of as rambles I by the molar mass which is, um, 55.8. And that's great from old then for offi we know that there are two boss eight hours. Sorry. So we're going to multiply two by 30.97 Nothing brings formal. We'll add that to our oxygen's and we see that we have eight Oxygen's here almost five 18 grand. We multiply that all up and added, Together we're going to get 357.49 and that's brands from all of the whole compound. I didn't shoot bossy. Then we will go ahead and cross. Most live that by what we know is the mass in grams, and that's going to give us the mold of this compound. So we know that we have 129.68 grams across most by that bythe mass. So for every one mole of this compound, you have 357 24 9 brands. So when you define 1 29.68 by 3 37.49 you're gonna end up with point every 6 to 8 moles of this iron too bossy


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