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Eutosane00 6 0 6 ,chem 211LabGavecMN MacHomeInserDratCesicnLavdutReterencesMailinggRevle %viewAcrcbatTeli MeShareComment:Callbi11LARAC-DAFELibO MF;bCc_c UspF -ifr AaBbi ~auui > aelira=MumNc ScadogSkisDicaiicCCrcicandshac Kcouc Aula PDC sumalu=3- We have crude mixture of the following compounds We are purifying the crude mixture by columm chromatography packed with silica gel. we decide tO use Hexanes and Ethyl acetate {1;1 ratio) as solvent; which compound would get eluted first? Which comp

Eutosane 00 6 0 6 , chem 211Lab Gavec MN Mac Home Inser Drat Cesicn Lavdut Reterences Mailingg Revle % view Acrcbat Teli Me Share Comment: Callbi 11 LARAC-DAFE LibO MF; bCc_c UspF -ifr AaBbi ~auui > aelira= Mum Nc Scadog Skis Dicaiic CCrcicandshac Kcouc Aula PDC sumalu= 3- We have crude mixture of the following compounds We are purifying the crude mixture by columm chromatography packed with silica gel. we decide tO use Hexanes and Ethyl acetate {1;1 ratio) as solvent; which compound would get eluted first? Which compound would get eluted last? Explain your answer in detail: (3 points) Sivmorn Hei 67 jnrcs Fnolish Wnited Satesl Fncus 150%



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The two compounds whose structures are depicted here are isomers. When derived from petroleum, they always occur mixed together. meta-Xylene is used in aviation fuels and in the manufacture of dyes and insecticides. The principal use of paraxylene is in the manufacture of polyester resins and fibers (for example, Dacron). Comment on the effectiveness of fractional distillation as a method of separating these two xylenes. What other method(s) might be used to separate them?

This problem introduces the concept of the mixing ratio, which was also introduced in problem 1 29. The mixing ratio is simply an expression of concentration as a function of the volume of the particular species per total volume in parts per million. It's a parts per million volume ratio in Problem 29. It defines the mixing ratio as the volume of the species in question at STP, divided by the volume of the total gas at STP times 10 to the six. So in the case of benzene to solve for the volume of benzene at STP, we need to take the mass of benzene and convert it to moles of benzene by dividing by the molar mass of benzene. 78.11 g per mole. Now that we have moles of benzene, we can use the conversion factor of 22.4 to convert the moles of benzene at STP to leaders of benzene at STP. Well, then, divide that by the volume of the entire gas at STP, were given the volume of the gas nodded STP as three leaders. This is the volume at a pressure and the temperature different from STP. So all we have to do is multiplied by the ratio of pressure over temperature in Kelvin and then convert it into the STP temperature and pressure my multiplied by the ratio of the temperature Um STP, which is zero Celsius or 2 73 Kelvin, divided by the standard pressure which is 7 60 tour or one atmosphere. This then will get us out of the volume at 7 48 Tour and 23 degrees Celsius and into the volume at STP. Then we simply multiply this by 10 to the sixth and we get 9.44 times 10 to the negative three parts per million volume. Now to calculate the number of moles of benzene that is president in a centimeter cubed of the entire gas mixture, we simply need thio again. Take the mass of benzene and converted into moles of benzene by dividing by the molar mass of benzene and then convert the moles of benzene into molecules have been seen by multiplying by alpha god rose number well, then divide that by the volume of the gaseous mixture, and it does not specify STP here, so it must be the three leaders that were sampled. So we'll calculate the molecules of benzene that air present per the volume specified three leaders. We can then convert the three leaders into Centimeters Cube, recognizing there are 1000 centimeters cube per leader. This then gives us 2.30 times 10 to the 11 molecules of benzene per centimeter cube of the gaseous mixture. The second part is to do the same thing for Paul you in. The only difference here is there's a different mass of tall you in and a different Moeller Mass for tall Ewing. But the set up otherwise, is exactly the same. So we will take the mass of valuing 153 10 of the negative nineties nano nano grams converted into moles by dividing by the molar Mass, and then multiply by 22.4 to get the moles into leaders at STP. Well, then divide by the volume of the gaseous mixture at STP and on the tail end. Multiply by 10 to the sixth and we get 1.37 times 10 to the negative two ppm V. Now, to calculate the molecules of Halloween that are present in a centimeter, cubed. We'll do what's similar toe what we did before. Take the mass of tall Ewing and nanograms converted into moles by dividing by the molar Mass and then multiplied by Allah. God rose number to get the molecules of valuing for the three leaders, which we then convert into Centimeters Cube by multiplying the three by 1000 and we get 3.33 times 10 to the 11 molecules of tall ewing per centimeter cubed of the gaseous mixture.

So in this question were given these substances and we're told that we placed 10 milliliters of each liquid and a small amount of each solid until graduated cylinder. And we're expected to predict what we see. So we're given the densities of all these different substances an important thing to remember what densities is. When you place admissible liquids together so that they don't dissolve, the highest density will sink to the bottom or the lowest density will float. So all the information that we needed is right here. We have the densities. So given this information, now, let's just go ahead and rank the densities. So we have. Will it rank them from highest to lowest? So with the highest density, Teflon has a density of 2.3 grams per cubic centimeter seconds. We have per floor axing with 1.669 and then we have PVC with 1.36 and then we have water with one http with 0.97 and heck, sane with 0.766 So if we place all these in a graduate cylinder would expect tough, want to be at the very bottom, followed by Per floor hacks eightfold by PVC and then so on and so forth. So let's go ahead and answer this question that started the bottom and go work our way to the top. So we expect six different Liq, six different layers as well, since these are all going to be admissible and they don't dissolve in one another. So we have 12345 Let's make one more room. Okay? And again, remember, here is the highest density and appear. We'll have the lowest density. So let's go ahead and put these in at the bar. Very bottom. We have Teflon with the highest density, followed by C six F 14 followed by PVC, followed by water, followed by HDP. Eat and finishing off with hexane. There we go. Here's the order that we will find the substances to be in. Oh, that was meant to be a square. Sorry,

In this problem, in this problem, the outside of the oxides of C 02 as to oh and S or two ah present, respectively. In this problem, the oxide of CO two H two, n s. O to our present, respectively. Therefore, according to the option, Option B. Agent, correct and said option B is correct. Answer for this problem.


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