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Li|b) (4 points) 'Your product 3 is much morC 'soluble in dichloromethane than in water: Based surprise- What is what you 've learned in general and ...

Question

Li|b) (4 points) 'Your product 3 is much morC 'soluble in dichloromethane than in water: Based surprise- What is what you 've learned in general and organic chemistry however; this could be it about your product: structural formula that might lead you believe that it would be more soluble in water than an does not share this fcature organic solid? Be specific. Hint: reactant with product 3.(6 points) Give an extraction procedure that will remove any unreacted _ triethylamine, and

Li| b) (4 points) 'Your product 3 is much morC 'soluble in dichloromethane than in water: Based surprise- What is what you 've learned in general and organic chemistry however; this could be it about your product: structural formula that might lead you believe that it would be more soluble in water than an does not share this fcature organic solid? Be specific. Hint: reactant with product 3. (6 points) Give an extraction procedure that will remove any unreacted _ triethylamine, and acetic acid from the remaining solution of product 3 in dichloromethane_ You need to specify important), and _ state which what aqueous solutions you Il use (their exact concentrations aren of these three undesired materials will be extracted into that solution_



Answers

Two of the substances listed here are highly soluble in water, two are only slightly soluble in water, and two are insoluble in water. Indicate the situation you expect for each one. (a) iodoform, $\mathrm{CHI}_{3}$ (b) benzoic acid, (c) formic acid, (d) 1-butanol, (e) chlorobenzene, (f) propylene glycol, $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}$

So here we have hypothetical ionic solids so we can dissolve them in water. So for A X. What we have is A. X. Will disassociate into A plus an X minus. Next we have a X. Two. List associates into a two plus and two X minus. Finally we've got AX. three which dissociates into a three plus And three x minor. So the model concentration of each compound does remain the same for part B. So we have not point to five miles. And so the concentration of the eight plus irons in each of the solutions will be different though because the stroke geometry of the reaction. And so the marlins liability is greatest for the compound which associates into the largest number of irons per formula unit. So that would be the final example here. A X. Three. And so taking a look at the mall of scalability of the compound A. X. Consult for the mola scalability of this A.X. which is 7.4 Times 10 to the -4 Mola X two. 5.2 times 10 to the -3 Molar. Finally X. Three It's 1.2 times 10 to the -2 molar.

So here we have a scientific scenario. And so looking at the first part that we need to be answering. According to our scalability rules, nitrates are water soluble. So therefore K. N. 03 is soluble in motor most sulphites except NH 4 to S In Group one. So far it's insoluble. So N. H four to S. And K two as a soluble. Whereas Mns is insoluble. So most chlorides except silver chloride, HD two cl two and B B C L two soluble. So A G C L. Isn't soluble in water. Most sulfates except Bso. For cso, for srs oh for G two, S. 04, H G two, S. 04 PBS. So for our for soluble, so Bso four will be insoluble. And so from our sustainability rules, it can be inferred that the beakers contain M. N. S. A. G. C. L. And B. S. 04 are observed as they still contain the solid. Where we have the precipitates of this material. It is M and S. A. G C. L. BA. S. 04. There are precipitates. So looking at part B. The K plus concentration Is equal to not .1 moles divided by one L. That gives us not .10 Mola. So the rest the concentration of the case plus irons for the solutions of Kono three and K two baths. Next we can just look at the concentration of cable science for the solution of K two And that is not .20 Mola where the concentration of capable sciences twice compared to capable science from K. N. 03. So amongst the three given we have K. Two S. N. H four to S and M N. S. Well known K two S and H and H four to s of water soluble, where's reminisces water insoluble. So the faster materials are completely ionized. So what we have This K two s gives us two K plus on S two minus The next example, NH four to S gives us to NH four plus That S 2-. On the other hand, Mns forms an insoluble precipitate in water. So what we have is um And FC at H 20. And then to Paula At S 2-. So the concentration of the attitude I am for the solutions of n H four to S and K two S can be calculated from the concentrations of the salt directly because we have complete ionization. So the concentration of SG minus for the solution of men S cannot be directly calculated because we do not have complete ionization. And so in order to calculate the concentration, the scalability product constant of Mns must be known. And so we have the KsB values for these equations. So for mns, Okay, SPS, he puts 2.5 Times 10 to the -10. A G C L KSP is equal to 1.8 Times 10 to the -10. Next we have B A S. 04 Where the Kspn because at 1.1 times 10 to the -10. And so what we can do is solve for the concentration of a G plus, Which is equal to 1.341, six Times 10 to the -5 malls political, which is the same concentration of chloride anagrams.

So in this fusion reaction inward rotation is key i. 303 Plus who killed plus six at cl gives kind of isil the tribes of Cashel. Cashel first ride of actual oh this is one more this is too all 20 ml. Office Tokyo solution is about 30 envelope and by 10-K solution, morality of kale polarity of care solution Physical to 13-1 and two Do I read by 29 to 10 until 2 10. That is the product three by 10 million. More than 15 million of the solution. 15 to 3 by And that is the total 15 million miles. Okay I left and they acted with the agenda of the solution is the car to go into the into one by 10. That is equal to 10 million miles. Okay idea. At reactor with the D. N. A 3 15 minor standard is required to five million miles of a G. N. Representing indiana three solution is required to five molecular weight of a Gmo free media. Not reasonable to 1 17 weight of a genetically in the solution is 5.2 10 respond my necessary -3 into 1 70 Is able to 0.850 g percentage of AA DNA sample. 0.850 divided by one in 225

This question will calculate the quotient, which is the expression will be th three Seo too C two, age five Initial h 20 initial. See each three CEO to H initial and C two h five old h initial quotient. Is the trial value always based on initial values? So for part, A will substitute into this formula, and we have all of the values that are given. So the quotient for a when two to leave units out all the units. Similarities 0.10 0.0 and 0.10 solving gives us a value of 220. We're gonna compare this to the actual value of key equilibrium constant given in the question. So in this case, the Q value the trial value Bush in 2 20 is greater than the equilibrium constant of 2.2. And because of this, um, this is going to be the result of the equilibrium shifting towards the, uh, sorry equilibrium shifting towards the reactant ce uh, equilibrium shifts towards the left and the polarity. Because of this, since each two O's on the product side and it gets used up similarity of H 20 decreases. We can talk about ability of h 20 because in this, uh, equation here it is in, um, it's in the gas fees. For a part. Be, um, sorry, we're not told us in the gas phase, but since I it is a saw, you two, we need to consider it, and that'll come up in part to death. At the end of this question, part B calculate the question with the values that are given point you two 0.20 00 to 0 and 0.10 Solving here gives us 2.2, and in this case, we make the comparison. Q, which is 2.2, is exactly equal to the equilibrium constant, which is 2.2, which means that the system is at equilibrium and the polarity of H 20 because it's already at equilibrium. It's constant, and it stays. Uh, it's very it stays the same. We'll see. Gin will calculate quotient. Look in the values expression was at the beginning of the question. 20 dates. What one to the numerator denominator go for four 6.0 quotient work, so 2.40 q for zero is less than the equilibrium constant of 2.2. And because of this, uh, since Q is less than K, the equilibrium ships right to favor the products. And because of this, the concentration of each to o increases due to the shift towards the wrecked last part. Um, two more parts party questions. What for? For putting foreign the numerator through when he dates 10.0 denominator works out to 2.2 this case again, Q. You point to the exactly equal to K just 2.2, similar to what we saw earlier. Ah, system is at equilibrium, which means that the reactors and products are equally favored. And then the polarity of H 20 stays the same part. E more calculation here. That cushion again, plug in values. Here, you've got 2.0. Um, sorry for parts key. We were asked to calculate the amounts of each to open. Uh, we'll bring up the formula. Uh, we're not gonna use cue this time. We're gonna use the equilibrium concept. My apologies. All right. The formula for us here see each three. Go to C two h five. It's you H three c o to H. C 2 85 a wage and were asked to calculate similarity of water at equilibrium. The equilibrium constants 2.2 plugging the values that were given, you know, no similarity of H 20 10 my 0.0 and all of these of polarities. But we're going to be about the units for simplicity. Solving here won't go through the math. But when you solve this, we'll get a minority vague stool, which is 0.55 Moeller left the answer for Park hee. Part f were asked, Why is water included in the equilibrium expression of this reaction? We're told this question. Um so water is included because water is a saw ute. Um, and because it is so, I saw you two not a solvent on. They saw you two must be included in the equilibrium expression. We're not told what the solvent is in this question here. We're just told that the, uh solved, you know, it's a non reacting solvent. It's not water. And since it's not water, water is part of the equilibrium expression. That is why we're including it. That's why we can solve for it. Ah, and use it in a question value for all of the parts A through E


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