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LetEMemgt ELLLIQuestionAdncto (arching the ecos Yoo much tCxt messabing group of teerapers He nas already shown that the 312 with 4ndard deviaticn pcvans Kng Jverag...

Question

LetEMemgt ELLLIQuestionAdncto (arching the ecos Yoo much tCxt messabing group of teerapers He nas already shown that the 312 with 4ndard deviaticn pcvans Kng Jveragc numbcr ol text mess ges sent ner day bt tcrnnsers i5 mani Mcseges scparste the the lowest 13%4 trom the highest 85% empling € strbution jeenaecrs, Usc th : Iable below:0,08 10.090dJ / 0,095,10,09340.092 0I15 0 1120,08 UUBC0i0n 00970.218

LetE Memgt ELLLI Question Adncto (arching the ecos Yoo much tCxt messabing group of teerapers He nas already shown that the 312 with 4ndard deviaticn pcvans Kng Jveragc numbcr ol text mess ges sent ner day bt tcrnnsers i5 mani Mcseges scparste the the lowest 13%4 trom the highest 85% empling € strbution jeenaecrs, Usc th : Iable below: 0,08 10.09 0dJ / 0,095,10,09340.092 0I15 0 112 0,08 UUBC 0i0n 0097 0.218



Answers

I low to we differcntiare betwecn $\mathrm{Fc}^{3}$ and $\mathrm{Cr}^{3}$ in group III? (a) by increasing NII ' ion conccntrarion (b) by decrcasing Oll ion conccarrarion (c) by atlding cxccss of $\mathrm{N} 11$ olt solurion (d) borh (a) and (b)

So for this problem, You have three different constitutionalist summers of a means with the structural formula of C3, age nine. And and you want to rationalize why these different structures have such variations in their boiling point. So remember that when you're thinking about boiling point, you have a higher boiling point is going to have stronger hydrogen bonding. And you're going to have this hydrogen bonding between and and hydrogen on the same molecule. So it's going to hydrogen bond with itself. So when you're looking at these structures, you have, In all three cases, you have been nitrogen in terms of this trim ethyl amine, there are no hydrogen for it's a hydrogen bond with. So because there are no hydrogen bonds, you know, it's going to be very easy to kind of break apart this structure, allowing for lower boiling point and by break apart the structure, I mean kind of break apart bonding or the attraction between one molecule of try muscle. I mean to another molecule try muscle, I mean, however, when you have Um problem in which is the boiling point of 48 And your ethyl methyl amine which has a boiling point of 37, they are considerably higher. And this is because both of these structures can form hydrogen bonds um within with themselves. However, because your proposal, I mean this is a primary, whoops primary mean, and this is the secondary I mean, and then this is a tertiary, you mean your primary mean is going to be less hysterically hindered. So because it's less hysterically hindered, it's going to better be able to form hydrogen bonds. So it's going to have a higher boiling point then your secondary mean? Which now has a little bit of Starik hindrance, but it still has a pretty high boiling point. But just know that you know, as you kind of get a more substituted I mean, so as you increase substitution, you're going to have a decrease in your boiling point.

So in this problem, the correct order of die carbonnel compound, the correct order of dying garbo nail come phones in decreasing order in decreasing up there in all, contain marriage follows, so I'm just writing the order here. Just look at it carefully, voice after that food, Then 3rd and then 2nd according to the option option B. H, correct answer.

Question is that decreasing order of nuclear Felicity among the nuclear files? Okay. McClure files strength increase as now negative charges, donation power will be increased. Okay means that negative charge should not be included in the resonance. Also not be highly electron negative and um correct. So we have molecules. First one is PFC. They were going to open a negative charge c. CN minus C H three S double bond double four minus. Okay. Now we can observe that this negative charge is contributed with carbon development and also Benji. And this negative charges also contributed with the to sell it for carbon oxygen. I only contributed highly fondue get the and this negative charge is the sp hybrid carbon have. Okay and here is density of negative charge increase because Al Qaeda group is the electron releasing group. So this molecule will be so be molecule the molecule will be the strongest meatloaf. I'll be dancing. But in the end entity in the area of some negative charges contributed with them. Here is C. S. Tree CS three. Okay, so in this area of some negative charges contributed with only carbon double bond oxygen, not into the complete molecule. That's why a molecule will be will be more than a glorified standard. The molecule the molecule is the highly concentrated with that to santa, fe, carbon oxygen. Okay. Thank you.

In this problem before adding before adding the re agents the A. Re agent of Group three of guru three Before adding the the Agent of Group three. The solution the solution H he did with It's heated with some concentrated some concentrated At 1003 solution HN or three solution in order in order to oxidize to oxidize FE two plus who? FE three plus, so Option C. It's correct here.


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