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NgstI"PreseniedJCCC & dafrmethr netwokFEpTEALILLLach Tuudthe house? tha door connecting eachcoon??Anetet...

Question

NgstI"PreseniedJCCC & dafrmethr netwokFEpTEALILLLach Tuudthe house? tha door connecting eachcoon??Anetet

Ngst I"Presenied JCCC & dafrme thr netwok FEpTEALILL Lach Tuud the house? tha door connecting each coon?? Anetet



Answers

Name each amine.

a. $\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}(\mathrm{NH}_2)\mathrm{CH}_3$

b. ($\mathrm{CH}_3\mathrm{CH}_2\mathrm{CH}_2\mathrm{CH}_2$)$_2$NH

This question says to use sigma notation to write for some. So you have your sigma and you really only have a beginning term and you're ending term to go on. They didn't give you anything in between. So if you kind of look at what's changing as you go from the first term to the last term, you've got to over an which is two men over end. So you can kind of see that this part up here is changing. And then you've got this minus two over an and it goes to minus to an over in so you can see that that numerator is changing. And so what you notice is not changing. Is this too over in and two of her end So we can kind of put that out here and not included in our summation. It wouldn't really matter if you included it or not. We need to pick an index estimation. I'll just pick a for this one. So where does case start and where does he end? So he's would be starting at a value of one and ending at a value of and because if you use your last term here, which, when there's variables involved, it's common to use the last time you can kind of see that you're plugging in something in that numerator. So if you think about this, you would have to win over an. So that's your two K over an and all that being cubed minus and then you could see that you had to end over end. So your index is changing, and that would be two K over end, and that would be your expression to write your son.

All right. Definitely one of the more interesting problems here. We have to see what's wrong with this truth, so to speak, which I could even put in air quotes. But let's just go ahead. In part by part, P event denotes a statement in any group of end cats. If one cat is black, then they are all black. That's definitely true. Friend is one cause in a group of one cats if or what? A group. It's not in a group with one cat. If it's black, obviously that whole group is black. So we're going to suppose p f k is true and show P f K plus one is also true, which is just mathematical induction. And they have that nice visual. But I'm just going to kind of summarize you've k plus one cats. We know that this tad polka is black. So we remove other Another cat called Sparky from the group left with Kay cats, one of whom is tadpoles black. So by induction, okay, of these cats are black. But now we're putting Sparky back in and we take Tad pull out again. We have a group of K cats, so we're basically saying that that group of K cuts with Sparky in without Sparky is exactly the same, and that is definitely one assumption that can't be made. We can't kind of say and I we can't say that I'll quit. Esther Sparky is the exact Shane as one, which is the exact same as T for Tadpole. We we can't really assume in this situation that both cats are quantifiable merely by a number, especially when we're talking about the color of the cat. I mean to say that they both equal one, and if you just remove it, then we're able to say that it's just does not work to say that they are exactly equal and on Lee, quantifiable by this number one because they're each one cat. And so that's the main problem. It's kind of like if you were to remove a number, say from the Fibonacci sequence, you can't assume it's one or you can't assume it's three or I mean eight is later on. You can't really assume that it's any number if you removed. I guess you just have to know that what you know about the Fibonacci numbers into kind of fudge is around and say OK, but they're just quantifiable by the number one and that's that is the main problem with this proof, which is obviously not true.

I have even so in this question they asked the second in addition energy of the carbon nitrogen option. And flooring atoms are such as is carbon far far greater than nitrogen far far greater than flooring for greater than carbon bees. Flooring greater than option greater than nitrogen greater than carbon C. Is carbon greater than night auction greater than nitrogen greater than deep. Please oxygen greater than floating greater than nitrogen greater than got a button so yeah. Option dis correct. Yeah. Option beads curry nine days, 9 days the second ionization energy of carbon nitrogen option flooring are yeah, The 2nd Innovation and Adultery of carbon nitrogen option and flooring are in the older Yes, yeah auction greater than flooring which is greater than nitro ocean. This is greater than carbon. Yeah. Yeah. Yeah. Yeah. Yeah. Mhm. Half fill two P three sub shell of a plus is highly stable, highly stable then stable compared to compared two to be force sub selloff. Yes. Yeah. Of a plus sold. I used to of folks who is young will be larger than f minus. Yeah. No Jeff then eight minus. Thanks lot.

In this problem I can write the reaction and CST CH two managed to in perchance of action or two 0-5° integrate will give this compound CS three CH 20 H. And this component to denso. PBL three will give CST CH two beyond and this compound in pageants of GCN we'll give CST CH two M. C. And this component presence of AL I am at full well finally give CST CH two NHCS three. This is compounded by this is component so according to the option, option C it correct here, option siege, correct answer.


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