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A"T3l 12T3/2 : 4-74a; andX ~Tud...

Question

A"T3l 12T3/2 : 4-74a; andX ~Tud

A"T3l 12T3/2 : 4-74a; andX ~Tud



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$$14,356-13,253$$

Okay, so we're asked his objective falling. So since one is just in to we need to borrow. So we get to and this is now 11. So 11 minus two is nine. And now, since two is us and eight, we have to borrow. So this is a five, and it turns into a 12 12. Minus eight is for Okay, So now since five minus 85 is less than five minutes borrowed, we get a one here. And in the 15 year 15 a minus eight is seven. And since when is less than seven, we again need to borrow. So this has turned into four. And now we have 11.7, which is for and in four minutes for is zero and two Do not equal to okay.

Okay, This problem is asking us which of these molecules between all four of these is my m ing going to naturally to tomorrow's into. So let's go ahead and define a couple terms here. First off to Tom Rice or to Tamar Ization is essentially I saw memorization, right? Because whatever we have right here in this case, we have an I mean, the Tomur, the Tanamor is going to be in equilibrium. It exists in equilibrium with my I mean so in order to exist in equilibrium, that means that we're going to have basically the same atoms as we do in this. I mean, but just in a different order, right, we're gonna have different connections. Basically another item, er which is going to be called the totem. Or in this case, so we're looking among all these molecules, Which one has the same atoms just in different connections. So over here we have an oxide. Does this have the same number of atoms in the same type of atoms? The answer is no, because we have an oxygen here that is simply not a cook cook component of Miami. So this is not going to be an option. Next step. We have a hydro zone. Is this considered to be the same atoms as we do in my me? The answer is no. Because we have another nitrogen, right? We have the extra nitrogen. That is not a option. Next up we have this semi carpet zone. This is obviously not the right answer because this one has way too many nitrogen, right? We have one too many. This is not gonna be an option. The last option is my enemy. So does this have the same number of atoms? Same type of atoms and just in different connections? The answer is yes. So it's going to be my enemy. And what is that? Because if we're to simply use a base, for example let's go. You had a user base. My bass is able to Deep protein ate the aesthetic proton associative right here. So this is analogous to my alfa hydrogen usually associated with carbon nails. So I'm gonna take off the hydrogen. Moving the electrons onto this carbon or another way I can show is to move these electrons onto the single bond to create a double bond. As I do that, I'll have to move these electrons up to this nitrogen to produce this intermediate in which I have a nitrogen with a single bond because it has a negative charge. My method group right there and then my double bond right there. Right. So this is very, very similar to my enemy, which I have drawn right here. The only other thing missing is the presence of the hydrogen, which can simply be used over here. We have HB used from the protein nation of be in the previous step that can simply get pregnant by HB to eventually form my enemy shown right here. So this is a Tanamor because this exists in equilibrium, right? It just depends on the reaction conditions. But normally we're going to have my maybe the most stable form that is going to be the one that is present in the greatest excess, right? So enemy and I mean those are taught Immers, and they're going to exist naturally.

In this question were being asked to add. So keep in mind that when we add, we go from or right, call him to our left, just in case you have any numbers that we need to carry over. So we were start here in our far right column, you would add seven plus two plus six plus eight plus zero. This gives us 23 now, since some Berries greater than nine, we need to make sure that we carry over this too. So you would put our three down here and who would k over our two and that we need to make sure that we added to and to our next call him. So we have to plus nine plus three plus two plus zero post one. That's as that to 17. Same thing. We would place our seven down here and carry over our one. Let me. So where next call him. We have one plus zero plus four plus one was five plus three, which adds up to 14. That's right in our seven here. I said to take this for right in our form and again carry over our one. Next we saw this column. We have one plus four plus three plus seven plus four plus eight, which adds up to 27 again place. Our seven here carry over our chill every so our final column, which is two plus five plus three plus nine plus six plus seven. This adds up to 32. And since this is our final call home, we do not need to carry over our three. So we just write in our 32. This gives us our final answer, which is 327,473.


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