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Question

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Answers

The product (s) obtained via oxymercuration $\left(\mathrm{HgSO}_{4}+\mathrm{H}_{2} \mathrm{SO}_{4}\right)$ of But-1-yne would give [IIT, 1999] a.
(b) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}-\mathrm{CHO}$ (c) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CHO}+\mathrm{HCHO}$ (d) $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{COOH}+\mathrm{HCOOH}$

First chemical reaction where we have each three people for an acid reacting with C a. O. H. Two calcium carbonate. This is just a simple acid base reaction. However, balancing becomes a little tricky because of the two plus charge of calcium and the three minus charge of the phosphate. So we're gonna need to count three. Calcium is for every to fall states. And then, of course, we produce water because of the acid base reaction that we're going to need three calcium hydroxide because you're three calcium to phosphoric acids. Because of the two frost states, which will then require six waters, the next one is potassium bicarbonate reacted with the acid nitric acid producing the salt, potassium nitrate and carbonic acid, which will fall apart into water and carbon dioxide.

So for this problem we're going to complete and balanced the reactions given. So for the first one we have a wound oxide reacting sulfuric acid. So this is a double displacement. So basically these two will switch partners. So we're going to have aluminum sulfate and also water because the hydrants pair with the oxygen. So when we go to balance this we can see that we have three sulfate compared with the aluminum because the aluminum has a charge of three plus and they self it has a charge of negative two. So then we'll put a three in front of the sulfuric acid because we need to have three sulfate islands. So now we see we have six hydrogen. So we're going to go to our water and put a three in front of that Because now we have 600 is here as well so that we can check the balance of the options. We have three oxygen's on the aluminum oxide and we have to aluminum's on both sides. So we're done with that. So for the second one, aluminium reacts with silver nitrate. This is a single displacement. So basically we're going to get the silver battle by itself and then aluminum nitrate. So all the women have the charge of three plus. So it's going to pair with three nitrate ions. So now that It's fair with three nitrate ions, we have to put a three in front of the silver nitrate to balance out the nitrates. So now we have three silver. So we just put a three and from the silver and then we check the aluminum that has already balanced so we're doubling down. Now. For C this is another devil displacement that just switching partners. So the lead has a charge of two plus because it's paired with two nitrates which has a charge of negative. So therefore there are two Iodide island compared with the lead. So then we're just going to balance up the iodide by putting into in front of sodium iodide. Now we have to sodium so put it to in front of sodium nitrate and we see that we already have two nitrates on the left and only one. Let's so we're done with that one. So for E this is a single displacement. So for this one we need to first of all take a look at the moment of so if we look at the aluminum, we can see that we have to, aluminums. Mhm. Yeah. Mhm. On the right side. So let's drive putting a two here. So after we put the two there, that was about field. So now we need to try to balance out the oxygen. So we have three here and we have four on the other side. So we're going to put a four in front of this. Then we know that we can't actually use that, you know, so we're going to put a in front of the so then You put a three in front of this because we have bounced out the oxygen And then we have nine Megane, so we just have to put a nine in front of that over here. We heat gallium hydroxide, we get gallium oxide and water. So the gallium has a charge of plus three. So we have to have two paired with three oxygen's. Now we have two gallons, we're putting a to and from the gallon hydroxide development galleries and then we need to count how many hae jin's we have. We have six hydrants, what are three from the water to balance out the hydrogen and then check the balance of the oxygen's. We have three oxygen's and three accidents about six. We have six over here, so.

The ocean is complete without formal institutions more balancing. We have already five little of the following reaction is how can we balance the chemical equation? Right believes that the condition because we have to it has a number of items. Okay, thank you. So first reaction is given in the question L plus three at three minus here the ST louis as minus. So first we are getting air which leads to heart block. Said so three more of those that were taken. I'm sure the gas bill Gates police three more of the best guess please because the first year channel and second reaction in BRT Just keep them cr two seven plus then it will convert into them six here. 3 14 three here looking here makes Then you know all of the three piece of work well gets warm plus six months of l. p. 0. 4 And 21 more of a street pure food. Nine more of br ko Yes six more of 66 more. Ocr beautiful. Like the balance the organization. I hope you are getting

Okay. Good day. Ladies and gentlemen, today we're looking at problem number 11 here. And the question is whether or not we can apply the domesticated of undetermined coefficients Thio this, uh, ordinary different show equation here. And so I'm not really gonna, um, go through all the different parts of the undetermined coefficients cause I think there's probably 5 to 6 distinct cases. But it is important, I think, for you two know each of those cases because they tell you how to go about solving, um, the, uh, ordinary differential equations. They'll tell you how to solve a bunch of cases of, or a bunch of, um, ordinary difference, your equations on particular ones with constant coefficients here. And so the first thing is to realize that, um oops, sorry about that. So if I take three different, distinct functions, I think one of the year of two of two here at three of tea Now I'm going to really look at this case by case in each case. So in the first case, this one here is, um and it is in fact, a proper form. It is one of the cases, and I'm not sure which But if you look, if you flip through, you'll see the thing cases. And this is in fact, one of the cases, um, in the 2nd 1 here again is also a case again. I don't know exactly what's important, but it is, in fact, one of the cases covered by the, um, undetermined coefficients. But the 3rd 1 is not and in particular one over tea is not a polynomial. It is Tito the negative first, and that is not, um, covered by any of the cases. So in particular than, um, since you have one I mean, really, you can't get rid of this one over t s o. The end of the final answer is that it's not applicable, and it's not applicable because there's no case that covers us. And the older way you could solve this. Using the undetermined coefficients is to break this into threes. In cases here, I'm solve each one and then applies the the superposition principle. And in this case, you can't because one of those, uh, you know, is not solvable using that method. Uh huh. But it doesn't mean that there's not other methods to solve it. It's just not solvable using this method really, all the collections after. So there's no need to go any further with. So, um, again, I would just mention that, um, it's a good idea to have in the back your mind. What thes, um the what the undetermined coefficients method involves and sort of go through each of the steps because it's it's actually fairly involved. There's quite a few different steps, and there's a bunch of different cases. So there kind of TVs, but still probably could know. Uh, okay, so that's it for this problem. Thank you very much. I haven't.


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