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Wrt: Ch: liritofthc Ricraxn sucn , lira 2i-1{cos-'{#i) + 8(+i} )urx o [+, 1l]a 67 delinite 'riegral.Deemine (22 + 3r + Wjz using %he lirrit ofike Ricrnar ...

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Wrt: Ch: liritofthc Ricraxn sucn , lira 2i-1{cos-'{#i) + 8(+i} )urx o [+, 1l]a 67 delinite 'riegral.Deemine (22 + 3r + Wjz using %he lirrit ofike Ricrnar %Jri, [in` J6N;

Wrt: Ch: liritofthc Ricraxn sucn , lira 2i-1{cos-'{#i) + 8(+i} )urx o [+, 1l]a 67 delinite 'riegral. Deemine (22 + 3r + Wjz using %he lirrit ofike Ricrnar %Jri, [in` J6N;



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The oxidarion srare of xenon arom in $\mathrm{X} \mathrm{c} \mathrm{F}_{s}$, $\mathrm{HX}_{\mathrm{c} \mathrm{O}}_{4}, \mathrm{Na}_{4} \mathrm{X} \mathrm{cO}_{6}$ are, rcspecrively (a) $+i,+6,+8$ (b) $+1,+6,+6$ (c) $+1,+6,+7$ (d) $+1,+5,+8$

Continuing with some redox chemistry. So that is the reduction and oxidation of different elements occurring within the reaction. Where reduction is the gain of electrons and oxidation is the loss of electrons. So here we have two different reactions. You've got part A and part B to consider. And so the equation we are using is E not is equal to E reduction minus E oxidation. And so for the fast example, we have seen A is equal to -1.229V At 1.953V. And so what we get is positive, not .7-4 volts. So these batteries you can pick up from tables quite easily. So the E not for this reaction is positive, not .7-4V. So we have another example to take a look at using the exact same equation. So we have Enoch is equal to the reduction part of the reaction, Which is positive, not .408V At no .534V. And so what we get is not .942 vaults.

First time drawing the diagram of 43 so just look at it carefully editing or This is compound one. This is the oxygen one. Here it is for There are two long pair. This is oxygen number two And this is oxygen number three, So former charge on oxygen one Is equal to 6 -2 -1 x two, multiplication 6, which is equal to plus one. one of my charge on oxygen to Is equal to 6 -4 -1 x two multiplication, for which is equal to zero. Former child on accidentally equal to 6 -6 -1 x two, multiplication to, So it is equal to -1. So according to the option, obstinacy is correct. Siege correct answer. I hope you understand the tradition.

So next we're looking at a Siri's off chemical transformations again. Where we are given are re agents here that I've underlined in green. And then we are drawing out the product. And where exactly are components off? What waas one of our starting materials ends up So as you can see in our first example, I've just labeled one. You can see that the proton added is added to the carbon that already had the most protons. So here was a double bond. So it adds duty carbon that was C. H. So moving on to our second example, we've got hydrogen iodide and the same rationale are proton adds to the area that already had the most protons in the starting material. So moving on, we've got h 20 on H two s 04 which adds a proton ended alcohol functional group. The Marco Viticultural again still stands here on we're adding our protons while least substituted carbon. This product is also relevant for e f. In our examples on lastly, we have h c l. Where again are Proton Jews to mark Kalashnikovs, Rule ads do the least substituted and where we already hard the most protons. It's true. Start with and then our chlorine adds to the other end off. What was our double bond?

Hi in this problem from the chapter Elkins and all kinds, some of them in complete chemical reactions of Elkins are given, we have to complete these equations. The first one is this here, the reactant is reacted with that seal and the product formed will have the structure as shown here in this reaction to the double bond, it's still molecule is added where one of the carbon atom, that's the chlorine and the other gets the hydrogen wanted to it and the product formed has the structure as shown here. Then we have the next reaction, we are again, the al Kane is reacted with water. The product form has the structure where one of the carbon atom of Elkin gets the negative part of the water, that is uhh part and the other gets etch that is positive part. And the product form is shown here. Then we have the third reaction where H. I. Is added to the reactant. In the product form has the structure CH three, Ch two units, which has repeated five times, then C. H. Having iodine, wanted to eat CH three. In the 4th reaction, the reactant is reacting with hydrochloric acid that seal. And the product farm has the structure. As shown here we are. This carbon atom has CH three wanted to it chlorine And Ch three again wanted to it. Then the next reaction is shown here where water is being added to the elkin. And the product form has a structure CH three, ch two, ch having O. H. Group CH two CH 3. This reaction occurs in the presence of mineral acid acting as a catalyst. The next reaction is a reaction between Elkin and water in the presence of mineral acid and the product formed is CH three Ch having Ohh group bonded to eight Ch two, Ch 2 Ch three. So these are the product form for the given incomplete reactions.


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