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Question

You did not answer Ina quesy You did not answer the ques Gha the parraril slulon lo 2u^ Yv - JchleGhue bhe ceneral goljon loAMAf)J-r (ri+rC% piulli Y sip(3z)-7-Cr blt)+1~Z+cJlxl{c+1sip(3-)P-13/blr) +PcNana citeJba UdmalldmaYou did not answer the quastion.Gne Ihe Gerntijlgolulon lo6rdE++obltI+outi-cDnastlons#)-(H)-= 6nt =(H)- ?You did not answer the quesiion econes cchi lamty Iparabolas centalain Arer aiche Doini _Give the lamik onocona(I-3+26+502-c (Y+312+20-S17-CoWri-COlne AboleS2-3-(x-3*+6+51*

You did not answer Ina quesy You did not answer the ques Gha the parraril slulon lo 2u^ Yv - Jchle Ghue bhe ceneral goljon lo AMAf) J-r (ri+rC % piulli Y sip(3z)- 7-Cr blt)+1 ~Z+c Jlxl {c+1 sip(3-) P-13/blr) +Pc Nana citeJba Udmalldma You did not answer the quastion. Gne Ihe Gerntijlgolulon lo 6rdE+ +obltI +outi-c Dnastlons #)-(H)-= 6nt =(H)- ? You did not answer the quesiion econes cchi lamty Iparabolas centalain Arer aiche Doini _ Give the lamik onocona (I-3+26+502-c (Y+312+20-S17-C oWri-C Olne Abole S2-3- (x-3*+6+51*-C (r+3+6-517_C None ol Ihe Jbove



Answers

A 7.nSO $_{4}$ solurion is boilcd wirh an NallCO $_{3}$ solution ro producc (a) $7 . n] 1 C O_{3} .7 . n 0$ (b) $7 n(011)_{2}$ (c) $7 \mathrm{nCO}_{3}$ (d) $\angle \mathrm{L} C O_{3} \cdot \mathrm{Na}_{2} \mathrm{SO}_{4}$

Hello and welcome to this video solution of numerous. Here we are given the rules of silver and copper are constantly reducing their scalability in So we have got option A. S case in which is the correct option. Now using the in the Synod process, what we do is okay let's let's try to find out what in what is happening in case of signing process. So let us take the Argentina, which is a G two S. Plus. We have Casey in now for this occasion this gives us a complex of silver. This is okay, ain't you? Him two plus it works now. This compound that is formed is soluble in the solution. He's so little now this is thing for copper also. Thus we can have a quick and select option skc I hope this is clear to you and have a very good thank you.

I'm writing the reaction which is really event to the micro's cosmic salt bead aged. So just look at it carefully and I think that taste the relieving reaction, then see, oh, we'll react with anne B or three, then the product form it. Any issue, be all four. And according to the problem, this is the answer. So option B is only got it answer for this problem.

So for a we know that we have the same exponent values, so we just have to add the coefficients. So that's gonna become 18 into the 10th. Same with be, except we're subtracting. That's gonna become 12 under the tent for see if we have these two. I'm gonna start by multiplying the coefficients of 15 day and three, which will give me 45 a and then ends the 10th time suspended tent will become end to the 20th. Since we have the same base, we just want to add the cliff values of the exponents for D. If we have 15 end to the 10th over three and to the 10th you can see on this 15 and three will cancel out to 501 and the end of the tent and under the 10th of the same. So they cancel out

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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