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3. (12 pcinis Lei &E"dDatarmie "lethzz Jc; gd are perpenclicelar:10...

Question

3. (12 pcinis Lei &E"dDatarmie "lethzz Jc; gd are perpenclicelar:10

3. (12 pcinis Lei & E"d Datarmie "lethzz Jc; gd are perpenclicelar: 10



Answers

I Icre, ycllow procipirare (S) is of (a) $\mathrm{PbCrO}_{4}$ (b) $\mathrm{PbCO}_{3}$ (c) $\mathrm{PbCl}_{2}$ (d) $\mathrm{Fc}(\mathrm{OII})_{3}$

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.

This problem we're gonna be using lap has transformed tables to evaluate the lack class. Transfermarkt t sweat minus three. T minus to you to the negative t sign three t. This is easily done by linearity. It means we only need to calculate the lack. Last transport squared, the last last transom of tea and the last last transform of each of the negative D sign three teams. Now, the last ransom of T squared is quite easy. It's equal to two over s cubes. House transom of T is equal toe one of the square on the blacklist transform of each of the negative. T sign three t is gonna be equal. Tow three over s plus one squared plus nine and that's it.

Problems three through 10 we're going to be calculating dot products. So first we're going to have negative to one third and negative 5 12. So we have negative two times negative five which gives us 10 plus one third times 12 gives us four. So we end up getting 14 of our answer. And for the next one we have negative too Times .7. So it's a negative 1.4 Plus three times 1.2 that's positive 1.6. So we get a point to it's our product And we have four times 6 which is 24. Um one times negative three is negative three and then we have 1/4 times a negative eight. That's a negative too. So we end up killing 19 as our product. The problem six we're going to have S times T. Which is S. T. Than to s. Times a negative T. Which is a negative to S. T. And then we have three S. And five piece that's going to be plus 15 S. T. So we end up getting um 16 -2 is going to be 14 s. T. For the final product. Now problems seven, we're going to have one times five is one. Um Plus a negative two times zero which is just zero and then three times nine is 27. So we end up getting 28 as our dot product. And we're on to eight. Almost done. We're going to have um for two times 0 which is zero four times four which will be 16. And that's going to be added to a negative three times six, which is negative 18. So we end up getting negative two is our God product. Now we're looking at um A and B. Where we have the magnitudes given to us. So we have magnitude of A. Which is six times the magnitude of the which is five. I'm code sign Have to Pi over three. That's kind of giving us a negative 15 a third our product and then far very last one. We have three en route six. So three times route six times the code sign of This case, it can be 45°. So we know that's going to give us route two Over two. So we're going to have three halves times the square root of 12, which we can simplify further.

In this problem, first time writing the accent. So just look at it carefully. Six came for f e CN six plus or three plus city as to so it will react to give the product ID six get to the FECN six plus six. He who add as the product. Therefore, according to the given option, option B age, correct and set absent. Be correct. Answer for this problem. I hope you understand the solution of this problem.


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