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Ilcrc, whire precipirare $\{\mathbb{R})$ obraincd on rrcarment wirh aqucous solurion of $\mathrm{BaCl}_{2}$ is of (a) $\mathrm{AgC}$. (b) $\mathrm{CaCl}_{2}$ (c) $\mathrm{BaSO}_{4}$ (d) $\mathrm{PbC}^{\prime} \mathrm{I}_{2}$

In this problem, I can write the reaction and just look at it carefully. After reading the comprehension, I conclude that the reaction happen. It's something like this. B. B and no. Three or two plus two K. I will react to give the productive The VI, two, B B. I do. This is compound C. And this is yellow PPT blood, you KN or three. Therefore, according to the option of some B, each correct answer option B. H. Correct answer for this problem.

Asking us to find the oxidation number of each oxygen within the compounds and then identify it as either in oxide a peroxide or a super oxide. And before even get started, The oxidation numbers are going to be what tell us whether the compound is an oxide of peroxide or a super oxide. If the charge on the oxygen is a minus one, then it's going to be a peroxide. If it's a minus half, it's going to be a super oxide, and if it's a minus two, it's going to be an oxide. So that being said, let's get started. Part A. We have a rubidium and an oxygen, and if we look at rubidium, it's in Group one. A. So each rubidium is going toe. Have a plus one charge. That means Theo Oxygen, because they're in the same issue, is going to have a minus one charge, and thus this is going to be a peroxide for part B. Calcium has a plus to charge, which means the oxygen is going toe. Have a minus two charge, and that's gonna let us know that B is an oxide with part C. CCM House I plus one charge. But if you look at the ratios, there are two oxygen's for one cc, Um, so to balance the oxidation numbers, each oxygen is going toe. Have a negative 1/2 charge, which means, see is a super oxide for Part D. Strong TM is a plus two, which means the oxygen is going to be a minus one. So D is a peroxide. And lastly, for part E. Calcium are sorry. That's not calcium. That's carbon. Carbon has a plus four charge, which means each oxygen is going toe, have a minus two charge, and thus this is an oxide.

K two Cr awful gives a yellow precipitated it's a yellow policy heated on reaction with on reaction with Be a two plus, as well as B B two plus, therefore, option C and option the uh, correct answer for this problem, Option C and option B. R. Correct answer for this problem.

All right, guys, for 29 he wants me to go ahead and simplify the expression, have currency to baby to the power of native to over to a B squared. So first unforeseen manages to beat the negative into all three pieces When I have to go negative ate the native to and the native all over two a b. So all that means is I'm gonna bring my entire numerator down to my denominator. So I'll now have four. I'm sorry. Let me write the 1st 1 first. So I'll have to a v squared times squared, which is for a squared. So now, since I'm gonna multiply down here on the bottom two times four gives me 81 day into a gives me three days to to give me four from my final answer Will be won over eight. Cubed over beat over eight a