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Ol (te Mbt br&r ccrb aJoucous IcluOnJoou 12CenpetrntFaentntJaconentres tor Il,0 67 ? Namizar JiuccncnugsDex...

Question

Ol (te Mbt br&r ccrb aJoucous IcluOnJoou 12CenpetrntFaentntJaconentres tor Il,0 67 ? Namizar JiuccncnugsDex

ol (te Mbt br&r ccrb a Joucous IcluOn Joou 12 Cenpetr ntFaentnt Jacon entres tor Il,0 67 ? Namizar Jiu ccnc nugs Dex



Answers

licre the ycllow and orangc precipirares arc, rcspectively (a) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}, \mathrm{~K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{-}$ (b) $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}, \mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}$ (c) $\mathrm{Na}_{2} \mathrm{CrO}_{4}, \mathrm{~K}_{2} \mathrm{CrO}_{4}$ (d) $\mathrm{Na}_{2} \mathrm{Cr}_{2} \mathrm{O}_{2}, \mathrm{~K}_{2} \mathrm{CrO}_{4}$

In this problem, the green solid is CR 203 This is the green solid which fueled with which filled with. I need to see you three to give yellow solution and the yellow solution. Age any to see ERo four. This is yellow solution and this is green solid. So according to the option option B h, correct answer here, I hope you understand the action of this problem.

The common name Age? The common name used Age Okay as me oh, C484 or six so option A two s b o C484 46 Age correct. Ben said for this problem, KSBOC four at 406 is the common name used, So option, age correct and said what this problem?

So here we have to do a matching The first option his plaster of paris stood off. That is we need to write down words, composites and orders chemical formula. This family is, yeah, so forth. Will play with the half. It's too next components oral cement or else email. It's correct answer is M G. O. Component of Carmelite, component of cardinal light and its composition is NGC two. The next is Sylvain and its completion is S E. L. So the character matching will be as follows here to S and B two Q, then our to see and see to her and then they took it. This is correct sequence.

From 88 of chapter three and chemistry Molecular approach. The chemist decomposed the samples of several compounds the massive their constituent elements are listed just like he appeared before him. A lot of each compound look closer of a 1.245 grams of nickel and 5.3, 5.381 grams of iodine. So we're going to who convert the moles? Nicole is gonna be 50 if the 8.7 iodine is going to be 1 27 70 point 0 to 1 mole, two moles, then fuck. 1381 About 1 27 zero 0.0 for two, four moles. So now we're gonna divide these by the lowest number of moments. So that by this by 0.21 to end your 0.0 12 there's gonna be equal to two. Who then there's gonna be equal to one. That means the empirical form is gonna be and, uh, aye, aye, too. All right. Next, 2.677 grams of barium and three point 115 grams of Romy and barium atomic weight. That's gonna be 1 37 transfer moon and girl mean it's going to be 80 grasshopper mole and there's gonna be 0.195 moles, and this is gonna be to your point 0389 miles. So we're going to buy this buy points of your own 195 It's gonna be ableto one, and this is going to be equal to two. So our empirical formula that's going to be B A. We are too, right. Next 2.12 a gram's beryllium and Sep seven point first seven grams I'm so for and 50 15.107 grams oxygen. So now we're gonna convert these two moles. So, beryllium, that is atomic weight of nine. So for weight of 32 and action atomic weights of 16 0.25 miles and 7.7532 you point point 24 malls. And I think I think one of seven develop 16. That's gonna be zero point nine Hormel's. You're going to divide all these by point by 0.24 So this person I'm gonna need someone that's gonna be ableto one, and it's approximately gonna be equal to four. So That means that our purest form is gonna be or B s o four. And that's the imperative formulas of her sample.


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