All right, so in this problem, we're forming an unknown compound, and we're gonna go through a series of tests to figure out what it iss. So first, in this solution, we know we have ammonia. We have potassium iodide, which will completely disassociate. And then we have chromium nitrate. So the first thing we're told, I should say, and all of that we form some compound A that we're trying to figure out. So first, if we have 0.105 g of a and we reacted with 02 we get 0.203 grams of c R 02 So excuse me 03 So what we're gonna do is figure out how many moles of chromium got. So let's see, Chromium has a Moeller mass of 51 0.996 grams per mole. So for this, I'm just gonna round these numbers to make it easier as I'm writing. So I'll just say, here we have 52 grams per mole doing the same thing with oxygen. Oh, and let me clarify. That is grams of chromium for oxygen. We have approximately 16 g of oxygen. Her mole so the total Moeller Mass here is going to be 52 plus two times 16. So that's just excuse me, plus three times 16. So that's going to give us 100 g of this compound her mole. And we know that chromium makes up 52 g of that. So by mass chromium is 52% so it's 52% chromium. So if we take 0.0, 203 g time zero point 52 we get 0.0, one 06 grams of chromium. So what that tells us is from the original compound compound A here that we have 0.106 rams of chromium, her 0.10 5 g, which means we have 10% by mass chromium and compound A. So next we're told that Compound A was tight traded, so we had 30 2.93 mL of 0.100 Moeller, h CEO to titrate completely the ammonia present. So that's going to say that So Moeller is the same as moles, her leader And so here we have. If we were to change this to leaders, we have 0.0, 32 93 Leaders time 0.100 Moles her leader, which is going to give us 0.0 32 93 Moles. And that's we're gonna say that's moles of ammonia. Now, to convert that to mass nitrogen is 14 g per mole approximately, and hydrogen is approximately 1 g per mole. That gives us 17 grams of ammonia thermal. So then, if we multiply the number of moles by 17 what we get is 0.0 56 grams of ammonia, and we're told that comes out of zero point 341 g of compound A. So that tells us that it is 16 point four, her son ammonia. So if we put that all together, we have 10% chromium, 16.4% ammonia and were given 73.5% iodine. So now that we have these, we're gonna assume since we know this is the mass percent that basically we have 10 g of chromium, 16.4 g of ammonia and 73.5 grams of iodine. So now we go through and divide by the molar mass, So we would divide by 52 here, divide by 17 and then lastly, we would divide by 127. So that would give us 0.19 moles of chromium, 0.96 moles of ammonia and, lastly, 0.58 moles of iodine. So if we start, for instance, with one mole of chromium. So if we divide the 0.96 by 0.19 that gives us a ratio of 1 to 5. So we'll say that we have three ammonia. And then if we do the same with iodine so 0.58 divided by 0.19 it's about three. So this is our empirical formula looks so we have that five there. So from there, what we can say is move this down. We have chromium three plus, and it's going to be six. Coordinate with ammonia and possibly iodide. Well, if we only have five ammonia, that means we need one iodide. And since this is crummy and three plus ammonia is neutral, iodide is minus one. That gives us two plus. But looking at that, obviously from our original formula that we calculated we have three iodine, so How is that? Well, instead of putting that two plus there we can balance the charge. We have counter irons with those other two iodine.