Using the information provided. Let's determine the chemical formulas of each of the compounds and then draw their Lewis structure. So we have five compounds which are listed A, B, C, de and E. So let's first cell for our compound d. Um, because we're told that Compound D is 43.7% nitrogen and 50 percent oxygen. So for the mass of hydrogen, it will be 100% minus 50 percent minus 43.7% which would be equal to 6.3% hydrogen. So we're also told that the density of Compound D is equal to 2.86 grounds per leader at STP. So no, let's go ahead and find the formula for de. We can find the moles of n which is going to be, um, before we do that, let's assume that we have 100 g of D. And so we have 43.7 grams of N 50 g of oh, and 6.3 g of hydrogen. Let's convert each of these two moles, uh, 14 g per mole, smaller amounts of nitrogen. This would give us all right? Uh, yeah, 3.12 malls on 3 16 g per mole. So 55 about 16 gives me 3.12 malls and times 1 g per mole. That's equal to 6.3 moles we're going to divide by have 3.12 This would give me 11 and two. So my empirical formula for Compound B would be called to, uh and oh h two and we can solve for the empirical mouse for D, which would be equal to. And that's very each to their N O H. Two. So 32.0 g per mole. We're told that the density is 2.86 grounds for leader. So, uh, to let's hope for the molar mass of D. We have 2.86 g per liter times 22.4 liters per mole and this would work out to leaders would counsel 2.86 telestrate. You play for 64.1 pants per mole eso for n which is our whole number ratio. Mm. Over e m 64.1 grounds per mole, divided by 32 g per mole. And so we killed the two. So the molecular formula for D would be equal to two times the n o H. Two, which should be equal the end to O to h two. We rearrange this, we can get a compound which is equal the n h four n 02 So there is Compound de and H four to. Okay, so now there's d. We can identify compound. See, we're told compound. See, it has one more oxygen full, then compounds mhm d. So this is equal to N H four and 03 for the formula of C. Um, Now let's move on to compound A. We're told that compound C and a half one iron in common eso has one I am in common with compounds. See on when is acting is a strong electrolyte. The solution is strongly acidic. Eso it's strongly acidic. If it is strongly acidic, a strong acid would be nitric acid. The ironing common would be the nitrate iron. So there's compound a Andi, we've got Compound B, which we can do next. And here we have the tie. Trish, in of Compound B, requires 20. So whatever compound B is, um, it's gonna be tight treated with hydrochloric acid for complete neutralization. So I know that this has to be a base so tight, traded with HCL and obtain The Moller mouse s 07 to 6 g of compound B is tight, treated with 21.98 There's a neutralization, their middle leaders of mhm one Moeller hcl The moles of HCL is equal to one Moeller times 0.2198 leaders and this is equal to a point 0 to 1 98 moles of HCL and what we're going to assume here is we're going to assume a We're gonna assume a 1 to 1 more ratio, so we're going to assume 1 to 1 more ratio. So we have 0.2198 moles of compounds be and there for the Moeller mass of be would be equal 2.726 g over 0.2198 moles And this would work out to 30 1976 33.0 grounds per mole is the Moller massive beat a compound Be, you know must be basic since there's a neutralization. So it has to contain an O H group and ohh the O. H. Group is 17 g per mole, so 33 grounds per mol minus 17 g per mole will leave us with 16 grounds for more and so on a compound B. Uh, since it must be basic and it has to be 16 g from all the 16 g per mole must be comprised of N H two. Therefore, Compound B is an H two oh each and there's the identity of Compound B. And lastly, Compound E. It's as commercially available concentration solutions of air normally 16 Mueller Um, yeah. So, uh, compound A, we're told, is h N 03 and commercially available A 16 Moeller the solution of Compound E, which is 15 Moeller Um Hmm. Which company? E. Which is 15 Moeller I mean on this Must be but be solution with commercially available. Um, all the comrades are strong electrolytes. If it is a strong electrolyte 15 Mueller, This would be n h four Oh h Now let's draw our Lewis structures for compound A component E was H and 03 and it's Louis structure for nitric acid is and plus double bond. Oh, oh, minus Yeah, yeah. Oh, and H There's a Louis structure for a for B way have N h 20 h, which is hydraulic saw aiming on Lewis structure here O h and lone pairs here or compound. See, we have ammonium they treats, which would be ammonium, the ammonium iron and nitrates, which is in Yeah. Oh, this would be a minus there on the O double bond. Oh, lone pairs ceremony. Nitrates, compound D or Sorry, ammonium nitrate in each four. No end to their Oh, no. Sorry for and through. They get it wrong. NH 403 Let's make this correction here. This is N H four n 02 Let's fix. Come. Um, this should be nitrates, not nitrate. I hear So nitrate would be and plus couple bond. Oh, lone pairs. Oh, whole minus minus And ammonium nitrate B n plus H h h h. And things should be and plus oh, minus on pairs Double bond. Oh, so ammonium nitrate and compound E waas ammonium hydroxide So and plus and a drug side Oh h minus sell ammonium hydroxide there. So those are the five compounds and their Lewis structures. Yeah,