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YoliaelcAmakah:Th WealonAclda and a8aWhut Utk molsky ot# 12,6"0 solution &l shum curbonite slh # &nsity 0 /, EmL'Whit Bnl'Renent of' pot...

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Ammonia and potassium iodide solutions are added to an aqueous solution of $\operatorname{Cr}\left(\mathrm{NO}_{3}\right)_{3} .$ A solid is isolated (compound A), and the following data are collected: i. When $0.105 \mathrm{g}$ of compound $\mathrm{A}$ was strongly heated in excess $\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}$ was formed. ii. In a second experiment it took $32.93 \mathrm{mL}$ of $0.100 \mathrm{MHCl}$ to titrate completely the $\mathrm{NH}_{3}$ present in $0.341 \mathrm{g}$ compound $\mathrm{A}$. iii. Compound A was found to contain $73.53 \%$ iodine by mass. iv. The freezing point of water was lowered by $0.64^{\circ} \mathrm{C}$ when $0.601 \mathrm{g}$ compound $\mathrm{A}$ was dissolved in $10.00 \mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.$ $\left.1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)$ What is the formula of the compound? What is the structure of the complex ion present? (Hints: $\mathrm{Cr}^{3+}$ is expected to be sixcoordinate, with $\mathrm{NH}_{3}$ and possibly $\mathrm{I}^{-}$ as ligands. The $\mathrm{I}^{-}$ ions will be the counterions if needed.)

All right. So, in this question, it involves the analysis of a tablet of potassium iodide H Right. You want to determine the mess of potassium iodide date in the tablet. So, first the tablet was dissolved in water and acidified and it was converted to this tri iodide iron. Then they try Knight Iron. I was reacted with this amounts of it by yourself. A solution. All right. So, We want to know their mass of potassium iodide eight in the tablet. So France. We need to know the most of potassium my date, but we need to start with the face off it. All right. So, we will start with that. And if you have zero the most of thai software, we must apply the malacca city by the volume of the by yourself eat in liters. 0.02261. Yeah. Yes. All right. So, we will find the most of the try try how you died. I am by using a more issue. Uh So, we have one more of the it's a date I am and to most of the thyself feet. Yeah. All right. So that would be the most of the tri iodide. Now, there's more asia between the tri iodide and the idea is this. Yeah. So, uh we'll have one more Yeah, of the a date and three malls of Mhm. Uh The miners. All right. So, that's give us the most of the uh I did in that tablet. Okay, so that gives us 3.967/10 Temper minus four moles of Yeah. Mhm. Okay, so, we will convert this to mess by using the miller mass of potassium. Are you dates? Which is 21 4.001g of Yeah. For more. Yeah. All right. So That gives us zero 08 Just 61 grabs in Milligram. Will multiply by 1000 to get 80.61 milligram of protests in. Yeah, I did. All right. So this is the mass of potassium a date in the tablets. Yeah. All right. Yeah.

Using the information given in each of the four parts. Let's determine the formula of the compounds and the structure of the complex I in so from part one, we can calculate the mass of chromium present. So the mass of chromium is equal to zero 203 g of C R 03 and in 100 grounds of C R 03 b, 52.0 g of chromium, this would yield 0.106 g of chromium on that's calculate. Next, the mess percent of chromium, which would be equal to 0106 g over the 0.105 g times 100%. And this gives us a mass percent of chromium of 10.1% chromium for part to let's use this information to calculate the mass, uh, mass of NH three present. So the mass of NH three is equal to 39 32.93 Middle leaders of HCL times more clarity here. 0.1 Milly. Mole of HCL per Miller Leader times one, Milly Mole of HCL to one million mole, NH three And what times This by the molar Mass one million mole of N. H. Three is 17.4 mg. Event age three, and we consult for the massive NH 3 mg works out to 56.1 milligrams of ammonia. Now we'll calculate the mass percent of ammonia, which is equal to 56.1 milligrams over 341 mg times 100% and this would yield 16.5 percent Ammonia now for part three, Let's, um, combine the mass percent values to check if other elements might be needed. So we have 74.53% which is the percent, um, of the iodine by mouse, plus the percent of the chromium, plus the percent of the ammonia. I'm working this out here. We solve 400 0.1%. So this tells us that we're at 100% no other elements, our present. So if we assume 100 g yeah, then we have 74 53 g iodine 5. 74 point five three. Yes. Um, let's make sure we got this right here. Grounds of iodine, 10.1 g of chromium and 16 5 g of NH. Three times. 126.9 g per mole. That's what we call to you. 0.587 moles on this would be 52 10 g per mole 0.194 and 17 grams per mole for mole, a massive ammonia 0.971 Divide by the least. We're gonna divide by 0.194 Give me one. So work out to three this workout to five. So our empirical formula here is chromium. Yeah, NH three five I three. So, in this case here, chromium three forms Octa hydro complexes. Okay, this means that the compound Yeah, yeah is made up of the Octa hydro chromium and H three five I two plus complex I in and to I minus irons as counter ions. Sore formula based on this is going to be C R N H 35 I Yeah, how to? We didn't need the results from part four for the formula, but let's use them. And so delta T f zika the i k f m's or morality. So let's solve for the morality. First, the morality here is, um, 0.601 g of the complex over 100 zero times 10 to the minus 2 kg of water times. Moeller Mouse is 517.9 g of the complex in one mole of the complex, and this would work out to a morality of 0.116 moles per kilogram. So it's calculator adults A. T F. Here theme. Um, the constant here is assuming complete association would be three k f years 186 degrees Celsius kilogram Permal, times of morality of 0.116 moles per kilogram and delta T f here works out to 65 degrees Celsius. Eso We have some confirmation here, since this value is close to the given value and given value here is 0.64 degrees Celsius. We can say that the formula that we've created in part three is correct. So we have some confirmation here with part four. Yeah,

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.

Here for the solution in this. The reverse off the following equation is this. That is C B C 02 sorry at B C O to pull a row at B plus or two head out of the S plus 70 close would her at BCU at people asi is Delta equal a particular goal. Now here we subtract equation to which is this situation one which is this now, By subjecting it, we get Delta's equal to 10 clothes on and they were the final equation that it has B 02 plus CEO GHB CEO plus or two in her there. Turgeon minus stand. Carrie. Now for the next to calculate the equilibrium constant here, we know that Delta t know tickle toe minus rt. Okay. Hair and cake all to magazine out upon rt. By putting the value in the equation by simplifying it here, we guard that I am cake will to 40.36 and here to find this here. We got the cake order into the power. 40 points. 36 Now here we get the final value off K is 3.38 or 10. I was to the power 17. And and that looks a round off value ease. 3.1 to 10. 3 powers and 17. And this is the our answer. This was the solution. Her was reversed off equations and calculated the value of G not. And we're here by subjecting. Mhm both equation two from equation. Where we go developed C note and her were the combined equation and the value of see it her to calculate the equilibrium Constand ID, which is k equilibrium constant. Here we where the equation formula toe find. And here we find the value of okay. And this is the work. Commit well, thank you.


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