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(10 pts) Ammonium nitrate dissolves in 100 _ ol water according the following equatlon: NHLNO; - If adding NHANO3 NH watercause' NO3 tematramune change Irom 2S...

Question

(10 pts) Ammonium nitrate dissolves in 100 _ ol water according the following equatlon: NHLNO; - If adding NHANO3 NH watercause' NO3 tematramune change Irom 2S0C to 220C, how much cnergy exothermlc? exchangcd? Is the soluuon precess endothermicor

(10 pts) Ammonium nitrate dissolves in 100 _ ol water according the following equatlon: NHLNO; - If adding NHANO3 NH watercause' NO3 tematramune change Irom 2S0C to 220C, how much cnergy exothermlc? exchangcd? Is the soluuon precess endothermicor



Answers

The solubility of ammonium formate, $\mathrm{NH}_{4} \mathrm{CHO}_{2},$ in $100 \mathrm{g}$ of water is $102 \mathrm{g}$ at $0^{\circ} \mathrm{C}$ and $546 \mathrm{g}$ at $80^{\circ} \mathrm{C} .$ A solution is prepared by dissolving $\mathrm{NH}_{4} \mathrm{CHO}_{2}$ in $200 \mathrm{g}$ of water until no more will dissolve at $80^{\circ} \mathrm{C}$. The solution is then cooled to $0^{\circ} \mathrm{C} .$ What mass of $\mathrm{NH}_{4} \mathrm{CH} \mathrm{O}_{2}$ precipitates? (Assume that no water evaporates and that the solution is not supersaturated.)

This problem is a little bit involved. We want to know the mass ammonium chloride that we need to add in order to reach a Ph of nine. So I'm again going to use Anderson House. A bolt equation Target pH is nine. That's going to be equal to the PK of the acid that's present, which is ammonium. PK of ammonium can be calculated by taking the P k K B of ammonia divided into kw that will give us a K a. And then take the negative log that in order to get PK and then plus the log of. So we originally have 500 milliliters at 5000.1 bowler ammonia, and then we're going to divide by mass of ammonium chloride, which we don't know that we need to add in order to reach this target. PH. But we need moles down here, not mass. So it'll be the mass divided by the molar mass of ammonium chloride to give us moles in the denominator that all we need to do is the algebra in order to solve for the mass, which is our X value. I'll take the negative log of this to get my 9.25 Move the 9.25 over getting negative 0.255 equal to this law. Value this log ratio well, then do 10 to the to the anti log of both sides and 10 to the negative. 100.255 point 555 Repeating that would be equal to the law goes away, and I'll have 0.5 up top and then x times They reciprocal of 53.9 point 491 is 1.87 10 to the negative, too. Then I can isolate my ex value and I get 4.81 grams ammonium fluoride.

The question states that an equal a solution is 7.51% by mass of ammonium nitrate. So we are asked to find one mass of ammonia, matri and what massive water will be needed if the solution is 1.25 kilograms. So let's write out what we have. We have seven point 51% by Maris ammonium my tree. So that's an age for and no. Three. So this means that 7.51% of the solution will be that ammonium nitrate, and we also have the total mass of the solution. So mass, um, solution is 1.25 kilograms and I've also gone ahead to read out the equation 4% by mass of ammonium nitrate, which is equal to the mass of ammonium nitrate divided by the total mass of the solution. So that includes theme s of ammonium nitrate and the mass of water, and that is all multiplied by 100. So we can start by plugging into the equation what we already have. So we have the percent by mass of NH four. Um, I know three or ammonium nitrate, so that is 7.51 person that's equal to the mass of ammonium nitrate and we don't have that value. So we're just going to write X as are unknown variable mass divided buying us of the solution which care of which is 1.25 kilograms and this is multiplied by 100. So, in order to Seoul for X, which is our massive ammonium right tree, we have to isolate for X and our first up to do that is to divide both sides by 100. As that will cancel this out. We divide both sides by 100. This term here gives us zero point 0751 which is equal to x over 1.25 kilograms. And the reason that you know we have to divide both sides by 100 is because in this equation this 7.51% was originally 0.571 but then was multiplied by 100 to get this. So if we divide by 100 we get 0.51 0.751 So, essentially we're working backwards and we can rearrange this equation to solve for x so x I was going to be equal to 0.7 51 multiplied by 1.25 kilograms can. That is equal to 0.0 93 875 But we can roam at up to 0.939 kilograms. So that is our kilograms, uh, and H four and 03 So that was our kilograms of ammonium retreat. But we are also asked to find the massive water. So we know that the total mass of the solution is the mass of the ammonium nitrate, plus the massive water. So we have the massive ammonium nitrate and we have the mass of the solution. So now we confined amounts of water, which will be the massive solution which is 1.25 kilograms minus the mass of ammonium nitrate, which is zero point 0939 kilograms. And that is equal to 1.16 kilograms of water. And you can go and double check your work by thinking, you know, you have your 1.16 kilograms of water and your 0.939 kilograms of ammonium nitrate, and he's two numbers added together should give you 1.25 kilograms, which is your total mass. I'm solution. So it was your answers.

To calculate the molar mass of a compound, we do the same thing that we did to determine the formula mass. However, instead of assuming the numbers on the periodic table have units of emu, we will now assume that they have units of grams per mole as opposed to a m u s per adam. Or I guess there'd be atoms in the case of the periodic table. So we'll go to the periodic table, recognize that there are two moles of nitrogen in one mole of ammonium nitrate. Multiply the molar mass of nitrogen by two, add that to four times the molar mass of hydrogen Because there are four moles of hydrogen for every mole of ammonium nitrate And then add that to three moles of oxygen, three times the Mueller massive oxygen, And we get 80.043 g per mole.

With 10 39. I'm only in They did reacts to form. No, I didn't. All frickin on model. This is an explosive reaction. And we have to find out how many grams of oxygen will be produced when one Casey off ammonium nitrate is the compost one. Casey off more united Izbica. Both all money grams off all season is released. Now, in order to find out this well treated the molar masses self my money and I did on guard oxygen. Now, one balancing this person will get two more in my great woman two and two on for its tour. Now, from this equation, you can see to most of all, united he couldn't do two in the 80 g produces 32 g of oxygen. So 1 g off more in Madrid. Next work in a dream generates 32 of all to indu 80 g or oxygen. Mhm. So one Casey will produce 32 of all two in the 80 in the one case is above you. It was you zero point to G or oxygen or 200 g off boxes. It


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