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What is the mass, in grams, of 2.50X 1O J mol of ammonium phosphate? (N+L), 744 [4+(4)Js+ 31+ (c) = Mrht 1.50 1lo $ x (141,) 0.373r (NHy } P44How many moles of chlo...

Question

What is the mass, in grams, of 2.50X 1O J mol of ammonium phosphate? (N+L), 744 [4+(4)Js+ 31+ (c) = Mrht 1.50 1lo $ x (141,) 0.373r (NHy } P44How many moles of chloride ions are in 0.2550 & of aluminum chloride? AICIs +(5.4932193.54 /eul 0.4550 $ Aili; 3~lci- -0057303 mi| Ci" 155.57 AIU; Adriver is driving at the speed of 85.0 kilomiles per hour: What $ his speed at mm per sec? mile 16093 Ipk 7 m(5 35.05 IW Jr min Ikm bonin inr 1l $ 85.0 X /XI0` 60 + 60 236ll.1 mrV/s

What is the mass, in grams, of 2.50X 1O J mol of ammonium phosphate? (N+L), 744 [4+(4)Js+ 31+ (c) = Mrht 1.50 1lo $ x (141,) 0.373r (NHy } P44 How many moles of chloride ions are in 0.2550 & of aluminum chloride? AICIs +(5.4932193.54 /eul 0.4550 $ Aili; 3~lci- -0057303 mi| Ci" 155.57 AIU; Adriver is driving at the speed of 85.0 kilomiles per hour: What $ his speed at mm per sec? mile 16093 Ipk 7 m(5 35.05 IW Jr min Ikm bonin inr 1l $ 85.0 X /XI0` 60 + 60 236ll.1 mrV/s



Answers

(a) What is the mass, in grams, of $2.50 \times 10^{-3}$ mol of ammonium phosphate? (b) How many moles of chloride ions are in $0.2550 \mathrm{~g}$ of aluminum chloride? (c) What is the mass, in grams, of $7.70 \times 10^{20}$ molecules of caffeine, $\mathrm{C}_{8} \mathrm{H}_{10} \mathrm{~N}_{4} \mathrm{O}_{2} ?$ (d) What is the molar mass of cholesterol if $0.00105 \mathrm{~mol}$ weighs $0.406 \mathrm{~g}$ ?

So there's four parts to this problem. In the first part we're asked to find the mass of 2.5 times 10 to the negative three malls of ammonium phosphate. So ammonium phosphate. Ammonium is an NH form Phosphate is P 04. The PS four has a three minus charge NH three as a plus one charge. So you would need three NH threes to balance out the minus three from the P. +04. So let's find the molar mass of ammonium phosphate. And then all we would have to do is multiply and so there's three nitrogen Has a molar mass of 14.01. There's three times for 12 hydrogen. 1.8. There's one phosphorus 30.97 and there's four oxygen's 16 point oh so we can multiply all these together three times 14.1 Plus 12 times 1.8 Plus 30.97 plus four times 16. And you would get a molar mass of 149 10 grams per mole. And now all you'd have to do is multiply by this original number of moles. So 2. 2.5 times 10 to the -3. And the mass of this sample would be zero 373 grams. Okay. Um Yeah. And we need three sick fix because the least number of sick fix we have are from the original Number of moles which has three significant figures here. Let's look at Part two. You are asked to find how many moles of chloride ions are in malls C l minus in 0.2550 grams of aluminum chloride, aluminium us three charge chloride has a minus. Yeah. And so you would need three chlorine to balance this out. So let's find the molar mass of aluminum chloride, Mueller mass of aluminum is 26 0.98. And there's one of them plus 33 times 35.45 for the Chlorine. 26.98 plus three times 35.45 Is 133 0.33 grams per mole. So we multiply this. Bye. I'm sorry. We take 0.255 programs And we multiply this by one sir and more. LCL three Per 133.3 three g of a L C L three. And then we multiply how many chloride ions per aluminum? Try chloride. That's three. So three moles of CL per one mole of a L C L three. These cancel these cancel. And this equals .225 divided by 133.33 times three. And you reject five point oh 63 Times 10 to the -3 moles of C l minus. Let's look at part C. In this part we're looking at the mass of seven point zero I entered the 20th molecules of caffeine. Okay two C eight each 10 And 402. So like always we need to first find the molar mass of the molecule we're looking at. So the molar mass of caffeine, There's eight carbons. Each one has a molar mass of 12.1 plus 10 hydrogen. Each one has a Miller massive 1.0 H plus four nitrogen. Each one has a mass of 14.1. And then to oxygen's which each have a molar masses 16.0 eight times 12.6 plus 10 times 1.8 Plus four times 14.1 Plus two times 16 Is 194.2 grams per mole. Okay, so now that we have the molar mass, We need to find out how many moles 7.70 times 10 to the 20th molecules corresponds to Yeah. Okay. And this we use avocados constant. So one mole 6.02, 2 times 10 to the 23rd molecules. Okay. And then no, these cancel. We have 194.2 g per mole. These cancel. And so if we multiply all of these together, we get the mass in grams. So 7.70 times 10 to the 20th, Divided by 6.02, 2 times 10 to the 23rd gives you The number of moles that you're dealing with times 194.2 g per mole equals zero point 248 g. And make sure the number of sick figs is correct. You always look at the one with the least number of sick figs when you're multiplying which in this case to 7.70 with three. So our answer is .248g. Finally, let's look at part D. We're asking for the mass of cholesterol. If zero 105 moles, it's a mass 0.46 g. So really for this one, we don't actually need to find the molar mass. All we need to do is remember what the definition of Mueller masses. So Mueller mass is that grams Per one mole of a compound. And so if we know how many grams are in 0.1 of five moles, Yeah, then we can set up a simple equation To find out there'd be XG in one mole. And so all you have to do is take .406 divided by .0010 five miles. And you would get 386 .66 repeating moles. Okay, we only have 366 here. So we ground this up to 387. Hence Permal. Yeah. And if you're curious, look up the Mueller mats of cholesterol And it turns out that it does in fact have a molar mass of 286.65. Which is very close to our number that we got before we rounded it up. So that's the problem

I even saw in this question. They are calculated Mosque in grams of the following E. 1.25 more calcium prospect yeah. 1.25 more calcium or spade. No we know that w that is waiting grams That is equal to an into em. So 1.2 for you are the camera phones into moller muscle calcium phosphate is 310.18. And therefore therefore mosque in grams. Moss in ground that is equal to 387 point 71 for you. Uh huh. Mhm grabs then the second one. Yeah. 0.600 micro moll op. C. For age 10. Micro more. Love see forage 10 C. For 8 10. So this is equal to yeah 0.600 in to challenge to minus six. So we know that w it is recalled to any into em and that is equal to point 600 in. To tenders to minus six into 15.12. And therefore moss Ingraham. Yeah that is equal to 3.49 into appendage to minus five g. Yeah then see fart. Yeah. Yeah 0.62 for you. Millie mola ram. Really Mall F. E. And that is equal to I am 62 for you in to tenders to minus three mola firing. So I believe he can to end into end and that is equal to 0.625 into english to minus three into 41.86. And therefore mass in grams notice equal took. Sure. Yeah. Find 15 11 6 to 5 grand. Then be part one point 45 malt ammonium carbonate fact. So and I believe he can do any into young and that is equal to 1 45 into moller nasa ammonium carbonate 96.9 And therefore Massing graham of ammonium carbonate. That is equal to 1 39.33 Yeah 29 23 grand. Think

Even so in this example how to calculate the mass from the malls They're given 6.14 into tenders minus four Moles of SL three. Then be 3.11 into the industry. Five moles of lead off site the 3.495 moles of serious Hillary Then d 2.44 you into tenders to minus eight moles of Treichl or a 10 then e 0.16 and more of a leverage and have a point point 26 more stopping copper chloride. So let us begin with this exercise. So, you know, uh, they're given 6.14 into tenders to minus four Moles Episode three Walls of s a tree So you have to use the formula and is equal to W upon young and therefore w is equal to any into, um Therefore you need 6.14 into 10 national minus four into molar mass of cell phone to exile. It is 80 80 80 and that is equal to 491.2 into 10 to minus four. Okay. And therefore loss. Ingraham soft grams of 6.14 in two tenders to minus four months of s 03 that is equal to 4.912 into tended to minus 2 g. So this one day and so, yeah. Then the second one. Okay. Mhm. Yeah, right. Yeah. Three point 11 in to tenders to fire you more. Also, preview debut? No, Once again. Yeah, he's he called to w upon, um, and therefore w is equal to Yemen too young, and that is equal to Jan is 3.11 into 10 x 25 into molar mass of laid upside. It is 223.2 that is equal to 694.152 into $10 to 5. And therefore and studies ladies loss Ingraham of 3.11 into it and next to five moles models of bbo that is equal to six planes. Uh huh. 94 152 in two tenders to 7 g. Mhm the next. See one. Yeah, that is 0.495. Also, flora form. No, you know, w is equal to and in two young Well, that fall probably is equal to wind. 495 into 119.38 and therefore w sequel therefore Marcin Grandma loss in Graham. Oh, why in 49 for you model. So chloroform that is equal to 59 point 0931 g. Mm. Yeah, the next divan. Yes. Yeah, yeah, yeah. B two point 40 for you in two tenders to minus eight mile sob. Right, Laurita, you tend That is C two h, three cm three. So we know that w is equal to and into em. Therefore, W is equal to 2.40 for you in to tenders to minus aid in 233.4. That is equal to 3. 26 point 83 into the national minus eight. You go around and therefore W is equal to three point 26 83 You do, then managed to minus six times. Mhm. Yeah, the next. Mm. Then e 0.167 More love. Ally. Wait. So we know that w is called too young and too young. Uh, therefore lovely is equal to 0.167 into molar mass. 23 points. 95. Uh huh. And therefore, W is equal to three point 99 six for you Grand. Yeah, Mhm mhm school. My next, uh, 5.26 months sub copper chloride Wallace of right copper chloride. Yes. Yeah. Copper chloride, that is. No, please call to any into Young. Therefore, the place he can't do. Yeah. I mean, it's 5.26 into 1. 34.446 and therefore doubly is equal to 707 point 18596 in grand. Thanks.

We have 375 grounds of a solution that is 1.51% ammonium chloride, which means that there are 1.51 grams of ammonium chloride in 100 g of solution by definition of the mass person. So to find the number of moles of ammonium chloride, we can just cross multiply here to cancel out grams of solution and we left with We want mhm. So 3, 75 times 1.51 divided by 100 is equal to five point 66 grands of an H four c l ammonium chloride are in this solution. Now, if you have 2.91% sodium chloride mhm, that's 2.91 grams of N I C. L in 100 grams of solution. So if you have 125 g of it, multiplying by that massive solution will give you the grams of any cl. Okay, 1 25 times 2.91 divided by 100 gives you 3.64 grams of and I see all For part C, we're working with 4.92% uh, potassium nitrate. So That's 4.92 grands. Okay. N 03 in 100 grams solution. If we have 1.31 kg of solution, we know that there are 1000 g in a kilogram so kilograms cancels with kilograms. Grams of solution will cancel the grams of solution and we'll be left with grams of potassium nitrate 1.31 times 1000 times 4.92 divided by 100. Gives you 64.5 mhm and our unit is grams of potassium nitrate K and oh three. And lastly, if you have 470 mg and we're working with 12.5% ammonium nitrates or 12.5 g of NH four n 03 in 100 grams of solution 407 8 mg and we know that 1 g is 1000 mg. Yeah, Mhm. Yeah. So then milligrams will cancel with milligrams, grams of solution cancels, grams of solution will be left with grams of ammonium nitrate. You want to 4 78 divided by 1000 times 12.5 divided by 100 gives you 0.0 598 rounded to three significant figures grams of ammonium nitrate. Mm


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