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1z03-MWso >0.175MWSZSTOthenew golution What is the molar concentrations 0f 8042- iorsQUESTION 120.420M0.175MW Oo"L0.675M0.370MWhat is the molar concentratio...

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1z03-MWso >0.175MWSZSTOthenew golution What is the molar concentrations 0f 8042- iorsQUESTION 120.420M0.175MW Oo"L0.675M0.370MWhat is the molar concentrations ofKt ions in the new solution?QUESTION 110.375 the new solution? ions Assuming that the volumes are additive, what is the molar concentrations of Na0.900 M K2SO4 with 30.0 mL of A solution is made by mixing 50.0 mL of 0.300 M NazS04 Use the following information to onswer questions ## 10-12QUESTION 10

1z 03-M Wso > 0.175M WSZSTO thenew golution What is the molar concentrations 0f 8042- iors QUESTION 12 0.420M 0.175M W Oo"L 0.675M 0.370M What is the molar concentrations ofKt ions in the new solution? QUESTION 11 0.375 the new solution? ions Assuming that the volumes are additive, what is the molar concentrations of Na 0.900 M K2SO4 with 30.0 mL of A solution is made by mixing 50.0 mL of 0.300 M NazS04 Use the following information to onswer questions ## 10-12 QUESTION 10



Answers

Indicate the concentration of each ion or molecule present in the following solutions: (a) $0.25 \mathrm{M} \mathrm{NaNO}_{3}$, (b) $1.3 \times 10^{-2} \mathrm{M} \mathrm{MgSO}_{4}$, (c) $0.0150 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$, (d) a mix- ture of $45.0 \mathrm{~mL}$ of $0.272 \mathrm{M} \mathrm{NaCl}$ and $65.0 \mathrm{~mL}$ of $0.0247 \mathrm{M}$ $\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}$. Assume that the volumes are additive.

We are going to be combining mixtures. Okay, so we'll start off with two different kinds of solutions of sodium hydroxide. Okay, so feel a stopped their concentrations involved. The first one has 0.17 moles And the second one is slightly higher at .4. Okay Then their volumes. 0.42 L with the other at 0.0 376 leaders. Now we can do the calculation of most is nope concentration times volume to figure out their malls And we will get zero 0071 for moles. And then this one would be 0.015 mi. Okay, so where you know the concentration of sodium will be equal to the concentration of hydroxide of it. And they're combining basically that we're combining the same Calyon and and I on. So well we have to do to get this constitution is trying to the total moles and we can do that by adding these two moles here. And that gives us 0.02-14 moles than to find the volume total concentration is for the total volume used and we'll add this second round here. What we get is 0.796 leaders. Okay. So then if we do that calculation, the final moles concentration of sodium and hydroxide Are equivalent and are 0.0278 multiple later. Yeah, that was just the first example. Next is a slightly bit more involved. Okay, so we'll take sodium sulfate with some salt, potassium. Quite okay. And if we mix come together, what will we do? Yes let's first. Right. Um Are givens. Okay, so here are the concentrations given followed by their volumes. Because up we converted that to leaders because that's actually what's most useful actually sort of this were over a bit to give us some room king. Now we need to calculate the malls simply by multiplying what's above. What we get is 0.0044 moles, sodium sulfate and 0.00375 moles of the potassium chlorate. Okay. Now for going to find concentration of the ions in this solution, we need the proper volume of this solution by adding a second role as we did above. And that gives us 0.069 leaders. Okay, so let's start off by isolating these concentrations. Okay, so we know that the concentration of potassium would be equivalent to that of the chloride. Okay. And we'll just take this these molds here and divided by the volume. Get the total volume like this. And what were ended up with a zero 05 43 moles per liter. Okay then that's going to be concentration for potassium and for the chloride as well. So following for the concentration of the sulfate, we'll take malls of the sulfate invited by The total. Okay, you get above in that will yield of 0.0 0.0637. That's per liter. Then the malls of Sudan will be double that. I think the final problem that we're going through, it's going to be an interesting one. We're going to take potassium chloride and mix it with some calcium chloride. So here we're going to have some additive additive properties as we saw in the very first example. So here, since we were given maths of 3.6 g, we're going to need to remember the molar mass. Okay. Which is that over there? 74 g per mole. Okay. And the malls will simply be mass over molar mass To give the serpent 0483 the walls of this salt. Okay, So then for the calcium chloride We had zero 24 We're sorry to five multiple causing quiet. Okay, so the only volume we had which will call vine total because here we just got a dry mass of the first cop out Will be 0.075 leaders. And the moles then will be fi multiple. What's above your europe 1875. All right. But actually moles of calcium chloride doesn't change. So we can use this to find the malls of the calcium. It will also be the exact same thing here because this one is actually not deluded at all because we're not adding any volume for the rest. We will need to do some work here. All right. So, if we try to calculate concentration of the potassium I am. What we need to do is take this smalls over here here .0483 And divide that by the total volume, which was 0.075 leaders. Okay. And what that gives us is you're 26 44 knows. Okay. So now for the calcium is going to be completely different case when we got in blue. So most of calcium will be the component from the hook passing for it As well as double the component from the calcium cord. Since it's going to associate into two and my arms. Okay. So if we do that addition, What we end up getting is zero 0858 months. Hey, So we would have added it that way and then malls of chloride will be you know that and the concentration Will be most over the total volume to lead 1.1 for four months. The leader. So that was tricky part. There was a need to remember that the chloride ion is coming from both components.

Let's go for the Mueller concentration of arsenic three in solution. Arsenic three is present in the each three a s 03 compound. Let's find the moles of the H three A s. Three. We're told that we have 22.35 mL of the K B R 03 It's convert to leaders. We know the mill Arat E here one malls K B R 03 per leader K p 03 one mole of K B R 03 has one more of B R 03 minus and our story geometry here. One more B R 03 minus 23 moles of the H three A s 03 That's what This out here. Okay, so we're kind of tipping 0.671 Let's keep all our sick fix here. Five moles of H three a s three. The morality of the H three a s 03 is equal 2.6705 Moles divided by the volume 50 mL into leaders is 500.500 leaders. When we find that the majority here is 0.13 for one Mueller Mhm

So in this case we will use the formula. And when we want equals to m to it too. So The two is equal to 50 plus 25, 75 millimeters. M two is equal to. And when we won over we too. So I am too will be equal to 0.2 moller Multiplied by 50 million l Divided by. So this M two is 4. Yeah. Can they see a low 3? Okay. Just to be sure. Divided by 75 mm. So this will be 0.133 moller and UCL. Okay, concentration of I need to and so forth. In this case, I am too will be again And when we won over them too. So this will be equal to zero coin to moller In 225 divided by 75 mm. This is equal to 0.066, six more. Not. Can you do it so far? The concentration of irons are mm hmm. Do you .133 Mueller? Any CIA lutely. And do you open to 6 6? six more notes? And you do s. 0. 4? So for any positive, it is 0.133 Smalling plus two in 2 0.6 266 smaller. Any positive? And this is the answer.

In this problem were asked to determine the polarity when varying amounts of Salyut are dissolved in various amounts of solution, Our first prompt involves 150 grams of sucrose being dissolved in 250 milliliters of solution. You can start this problem by taking our 150 grams of sucrose and dividing that by the molar mass. By doing this, we determine our movements of sucrose that we have and we find that we have 0.438 moles of sucralose and we were given a volume in terms of milliliters. So we know that we have 250 milliliters of solution. We can do a quick unit conversion here, Tito leaders so that we can write our polarity improper units. So we have, ah, volume off 0.250 leaders. So to find similarity, we're just going to do the moles that we calculated divine by our volume amount that we found through a unit conversion. And by doing this, we find that our solution is 1.75 Moeller. In this next scenario, we're told that we have a certain massive Yuria with 97.9% purity and the serious being dissolved in five milliliters oven angriest solution. We can start solving this by taking our mass that we have, which is 98.3 milligrams. And we're told that we have a certain percent purity so we can multiply by that percentage, which is 97.9%. So we find that the actual mass of our Yuria that we have is equal to 96 0.24 milligrams. We can then find our moles of Yuria by taking our actual mass that we found through our previous calculation. And we can translate that wheat and two grams by doing a unit conversion. And we condone divide by the molar mass of Yuria to find our Mel's. And by doing this we find that 0.16 moles of Yuria, our president. Our volume was given to us in milliliters so we can do a unit conversion on our volume. We know that there are 1000 milliliters in every leader. So when we do that math, we find that we have a volume of 0.5 leaders and to find our polarity, we're going to take the moles of Yuria that we found in our calculation above and divide by the volume that we found in leaders. And by doing this calculation, we find that our polarity is equal. 2.32 Moller in this problem were provided with a certain amount of methanol in Melo. Leaders that is dissolved in water to make 15 liters of solution were also given the density of methanol. Here we consult for our moles of methanol by taking the volume of methanol that were given in the problem statement. I'm multiplying by the density that we were also given. So this translates our number into grands of methanol rather than milliliters of methanol. And from here we can divide by the molar mass of methanol to go from grams to moles. By doing this math, we find that we have 3.9 moles of methanol present and we know that we have 15 liters of solution so we can take our three moles of methanol and divide by our 15 leaders. And through this calculation we find that arm polarity is equal 2.206 Mohler


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