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Bradon tia sanblly ruks wtkh cthe folloning should be sotble Inwate7 HgzchzNazsAg2co3Ag2SBaSO4QUESTION 2make 3,1 102 mLofa0.41 Msolution? Determine how many Erjms ...

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Bradon tia sanblly ruks wtkh cthe folloning should be sotble Inwate7 HgzchzNazsAg2co3Ag2SBaSO4QUESTION 2make 3,1 102 mLofa0.41 Msolution? Determine how many Erjms . potassium Iichromate (K2CrzO7) solutes would be @eededQUESTIONWhat volune (In units of ml) 0f 3.7 MKOH IS required t0 prepareof 1,22 MKOH?and subimit Click Sallc All Annrersull anauGiLsat and Submit [0

Bradon tia sanblly ruks wtkh cthe folloning should be sotble Inwate7 Hgzchz Nazs Ag2co3 Ag2S BaSO4 QUESTION 2 make 3,1 102 mLofa0.41 Msolution? Determine how many Erjms . potassium Iichromate (K2CrzO7) solutes would be @eeded QUESTION What volune (In units of ml) 0f 3.7 MKOH IS required t0 prepare of 1,22 MKOH? and subimit Click Sallc All Annrers ull anau GiLsat and Submit [0



Answers

Potassium hydroxide is used to precipitate each of the cations from their respective solution. Determine the minimum concentration of $\mathrm{KOH}$ required for precipitation to begin in each
case. $\begin{array}{l}{\text { a. } 0.015 \mathrm{M} \mathrm{CaCl}_{2}} \\ {\text { b. } 0.0025 \mathrm{M} \mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2}} \\ {\text { c. } 0.0018 \mathrm{M} \mathrm{M} \mathrm{gBr}_{2}}\end{array}$

We're looking to mix the solution of Precor inc acid with sodium hydroxide. Okay. And we're looking for the concentration of acid at equivalence point. We just need to remember that moles of acid and moles of based equal each other at equivalents. Okay, So then if we write that out fully, you know that C. A. So concentration of us at times while we have acid equals concentration of base times volume of space. So let's just rearrange and we're absolutely looking for volume. It's changed out there. You have volume or concentration of acid. They're looking for volume. So we'll use this relationship here. Okay? So if we just multiply what we were given and divided by penetration of acid, Okay, we end up with here is 0.0 3 8 leaders of acid is required to get this solution to equivalents. Next example we have a solution of some magnesium hydroxide and you're mixing it or not a solution. A mass. We have 2.87 g of this stuff And we're going to put it into a 0.128 smaller concentration of some hydrochloric acid. Yeah. So we'll start off by recalling magnesium hydroxide, Molar mass which is 38 g promote approximately. And that will get us Um Malls of 0.0492. Yes. Okay. So if we're looking for the volume to get to equivalence, we'll just use this relationship and free. But so you just plug in all the components here. So understanding is zero 7687 leaders will take us to equivalents. Next up we will look at another reaction. Okay, We're going to take silver nitrate and mix it with some potassium chloride until we get some silver chloride as precipitate. Yes. What in fact do we have? So you know the volume of the silver nitrate solution? 0.0258 leaders. And we actually have some solid potassium chloride. It's 0.785 g. Okay, so again, we were well um search up the molar mass of potassium chloride to be 74.551 g per mole yielding a number of moles Of 0.01053. All right, so now we have malls. We have volume. We can solve for a concentration kept by just taking this these ones here, over the volume here. And what we end up Getting is zero Actually 0.48 40813 most per liter. Lastly we have another I'm not equilibrium kind of a reaction actually neutralization with as recalled acid. Okay. And Saudi or potassium hydroxide. Okay, so the concentration we have of the asset Is your .10 eight months And the volume is your .0453 leaders case. That gives us a mold. If you just multiply the two Of 0.00489 moles. All right. So the only thing we know about the potassium hydroxide here is that it's smaller mass 56 g per mole. Yes, but we do have moles because we know, as I said earlier, most of acid equal smells of base. So then if we take our more mass multiplied by moles will get actually the mass That is required to reach equivalent point which is 0.27 four or 5 g. I just want to play those numbers there.

I haven't even so in this question they asked how many grams of so little needed to make each of the following solutions. 8 50 ml a 500.100 smaller operations Selfridge. So eight bucks. We know that N is equal to M into V. And that is equal to points 100 into 2 50 point thousands of 50.0.2 15. And that is equal to 0.0 25 malls. Then we know that W is equal to number of months into moller Maas, a number of months, 0.25 into moller Mas of bluish himself ain't 1 74.26. And that is equal to four point 34 U seven and therefore W o r k two. So four that is equal to 4357 grams then be part and is equal to come into V. And that is equal to 0.2 50 in to to keep the upon 1000 that is 10000.2 50 that is equal to 0.625 moles. Now we know that W is equal to and into em. So that is equal to the number of moles, europe and 0625 into smaller mass of I ran three chlorides 90 is 1 62.2 and that is equal to 10.138 g. And therefore W five official three 10.138 g. My next see part N is equal to I'm in T V. And that mhm. Therefore, and is equal to no 0.400 into 400 upon 1000 0.500 that is equal to point two months. Now, any sequel to W upon em. Therefore, W is equal to and into um that is equal to 0.2 into 255 points 43. That is equal to 51 point 0 86. And their forward Yes. W Off medium accident. Bad idea. I see that. That is equal to 51.0 86 graham. Thanks

For optioning to calculate the volume off etc. Alot for needed to neutralize the enemy which we use. The formula m one V one equals m toe vetoed We're em one and we won corresponds toe Etch Cielo for and em to veto corresponds to the values off any which the volume off it. Substituting in the values The volume off at Cielo four turns out to be 38 point the women to calculate the moles off mg which toe solution we first we need to determine the molar mass off MD which which turns out to be 58.32 grams. Using the formula malls equals weight divided by the molecule months. The number off mole turn out to be 0.49 more writing the balance chemical equation between its seal on MD which one more off MD which, neutralized by two moons off its sale, does 0.49 moles off mg a witch too Nuke is neutralized by 0.1 mole off etc. To calculate the volume off etc. L needed to neutralize mg which took the volume. Don't know Toby 769 Emily for the we need to calculate the moves off case he'll So the molar mass off Casey list 74 points 54 grams To determine the moon. We use the formula. Moles equals mass divided by more like llamas and the number off mortal Now. Toby 0.105 More writing the balance chemical equation between Casey Ln India No. Three one Moon off Casey and needed to precipitate one moon off Eiji Positive. I'm for then calculating the polarity off 80 and No. Three, the polarity turns out to be 0.407 em calculating the moles off its shell solution from polarity. The arranging the polarity formula. The more turn out to be four point 89 in 10 days to turn negative three more writing the balance chemical equation between ETC. L and K which one mole off its seal neutralized with one more off a witch

This question. First of all we'll have to write about a couple correctional for the first one we see the reaction once one mole ratio. So we have the polarity of the sodium hydroxide multiplied by the volume of sodium hydroxide in leaders. So do you have familiar solutions about 5000 and then covered from moles of sodium hydroxide two moles of for caloric acid. Then you will need to use the polarity of the acid to convert from moles of acid to leaders of acid. So we're just reverse. We're just flipping it over to uses courage factor. Then we have the volume and leaders and we can also convert it to middle leaders want. So the next one we have to write the balanced chemical reaction between hcl and mg Oh h two. So before that's balanced and we need to put two information to and from the water. So then we have the mass of the magnesium hydroxide. Bye bye. Mueller mask which you get from the periodic table and atomic masses and we get the moles of potassium. It jobs then from the balanced chemical reaction we see that two moles basically react with one mole of magnesium hydroxide. So we convert two moles of hcl. Then use the polarity of hcl convert to leaders of hcl. We get the volume of hcl needed and we can convert that to milliliters if you want. So now for the next one right the balance couple correction like we've been doing. Then we start with milligrams of K. C. L. Divide by 1000 to get grams of K. C. L. Divided by the molar mass K C. L. To get molds for K C. L. Then convert from altercation multiples of silver nitrate. Using this political correction, mm divide by the volume of silver nitrate and leaders. Did we get the polarity? We lost one right, the balance coming correction. Take the polarity of a seal, multiplied by the volume, then one more of your silk reactive. One more college. And then use the molar mass of college to convert to mask college. And we get the massive college that must be present in solution.


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