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Its0W Gi_hAtelution Hl utdY ncllilun "hleIn3 ru-larityulur- Vinl wlutunItml4coneentralion when 5O WmL ofwalct / addcd t 725AallElculilcGc ritrlc diluli g 1 sto...

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Its0W Gi_hAtelution Hl utdY ncllilun "hleIn3 ru-larityulur- Vinl wlutunItml4coneentralion when 5O WmL ofwalct / addcd t 725AallElculilcGc ritrlc diluli g 1 stock makc 0 S79 Lof,150 M SupposJnu wished mLof thc Sfock solution wuld Yon Rced f0 solution 01.675 E siket nirrnGe. How" many FElatis diluted with Walef t0 make 16s ML stock solutian of suerosc If55.0 mLola 2,45 Lhc mlority ofhe Auoa] ~olutuan? Rucroy toluion; Mhr

Its0W Gi_hAtelution Hl utdY ncllilun "hleIn3 ru-larityulur- Vinl wlutun Itml4 coneentralion when 5O WmL ofwalct / addcd t 725 AallE lculilc Gc ritrlc diluli g 1 stock makc 0 S79 Lof,150 M SupposJnu wished mLof thc Sfock solution wuld Yon Rced f0 solution 01.675 E siket nirrnGe. How" many FElat is diluted with Walef t0 make 16s ML stock solutian of suerosc If55.0 mLola 2,45 Lhc mlority ofhe Auoa] ~olutuan? Rucroy toluion; Mhr



Answers

Given a $0.250 \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}$ stock solution, describe how you would prepare a solution that is $0.0125 \mathrm{M}$ $\mathrm{K}_{2} \mathrm{CrO}_{4} .$ That is, what combination(s) of pipet and volumetric flask would you use? Typical sizes of volumetric flasks found in a general chemistry laboratory are $100.0,250.0,500.0,$ and $1000.0 \mathrm{mL},$ and typical sizes of volumetric pipets are 1.00,5.00,10.00 $25.00,$ and $50.00 \mathrm{mL}$.

I even saw in this question dusk calculate number of grams of each salute. That has to be taken in the following a 1 25 ml of 250.200 Mueller initial So a part and is equal to M into V. And that is equal to point 200 into 1 25 Miles. Divided by 1000 nothing but 10000.1 25 liter. That is equal to zero and 0 to 5 mall. Yeah. No, we know that w sequel to number of moles into moller MAs. So number of moles 0.25 into mama's 58.5. And that is equal to one point 463 g. And that forward. Yeah. Yeah. Okay. Uh huh. W Financial that is equal to 1.4 16 grab then be part. Yeah. N is equal to M. And T. V. And that is equal to 0.2 50 into 0.3 16. That is equal to 0.9 months. So N is equal to doubly upon M Therefore lovely is equal to and into em And that is equal to 0.9 into one entity 19. Currently 16 points two. And therefore the blue of C six. H 12 06 Mhm. Oh all six. That is equal to 16.2 g then see part. Okay. Any sequel to em. Yes. Yeah. Yeah. And do we That is equal to find 250 into what you means to keep the ml divided by 1000. So it is 0.250 That is equal to 0.625 moles. Then W is equal to and into em. And that is equal to 0.625 into 98. And then forward. W Aww it should be a softball. That is equal to 6.1 25 grab. Okay. Yeah.

Okay, so this problem is a simple AM on few on equal toe MTV to formula was started. The first bird you first. It's in the first part our, ah, we can consider the stock solution concentration as M one, which is the stock's risen Contreras six. Moller. And if we considered the falling on the stock solution required as few one, that's what we have to find out and came to hear is on the final or the resulting concentration which is point five Mueller and the resulting or the final four volume. Here is the volume of the resulting solution which is one hundred ten milliliters. So you have to find out the fuel on. So just put these values in this formal or were just rearrange it as em to veto over m one. That witch has put the values so himto. Here is your plan Fine Mourner Times The veto is hundreds, captain. Many later someone is six smaller from here are very one bell will be nine point one six militar. So this is the falling on the stocks solution we require Teo prepare hundred animal off point five Mueller Patrick is a solution now in the second part, we have to dilute the animal of the stock solution to a final volume of point two five three litre. If you find out, concentrate on the night. Your solution of the final solution again here or any one is a stock solution. Concentration one six Mueller V. One is what we're taking from the stock solution which is ten millimeter and our aim toe is the final concentration with you have to find out. And our veto, um is the final volume which is point two five zero liter. No, we have wanted Millie turn one of letters. We have to convert either one. So come back later, toe to fifty million liter. I have to find a veil off em too. So if we just really is the formal, I'm too will be em. One view on over V two. So everyone here is The stock's a different concentration. Okay, so it is not point six. If six smaller, it has six smaller. All right, so we have him six times. Then he shall follow him for the volume of the stock solution we're taking, which is ten million later. And the feet too is the final volume order or the volume the resulting solution which is two. Fifty militar. It will give us the final concentration, which is zero point two four Mueller. So this is the concentrate on the diluted solution if we dial ten militar off the stock solution to produce a solution of a bowling off point two five zero later or two fifty militar.

This question is moderately lengthy with multiple parts. You have a five mil leader stock solution of H two S. 04 for which you want to determine a concentration. You're going to dilute five million liters of it to 18 ml. This diluted solution, if you know its concentration could be used to calculate the concentration of the stock solution. But you don't know the concentration of the diluted solution. But you are given information to calculate the concentration of the diluted solution To this diluted solution is added 30 ml of 3.5 moller calcium chloride and 7.1 g of calcium sulfate is precipitated. So you can use this information to figure out the concentration of the sulfuric acid solution that you diluted. Let's start with the balanced chemical reaction, one mole H two S. 04 reacts with one mole calcium chloride to produce one mole calcium sulfate and two moles hydrochloric acid. We're going to assume that the sulfate reacts completely to form calcium sulfate, the sulfate from sulfuric acid. The concentration of sulfuric acid is to find as the mold sulfuric acid divided by the leaders. This is now the concentration of the sulfuric acid in the diluted solution that reacted with calcium chloride. To calculate the moles of sulfuric acid, we can start with the moles of the grams of the calcium sulfate that was produced because all the sulfate in calcium sulfate came from sulfuric acid. So if we convert the grams of calcium sulphate into moles calcium sulfate by dividing by the molar mass calcium sulfate, we will have moles calcium sulfate. Then knowing moles calcium sulfate with the 1-1 relationship, we can calculate what we desire. One mole sulfuric acid per one mole calcium sulfate. We get mold sulfuric acid. We now need to divide that by the volume of the solution. Well the solution of sulfuric acid that was That reacted with the 30, ml 35 Mueller calcium chloride with 18 mm. So we simply convert the 18 ml into leaders by dividing by 1000. This ratio here then allows us to calculate the concentration of sulfuric acid in that diluted solution 29 moller. Then we'll use M one V one equals MTV two to calculate the concentration of sulfuric acid in the five mil leader stock solution. Rearrangement of this equation Gives us M. one a concentration in the stock solution will be equal to M. To the concentration of the diluted. We just calculated, multiplied by its volume, 18 ml divided by the volume of the stock solution. Used to make the diluted solution that was five ml Plugging in these values, then gives us a concentration of the stock solution at 10.43 moller. So this is one part to the question. It also wants you to calculate the concentration of calcium chloride that's left over after the reaction occurs. This one is a little bit more challenging. The concentration of calcium chloride after the reaction is left after the reaction is complete will be equal to the moles of calcium chloride leftover divided by the new volume. To calculate the moles of calcium chloride leftover, let's first calculate the moles of calcium chloride we have before the reaction occurs, We have 3.5 moller And 30 ml. 30 ml is .2030 leaders. If we take polarity times volume and leaders, we will get the moles of calcium chloride. We start with well then subtract off the moles of calcium chloride that reacted. We can calculate the moles of calcium chloride that reacted by calculating the or by using the mass of calcium sulfates that formed every time we form a calcium sulfate, we used a calcium chloride. So we'll take 7.1 g calcium sulfate divided by its smaller mass to get moles calcium sulfate and then use the 1 to 1 relationship to get molds of calcium chloride that reacted to produce this amount of calcium sulfate. So here are the molds calcium chloride. We started with these are the molds that reacted to form our calcium sulfate. That difference gives us the moles of calcium chloride left in solution after the reaction occurred. We then divide by the new volume. We added 18 mil leaders to 30 mil leaders. So assuming volumes are additive, we will some these up to get 48 mil leaders as our new volume And then we'll convert the mill leaders into leaders by dividing by 1000. This setup then gives us the concentration moles calcium chloride per liter solution After the reaction has occurred, 1.10 Mueller calcium chloride

So here we are asked for the amount of moles of acid required for equivalence for certain compounds. So equivalence occurs when the number of moles of acid added is equal to the number of moles of base. However, we have to account for contributions, such as whether we have two or three moles of hydroxide, which contributes double or triple, respectively. However, the number of moles of acid is always going to be M A V A. Since our strong asked, we're using in this case, it's just hcl which is monochromatic, which also is equivalent to the number of moles of base. Uh the polarity of the base volume in the base. This value here depends on the number of hydroxide or basically the contribution total of a certain base, whether it's monochromatic die product or try product etcetera. So in our first case, we have So are more clarity of assets. Initially given a 0.150 molars Were given in our first case that we have a polarity of base of 0.175 molars And the volume base of 25.0 mm. Since it's potassium hydroxide, there is only one hydroxide contribution. So the row factor is just one in this case. And now we can substitute everything into our equation here. Okay, okay. And we can find that the volume required is equivalent to $29.2 million. For our 1st part. For our second part we have once again we have modern protic acid mono product base since we have ammonia Which can only accept one proton. So our row factors once again, one volume base is 15 millimeters Similarity is six molars mhm. And by substituting in all our information here, we can find that the volume of acid required in this case Is 600. And our third case, instead of given directly the polarity Were given information that we have a density of 0.712 g per million. And in this case, once again, we're given monochromatic base which is similar to ammonia. So we're given that we have 15 ml this base here. So we would have to convert this in terms of mass. Using basically the fact that tends these mass over volume. We can find that the mass of this substance here mm Is 10.68 g. However, we have to now convert everything we have here. Two more clarity. So our compound CH three CH two, CH two, NH two. So it'll be 36 Uh plus nine hydrogen plus one nitrogen. So the molar mass is 59 g per mole. And we can find the number of moles then divided by the volume, which I'll just do it in one step here and we need to use for concentration. We have to use the leaders. Okay. And as a result of doing all of this, we essentially can find that the more clarity Of our base in this case is about 12.1 molars. And now we have all the information. We need to solve this and we can find that the volume required is 12,011,210. In our last part we have a compound that has to higher oxides. So our role factors equivalent to two in this case And we have 0.0050 molars of the substance. So this is the polarity based the volume of basis. 40. Yeah. Okay. Mhm. And by substituting our information in here, We can find that the volume of required is about two seven mL. And these are final answers.


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