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A student obtained the following data based on the procedure inthe lab manual. What is the percent difference between the twotrials? Report your answer with two pla...

Question

A student obtained the following data based on the procedure inthe lab manual. What is the percent difference between the twotrials? Report your answer with two places after the decimal.Mass of empty evaporating dish and filter paper: 15.282 gMass of evaporating dish plus dried product, first drying:15.982 gMass of evaporating dish plus dried product, second drying:15.874 g

A student obtained the following data based on the procedure in the lab manual. What is the percent difference between the two trials? Report your answer with two places after the decimal. Mass of empty evaporating dish and filter paper: 15.282 g Mass of evaporating dish plus dried product, first drying: 15.982 g Mass of evaporating dish plus dried product, second drying: 15.874 g



Answers

A student calculated the theoretical yield of barium sulfate in a precipitation experiment to be $1.352 \mathrm{g}$ When she filtered, dried, and weighed her precipitate, however, her yield was only $1.279 \mathrm{g}$. Calculate the student's percent yield.

For this problem. There's multiple steps involved, although it's a very short problem in terms of its words. If we have 1.271 g of sodium sulfate, we can convert this into molds of sodium sulfate. Then what we have to do is determine how many moles of sodium sulfate became sodium sulfate. Decca hydrate. To do that will take the math that was gained 0.387 g and recognize that this mass was gained when exposed to the atmosphere due to the acquisition of water. So this 0.387 g is completely water. If we take the mass that increased from water and convert the mass into moles of water, we can then figure out how many moles of sodium sulfate Decca hydrate were formed. Bye. Recognizing 10 miles of water are required to form one mole of sodium sulfate. Decca hydrate. So if this were the moles of sodium sulfate we started with, and this is now the moles of sodium sulfate Decca hydrate that were, uh, created subtracting those two will tell us how many moles of sodium sulfate are left in the mixture. Then, if we take the mass of water converted into moles of water and then moles of water into moles of sodium sulfate. Decca Hydrate. This will be the moles of sodium sulfate Decca hydrate in the mixture, but they don't want a mole percent. They want a mass percent. So now that we know the moles of each in the mixture, we need to take the moles of each and convert them into a mass by using, in this case the Moller massive sodium sulfate and the moles of sodium sulfate to get the massive sodium sulfate 0.9658 g. Then we'll take the moles of sodium sulfate, Decca Hydrate and the Mueller massive Sodium sulfate Decca Hydrate 3 22.2 and get the mass of sodium sulfate Decca Hydrate. Then, to get the mass percent of each of them, will take the mass of each of them massive sodium sulfate divided by the total mass of the newly created mixture, which is going to be the some of these two values, which should also be equal to the 1.271 g of the sample, plus the 0.387 g of water that absorbed that some should be equal to this some, which is 1.658 g. Multiply that by 100 we get 58.24%. So subtract 58.24% from 100 you'll get the rest of the percent. That is the sodium sulfate Decca hydrate, which will be 41.76% sodium sulfate Decca Hydrate.

Yeah, a student was given a 1.640 g sample of a mixture of sodium nitrate and sodium chloride and was asked to find the percent of each compound in the mixture. We're told that dissolving a sample and then added solution containing excess of silver nitrate the silver I'm precipitated out the Claritin and the mixture was filter. Dr. Wade in its mouse was 2056 g. Starting from the precipitate, the 2.56 g of the silver chloride. Let's find the mass of sodium chloride. So 143.4 g it's a Mueller massive gcl, one mole of a gcl thio to one more of our balanced equation Is that Z way didn't write a balanced equation for the formation of silver chloride? Uh, we have an A C l plus silver nitrates to produce silver chloride, and this is our precipitate and sodium nitrates. Lee Quis Hey, Chris Equus. This is a 1 to 1 more ratio. This will be one more of the A. C l. And in a c l. One more is 58.44 g and this will be equal to 0.8379 g of N a c l. Now, to find the mass of the sodium nitrates, we could take the mass of the sample entire sample minus the mouse of sodium chloride. So the mass of the sample was 1 64 0 g, minus the mass of the 0.8379 g, and we confined the mass of the sodium nitrate to be 0.7861 g. And now we can calculate the percent of both components in the mixture. So the percent of and a C L would be the mass of N a. C l 0.8379 g divided by the mass of the sample, 1.6240 g times 100% and this would be equal to 51.59% would be the percent of N A C l. And the percent of the sodium nitrate would be 0.7861 g is the mass of sodium nitrate over the mass of the sample, and this would be equal to 48.41%. So here we have our percent of any CL in the mixture percent of sodium nitrate in the mixture

There's an equation that allows you to calculate the percent yield from it and it's rearrangement. You can calculate other things such as theoretical yield and actual yield percent yield is equal to the actual yield. This will be the amount that you get in the laboratory. You don't calculate this amount. You measure this amount divided by the theoretical yield, which is the value you calculate from the spike geometry of the balanced chemical reaction, Multiplied by 100. So for this problem, it wants us to calculate the actual yield rearrangement of this equation allows you to calculate the actual yield that you're going to get in the laboratory, that you're going to measure in the laboratory knowing the theoretical yield And that's going to be equal to the % yield, which was 40 multiplied by the theoretical yield, Divided by 100. So if we put in our 40 multiplied by the theoretical yield, that would have been calculated at 12.5, divided by 100 we get 5.0 g

It's in a laboratory experiment, we have 15 mL of a K. C. L solution and it's poured into an evaporating dish with a massive 24.10 g. The combined mass of the evaporating dish and the k. C. L solution is 41.50 g. So the difference between the mass of the evaporating dish and the mass of the evaporating dish with the K. C. L solution will be the mass of the K C. L solution that was added. So 41.50 minus 24.10 gives us the mass of the K. C. L solution, 17.4 g. Then it tells us that after heating the evaporating dish to dryness, we end up with dry potassium chloride stuck to the bottom of the evaporating dish with a combined mass of the dry potassium chloride and the evaporating dish of 28.28. So the mass of just the chloride that was in the 15 mil leader potassium chloride solution would be the mass of the evaporating dish with the dry potassium chloride minus again, the mass of the empty evaporating dish 24.10 g. So the mass of the potassium chloride in the potassium chloride solution would be 4.18 g. With this information, you should be able to answer the three parts to this question. The first part is what is the mass percent mass potassium chloride per mass potassium chloride solution. And because it's a percent, we need to multiply by 100. So to calculate this mass percent potassium chloride in the potassium chloride solution will take the mass of potassium chloride divided by the mass of the potassium chloride solution and multiplied by 100 and we get 24% potassium chloride mass to mass or 24% 24 mass percent potassium chloride. Then, for part B, it asks what is the polarity of the potassium chloride solution? Well, polarity is defined as moles of the solute. In this case potassium fluoride divided by the leaders of the solution. We can calculate the moles of potassium chloride knowing the grams of potassium chloride, which is what we calculated up here. If we have 4.1 g of potassium chloride, we can divide by the molar mass potassium chloride, which is 74.55 g and get moles potassium chloride. We now need to divide by the leaders of the potassium chloride solution. We were not given leaders but we were given the volume in millilitres of the potassium chloride solution, 15 mL. So if we divide that by 1000 we will now have in our denominator the leaders of the solution. So moles potassium chloride divided by leaders of solution is more clarity and we get 3.74 Mueller. For the last part, it asks if water is added to 10 mL of the potassium chloride solution to give a final volume of 60 mL. What is the polarity of the diluted solution? If we take the dilution equation, M one, V one equals MTV two and rearrange it to solve for M two, the polarity of the diluted solution, it will be equal to V one, which is the volume of the potassium chloride solution. Were diluting 10 mL multiplied by M one, which is the concentration of the potassium chloride solution that we determined appear divided by V two. V two will be the new final volume to which the solution was diluted, which is 60 millimeters. And this gives us a final concentration of the diluted solution to be 0.6 to three moller potassium chloride


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