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An aquatic plant sample containing copper (II) was analyzed byflame atomic absorption spectrophotometry. 0.2495 g of the samplewere taken and after the digestion pr...

Question

An aquatic plant sample containing copper (II) was analyzed byflame atomic absorption spectrophotometry. 0.2495 g of the samplewere taken and after the digestion process the solution was broughtup to the 50.0 mL mark. This solution had to be diluted 10 timesand then its absorbance was measured as 0.182. The least squaresregression of the calibration data resulted in the followingequation: y = 1.12 x (in ppm Cu) + 0.0412. Calculate the copperconcentration, in ppm, in the plant sample.

An aquatic plant sample containing copper (II) was analyzed by flame atomic absorption spectrophotometry. 0.2495 g of the sample were taken and after the digestion process the solution was brought up to the 50.0 mL mark. This solution had to be diluted 10 times and then its absorbance was measured as 0.182. The least squares regression of the calibration data resulted in the following equation: y = 1.12 x (in ppm Cu) + 0.0412. Calculate the copper concentration, in ppm, in the plant sample.



Answers

In the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing $10.0,25.0,50.0,75.0,$ and $100 .$ ppm of copper from a commercially produced 1000.0 -ppm solution? Assume each solution has a final volume of $100.0 \mathrm{mL}$. (See Exercise 123 for definitions.)

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

For this question, we need to recognize that there is a direct relationship between absorbent and concentration according to Beer's law. So if we do a delusion calculation, a change in absorb INTs, the percent change in a verb Unz will directly correspond to the percent change in concentration. So it says that the concentration decreases by 45% if it decreased by 45% than it became 55%. So that's why I have point by five of the original concentration at one Moeller being the new concentration. The only thing then that I have left to calculate is the final volume. The final volume is the initial volume 15 milliliters plus, however much I added to carry out the delusion toe have it decreased by 45%. So I plug everything into the delusion equation I saw for my ex is 12.27 milliliters, so that means I simply need to add 12.27 or about 12.3 mil leaders to the original 15 milliliters in order to see the reduction by 45%

This video, we're gonna go over question 136 from Chapter four, which says in the spectroscopic analysis of many substances, a series of standard solutions of known concentration are measured to generate a calibration curve. How would you prepare standard solutions containing 10 25 50 75 in 100 parts per 1,000,000 of copper from a comfort from a commercially produced 1000 parts per 1,000,000 solution? Assume each solution has a final volume of 100 no leaders. Um, so basically, we need to do a series of dilutions to get these more dilute. 10 25 50 75 1 100 parts per 1,000,000 solutions from this more concentrated 1000 parts per 1,000,000 solution. Um, so once we do these delusions, you can measure whatever property you need thio about these from these solutions so that you can determine the concentration of an unknown solution based on where it falls on your calibration curve. Let's go ahead and do our delusions. So if we want a solution, we want 100 milliliters of a 10 part per 1,000,000 10.0 part per 1,000,000 solution on from 1000 parts per 1,000,000. Then we're going to need to do a dilution we're going to use C one V one equals C to be too where c wouldn't be one are the concentration and volume of the more concentrated solution that you're diluting and then see to envy To are your desired concentration at the end of your dilution and the volume of your solution that you're making eso In this case, C one is 1000 parts per 1,000,000. V one is actually our unknown. We want to know what volume of our 1000 parts 1,000,000 solution we need in order. Thio, in order to make this 100 leaders of 10 parts per 1,000,000 solution S O C two is our tens parts per 1,000,000 and V two is our 100 no leaders. So if I sold this for B one, what I get is one millimeter 1.0 millilitres. I'm so in order to make the solution, you would take 10 take one mil leader of the 1000 parts per 1,000,000 solution on, then add enough water to equal 100 old leaders Next we want, um 25 ppm. So again we have C one V one is equal to see to be too. Um and now see, one is still our 1000 ppm solution. Everyone is a gun are unknown. C two is now 25.0 ppm and V two is our 100 milliliters s o. We solve this for viewing. What we get is 2.50 millilitres. So we would, uh we would measure out 2.50 millilitres of the 1000 parts per 1,000,000 million solution on then out enough water to equal 100 milliliters s o the next. A solution we want to make is a 50 parts per 1,000,000 solution. So again see won we won is equal to see to be too C one is 1000 ppl. The one is our unknown. C two is 50 ppm and V two is our 100 milliliters. It's how much of our solution we want to make s o. We saw this for view one. What we get is 5.0 millilitres. So we would measure out five milliliters of the 1000 parts per 1,000,000 solution on and then add enough water to equal 100 milliliters. So next we have 75 ppm. So again, C one everyone is equal to see to V two. C one is our 1000 ppm. V one is our unknown. C two is 75 ppm and V two is 100 milliliters. Eso we solve this for everyone and what we get is 7.5 zero milliliters Noticed that since I gave this volume in milliliters the volume I get is my answers and the leaders eso again. We would measure out 7.50 millilitres of the 1000 parts per 1,000,000 solution and then add enough water to equal 100 milliliters. Finally, the last solution we want to make is the 100 Um, the 100 ppm solution. Um so again, C one V one is equal to see to you too. C one is 1000 ppm. Be one is the unknown. C two is 100 ppm and V two is 100 milliliters. So we saw this for everyone and what we get is 10 milliliters. So you would measure out 10 milliliters of the 1000 parts per 1,000,000 solution and then add enough water to dilute it to 100 milliliters.

We're given here we have essentially basically a process for electrolysis. So the process for electrolysis we also accordingly have a solution of interest that we're trying to basically perform electrolysis on. So we have a copper sulfate acquis solution. So this is soluble to this just forms copper which is basically the iron of our interest. And were given information that the copper concentration decreases. So essentially the important thing to recognize here is that essentially the reduction of copper to essentially copper solid. This reaction itself is already a favorable reaction which has an N. A. Basically an E not of creator than zero. So this reaction itself will basically if it has the opportunity to essentially basically have electrons be transferred to this reaction. This reaction will even proceed with an basically coupled with an unfavourable reaction. So here we're essentially given information that also the concentration of hydro knee um increases. Uh huh. Which means that we have to have some factor that's producing essentially more and more hydrogen ions. And also were given information that mass of one electorate increases. Which likely implies that essentially since basically you're losing see you two plus. This essentially means you're forming more copper solid is forming. And on the other hand, essentially we also have some type of gas formation. So based on these clues, essentially we have to determine a set of reactions. And even for electrolysis, we're looking for systems that essentially minimize essentially basically the use of energy. So are you basically trying to find systems that basically have the least use of energy as possible? Since even for unfavorable reactions, we have to look for the least unfavorable reaction that's going to. So in this case, essentially the possible reactions based on the information given. So essentially the cathode we have the reduction of copper. So this checks off the fact that there is the increase in mass of something. And this checks off the fact that the concentration of C. U. two plus is decreasing Since you two plus is consumed. And basically C plus concentration basically see amount copper solid amount increasing increases. Which means that there is the possibility for see you to be plated essentially on the cathode on basically one of the basically under basically captured in their plates. And on the other hand, in the an ode, The reaction that occurs essentially is to H20 yields essentially oxygen gas plus hydro knee um plus 40 minus. So this reaction has an E. Not basically less than zero and basically the overall may not sell as a result of the entered reaction will be less than zero. So this satisfies also the conditions for electrolysis and basically this checks off the conditions for gas formation. We have oxygen gas and with hydro any information this checks off this condition. So this set of reactions essentially matches all the observational information given in the problem


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