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For the following data set; calculate mean and SD (Standard deviation}: points:SamleEreshuciehu lel 0.25 0.33 0.330.22DMSO (dimethyl sulfoxide) liquid that can make...

Question

For the following data set; calculate mean and SD (Standard deviation}: points:SamleEreshuciehu lel 0.25 0.33 0.330.22DMSO (dimethyl sulfoxide) liquid that can make cells more permeable compounds You are asked t0 prepare mL each of a 590 (vlv) 107 (vfv); 15% (vlv); and 2070 (vfv) DMSO in Water Explain precisely how you would do this. (5 points).You need make buffer for the restriction endonuclease Smal: This enzyme requires 20 mM KCI; 10 mM Tris-HCI Iph O1, 10 mM MEClz and mM dithiothreitol (DTT

For the following data set; calculate mean and SD (Standard deviation}: points: Samle Ereshuciehu lel 0.25 0.33 0.33 0.22 DMSO (dimethyl sulfoxide) liquid that can make cells more permeable compounds You are asked t0 prepare mL each of a 590 (vlv) 107 (vfv); 15% (vlv); and 2070 (vfv) DMSO in Water Explain precisely how you would do this. (5 points). You need make buffer for the restriction endonuclease Smal: This enzyme requires 20 mM KCI; 10 mM Tris-HCI Iph O1, 10 mM MEClz and mM dithiothreitol (DTT) Calculate exactly how You would make mL of this Smal buffer using the following stock solutions (0.5 M KCI; M Tris-HCI (pH 01, 250 mM MeIz MDTT): Show all calculations. Write answers below: (5 pointsk



Answers

Determine the range and sample standard deviation for each of the data sets. For the sample standard deviation, round each answer to one more decimal place than that used for the observations. An article by D. Schaefer et al. (Journal of Tropical Ecology, Vol. $16,$ pp. $189-207$ ) reported on a longterm study of the effects of hurricanes on tropical streams of the Luquillo Experimental Forest in Puerto Rico. The study showed that Hurricane Hugo had a significant impact on stream water chemistry. The following table shows a sample of 10 ammonia fluxes in the first year after Hugo. Data are in kilograms per hectare per year. $$\begin{array}{rrrrr} \hline 96 & 66 & 147 & 147 & 175 \\ 116 & 57 & 154 & 88 & 154 \\ \hline \end{array}$$

So we're being giving a buffer solution here and we're going to try to find its ph and then find the ph after adding some strong acid or some strong base. So first let's go ahead and look at the two components and find the moles of each. Don't make it easier for us to work with them. So I've got the polarity and the leaders. If I multiply those I'll get my moles 447 moles. Mhm. This is the H. C. Four age four oh six which is our weak acid. Mhm. And then our weak bases, its conjugate weak base. So we'll take its polarity times its volume and make sure you're in leaders. And we'll get the malls 0395 malls. Well this conjugate face. Mhm. Mhm. So we're gonna we definitely have a buffer. Right? We have a weak ass and its conjugate base. So we're going to try to find the ph of that. So I'm going to use H plus is K. A. Times the concentration of the acid over the base. But since I have moles, I'm gonna go ahead and rewrite that as ca the moles of my acid over the malls of my base. And we can do that because concentration is moles per liter, so moles per liter over moles per liter. You can see that the volumes are going to cross out and we're just left with moles. So we'll go ahead and plug in here. We'll use the K. A. For our weak acid right here, which is 455 Who was given to us on the problem, Time to send the negative five and then we'll multiply this by moles of acid, which is point 0447 Over base was 0395 So this will give us 5.15 times 10. The negative five. And this is the molar itty of R. H. Plus. So if we just do minus the log of that number will get our ph which is 4.288 So that's the ph of our buffer before we've added anything to it. Mhm. In part B though we're going to add some strong acid. Well the strong acid is going to react with the weak base so the H plus will react with our weak base. See for H 406 to minus. And it will make some weak acid. Okay, it's congregant weak acid here. So we'll use up some of the week base and will produce some more weak acid. So I'm gonna just write a table here where I do the initial change and final. So I'm adding 0.25 moles of H plus to these malls that we found above. Okay, so we have both of these to start. These are all in malls here. I hope you can see that the H. Pluses are limiting reactant. So we're gonna use up all of this. So we must use the same amount of our weak base and produce the same amount of our weak acid because it's all in a 1 to 1 to one ratio. So I'll end up with none of this. And I went 1.145 walls here. Mm 0.697 moles here. Mm Well we have a weak base and its conjugate weak acid. So we have a buffer. So again we'll go ahead and plug into our buffer equation. Or H plus is R K A 455 I'm still negative five. And then we're gonna do moles of acid over moles of base. So point 0697 moles over 6970.145 moles. You multiply that out. We've got 2.19 I'm telling negative four. And this is the molar itty of R. H. Plus. So if we take minus the log of that, we'll get our ph so minus the log of that number. Gives us 3.660 in the final step here, we're adding hydroxide. So a strong base. So that's going to react with our weak acid. So let's write that reaction. We've got the hydroxide reacting with our weak acid. Mhm. So that's going to make water and its conjugate base. So we're gonna use up some of our acid and produce some more base. So let's keep track of our initial change and final conditions here, I'm adding 0.25 moles of O. H minus to our original moles of our two substances here in our buffer. Okay 0395 We can see that the O. H. Minuses are limiting reactant. So this is all going to react. So I'll react the same amount of my weak acid and produce that same amount of my week base since the ratios are all 1 to 1 to one. So I'll end up with none of this and 0.197 malls of my weak acid 1.645 moles of my week base. So this is a weak acid and this is its conjugate weak base. So we have a buffer. So let's go ahead and plug in. Our equation H plus is R. K A. Which is 455 I'm sending negative five and then we're going to do the moles of acid over base with 0197 moles divided by 0.645 moles. So this will give us 1.39 I'm sending negative five. This is the molar itty of r h plus. So if we just take minus the log of that number, we'll get the ph The ph is 4.8 five seven.

Okay. So we've got a buffer solution here. Okay, it's made of ammonium ion and ammonia. So we've got the ph Ph is 10. Therefore the hydrogen ion concentration must be one times 10: -10 More. Okay. And we have the K. For the acid part of this which is the ammonium ion Which is 5.6 Time some -10. So we'll use our equation for buffers. H plus there's a K. A. Times the concentration of the acid which is ammonium ion divided by the concentration of our base which is ammonia. Since we have H plus in K. We'll go ahead and plug those in one Time. To turn the negative 10 Equals 5.6. I'm saying the negative 10 times are ratio. But so we're gonna go ahead and solve for that ratio now the ratio of our acid compared to our base and that's going to equal 0.18 21 mm. So we're gonna use that ratio now to answer some other questions. So in the second one we've given some, we've been given some information about ammonia. I know of 1.24 moller And the volume was .465 L. So I have 5, 77 moles Of the NH three. But we can use that ratio. We just found to answer some questions since concentration is moles per liter. And so we're doing moles per liter over moles per liter. Well, the leaders have to be the same. So we can also write this as the malls of ammonium ion divided by the moles of pneumonia, Equal that same ratio of .18. So we'll go ahead and plug in what we know here. Okay, we're looking for the moles of ammonia mayan and we'll divide that by the malls of ammonia. Set that equal to our ratio to our moles of ammonium ion. This problem Is going to be zero .10 moles. Okay, in our next example, Were given 2.08 g of ammonium ion. Well ammonium chloride. Right? So we'll go ahead and change grams to moles. This is ammonium chloride. So we use the molar mass, which is 53.5. So that will give us .0390 moles of ammonium chloride, which is the same as the ammonium. Bye on. And then we'll go ahead and use our ratio again. We've got moles of our acid divided by Moles of our base. So moles of NH three ph hasn't changed. So we need to keep this the same ratio So we can solve for moles of NH three here we'll get .216 Malls. But we've been asked for what volume of that. But they gave us the polarity. So if we divide by arm polarity here, .236 moles per liter. We divide that out. We'll get our leaders which is .915 L. Okay. Or 9.2 Times 10 of the Second Mill leaders about 920. So on our final example, we've got the polarity and the volume of our ammonium ion solution. So we'll go ahead and multiply more clarity times leaders to get moles. We've got .043 moles of ammonium ion NH four plus. So we'll go ahead. Use our same ratio moles of our acid Divided by moles of our base, which is NH three Most equal .18. So if we multiply that out, we'll get malls of ammonia that's going to come out to be .239 moles mm. And again they are asking us for the volume that we need. So we're gonna go ahead and divide it by the polarity Just .499 Moller, which will give us points 479 L, four, Times 10 of the Second Mill leaders if we want to use proper sig figs.

Okay. So we've got a buffer solution here. Okay, it's made of ammonium ion and ammonia. So we've got the ph Ph is 10. Therefore the hydrogen ion concentration must be one times 10: -10 More. Okay. And we have the K. For the acid part of this which is the ammonium ion Which is 5.6 Time some -10. So we'll use our equation for buffers. H plus there's a K. A. Times the concentration of the acid which is ammonium ion divided by the concentration of our base which is ammonia. Since we have H plus in K. We'll go ahead and plug those in one Time. To turn the negative 10 Equals 5.6. I'm saying the negative 10 times are ratio. But so we're gonna go ahead and solve for that ratio now the ratio of our acid compared to our base and that's going to equal 0.18 21 mm. So we're gonna use that ratio now to answer some other questions. So in the second one we've given some, we've been given some information about ammonia. I know of 1.24 moller And the volume was .465 L. So I have 5, 77 moles Of the NH three. But we can use that ratio. We just found to answer some questions since concentration is moles per liter. And so we're doing moles per liter over moles per liter. Well, the leaders have to be the same. So we can also write this as the malls of ammonium ion divided by the moles of pneumonia, Equal that same ratio of .18. So we'll go ahead and plug in what we know here. Okay, we're looking for the moles of ammonia mayan and we'll divide that by the malls of ammonia. Set that equal to our ratio to our moles of ammonium ion. This problem Is going to be zero .10 moles. Okay, in our next example, Were given 2.08 g of ammonium ion. Well ammonium chloride. Right? So we'll go ahead and change grams to moles. This is ammonium chloride. So we use the molar mass, which is 53.5. So that will give us .0390 moles of ammonium chloride, which is the same as the ammonium. Bye on. And then we'll go ahead and use our ratio again. We've got moles of our acid divided by Moles of our base. So moles of NH three ph hasn't changed. So we need to keep this the same ratio So we can solve for moles of NH three here we'll get .216 Malls. But we've been asked for what volume of that. But they gave us the polarity. So if we divide by arm polarity here, .236 moles per liter. We divide that out. We'll get our leaders which is .915 L. Okay. Or 9.2 Times 10 of the Second Mill leaders about 920. So on our final example, we've got the polarity and the volume of our ammonium ion solution. So we'll go ahead and multiply more clarity times leaders to get moles. We've got .043 moles of ammonium ion NH four plus. So we'll go ahead. Use our same ratio moles of our acid Divided by moles of our base, which is NH three Most equal .18. So if we multiply that out, we'll get malls of ammonia that's going to come out to be .239 moles mm. And again they are asking us for the volume that we need. So we're gonna go ahead and divide it by the polarity Just .499 Moller, which will give us points 479 L four, Times 10 of the Second Mill leaders if we want to use proper sig figs.

So for a ever going to use a Henderson house a black equation. Um, the volume totaled 100 milliliters of K is 14 miles PKB, which is equal the 9.26 So, using that information, NH three here is going equal 36 times. Their point to over 100 is equal to zero plane, 07 to 0 rolled color and then in age for plus legal. The 64 times zero went to over 100 which is equal Teoh One hopes 0.1 to a fuller. So then p age is equal to 9.26 Plus the log of those two do that to their and this is all equal to nine points here said It's for B, um so H minus is equal to 10 to the negative, More so for diluting it to one leader. The, um in ages three hopes been able to 7.2 10 cents and I get third above the K A value and saying for NH four plus equals one point 8/10. And then I get a second much bigger than K. Um, so Henderson Hospital question is valid if it's diluted. Teoh this is, Ah, one leader, but it's a letter to 1000 leaders. There's no distinguishable fact of impure water. So now we want to see adding this h it won't affect my change in volume is because Delta age 30 plus equals 0.2 times Warren, which is equal to 0.2 normals of HBO plus. So we set up an equation. Nothing there. This is 7.2. My syrup went to zero. This is 12.8 plus point to so pH would equal 9.26 plus log of 7/13 which is equal to 8.99 And lastly, we have d so have an age three plus h 30 plus equal to many four plus eight. Sure, nothing There no 0.8 points x 7.2 minus eggs. PH equals 8.9, which would then equal 9.26 a minus log 12.8 plus x over 1.2 minus bags, so X equals 1.1 a. Miller Moles of age 30 plus


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