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DATA TABLE Part I: Experimental Data CollectionRWunCrystal viclet concentration M Vorume used (tnl) NaOH concentralion M Volume used (tnl) TOIAL Volumne (m)) Slope ...

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DATA TABLE Part I: Experimental Data CollectionRWunCrystal viclet concentration M Vorume used (tnl) NaOH concentralion M Volume used (tnl) TOIAL Volumne (m)) Slope frorn DUSI Iit straight line1 *ioLsxlo $65400""[0 M0.0100+020 K2 . Heov I0 YatI0 ml20 IAta ({nCALCULATIONSWhat is the order with respect to CV ? Eib MLCalculate the initial concentration of CV- tcactiun murlum and place in lhe Klve Table below. Tu-] Wl [r" Dr k [cv * J-Vi el ble Kiionl 5# [ou _ r/o LevlnI 0Sk %*44*20 [0

DATA TABLE Part I: Experimental Data Collection RWun Crystal viclet concentration M Vorume used (tnl) NaOH concentralion M Volume used (tnl) TOIAL Volumne (m)) Slope frorn DUSI Iit straight line 1 *io Lsxlo $ 65400"" [0 M 0.010 0+020 K 2 . Heov I0 Yat I0 ml 20 I Ata ( {n CALCULATIONS What is the order with respect to CV ? Eib ML Calculate the initial concentration of CV- tcactiun murlum and place in lhe Klve Table below. Tu-] Wl [r" Dr k [cv * J-Vi el ble Kiionl 5# [ou _ r/o LevlnI 0Sk %*44*20 [01, = [cv*J, Jowa | [cv J [04" ] 2*/" [ev* Jo 3-10 of OH The reichon MLIQMIC and place Calculate the initial concentration Table below. (OHJa (CV Jo 3710 Vilo Irial Trial Trial the order with respect to OH . Delenine @u]- rale law. [w" Write the overall



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Two concentration cells are prepared, both with 90.0 $\mathrm{mL}$ of 0.0100 $\mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}$ and a Cu bar in each half-cell.
(a) In the first concentration cell, 10.0 $\mathrm{mL}$ of 0.500 $\mathrm{M} \mathrm{NH}_{3}$ is added to one half-cell; the complex ion $\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}$ forms, and $E_{\mathrm{cell}}$ is 0.129 $\mathrm{V}$ . Calculate $K_{\mathrm{f}}$ for the formation of the complex ion.
(b) Calculate $E_{\text_{cell}$ when an additional $10.0 \mathrm{mL}}$ of 0.500 $\mathrm{M} \mathrm{NH}_{3}$ is added.
(c) In the second concentration cell, 10.0 $\mathrm{mL}$ of 0.500 $\mathrm{M} \mathrm{NaOH}$ is added to one half-cell; the precipitate $\mathrm{Cu}(\mathrm{OH})_{2}$ forms $\left(K_{\mathrm{sp}}=\right.$ $2.2 \times 10^{-20}$ ). Calculate $E_{\mathrm{cell}}^{\circ}$
(d) What would the molarity of NaOH have to be for the addition of 10.0 $\mathrm{mL}$ to result in an $E_{\mathrm{cell}}^{\circ}$ of 0.340 $\mathrm{V}$ ?

The solution of the answer is there'd the dissociation off cso. Fourth is as that CSO for irreversible CIA to positiveness s afford to narrative the sole ability Expression for the reaction can be written as follows that KSB is equal talks here too positive as opposed to negative and we consider it as a question too. So in a saturated solution, cierto positive will be equal to a supporter. Negative now from the daughter the sample off as to positive is added to the base and the ohh informed sodium iron and water molecules has the reaction for that that is too positive plus any wedge which gives twist to hopeless and a positive first, calculate the number off the malls off the esto as follows here that 100 ml sample multiplied by 8.25 mln neo activated by 10.0 amel sample multiplied by 0.105 mall and well divided by one later multiplied by 100,000 ml by one leader multiplied by one Emmel and Austria positive divided by one more off and there was which will be equal to by calculating this 8.66 multiply by 10 to the power minus four more now using the moles off s three a positive and value off The simple critical clear the concentration off CIA to positive as follows by putting the values in the equation and by simplifying the heavy gear the value is 0.173 moles per liter. These are the This is the concentration off See a too positive European 0173 moles per liter. Therefore the concentration of CIA operative in 0.173 AM But the But we know that in a saturated solution see it to positive physical to a support to negate him. That's the concentration off before too negative. There's a a call to 0.173 a. M. Now substitute the concentration off Cito positive and as off to negative in the equation toe as follows that we know the question towards does okay equal to see two positive and as a Fortunati here by putting the values and we get final value is 3.0, multiplied by 10 to the power minus four. Therefore, the KSB off the kills himself here that this year support is 3.4. Multiply white into par, minus port. So this is a complete solution of the answer. Please go through this.

To calculate the concentration of acetic acid. We need to know the moles of acetic acid and we need to know the volume of the acetic acid solution. The volume of the acetic acid solution is provided as 25 mil leaders but we need to know it in units of leaders. So we'll divide by 1000 and we get .050 leaders. The moles of acetic acid can be calculated by using the average volume of sodium hydroxide, Which is 1983 ml. This volume of sodium hydroxide required to neutralize the acidic acid can be converted into Leaders by dividing by 1000 and then converted from leaders into moles by multiplying by the concentration of sodium hydroxide. This will give us units of moles sodium hydroxide, one mole sodium hydroxide having one hydroxide will react with one mole acetic acid, having one hydrogen ion to donate. This then gives us molds acetic acid, which is 240 times 10 to the negative three. So we then calculate the concentration acetic acid by taking the moles acetic acid Divided by the volume of citric acid, and leaders, and we get .20960 molar acetic acid.

In this first example we're combining two solutions with the same compound. So the resulting polarity can be found by taking the total moles and dividing it by the total volume. So let's start by finding the total the moles in each one. Some moles. Osmolarity times leaders 2.17 Moller Times .4-0 leaders will give us 0.714 moles of nuh. From that first solution 0.4 moller .0376 L. Give us With 015 moles. Okay so the polarity is going to be our total moles which is .00714 Plus .015. And now that total volume as we add those two leaders together and we end up with .0796 L. So malaria malaria he comes out to be .278 moller in a way. Which but we want the concentration of the ions. Okay so the concentration of the N. A plus will be the same as a concentration of the O. H minus because it's all 1 to 1 there and those will equal .278. More. Okay so our second example we're going to take a solution and we're gonna dilute them both. So two solutions are gonna get diluted. So we're gonna want to use our dilution formula and one V. One equals M two V. Two. So our new polarity equals the original more clarity times its volume over the over the new volume which is the total volume. Okay so we used the total volume there. So for our first solution and two is .1 moller Times The 44 ml Over a total of 69 million liters. So that will give us .0638 moller where are sodium sulfate? So the concentration of our sodium ion Since there's two of them in there will double this with 0638. So that's gonna be .128 moller sodium ion. And the concentration of our sulfate ion Is simply going to be the .0638 Mueller. And there's your sulfate. Okay, but we have another solution now So we'll do the same thing. So IM2 is its polarity .150 moller Times 25 million liters Over the total of 69. So that's going to give us a more clarity of .05 43 molar K. C. L. So the concentration of the potassium ion, it will be the same as the chloride ion because there's just one of each And that will be equal to 05 43 more. And this third example we're gonna add 3.6 g of some K. C. L. Let's go ahead and find the moles of that. So one over 74.55. So that's gonna give us .048 three malls of the K. C. L. So if we divide that by the volume With 75 l We'll get the polarity is .644 Mueller K. C. L. So the concentration of the potassium ion Is going to be .644 moller. We can't do the chloride yet because we're getting chloride from another source as well. So let's find the moles of the calcium chloride. We'll take the polarity .25 Moller Times .075 l And that will give us .0188 moles of calcium chloride. Which means we've got .0188 moles of our calcium ion and we have two times that of our chloride ion. Okay so to start with we can pretty easily do the concentration of the calcium ion because that was given to us. That's just gonna be the .25 more. That was the mole aren t of the calcium chloride therefore it's the polarity of the calcium ion. And then the concentration of your chloride ion is going to be the moles from the first part from the first compound, plus the moles from the second compound. And then we'll divide it by the volume of solution With 0750 L. So that will give us 1.14 more C. L minus. So we've got all of our concentrations now

To answer this question and determine the approximate concentration of a after 110 seconds for the zero first and second order reactions, we need to know what the rate constant is. So if you've done problem 27 you've plotted everything, you will get the rate constant for each of them from the slope. So Experiment one is first order its rate constants point to one experiment to zero order. Its rate constant is 00.1 an experiment three of second order and its rate constant is essentially 30.1 also so to determine the concentration at 110 seconds for the zero order reaction. Well, it goes to zero at 100 seconds. So if we're at 110 it's still nothing for B consulates. For first order, we use the first order integrated rate law, natural log of concentration at 110 seconds over, natural over a natural log of concentration at 110 seconds, divided by the concentration at time zero will be equal to negative K multiplied by t 110 seconds. Doing a little bit of algebra will find out that the natural log of concentration is negative 1.1 or 0.33 Moller. And that should make sense because here we see that at 100 seconds it's 1000.37 So it should be just a little bit less than that. At 100. And 10 0.33 is reasonable for the second order. Reaction will use the second order Integrated rate law one over. Concentration at time, zero minus one over concentration at time T equals positive. I'm sorry. Negative, Katie. I'm doing a little bit of algebra. Will see that one over. Concentration 110 seconds equals 2.9. Take the inverse of 2.9 and we get 0.48 Moller. And at 100 seconds for the second order we see it's 0.5. So it should be just a little bit less than that. 2.48 makes sense at 110 seconds.


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