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Aooo T-Mobile LTE 5.58 PM 28% organic cc stonybrookedu The data for the absorbance measurements the reaction mixture at 498 nm (the analutica wavelength of the dye)...

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Aooo T-Mobile LTE 5.58 PM 28% organic cc stonybrookedu The data for the absorbance measurements the reaction mixture at 498 nm (the analutica wavelength of the dye) is given below Initial readings for both runs were taken at approximately after mixing Additional readings were then measured approximately every 120 (tor the lover concentra Ition of OH-) and approximately every 60 (for the higher concentration OH-)Data Run Haroroinen0lsorWeh0.414 072iWeua0 176Data for Run AbrorbancoMeue0.938 0,175

Aooo T-Mobile LTE 5.58 PM 28% organic cc stonybrookedu The data for the absorbance measurements the reaction mixture at 498 nm (the analutica wavelength of the dye) is given below Initial readings for both runs were taken at approximately after mixing Additional readings were then measured approximately every 120 (tor the lover concentra Ition of OH-) and approximately every 60 (for the higher concentration OH-) Data Run Haroroinen 0lsor Weh 0.414 072i Weua 0 176 Data for Run Abrorbanco Meue 0.938 0,175 Uehre 0 574 DeY This data produced the plots below.



Answers

A colored dye compound decomposes to give a colorless product. The original dye absorbs at $608 \mathrm{nm}$ and has an extinction coefficient of $4.7 \times 10^{4} \mathrm{M}^{-1} \mathrm{~cm}^{-1}$ at that wavelength. You perform the decomposition reaction in a $1-\mathrm{cm}$ cuvette in a spectrometer and obtain the following data: $$ \begin{array}{rl} \hline \text { Time (min) } & \text { Absorbance at } 608 \mathrm{nm} \\ \hline 0 & 1.254 \\ 30 & 0.941 \\ 60 & 0.752 \\ 90 & 0.672 \\ 120 & 0.545 \end{array} $$ From these data, determine the rate law for the reaction "dye $\longrightarrow$ product" and determine the rate constant.

For this problem we're going to be plotting absorbent versus concentration were given some data. Let's get caught up here. We're going to. But I did my the slope in the, um, graphing calculator. I don't know. I think I tried to do a plot, but it was pretty poorly done. There it is. So I just took pictures of my calculator and the data are listed right there where you see me writing Absorb INTs. That's the absorb INTs at 4 75 nanometers, and the dye concentration is listed as L two. There's the dime Alaric E. I plug those numbers into the calculator, and there you can see that Y equals X plus B. There's the slope and there's the Y intercept. So we were asked to conduct construct a calibration plot, which I put the points on there for you. Excuse that ugly, ugly graph. I planted a zor observance against concentration, and there's like a line pretty straight and then were asked, What is the dye concentration in a solution? What is the dye concentration in a solution with absorb INTs? A 0.52 Okay, so the slope equals 11.15 1.2 times 10 to the fifth. The Y intercept is 0.1785 0.18 So we plotted. We've got the slope. You've got the intercept. I forgot the tien intercept. They're getting the units on things. Here's the last part. If the absorb INTs, the why is 0.52 I think you can see where this is going to go. Find X. So we're simply going to put that into our equation. 1.15 times 10 to the minus fifth. Or that should be positive. Five x I feel like that should be positive. Five. My bad. But it's 1.15 I'm just looking at things here. Let's wait and get this done. 0.15 x. I keep doing minus fifth there. I should be doing 50. So change that to five. Sometimes I copied is down wrong. I'm doing a voice over on this one there. I've got fifth. Finally there, fixed it and we get 3.0 times. 10 to the minus six, the polarity. And that's the answer

Problem. party we know that equals to extinction copies and E B C. Okay. Sorry, extinction coefficient. Asylum not being to C. C. Is the concentration from equation 14.5. No value of a Given the question that is zero point 605 Value of extinction coefficient. E is given a question that is 5.60 into 10 to the power three centimetre per centimeter. Mm. Okay. And value of B is also given question that is one centimetre. Now use this equation. We can see that concentration equals two A divided by extinction coefficient divided by B. Now simply put the values That is value of 8 0.605 value of extinction coefficient. That is 5.60 into 10 to the power three per centimeter. But boy multiply the value of the dirties. One centimetres. No After calculating we get one Now we can get rid of forgetting the concentration is one 080 into 10 to the power minus poor. And this is the answer for party. So this is the initial concentration of coloured reaction. If the observance is G 0.605 at beginning of reaction. Now in part B we need to calculate concentration at time t using beers, love. Okay so we calculated see not in part A. Okay we use. Okay now we can see that A At time 30 seconds it calls to extinction copies in and to be the concentration that her this again, We can say that concentration at 32nd her teeth second. It goes to Value up a noticed participant that is zero 250. Given questions they've had it where Value of extinction copy isn't also given a question that is 5.60 into 10 to the power -3 part centimeter per and multiply by Multiplied by the value of b. That is also given question one centimetre. After calculating we get Concentration at 32nd equals two. Poor point port 64 into 10. To the power minus five. M. No from a cushion for pinpoint 1 3. We can say that Ellen CT equals 2- Katie plus LNC. Not Okay. No We can see that key equals two rearranging this equation. LNC not minus. Alan L N C D divided by T. So we can say that Kay calls to LNC. Not so he Well you upsy not already calculated, calculated that is one point already calculated part first initial concentration. So Ellen 1.080 into 10 to the power minus food minus concentration at time. T. We already calculated here that is alan poor point pull six, pull into 10 to the power minus pipe and time Time given a 30 minute. So it's converting a second after multiplying 60 seconds, so one 1800 seconds were divided by 1800 seconds. After calculating we get K equals two 4.9, 1, 0 into 10 to the Power -4. That is approximately equals two. 4.9, 1 into 10 to the power minus pool per second. This is a rate constant for this reaction. So this is final answer. Hard part B. No for part C. Our first order reaction, the hub lie because to 0.69 three divided by key, now simply put the value 0.69 three helicopter already calculated here, That is poor .91 into 10 to the power minus four per second. After calculating we get one point poured 11 into 10 to the power three second. That is supposed to 23 0.5. I mean that this is and Singapore part C. So this is half Leiper. This reaction now in part D. In part day concentration of a time. T Calls to 0.100. So now we need to calculate city using beers, love then time from first order integrated rate equation. So first we need to calculate concentration at time. T it calls to a divided by extinct extinct X. Things and coefficient. And to be now simply put the value value up. Given the question that is 0.100, divided by the value of extinction coefficient is 5.60 into 10 to the power three centimetre universe. I am universe into one centimeter. Okay, After calculating we get 1.786 and two 10 to the power minus pipe, that is approximately equals two. 1.79 Into 10 to the Power -5 in. Okay. No okay we use this formula so we can see that time equals two. Ellen concentration of C note minus. Ellen concentration of city divided by key. Now simply put the values so Alan 1.080 into 10 to the power minus poor minus. Alan consider attempting already calculated that is 1.786 into 10 to the power minus pipe divided where time given question put Time given question support .9 10 into 10 to the power minus poor. Sorry sorry sorry this is Hello kitty. Well you already calculated here. Okay That is 4.19, to the power -4 2nd universe calculating we get Time equals to 3.66 six into 10 to the power three. That is equal to 3.67 Into 10 to the power three seconds. That is equal to 61 one. You mean that? Mhm. So this is an support are. Today, we can see that 61.1 minutes Long does it take for observance to fall to zero 100 So this answered final answer for party.

Problem. 23.56 Question is asking. What color would you expect for the complex nickel by period ing three two plus All right, the question tells you that the absorption maximum occurs for the complex at about 520 nanometers. So we know that if a substance absorb Allied's two wore a region 520 is about in this region right here. Green collar. So the color observed is toward the complimentary color and the complimentary color of this green. It would be red. So the answer to this question is the color we observe. ISS Red Party. Based on this data, where would you put bye purity in spectral pickled? Serious? All right, So the data on the table have Nicole Water six two plus. The absorption occurs at 729 meters. Nickel and age 36 two plus occurs at about 570 nana meters, and they call and 32 plus a curse on 545 nano meters, and the last one, Nicole Bye purity 32 plus the maximum absorption occurs at 520 nada meters, so the shorter the wavelength the light absorb, the greater the value ofthe a and the stronger the league infield. So the order for increasing llegan fill strings is the order of decreasing way Flynn strips. So we will write the order here. The water will be smaller than the image three smaller than a theme dia mean group and smaller than the by period in group.


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