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5. The following results are obtain for the absorbance values. Cuvet 1,2, 3, 4 and 5 (0.50 M) are 0.422, 0.792, 1.133, 1.395 and 1.571 respectively: The unknown abs...

Question

5. The following results are obtain for the absorbance values. Cuvet 1,2, 3, 4 and 5 (0.50 M) are 0.422, 0.792, 1.133, 1.395 and 1.571 respectively: The unknown absorbance is 1.268. Use the results to calculate the unknown concertation Use beer's law relationship but a more manageable equation. Show all your work!Discuss beer' s law using the above data as an example?

5. The following results are obtain for the absorbance values. Cuvet 1,2, 3, 4 and 5 (0.50 M) are 0.422, 0.792, 1.133, 1.395 and 1.571 respectively: The unknown absorbance is 1.268. Use the results to calculate the unknown concertation Use beer's law relationship but a more manageable equation. Show all your work! Discuss beer' s law using the above data as an example?



Answers

Spectrometry. The absorbance $A$ of a solution is defined as $$A=\log _{10}\left(I_{0} / I\right)$$ in which $I_{0}$ is the incident-light intensity and $I$ is the transmitted-light intensity. The absorbance is related to the molar absorption coefficient (extinction coefficient) $\varepsilon$ (in $\mathrm{M}^{-1}$ $\mathrm{cm}^{-1}$ ), concentration $c(\text { in } \mathrm{M}),$ and path length $l(\text { in } \mathrm{cm}$ ) by $$A=\varepsilon l c$$ The absorption coefficient of myoglobin at $580 \mathrm{nm}$ is $15,000 \mathrm{M}^{-1} \mathrm{cm}^{-1} .$ What is the absorbance of a $1 \mathrm{mg} \mathrm{ml}^{-1}$ solution across a $1-\mathrm{cm}$ path? What percentage of the incident light is transmitted by this solution?

Alarm but low. We have locked. I not over. I equals E C l for the absorption coefficient, the concentration on the path length so e and L a constant for specific Anna light and specific measurements because the Bear Lambert law will be invalidated when a linear plot is obtained After plotting concentration verses absorbent where we have the following data, we have transmission percent concentration micrograms per mil. Anita Trans and a U on absorb INTs in a you. So we have 66.8, 44.7 29.2, 19.9 and 13.3. The concentrations are 1234 and five. The Trans Mittens is not 0.668 No point for for seven not 0.292 no 0.19 nine on not 90.133 So our absorb Ince's are not 0.17522 no 0.34969 Next, we have no 0.53462 seconds. Last does not 0.70115 And lastly, we have no 0.87615 So we plot that on our graph of course, and so the linearity off the plot indicates that the bear lamp but law is valid for the following data. So the solution that has 35% transmitting to address that the absorb its negative log multiplied by 9.35 which is not 0.46 So the above linear plot suggests another 0.46 absorbing corresponds to the concentration of 2.6 micro grounds for milita, So the concentration of vitamin A is 2.6 micrograms per mil Elita.

So we have law and I over I zero equals who are fussy I So we'll have I hear eco's thio zero times e to the power off See, I do You have the i o our the X just equals Teoh a zero e r for C I times are fussy equals thio I times R c so just equal to minus k I so okay, here just eek also my minus r C which is

Okay. Were given the intensity of light, uh, labeled as I satisfies a bier Lambert law and were given that Ellen of Eye over. Why not? You go to our first. He acts where I know it is in the show. Intensity okay. And exit the distance Traveled by light travel light. Okay. And I were asked to show that I satisfies the eye over t X recruiting native Che I for some k. Okay, let's start. So we're giving our equation. Oh, and I over I not Is he good to Alfa? See axe every take he of both sides get I over. I not they could to e offa see axe. Okay, now moving our not so multiplying by eye not on both sides You get eye is equal to I know e l f a c x I wanted to isolate I because if you see it back here, we're taking to the derivative of eye with with with respect to X Well, that So now that we have this, let's take degenerative with expect X What do we get? We get the eye overthe e x is equal to I not Alfa. See each hour of Alpha C X. Okay, if you recall. He said I Is he going to I not e Alfa. See, x So I go to I not I not e repair off a c x sort of shooting here indigo to I Okay. And we consider that K is equal to negative Alfa because the neg Okay, Is it good to out for C? So if you plug those in, we get Do you? I over t X is equal to neighbor. Okay, I and looking back at what we wanted, we're asked to show that I satisfies the eye over the exit code in there. Okay. And this is what we showed over here.

Mhm. Okay so this problem actually has our understanding on the bier lambert store. So that's first right down the beer lambert store. The beer lambert law states at the adoption right equals to the extinction coefficient times the concentration times the past lands of the solution sell. And if you look at the question the question gives us true as orphans and gives us a one solution corresponding one opens and we need to calculate the constitution of another adornments. So based on that we can write down some equivalence right? So As I said it gives us one Roberts and it also gives us the concentration right corresponding to the endorphins which is given and another one is it gives another door opens to right? But this time the concentration is unknown which is what we want to calculate, right? So we can actually combined this true clearance. We just used one divided by two. And you were fined right? The romans, one over the romans, two were equal to the concentration one over concentration to and the concentration to is what we want to calculate and now what we need to do is just plug in the numbers. So for the robins one Actually is .72 right? And the concentration correspond to the .72 is also given which is 2.5 times 10- -3. Mhm. And for the adornments to it's also given one And the say two is what we want calculate. So now what we need to do is do the calculation on this Right we can do arrangement, you will find the say two equals 2 right 2.5 times 10 to minus three over 30.72 And the final answer will be 3.472 times 10 to -3 more per liter. And this is the answer.


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