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Calculate Extinction coefficient of substance A that has a concentration of 12 micrograms/ml, A= 0.057 and path length is 1cm
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- 6. Blue Blue dye stock solution 0.293 M Absorbance at 630 nm 0.00265 Calibration curve y = 0.0833x A solution is prepared by diluting 2.79 mL of the blue dye stock solution to 25.00 mL. The measured absorbance for the prepared solution is listed in the data table. (a) What is the theoretical molar concentration? [Blue]theoretical x 10 |M (b) What is the experimental molar concentration? [Blue]experimental x 10 M (c) What is the percent error? Percent error (blue) = %Substance P has an extinction coefficient of 72.22 mM-1cm-1 at 420nm. 2 μl of Substance P was added to 998 μl water and the Absorbance at 420nm was 0.820. Calculate the mM concentration of the original Substance P solution3. The chemical sunset yellow, also known as yellow #6, looks yellow. What color(s) must be absorbed by this chemical? RED YELLOW GREEN WHITE CYAN MAGENTA RGB System (Transmitted light) Additive color syslem colored light Rendered with Corel Dro RGB pollete BLUE-VIOLET 4. Sunset yellow has an extinction coefficient of 20,000 M-1 cm-l at a lambda max of 470 nm. What is the concentration of a solution that has an absorbance of 0.893?
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- 45.0 mL of an unknown FeCl3 solution is diluted to a total volume of 230.0 mL. The diluted FeCl3 solution measured an absorbance of 0.223. Calculate the molarity of the undiluted unknown FeCl3 solution if the molar extinction coefficient is 230.0 M-1⋅cm-1 and a path length of 1.00 cm.The Ksp of M2CrO4 is 7.11 x 10-15. How many moles of Mt will dissolve in 744 mL of solution in which [CrO,2] = 0.056 M? 2.4 x 10-5 mol M* 3.6 x 107 mol M* 3.6 x 10-7 mol M+ 2.7 x 10-7 mol M* None of these 6.2 x 10-6 mol M*7. If the equilibrium constant of binding of NADH to an enzyme [E] is 2.7=2,7 v Ko = [NADH][E]/[E-NADH] = 1.5 x 10-6 M, and solution concentrations of [E] and [NADH] are 10 nM and 1.0 µM, respectively, what is the concentration of the [E-NADH] complex? %3D 51210 (a) 16 µM0.:0 (b) 0.62 x 109 M 1.5 E-NADH (c) 62 nM .0000000 (d) None of the above. 5 -2.51x
- (Q53) A solution is prepared this is 1.38 x 104 M in magnesium nitrate and 1.76 x 104 M in sodium fluoride at the moment of mixing. If, by chance, a precipitate were to form, it would have a Ksp value of 5.16 x 1011. Given this information, will a precipitate form? Yes, because Q = 6.65 x 10-8 O Yes, because Q = 4.27 x 10-12 Both sodium nitrate and magnesium fluoride will precipitate out of solution O No, because Q = 4.27 x 10-12 O No, because Q = 6.65 x 10-8An antibody (Ab) can bind to its antigen binding site (B) with a high affinity (large negative AGO). The antibody and the antigen are mixed at low concentrations and allowed to reach equilibrium. The concentrations measured are [Ab] = 1 nM, [B] = 40 %3D nM, and [Ab-B] = 30 nM, for the equilibrium Ab + B= Ab-B. A. What is the equilibrium constant (association constant), Ką, in M1? B. What is Kd = 1/K, in nM? C. What is AG° in kgT/molecule? Use the equation AG° = -kgTln(Ka) = kgTln(Ka) D. Imagine that this binding were due entirely to hydrogen bonds. Referring to question 3C above, how many hydrogen bonds would form between Ab and B upon binding? E. If you were going to manufacture a coronavirus test kit with an antibody that tightly binds to an antigen found in the virus, would you want the antibody/antigen binding to have a Ka in the nM, µM, or mM range? Why?Equilibrium Solutions Volume (ml) Trial # 2.00 x 10-3 M Fe(NO3)3 (ml) in 5.00 M HNO3 2.00 x 10-3 M HSCN(ml) in 5.00 M HNO3 0.5000M HNO3(ml) Absorbance [H+] 1 5.00 ml 1.00 ml 4.00 ml 0.085 0.500 2 5.00 ml 2.00 ml 3.00 ml 0.195 0.500 3 5.00 ml 3.00 ml 2.00 ml 0.345 0.500 4 5.00 ml 4.00 ml 1.00 ml 0.432 0.500 5 5.00 ml 5.00 ml 0.00 ml 0.506 0.500 What is the [Fe3+]initial (moles /L x 10 -4) for trial 1? What is the [HSCN]initial (moles /L x 10 -4) for trial 1?