Please help. I\'m having trouble with this pre-lab. Define A, A_0, and A_t omega
ID: 894864 • Letter: P
Question
Please help. I'm having trouble with this pre-lab.
Define A, A_0, and A_t omega as they apply to this experiment. Two drops of a 0.20 M imidazole solution were added to a solution in a cuvette containing ferrihemoglobin. The mass of the ferrihemoglobin solution in the cuvette was measured before the addition of the imidazole, and then the mass of the total final solution in the cuvette was measured after adding the imidazole. The results are recorded in the following table Assuming that the densities of all solutions are identical, calculate the concentration (in molarity) of imidazole in the final solution after the 2 drops of imidazole were added. Provide answer with correct units and significant figures. (Since the experiment has been designed to ensure that imidazole is the excess reagent, the concentration calculated can be assumed to be the equilibrium concentrations as well.) The K_eq for imidazole binding to hemoglobin can be determined from the following linear equation: 1/Delta A_100 K_eq Times 1/[Im] + 1/Delta A_100 = (1/Delta A) For the plot of the reciprocal of the change in absorbance versus the reciprocal of the imidazole concentration, identify the equivalent of the slope and the y-intercept in the above equation. Copy the equation in your notebook, and clearly mark a circle around the term equivalent to slope and a square around the y-intercept. If the best-fit line determined for a set of data gives a slope of 0.057 M and an intercept of 21, what is the value of Delta A_100 and K_eq? For the reaction, what does the calculated value of K_eq from question 3b predict for the favorability of reactants or products?Explanation / Answer
Given: The densities of two solutions are same
let it be 1g / mL
imidazole added = 0.086g
0.086g imidazole / 1g/ml = 0.086ml imidazole added
0.086ml = 8.6x10^-2ml = 8.6x10^-5L
So moles of imidazole added
8.6x10^-5L x 0.2M = 1.72x10^-5moles imidazole added
Amount of ferrihaemogolobin = 2.81 g
2.81g / 1g/ml = 2.81ml = 0.00281L = volume of ferrihemoglobin
so
total volume = 2.81ml + 0.086ml = 2.896ml = 0.002896L
1.72x10^-5moles imidazole / 0.002896L = 0.0059M
3) absorbance not given