For dissociation of a complex insolving a simple equilibrium: ab a+ b, the equilibrium constant, k- tab is equal to the ratio of forward and reverse rate constants: kp kotl/ kon for avidin-biotin, kp- 10-15. suppose kon were as fast as the diffusion limit, -109 ms1. (a) what is the value of ko? (b) what would be the half-life of the avidin-biotin complex? (c) suppose kon for avidin-biotin were 10-m s. what would be the half-life of the complex?

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o

hypostyle hall;