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The half life period of a subtance in a certain enzyme-catalysed reaction is 138s. The time required for the concentration of the substance to fall from 1.28 mg`L^(-1)` is:A. 414sB. 552 sC. 690 sD. 276 s |
Answer» Correct Answer - C c) Amount left after n half life peridos `[A]_(0)` `=[A]_(0)/(2^(n))` `2^(n) = [A]_(0)/[A]= (1.28 mg L^(-1))/(0.04 mg L^(-1))` `[2]^(n) = 32=[2]^(5)` No. of `t_(1//2)` periods =5 Time required = `138 xx 5=690s` |
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