1.

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 mgL^(-1) is:

Answer»

414s
552 s
690 s
276 s

Solution :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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