The half-life of a substance in certain enzyme catalyzed reaction is 138 s. The time(in sec.) required for the concentration of the substance to fall from 1.28 mg L-to 0.04mg L-is:

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    Solution:

    n = no. of half - life periods is given as:

        0.04mgL1.28mgL=132;12n=125  or  n=5

      Time required 

    =n  x t1/2 =5 x 138 s=690 s.

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