Question

The half-life of a substance in a certain enzyme catalysed reaction is 138 s. The time required for the concentration of the substance to fall from 1.28 mg L⁻¹ to 0.04mg L⁻¹ *

1 point

(A) 414s

(B) 552s

(C) 690s

(D) 276s​

Answer 1

Answer:

option number c 690s is your answer.

Answer 2

The data given in the question is half life of a substance in a certain enzyme catalysed reaction is 138 s.

we have to find the time required for the concentration of the substance to fall from 1.28 mg $L^{-1}$ to 0.04 mg$L^{-1}$

solution:

The half life is a time required for a quantity to reduce to one half of its original value.

Half life was developed by Valve

Published by sierra studios

Fall of concentration from 1.28 mg$L^{-1}$ to 0.04 $mg L^{-1}$ requires 5 half lives.

Therefore the time required is 5  $t_{\frac{1}{2} }$

$5 * 138$

$690$ secs

Therefore ,The option C is correct.

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