1.

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 "0.04" mg L"^(-1), is

Answer»

414 s
552 s
690 s
276 s

Solution :AMOUNT LEFT after n HALF lives `=([A]_(0))/(2^(n))`
`0.04 = (1.28)/(2^(n)) or 2^(n) = 32 = 2^(5) therefore n = 5`
Time taken ` = 5 ` half lives`= 5 xx 138 s = 690 s.`


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