1.

What fraction of a reactant showing first order remains after 40 miute if t_(1//2) is 20 minute ?

Answer»

`1//4`
`1//2`
`1//8`
`1//6`

Solution :`[A]_t/[A]_0= 1/2^N " and Half lives "n = 1/t_(1//2)= 40/ 20 =2 `
`:. [A]_t= [A]_0/2=[A]_0/4= 1/4`


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