The half life period of a first order reaction is 60 min. What percentage will be left after 240 min.

Explanation

`k=(0.693)/(t_(1//2)) = (0.693)/60`
`t=(2.303)/klog.([A_0])/([A_t])`
`240=2.303/(((0.693)/60))log.([A_0])/([A_t])`
`(4xx0.693)/2.303=log.([A_0])/([A_t])`
`A_t=A_0/16=0.0625[A_0]`
`[A_t]=6.25% "of " [A_0]`