Question

Question #c81c4

Answer 1

`t=248791.2 sec = 69.1 t`

Explanation:

`rate=-(dA)/(dt)=k[A]^2`

Integrated rate Law:
`1/([A]_t)=k*t+1/([A]_0)`

0,249 M of A is used = `95.77%`
0.011 M of A is not used = `4.23%`

fraction of A not used: `4.23/100=1/23.64`

`[A]_t=([A]_0)/23.64`

and `t=t_(1/23.64)`

`1/(([A]_0)/23.64)=k*t_(1/23.64)+1/[A]_0`

`t_(1/23.64)=1/k(23.64/[A]_0-1/[A]_0)=1/(3.5*10^(-4))(22.64/0.260)`

`t_(1/23.64)=248791.2 sec = 69.1 t`