1.

The half-life period of a order reaction is 15 minutes. The amount of substance left after one hour will be:

Answer»

`(1)/(4)` of the ORIGINAL amount
`(1)/(8)` of the original amount
`(1)/(16)` of the original amount
`(1)/(32)` of the original amount

Solution :Given `t_((1)/(2)=15`minutes
Total time (T)=1hr=60min
From `T=ntimest_((1)/(2))`
`n=(60)/(15)=4`
Now from the formula `(N)/(N_(0))=((1)/(2))^(n)=((1)/(2))^(4)=(1)/(16)`
Where `N_(0)=` INITIAL amount
N=amount LEFT after times t
hence the amount of substance left after 1 hour will be `(1)/(16)`


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