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$75 \%$ of a zero order reaction complete in $4 \mathrm{~h}$ $87.5 \%$ of the same reaction completes in
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Verified Answer
The correct answer is:
$6 \mathrm{~h}$
Using the relation,
$t_{87.5 \%}=\frac{3}{2} t_{75 \%}$
$t_{87.5 \%}=\frac{3}{2} \times \frac{4}{1}=6 \mathrm{~h}$
$t_{87.5 \%}=\frac{3}{2} t_{75 \%}$
$t_{87.5 \%}=\frac{3}{2} \times \frac{4}{1}=6 \mathrm{~h}$
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