$t_{1 / 2}=\frac{12}{5} min =2.4$
$\approx 2$
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$\mathrm{A}+\mathrm{B} \rightarrow \mathrm{C}$ (Reaction 1$)$
$\mathrm{P} \rightarrow \mathrm{Q}$ (Reaction $2$)
The ratio of the half life of Reaction $1$ : Reaction $2$ is $5: 2$. If $t_1$ and $t_2$ represent the time taken to complete $2 / 3^{\text {dd }}$ and $4 / 5^{\text {dd }}$ of Reaction $1$ and
Reaction $2$, respectively, then the value of the ratio $\mathrm{t}_1: \mathrm{t}_2$ is . . . .$\times 10^{-1}$ (nearest integer).
[Given: $\log _{10}(3)=0.477$ and $\log _{10}(5)=0.699$ ]
The $E _{\text {ceil }}$ for the given cell is $0.1115\,V$ at $298\,K$ when $\frac{\left[ M ^{+}( aq )\right]}{\left[ M ^{3+}( aq )\right]}=10^{ a }$
The value of a is
Given : $E _{ M ^{3+} / M ^{+}}=0.2\,V$
$\frac{2.303 RT }{ F }=0.059\,V$
$\left[ K _{ f } \text { of water }=1.86 \ K \ kg \ mol ^{-1}\right]$
