$K = \frac{{2.303}}{t}\log \frac{a}{{a - x}}$
Given:
$a = \frac{1}{{10}} = .1\,m$;
$a - x = \frac{1}{{100}} = .01\,m$;
$ t = 500 $ $\sec$
$\therefore \;\;K = \frac{{2.303}}{{500}}\log \frac{{.10}}{{.01}} = \frac{{2.303}}{{500}}\log \,10$
$ = \frac{{2.303}}{{500}} = 0.004606 = 4.6 \times {10^{ - 3}}\,{\sec ^{ - 1}}$.
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Given :
Molar mass $N =14\,g\,mol ^{-1} ; O =16\,g\,mol ^{-1} ; C =12\,g\,mol ^{-1} ; H =1\,g\,mol ^{-1}$;
$2 Fe _{( s )}+ O _{2( g )}+4 H _{( aq )}^{+} \rightarrow 2 Fe _{( aq )}^{2^{+}}+2 H _2 O ( l ) \quad E ^{\circ}=1.67 V$, At $\left[ Fe ^{2+}\right]=10^{-3} M , P \left( O _2\right)=0.1$ atm and $pH =3$, the cell potential at $25^{\circ} C$ is
