- A$3300 $
- B$2200$
- ✓$1100$
- D$4400 $
${10^{ - 3}} = \frac{{2.303}}{t}\log \frac{a}{{(a - \frac{{2a}}{3})}}$,
${10^{ - 3}} = \frac{{2.303}}{t}\log \,3$
${10^{ - 3}} = \frac{{2.303}}{t} \times 0.4771,$
$t = \frac{{2.303 \times 0.4771}}{{{{10}^{ - 3}}}}$
$ =1100\,\sec $
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
${H_2}O\left( g \right) + C\left( s \right) \to CO\left( g \right) + {H_2}\left( g \right);\Delta H = 131\,kJ$
$CO\left( g \right) + \frac{1}{2}{O_2}\left( s \right) \to C{O_2}\left( g \right);\Delta H = - 282\,kJ$
${H_2}\left( g \right) + \frac{1}{2}{O_2}\left( s \right) \to {H_2}O\left( g \right);\Delta H = - 242\,kJ$
$C(s) + O_2(g) \to CO_2(g); \Delta H = X\,kJ$
The value of $X$ will be......$kJ$
Reason : On increasing dilution, degree of ionisation of weak electrolyte increases and molality of ions also increases.
$A$. Compound ' $B$ ' is aromatic
$B$. The completion of above reaction is very slow
$C$. '$A$' shows tautomerism
$D$. The bond lengths $C - C$ in compound $B$ are found to be same
Choose the correct answer from the options given below :
