- A$1 \times {10^{ - 20}}$
- ✓$1 \times {10^{ - 12}}$
- C$1 \times {10^{ - 14}}$
- D$1 \times {10^{ - 16}}$
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${N_2}\left( g \right) + {O_2}\left( g \right)\underset{{{k_2}}}{\overset{{{k_1}}}{\longleftrightarrow}}2NO\left( g \right)$
$C_0 = Ce^{-2.1×10^{-3}\ t}$ for the forward reaction and
$C_0'= C'e^{-4.2×10^{-4}\ t}$ for the backward reaction, hence $K_c$ for the above equilibrium is
| List-$I$ (Molecule) | List-$II$(Shape) |
| $A$ $\mathrm{NH}_3$ | $I$ Square pyramid |
| $B$ $\mathrm{BrF}_5$ | $II$ Tetrahedral |
| $C$ $\mathrm{PCl}_5$ | $III$ Trigonal pyramidal |
| $D$ $\mathrm{CH}_4$ | $IV$ Trigonal bipyramidal |
Choose the correct answer from the option below :
$C{{H}_{3}}-\underset{C{{H}_{3}}}{\mathop{\underset{|\,\,\,\,\,}{\mathop{\overset{OH}{\mathop{\overset{|\,\,\,\,}{\mathop{C-}}\,}}\,}}\,}}\,C{{H}_{2}}-\underset{C{{H}_{3}}\,}{\mathop{\underset{|\,\,\,\,\,\,\,}{\mathop{CH-}}\,}}\,C{{H}_{3}}$
$(A)$ the reaction of $Al _2 O _3$ with coke ($C$) at a temperature $>2500^{\circ} C$.
$(B)$ the neutralization of aluminate solution by passing $CO _2$ gas to precipitate hydrated alumina $\left( Al _2 O _3 .3 H _2 O \right)$
$(C)$ the dissolution of $Al _2 O _3$ in hot aqueous $NaOH$.
$(D)$ the electrolysis of $Al _2 O _3$ mixed with $Na _3 AlF _6$ to give $Al$ and $CO _2$.