MCQ
$0.01\,\,M\,HA\,(aq.)$ is $2\%$ ionized $; [OH^-]$ of solution is
- A$2 \times 10^{-4}$
- B$ 10^{-8}$
- ✓$5 \times 10^{-11}$
- D$5 \times 10^{-12}$
$\Rightarrow \frac{\left[\mathrm{H}^{+}\right]}{[\mathrm{HA}]} \times 100=2$
$\Rightarrow \frac{\left[\mathrm{H}^{+}\right]}{0.01} \times 100=2$
$\Rightarrow\left[\mathrm{H}^{+}\right]=\frac{0.02}{100}=2 \times 10^{-4} \mathrm{M}$
$\therefore\left[\mathrm{OH}^{-}\right]=\frac{10^{-14}}{2 \times 10^{-4}}=0.5 \times 10^{-10} \mathrm{m}$
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$CaCO _3( s ) \rightleftharpoons CaO ( s )+ CO _2( g )$
For this equilibrium, the correct statement(s) is (are)
$(A)$ $\Delta H$ is dependent on $T$
$(B)$ $K$ is independent of the initial amount of $CaCO _3$
$(C)$ $K$ is dependent on the pressure of $CO _2$ at a given $T$
$(D)$ $\Delta H$ is independent of the catalyst, if any
$1.\,\,(CH_3)_2 - \mathop C\limits^ + - CH_2 - CH_3$
$2.\,\,(CH_3)_3 - \mathop C\limits^ + $
$3.\,\,(CH_3)_2 - |\mathop C\limits^ + H|$
$4.\,\,CH_3 - \mathop C\limits^ + H_2$
$5.\,\,\mathop C\limits^ + H_3$