MCQ
$pH$ of $10^{-8}\, M$ $Ba(OH)_2$ solution will be
- A$7.4$
- B$6.92$
- ✓$7.08$
- D$7.7$
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$\begin{array}{*{20}{c}}
{C{H_3} - C{H_2} - C - COOH} \\
{\,\,\,\,\,\,\,\,||} \\
{\,\,\,\,\,\,\,\,\,\,\,\,C{H_2}}
\end{array}$
is :
| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{Bi}^{3+} \longrightarrow(\mathrm{BiO})^{+}$ | $(P)$ Heat |
| $(B)$ $\left[\mathrm{AlO}_2\right]^{-} \longrightarrow \mathrm{Al}(\mathrm{OH})_3$ | $(Q)$ Hydrolysis |
| $(C)$ $\mathrm{SiO}_4^{4-} \longrightarrow \mathrm{Si}_2 \mathrm{O}_7^{6-}$ | $(R)$ Acidification |
| $(D)$ $\left(\mathrm{B}_4 \mathrm{O}_7^{2-}\right) \longrightarrow\left[\mathrm{B}(\mathrm{OH})_3\right]$ | $(S)$ Dilution by water |
$2 NO _{( g )}+ O _{2}( g ) \rightleftarrows 2 NO _{2}( g )$
The reaction occurring as above comes to equilibrium under a total pressure of 1 atom. Analysis of the system shows that $0.6 mol$ of oxygen are present at equilibrium. The equilibrium constant for the reaction is $.........$(Nearest integer).
(given $R = \frac{{8.3\,J}}\,{{mol - K}}$, $1$ $lit$ -$atm$ $=$ $100$ $J$ $ln$ $2$ $=$ $0.693$)