MCQ
The hydrogen ion concentration in a given solution is $6 \times {10^{ - 4}}.$ Its $pH$ will be
- A$6$
- B$4$
- ✓$3.22$
- D$2$
$pH = - \log \,\,[{H^ + }]$ $ = - \log \,\,[6 \times {10^{ - 4}}] = 3.22$.
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$\mathrm{SO}_3(\mathrm{~g}) \rightleftharpoons \mathrm{SO}_2(\mathrm{~g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})$
is $\mathrm{K}_{\mathrm{C}}=4.9 \times 10^{-2}$. The value of $\mathrm{K}_{\mathrm{C}}$ for the reaction given below is
$2 \mathrm{SO}_2(\mathrm{~g})+\mathrm{O}_2(\mathrm{~g}) \rightleftharpoons 2 \mathrm{SO}_3(\mathrm{~g})$ is
$C_{(graphite)} +O_2(g) \rightarrow CO_2(g)\,;$
$\Delta _rH^o=-395.5 \, kJ\,mol^{-1}$
$H_2 (g) + \frac{1}{2} O_2 (g) \rightarrow H_2O(l)\,;$ $\Delta _rH^o =-285.8\, kJ\, mol^{-1}$
$CO_2(g) + 2H_2O(l) \rightarrow CH_4(g) + 2O_2(g)\,;$
$\Delta _rH^o = + 890.3\, kJ\, mol^{-1}$
Based on the above thermochemical equations, the value of, $\Delta _rH^o $ at $298\, K$ for the reaction
$C_{(graphite)} + 2H_2(g) \rightarrow CH_4(g) $ will be for $\Delta_{r} H^{\circ}$ ........... $kJ \,mol^{-1}$
Assertion $(A):$ The ionic radii of $O ^{2-}$ and $Mg ^{2+}$ are same.
Reason $( R )$ : Both $O ^{2-}$ and $Mg ^{2+}$ are isoelectronic species.
In the light of the above statements, choose the correct answer from the options given below