MCQ
$pH$ of $0.1\,\, M$ $N{H_3}$ aqueous solution is $({K_b} = 1.8 \times {10^{ - 5}})$
- ✓$11.13$
- B$12.5$
- C$13.42$
- D$11.55$
${K_b} = C{\alpha ^2}$ ; $\frac{{1.8 \times {{10}^{ - 5}}}}{{.1}} = {\alpha ^2}$;$\alpha = 1.34 \times {10^{ - 3}}$
$[O{H^ - }] = \alpha \;.\;C$$ = 1.34 \times {10^{ - 3}} \times .1$
$pOH = \log 10\frac{1}{{1.34 \times {{10}^{ - 4}}}}$; $pOH = 2.87$
$pH + pOH = 14$; $pH + 2.87 = 14$
$pH = 14 - 2.87$; $pH = 11.13$
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[Molar mass of glucose in $\mathrm{g} \mathrm{mol}^{-1}=180$ ]
$\mathrm{A}(l) \rightarrow 2 \mathrm{B}(\mathrm{g})$
$\Delta \mathrm{U}=2.1\; \mathrm{kcal}, \Delta \mathrm{S}=20\; \mathrm{cal} \mathrm{K}^{-1}$ at $300\; \mathrm{K}$
Hence $\Delta \mathrm{G}$ in $\mathrm{kcal}$ is
$H _{2( g )}+ Br _{2( g )} \rightarrow 2 HBr _{( g )}$
Given that bond energy of $H _{2}$ and $Br _{2}$ is $435 \;kJ mol ^{-1}$ and $192 \;kJ mol^{-1}$, respectively, what is the bond energy (in $kJ mol ^{-1}$ ) of $HBr?$