MCQ
Silver nitrate produces a black stain on skin due to
- ABeing a strong reducing agent
- BIts corrosive action
- CFormation of complex compound
- ✓Its reduction to metallic silver
$2AgN{O_3}\, \to \,\mathop {2Ag}\limits_{{\rm{black stain}}} + {N_2} + 3{O_2}$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| List-$I$ | List-$II$ |
| $\mathrm{P}_2 \mathrm{O}_3+3 \mathrm{H}_2 \mathrm{O} \rightarrow$ | $\mathrm{P}(\mathrm{O})\left(\mathrm{OCH}_3\right) \mathrm{Cl}_2$ |
| $\mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \rightarrow$ | $\mathrm{H}_3 \mathrm{PO}_3$ |
| $\mathrm{PCl}_5+\mathrm{CH}_3 \mathrm{COOH} \rightarrow$ | $\mathrm{PH}_3$ |
| $\mathrm{H}_3 \mathrm{PO}_2+2 \mathrm{H}_2 \mathrm{O}+4 \mathrm{AgNO}_3 \rightarrow$ | $\mathrm{H}_3 \mathrm{PO}_4$ |
$\Delta_f G^0[\mathrm{C}(\text { graphite })]=0 \mathrm{kJmol}^{-1}$
$\Delta_f G^0[\mathrm{C}(\text { diamond })]=2.9 \mathrm{kJmol}^{-1}$
The standard state means that the pressure should be $1$ bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2 \times 10^{-6} \mathrm{~m}^3 \mathrm{~mol}^{-1}$. If $\mathrm{C}$ (graphite) is converted to $\mathrm{C}$ (diamond) isothermally at $\mathrm{T}=298 \mathrm{~K}$, the pressure at which $C$ (graphite) is in equilibrium with C(diamond), is
[Useful information: $1 \mathrm{~J}=1 \mathrm{~kg} \mathrm{~m}^2 \mathrm{~s}^{-2} ; 1 \mathrm{~Pa}=1 \mathrm{~kg} \mathrm{~m}^{-1} \mathrm{~s}^{-2} ; 1 \mathrm{bar}=10^5 \mathrm{~Pa}$ ]
$CN ^{-}, NO ^{+}, O _{2}, O _{2}^{+}, O _{2}^{2+}$