MCQ
When lead nitrate is heated, it gives
- ✓$N{O_2}$
- B$NO$
- C${N_2}{O_5}$
- D${N_2}O$
$2Pb{(N{O_3})_2} \to 2PbO + 4N{O_2} + {O_2}$
So nitric oxide ${N_2}O$ is produced.
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|
List$-I$ (Parameter) |
List$-II$ (Unit) |
| $(a)$ Cell constant | $(i)$ $\mathrm{S}\, \mathrm{cm}^{2} \,\mathrm{~mol}^{-1}$ |
| $(b)$ Molar conductivity | $(ii)$ Dimensionless |
| $(c)$ Conductivity | $(iii)$ $\mathrm{m}^{-1}$ |
| $(d)$ Degree of dissociation of electrolyte | $(iv)$ $\Omega^{-1} \,\mathrm{~m}^{-1}$ |
Choose the most appropriate answer from the options given below :
$Mn_{(S)} | Mn^{+2}_{(aq)} (0.4\,M) | | Sn^{+2}_{(aq)} (0.04\,M)| Sn_{(S)}$,
Calculate free energy change $\left( {\Delta G} \right)$ at $298\, K.$ .......... $\mathrm{kJ}$
Given : $E_{M{n^{ + 2}}/Mn}^o = - 1.18\,V ; \,E_{S{n^{ + 2}}/Sn}^o = - 0.14\,{\rm{volt}}\,\frac{{2.303RT}}{F}=0.06$

$[Y]$ is