MCQ
Alkyl halides can be obtained by all methods except
- A$CH_3 CH_2OH + HCl/ZnCl_2 \to$
- B$C{{H}_{3}}-C{{H}_{2}}-C{{H}_{3}}-C{{H}_{2}}\xrightarrow{C{{l}_{2}}/UV\,light}$
- ✓$C_2H_5OH +NaCl \to$
- D$CH_3COOAg + Br_2/CCl_4 \to$

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$\mathrm{N}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{g})$
the correct option is
(given, $\frac{ F }{ R }=11500 K V ^{-1}$, where $F$ is the Faraday constant and $R$ is the gas constant, The value of $\ln (10)=2.30$ )