MCQ
Concentrated ${H_2}S{O_4}$ cannot be used to prepare $HBr$ from $NaBr$, because it
- AReduces $HBr$
- ✓Oxidises $HBr$
- CDisproportionates $HBr$
- DReacts slowly with $NaBr$
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Product in above reaction is
$C{O_{(g)}} + \frac{1}{2}{O_{2(g)}} \to C{O_{2(g)}}$
How are $\Delta E$ and $\Delta H$ releated for the reaction?
$C{H_3} - \mathop {\mathop {C = }\limits_{|\,\,\,\,} }\limits_{C{l_{}}\,} \mathop {\mathop {C\, - }\limits_{|\,\,\,\,\,\,} }\limits_{C{H_3}\,} \mathop {\mathop {CH - \,}\limits_{|\,\,\,\,\,\,\,\,\,} }\limits_{{C_2}{H_5}\,} C{H_2} - C \equiv CH$