MCQ
When $KBr$ is treated with concentrated ${H_2}S{O_4}$ redish brown gas evolved, gas is
- AMixture of bromine and $HBr$
- B$HBr$
- ✓Bromine
- DNone of these
$2KHS{{O}_{4}}+MnS{{O}_{4}}+2{{H}_{2}}O+B{{r}_{2}}$
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Given that $\left[ {\frac{{{{\left[ B \right]}_t}}}{{{{[C]}_t}}} = \frac{{16}}{9}} \right]$
$A\,\xrightarrow{{{K_1}\, = \,2\, \times \,{{10}^{^{ - 3}\,}}{S^{ - 1}}}}4B$
$A\to C$
Reason : Hydrogen is present above $Cu$ in the reactivity series.
