MCQ
Conjugate base of $HPO_4^{2 - }$ is
- ✓$PO_4^{3 - }$
- B${H_2}PO_4^ - $
- C${H_3}P{O_4}$
- D${H_4}P{O_3}$
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$2NO + Br \to 2NOBr$ is
$NO + Br_2 \rightleftharpoons NOBr_2$ (Fast)
$NOBr_2 + NO \to 2NOBr$ (Slow)
The rate law expression is
$\mathop {Ca{C_2}}\limits_{\left( A \right)} \xrightarrow{{{H_2}O}}B\xrightarrow[{HgS{O_4}}]{{{H_2}O,{H_2}S{O_4}}}C$
Reason : In presence of $H_2SO_4$, $HNO_3$ acts as a base and produces $NO_2^+$ ions.