MCQ
$S{O_2} + {H_2}S \to $ product. the final product is
- ✓${H_2}O + S$
- B${H_2}S{O_4}$
- C${H_2}S{O_3}$
- D${H_2}{S_2}{O_3}$
$S{O_2}$ oxidises ${H_2}S$ into $S$.
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Product $(A)$ in above reaction is
$0.48 + \log \left\{ { - \frac{{d[A]}}{{dt}}} \right\} = \log \left\{ { + \frac{{d[B]}}{{dt}}} \right\} + 0.7$
Then, $x : y$ is