MCQ
Molarity of $0.2\,N\,\,{H_2}S{O_4}$ is
- A$0.2$
- B$0.4$
- C$0.6$
- ✓$0.1$
i.e., $0.2=$ molarity $\times $ $ 2$
Molarity $ = 0.2/2 = 0.1$
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$K = \left( {4.5 \times {{10}^{11}}{S^{ - 1}}} \right){e^{ - \left( {28000\frac{{{K^2}mol}}{J}} \right)\frac{R}{T}}}$
The activation energy of reaction (in $J/mol$) is
