MCQ
Difference between two types of sulphur atoms in $Na_2S_4O_6$ will be
- ✓$5$
- B$3$
- C$2.5$
- D$6$
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$2N_2O_5 \rightarrow 4NO_2 + O_2$ can be written in three ways.
$\frac{-d[N_2O_5 ]}{dt} = k[N_2O_5]$
$\frac{d[NO_2 ]}{dt} = k'[N_2O_5]\,;$ $\frac{d[O_2 ]}{dt} = k"[N_2O_5]$
The relationship between $k$ and $k'$ and betweenk and $k''$ are
$Xe{F_6}\,\xrightarrow{{excess\,\,{H_2}O}}\,'X'\, + \,HF$
$Xe{F_6}\,\xrightarrow{{2\,moles\,{H_2}O}}\,'Y'\, + \,HF$