MCQ
Correct order of acidic strength
- A$HF > HCl > HBr > HI$
- ✓$H_3PO_2 > H_3PO_3 > H_3PO_4$
- C$B(OH)_3 > H_2CO_3 > HNO_3$
- D$H_2O > H_2S > H_2Se > H_2Te$
$O _3$ can acts as only oxidizing agent due to its unstable nature and decomposes to give nascent oxygen.
$HNO _3$ : Nitrogen is present in its highest oxidation state i.e., $+5$ so it can act as only oxidizing agent.
$SO _2$ :Sulphur is present in $+4$ oxidation state so it can act as both oxidizing as well as reducing agent.
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Given : $K _{ a }\left( CH _{3} CH _{2} COOH \right)=1.3 \times 10^{-5}$
The above reaction is carried out in a vessel starting with partial pressure $\mathrm{P}_{\mathrm{SO}_{2}}=250\, \mathrm{~m}$ $bar,$ $\mathrm{P}_{0_{2}}=750 \,\mathrm{~m}$ $bar$ and $\mathrm{P}_{\mathrm{SO}_{3}}=0 \,\mathrm{bar}$. When the reaction is complete, the total pressure in the reaction vessel is $.....\mathrm{m}$ $bar.$ (Round off to the Nearest Integer).