MCQ
$HNO_3 + P_4O_{10} \to 4HPO_3 + X$ . $X$ is
- ✓$N_2O_5$
- B$NO_2$
- C$NO$
- D$N_2O_3$
$4 HNO _3+ P _4 O _{10} \rightarrow 4 HPO _3+2 N _2 O _5$
The product, $x$ is dinitrogen pentaoxide $\left( N _2 O _5\right)$.
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$A \to B$ ${K_1} = {10^{15}}{e^{ - 25000/8.314\,T}}$
$C \to D$ ${K_2} = {10^{14}}{e^{ - 15000/8.314\,T}}$

$Al_{(s)}\, | Al^{+3}\, (0.1\,M)||Fe^{+2}\, (0.001\,M)|Fe_{(s)}$
If $E_{A{l^{ + 3}}/Al}^o = - 1.66\,V$ and $E_{Fe/F{e^{ + 2}}}^o = + 0.44\,V$
