The unbalanced chemical reactions given in List I show missing reagent or condition (?) which are provided in List II. Match List I with List II and select the correct answer using the code given below the lists :
List I | List II | ||
---|---|---|---|
P. | $$Pb{O_2} + {H_2}S{O_4}\buildrel ? \over \longrightarrow PbS{O_4} + {O_2} + Other\,products$$ |
1. | NO |
Q. | $$N{a_2}{S_2}{O_3} + {H_2}O\buildrel ? \over \longrightarrow NaHS{O_4} + Other\,products$$ |
2. | $${I_2}$$ |
R. | $${N_2}{H_4}\buildrel ? \over \longrightarrow {N_2} + Other\,products$$ |
3. | Warm |
S. | $$Xe{F_2}\buildrel ? \over \longrightarrow Xe + Other\,products$$ |
4. | $$C{l_2}$$ |
Match the chemical conversions in List I with the appropriate reagents in List II and select the correct answer using the code given below the lists :
Let $\omega=\frac{\sqrt{3}+i}{2}$ and $P=\left\{\omega^n: n=1,2,3, \ldots\right\}$. Further
$\mathrm{H}_1=\left\{z \in \mathrm{C}: \operatorname{Re} z<\frac{1}{2}\right\}$ and
$\mathrm{H}_2=\left\{z \in \mathrm{C}: \operatorname{Re} z<\frac{-1}{2}\right\}$, where C is the
set of all complex numbers. If $z_1 \in \mathrm{P} \cap \mathrm{H}_1, z_2 \in$ $\mathrm{P} \cap \mathrm{H}_2$ and O
represents the origin, then $\angle z_1 \mathrm{O} z_2=$
Area of S =