\[N{{H}_{4}}N{{O}_{2\left( aq \right)}}\xrightarrow{\text{heat}}{{N}_{2\left( g \right)}}+2{{H}_{2}}{{O}_{\left( l \right)}}\]

No explanation is given for this question. Let's discuss the answer together.

Dinitrogen pentaoxide $$\left( {{N_2}{O_5}} \right)$$  is prepared by dehydrating the concentrated nitric acid with phosphorus pentoxide.
$$4HN{O_3} + {P_4}{O_{10}} \to 2{N_2}{O_5} + HP{O_3}$$
So, $${{N_2}{O_5}}$$  is regarded as anhydride of $$HN{O_3}.$$

Helium is obtained during radioactive decay.

Water gas is $$CO + {H_2}$$

Carbon is a non-metal.

\[\underset{\begin{smallmatrix} \text{Thiosulphate} \\ \text{ion} \end{smallmatrix}}{\mathop{2{{S}_{2}}O_{3}^{2-}}}\,+{{I}_{2}}\to \underset{\begin{smallmatrix} \text{Tetrathionate} \\ \text{ion} \end{smallmatrix}}{\mathop{{{S}_{4}}O_{6}^{2-}}}\,+2{{I}^{-}}\]

In the formation of $$Xe{F_2},s{p^3}d$$   hybridisation occurs which gives the molecule a trigonal bipyramidal structure.

In the formation of $$Xe{F_4},s{p^3}{d^2}$$   hybridization occurs which gives the molecule an octahedral structure.

In the formation of $$Xe{F_6},s{p^3}{d^3}$$   hybridization oocurs which gives the molecule a pentagonal bipyramidal structure.

$$HF + Xe{F_6} \to XeF_5^ + + HF_2^ - $$