On heating copper nitrate strongly,$.......$ is finally obtained.

The presence of $Si$ in steel gives it

$SiCl_4$ on hydrolysis forms '$X$' and $HCl$. Compound '$X$' loses water at $1000^{\circ} C$ and gives '$Y$'. Compounds '$X$' and '$Y$' respectively are:

Consider the following route of reactions $R_2SiCl_2 + H_2O \to (A) \xrightarrow{\text{Polymerisation}} (B)$. Compound $(B)$ in the above reaction is:

How is dry ice produced? Mention its uses.

Assertion : $PbCl_2$ is more stable than $PbCl_4$.
Reason : $PbCl_4$ is a powerful oxidising agent.