$$2Al + 3{H_2}S{O_4} \to $$    $$A{l_2}{\left( {S{O_4}} \right)_3} + 3{H_2}$$
$$2Al + 2NaOH + 2{H_2}O \to $$      $$2NaAl{O_2} + 3{H_2}$$
Thus, ratio of volumes of hydrogen evolved is 1 : 1.

Non-metal hydrides are covalent hydrides.

Heavy water freezes at $$3.8{\,^ \circ }C.$$

$$Ca{\left( {HC{O_3}} \right)_2} + Ca{\left( {OH} \right)_2} \to $$       $$\mathop {2CaC{O_3}}\limits_{{\text{(Insoluble)}}} + 2{H_2}O$$

The water has maximum density at $${4^ \circ }C$$  when it is converted to ice, an open cage three dimensional structure is formed therefore the density decreases.

\[\underset{\begin{smallmatrix} \text{Sodium} \\ \text{peroxide} \end{smallmatrix}}{\mathop{N{{a}_{2}}{{O}_{2}}}}\,+{{H}_{2}}S{{O}_{4\left( \text{dil}\text{.} \right)}}\to \]     \[\underset{\begin{smallmatrix} \text{Sodium} \\ \text{sulphate} \end{smallmatrix}}{\mathop{N{{a}_{2}}S{{O}_{4}}}}\,+\underset{\begin{smallmatrix} \text{Hydrogen} \\ \text{peroxide} \end{smallmatrix}}{\mathop{{{H}_{2}}{{O}_{2}}}}\,\]

Two ice cubes stick to each other due to $$H$$ - bonding

$${H^ - }\left( {aq} \right) + {H_2}O\left( l \right) \to O{H^ - } + {H_2}.$$        Since $${H^ - }$$  is a strong base it will abstract $${H^ + }$$  to form $${H_2}O.$$

$$N{a_2}{O_2} + {H_2}S{O_4}\left( {20\% \, \,{\text{ice}}\,{\text{cold}}} \right) \to N{a_2}S{O_4} + {H_2}{O_2}$$

$$Mn{O_2},Pb{O_2}$$   and $$BaO$$  will not give $${H_2}{O_2}$$  with $$HCl.Mn{O_2}$$   and $$Pb{O_2}$$  will give $$C{l_2}$$  and $$BaO$$  will react with $$HCl$$  to give $$BaC{l_2}$$  and water.