The formula of a metallic oxide is $MO$. What will be the formula of its phosphate?

$300 \ mL$ of a gaseous hydrocarbon when burnt in excess of $O_2$ gave $2.4 \ L$ of $CO_2$ and $2.7 \ L$ of water vapour under same conditions. The molecular formula of the hydrocarbon is:

$18 \ L$ mixture of $N_2$ and $H_2$ gives maximum $6 \ L$ of $NH_3$ at same temperature and pressure. What will be the ratio of $N_2$ and $H_2$ initially taken?

What volume in liters is required to prepare a $0.1 \ N$ $HNO_3$ solution from $6.3 \ g$ of $HNO_3$?

In the balanced combustion reaction of $1 \, mol$ of $C_xH_{2n-2}$ hydrocarbon,the coefficients of $O_2$,$CO_2$,and $H_2O$ will be respectively ...

The volume of air needed for complete conversion of $100 \ cm^3$ of $NO$ to $NO_2$ will be (assuming $20 \%$ $O_2$ by volume in air).
$2 NO_{(g)} + O_{2(g)} \rightarrow 2 NO_{2(g)}$ (in $cm^3$)