To $50 \ mL$ of $0.1 \ N \ Na_2CO_3$ solution,$150 \ mL$ of water is added. What is the molarity of the resultant solution?

$CaCO_3$ reacts with $HCl$ to produce $CaCl_2, CO_2$ and $H_2O$. The approximate mass (in $g$) of $CaCO_3$ required to react completely with $25 \ mL$ of $0.75 \ M \ HCl$ is (atomic mass of $Ca=40, C=12, O=16, Cl=35.5$ and $H=1$).

At $400 \ K$,$1 \ mol$ of a hydrocarbon is completely burned. It produces $132 \ g$ of a gas along with $72 \ g$ of water vapour. The hydrocarbon is:

The number of chlorine atoms in $20 \, mL$ of chlorine gas at $STP$ is ..... $\times 10^{21}$. (Round off to the Nearest Integer).
[Assume chlorine is an ideal gas at $STP$,$R = 0.083 \, L \, bar \, mol^{-1} \, K^{-1}$,$N_{A} = 6.023 \times 10^{23}$]

$1.44 \ g$ of titanium (At. wt. $= 48$) reacted with excess of $O_2$ to produce $x \ g$ of a non-stoichiometric compound $Ti_{1.44}O$. The value of $x$ is:

$12 \ g$ of $Mg$ (atomic mass $24$) will react completely with acid to give: