The amount of $50\% \ (w/w)$ solution of hydrochloric acid required to react with $200 \ g$ of $CaCO_3$ would be (in $g$)

$KMnO_4$ reacts with oxalic acid according to the equation,$2MnO_4^- + 5C_2O_4^{2-} + 16H^{+} \to 2Mn^{2+} + 10CO_2 + 8H_2O$. Here,$20 \ mL$ of $0.1 \ M$ $KMnO_4$ is equivalent to:

The ratio of the molar amounts of $H_2S$ needed to precipitate the metal ions from $20 \ mL$ each of $1 \ M \ Cd(NO_3)_2$ and $0.5 \ M \ CuSO_4$ is

At sea level, the dry air mass percentage composition is given as nitrogen gas: $70.0$, oxygen gas: $27.0$, and argon gas: $3.0$. If the total pressure is $1.15 \ atm$, calculate the ratio of the following respectively:
$(i)$ partial pressure of nitrogen gas to partial pressure of oxygen gas
$(ii)$ partial pressure of oxygen gas to partial pressure of argon gas
(Given: Molar mass of $N_2$, $O_2$, and $Ar$ are $28$, $32$, and $40 \ g \ mol^{-1}$ respectively)

The decomposition of $CaCO_3$ produces $11.2 \, dm^3$ of $CO_2$ gas at $STP$. What mass of $KOH$ in grams is required to completely neutralize this gas?

In the following reaction sequence,the $\%$ yield for reaction $I$ is $60 \%$ and that of reaction $II$ is $50 \%$. The overall yield of the complete reaction is $..........\%$ [nearest integer].