In Dumas method for estimation of nitrogen,$0.3 \ g$ of an organic compound gave $41.9 \ mL$ of nitrogen gas at $\text{STP}$. What is the percentage of nitrogen (in $.5$)?

In phosphorus estimation,$0.31 \ g$ of an organic compound gave $0.444 \ g$ of magnesium pyrophosphate (molar mass $= 222 \ g \ mol^{-1}$). The percentage of phosphorus in the compound is

$0.53 \ g$ of an organic compound $(x)$ when heated with excess of nitric acid (concentrated) and then with silver nitrate gave $0.75 \ g$ of silver bromide precipitate. $1.0 \ g$ of $(x)$ gave $1.32 \ g$ of $CO_2$ gas on combustion. The percentage of hydrogen in the compound $(x)$ is . . . . . . $\%$. [Nearest Integer]
[Given : Molar mass in $g \ mol^{-1} H : 1, C : 12, Br : 80, Ag : 108, O : 16$; Compound $(x) : C_xH_yBr_z$] (in $\%$)

What is the reason for passing dry air through the combustion tube during the estimation of $C$ and $H$?

In Kjeldahl's method for estimation of nitrogen,$CuSO_4$ acts as :

In the Kjeldahl method,$0.35 \ g$ of an organic compound produces ammonia,which is absorbed in $100 \ mL$ of $0.1 \ M \ H_2SO_4$. The excess $H_2SO_4$ requires $154 \ mL$ of $0.1 \ M \ NaOH$ for titration. Calculate the percentage of nitrogen in the compound. (Given atomic masses: $C=12, H=1, O=16, N=14, S=32$)