If 1 , L of N_2 and 7/8 , L of O_2 gases are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) For the same temperature $T$ and pressure $P$,according to Avogadro's law,$n \propto V$. $	herefore \frac{n_{N_2}}{n_{O_2}} = \frac{V_{N_2}}{V_{O

Vedclass · Accepted Answer

$m(N_2) = m(O_2)$

Vedclass · Answer

(C) For the same temperature $T$ and pressure $P$,according to Avogadro's law,$n \propto V$. $	herefore \frac{n_{N_2}}{n_{O_2}} = \frac{V_{N_2}}{V_{O_2}} = \frac{1}{7/8} = \frac{8}{7}$. Since $n = \frac{m}{M}$,where $m$ is mass and $M$ is molar mass: $\frac{m(N_2) / M(N_2)}{m(O_2) / M(O_2)} = \frac{8}{7}$. Given $M(N_2) = 28 \, g/mol$ and $M(O_2) = 32 \, g/mol$: $\frac{m(N_2) / 28}{m(O_2) / 32} = \frac{8}{7}$. $\frac{m(N_2)}{m(O_2)} = \frac{8}{7} 	imes \frac{28}{32} = \frac{8}{7} 	imes \frac{7}{8} = 1$. Therefore,$m(N_2) = m(O_2)$.

If $1 \, L$ of $N_2$ and $7/8 \, L$ of $O_2$ gases are mixed at the same temperature and pressure,what is the relationship between their masses in the mixture?

Similar Questions

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?

How many gram-equivalents of $NaOH$ are required to neutralize $25 \ cm^{3}$ of a decinormal $HCl$ solution?

If $0.45 \ g$ of a gas with molecular weight $60$ and $0.22 \ g$ of another gas with molecular weight $44$ are present,and the total pressure exerted by the mixture is $75 \ cm \ Hg$,then the partial pressure of the second gas is .......... $cm$ of $Hg$.

For the combustion reaction $C_2H_{6(g)} + nO_{2(g)} \to CO_{2(g)} + H_2O_{(l)}$,what is the ratio of the stoichiometric coefficients of $CO_2$ and $H_2O$ in the balanced equation?

The volume of hydrogen liberated at $STP$ by treating $2.4 \ g$ of magnesium with excess of hydrochloric acid is $x \times 10^{-2} \ L$. Find the value of $x$. Given: Molar volume of gas is $22.4 \ L$ at $STP$. Molar mass of magnesium is $24 \ g \ mol^{-1}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module