7.0 , g of N_2,8.0 , g of O_2,and 34.0 , g of | vedclass

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(A) Active mass is defined as molar concentration,which is $\frac{	ext{moles}}{	ext{volume in liters}}$.$(a)$ For $N_2$: $	ext{Moles} = \frac{7.0 \

Vedclass · Accepted Answer

$0.25, 0.25, 2.0$

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(A) Active mass is defined as molar concentration,which is $\frac{	ext{moles}}{	ext{volume in liters}}$.$(a)$ For $N_2$: $	ext{Moles} = \frac{7.0 \, g}{28 \, g \, mol^{-1}} = 0.25 \, mol$. Active mass $= \frac{0.25 \, mol}{1 \, L} = 0.25 \, mol \, L^{-1}$.$(b)$ For $O_2$: $	ext{Moles} = \frac{8.0 \, g}{32 \, g \, mol^{-1}} = 0.25 \, mol$. Active mass $= \frac{0.25 \, mol}{1 \, L} = 0.25 \, mol \, L^{-1}$.$(c)$ For $NH_3$: $	ext{Moles} = \frac{34.0 \, g}{17 \, g \, mol^{-1}} = 2.0 \, mol$. Active mass $= \frac{2.0 \, mol}{1 \, L} = 2.0 \, mol \, L^{-1}$.Thus,the active masses are $0.25, 0.25, 2.0$.

$7.0 \, g$ of $N_2$,$8.0 \, g$ of $O_2$,and $34.0 \, g$ of $NH_3$ are present in a $1 \, L$ vessel. The active masses in $mol \, L^{-1}$ are respectively:

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