Question #bde13

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Question #bde13

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Answer 1 · 2016-10-22T15:44:14

Showing calculation only

given

$w = 2.08 g$ $w=2.08g$

$M = 28 g/mol \to for N_{2}$ $M=28" g/mol "->"for "N_2$

$R = 0.08206 L a t m K^{- 1} m o l^{- 1}$ $R=0.08206LatmK^-1mol^-1$

$T = 50 + 273.15 = 323.15 K$ $T=50+273.15=323.15K$

$V = 0.285 L$ $V=0.285L$

$P = \frac{w}{M} \frac{R T}{V} = \frac{2.08}{28} \times \frac{0.08206 \times 323.15}{0.285} a t m \approx 6.9 a t m$ $P=w/M(RT)/V=2.08/28xx(0.08206xx323.15)/0.285atm~~6.9atm$

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