The pressure exerted by a non-reactive gaseou | vedclass

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(D) First,calculate the number of moles of each gas:$n_{CH_4} = \frac{6.4 \ g}{16 \ g \ mol^{-1}} = 0.4 \ mol$$n_{CO_2} = \frac{8.8 \ g}{44 \ g \ mol^

Vedclass · Accepted Answer

$150$

Vedclass · Answer

(D) First,calculate the number of moles of each gas:$n_{CH_4} = \frac{6.4 \ g}{16 \ g \ mol^{-1}} = 0.4 \ mol$$n_{CO_2} = \frac{8.8 \ g}{44 \ g \ mol^{-1}} = 0.2 \ mol$Total moles,$n = 0.4 + 0.2 = 0.6 \ mol$Convert volume to $m^3$:$V = 10 \ L = 10 	imes 10^{-3} \ m^3 = 0.01 \ m^3$Convert temperature to Kelvin:$T = 27 + 273 = 300 \ K$Using the ideal gas equation $PV = nRT$:$P = \frac{nRT}{V} = \frac{0.6 \ mol 	imes 8.314 \ J \ mol^{-1} K^{-1} 	imes 300 \ K}{0.01 \ m^3}$$P = 149652 \ Pa = 149.652 \ kPa$Rounding to the nearest integer,we get $150 \ kPa$.

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