A gas (Molar mass = 280 g mol^{-1}) was bur | vedclass

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(A) Let $x \ g$ be the mass of the gas burnt.Moles of gas $= \frac{x}{280} \ mol$.Heat released by the combustion $= 	ext{Heat capacity} 	imes \Delt

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(A) Let $x \ g$ be the mass of the gas burnt.Moles of gas $= \frac{x}{280} \ mol$.Heat released by the combustion $= 	ext{Heat capacity} 	imes \Delta T = 2.5 \ kJ \ K^{-1} 	imes (298.45 - 298.0) \ K = 2.5 	imes 0.45 \ kJ = 1.125 \ kJ$.Since the combustion occurs in a constant volume calorimeter,the heat released is equal to the internal energy change,$\Delta U$.Given the enthalpy of combustion $\Delta H = 9 \ kJ \ mol^{-1}$,and assuming $\Delta H \approx \Delta U$ for this calculation,we have:$\Delta U 	ext{ per mole} = 9 \ kJ \ mol^{-1}$.Total heat released $= 	ext{moles} 	imes \Delta U 	ext{ per mole}$.$1.125 \ kJ = (\frac{x}{280} \ mol) 	imes 9 \ kJ \ mol^{-1}$.$x = \frac{1.125 	imes 280}{9} = 35 \ g$.

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