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SVEN [57.7K]
3 years ago
14

Moles in 9.8g of calcium

Chemistry
1 answer:
Nataly [62]3 years ago
8 0
(9.8/1) x 1mol/40g = 0.245 mol
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Answer : The vapor pressure of propane at 25.0^oC is 17.73 atm.

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The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of propane at 25.0^oC = ?

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T_1 = temperature of propane = 25.0^oC=273+25.0=298.0K

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\ln (\frac{1atm}{P_1})=\frac{24540J/mole}{8.314J/K.mole}\times (\frac{1}{298.0K}-\frac{1}{230.96K})

P_1=17.73atm

Hence, the vapor pressure of propane at 25.0^oC is 17.73 atm.

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