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Chemistry 4 Online
OpenStudy (anonymous):

HELP! Liquid ammonia boils at –33.4ºC and has a heat of vaporization of 23.5 kJ/mol. What is its vapor pressure at –50.0ºC? (R = 8.314 J/K•mol)

OpenStudy (anonymous):

ln (P1 / P2) = - (ΔH / R) (1/T1 - 1/T2)

OpenStudy (anonymous):

solve for p1. change of H is 23.5

OpenStudy (anonymous):

the vapor pressure of a substance at its normal boiling point is 760 mmHg

OpenStudy (anonymous):

so use that as your p2 @bs46shal

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