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anonymous
 4 years ago
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)
anonymous
 4 years ago
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)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ln (P1 / P2) =  (ΔH / R) (1/T1  1/T2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0solve for p1. change of H is 23.5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the vapor pressure of a substance at its normal boiling point is 760 mmHg

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so use that as your p2 @bs46shal
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