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bs46shal
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)
ln (P1 / P2) = - (ΔH / R) (1/T1 - 1/T2)
solve for p1. change of H is 23.5
the vapor pressure of a substance at its normal boiling point is 760 mmHg
so use that as your p2 @bs46shal