Answer:
Constants: 1 atm = 760 mm Hg = 760 torr = 1.01325 bar = 101.325 kPa R = 8.314 J·mol-1K-1= 0.08206 L·atm·mol-1K-11 Calorie = 1 kcal = 4.184 kJ Data:Properties of liquid water: b.p. (at 1 atm) = 100.00°CCH2O(l)= 4.184 J·g-1·K-1ΔHovap= 40.7 kJ·mol-1dH2O(l)(at 25°C) = 1.00 g·mL-1 KfH2O= 1.86 °C·kg·mol-1KbH2O= 0.52 °C·kg·mol-1Po(298 K)= 23.8 mm Hg Properties of ice: m.p. (at 1 atm) = 0.00°CCH2O(s) = 2.06 J·g-1·K-1 dH2O(s)= 0.917 g·mL-1 ΔHofus= 6.02 kJ·mol-1Formulae: ΔGo= ΔHo-TΔSo ΔG = ΔGo+ RT lnQΔGo= -RT lnKeqC = k P (or, S = k P) P = χPoΔT = KmπV = nRT [A]t= -k t + [A]oln[A]t= -k t + ln[A]o1/[A]t= k t + 1/[A]o
I am not sure If that helps im sorry I'm only in middle school :P.
