Answer to Question #144429 in General Chemistry for martha

Question

the freezing point of mercury is -38.8 degree Celsius . what quantity of energy in joules is released to the surrounding if 1.oomL of mercury is cooled from 23.0 degree Celsius to -38.8 degree Celsius and then frozen to a solid?

Answer 1

Q=C x m x deltaT

deltaT=23+38.8=61.8°

С(Hg)=140 J/kg x C°

m=n/M

V=n/22.4

M (Hg)=200.59 g/mol

V (Hg)=1 ml=0.001 L

n (Hg)=Vx22.4=0.001 x 22.4 = 0.0224 mol

m (Hg) = 0.0224/200.59=0.0001 g = 1 x 10^-7 kg

Q=140 x 10^-7 x 61.8 = 10^-3 J