The normal boiling point of mercury (Hg) is 356.7 °C. What is the vapor pressure of mercury at 333 °C in atm? (∆Hvap = 58.51 kJ/mol

Here you can use the Clausis Clayperon equation: ln P1/P2=-Ea/R-(1/T1 - 1/T2)

where P1 is the pressure at standard condition: 760 mm Hg

P2 is the variable we need to solve

Ea is the activation energy, which in this case is delta H vaporisation: 56.9 kJ/mol

R is the gas constant 8.314 J/mol or 8.314 J/mol /1000 to convert to kJ

T1 is the normal boiling point 356.7 C, but converted to Kelvin: 629.85K

T2 is room temperature 25 C, but converted to Kelvin: 298.15 K

Once you plug everything in, you should get 4.29*10^-3 mmHg

Explanation:

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b