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:

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.