The normal boiling point of bromine is 58.8°C, and its enthalpy of vaporization is 30.91 kJ/mol. What is the approximate vapor pressure of bromine at 10.0°C?

Answer : The vapor pressure of bromine at is 0.1448 atm.

Explanation :

The Clausius- Clapeyron equation is :

where,

= vapor pressure of bromine at = ?

= vapor pressure of propane at normal boiling point = 1 atm

= temperature of propane =

= normal boiling point of bromine =

= heat of vaporization = 30.91 kJ/mole = 30910 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

Hence, the vapor pressure of bromine at is 0.1448 atm.

there is none because earth is stationary and flat.

the density of gold = 19.3 g/cm3

the relation between density and mass and volume is

density = mass / volume

given :

mass of gold = 13 g

density = 13g / volume = 19.3

volume = 13 /19.3 = 0.67 cm^3

thus volume of gold nugget = 0.67 cm^3