A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature rose by 5.72°C. If the heat capacity of the bomb plus water was 9.47 kJ / °C, calculate the molar heat of combustion of methanol.

872.28 kJ/mol

Explanation:

The heat released is:

ΔH = C*ΔT

where ΔH is the heat of combustion, C is the heat capacity of the bomb plus water, and ΔT is the rise of temperature. Replacing with data:

ΔH =  9.47*5.72 = 54.1684kJ

A quantity of 1.922 g of methanol in moles are:

moles = mass / molar mass

moles = 1.992/32.04 = 0.0621 mol

Then the molar heat of combustion of methanol is:

ΔH/moles = 54.1684/0.0621 = 872.28 kJ/mol

the only liquid elements at standard temperature and pressure are bromine (br) and mercury (hg). although, elements caesium (cs), rubidium (rb), francium (fr) and gallium (ga) become liquid at or just above room temperature.

explanation: