When 8.70 kj of thermal energy is added to 2.50 mol of liquid methanol, it vaporizes. determine the heat of vaporization in kj/mol of methanol.

6.4 cm (max) - 25 cm (min)

the number of valence electrons for each molecule or ion is shown beneath the structure. remember hydrogen will not have more than two electrons. this hydrogen is part of a covalent bond (sharing two electrons). a double bond here would cause hydrogen to share four electrons with phosphorus.

The heat of vaporization in kJ/mol of methanol

= 3.48 kJ/mole

Explanation:

Here,the heat of vaporization in kJ/mol is asked to calculate.This means it is asked to calculate heat required for 1 mole of methanol. Since the substance vaporizes , so this is called heat of vaporization.

Divide the thermal energy with the number of moles .

This is simple mathematics :

If 2.50 mol of liquid require = 8.70 kJ of energy

Then, 1 mole will need =

= 3.48 kJ/mole

Follow the units :

The heat of vaporization in kJ/mol of methanol is asked.

Unit = kJ /Mole

The heat of vaporization in kJ/mol of methanol

= 3.48 kJ/mole

Explanation:

Here,the heat of vaporization in kJ/mol is asked to calculate.This means it is asked to calculate heat required for 1 mole of methanol. Since the substance vaporizes , so this is called heat of vaporization.

Divide the thermal energy with the number of moles .

This is simple mathematics :

If 2.50 mol of liquid require = 8.70 kJ of energy

Then, 1 mole will need =

= 3.48 kJ/mole

Follow the units :

The heat of vaporization in kJ/mol of methanol is asked.

Unit = kJ /Mole