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.

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

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.