Astudent determined that it requires 106220 j of energy to vaporize 47g of water. is the student is right

He is right that the energy of vaporization of 47 g of water s 106222 j.

Enthalpy of vaporization or heat of vaporization is the amount of energy which is used to transform one mole of liquid into gas.

In case of water it is 40.65 KJ/mol. And 18 g of water is equal to one mole.

It means for vaporizing 18 g, 40.65 kJ energy is needed.

So for energy 47 g of water = 47/18 * 40.65 = 106.1 KJ

Hence the student is right about the energy of vaporization of 47 g of water.

it would be b. 6co2 + 6h2o --> c6h12o6 + 6o2