the molar heat of vaporization for water is 10.79 kj/mol. how much energy must be absorbed by 100 grams
of water at 100 °c in order to convert it to steam at 100 °c?
a 4079 0
b 140.79 k)
c 7.35 k)
d 226 kj

59.878 Kilo joules

We are given;

We are required to calculate the amount of energy absorbed to convert water to ice.

In this case;

Mass of water = 100 g

But, molar mass of water = 18.02 g/mol

Moles of water = 100 g ÷ 18.02 g/mol

                         = 5.549 Moles

Therefore;

Q = 5.549 moles × 10.79 kJ/mol

   = 59.878 Kilo joules

the answer is option a.

the cold front replaces the warmer air at the ground level changing the wind to be breezy and denser.there is also an immediate drop in the temperature.