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?

59.87 kJ

In the question , we are given;

We are required to calculate the amount of heat energy absorbed to convert water at 100°C to steam at 100°C.

Moles = Mass ÷ Molar mass

Molar mass of water = 18.02 g/mol

Moles of water = 100 g ÷ 18.02 g/mol

                        = 5.549 moles

But;

Q = n × ΔHv , where n is the number of moles and ΔHv molar heat of Vaporization.

Therefore;

Q = 5.549 moles × 10.79 kJ/mol

   = 59.874 kJ

    = 59.87 kJ

Thus, the amount of heat absorbed is 59.87 kJ

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.