The molar heat of vaporization for liquid water is 40.6 kJ/mole.
How much energy is required to change 3.4 g of liquid water to steam if the water is already at 100oC?

7670 joules

Explanation:

Molar heat → 40.6 kJ / mol

1 mol of water occupies 18 g, but we have only 3.4 g. Let's make a rule of three:

18 g of H₂O release 40.6 kJ of heat

Then, 3.4 g will release (3.4 . 40.6) / 18 = 7.67 kJ

If we convert the kJ to J → 7.67 kJ . 1000 J / 1 kJ = 7670 Joules

We can also do this:

If water is already at 100°C, we apply the latent heat (heat that is relased in a change of state)

Q = 2256 J/g . 3.4 g → 7670.4 J

i think it is true since you can test the answer and can use observations to complete it.

1 is the answer to this problem