Question

65.0 mL of water is heated to its boiling point. How much heat in Kj is required to vaporize it? (Assume a density of 1.00 g/mL)

Answer 1

q = mass x heat vaporization

Answer 2

This doesn't help me.

Answer 3

Then tell me what you don't understand. All you need to do is to substitute the mass of 65.0 mL water (65.0 grams) and surely you have a table giving the heat vaporization of water. That calculates q, the heat required to vaporize water at it's boiling point. How did I get the mass of 65.0 grams?
mass = density x volume
mass = 1.00 g/mL x 65.0 mL = 65.0 grams.

Answer 4

Can someone please answer this.

Answer 5

use dimensional analysis and the heat vaporization table.

Water = 40.7 kj/mol.

Start with the information that you were given and make sure that all of your units cancel out!

(65.0mL)(1.00g/1mL)(1 mole H2O/18.0g H2O)(40.7kj/1 mole H2O) = 146.9 kj.

There are 3 significant figures, so 147kj is your answer.