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)
5 answers
q = mass x heat vaporization
This doesn't help me.
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.
mass = density x volume
mass = 1.00 g/mL x 65.0 mL = 65.0 grams.
Can someone please answer this.
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.
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.