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)
chemistry - DrBob222, Friday, September 10, 2010 at 9:49pm
q = mass x heat vaporization
chemistry - jessa, Friday, September 10, 2010 at 10:09pm
This doesn't help me.
chemistry - DrBob222, Friday, September 10, 2010 at 10:51pm
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.
chemistry - hrashid, Monday, June 10, 2013 at 8:12pm
Can someone please answer this.
chemistry - Alison, Tuesday, July 9, 2013 at 8:43pm
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.