a 50.0 ml sample of water is heated to its boiling point. how much heat (in KJ) is required to vaporize it? (assume a density of 1.00g/ml).
3 answers
Read the other problem you posted about moving T of steam at 135 C to ice at -45 and pull out the one part that deals with condensing the steam to liquid at 100 C.
You have to change 50 mL into moles by taking 50mLx1.00g/mL
mL will cancel and now you have 50 grams. Turn grams into moles.
50/18.01 to get moles
then times by deltaHvap at boiling point which is 40.7
mL will cancel and now you have 50 grams. Turn grams into moles.
50/18.01 to get moles
then times by deltaHvap at boiling point which is 40.7
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!
(50.0mL)(1.00g/1mL)(1 mole H2O/18.0g H2O)(40.7kj/1 mole H2O) = 113.0 kj.
There are 3 significant figures, so 113kj 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!
(50.0mL)(1.00g/1mL)(1 mole H2O/18.0g H2O)(40.7kj/1 mole H2O) = 113.0 kj.
There are 3 significant figures, so 113kj is your answer.