Asked by reese
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).
Answers
Answered by
DrBob222
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.
Answered by
Anonymous
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
Answered by
Alison
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.
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