Ok so I knew how to do this when I took the exam but now I forget what I did.
Question:
Calculate the quantity of heat that is absorbed in kJ when 1.00 mol of liquid water forms water vapor at 100 degrees Celsius. Heat of vaporization = 2258 J/g.
Answer: 40.7 kJ
If you can tell me how I got this with steps i'd really appreciate it! Thanks!
Question:
Calculate the quantity of heat that is absorbed in kJ when 1.00 mol of liquid water forms water vapor at 100 degrees Celsius. Heat of vaporization = 2258 J/g.
Answer: 40.7 kJ
If you can tell me how I got this with steps i'd really appreciate it! Thanks!
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heat = M*vap
multiply mass by heat of vaporization and make sure u have all the correct units.
1.00mol of water is 18g/mol, so you have 18g of H2O.
you want the heat energy in kJ so convert 2258J/g to kJ (conversion 1kJ=10^3 J) so 2.258 kJ/g
18g*2.258kJ/g = 40.644 kJ
it helps if you write out units and they should cancel out and leave correct units for the answer, if it doesn't match up you know u prolly did something wrong.
multiply mass by heat of vaporization and make sure u have all the correct units.
1.00mol of water is 18g/mol, so you have 18g of H2O.
you want the heat energy in kJ so convert 2258J/g to kJ (conversion 1kJ=10^3 J) so 2.258 kJ/g
18g*2.258kJ/g = 40.644 kJ
it helps if you write out units and they should cancel out and leave correct units for the answer, if it doesn't match up you know u prolly did something wrong.
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that girl who got best answer is pretty much amazing :)
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