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anonymous
 4 years ago
What is the enthalpy change when 13.8 grams of H2O changes from a liquid to a solid at zero degrees Celsius? ΔHcondensation = −40.65 kJ/mol ΔHfreezing = −6.03 kJ/mol
−4.62 kJ
−83.2 kJ
−7.87 kJ
−31.3 kJ
anonymous
 4 years ago
What is the enthalpy change when 13.8 grams of H2O changes from a liquid to a solid at zero degrees Celsius? ΔHcondensation = −40.65 kJ/mol ΔHfreezing = −6.03 kJ/mol −4.62 kJ −83.2 kJ −7.87 kJ −31.3 kJ

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JFraser
 4 years ago
Best ResponseYou've already chosen the best response.1how many moels of water is in 13.8g of H2O?

JFraser
 4 years ago
Best ResponseYou've already chosen the best response.1no. 1mol of H2O is 18g. You've only got 13.8g. Flip the fraction upside down

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay so the answer would −7.87 kJ ?

JFraser
 4 years ago
Best ResponseYou've already chosen the best response.1your mole conversion should look like\[13.8g H_2O * (\frac{1mol H_2O}{18gH_2O}) = 0.767 moles H_2O\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so is my answer wrong?

JFraser
 4 years ago
Best ResponseYou've already chosen the best response.1mole of water "costs" 40.65kJ of energy to freeze. You've got only 76.7% of 1 mole. The heat of fusion for .767mol of H2O will be \[0.767mol H_2O * (\frac{40.65kJ}{1mol H_2O}) = 31.17kJ\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh okay . −31.3 kJ so this would be the final result?

JFraser
 4 years ago
Best ResponseYou've already chosen the best response.1@mos1635 is correct, replace the 40.65kJ with 6.03kJ. I missed the condensation vs freezing.

mos1635
 4 years ago
Best ResponseYou've already chosen the best response.04.623\[\approx\] 4.62
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