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Hayatcakes
Group Title
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
 2 years ago
 2 years ago
Hayatcakes Group Title
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
 2 years ago
 2 years ago

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

Hayatcakes Group TitleBest ResponseYou've already chosen the best response.0
there is 1.22 moles
 2 years ago

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

Hayatcakes Group TitleBest ResponseYou've already chosen the best response.0
okay so the answer would −7.87 kJ ?
 2 years ago

Hayatcakes Group TitleBest ResponseYou've already chosen the best response.0
or am i wrong?
 2 years ago

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

Hayatcakes Group TitleBest ResponseYou've already chosen the best response.0
so is my answer wrong?
 2 years ago

JFraser Group TitleBest ResponseYou've already chosen the best response.1
mole 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 \[0.767mol H_2O * (\frac{40.65kJ}{1mol H_2O}) = 31.17kJ\]
 2 years ago

mos1635 Group TitleBest ResponseYou've already chosen the best response.0
40.65 or 6.03 ????
 2 years ago

Hayatcakes Group TitleBest ResponseYou've already chosen the best response.0
Oh okay . −31.3 kJ so this would be the final result?
 2 years ago

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

mos1635 Group TitleBest ResponseYou've already chosen the best response.0
4.623\[\approx\] 4.62
 2 years ago
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