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Hayatcakes

  • 2 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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  1. JFraser
    • 2 years ago
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    how many moels of water is in 13.8g of H2O?

  2. Hayatcakes
    • 2 years ago
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    there is 1.22 moles

  3. JFraser
    • 2 years ago
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    no. 1mol of H2O is 18g. You've only got 13.8g. Flip the fraction upside down

  4. Hayatcakes
    • 2 years ago
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    okay so the answer would −7.87 kJ ?

  5. Hayatcakes
    • 2 years ago
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    or am i wrong?

  6. JFraser
    • 2 years ago
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    your mole conversion should look like\[13.8g H_2O * (\frac{1mol H_2O}{18gH_2O}) = 0.767 moles H_2O\]

  7. Hayatcakes
    • 2 years ago
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    so is my answer wrong?

  8. JFraser
    • 2 years ago
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    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 \[0.767mol H_2O * (\frac{40.65kJ}{1mol H_2O}) = 31.17kJ\]

  9. mos1635
    • 2 years ago
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    40.65 or 6.03 ????

  10. Hayatcakes
    • 2 years ago
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    Oh okay . −31.3 kJ so this would be the final result?

  11. JFraser
    • 2 years ago
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    @mos1635 is correct, replace the 40.65kJ with 6.03kJ. I missed the condensation vs freezing.

  12. mos1635
    • 2 years ago
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    -4.623\[\approx\] -4.62

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