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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

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  1. JFraser Group Title
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    how many moels of water is in 13.8g of H2O?

    • 2 years ago
  2. Hayatcakes Group Title
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    there is 1.22 moles

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

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

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

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

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

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

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

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

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

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

    • 2 years ago
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