anonymous
  • anonymous
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
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
JFraser
  • JFraser
how many moels of water is in 13.8g of H2O?
anonymous
  • anonymous
there is 1.22 moles
JFraser
  • JFraser
no. 1mol of H2O is 18g. You've only got 13.8g. Flip the fraction upside down

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
okay so the answer would −7.87 kJ ?
anonymous
  • anonymous
or am i wrong?
JFraser
  • JFraser
your mole conversion should look like\[13.8g H_2O * (\frac{1mol H_2O}{18gH_2O}) = 0.767 moles H_2O\]
anonymous
  • anonymous
so is my answer wrong?
JFraser
  • JFraser
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 \[0.767mol H_2O * (\frac{40.65kJ}{1mol H_2O}) = 31.17kJ\]
mos1635
  • mos1635
40.65 or 6.03 ????
anonymous
  • anonymous
Oh okay . −31.3 kJ so this would be the final result?
JFraser
  • JFraser
@mos1635 is correct, replace the 40.65kJ with 6.03kJ. I missed the condensation vs freezing.
mos1635
  • mos1635
-4.623\[\approx\] -4.62

Looking for something else?

Not the answer you are looking for? Search for more explanations.