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Topic: Calorimetry Help  (Read 28692 times)

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Offline Mitch

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Re:Calorimetry Help
« Reply #15 on: March 23, 2004, 06:22:45 PM »
The delta H is equal to the heat of neutralization divided by moles of the limiting reagent. :)
« Last Edit: March 23, 2004, 06:23:47 PM by Mitch »
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meme

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Re:Calorimetry Help
« Reply #16 on: March 23, 2004, 06:30:05 PM »
okay... lemme just make sure I got this all straight:

  • Heat evolved:

.05 mol NaOH x (38.99g/1mol) = 1.95g NaOH
heat evolved: (4.18)(1.95g)(6.9C) = 56.2J
  • Amount of OH- reacted:
    so... since it's the same as the amount of NaOH - its .05 mol OH- right?
  • Amount of H2O formed:
    1:1 mole ratio - therefore .05 mol water
  • Heat evolved per mole of H2O, (heat of neutralization):

56.2 J/ .05 mol = 1124.5 J/mol ?? [/list]

*squeezes eyes shut, crosses fingers, and hopes all that's right*

Offline Mitch

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Re:Calorimetry Help
« Reply #17 on: March 23, 2004, 07:56:16 PM »
Nope you were close though. :P

You have the right answer but the .05moles at the last step isn't coming from the moles of water but is the moles of hydroxide. It is the limiting reagent that matters when you divide not the water. :)

But, yeah it all looks right to me.
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meme

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Re:Calorimetry Help
« Reply #18 on: March 23, 2004, 08:05:12 PM »
56.2 J/ .05 mol HYDROXIDE = 1124.5 J/mol

so everything is right now, right?  :P

anyway, thanks a million! It would have taken me forever without your help.  :)

Offline Mitch

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Re:Calorimetry Help
« Reply #19 on: March 23, 2004, 08:13:32 PM »
looks good. :)
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meme

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Re:Calorimetry Help
« Reply #20 on: March 23, 2004, 11:44:12 PM »
Is there any way I can donate to the site?
« Last Edit: March 24, 2004, 01:07:54 PM by meme »

Offline Mitch

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Re:Calorimetry Help
« Reply #21 on: March 24, 2004, 04:52:58 PM »
We might start selling merchandise in the coming month, like little periodic tables and the like. But we don't accept straight donations. Although, the sooner Greg makes the logo, the sooner we can start selling some stuff. :P
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