anonymous
  • anonymous
for the reaction c(graphite)+1/2o2=co at 298k and 1 atm ,H=-26416cal.if the molar vol. of graphite is 0.0053litre,calculate E?
Chemistry
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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hba
  • hba
Yes Ruchi
anonymous
  • anonymous
wait
anonymous
  • anonymous
for the reaction \[c(graphite)+1/2o2=co\] at 298k and 1 atm ,H=-26416cal.if the molar vol. of graphite is 0.0053litre,calculate E?

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anonymous
  • anonymous
@hba @rambo2210
anonymous
  • anonymous
@timo86
anonymous
  • anonymous
@hitten101 @.Sam.
anonymous
  • anonymous
@SinginDaCalc2Blues
anonymous
  • anonymous
oh no...not hba again! haha
hba
  • hba
Hello :)
anonymous
  • anonymous
What is the question Ruchi?
hba
  • hba
Are You Blind ?
hba
  • hba
for the reaction c(graphite)+1/2o2=co at 298k and 1 atm ,H=-26416cal.if the molar vol. of graphite is 0.0053litre,calculate E?
anonymous
  • anonymous
Nope...not blind...just wanted you to do the dirty work for me hba! sucker! haha
anonymous
  • anonymous
Because I know you can't resit putting me down!
anonymous
  • anonymous
hey hlp me out
anonymous
  • anonymous
Hey Ruchi, what book do you use?
hba
  • hba
Stop Crying Girl
anonymous
  • anonymous
do you have to have energy in Calories, Kilocalories, Joules, or Kilojoules?
anonymous
  • anonymous
ARE YOU STILL HERE?
hba
  • hba
In Cal
anonymous
  • anonymous
okay, well if you see this, as far as I can gather, the question looks like it is asking how much heat(E) is evolved from the reaction of carbon and o2. tHE -26416 cal is the Hrxn (Heat of the reaction). There is only 1 mole of carbon, and the molar mass of carbon is 12.01grams... so we have 1 mol C= 12.01g and 1 mol C: -26416 cal (from the reaction) so set up this equation: (1 mol C/12.01 g C) x (-26416 cal/1 mol C) = -2199.5cal I think this is correct.
anonymous
  • anonymous
The logic seems correct as the energy given off is negative, as it should be for an exothermic combustion reaction, and it is much smaller than your Heat of reaction...I think this is correct...hope that helps!
anonymous
  • anonymous
bye hba!!! haha :o)~
hba
  • hba
Bye :)
anonymous
  • anonymous
@TuringTest @sauravshakya hlp me out

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