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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?
 one year ago
 one year ago
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?
 one year ago
 one year ago

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Ruchi.Best ResponseYou've already chosen the best response.0
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?
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
oh no...not hba again! haha
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
What is the question Ruchi?
 one year ago

hbaBest ResponseYou've already chosen the best response.0
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?
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
Nope...not blind...just wanted you to do the dirty work for me hba! sucker! haha
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
Because I know you can't resit putting me down!
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
Hey Ruchi, what book do you use?
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
do you have to have energy in Calories, Kilocalories, Joules, or Kilojoules?
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
ARE YOU STILL HERE?
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
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.
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
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!
 one year ago

SinginDaCalc2BluesBest ResponseYou've already chosen the best response.0
bye hba!!! haha :o)~
 one year ago

Ruchi.Best ResponseYou've already chosen the best response.0
@TuringTest @sauravshakya hlp me out
 one year ago
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