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anonymous
 4 years 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?
anonymous
 4 years 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?

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for 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
 4 years ago
Best ResponseYou've already chosen the best response.0oh no...not hba again! haha

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What is the question Ruchi?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Because I know you can't resit putting me down!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hey Ruchi, what book do you use?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you have to have energy in Calories, Kilocalories, Joules, or Kilojoules?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay, 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
 4 years ago
Best ResponseYou've already chosen the best response.0The 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
 4 years ago
Best ResponseYou've already chosen the best response.0@TuringTest @sauravshakya hlp me out
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