Question

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?

Answer 1

Yes Ruchi

Answer 2

wait

Answer 3

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?

Answer 4

@hba @rambo2210

Answer 5

@timo86

Answer 6

@hitten101 @.Sam.

Answer 7

@SinginDaCalc2Blues

Answer 8

oh no...not hba again! haha

Answer 9

Hello :)

Answer 10

What is the question Ruchi?

Answer 11

Are You Blind ?

Answer 12

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?

Answer 13

Nope...not blind...just wanted you to do the dirty work for me hba! sucker! haha

Answer 14

Because I know you can't resit putting me down!

Answer 15

hey hlp me out

Answer 16

Hey Ruchi, what book do you use?

Answer 17

Stop Crying Girl

Answer 18

do you have to have energy in Calories, Kilocalories, Joules, or Kilojoules?

Answer 19

ARE YOU STILL HERE?

Answer 20

In Cal

Answer 21

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.

Answer 22

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!

Answer 23

bye hba!!! haha :o)~

Answer 24

Bye :)

Answer 25

@TuringTest @sauravshakya hlp me out