Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
Ruchi.
Group Title
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
Ruchi. Group Title
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

This Question is Closed

Ruchi. Group TitleBest 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

Ruchi. Group TitleBest ResponseYou've already chosen the best response.0
@hba @rambo2210
 one year ago

Ruchi. Group TitleBest ResponseYou've already chosen the best response.0
@hitten101 @.Sam.
 one year ago

Ruchi. Group TitleBest ResponseYou've already chosen the best response.0
@SinginDaCalc2Blues
 one year ago

SinginDaCalc2Blues Group TitleBest ResponseYou've already chosen the best response.0
oh no...not hba again! haha
 one year ago

SinginDaCalc2Blues Group TitleBest ResponseYou've already chosen the best response.0
What is the question Ruchi?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Are You Blind ?
 one year ago

hba Group TitleBest 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

SinginDaCalc2Blues Group TitleBest 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

SinginDaCalc2Blues Group TitleBest ResponseYou've already chosen the best response.0
Because I know you can't resit putting me down!
 one year ago

Ruchi. Group TitleBest ResponseYou've already chosen the best response.0
hey hlp me out
 one year ago

SinginDaCalc2Blues Group TitleBest ResponseYou've already chosen the best response.0
Hey Ruchi, what book do you use?
 one year ago

hba Group TitleBest ResponseYou've already chosen the best response.0
Stop Crying Girl
 one year ago

SinginDaCalc2Blues Group TitleBest ResponseYou've already chosen the best response.0
do you have to have energy in Calories, Kilocalories, Joules, or Kilojoules?
 one year ago

SinginDaCalc2Blues Group TitleBest ResponseYou've already chosen the best response.0
ARE YOU STILL HERE?
 one year ago

SinginDaCalc2Blues Group TitleBest 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

SinginDaCalc2Blues Group TitleBest 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

SinginDaCalc2Blues Group TitleBest ResponseYou've already chosen the best response.0
bye hba!!! haha :o)~
 one year ago

Ruchi. Group TitleBest ResponseYou've already chosen the best response.0
@TuringTest @sauravshakya hlp me out
 one year ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.