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Ruchi.

  • 2 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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  1. hba
    • 2 years ago
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    Yes Ruchi

  2. Ruchi.
    • 2 years ago
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    wait

  3. Ruchi.
    • 2 years ago
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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?

  4. Ruchi.
    • 2 years ago
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    @hba @rambo2210

  5. Ruchi.
    • 2 years ago
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    @timo86

  6. Ruchi.
    • 2 years ago
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    @hitten101 @.Sam.

  7. Ruchi.
    • 2 years ago
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    @SinginDaCalc2Blues

  8. SinginDaCalc2Blues
    • 2 years ago
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    oh no...not hba again! haha

  9. hba
    • 2 years ago
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    Hello :)

  10. SinginDaCalc2Blues
    • 2 years ago
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    What is the question Ruchi?

  11. hba
    • 2 years ago
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    Are You Blind ?

  12. hba
    • 2 years ago
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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?

  13. SinginDaCalc2Blues
    • 2 years ago
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    Nope...not blind...just wanted you to do the dirty work for me hba! sucker! haha

  14. SinginDaCalc2Blues
    • 2 years ago
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    Because I know you can't resit putting me down!

  15. Ruchi.
    • 2 years ago
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    hey hlp me out

  16. SinginDaCalc2Blues
    • 2 years ago
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    Hey Ruchi, what book do you use?

  17. hba
    • 2 years ago
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    Stop Crying Girl

  18. SinginDaCalc2Blues
    • 2 years ago
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    do you have to have energy in Calories, Kilocalories, Joules, or Kilojoules?

  19. SinginDaCalc2Blues
    • 2 years ago
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    ARE YOU STILL HERE?

  20. hba
    • 2 years ago
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    In Cal

  21. SinginDaCalc2Blues
    • 2 years ago
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    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.

  22. SinginDaCalc2Blues
    • 2 years ago
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    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!

  23. SinginDaCalc2Blues
    • 2 years ago
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    bye hba!!! haha :o)~

  24. hba
    • 2 years ago
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    Bye :)

  25. Ruchi.
    • 2 years ago
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    @TuringTest @sauravshakya hlp me out

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