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RE: #Chemistry Challenge 5
1. Your equation is correctly balanced, but I would have preferred if you use the smallest possible whole numbers as factors in front of the element symbols. I count your answer as correct though.
2. Correct answer!
3. Why didn't you use the given (rounded) molar masses? In that case we would get 'nicer' numbers as results and I wouldn't need to verify every single of your calculations.
Your mass of isooctane is correct. Also n(C8H18) and n(CO2), but the mass of CO2 (even with not rounded molar masses) is a little bit too small. What did you use as M(CO2)? The volume (calculated with your numbers) is correct again (I guess you rounded R different than given).
4. The speed of chemical reactions isn't measured by volume / time!
I am just lazy so I use chemdraw to generate the molecular weigh LOL
More explanation on
Q3)
mole of CO2 = 1596.78 mole
implies mass of CO2 = 1596.78 * 44 = 70258 g = 70.26 kg
I use the given R btw
Q4
I thought the v(C8H18) where the v means volume LOL, but now i guess probably you mean speed
I would take rate of reaction as concentration over time usually, but I don't have concentration here, so I guess you are looking for change of mole per unit time?
If this is the case, here would be my answer:
5 L/ hr implies 32.89 mol/hr
Yes, right, but first you had another volume and also another mass of CO2. Better is to add a new comment if you change any parts of your solutions.
Concerning part 4: if any information seems to be missing you can use every given detail from parts 1 to 3 to find it out.
So far your answer is wrong.
How about:
v(C8H18) = - 32.89 mol/hr
NOT correct! :-)
But apart from that so far you did well ... :)
Edit: by the way, the reaction speed v is not negative (even if ∆c is negative), but that's not the point: your number is wrong, too.
humm... if you are expecting the answer in concentration:
[CO2] = 4.734 x 10-2 mol dm-3
implies rate of CO2 production = 6.7 x 10-3 mol dm-3 hr-1
implies speed of dissipating isooctane = 8.37 x 10-4 mol dm-3 hr-1
Thanks for letting us know about the concentration of carbon dioxide.
How exactly do you get
c(CO2) = 4.734 x 10-2 mol/dm3?
Nah.... sorry for getting so many typo
[CO2] = 1596.78 mol / 3.92 x 10 4 dm3
= 4.0734 x 10-2 mol dm-3
But the rest should be the same
:)
This result is the (lower) concentration of CO2 when it was emitted as a gas already ... but gasoline is a liquid ...