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WarrenBuffet
01-30-2008, 10:31 PM
*nothing Happened Here

bosse de nage
01-30-2008, 10:48 PM
i need one.
my reps are too low.
ahhh

WarrenBuffet
01-30-2008, 10:49 PM
well, if no one higher than 9 posts.... for a while, you'll be in luck

bosse de nage
01-30-2008, 10:54 PM
hold on....i'll give some stuff away real quick and get it up to 10.
:geptard:

traa
01-30-2008, 10:58 PM
i need TL invite.
Can you give it to me?

etrade
01-30-2008, 10:59 PM
how can it be that +55 need a TL invite :) weird!

jappere
01-30-2008, 11:15 PM
how can it be that +55 need a TL invite :) weird!
he is giving the invite

Definite
01-30-2008, 11:19 PM
Today is opposite day. :P

FaragelloOo
01-30-2008, 11:21 PM
good luck all i have acc + invite

good man and goodluck all
.

WarrenBuffet
01-30-2008, 11:36 PM
Nothing Happened Here

hasnainbk
01-30-2008, 11:37 PM
can I plz have it?

Answer:
2co+o2-> 2co2

bosse de nage
01-30-2008, 11:43 PM
2CO(g) O 2(g) y 2CO 2(g)

hey he copied my answer

hasnainbk
01-30-2008, 11:48 PM
nopes bosse de nage...look at your equation...you cant even write it correct.

Definite
01-30-2008, 11:49 PM
Approximately 7.5708236 liters.

bosse de nage
01-30-2008, 11:52 PM
it's 3 liters for 6

pdj2
01-31-2008, 12:03 AM
3 meters?

bosse de nage
01-31-2008, 12:04 AM
oh well...i went to liberal arts college.

wrUmm568
01-31-2008, 12:06 AM
6.7 Liters:
2CO + O2 -> 2CO2
4.8g @ 32g/mol
0.30 0.15mol

PV = nRT -> V = nRT/P
STP: 1 atm, 273 K

or just use molar gas vol at 22.4 L /mol

hasnainbk
01-31-2008, 12:22 AM
if 16 grams of oxygen react with 28 gram of CO;
4.8g of O2 will combine with 8.4g of CO;
2CO(g) + O2(g) -> 2CO2(g)
Because 28 g of CO volumes 22.4 litres
8.4 gram will occupy 6.72
producing 6.48 litres of CO2.

==> thats the correct solution.

WarrenBuffet
01-31-2008, 12:25 AM
Nothing Happened Here

wrUmm568
01-31-2008, 12:31 AM
if 16 grams of oxygen react with 28 gram of CO;
4.8g of O2 will combine with 8.4g of CO;
2CO(g) + O2(g) -> 2CO2(g)
Because 28 g of CO volumes 22.4 litres
8.4 gram will occupy 6.72
producing 6.48 litres of CO2.

==> thats the correct solution.
This is incorrect, the ideal gas law (PV = nRT) says volume is proportional to n (number of molecules) not the mass.

WarrenBuffet
01-31-2008, 12:31 AM
Nothing Happened Here

bosse de nage
01-31-2008, 12:32 AM
thanks for the chance.

yep..it's n not m..but equation is right...as i demonstrated 1st..
ha

hasnainbk
01-31-2008, 12:40 AM
thats the way of converting grams into liters which didn't really mattered with the solution.you are right about this formula and it implements the same as I did.

wrUmm568
01-31-2008, 12:43 AM
hmm....... O reely? can someone verify this?

further more this:



2CO(g) + O2(g) -> 2CO2(g)
Because 28 g of CO volumes 22.4 litres
is wrong because this

2CO(g) + O2(g) -> 2CO2(g) equation references amount (ie moles) not mass therefore it takes 2 moles of CO for every 1 mol of O2. Also the question asked for the amount of gas in Liters and not just an equation

TP635
01-31-2008, 12:44 AM
How many liters of carbon monoxide (at STP) are needed to react with 4.8g of oxygen gas to produce carbon dioxide?Do you realized that you ask a bad/wrong question? Read it again.
Any positive number is a correct ans to that question.

wrUmm568
01-31-2008, 12:53 AM
How many liters of carbon monoxide (at STP) are needed to react with 4.8g of oxygen gas to produce carbon dioxide?Do you realized that you ask a bad/wrong question? Read it again.
Any positive number is a correct ans to that question.
The question is correct and has only one right answer if you interpret it as:
How many liters of carbon monoxide (at STP) are needed to fully react with 4.8g of oxygen gas such that there is none left over to produce carbon dioxide?

hasnainbk
01-31-2008, 12:56 AM
zaa730: I guess wrUmm568 deserves it more.He has really done some hard work.I am not sure if he is right or wrong but I appreciate him and would ask you to cancel the invite you sent to me and send it again to wrUmm568.

TP635
01-31-2008, 01:02 AM
Do you realized that you ask a bad/wrong question? Read it again.
Any positive number is a correct ans to that question.
The question is correct and has only one right answer if you interpret it as:
How many liters of carbon monoxide (at STP) are needed to fully react with 4.8g of oxygen gas such that there is none left over to produce carbon dioxide?

The original question is ambiguous. The CORRECT question is what you stated above. You should be given the invite for getting the right question and ans.

mrnobody
01-31-2008, 01:05 AM
How many liters of carbon monoxide (at STP) are needed to react with 4.8g of oxygen gas to produce carbon dioxide?

2CO (g) + O2 (g) <=> 2CO2

Mole of O2 = 4.8 g/ (16 g/mol ) = # of Moles of O2

Moles of CO = 2 * (# of Moles of O2)

Then, you can go to liters simply by multiplying # of Moles of CO with 22.4 liters (coz the reaction is at STP)

or use classic PV = nTR

=> V = nTR / P

where,

P= pressure (1 atm at STP)

V = Volume

n = # of moles

T = Temperature in kelvin (0 degree celcius in STP)

R = constant (forgot the value lol)

P.S. me not want an invite...was barely showing http://www.pnwriders.com/forum/images/smilies/epenis.gif

WarrenBuffet
01-31-2008, 01:24 AM
NOTHING HAPPENED HERE

kthxbye.

CLOSE TIME