View Full Version : Chemist smarties answer this
Marklar MM
11-20-2006, 08:07 PM
A 4.40-g piece of solid CO2(dry ice) is allowed to vaporize (change from CO2(s)to CO2(g)) in a balloon. The final volume of the balloon is 1.00 L at 300. K. What is the pressure of the gas in atmospheres?
Friend needs help.
resistanze
11-20-2006, 08:29 PM
Probably will get this shit wrong but I don't wanna complete my philosophy essay.
Converted CO2 to moles:
(4.40g)(1mol/12.011) = 0.366 mol
PV = nRT => P = nRT/V
P = (0.366mol)(300K)(0.08206 L atm/mol K)/1.00L
P = 9.01 atm
Marklar MM
11-20-2006, 08:36 PM
ty ty ty. :toast :makeout
LaMarcus Bryant
11-20-2006, 09:03 PM
that is such a fundamental equation
your friend is just lazy
Marklar MM
11-20-2006, 09:15 PM
that is such a fundamental equation
your friend is just lazy
He is not veddy good at chem. :) Then again, neither am I.
I compeletly forgot all that. I had Chem my Sophomore year and 2 years erased everything I know about that class.
CosmicCowboy
11-21-2006, 09:11 AM
Maybe your friend and resistanze should swap homework assignments. Is he any good at philosophy since he sucks at chemistry?
resistanze
11-21-2006, 11:44 AM
Maybe your friend and resistanze should swap homework assignments. Is he any good at philosophy since he sucks at chemistry?
:lol
Well to late I already handed in my shitty 40%-of-my-final-grade paper.
Phenomanul
11-21-2006, 04:17 PM
Probably will get this shit wrong but I don't wanna complete my philosophy essay.
Converted CO2 to moles:
(4.40g)(1mol/12.011) = 0.366 mol
PV = nRT => P = nRT/V
P = (0.366mol)(300K)(0.08206 L atm/mol K)/1.00L
P = 9.01 atm
The molecular weight of CO2 is 12.011 + (15.9994 x 2) = 44.0098
Therefore:
Converted CO2 to moles:
(4.40g) (1 mol/44.0098g) = 0.099978 mol
PV = nRT => P = nRT/V
P = (0.099978 mol) (0.08206 L atm/mol K) (300 K) / 1.00 L
P = 2.46 atm
resistanze
11-21-2006, 04:33 PM
^^ Bwahaha, touche.
ALVAREZ6
11-21-2006, 04:38 PM
I fucking hate chem, I actually had a chem test today...hope I did ok.
Zunni
11-21-2006, 08:53 PM
Remember: if you're not a part of the solution, you're part of the precipitate.
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