Chemist smarties answer this

[Marklar MM]

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]

Probably will get this 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]

ty ty ty.

[LaMarcus Bryant]

that is such a fundamental equation
your friend is just lazy

[Marklar MM]

that is such a fundamental equation
your friend is just lazy

He is not veddy good at chem. Then again, neither am I.

[E20]

I compeletly forgot all that. I had Chem my Sop re year and 2 years erased everything I know about that class.

[CosmicCowboy]

Maybe your friend and resistanze should swap homework assignments. Is he any good at philosophy since he sucks at chemistry?

[resistanze]

Maybe your friend and resistanze should swap homework assignments. Is he any good at philosophy since he sucks at chemistry?

Well to late I already handed in my ty 40%-of-my-final-grade paper.

[Phenomanul]

Probably will get this 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]

^^ Bwahaha, touche.

[ALVAREZ6]

I ing hate chem, I actually had a chem test today...hope I did ok.

[Zunni]

Remember: if you're not a part of the solution, you're part of the precipitate.