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Here are the incorrect answers so do not submit these... a) 4.90 atm b) 2.23 L c) 5.35 atm

An aerosol spray can with a volume of 456 mL contains 3.99 g of propane gas (C3H8)as a propellant.

(a) If the can is at 23°C, what is the pressure in the can?
atm

(b) What volume would the propane occupy at STP?
L

(c) The can says that exposure to temperatures above 130°F maycause the can to burst. What is the pressure in the can at thistemperature?
atm

a) 4.90 atm b) 2.23 L c) 5.35 atm

Explanation / Answer

a) PV= nRT         P = [(3.99/44)mol *0.0821L.atm/mol.K*296K]/ 0.456L       = 4.83atm b) 2.031L c) P= 5.35atm

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