posted by Samantha .
how do i ansewr this question?
An aerosol spray can with a volume of 400. mL contains 2.25 g of propane gas (C3H8) as a propellant
(a) If the can is at 31°C, what is the pressure in the can?
(b) What volume would the propane occupy at STP?
(c) The can's label says that exposure to temperatures above 130.°F may cause the can to burst. What is the pressure in the can at this temperature?
Use PV = nRT
n = grams/molar mass
i got 1.28 but its wrong
You have three questions here. Which gave you an answer of 1.28? That's 1.28 WHAT unit? Show your work. I can't find errors with an answer only.
Perhaps I can with just an answer. You used P = nRT. You never used the 400 mL. And you must use 0.400 L; P will be in atmospheres. T must be in kelvin = 304K.
that was my answer for letter a and i did all of those conversions first and i still got that anwerrs which is wrong. can you do the problem and tell me what you get?
PV = nRT
P = ?
V = 0.400L
n = grams/molar mass = 2.25/44 = 0.0511
T = 273 + 31 = 304K
P = 0.0511 x 0.08206 x 304/0.400 = 3 something.
Tell me what you did wrong. I think you did not divide by 0.400.
thanks! for b i used that equation and got the volume to be .400 buts its wrong. do i use the presure from a or do i just use 1atm?