All Chapters
Ch.1 - Intro to General Chemistry
Ch.2 - Atoms & Elements
Ch.3 - Chemical Reactions
BONUS: Lab Techniques and Procedures
BONUS: Mathematical Operations and Functions
Ch.4 - Chemical Quantities & Aqueous Reactions
Ch.5 - Gases
Ch.6 - Thermochemistry
Ch.7 - Quantum Mechanics
Ch.8 - Periodic Properties of the Elements
Ch.9 - Bonding & Molecular Structure
Ch.10 - Molecular Shapes & Valence Bond Theory
Ch.11 - Liquids, Solids & Intermolecular Forces
Ch.12 - Solutions
Ch.13 - Chemical Kinetics
Ch.14 - Chemical Equilibrium
Ch.15 - Acid and Base Equilibrium
Ch.16 - Aqueous Equilibrium
Ch. 17 - Chemical Thermodynamics
Ch.18 - Electrochemistry
Ch.19 - Nuclear Chemistry
Ch.20 - Organic Chemistry
Ch.22 - Chemistry of the Nonmetals
Ch.23 - Transition Metals and Coordination Compounds

Solution: An aerosol spray can with a volume of 250 mL contains 2.30 g of propane gas (C3H8) as a propellant.The can says that exposure to temperatures above 130 oF may cause the can to burst. What is the press

Solution: An aerosol spray can with a volume of 250 mL contains 2.30 g of propane gas (C3H8) as a propellant.The can says that exposure to temperatures above 130 oF may cause the can to burst. What is the press

Problem

An aerosol spray can with a volume of 250 mL contains 2.30 g of propane gas (C3H8) as a propellant.

The can says that exposure to temperatures above 130 oF may cause the can to burst. What is the pressure in the can at this temperature?

Solution
  • Use the ideal gas equation to find the pressure where we have to convert the provided mass into moles
  • Ideal gas equation will appear as:

  • Calculating for moles (n):

MM of C3H8:

C - 12.01 (3) = 36.03

H - 1.01 (8) = 8.08

36.03 + 8.08 = 44.11 g/mol

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