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: If 250 mL of methane, CH4, effuses through a small hole in 28 s, the time required for the same volume of helium to pass through the hole under the same conditions will be 

Solution: If 250 mL of methane, CH4, effuses through a small hole in 28 s, the time required for the same volume of helium to pass through the hole under the same conditions will be 

Problem

If 250 mL of methane, CH4, effuses through a small hole in 28 s, the time required for the same volume of helium to pass through the hole under the same conditions will be 



Solution

We’re being asked to calculate the time required for 250 mL of helium (He) gas to effuse through a hole. The same volume of methane (CH4) gas took 28 s to effuse through the same hole.


Recall that Graham's Law of Effusion allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.



This means that when comparing two gases:



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