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: It takes 125 minutes for 10 mL of Ar gas to effuse through a porous barrier. How many minutes would it take the same amount of Ne gas to effuse through the same barrier?A. 108.4 min B. 70 min C. 72.5 min D. 85.3 min E. 88.8 min

Problem

It takes 125 minutes for 10 mL of Ar gas to effuse through a porous barrier. How many minutes would it take the same amount of Ne gas to effuse through the same barrier?

A. 108.4 min 

B. 70 min 

C. 72.5 min 

D. 85.3 min 

E. 88.8 min

Solution

We’re being asked to identify the unknown gas given the rate of effusion of O2.


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.


rate=1MMgas


This means that when comparing two gases:


rategas 1rategas 2=MMgas 2MMgas 1


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