Ch.13 - Chemical KineticsWorksheetSee all chapters
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: The decomposition of dinitrogen pentoxide is described by the chemical equation 2N2O5(g) → 4NO2(g) + O2(g) If the rate of appearance of NO 2 is equal to 0.78 mol/min at a particular moment, what is the rate of appearance of O2 at that moment? 

Solution: The decomposition of dinitrogen pentoxide is described by the chemical equation 2N2O5(g) → 4NO2(g) + O2(g) If the rate of appearance of NO 2 is equal to 0.78 mol/min at a particular moment, what is th

Problem

The decomposition of dinitrogen pentoxide is described by the chemical equation 

2N2O5(g) → 4NO2(g) + O2(g) 

If the rate of appearance of NO 2 is equal to 0.78 mol/min at a particular moment, what is the rate of appearance of O2 at that moment? 


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