Explanation Given: The rate constant is 0.00752 s − 1 , the temperature is 20 ° C , the volume is 150.0 mL , the mass percentage of H 2 O 2 is 30%, density is 1.11 g/mL , time is 85.0 s , and barometric pressure is 742.5 mmHg . Explanation: Calculate mass as follows: Mass = d V Here, d is density and V is volume. Substitute 1.11 g/mL for d and 150.0 mL for V in the above equation as follows: Mass = ( 1.11 g/mL ) ( 150.0 mL ) = 166.5 g Mass of H 2 O 2 will be as follows: Mass of H 2 O 2 = 30 % of mass Substitute 166.5 g for mass in the above equation as follows: Mass of H 2 O 2 = 30 % ( 166.5 g ) =49 .95 g Calculate the number of moles of H 2 O 2 ( n ) as follows: n = Mass of H 2 O 2 Molar mass of H 2 O 2 Substitute 49.95 g for mass and 34 g/mol for molar mass in the above equation as follows: n = 49.95 g 34 g/mol = 1.47 mol Calculate initial concentration ( A ∘ ) as follows: A ∘ = n V Substitute 1.47 mol for n and 150.0 mL for V in the above equation as follows: A ∘ = 1.47 mol 150.0 mL ( 1000 mL 1 L ) = 9.8 mol/L =9 .8 M The relation between initial concentration ( A ∘ ) and final concentration ( A ) will be as follows: A = A ∘ e − k t Here, k is rate constant and t is time. Substitute 9.8 M for A ∘ , 0.00752 s − 1 for k and 85.0 s for t in the above equation as follows: A = ( 9.8 M ) e − ( 0.00752 s − 1 ) ( 85.0 s ) = 5.1716 M Calculate remaining moles of H 2 O 2 ( n r ) as follows: n r = A V Here, A is final concentration and V is volume

Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

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ISBN: 9780321955517

Author: Nivaldo J. Tro

Publisher: PEARSON

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Chapter 13, Problem 92E

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To determine: The volume of the O2.

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Chapter 13, Problem 91E

Chapter 13, Problem 93E

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Chemistry: A Molecular Approach & Student Solutions Manual for Chemistry: A Molecular Approach, Books a la Carte Edition Package

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Chapter 13 Solutions