Chemical equation used: 2 MnO4 + 5 H2O2 + 6 H* 2 Mn2+ + 5 O2 + 8 H2O [MnO4] = 0.0090 M Volume H2O2 used: 1.0 mL Volume H2SO4 used: 7.0 mL Trial 1: 41.9 mL MnO4 used Trial 2: 40.4 mL used Trial 3: 40.01 mL used Remember you want to find the M (mol/L) of H2O2. You must use stoichiometry to find moles of H2O2 and the volume of hydrogen peroxide used to get liters. Calculate the concentration for each trial and average the results.
Chemical equation used: 2 MnO4 + 5 H2O2 + 6 H* 2 Mn2+ + 5 O2 + 8 H2O [MnO4] = 0.0090 M Volume H2O2 used: 1.0 mL Volume H2SO4 used: 7.0 mL Trial 1: 41.9 mL MnO4 used Trial 2: 40.4 mL used Trial 3: 40.01 mL used Remember you want to find the M (mol/L) of H2O2. You must use stoichiometry to find moles of H2O2 and the volume of hydrogen peroxide used to get liters. Calculate the concentration for each trial and average the results.
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![Decomposition of H202 Data (1) - Word
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12 A A Aa
、 |A|三▼三▼i、|EE|2↓T
Po
lal
AaBbCcD AABBCCD AaBbC AABBCCC Aa
I U - ab x, x A A-
|■===|=。12-田
T No Spac.. Heading 1
T Normal
Heading 2
Titl
Font
Paragraph
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Chemical equation used:
2 Mn04 + 5 H2O2 + 6 H* →2 Mn2+ + 5 02 + 8 H2O
[MnO4] = 0.0090 M
Volume H2O2 used: 1.0 mL
Volume H2S04 used: 7.0 mL
Trial 1: 41.9 mL MnO4 used
Trial 2: 40.4 mL used
Trial 3: 40.01 mL used
Remember you want to find the M (mol/L) of H2O2. You must use stoichiometry to find
moles of H2O2 and the volume of hydrogen peroxide used to get liters. Calculate the
concentration for each trial and average the results.
Thar nocho mictrv](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc052376c-7813-42b5-b3df-5a028cf07b80%2F691c37d1-d697-452a-aae8-50fca3abd7f2%2F9jj38o_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Decomposition of H202 Data (1) - Word
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12 A A Aa
、 |A|三▼三▼i、|EE|2↓T
Po
lal
AaBbCcD AABBCCD AaBbC AABBCCC Aa
I U - ab x, x A A-
|■===|=。12-田
T No Spac.. Heading 1
T Normal
Heading 2
Titl
Font
Paragraph
Styles
Chemical equation used:
2 Mn04 + 5 H2O2 + 6 H* →2 Mn2+ + 5 02 + 8 H2O
[MnO4] = 0.0090 M
Volume H2O2 used: 1.0 mL
Volume H2S04 used: 7.0 mL
Trial 1: 41.9 mL MnO4 used
Trial 2: 40.4 mL used
Trial 3: 40.01 mL used
Remember you want to find the M (mol/L) of H2O2. You must use stoichiometry to find
moles of H2O2 and the volume of hydrogen peroxide used to get liters. Calculate the
concentration for each trial and average the results.
Thar nocho mictrv
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