INTO TO CHEMISTRY (EBOOK ACCESS CODE)
5th Edition
ISBN: 9781307892864
Author: BAUER
Publisher: MCG
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Chapter 15, Problem 56QP
Interpretation Introduction
Interpretation:
The time taken to decrease the radiation intensity by
Concept Introduction:
The half-life of a nuclide is always the same irrespective of the amount of nuclides, temperature, or any other conditions. During each half-life, one-half of the nuclides present at the beginning of the time period undergo decay.
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2CIO2 + 20H-1 CIO31 + CIO2 + H2O
Experiment
[CIO2], M
[OH-1], M
1
0.0500
0.100
23
2
0.100
0.100
3
0.100
0.0500
Initial Rate, M/s
0.0575
0.230
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...
Given this date, calculate the overall order of this reaction.
2
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.(be)_[Ɔ+(be)_OI ← (b²)_IƆO+ (be)_I
Experiment
[1-] M
0.005
[OCI-]
0.005
Initial Rate M/min
0.000275
0.0025
0.005
0.000138
0.0025
0.0025
0.000069
4
0.0025
0.0025
0.000140
Calculate the rate constant of this reaction using the table data.
1
2
3
4
I(aq) +OCl(aq) → IO¯¯(aq) + Cl¯(aq)
Experiment
[I-] M
0.005
[OCI-]
0.005
Initial Rate M/min
0.000275
0.0025
0.005
0.000138
0.0025
0.0025
Calculate the overall order of this reaction using the table data.
0.0025
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Chapter 15 Solutions
INTO TO CHEMISTRY (EBOOK ACCESS CODE)
Ch. 15 - Prob. 1QCCh. 15 - Prob. 2QCCh. 15 - Prob. 3QCCh. 15 - Prob. 4QCCh. 15 - Prob. 5QCCh. 15 - Prob. 6QCCh. 15 - Prob. 1PPCh. 15 - Prob. 2PPCh. 15 - Prob. 3PPCh. 15 - Prob. 4PP
Ch. 15 - Prob. 5PPCh. 15 - Prob. 6PPCh. 15 - Prob. 7PPCh. 15 - Prob. 8PPCh. 15 - Prob. 9PPCh. 15 - Prob. 10PPCh. 15 - Prob. 11PPCh. 15 - Prob. 1QPCh. 15 - Prob. 2QPCh. 15 - Prob. 3QPCh. 15 - Prob. 4QPCh. 15 - Prob. 5QPCh. 15 - Prob. 6QPCh. 15 - Prob. 7QPCh. 15 - Prob. 8QPCh. 15 - Prob. 9QPCh. 15 - Prob. 10QPCh. 15 - Prob. 11QPCh. 15 - Prob. 12QPCh. 15 - Prob. 13QPCh. 15 - Prob. 14QPCh. 15 - Prob. 15QPCh. 15 - Prob. 16QPCh. 15 - Prob. 17QPCh. 15 - Prob. 18QPCh. 15 - Prob. 19QPCh. 15 - Prob. 20QPCh. 15 - Prob. 21QPCh. 15 - Prob. 22QPCh. 15 - Prob. 23QPCh. 15 - Prob. 24QPCh. 15 - Prob. 25QPCh. 15 - Prob. 26QPCh. 15 - Prob. 27QPCh. 15 - Prob. 28QPCh. 15 - Prob. 29QPCh. 15 - Prob. 30QPCh. 15 - Prob. 31QPCh. 15 - Prob. 32QPCh. 15 - Prob. 33QPCh. 15 - Prob. 34QPCh. 15 - Prob. 35QPCh. 15 - Prob. 36QPCh. 15 - Prob. 37QPCh. 15 - Prob. 38QPCh. 15 - Prob. 39QPCh. 15 - Prob. 40QPCh. 15 - Prob. 41QPCh. 15 - Prob. 42QPCh. 15 - Prob. 43QPCh. 15 - Prob. 44QPCh. 15 - Prob. 45QPCh. 15 - Prob. 46QPCh. 15 - Prob. 47QPCh. 15 - Prob. 48QPCh. 15 - Prob. 49QPCh. 15 - Prob. 50QPCh. 15 - Prob. 51QPCh. 15 - Prob. 52QPCh. 15 - Prob. 53QPCh. 15 - Prob. 54QPCh. 15 - Prob. 55QPCh. 15 - Prob. 56QPCh. 15 - Prob. 57QPCh. 15 - Prob. 58QPCh. 15 - Prob. 59QPCh. 15 - Prob. 60QPCh. 15 - Prob. 61QPCh. 15 - Prob. 62QPCh. 15 - Prob. 63QPCh. 15 - Prob. 64QPCh. 15 - Prob. 65QPCh. 15 - Prob. 66QPCh. 15 - Prob. 67QPCh. 15 - Prob. 68QPCh. 15 - Prob. 69QPCh. 15 - Prob. 70QPCh. 15 - Prob. 73QPCh. 15 - Prob. 74QPCh. 15 - Prob. 75QPCh. 15 - Prob. 76QPCh. 15 - Prob. 77QPCh. 15 - Prob. 78QPCh. 15 - Prob. 79QPCh. 15 - Prob. 80QPCh. 15 - Prob. 81QPCh. 15 - Prob. 82QPCh. 15 - Prob. 83QPCh. 15 - Prob. 84QPCh. 15 - Prob. 85QPCh. 15 - Prob. 86QPCh. 15 - Prob. 87QPCh. 15 - Prob. 88QPCh. 15 - Prob. 89QPCh. 15 - Prob. 90QPCh. 15 - Prob. 91QPCh. 15 - Prob. 92QPCh. 15 - Prob. 93QPCh. 15 - Prob. 94QPCh. 15 - Prob. 95QPCh. 15 - Prob. 96QPCh. 15 - Prob. 97QPCh. 15 - Prob. 98QPCh. 15 - Prob. 99QPCh. 15 - Prob. 100QPCh. 15 - Prob. 101QPCh. 15 - Prob. 102QPCh. 15 - Prob. 103QPCh. 15 - Prob. 104QPCh. 15 - Prob. 105QPCh. 15 - Prob. 106QPCh. 15 - Prob. 107QPCh. 15 - Prob. 108QPCh. 15 - Prob. 109QPCh. 15 - Prob. 110QPCh. 15 - Prob. 111QPCh. 15 - Prob. 112QPCh. 15 - Prob. 113QPCh. 15 - Prob. 114QPCh. 15 - Prob. 115QPCh. 15 - Prob. 116QPCh. 15 - Prob. 117QPCh. 15 - Prob. 118QPCh. 15 - Prob. 119QPCh. 15 - Prob. 120QPCh. 15 - Prob. 121QPCh. 15 - Prob. 122QPCh. 15 - Prob. 123QP
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- H2O2(aq) +3 I¯(aq) +2 H+(aq) → 13(aq) +2 H₂O(l)· ••• Experiment [H2 O2]o (M) [I]o (M) [H+]。 (M) Initial rate (M/s) 1 0.15 0.15 0.05 0.00012 234 0.15 0.3 0.05 0.00024 0.3 0.15 0.05 0.00024 0.15 0.15 0.1 0.00048 Calculate the overall order of this reaction using the table data.arrow_forwardThe U. S. Environmental Protection Agency (EPA) sets limits on healthful levels of air pollutants. The maximum level that the EPA considers safe for lead air pollution is 1.5 μg/m³ Part A If your lungs were filled with air containing this level of lead, how many lead atoms would be in your lungs? (Assume a total lung volume of 5.40 L.) ΜΕ ΑΣΦ = 2.35 1013 ? atoms ! Check your rounding. Your final answer should be rounded to 2 significant figures in the last step. No credit lost. Try again.arrow_forwardY= - 0.039 (14.01) + 0.7949arrow_forward
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