Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces: 一+ rate = 4.85 X 10-2 dayP What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M O Rate = 4.85x 102 day/[0.0033 M] = 15 mol/L/d O Rate 4.85 x 10² day{0.0033 M? = 5.3× 107 mol/L/d O Rate = 4.85 x 10 day"{0.0033 M] = 1.6 × 10 mol/L/d
Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces: 一+ rate = 4.85 X 10-2 dayP What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M O Rate = 4.85x 102 day/[0.0033 M] = 15 mol/L/d O Rate 4.85 x 10² day{0.0033 M? = 5.3× 107 mol/L/d O Rate = 4.85 x 10 day"{0.0033 M] = 1.6 × 10 mol/L/d
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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![Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are
components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is
bound in) can be detected from the electrons (beta particles) it produces:
即→+6
rate= 4.85 × 10-2 day-"P
day" [*P]
15
10
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M
O Rate = 4.85 x 102 day/(0.0033 M| = 15 mol/L/d
%3D
O Rate 4.85x 102 day (0.0033 M? = 5.3 × 107 mol/L/d
%3D
O Rate = 4.85x 10 day(0.0033 M] = 1.6 x 10 mol/L/d](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F21602a64-d68c-457a-9689-0f6f322dc5be%2F5bcb44a6-d5f0-4019-99f8-cf4d298155a8%2Fk33yapf_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are
components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is
bound in) can be detected from the electrons (beta particles) it produces:
即→+6
rate= 4.85 × 10-2 day-"P
day" [*P]
15
10
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M
O Rate = 4.85 x 102 day/(0.0033 M| = 15 mol/L/d
%3D
O Rate 4.85x 102 day (0.0033 M? = 5.3 × 107 mol/L/d
%3D
O Rate = 4.85x 10 day(0.0033 M] = 1.6 x 10 mol/L/d
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