Explanation Explanation The total counts of radioactivity is calculated by the formula, Total counts = Volume of the sample × Counts per minute per millilitre (1) The counts per minute per millilitre is 5 × 10 3 cpm/mL . Volume of the sample is 0.10 cm 3 . The conversion of cm 3 to mL is done as, 1 cm 3 = 1 mL Therefore, the conversion of 0.10 cm 3 into mL is, 0.10 cm 3 = 0.10 × 1 mL = 0.10 mL Hence, the volume of the sample is 0 .10 mL . Substitute the value of volume of sample and counts per minute per milliliter in the above equation

9th Edition

ISBN: 9781133611097

Author: Steven S. Zumdahl

Publisher: Cengage Learning

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Question

Chapter 19, Problem 81CP

Interpretation Introduction

Interpretation: Injection of a sample solution of a radioactive nuclide in a rat and its counts per minute per milliliter is given. The removal of 1 cm3 blood after several minutes is given. The volume of blood in the rat is to be stated.

Concept introduction: A process through which an unstable nuclide looses its energy due to excess of protons or neutrons is known as radioactive decay. Counts per minute rate determine the decay events per minute.

To determine: The volume of the blood in the rat.

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Chapter 19 Solutions

Chemistry

Ch. 19 - Prob. 1RQ Ch. 19 - Prob. 2RQ Ch. 19 - Prob. 3RQ Ch. 19 - Prob. 4RQ Ch. 19 - Prob. 5RQ Ch. 19 - Prob. 6RQ Ch. 19 - Prob. 7RQ Ch. 19 - Prob. 8RQ Ch. 19 - Prob. 9RQ Ch. 19 - Prob. 10RQ

Ch. 19 - Prob. 1Q Ch. 19 - Prob. 2Q Ch. 19 - Prob. 3Q Ch. 19 - Prob. 4Q Ch. 19 - Prob. 5Q Ch. 19 - Prob. 6Q Ch. 19 - Prob. 7Q Ch. 19 - Prob. 8Q Ch. 19 - Prob. 9Q Ch. 19 - Prob. 10Q Ch. 19 - Prob. 11E Ch. 19 - Prob. 12E Ch. 19 - Prob. 13E Ch. 19 - Prob. 14E Ch. 19 - Prob. 15E Ch. 19 - Prob. 16E Ch. 19 - Prob. 17E Ch. 19 - Prob. 18E Ch. 19 - Prob. 19E Ch. 19 - Prob. 20E Ch. 19 - Prob. 21E Ch. 19 - Prob. 23E Ch. 19 - Prob. 24E Ch. 19 - Prob. 26E Ch. 19 - Prob. 27E Ch. 19 - Prob. 28E Ch. 19 - Prob. 29E Ch. 19 - Prob. 30E Ch. 19 - Prob. 31E Ch. 19 - Prob. 32E Ch. 19 - Prob. 33E Ch. 19 - Prob. 34E Ch. 19 - Prob. 35E Ch. 19 - Prob. 36E Ch. 19 - Prob. 37E Ch. 19 - Prob. 38E Ch. 19 - Prob. 39E Ch. 19 - Prob. 40E Ch. 19 - Prob. 41E Ch. 19 - Prob. 42E Ch. 19 - Prob. 43E Ch. 19 - Prob. 44E Ch. 19 - Prob. 46E Ch. 19 - Prob. 47E Ch. 19 - Prob. 48E Ch. 19 - Prob. 49E Ch. 19 - Prob. 50E Ch. 19 - Prob. 51E Ch. 19 - Prob. 52E Ch. 19 - Prob. 53E Ch. 19 - Prob. 54E Ch. 19 - Prob. 55E Ch. 19 - Prob. 56E Ch. 19 - Prob. 57E Ch. 19 - Prob. 58E Ch. 19 - Prob. 59AE Ch. 19 - Prob. 60AE Ch. 19 - Prob. 61AE Ch. 19 - Prob. 62AE Ch. 19 - Prob. 63AE Ch. 19 - Prob. 64AE Ch. 19 - Prob. 65AE Ch. 19 - Prob. 66AE Ch. 19 - Prob. 67AE Ch. 19 - Prob. 68AE Ch. 19 - Prob. 69AE Ch. 19 - Prob. 70AE Ch. 19 - Prob. 71AE Ch. 19 - Prob. 72AE Ch. 19 - Prob. 73CWP Ch. 19 - Prob. 74CWP Ch. 19 - Prob. 75CWP Ch. 19 - Prob. 76CWP Ch. 19 - Prob. 77CWP Ch. 19 - Prob. 78CWP Ch. 19 - Prob. 79CP Ch. 19 - Prob. 80CP Ch. 19 - Prob. 81CP Ch. 19 - Prob. 82CP Ch. 19 - Prob. 83CP Ch. 19 - Prob. 84CP Ch. 19 - Prob. 85CP Ch. 19 - Prob. 86CP Ch. 19 - Prob. 87IP Ch. 19 - Prob. 88IP

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