OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
5th Edition
ISBN: 9781285460369
Author: STANITSKI
Publisher: Cengage Learning
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Question
Chapter 18.4, Problem 18.7PSP
Interpretation Introduction
Interpretation:
Age of sealed sample of Scotch whiskey has to be estimated if it contains tritium content 0.60 time that of water.
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The amount of 14C in an old, isolated sample of vegetation is 0.770 that expected from a present day sample with an equivalent carbon content. Calculate the age of the old sample in years.
Scientists can determine the age of ancient objects by the method of radiocarbon dating. The bombardment of the upper atmosphere by cosmic rays converts nitrogen to radioactive isotope of carbon, 14C, with
a half-life of about 5730 years. Vegetation absorbs carbon dioxide through the atmosphere and animal life assimilates 14C through food chains. When a plant or animal dies, it stops replacing its carbon and the
amount of ¹4C begins to decrease through radioactive decay. Therefore the level of radioactivity must also decay exponentially. A parchment fragment was discovered that had about 74% as much 14C
radioactivity as does plant material on earth today. Estimate the age of the parchment. (Round your answer to the nearest whole number.)
yr
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The age of water samples or aqueous solutions can be determined by
determining the radioactive tritium. Calculate the age of the cognac sample that
had 10 times less radioactivity than the freshly prepared sample. The half-life of
tritium is 12.5 years.
Chapter 18 Solutions
OWLV2 FOR MOORE/STANITSKI'S CHEMISTRY:
Ch. 18.2 - Prob. 18.1PSPCh. 18.2 - Prob. 18.1ECh. 18.2 - Prob. 18.2PSPCh. 18.2 - Prob. 18.2ECh. 18.3 - Prob. 18.3PSPCh. 18.3 - Prob. 18.3ECh. 18.3 - Prob. 18.4CECh. 18.4 - Prob. 18.4PSPCh. 18.4 - Prob. 18.5ECh. 18.4 - Prob. 18.5PSP
Ch. 18.4 - Prob. 18.6PSPCh. 18.4 - Prob. 18.7PSPCh. 18.4 - Prob. 18.6ECh. 18.4 - Prob. 18.7CECh. 18.5 - Prob. 18.8ECh. 18.5 - Prob. 18.9CECh. 18.6 - Prob. 18.10ECh. 18.6 - Prob. 18.11ECh. 18.7 - Prob. 18.12ECh. 18.8 - Prob. 18.13ECh. 18.8 - Prob. 18.14ECh. 18.9 - Prob. 18.15ECh. 18 - Prob. 1SPCh. 18 - Prob. 2SPCh. 18 - Prob. 3SPCh. 18 - Prob. 4SPCh. 18 - Prob. 5SPCh. 18 - Prob. 1QRTCh. 18 - Prob. 2QRTCh. 18 - Prob. 3QRTCh. 18 - Prob. 4QRTCh. 18 - Prob. 5QRTCh. 18 - Prob. 6QRTCh. 18 - Prob. 7QRTCh. 18 - Prob. 8QRTCh. 18 - Prob. 9QRTCh. 18 - Complete the table.Ch. 18 - Prob. 11QRTCh. 18 - Prob. 12QRTCh. 18 - Prob. 13QRTCh. 18 - Prob. 14QRTCh. 18 - Prob. 15QRTCh. 18 - Prob. 16QRTCh. 18 - Prob. 17QRTCh. 18 - Prob. 18QRTCh. 18 - Prob. 19QRTCh. 18 - Prob. 20QRTCh. 18 - Prob. 21QRTCh. 18 - Prob. 22QRTCh. 18 - Prob. 23QRTCh. 18 - Prob. 24QRTCh. 18 - Prob. 25QRTCh. 18 - Prob. 26QRTCh. 18 - Prob. 27QRTCh. 18 - Prob. 28QRTCh. 18 - Prob. 29QRTCh. 18 - Prob. 30QRTCh. 18 - Prob. 31QRTCh. 18 - Prob. 32QRTCh. 18 - Prob. 33QRTCh. 18 - Prob. 34QRTCh. 18 - Prob. 35QRTCh. 18 - Prob. 36QRTCh. 18 - Prob. 37QRTCh. 18 - Prob. 38QRTCh. 18 - Prob. 39QRTCh. 18 - Prob. 40QRTCh. 18 - Prob. 41QRTCh. 18 - Prob. 42QRTCh. 18 - Prob. 43QRTCh. 18 - Prob. 44QRTCh. 18 - Prob. 45QRTCh. 18 - Prob. 46QRTCh. 18 - Prob. 47QRTCh. 18 - Prob. 48QRTCh. 18 - Prob. 49QRTCh. 18 - Prob. 50QRTCh. 18 - Prob. 51QRTCh. 18 - Prob. 52QRTCh. 18 - Prob. 53QRTCh. 18 - Prob. 54QRTCh. 18 - Prob. 55QRTCh. 18 - Prob. 56QRTCh. 18 - Prob. 57QRTCh. 18 - Prob. 58QRTCh. 18 - Prob. 59QRTCh. 18 - Prob. 60QRTCh. 18 - Prob. 61QRTCh. 18 - Prob. 62QRTCh. 18 - Prob. 63QRTCh. 18 - Prob. 64QRTCh. 18 - Prob. 65QRTCh. 18 - Prob. 66QRTCh. 18 - Prob. 67QRTCh. 18 - Prob. 68QRTCh. 18 - Prob. 69QRTCh. 18 - Prob. 70QRTCh. 18 - Prob. 71QRTCh. 18 - Prob. 72QRTCh. 18 - Prob. 73QRTCh. 18 - Prob. 74QRTCh. 18 - Prob. 75QRTCh. 18 - Prob. 76QRTCh. 18 - Prob. 77QRTCh. 18 - Prob. 78QRTCh. 18 - Prob. 79QRTCh. 18 - Prob. 80QRTCh. 18 - Prob. 81QRTCh. 18 - Prob. 82QRTCh. 18 - Prob. 83QRTCh. 18 - Prob. 84QRTCh. 18 - Prob. 85QRTCh. 18 - Prob. 86QRTCh. 18 - Prob. 87QRTCh. 18 - Prob. 88QRTCh. 18 - Prob. 89QRTCh. 18 - Prob. 91QRTCh. 18 - Prob. 92QRTCh. 18 - Prob. 93QRTCh. 18 - Prob. 94QRTCh. 18 - Prob. 95QRTCh. 18 - Prob. 96QRTCh. 18 - Prob. 18.ACPCh. 18 - Prob. 18.BCPCh. 18 - Prob. 18.CCPCh. 18 - Prob. 18.DCPCh. 18 - Prob. 18.ECP
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Similar questions
- Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of 109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is F918O188+e+10 Physicians use 18F to study the brain by injecting a quantity of ?uoro-substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment. (a) What is the rate constant for [lie decomposition of ?uorine-18? (b) If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h? (c) How long does it take for 99.99% of the 18F to decay?arrow_forwardCarbon-14 (C-14) with a half-life of 5730 years decays to nitrogen-14 (N-14). A sample of carbon dioxide containing carbon in the form of C-14 only is sealed in a vessel at 1.00-atmosphere pressure. Over time, the CO2 becomes NO2 through the radioactive decay process. The following equilibrium is established: 2NO2(g)N2O4(g) If the equilibrium constant for the equation as written is 1105 , what is the pressure of N2O4 after 20,000 years? Assume that the CO2 does not participate in any chemical reactions.arrow_forward
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