PHYSICAL CHEMISTRY-WEBASSIGN
11th Edition
ISBN: 9780357087411
Author: ATKINS
Publisher: CENGAGE L
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Question
Chapter 13, Problem 13C.9P
Interpretation Introduction
Interpretation:
The expression for the root mean square energy has to be stated in terms of partition function. The expression for root mean square for root mean square deviation has to be stated. The expression for root mean square for root mean square deviation for harmonic oscillator has to be stated.
Concept introduction:
Statistical
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Chapter 13 Solutions
PHYSICAL CHEMISTRY-WEBASSIGN
Ch. 13 - Prob. 13A.1STCh. 13 - Prob. 13B.1STCh. 13 - Prob. 13B.2STCh. 13 - Prob. 13C.1STCh. 13 - Prob. 13F.1STCh. 13 - Prob. 13F.3STCh. 13 - Prob. 13A.1DQCh. 13 - Prob. 13A.2DQCh. 13 - Prob. 13A.3DQCh. 13 - Prob. 13A.4DQ
Ch. 13 - Prob. 13A.1AECh. 13 - Prob. 13A.1BECh. 13 - Prob. 13A.2AECh. 13 - Prob. 13A.2BECh. 13 - Prob. 13A.3AECh. 13 - Prob. 13A.3BECh. 13 - Prob. 13A.4AECh. 13 - Prob. 13A.4BECh. 13 - Prob. 13A.5AECh. 13 - Prob. 13A.5BECh. 13 - Prob. 13A.6AECh. 13 - Prob. 13A.6BECh. 13 - Prob. 13A.1PCh. 13 - Prob. 13A.2PCh. 13 - Prob. 13A.4PCh. 13 - Prob. 13A.5PCh. 13 - Prob. 13A.6PCh. 13 - Prob. 13A.7PCh. 13 - Prob. 13B.1DQCh. 13 - Prob. 13B.2DQCh. 13 - Prob. 13B.3DQCh. 13 - Prob. 13B.1AECh. 13 - Prob. 13B.1BECh. 13 - Prob. 13B.2AECh. 13 - Prob. 13B.2BECh. 13 - Prob. 13B.3AECh. 13 - Prob. 13B.3BECh. 13 - Prob. 13B.4AECh. 13 - Prob. 13B.4BECh. 13 - Prob. 13B.7AECh. 13 - Prob. 13B.7BECh. 13 - Prob. 13B.8AECh. 13 - Prob. 13B.8BECh. 13 - Prob. 13B.9AECh. 13 - Prob. 13B.9BECh. 13 - Prob. 13B.10AECh. 13 - Prob. 13B.10BECh. 13 - Prob. 13B.11AECh. 13 - Prob. 13B.11BECh. 13 - Prob. 13B.12AECh. 13 - Prob. 13B.12BECh. 13 - Prob. 13B.4PCh. 13 - Prob. 13B.5PCh. 13 - Prob. 13B.6PCh. 13 - Prob. 13B.7PCh. 13 - Prob. 13B.8PCh. 13 - Prob. 13B.10PCh. 13 - Prob. 13C.1DQCh. 13 - Prob. 13C.2DQCh. 13 - Prob. 13C.1AECh. 13 - Prob. 13C.1BECh. 13 - Prob. 13C.6AECh. 13 - Prob. 13C.6BECh. 13 - Prob. 13C.7AECh. 13 - Prob. 13C.7BECh. 13 - Prob. 13C.3PCh. 13 - Prob. 13C.7PCh. 13 - Prob. 13C.8PCh. 13 - Prob. 13C.9PCh. 13 - Prob. 13D.1DQCh. 13 - Prob. 13D.2DQCh. 13 - Prob. 13D.3DQCh. 13 - Prob. 13D.4DQCh. 13 - Prob. 13D.1AECh. 13 - Prob. 13D.1BECh. 13 - Prob. 13D.1PCh. 13 - Prob. 13D.2PCh. 13 - Prob. 13E.1DQCh. 13 - Prob. 13E.2DQCh. 13 - Prob. 13E.3DQCh. 13 - Prob. 13E.4DQCh. 13 - Prob. 13E.5DQCh. 13 - Prob. 13E.6DQCh. 13 - Prob. 13E.1AECh. 13 - Prob. 13E.1BECh. 13 - Prob. 13E.2AECh. 13 - Prob. 13E.2BECh. 13 - Prob. 13E.3AECh. 13 - Prob. 13E.3BECh. 13 - Prob. 13E.4AECh. 13 - Prob. 13E.4BECh. 13 - Prob. 13E.5AECh. 13 - Prob. 13E.5BECh. 13 - Prob. 13E.6AECh. 13 - Prob. 13E.6BECh. 13 - Prob. 13E.7AECh. 13 - Prob. 13E.7BECh. 13 - Prob. 13E.8AECh. 13 - Prob. 13E.8BECh. 13 - Prob. 13E.9AECh. 13 - Prob. 13E.9BECh. 13 - Prob. 13E.1PCh. 13 - Prob. 13E.2PCh. 13 - Prob. 13E.3PCh. 13 - Prob. 13E.4PCh. 13 - Prob. 13E.7PCh. 13 - Prob. 13E.9PCh. 13 - Prob. 13E.10PCh. 13 - Prob. 13E.11PCh. 13 - Prob. 13E.14PCh. 13 - Prob. 13E.15PCh. 13 - Prob. 13E.16PCh. 13 - Prob. 13E.17PCh. 13 - Prob. 13F.1DQCh. 13 - Prob. 13F.2DQCh. 13 - Prob. 13F.3DQCh. 13 - Prob. 13F.1AECh. 13 - Prob. 13F.1BECh. 13 - Prob. 13F.2AECh. 13 - Prob. 13F.2BECh. 13 - Prob. 13F.3AECh. 13 - Prob. 13F.3BECh. 13 - Prob. 13F.3PCh. 13 - Prob. 13F.4PCh. 13 - Prob. 13F.5PCh. 13 - Prob. 13F.6PCh. 13 - Prob. 13.1IA
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- Part II. two unbranched ketone have molecular formulla (C8H100). El-ms showed that both of them have a molecular ion peak at m/2 =128. However ketone (A) has a fragment peak at m/2 = 99 and 72 while ketone (B) snowed a fragment peak at m/2 = 113 and 58. 9) Propose the most plausible structures for both ketones b) Explain how you arrived at your conclusion by drawing the Structures of the distinguishing fragments for each ketone, including their fragmentation mechanisms.arrow_forwardPart V. Draw the structure of compound tecla using the IR spectrum Cobtained from the compound in KBr pellet) and the mass spectrum as shown below. The mass spectrum of compound Tesla showed strong mt peak at 71. TRANSMITTANCE LOD Relative Intensity 100 MS-NW-1539 40 20 80 T 44 55 10 15 20 25 30 35 40 45 50 55 60 65 70 75 m/z D 4000 3000 2000 1500 1000 500 HAVENUMBERI-11arrow_forwardTechnetium is the first element in the periodic chart that does not have any stable isotopes. Technetium-99m is an especially interesting and valuable isotope as it emits a gamma ray with a half life ideally suited for medical tests. It would seem that the decay of technetium should fit the treatment above with the result In(c/c) = -kt. The table below includes data from the two sites: http://dailymed.nlm.nih.gov/dailymed/druginfo.cfm?id=7130 http://wiki.medpedia.com/Clinical: Neutrospec_(Technetium_(99m Tc)_fanolesomab). a. b. C. Graph the fraction (c/c.) on the vertical axis versus the time on the horizontal axis. Also graph In(c/c.) on the vertical axis versus time on the horizontal axis. When half of the original amount of starting material has hours fraction remaining disappeared, c/c = ½ and the equation In(c/c.) = -kt becomes In(0.5) = -kt1/2 where t₁₂ is the half life (the time for half of the material to decay away). Determine the slope of your In(c/c.) vs t graph and…arrow_forward
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