Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
11th Edition
ISBN: 9781285869759
Author: Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher: Cengage Learning
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Textbook Question
Chapter 9.5, Problem 9.6P

Problem 9-6

A radioactive isotope in a 9.0-mL vial has an intensity of 300. mCi. A patient is required to take 50. mCi intravenously. How much liquid should be used for the injection?

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Chapter 9 Solutions

Introduction to General, Organic and Biochemistry

Ch. 9 - Prob. 9.11PCh. 9 - Prob. 9.12PCh. 9 - Prob. 9.13PCh. 9 - 9-14 Write the symbol for a nucleus with the...Ch. 9 - Prob. 9.15PCh. 9 - Prob. 9.16PCh. 9 - Prob. 9.17PCh. 9 - Prob. 9.18PCh. 9 - Prob. 9.19PCh. 9 - Prob. 9.20PCh. 9 - Prob. 9.21PCh. 9 - Prob. 9.22PCh. 9 - Prob. 9.23PCh. 9 - Prob. 9.24PCh. 9 - Prob. 9.25PCh. 9 - Prob. 9.26PCh. 9 - Prob. 9.27PCh. 9 - Prob. 9.28PCh. 9 - Prob. 9.29PCh. 9 - Prob. 9.30PCh. 9 - Prob. 9.31PCh. 9 - Prob. 9.32PCh. 9 - Prob. 9.33PCh. 9 - Prob. 9.34PCh. 9 - Prob. 9.35PCh. 9 - Prob. 9.36PCh. 9 - Prob. 9.37PCh. 9 - Prob. 9.38PCh. 9 - 9-39 If you work in a lab containing radioisotopes...Ch. 9 - Prob. 9.40PCh. 9 - Prob. 9.41PCh. 9 - Prob. 9.42PCh. 9 - Prob. 9.43PCh. 9 - Prob. 9.44PCh. 9 - Prob. 9.45PCh. 9 - Prob. 9.46PCh. 9 - Prob. 9.47PCh. 9 - Prob. 9.48PCh. 9 - Prob. 9.49PCh. 9 - Prob. 9.50PCh. 9 - Prob. 9.51PCh. 9 - Prob. 9.52PCh. 9 - Prob. 9.53PCh. 9 - Prob. 9.54PCh. 9 - Prob. 9.55PCh. 9 - Prob. 9.56PCh. 9 - Prob. 9.57PCh. 9 - Prob. 9.58PCh. 9 - Prob. 9.59PCh. 9 - Prob. 9.60PCh. 9 - Prob. 9.61PCh. 9 - Prob. 9.62PCh. 9 - Prob. 9.63PCh. 9 - Prob. 9.64PCh. 9 - Prob. 9.65PCh. 9 - Prob. 9.66PCh. 9 - Prob. 9.67PCh. 9 - Prob. 9.68PCh. 9 - Prob. 9.69PCh. 9 - Prob. 9.70PCh. 9 - Prob. 9.71PCh. 9 - Prob. 9.72PCh. 9 - Prob. 9.73PCh. 9 - Prob. 9.74PCh. 9 - Prob. 9.75PCh. 9 - Prob. 9.76PCh. 9 - Prob. 9.77PCh. 9 - Prob. 9.78PCh. 9 - Prob. 9.79PCh. 9 - Prob. 9.80PCh. 9 - Prob. 9.81PCh. 9 - Prob. 9.82PCh. 9 - Prob. 9.83PCh. 9 - Prob. 9.84PCh. 9 - Prob. 9.85PCh. 9 - 9-86 is effective in prostate cancer therapy when...Ch. 9 - Prob. 9.87PCh. 9 - Prob. 9.88PCh. 9 - Prob. 9.89PCh. 9 - Prob. 9.90PCh. 9 - Prob. 9.91PCh. 9 - Prob. 9.92PCh. 9 - Prob. 9.93P
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  • A worker receives a dose of 8.5 units of radiation at a distance of 5.0m from the source. What will the dose be at a distance of 5m? (Estimated Answer 100)
    A sealed flask contains saturated sodium iodide with excess solid at the bottom of the flask. If a sample of radioactively labelled NaI(s) is added to the system, the most likely observation after 24 hours will be Select one: A. neither the NaI(s) nor the only the NaI(aq) will register as radioactive B. both the NaI(s) and the NaI(aq) will register as radioactive C. only the NaI(aq) will register as radioactive D. only the NaI(s) will register as radioactive    Use the following information to answer the next question.     Crystals of magnesium phosphate and glucose, C6H12O6(s), were dissolved together in a beaker of distilled water.    All the entities that will be present when the two compounds are added to water are Select one: A. Mg2+(s), PO43-(s), C6H12O6(aq), H2O(l) B. Mg2+(aq), PO43-(aq), C6H12O6(s), H2O(l) C. Mg3(PO4)2(aq), C6H12O6(s), H2O(l) D. Mg3(PO4)2(s), C6H12O6(aq), H2O(l)
    How did you get this? k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs   0.693 a 102.0 nanogram(ng) sample of sodium-24 was stored in a lead-lined cabinet for 2.5 days How much sodium-24 remained? The half-life os sodium -24 is 15 hours    _________ ng sodium -24 remain  check_circle Expert Answer thumb_up   thumb_down Step 1 Radioactive decay is a first order reaction. Conversion of days into hours: 1 day2.5 days===24 hrs2.5×2460 hrs1 day=24 hrs2.5 days=2.5×24=60 hrs Calculation of k: k===0.693t1/20.69315 hrs0.0462 hrs−1k=0.693t1/2=0.69315 hrs=0.0462 hrs-1 Step 2 Radioactive decay is a first order reaction, therefore: [A]=[Ao] e−kt[A]=[Ao] e-kt where [A] is the final amount remained at time t, [Ao] is the initial amount. Putting values: [A]====102.0 ng×e−0.0462 hrs−1×60 hrs102.0 ng×e−2.772102.0 ng×0.062546.38 ng[A]=102.0 ng×e-0.0462 hrs-1×60 hrs=102.0 ng×e-2.772=102.0 ng×0.06254=6.38 ng Thus 6.38 ng of sodium-24 remained.
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