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owLV2 | Online teachinc X FCNot secure sjc.cengagenow.com/ilrn/takeAssignment/takeCovalentActivity.do?locator assignment-take&takeAssignmentSessionLocator-assignment-take; AppsCanvas : Cengagerow on MyMathLab Pearso A webadvisor @ Pearson REVEL D English-101c a Rentals Cart D Formularna me searcWolfram Alpha 2. Nuclear Reaction: Identity Decay R...3 pts M Some Radioactive Isotopes Useful in Medical Imaging Mode of Decay Hf-life Use in Medical Imaging 3. Nuclear Reaction: Predict Decay Pr.. 3 ptsM Isotope 20.3 mrai scan to trace glucose metabolism 109 m 14.3 d 27.7 d 44.5 d 78.3 h 118d 13.3 Lung ventilation scan 22.2 m Sca for bone diseases, including cancer 6.01 h ri ver, kidney, bone scans; diagnosis of damaged heart muscle 8.04 d 641 h 3.05 d Carbon-11 Fluorine-18 Phosphorus-32 4. Half-Life Calculations 4 pts Brain scan to trace glucose metabolism Detect eye tumors Diagnose albinism, image the spleen and gastrointestinal tract Bone marrow function, diagnose anemias whole-body scan for tumors Pancreas scan 5. Radionuclides in Medical Imaging 2 pts 2req Cr Chromium-51 Iron-59 Gallium-67 Selenium-78 rypton-81m EC..Y , EC., E.C., Question Strontium-81 5 pts Sr XTe 6. Binding Energy Technetium-99m Iodine-131 , EC..y EC., Diagnosis of thyroid malfunction Hg Mercury-197 T Thallium-201 Kidney scan Heart scan and exercise stress test The radioactive isotope mercury-197 is used in medical imaging as indicated on the table above. If 68.2 milligrams of mercury-197 is administered to a patient, how many milligrams are left in the body after 128.2 hours? mg Submit Answer Retry Entire Group 5 more group attempts remaining Progress 4/6 groups Due Dec 9 at 11:00 PM Previous ext 1:29 PM 12/8/2017 Type here to searchExplanation / Answer
we have:
Half life = 64.1 h
use relation between rate constant and half life of 1st order reaction
k = (ln 2) / k
= 0.693/(half life)
= 0.693/(64.1)
= 1.081*10^-2 h-1
we have:
[Hg]o = 68.2 mg
t = 128.2 h
k = 1.081*10^-2 h-1
use integrated rate law for 1st order reaction
ln[Hg] = ln[Hg]o - k*t
ln[Hg] = ln(68.2) - 1.081*10^-2*128.2
ln[Hg] = 4.2224 - 1.081*10^-2*128.2
ln[Hg] = 2.8364
[Hg] = 17.05 mg
Answer: 17.05 mg
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