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Answers CHEM 101 Study Guide on Nuclear Chemistry & Balancing Redox Summer 2005

  1. Find the nuclear binding energy per mole for lithium-7 (at. mass = 7.01600 g/mol); chlorine-35 (at. mass = 34.95952 g/mol); and bismuth-209 (at. mass = 208.9804 g/mol). Then find the nuclear binding energy per atom, and finally, the nuclear binding energy per nucleon.

    total particle mass= 3(mass of a proton)+3(mass of electron)+4(mass of neutron)
    total particle mass = 3(1.67252x10-24g)+3(9.1095x10-28g)+4(1.67495x10-24g)
    total particle mass = 1.172 x 10-23 g/atom
    total particle mass per mole =(1.172 x 10-23g/atom)(6.022 x 1023 atoms/mol) total particle mass per mole = 7.05784 g/mol
    mass defect = 7.05784 g/mol - 7.01600 g/mol = 0.04184 g/mol
    mass defect in kg/mol = 0.04184 g/mol x 1 kg/1000 g = 4.184 x 10-5 kg/mol

    E = mc2, c = speed of light = 3.00 x 108 m/s
    E/mol = (4.184 x 10-5 kg/mol)(3.0 x 108m/s)2 = 3.766 x 1012 J/mol
    binding energy per atom = 6.254 x 10-12 J/atom (above divided by Avagadro's number, 6.022 x 1023)
    binding energy per nucleon = 8.934 x 10-13 J/nucleon (nuclear binding energy per atom divided by 7, the total number of protons and neutrons)

    All other problems are set up similarly

    For Cl-35 binding energy per mole = 2.951 x 1013 J/mol
    nuclear binding energy per atom = 4.900 x 10-11 J/atom
    nuclear binding energy per nucleon = 1.400 x 10-12 J/nucleon

    For Bi-209 binding energy per mole = 1.577 x 1014 J/mol
    nuclear binding energy per atom = 2.619 x 10-10 J/atom
    nuclear binding energy per nucleon = 1.25 x 10-12 J/nucleon

  2. The half-life of 239Pu is 24,000 yr. What fraction of the plutonium waste present today will have decayed in 1000 yrs?

    Number of half-lives = 1000/24,000 = 0.0416666
    Amount left = [(0.5)0.0416666 ]100% = 97.15%
    Amount decayed = 100.00% - 97.15 = 2.85%

  3. A 0.500 g sample of strontium-90 diminishes to 0.393 g in 10.0 yrs. What is the half-life of strontium-90?

    ln(C0/Ct) = kt
    k = [ln(C0/Ct)]/t = [ln(0.500g/0.393g)]/10.0y = 0.024079848 1/y
    t1/2 = ln 2/0.024079848 = 28.8 y

  4. Balance the nuclear equations below by supplying the missing reactant or product.

    1. 94Be + 42He ------------------> 10n + 126C

    2. 24395Am + 42He --------------->24697Bk + 10n

    3. 42He + 136C ---------------> 168O + 10n

    4. 136C --------------------> 42He + 84Be + 10n



  5. How many neutrons are in the nucleus of the isotopes below?

    1. 105B, 5 n

    2. 146C, 8 n

    3. 7935Br, 44 n

    4. 6529Cu, 36 n

    5. 20983Bi, 126 n

    6. 23592U, 143 n

    7. 23390Th, 143 n

    8. 4019K, 21 n

    9. 10847Ag, 61 n

    10. 9038Sr, 52 n

  6. Balance the redox equations below:

    1. Fe + Cl2 ------> FeCl3

      balance by inspection 2 Fe + 3 Cl2 -----> 2 FeCl3