University:
Glendale Community College
Course:
CHM152 General Chemistry II
Academic year:

2020
Views:

81

Pages:

2
Author:

onrijp69

CHM152LL: Nuclear Chemistry Practice Worksheet Key Exercise 1: Complete the following table: Isotope Mass Number # of Protons # of Neutrons # of Electrons Strontium - 90 90 38 52 38 222 86 136 86 222 Rn Exercise 1: Identify the unknown element in each to complete the following nuclear equations: a. Argon-40 is bombarded with a proton. 40 18 Ar  H  1 1 n 1 0  _____ 40 19 K b. Uranium-238 has a half-life of about 4.5 billion years and is used to date very old rocks. 238 92 U   6 0 -1 e  8 24  _____ 206 82 Pb Exercise 2: Write complete nuclear equations for the following processes: a. Uranium-234 is produced when a radioactive isotope undergoes alpha decay. 238 94 Pu 42 α  234 92 U b. Cobalt-60 is produced when a radioactive isotope undergoes beta decay. 60 26 Fe  01β  60 27 Co Exercise 3: The inhalation of radon-222 and its decay to form other isotopes poses a health hazard. Write balanced nuclear equations for the decay of radon-222 to lead-206 in eight steps. a. 222 86 Step 1: Radon-222 decays by alpha emission. (Radon has the element symbol Rn.) Rn  42 α  218 84 Po b. Step 2: The daughter product in part a decays by alpha emission. 218 84 c. 214 82 Po  42 α  214 82 Pb Step 3: The daughter product in part b decays by beta and gamma emissions. Pb  00  -01β  214 83 Bi d. Step 4: The daughter product in part c decays by beta and gamma emissions. 214 83 Bi  00  -01β  214 84 Po p. 1 of 2 e. 214 84 Step 5: The daughter product in part d decays by beta emission. Po  -01β  f. 214 85 At Step 6: The daughter product in part e decays by alpha emission. At  42 α  g. 210 83 214 85 210 83 Bi Step 7: The daughter product in part f decays by beta and gamma emissions. Bi  00  -01β  210 84 Po h. Step 8: The daughter product in part g decays by alpha and gamma emissions. 210 84 Po  00  42  206 82 Pb The final stable isotope is lead-206. III. Half-Life and the Amount of Sample Left Exercise 1: a. What is the half-life for Thorium-230? _8000 y__ b. How many half-lives have passed for Thorium-230 after 32,000 years? __4__ half-lives (HL) c. How much of an 95.6-mg sample of Thorium-230 would remain after 32,000 years? 95.6 mg ÷ 2 ÷ 2 ÷ 2 ÷ 2 = 5.98 mg or Amount remaining = initial amount x (0.5)n = 95.6 mg x (0.5)4 = 5.98 mg d. What is the value of the rate constant, k, for Th-230? k = 0.693 / 8000 y = 8.66 x 10-5 y-1 e. How much of a 35.8-mg sample of Thorium-230 would remain after 18500 years? ln [At /35.8] = -(8.66 x 10-5 y-1)*18500 y [At /35.8] = e -1.60256 = 0.2014 At = (0.2014)(35.8) = 7.21 mg f. How much time will it take for a 280 mg sample of Thorium-230 to decay to 11.0 mg? ln [11.0 / 280] = (-8.66 x 10-5 y-1)t t = (-3.237)/(-8.66 x 10-5 y-1)= 37378 yrs = 3.7x104 yrs p. 2 of 2