Unit 1 Honors Assignment Key

Honors Chemistry Name: ______________________per:_____ KEYHonors Chemistry Assignment Sheet- Unit 1 Learning Objectives: Assign #1: perform various calculations related to the percent abundance of each isotope for a given element. Assign #2: Perform calculations involving ½ lives. Assignments Due the day of Unit 1 Test Use your textbook & reliable internet resources to learn material Answer Keys are posted on the WHS chemistry website (see “important handouts”) Honors Assign # 1- % Abundance (show work) 1. Nitrogen is made up of two isotopes, N-14 (mass of 14.003074 amu) and N-15 (15.000108 amu) . Given nitrogen's average atomic weight of 14.007amu, what is the percent abundance of each isotope? 14.003074 (x) + 15.000108 (1-x) = 14.007 14.003074 x + 15.000108 - 15.000108 x= 14.007 Please let use know if you find any errors on this key  X= .997034 N-14: 99.6% N-15: 0.394% 2. Copper is made up of two isotopes, Cu-63 (62.9296 amu) and Cu-65 (64.9278 amu). Given copper's atomic weight of 63.546 amu, what is the percent abundance of each isotope? 3. 50.54% of the naturally occurring isotopes of bromine have an atomic mass of 78.92 amu while 49.46% of the naturally occurring isotopes of bromine have an atomic mass of 80.92 amu. Calculate the average atomic mass of bromine. (78.92)(.5054) + (80.92)(.4946)=avg atomic mass 39.886168 + 40.023032 = 79.9092 amu 4. Magnesium has 3 Isotopes: 24 Mg (78.70 % abundance), 25 Mg (10.13 % abundance), & 26 Mg (11.17 % abundance). Using the following data, calculate the average atomic mass of magnesium. 5. Chlorine-35 has a mass of 34.968852 amu and an abundance of 75.77 %. Chlorine-37 has a percent abundance of 24.23. Calculate the actual mass of Chlorine-37. (34.968852)(.7577) + (x)(.2423) = 35.45 amu … 35.45 came from the periodic table X = 36.9546052 6. In a sample of 400 Li atoms, it is found that 30 atoms are Li-6 (6.015 amu) and 370 are Li-7 (7.016 amu). Calculate the avg mass of Li. (get % and then calculate mass) Honors Assign. # 2- 1/2 Life for Radioactive Substances See Practice problems Pg 709 for help & also reliable internet links 1. The half-life of radon-222 is 3.824 days. After what time will one-fourth of a given amount of radon remain? 1 -> ½-> ¼ (3.824)(2) = 7.648 2. The half-life of cobalt-60 is 10.47 min. How many milligrams of cobalt-60 will remain after 104.7 min if you start with 10.0 mg? 3. The half-life of thorium-227 is 18.72 days. How many days are required for three-fourths of a given amount to decay? 1-> ½ -> ¼ (18.72)(2) = 37.44 days ½ Life equation- For those times when the numbers aren’t so simple  The equation will be provided on the test. (starting amount) x (1/2)# of half-lives = ending amount (sometimes remaining rather than ending is used) “# of half-lives” elapsed can be expressed as (total time elapsed ÷ length of half-life) Example Problem using formula: The half-life of Zn-71 is 2.4 minutes. If one had 100.0 g at the beginning, how many grams would be left after 7.2 minutes has elapsed? Answer: 7.2 / 2.4 = 3 half-lives (1/2)3 = 0.125 (the amount remaining after 3 half-lives) 100.0 g x 0.125 = 12.5 g remaining Complete the following problems using the half life equation. 4. Pd-100 has a half-life of 3.6 days. If one had 6.02 x 1023 atoms at the start, how many atoms would be present after 20.0 days? 5. Os-182 has a half-life of 21.5 hours. How many grams of a 10.0 gram sample would have decayed after exactly three half-lives? (10.0g) (1/2)3 = ending amount = 1.25 g *Since there is 1.25 g left, 8.75g have decayed