Chem39Nuclear - Calculations In Chemistry ***** Module 39...

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Unformatted text preview: Calculations In Chemistry ***** Module 39 – Nuclear Chemistry Module 39 – Nuclear Chemistry................................................................................ 1221 Lesson 39A: Lesson 39B: Lesson 39C: Lesson 39D: The Nucleus - Review .................................................................................... 1221 Radioactive Decay Reactions ......................................................................... 1226 Fission and Fusion........................................................................................... 1231 Radioactive Half-Life Calculations ............................................................... 1234 For additional lessons, visit www.ChemReview.Net © 2010 www.ChemReview.Net v.1g Page i Module 39 – Nuclear Chemistry Module 39 – Nuclear Chemistry Prerequisites: This module may be started at any point after Lessons 6A and 6B. Timing: This module covers alpha decay, beta decay, fission, and fusion reactions. If you are assigned calculations involving the half-life of radioactive isotopes, you should also complete the sections on first-order reactants in Lesson 27H. ***** Lesson 39A: The Nucleus – Review Pretest: If you recall the rules for the structure of the nucleus, try the problems in the practice set at the end of this lesson. If you can do those problems, you may skip this lesson. ***** Chemistry and Nuclear Reactions The rules for nuclear reactions are quite different from those that govern chemistry. For example, • During chemical reactions, the nucleus is not changed in any of the reacting atoms. Since the number of protons in the nucleus determines the identity of the atom, the number and kind of atoms must remain the same during chemical reactions. In nuclear reactions, nuclei can combine or divide to form different atoms. • In nuclear reactions, the amount of energy added or released per atom is nearly always much higher in chemical reactions. • On earth, chemical reactions occur with relative ease. In every biological organism, millions of chemical reactions occur each minute. Nuclear reactions occur naturally during radioactive decay, and may occur when cosmic rays enter our atmosphere, but on earth, compared to chemical reactions, nuclear reactions are rare. (In a star, nuclear reactions are common.) Because the rules are so different, the detailed study of nuclear reactions is generally assigned to physics rather than chemistry. We make an exception for three nuclear reactions: radioactive decay, fission, and fusion. In nuclear chemistry, radioactive decay is a powerful tool in the study of chemical reactions. Together, decay, fission, and fusion explain how the atoms that we study in chemistry came to be formed. A Model for the Nucleus Science has only a partial understanding of the nature of the atomic nucleus. When understandings in science are limited, models are developed that may be simplified, incomplete, or even speculative, but allow us to predict how systems will behave. To explain the nuclear reactions that are important in chemistry, the model for the nucleus that we utilize in chemistry is simplified compared to the models of physics, but this simplified model can nearly always predict the impact of the structure of the nucleus on processes of interest in chemistry and biology. © 2010 www.ChemReview.Net v.1g Page 1244 Module 39 – Nuclear Chemistry Our chemistry model for the atom and its nucleus was introduced in Lesson 6B. To briefly review: 1. Nuclear Structure Atoms are composed three subatomic particles. Chemical reactions involve changes to the outer shell electrons in atoms. Nuclear reactions involve changes in the number or state of the protons and neutrons in the nucleus of the atom. • Protons o o The number of protons is the atomic number of a nucleus or atom. The number of protons determines the name (and thus the symbol) of a nucleus or atom. The number of protons determines the nuclear charge of an atom’s nucleus. o The number of protons is a major factor in the atom’s behavior. o • Protons have a +1 electrical charge (1 unit of positive charge). Each proton has an mass of approximately 1.0 amu (atomic mass units) which is equivalent to 1.0 grams per mole. The number of protons in an atom is never changed by chemical reactions, but can change during nuclear reactions. Neutrons o Neutrons have an electrical charge of zero. A neutron has about the same mass as a proton: 1.0 amu. o Neutrons are believed to act as the glue of the nucleus: the particles that keep the repelling protons from flying apart. o Neutrons, like protons, are never gained or lost in chemical reactions, but the number of neutrons and protons in an atom can change in a nuclear reaction. o Unlike the number of protons, the number of neutrons in an atom has little to no influence on the types of chemical reactions that substances containing that atom will undergo. However, nuclei with the same number of protons but different numbers of neutrons will undergo different nuclear reactions. In addition, atoms with the same numbers of protons but differing numbers of neutrons in their nucleus will have different masses. As a result, some physical properties of the particles of a substance, such as their densities and the relative speeds at which the particles move, may differ measurably if its atoms have differing numbers of neutrons. 2. The Nucleus All of the protons and neutrons in an atom are found in the nucleus at the center of the atom. The diameter of a nucleus is roughly 100,000 times smaller than the effective diameter of most atoms. However, the nucleus contains all of an atom’s positive charge and nearly all of its mass. Electrons are located outside the nucleus and are much lighter than protons and neutrons. © 2010 www.ChemReview.Net v.1g Page 1245 Module 39 – Nuclear Chemistry 3. Types of Nuclei Only certain combinations of protons and neutrons form a nucleus that is stable. In a nuclear reaction, if a combination of protons and neutrons is formed that is unstable, the nucleus will decay. In terms of stability, nuclei can be divided into three types. • Stable nuclei are combinations of protons and neutrons that do not change in a planetary environment such as Earth over many billions of years. • Radioactive nuclei are somewhat stable. Some radioactive nuclei exist for only a few seconds, and others exist on average for several billion years, but they fall apart (decay) at a constant and characteristic rate. • Unstable nuclei, if formed in nuclear reactions, decay within a few seconds. Nuclei that exist in the earth’s crust include all of the stable nuclei plus some radioactive nuclei. All atoms with between one and 82 protons [except technetium (Tc) with 43 protons] have at least one nucleus found in the earth’s crust that is stable. Atoms with 83 to 92 protons exist in the earth’s crust but are always radioactive. Atoms with 93 or more protons exist on earth only when they are created in manmade nuclear reactions. Radioactive atoms comprise a very small percentage of the matter on earth. Over 99.99% of the earth’s atoms have stable nuclei that have not changed since their atoms came together to form the earth billions of years ago. 4. Terminology Protons and neutrons are termed the nucleons. The combination of a certain number of protons and neutrons is called a nuclide. The set of nuclides that have the same number of protons (so they are the same atom) but differing numbers of neutrons are called the isotopes of the atom. Isotopes Some atoms have only one stable nuclide; others have as many as 10 stable isotopes. Examples: All atoms with 1 proton are called hydrogen. Two kinds of hydrogen nuclei are stable: those with • 1 proton; and • 1 proton and 1 neutron, an isotope that is often referred to as deuterium or heavy hydrogen. Most hydrogen atoms found on earth are the isotope containing one proton and no neutrons: only 1 H atom in about 6,400 contains a deuterium nucleus. However, deuterium can be separated from the majority isotope, and it has many important uses in chemistry. As a result, deuterium is often represented in substance formulas by its own “atomic” symbol: a D. When both hydrogens in a water molecule contain a deuterium nucleus, its formula may be written as D2O (instead of H2O). Water that has a substantially higher percentage of deuterium than normal is termed heavy water. An isotope of hydrogen consisting of one proton and two neutrons, called tritium, is not found in the earth’s crust, but it can be isolated in measurable amounts from © 2010 www.ChemReview.Net v.1g Page 1246 Module 39 – Nuclear Chemistry products in nuclear reactors. Unlike deuterium, tritium is radioactive: half of the nuclei in a sample of tritium will decay in 12 years. Nuclide Symbols Each nuclide has a mass number which is the sum of its number of protons and neutrons. Mass Number of a nucleus = Protons + Neutrons Example: All nuclei with 6 protons are carbon. If a carbon nucleus has 8 neutrons, the mass number of the carbon isotope is 14. A nuclide can be identified in two ways, • by its number of protons and number of neutrons, or • by its nuclide symbol (also termed its isotope symbol). The nuclide symbol for an atom has two required parts: the atom symbol and the mass number. The mass number is written as a superscript in front of the atom symbol. Example: The three isotopes of hydrogen can be represented as • 1 proton + no neutrons or as 1H (a nuclide named “hydrogen-1”); • 1 proton + 1 neutron or as 2H (termed hydrogen-2 or deuterium); and • 1 proton + 2 neutrons or as 3H (called hydrogen-3 or tritium). Uranium has two isotopes that are important commercially and historically. • 238U is the most common naturally occuring isotope, containing 92 protons and 146 neutrons. • 235U , the isotope that is split in atomic bombs and nuclear power plants, contains 92 protons and 143 neutrons. Using a table of atoms that includes atomic numbers, try this question. Q. A nuclide with 49 protons and 62 neutrons would have what nuclide symbol? ***** A. Atoms with 47 protons must be named silver, symbol Ag. The mass number of this nuclide is 47 protons + 62 neutrons = 109 . This isotope is called silver-109 and its symbol is 109Ag . Nuclide symbols may also be written with the nuclear charge below the mass number. This is called A-Z notation, illustrated for tritium at the right. A is the symbol for mass number and Z is the symbol for nuclear charge. 3 1H Any nucleus that includes protons is by definition an atom, and since the atom symbol also identifies the number of protons in the nucleus, Z values are not required to identify a nucleus. However, for many subatomic particles, the nuclear charge is not the same as the number of protons, and showing the nuclear charge will be helpful in clearly identifying these particles. In addition, for problems in which we must balance nuclear reactions, knowing the nuclear charge (Z) is necessary, and showing Z will be helpful. © 2010 www.ChemReview.Net v.1g Page 1247 Module 39 – Nuclear Chemistry Practice: First learn the rules above, then complete these problems. 1. The charge on a nucleus is determined by its number of _________________. 2. The mass number of a nucleus is determined by its number of _____________________. 3. Isotopes have the same number of ____________ but different numbers of ___________. 4. Of the three sub-atomic particles, the two with the highest mass are _________________ and _________________. 5. Write the nuclide (isotope) symbol for a single proton using A-Z notation. 6. Consulting a table of atoms or periodic table, fill in the blanks below. Atom Name Atom Symbol Protons Neutrons Helium Atomic Number Mass Number Nuclide Symbol 2 Au 118 82 206 242Pu ANSWERS 1. Protons 2. Protons + Neutrons 4. Protons and Neutrons. 5. 3. Same number Protons, different number of Neutrons 1 1H (A particle with one proton is always given the symbol H.) 6. Atom Name Atom Symbol Protons Neutrons Atomic Number Mass Number Nuclide Symbol Helium He 2 2 2 4 4He Gold Au 79 118 79 197 197Au Lead Pb 82 124 82 206 206Pb Plutonium Pu 94 148 94 242 242Pu In writing the nuclide or isotope symbols for atoms, the nuclear charge below the mass number is optional. ***** © 2010 www.ChemReview.Net v.1g Page 1248 Module 39 – Nuclear Chemistry Lesson 39B: Radioactive Decay Reactions Pretest: If you have previously studied nuclear reactions, try the problems in the practice set at the end of this lesson. If you can do those problems, you may skip this lesson. ***** Stable Nuclei Protons have a positive electrical charge. If there is more than one proton in a nucleus, the like charges of the protons repel and the nucleus will have a tendency to fly apart. Neutrons are neutral: they have a zero electrical charge and they do not repel other particles or each other. Neutrons act in some way as the “glue” of the nucleus: if the right number of neutrons is mixed with the protons, the repelling protons remain in the nucleus, and the nucleus is stable. For small nuclei, the neutron to proton ratio that results in a stable nucleus is about one to one. As the number of protons in nuclei increases, the number of neutrons needed to form a stable nucleus increases slightly faster: the n0/p+ ratio gradually increases. For example: • All stable helium nuclei have 2 protons and 2 neutrons. • Chlorine has two stable nuclei: both have 17 protons, while one has 18 and the other 20 neutrons. • All lead have 82 protons. The four stable isotopes of lead have 122, 124, 125, or 126 neutrons. Radioactive Nuclei An unstable nucleus does not have a stable ratio of neutrons to protons. Making a nucleus stable is more complex than just adding more neutrons as “glue” or being in the right range of ratios. Certain combinations of protons and neutrons are stable, but others are not. A nucleus that is radioactive is between stable and unstable: it will have a tendency to gradually expel particles until a stable neutron and proton combination is achieved. The process of expelling particles from the nucleus is termed radioactive decay. Depending on the nucleus, radioactive decay may occur on average in seconds or gradually for up to billions of years. Radioactive decay can be a powerful tool in the study of chemical reactions. For example, for most atoms with less than 84 protons, stable isotopes exist, but radioactive isotopes also exist: the radioactive nuclei can either be found in nature or synthesized in nuclear reactors. A radioactive atom will undergo decay that we can detect, but the radioactive and non-radioactive forms of the atom have essentially the same behavior in chemical reactions. By substituting a radioactive nucleus for a stable nucleus, we can “tag” the atoms in substances. Because we can detect the location of radioactive nuclei as they decay, we can track where they go and how they behave during chemical reactions, including © 2010 www.ChemReview.Net v.1g Page 1249 Module 39 – Nuclear Chemistry reactions in biological systems. The use of radioactive dyes in medical imaging is one example of the importance of nuclear chemistry. There are several types of radioactive decay, but the two types encountered most often in chemistry are alpha (α ) decay and beta (β ) decay. Alpha Decay In alpha decay, a particle with 2 protons and 2 neutrons is ejected from the nucleus. Such a particle is termed an alpha particle. Because it has 2 protons and 2 neutrons, an alpha particle has the same structure as a helium-4 nucleus, and it is given the same isotopic symbol as an He-4 nucleus. Symbol for alpha particle = Example: α particle 4 = 2 He = 4 He The isotope U-238 undergoes radioactive decay by emitting an alpha particle. This nuclear reaction is written as 238 92 U α 234 ⎯ → 90Th ⎯ + 4 He 2 Alpha decay always lowers the atomic number (and nuclear charge) of a nucleus by 2 and its mass number by 4. Balancing Nuclear Reactions Nuclear reactions balance differently than chemical reactions, but they balance relatively easily. The rule is: In nuclear reactions, mass numbers and nuclear charge must be conserved. In a balanced nuclear reaction, on both sides of the arrow, • The sum of the mass numbers (A, on top) must be the same, and • The sum of the nuclear charges (Z, on the bottom) must be the same. The result is that nuclear reactions can be balanced by simple addition and subtraction. Q. Use the nuclear balancing rule to write below the symbol for the nucleus remaining after the alpha decay of plutonium-242. α 242 ⎯ 94 Pu ⎯ → ___________ + 4 He 2 ***** The mass numbers on top add up to 242 on both sides. The nuclear charges on the bottom total 94 on both sides. The nuclear charge must be 92, and that means the atom is uranium. The mass number of the nucleus must be 238. The balanced nuclear reaction is: α 242 ⎯ 94 Pu ⎯ → 238 92 U + 4 2 He Try one more. © 2010 www.ChemReview.Net v.1g Page 1250 Module 39 – Nuclear Chemistry Q. Which isotope is produced by the alpha decay of radium-226? ***** A key to balancing nuclear reactions is to write the isotope symbols in A-Z notation. Start there for Ra-226. ***** Radium by definition has a nucleus with 88 protons, so this reaction begins α 226 ⎯ 88 Ra ⎯ → Fill in the missing symbols. ***** In alpha decay, one product is always an alpha particle. Add its symbol on the right. ***** α 226 ⎯ 88 Ra ⎯ → 4 2 He + Use the balancing rule to write the isotopic formula for the remaining particle: the nucleus left behind after the alpha particle is expelled. ***** α 226 ⎯ 88 Ra ⎯ → 4 2 He + 222 86 Rn After the decay, the nucleus has 86 protons, so it must be radon (Rn). For the mass numbers to balance, the Rn nucleus must have a mass number of 222. Practice A 1. Write a balanced equation for the alpha decay of radon-219. 2. How many protons and how many neutrons are in Rn-219? 3. How many protons and how many neutrons are in the nucleus left behind after the alpha decay of Rn-219? How many protons and neutrons are lost in the decay? 4. Lead-206 can be formed by the alpha decay of which radioactive isotope? Beta Decay Beta decay is another type of radioactive decay. In beta decay, a neutron decays into a proton and an electron, and the electron is expelled from the nucleus at high speed. An electron formed in this manner is termed a beta particle. • Because an electron has no protons and neutrons, its mass number is zero. • Because an electron has a negative charge, when it is formed in the nucleus its “nuclear charge” is negative one. In nuclear reaction equations, a beta particle can be represented in these ways: Symbol for beta particle = © 2010 www.ChemReview.Net v.1g β particle = 0 −1 β or 0 −1 e Page 1251 Module 39 – Nuclear Chemistry Normally, in the nucleus of an atom, protons are the only particles with a charge. However, on a subatomic particle, the charge may be positive, zero, or negative. In the special case of an electron formed in the nucleus by beta decay, briefly, before the electron is expelled, it is a nuclear particle with a negative one charge. In beta decay, the number of neutrons in a nucleus decreases by one, but the number of protons increases by one, so the mass number of the isotope stays the same. Example: The equation for the beta decay of the radioactive isotope lead-210 can be written as β 210 ⎯ 82 Pb ⎯ → 0 −1 e+ 210 83 Bi Is the above equation balanced? ***** Before and after the reaction, the mass numbers total 210 and the nuclear charges total 82, so this is a balanced nuclear equation. Using the rules for nuclear balancing, we can predict the structure and symbol for the products of beta decay. Apply the rules to this question. Q. The isotope carbon-14 undergoes beta decay. Write the isotope symbols for the two nuclear particles formed in this reaction. ***** In nuclear balancing, begin by converting to A-Z notation: β 14 ⎯ 6C ⎯ → ***** One product must be a beta particle. As its symbol, you may use a β or an e . β 14 ⎯ 6C ⎯ → ***** 0 −1 e+ β 14 ⎯ 6C ⎯ → Complete the balancing. 0 −1 e+ 14 7N The isotope formed by the beta decay of carbon-14 is nitrogen-14. Practice B 1. From memory, write the symbols for an alpha particle and a beta particle. 2. Write balanced equations for these decay reactions. a. 40 K ⎯β → ⎯ b. 239 Pu ⎯α → ⎯ 3. The isotope iodine-131 is used for the treatment of hyperthyroidism: a condition in which the thyroid gland produces too much thyroid hormone. In the body, iodine is absorbed by the thyroid gland. If I-131 is administered to a patient, its beta decay kills © 2010 www.ChemReview.Net v.1g Page 1252 Module 39 – Nuclear Chemistry cells in the thyroid. The result is a reduced level of thyroid hormone without surgery. Write the symbol for the nucleus produced by the beta decay of I-131. 4. Lead-206 can be produced by the beta decay of which nucleus? 5. In a part of what is termed a radioactive decay series, nucleus A with 88 protons and 140 neutrons can beta decay to form nucleus B. B can emit a high speed electron to form nucleus C, which can α decay to form nucleus D. Write isotopic symbols for A, B, C, and D. 6. Radioactive atom A with 82 protons and 132 neutrons emits a beta particle to become atom B. Atom B emits an alpha particle to become atom C, which can emit a high speed electron from the nucleus to form Atom D. Write the isotopic formulas for A, B, C, and D. ANSWERS Practice A 1. α ⎯→ ⎯ 219 86 Rn 4 2 He + 215 84 Po 2. 86 Protons and 133 neutrons. 3. 84 protons and 131 neutrons: 2 protons and 2 neutrons are always lost in alpha decay. 4. α 4 ⎯ → 2 He ⎯ 210 84 Po + 206 82 Pb Practice B 1. α particle β particle β 0 + 40 20 Ca 2b. β 131 0 ⎯ 53 I ⎯ → −1 e+ 131 54 Xe 4. e+ 228 89 Ac 40 4 ⎯ 2a. 19 K ⎯ → −1 e 3. 5. 228 88 Ra ⎯β → ⎯ 0 −1 A 6. β = 2 He ; 214 82 Pb 0 = −1 or ⎯β → ⎯ 0 −1 0 −1 α 239 ⎯ 94 Pu ⎯ → 206 81Tl e+ B β 0 ⎯ → −1 e ⎯ + A 214 83 Bi B e β ⎯→ ⎯ 4 2 He 0 −1 + 235 92 U e+ 206 82 Pb α 228 4 ⎯→ 2 He 90Th ⎯ + C α 4 ⎯ → 2 He ⎯ + 210 Tl 81 C 224 88 Ra D β 0 ⎯ → −1 e ⎯ + 210 82 Pb D ***** © 2010 www.ChemReview.Net v.1g Page 1253 Module 39 – Nuclear Chemistry Lesson 39C: Fission and Fusion Pretest: If you have previously studied fission and fusion, try the problems in the practice set at the end of this lesson. If you can do those problems, you may skip this lesson. ***** Fission Nuclear fission is a reaction in which a large nucleus divides into two smaller nuclei, both of which contain more than two protons. If a fission reaction is accompanied by the creation of free neutrons, those neutrons can collide with other nearby fissionable nuclei and cause them to split. If this “splitting of atoms” begins in a sample of fissionable nuclei that is large enough to have critical mass, the result can be a chain reaction which releases large amounts of energy. If the chain reaction is not controlled, the result is an atomic bomb. In a nuclear power plant (a type of nuclear reactor), a chain reaction is controlled by adding materials that absorb some of the free neutrons. The large amount of energy produced by a chain reaction is then released gradually, and the resulting heat can be harnessed to drive turbines that produce electricity. The example of nuclear fission encountered most often is the splitting of uranium-235. A typical reaction is 235 92 U 1 + 0n 236 92 U 140 54 Xe + 94 38 Sr +2 1 0 n + 2 x 1010 kJ/mol In this reaction, the U-235 nucleus is struck by a free neutron. The neutron is at first absorbed, but this unstable nucleus then splits into two smaller nuclei plus 2 free neutrons. Those two neutrons can collide with other fissionable nuclei to create a chain reaction. These reactions produce amounts of energy per mole that are millions of times larger than that produced by chemical reactions such as the burning of fossil fuels. A disadvantage of using fission for electricity generation is that the products include highly radioactive isotopes. Exposure to the radiation released by radioactive decay can cause cancer, and some of the waste products of fission remain significantly radioactive for thousands of years. A major issue in nuclear power generation is how to store the waste products so that they will not escape into the earth’s biological environment. Isotopic Separation Both U-235 and U-238 can be split, but in practice only U-235 is an effective fuel for chain reactions. For use in nuclear power plants or weapons, naturally occurring uranium must enriched: meaning that the percentage of nuclei that are U-235 must be increased. In mined uranium ore, 99.3% of nuclei are U-238 and 0.7% are U-235. For nuclear weapons, uranium must be enriched to 20-80% U-235, and over 50 kilograms of this “weapons grade” uranium must be collected. Since all isotopes, including U-235 and U-238, have the same tendency to react chemically, chemical reactions cannot effectively separate isotopes. However, particles with the lighter isotopes will be less dense, and they will move faster at a given temperature. Since atoms © 2010 www.ChemReview.Net v.1g Page 1254 Module 39 – Nuclear Chemistry containing U-235 are lighter than atoms with U-238, in gaseous substances containing uranium the particles containing U-235 atoms diffuse slightly faster. Gaseous diffusion is therefore one method that is used to separate uranium isotopes. Because particles that are less dense tend to be moved toward the inside when spun in a circle at high speed, use of a centrifuge is another way to separate uranium isotopes. Both gaseous diffusion and centrifugation of uranium-containing substances are slow and expensive processes. This makes it difficult (thankfully) to obtain the amount of highly enriched uranium needed to build a nuclear weapon. Practice A 1. From memory, write the isotopic symbol for a free neutron. 2. Briefly describe the difference between fission and fusion. 3. Fill in the one missing isotopic symbol in this nuclear fission reaction. 235 92 U 1 + 0n 87 35 Br + +6 1 0 n 4. Fill in the missing isotopic symbol that is the first reactant in this fission equation. 1 + 0n 91 38 Sr + 146 56 Ba +3 1 0 n Fusion Nuclear fusion is a reaction that combines two small nuclei to make a larger one. When a small nucleus such as helium is the product of fusion, this reaction produces large amounts of energy. An example of a fusion reaction is 2 1H + 3H 1 4 2 He + 1 0 n+ 2 x 109 kJ/mol In this reaction, two hydrogen nuclei are fused, and one product is a heavier helium nucleus. Fusion is a reaction that takes place in stars, including our sun, and in hydrogen bombs. The primary reaction that causes stars to “burn” (release energy) is the conversion from hydrogen to helium. At the extremely high temperatures and pressures found in stars, lighter nuclei can fuse to form heavier nuclei, and those nuclei can undergo successive fusion reactions. After long periods of making heavier nuclei, some stars become unstable and explode. The atoms scattered into space from exploded stars can accumulate over time due to gravitational attraction to form new stars and planets. The atoms that coalesced to form our own planet billions of years ago are nuclei, or the decay products of nuclei, that were originally formed by fusion in a star. Fusion can use hydrogen as its fuel, and the products of hydrogen fusion are generally stable nuclei rather than long-lived radioactive isotopes. A nuclear reactor that could slow and control the fusion of hydrogen would therefore be a source of inexpensive and clean energy. To produce the energy needed for our society, such fusion reactors could replace © 2010 www.ChemReview.Net v.1g Page 1255 Module 39 – Nuclear Chemistry the burning of fossil fuels and current nuclear power plants, both of which form products that can harm our environment. However, while nuclear reactors can control the rate of nuclear fission, currently no way has been discovered to engineer the gradual release of the energy of nuclear fusion. Summary: To balance nuclear reactions, the rules you need in memory are α particle 1. 4 = 2 He ; β particle 0 = −1 β or 0 −1 e 1 , neutron = 0 n 2. Always use A-Z notation: mass number on top, nuclear charge on the bottom. 3. The mass numbers and nuclear charges must be conserved: both must add to give the same number on both sides of the arrows. 4. Fission splits a nucleus, fusion combines nuclei. Practice B 1. If a single nucleus is formed as the product of this reaction, write its isotope symbol. 2 1H + 2H 1 2. In stars that are red giants, helium-4 can fuse with beryllium-8 to form a single nucleus. Write the equation for this reaction. 3. Assuming that one isotope symbol is missing from these equations, fill in the missing isotope, then write the name for this type of reaction. Type: a. 235 U+ b. 238 c. 1 d. 234 1 0 92 n 4 U H+ 3 He +3 1 0 + n _________ _________ _________ H 0 −1 Th Kr + e+ _________ ANSWERS Practice A 1. 3. 1 0 n 235 92 U 2. Fission splits a nucleus, fusion combines nuclei. + 1 0 n 87 35 Br © 2010 www.ChemReview.Net v.1g + 143 57 La +6 1 0 n Page 1256 Module 39 – Nuclear Chemistry 4. 239 94 Pu 1 + 0n + 146 56 Ba 2. 91 38 Sr 4 2 He +3 1 0 n Practice B + 2H 1 1. 2 1H 3a. 235 92 U 3b. 1 3d. 1 + 0n + 234 90Th 92 36 Kr 4 2 He 238 92 U 3c. 1 H 4 2 He + + 234 90Th 0 −1 e+ 8 4 Be +3 1 0 12 6C n _fission_ _ alpha decay_ _ fusion_ 4 2 He 3 1H 141 56 Ba + 234 91 Pa beta decay_ ***** Lesson 39D: Radioactive Half-Life Calculations Timing: Complete this lesson when you are asked to solve radioactive half-life calculations or half-life calculations for first-order reactants. Prerequisites: Before starting this lesson you must complete Lesson 9A (on fractions and percentages) and Lessons 27D and 27E (on base 10 and base e logarithm calculations). If you have not already done so, do those lessons now. ***** Radioactive Decay and First-Order Kinetics In a reaction, if a reactant is used up at a rate that is proportional to its concentration, the rate of its reaction can be described by a differential rate law equation as Rate = k[A]1 = k[A] The reactant is said be “first order” due to the exponent 1 in the rate equation. If this first-order rate law is expressed in terms of time, the reaction rate can also be expressed by the following mathematically equivalent equation, termed the integrated rate law for a first-order reactant: ln[A]t = ─kt + ln[A]0 In the above equations, t represents time in any units, 0 is the time at the start of the measurements, and k is the symbol for the rate constant for the reaction. Radioactive decay, a nuclear reaction, is always first-order. However, reactants in standard chemical reactions can also be first order as well. In this lesson, we will learn to solve half-life calculations for radioactive nuclei, but the same equations can be applied to standard chemical reactions that involve first-order reactants. © 2010 www.ChemReview.Net v.1g Page 1257 Module 39 – Nuclear Chemistry Half-Life The half-life of any reactant (symbol t1/2) is the time that is required for half of the particles of the reactant to be used up in a reaction. First-order reactants are a special case in which the half-life of the reactant is not dependent on its initial concentration. In chemical reactions, at a given temperature, the a first-order half-life is constant. In chemical reactions, the rate of reaction changes with temperature, and the half-life of a reactant changes as well. However, in radioactive decay, a nuclear reaction, the time of a half-life does not change significantly at temperatures up to several thousand kelvins. This means that each radioactive nucleus has a characteristic half-life. Some radioactive nuclides have a half-life of a few seconds; others have a half-life of billions of years. It is not possible to predict when any one nucleus will decay. However, in any sample of more than a few hundred of a given nucleus, the time in which half of the nuclei decay is always the same. If we can calculate (or look up) the half-life, we know how long it will take for half of the nuclei to decay. By solving half-life equations, we can also calculate how long it will take for any percentage of the nuclei in a sample to decay. Half-Life Calculations For Simple Multiples In calculations involve half-life, the two variables will generally be the time over which a given nuclide in a sample decays and the percentage of those nuclei that remain. If the time period for the decay is equal to either the half-life or a simple multiple of the half-life, answers can be calculated by mental arithmetic. • If a sample of a given nucleus has decayed for a time equal to one half-life, 1/2 of the original nuclei have decayed and half remain. If the nuclei are in a sample that has a constant volume (which should be assumed unless other conditions are stated), 1/2 of the original concentration of the reactant also remains after one half-life. • After two half-lives (double the time of the half-life), the number of nuclei remaining is half of the half that remained after the first half-life: half of 1/2 = 1/4 (25%) of the original nuclei remain and 75% have decayed. • At triple the half-life, 1/2 of 1/4 = 1/8 (12.5%) of the original nuclei remain. Apply the logic above to the following problem. Q. Fluorine-18, a radioactive isotope used in nuclear medicine, has a 1.8 hour half-life. How long will it take for 87.5% of the F-18 nuclei in a sample to decay? * ** * * A. If 87.5% has decayed, 12.5% remains. How many half-lives are required? How much time would this be? * ** * * 12.5% remains at 3 half-lives; 3 x 1.8 hours = 5.4 hours To solve radioactive decay calculations for these “simple multiple” cases, given any one of headings in the table below, you will need to be able to fill in the rest of the table from © 2010 www.ChemReview.Net v.1g Page 1258 Module 39 – Nuclear Chemistry memory. This should not be difficult: note that in the two middle columns, each number is simply half of the one above. For Radioactive Nuclei, at time = Fraction Remaining Percent Remaining Percent Decayed 1 100% 0% One half-life 1/2 50% 50% Two half-lives 1/4 25% 75% Three half-lives 1/8 12.5% 87.5% 0 For calculations that are not easy multiples, we will use these rules to make estimates of answers. Practice A: Write the table above until, given the top row, you can fill in four rows below from memory. Then complete these problems. 1. The nucleus of Pu-239 undergoes first-order radioactive decay with a half-life of 24,400 years. In a sample of constant volume containing Pu-239, a. After how many years will 25% of the original Pu-239 nuclei remain? b. After how many half-lives will the [Pu-239] be 1/16th of its original concentration? c. What percentage of the Pu-239 has decayed after exactly 4 half-lives? Rate Constants for Radioactive Decay In half-life calculations that do not involve simple multiples, we can solve using rate equations and the math of natural logs. Each radioactive nucleus has a rate constant for decay (k) that is characteristic: a value that is constant. Different radioactive nuclei will have different values for k . One way to write the equation that predicts the decay rate for radioactive nuclei is ln [A]t = ─kt which can be abbreviated as ln(fraction remaining) = ─kt [A]0 In the first equation, [A]t / [A]0 is the fraction of nuclei remaining. For example: after one half-life, half (50%) of the original sample remains and the fraction remaining is 0.50 . After two radioactive half lives, the fraction remaining would be? ***** After two half-lives, the percentage remaining is 25% (1/4 ), and the fraction remaining is therefore 0.25 . © 2010 www.ChemReview.Net v.1g Page 1259 Module 39 – Nuclear Chemistry Since the reactant is being used up over time, the value of [A] after time = t will be less than it was at time = 0, and the value of the fraction will be less than one. The decimal equivalent of the fraction will always have the form 0.xxx . The units used to calculate the fraction can be concentration or any consistent units that are proportional to concentration, including the mass or number of particles in a sample that has a fixed volume. Radioactive half-life calculations often involve fractions or percentages, and in those cases the form of the equation above that includes (fraction remaining ) will be the most convenient to use. However, to use the equation with the term (fraction remaining), you must calculate using fractions or the decimal equivalent of fractions, rather than percentages. If a percentage is WANTED, you will need to solve for the fraction first. If a percentage is given, you will need to convert to its decimal equivalent to use in the equation. When using fractions and percentages, recall that • A fraction can be expressed as x/y or as a decimal equivalent value that is x divided by y. • Percentage = fraction x 100% • Percentage remaining = 100% ─ percentage decayed • Fraction remaining = 1.000 ─ fraction decayed • In decay calculations, the fractions will have a value between 1.00 and 0 (such as 0.25). and decimal equivalent = percent/100% (For practice in converting between percentages and fractions, see Lesson 9A.) Practice B: Commit to memory the equation above that includes (fraction remaining), the complete each of these. 1. If 90% of a sample has decayed, what is the fraction remaining? 2. If the fraction of a sample that has decayed is 0.40, what percent remains? 3. If the value for ln(fraction of sample remaining) is ─ 1.386 , a. What is the fraction of the sample remaining? b. What percentage has decayed? 4. For the decay of a radioactive nucleus, if the rate constant of the reaction is k = 0.04606 hours─1 , what percentage remains after 50.0 hours? © 2010 www.ChemReview.Net v.1g Page 1260 Module 39 – Nuclear Chemistry Half-Life Calculations For Non-Simple-Multiples For a radioactive nucleus, after a time equal to one half-life ( t1/2 ), half of a sample has decayed and half remains. Substituting into the equation ln(fraction remaining) = ─kt we can write ln(1/2) = ─k t1/2 at a time equal to one half-life, Solve this equation in symbols for half-life. * * ** * One way of several to write the equation is t1/2 ≡ ─ ln(1/2) / k This equation is one way to define radioactive half-life. In the above equations, the rate constant (k) and half-life ( t1/2 ) are variables: their numeric values will differ for different radioactive nuclei. The term ln(1/2) is a constant: it can be converted to a number. Use your calculator to find its fixed decimal value. ln(1/2) = __________________ * * ** * ln(1/2) = ln(0.500) = ─ 0.693 Substitute this numeric value into the equation above defining half-life, thensimplify. * * ** * t1/2 = ─ (─0.693) / k which simplifies to t1/2 ≡ 0.693 / k This last equation above is often listed in textbooks as a definition of radioactive half-life. From this form, it is clear that if you know the half-life, you can find the rate constant k, and if you know the rate constant you can find the half-life. To solve decay calculations, we need equations that relate the fraction remaining, half-life, time, and k . Several combinations of the equations above can be used, but the best equations may be those that are easy to remember. In these lessons, we will solve using what we will call the Radioactive Decay Prompt If a radioactive decay calculation includes or half-life and a fraction or percentage of a sample, and the answer cannot be calculated using simple multiples, write in the DATA: ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Note that the second equation is simply a special case of the first: when the fraction remaining is 1/2, the time is equal to the half-life. Commit the radioactive decay prompt to memory, then apply it to solve this problem. Q. Iodine-131, a radioactive isotope used to treat thyroid disorders, has a half-life of 8.1 days. What percentage of an initial [I-131] remains after 48 hours? ***** 2010 www.ChemReview.Net v.1g Page 1261 Module 39 – Nuclear Chemistry WANT: Percent [I-131]48 hrs. = % remaining DATA: 8.1 days = radioactive half-life = t1/2 48 hours = 2.0 days = t (choose any consistent time unit) Strategy: Write the equations that relate the symbols in the problem. For radioactive decay calculations that include half-life and fraction or percentage, write and use ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Percentage remaining = fraction remaining x 100% If needed, adjust your work and solve from here. * * ** * The variable that links the prompt equations is k . Since we know the half-life, the second equation will find k . Knowing k and t, the first prompt equation will find ln(fraction remaining). Knowing ln(fraction remaining), the fraction remaining can be found using Fraction = eln(fraction) and Percentage = Fraction x 100% Apply those steps and solve. ***** ln(1/2) = ─k t1/2 , has two variables, k and t1/2 , and we know t1/2 . k = ─ ln(1/2) / t1/2 = ─ (─ 0.693) / t1/2 = + 0.693 / 8.1 days = 0.0856 day─1 = k ln(fraction remaining) = ─kt = ─ ( 0.0856 day─1)(2.0 days) = ─ 0.171 Fraction = eln(fraction) = e─0.171 = 0.843 = 84 % I-131 remains after 2 days Practice C: If you are unsure of the answer to a part, check it before doing the next part. 1. The rate constant for the decay of the tritium isotope of hydrogen is 0.0562 years─1 . Calculate the half-life of tritium. 2. Strontium-90 is a radioactive nuclide found in fallout: dust particles in the cloud produced by the atmospheric testing of nuclear weapons. In chemical and biological systems, strontium behaves much like calcium. If dairy cattle consume crops exposed to dust or rain containing fallout, dairy products containing calcium will also contain Sr-90. Similar to calcium, Sr-90 will be deposited in the bones of dairy product consumers, including children. In part for this reason, most (but not all) nations conducting nuclear tests signed a 1963 treaty which banned atmospheric testing. © 2010 www.ChemReview.Net v.1g Page 1262 Module 39 – Nuclear Chemistry Strontium-90 undergoes beta decay with a half-life of 28.8 years. What percentage of an original [Sr-90] in bones will remain after 40.0 years? a. Estimate the answer. b. Calculate the answer. 3. The element Polonium was first isolated by Dr. Marie Sklodowska Curie and named for her native Poland. Radioactive Po-210 is found in significant concentrations in tobacco. If 20.0% of Po-210 remains in a sample after 321 days of alpha decay, a. estimate the half-life for Po-210. b. Calculate a precise half-life of Po-210. Compare it to your Part A estimate. 4. If 10.0% of a sample of Rn-222 remains after 12.6 days, a. estimate the half-life for Rn-222. b. Calculate a precise half-life of Rn-222. Compare it to your Part A estimate. 5. If the half-life of carbon-14 is 5,730 years, what fraction of the original carbon-14 in a sample has decayed after 1650 years? Estimate, then calculate. ANSWERS Practice A 1a. First-order half-life is constant. Half remains after one half-life, half of that half (25%) remains after two half-lives. Two half-lives = 2 x 24,400 years = 48,800 years. 1b. Half remains after one half-life, 1/4th after two, 1/8th after three, 1/16th after four half-lives. 1c. Half remains after one half-life, 1/4th after two, 1/8th after three; 1/16th after four half-lives. 1/16 = 0.0625 = 6.25% remains, so 93.75% has decayed. Practice B 1. If 90% has decayed, 10% remains, and fraction remaining = 10% / 100% = 0.100 2. If fraction decayed = 0.40, percentage decayed = 0.40 x 100% = 40%, and percentage remaining = 100% ─ 40% = 60% 3a. WANT: DATA: fraction of sample remaining ln(fraction remaining) = ─ 1.386 Knowing a value for ln(fraction remaining), to find fraction remaining, use Fraction remaining = eln(fraction remaining) * * * ** = e─1.386 = 0.250 3b. If the fraction remaining is 0.250, the percentage remaining is 25.0%, and the percentage decayed is 75.0%. © 2010 www.ChemReview.Net v.1g Page 1263 Module 39 Nuclear Chemistry 4. WANT: DATA: % of sample remaining . To find percentage, find fraction first. 0.04606 hours─1 = k 50.0 hours = t Strategy: The equation that relates the three terms in the data is ln(fraction remaining) = ─kt Knowing k and t, ln(fraction remaining ) and then fraction remaining can be calculated. * * * ** ln( fraction remaining ) = ─kt = ─ (0.04606 hours─1 )( 50.0 hours ) = ─ 2.303 Fraction remaining = eln(fraction remaining) = e─2.303 = 0.100 The percentage remaining is fraction x 100% = 0.100 x 100% = 10.0% Practice C 1. WANT: t1/2 for tritium DATA: 0.0562 years─1 = k Strategy: In radioactive decay calculations that include half life and fraction or percentage, write ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 In this problem, only the second equation is needed to relate the symbols in the WANTED and DATA. SOLVE: t1/2 = ln(1/2) / ─k = (─ 0.693) / ─ 0.0562 years─1 = 12.3 years ( 1/ years─1 = (years─1) ─1 = years ) 2a. WANT: Estimate of % [Sr-90] remaining after 40 years. If one half life is about 30 years and 50% remains, and two half-lives is about 60 years and 25% remains, then at 40 years, about…. 40% remains? 2b. WANT: % [Sr-90]40.0 yrs. remaining DATA: 28.8 yrs. = t1/2 40.0 yrs = t In radioactive decay calculations that include half life and fraction or percentage, write ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 From half-life, k can be found. From k and t, ln(fraction) and then fraction can be found. Percentage = Fraction x 100% SOLVE: Since ln(1/2) = ─k t1/2 , k = ─ ln(1/2) / t1/2 = ─ (─ 0.693) / t1/2 = 0.693 / 28.8 yrs. = 0.02406 yrs.─1 = k © 2010 www.ChemReview.Net v.1g Page 1264 Module 39 – Nuclear Chemistry ln(fraction remaining) = ─kt = ─ ( 0.02406 yrs.─1 )( 40.0 yrs. ) = ─ 0.9624 = ln(fraction) Fraction = eln(fraction) = e─0.9624 = 0.382 = 38.2 % Sr-90 remains after 40 years Compare this to the estimate. 3a. Estimate 20% remains after 321 days. 50% remains after one half-life, and 25% after 2 half -lives. At about 300 days, 25% would remain, which is 2 half-lives, so one half-life is about ….. * * * ** About 150 days. 3b. WANT: t1/2 DATA: 20.0% Po-210 remains fraction remaining = 20% / 100% = 0.200 t = 321 days. See radioactive decay, half life, and fraction or percentage? Write ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Percentage = Fraction x 100% Knowing the fraction and t, k can be found from the first prompt equation. Half-life can then be found from the second equation. Solving for k in symbols first: k = ─ ln(fraction) }/ t = ─ {ln(0.200) }/ (321 days) = ─ (─1.61) / (321 days) = 5.01 x 10─ 3 days─1 t1/2 = ─ln(1/2) / k = + 0.693 / 5.01 x 10─ 3 days─1 = 138 days = half-life of Po-210 Is this answer close to the estimate in Part A? 4a. Estimate 10% remains after 12.6 days. 12.5% remains after three half –lives, which is close to 10%. If three half-lives is about 12 days, then one half life would be…. * * * ** About 4 days. 4b. WANT: DATA: t1/2 90.0% Rn-222 has decayed, so 10.0% remains , and fraction remaining = 0.100 t = 12.6 days See radioactive decay, half life, and fraction or percentage? Write ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 Fraction remaining = 10% / 100% = 0.100 Knowing the fraction and t, k can be found from the first equation. Half-life can then be found from the second equation. © 2010 www.ChemReview.Net v.1g Page 1265 Module 39 – Nuclear Chemistry k = ─ ln(fraction) / t = ─ ln(0.100) / (12.6 days) = ─ (─2.30) / (12.6 days) = 0.183 days─1 t1/2 = ─ ln(1/2) / k = +0.693 / 0.183 days─1 = 3.79 days = half-life of Rn-222 Close to the estimate in Part A? 5. Estimate: 0.50 is the fraction gone after about 6,000 years, so maybe 0.15 is the fraction gone in about a third of that time? Calculate: WANT: fraction [C-14] decayed = 1.000 ─ fraction remaining DATA: t1/2 = 5,730 yrs. t = 1,650 yrs. ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 From half-life, k can be found. From k and t, ln(fraction) and then fraction can be found. SOLVE: k = ─ ln(1/2) / t1/2 = ─ (─ 0.693) / t1/2 = + 0.693 / 5730 yrs. = 1.21 x 10─ 4 years─1 ln( fraction remaining ) = ─kt = ─ (1.21 x 10─ 4 yr─1 )( 1,650 yrs. ) = ─ 0.1996 Fraction remaining = eln(fraction remaining) = e─0.1996 = 0.819 fraction C-14 remaining If 0.819 = fraction remaining, 1.000 ─ 0.819 = 0.181 = fraction C-14 decayed ***** © 2010 www.ChemReview.Net v.1g Page 1266 Module 39 – Nuclear Chemistry Summary: Nuclear Chemistry 1. α particle β particle 4 = 2 He ; 0 = −1 β or 0 −1 e 1 , neutron = 0 n 2. To balance nuclear reactions, a. Always use A-Z notation: mass number on top, nuclear charge on the bottom. b. Both mass numbers and nuclear charge must be conserved; each must add to give the same number before and after the reaction. 3. Fission splits a nucleus, fusion combines nuclei. 4. In the decay of radioactive isotopes, the half-life is constant. • At the half-life, 1/2 of the original number of nuclei remain; • At double the time of the half-life, 1/4 of the original nuclei remain; • At triple the half-life, 1/8 of the original nuclei remain. 5. Radioactive Decay Prompt If a radioactive decay calculation includes half-life and a fraction or percentage of a sample, and the answer cannot be calculated using simple multiples, write in the DATA: ln(fraction remaining) = ─kt and ln(1/2) = ─k t1/2 and use the math of natural logs to solve. 6. When using the radioactive decay prompt, • To use the equation with (fraction remaining), you must work in fractions, not percentages. • Fraction = percent/100% • Percentage remaining = 100% ─ percentage decayed • Fraction remaining = 1.000 ─ fraction decayed • The fractions in decay calculations will have a value between 1.00 and 0 (such as 0.25). and Percentage = fraction x 100% ##### © 2010 www.ChemReview.Net v.1g Page 1267 The ATOMS – The third column shows the atomic number: The protons in the nucleus of the atom. The fourth column is the molar mass, in grams/mole. For radioactive atoms, ( ) is the molar mass of most stable isotope. Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Boron Bromine Cadmium Calcium Californium Carbon Cerium Cesium Chlorine Chromium Cobalt Copper Curium Dysprosium Erbium Europium Fermium Fluorine Francium Gadolinium Gallium Germanium Gold Hafnium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Ac Al Am Sb Ar As At Ba Bk Be Bi B Br Cd Ca Cf C Ce Cs Cl Cr Co Cu Cm Dy Er Eu Fm F Fr Gd Ga Ge Au Hf He Ho H In I Ir Fe Kr La Lr Pb Li 89 13 95 51 18 33 84 56 97 4 83 5 35 48 20 98 6 58 55 17 24 27 29 96 66 68 63 100 9 87 64 31 32 79 72 2 67 1 49 53 77 26 36 57 103 82 3 (227) 27.0 (243) 121.8 40.0 74.9 (210) 137.3 (247) 9.01 209.0 10.8 79.9 112.4 40.1 (249) 12.0 140.1 132.9 35.5 52.0 58.9 63.5 (247) 162.5 167.3 152.0 (253) 19.0 (223) 157.3 69.7 72.6 197.0 178.5 4.00 164.9 1.008 114.8 126.9 192.2 55.8 83.8 138.9 (257) 207.2 6.94 Lutetium Magnesium Manganese Mendelevium Mercury Molybdenum Neodymium Neon Neptunium Nickel Niobium Nitrogen Nobelium Osmium Oxygen Palladium Phosphorus Platinum Plutonium Polonium Potassium Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Rubidium Ruthenium Samarium Scandium Selenium Silicon Silver Sodium Strontium Sulfur Tantalum Technetium Tellurium Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium Lu Mg Mn Md Hg Mo Nd Ne Np Ni Nb N No Os O Pd P Pt Pu Po K Pr Pm Pa Ra Rn Re Rh Rb Ru Sm Sc Se Si Ag Na Sr S Ta Tc Te Tb Tl Th Tm Sn Ti W U V Xe Yb Y Zn Zr 71 12 25 101 80 42 60 10 93 28 41 7 102 76 8 46 15 78 94 84 19 59 61 91 88 86 75 45 37 44 62 21 34 14 47 11 38 16 73 43 52 65 81 90 69 50 22 74 92 23 54 70 39 30 40 175.0 24.3 54.9 (256) 200.6 95.9 144.2 20.2 (237) 58.7 92.9 14.0 (253) 190.2 16.0 106.4 31.0 195.1 (242) (209) 39.1 140.9 (145) (231) (226) (222) 186.2 102.9 85.5 101.1 150.4 45.0 79.0 28.1 107.9 23.0 87.6 32.1 180.9 (98) 127.6 158.9 204.4 232.0 168.9 118.7 47.9 183.8 238.0 50.9 131.3 173.0 88.9 65.4 91.2 ...
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This note was uploaded on 11/16/2011 for the course CHM 1025 taught by Professor J during the Summer '09 term at Santa Fe College.

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