CHAPTERS 19, 20, 21 CHEM 162-2007 POST-EXAM III PRACTICE PROBLEMS CHAPTER 19 - THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS RADIOACTIVE DECAY KINETICS NUCLEAR BINDING ENERGY MISCELLANEOUS CHAPTER 20 & 21A - GROUPS 1A THROUGH 4A GROUP 1A ELEMENTS GROUP 2A ELEMENTS GROUP 3A ELEMENTS GROUP 4A ELEMENTS CHAPTER 21B - GROUPS 5A THROUGH 8A GROUP 5A ELEMENTS GROUP 6A ELEMENTS GROUP 7A ELEMENTS GROUP 8A ELEMENTS E. Tavss, PhD CHAPTER 18 - THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Radioactive decay particles and rays 52. Radioactivity has been used for all of the following except: A. sterilizing food B. tracing biological processes C. generation of power D. creating a new, stable element E. dating archeological sites All of the elements that radioactivity has produced are either unstable or are already naturally available. Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Radioactive decay particles and rays 53. Which of the following are assumptions of 14C dating? X. 14C content ceases to be renewed when the plant is harvested or dies. Y. The 14C/12C ratio in a living organism reaches a steady state. Z. Nothing can change the half life of the decay. A. All are assumptions B. Z only C. Y only D. X and Y E. X only X. 14C content ceases to be renewed when the plant is harvested or dies. Y. The 14C/12C ratio in a living organism reaches a steady state. Z. Nothing can change the half life of the decay. X. Through transpiration (in plants) or transpiration (in animals) 14C is constantly replaced by fresh 14C from the atmosphere. When the plant dies, transpiration stops and the 14C is no longer replaced. Y. I?m not so sure that I agree with the verbiage of ?reaches? a steady state. I believe that the 14C/12C ratio begins at a steady state and is maintained at a steady state throughout the lifetime of the organism. But the intentions of the meaning of this statement is true. Z. True. Here, also, I?m not so certain that under conditions of enclosing the radiation, such as entombing the organism in a lead shield, that the rate of half-life decay cannot be slowed down. But this has never been demonstrated as being true, to my knowledge. Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Radioactive decay particles and rays 54. The purpose of control rods in a nuclear reactor is A. to contain radioactivity in the event of a leak from the reactor. B. to produce neutrons to speed up the rate of fission. C. to absorb neutrons to slow the rate of fission. D. to convert the steam to electrical power. E. to cool the steam generated by the reactor. Control rods, composed of substances that absorb neutrons, are used to regulate the power level of the reactor. The reactor is designed so that should a malfunction occur, the control rods are automatically inserted into the core to stop the rreaction. Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Radioactive decay particles and rays 55. When a nucleus splits into smaller nuclei by fission, high energy neutrons are released. Which statement best describes supercritical mass? A. The material is massive enough to cause fusion to occur. B. The material is packed densely enough to allow exactly one neutron to cause another nucleus to split. C. There are too few nuclei to sustain the continued fission. D. The process occurs in a manner that releases more neutrons. E. The material is packed densely enough that the released neutrons cause more than one additional nucleus to split. A. The material is massive enough to cause fusion to occur. This is not releated to supercritical mass. The issue is fission, not fusion. Also, the issue is mass density, not having enough mass. B. The material is packed densely enough to allow exactly one neutron to cause another nucleus to split. Having the exact neutron emission necessary to minimally sustain a chain reaction requires a ?critical mass?, not a ?supercritical mass?. C. There are too few nuclei to sustain the continued fission. This means that there is a ?subcritical mass?. D. The process occurs in a manner that releases more neutrons. This is simply a chain reaction occurring with a critical mass. E. The material is packed densely enough that the released neutrons cause more than one additional nucleus to split. This is an accelerating chain reaction, and requires a supercritical mass of fissionable material in order to occur. Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Radioactive decay particles and rays 28. Which mode of nuclear decay can result in an increase in the atomic number? A. ?-emission B. ?-emission C. electron capture D. positron emission E. ?-emission A. ?-emission X ? 42He + -4-2Y; i.e., a decrease in atomic number B. ?-emission X ? 00? + 00Y; i.e., no change in atomic number C. electron capture X +0-1e ? 0-1Y ; i.e., a decrease in atomic number D. positron emission X ? 0+1e + 0-1Y; i.e., a decrease in atomic number E. ?-emission X ? 0-1e + 0+1Y; i.e., an increase in atomic number 29. Chem 162-2005 Final Exam + Answers Chaper 18 - The Nucleus Radioactive decay particles and rays The biological effects of a particular source of radiation depend on: A. The ionizing ability of the radiation B. Choose this choice if all the others are correct C. The penetrating ability of the radiation. D. The energy of the radiation. E. The chemical properties of the radiation source. See page 902 of Zumdahl. A. True. Extraction of electrons from biomolecules to form ions is particularly detrimental to their functions. C. True. The particles and rays produced in radioactive processes vary in their abilities to penetrate human tissue. ? rays are highly penetrating. D. True. The higher the energy content of the radiation, the more damage it can cause. E. True. When a radioactive nuclide is ingested into the body, its effectiveness in causing damage depends on its residence time. Strontium, being chemically similar to calcium, can collect in bones, where it may cause leukemia and bone cancer. 34. Chem 162-2005 Final Exam + Answers Chaper 18 - The Nucleus Radioactive decay particles and rays What type of particle is captured and what type is emitted in the nucleosynthesis below? 54Fe 54Mn A. Neutron capture and proton emission occurs B. Neutron capture and particle emission C. Only neutron capture occurs D. Neutron capture and positron emission occurs E. Electron capture and particles emission occurs 54Fe 54Mn 5426Fe 5425Mn A. 5426Fe + 10n 5425Mn + 11H B. 5426Fe + 10n 5425Mn + 0-1e C. 5426Fe + 10n 5425Mn D. 5426Fe + 10n 5425Mn + 0+1e E. 5426Fe + 0-1e 5425Mn + 0+1e The only option that is algebraically acceptable is ?A?. 44. Chem 162-2005 Final Exam + Answers Chaper 18 - The Nucleus Radioactive decay particles and rays Beta particles are identical to: A. hydrogen atoms = 11H + 0-1e B. electrons = 0-1e = ? particles C. helium nuclei = 42? = ? particles D. neutrons = 10n E. protons = 11H 47. Chem 162-2005 Final Exam + Answers Chapter 18 - The Nucleus Radioactive decay particles and rays The stable isotopes of lead are Pb-204, Pb-206, and Pb-208. The unstable isotope Pb-214 is most likely to undergo A. electron capture B. neutron emission C. particle emission D. positron emission E. particle emission The difference between Pb-214 and the other Pb isotopes is the number of neutrons. The greater number of neutrons in Pb-214 makes it unstable. Hence, reducing the number of neutrons should help stabilize the atom. 21482Pb ? 6 10n + 20882Pb RADIOACTIVE DECAY KINETICS Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Radioactive decay kinetics 38. How long does it take for a 2.78 g sample of 10143Tc to decay to 0.0l0lg if its half-life is 14 min.? A. 24.1 h B. 178 h C. 4.40 h D. 1.89 h E. 4.40 s t1/2 = 0.693/k k = 0.693/14 = 0.0495 ln(At/Ao) = -kt ln(0.0101/2.78) = -0.0495 x t t = 113.5 min 113.5 min x 1 hr/60 min = 1.89 h NUCLEAR BINDING ENERGY Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Nuclear binding energy 47. What is the binding energy of a nucleus of 23290Th if its nuclear mass is 232.0382 amu?* (*The actual question said ?atomic mass?, not ?nuclear mass?, but nuclear mass was intended. Data and conversions were on last page: Mass of proton = 1.0073 amu; mass of neutron = 1.0087 amu; 1 amu = 931 MeV; 1 eV = 1.602 x 10-19 J) A. 3.43 x 1012 J B. 2.76 x l0-10 J C. 2.86 x 10-21 J D. 3.98 x 10-7 J E. 3.43 x 1015 J 23290Th = 90 protons + 142 neutrons Massprotons + Massneutrons = Massnucleus + mass defect (90 x 1.0073) + (142 x 1.0087) = 232.0382 + mass defect Mass defect = 1.8542 amu 1.8542 amu/(6.022 x 1023 amu/g) = 3.079 x 10-24 g = mass defect E = mc2 E = 3.079 x 10-24 g x (1 kg/1000 g) x (3 x 108m/s)2 E = 2.77E-10 J Another way to convert the mass defect into J: 1.8542 amu x (931 MeV/amu) x (1 x 106eV/MeV) x (1.602 x 10-19 J/eV) = 2.77E-10 J Chem 162-2003 Final exam + answers Chapter 18-The Nucleus Nuclear binding energy 51. Which is true about nuclear fusion? X. It requires very high temperatures. Y. It requires nuclei to be forced very close together. Z. It is the energy source of stars. A. X only B. Y and Z only C. All are true D. Z only E. X and Z only X. It requires very high temperatures. The protons are identically charged and repel each other electrostatically. The temperature must be very high to give the protons the velocity necessary to overcome the electrostatic repulsion. Y. It requires nuclei to be forced very close together. They must be very close together for the two nuclei to merge. Z. Stars produce their energy through nuclear fusion. Our sun gives off vast quantities of energy from the fusion of protons to form helium. CHAPTER 18 ? THE NUCLEUS MISCELLANEOUS CHAPTER 19 - GROUPS 1A THROUGH 4A GROUP 1A ELEMENTS Chem 162-2003 Final exam + answers Chapter 19-Groups 1A through 4A Group 1A Elements 39. Which of the following are produced when alkali metals react with water? X. H2 Y. M+ Z. A. X, Y and Z B. Y, Z and W C. X and Y D. Y and Z E. X, Y and W 2H2O + 2e- ? H2 + 2OH- -0.83 2(Na ? Na+ + e-) +2.71 2M(s) + 2H2O(l) ? 2M+(aq) + 2OH-(aq) + H2(g) +1.88 GROUP 2A ELEMENTS GROUP 3A ELEMENTS GROUP 4A ELEMENTS Chem 162-2003 Final exam + answers Chapter 19-Groups 1A through 4A Group 4A elements 42. Which one of the following statements is false? A. Carbon dioxide is gas whereas silica is solid. True B. The silicon 3p valence orbitals overlap very effectively with the smaller oxygen 2p orbitals to form ? bonds. False C. The structure of silica is based on SiO4 tetrahedra with Si ? O ? Si bridges. True. D. Carbon dioxide is composed of discrete CO2 molecules with Lewis structure . . . . : O = C = O : True E. Discrete SiO2 molecules with the Lewis structure . . . . : O = Si = O : are not stable. True. There isn't such a molecule as Si double bonded to two oxygen atoms. See Hill and Petrucci text p. 892 including figure 21.5. The overlap of the 3p orbital of silicon and the 2p orbital of oxygen is too weak to form a double bond. Hence, this double bond doesn't form. Instead, the silicon forms single bonds to four oxygen atoms, forming a solid complex lattice. CHAPTER 20 - GROUPS 5A THROUGH 8A GROUP 5A ELEMENTS Chem 162-2003 Final exam + answers Chapter 20-Groups 5A through 8A Group 5A elements 33. Which of the following is not associated with the nitrogen cycle? A. Decomposition of plant material. B. Nitric acid in acid rain. C. Lightning. D. The Haber process for the synthesis of ammonia. E. The stabilization of nitroglycerin with silica. The nitrogen cycle involves going from N2 in the atmosphere to nitrogen-containing compounds used by plants and animals, and then decay back to N2 in the atmosphere. Stabilization of any nitrogen-containing compound inhibits its return to N2, thusly breaking the cycle. 52. Chem 162-2005 Final Exam + Answers Groups 5A through 8A Group 5A Elements The process of transforming N2 to a form usable by animals and plants is called A. nitrogen fixation. B. fertilization C. denitrification D. the Ostwald process E. nitrogenation A. Nitrogen fixation is the process of transforming N2 to nitrogen-containing compounds useful to plants and animals. B. Fertilization is the process of making fertile, by spreading nitrogen containing materials on land. C. Denitrification is the return of nitrogen from decomposed matter to the atmosphere by bacteria that change nitrates to nitrogen gas. D. The Ostwald process is a commercial process for producing nitric acid by the oxidation of ammonia. E. I don?t know what nitrogenation is. It is probably simply the process of adding nitrogen to something. 54. Chem 162-2005 Final Exam + Answers Groups 5A through 8A Group 5A Elements Which compound is not associated with acid rain? A. NH3 B. SO2 C. SO3 D. Choose this answer if they are all associated with acid rain. A. NH3 + H2O ? NH4OH, a base, therefore not associated with acid rain. B. Oxides in the upper right of the periodic table form acids. SO2 + H2O ? H2SO3 C. SO3 + H2O ? H2SO4 + H2O ? 2HNO3 Chapter 20-Groups 5A through 8A GROUP 6A ELEMENTS Chapter 20-Groups 5A through 8A GROUP 7A ELEMENTS GROUP 8A ELEMENTS Chem 162-2003 Final exam + answers Chapter 20-Groups 5A through 8A Group 8A elements 48. Which compound is correctly associated with its structure? A. XeO2F2 Distorted tetrahedron B. XeO4 Square planar C. XeO3F2 Square pyramidal D. XeO3 Trigonal planar Angular ("Bent") VSEPR structure: (a) XeO2F2 . . : O : . . . . | . . : F ?Xe ? F : . . | . . : O : . . Distorted tetrahedron (also known as ?See-saw?) (b) XeO4 . . : O : . . | . . : O ?Xe ? O : . . | . . : O : . . Tetrahedral (c) XeO3F2 . . . . : O : : O : . . . . : O ? Xe ? F : . . | . . : F : . . Trigonal bipyramidal (d) XeO3 . . : O : . . | . . : O ? Xe ? O : . . . . . . Trigonal pyramidal (e) XeF2 . . . . . . . . : F ? Xe ? F : Linear . . . . . . Given structures of b, c, d and e are not consistent with the actual structures. I don?t know about ?a? because I don?t know the structure of AX4E1 geometry. 32 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 18 ? THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS An isotope with a high neutron to proton (N/Z) ratio will tend to decay through A. ?-decay B. ?-decay C. ?-decay D. electron capture E. positron emission An isotope with a high neutron to proton ratio will tend to form a stable nuclide by decreasing the neutrons or increasing the protons. Losing an electron (?-decay) will do this. 10n ? 11H + 0-1e Capturing a positron will also do this. 10n + 0+1e ? 11H 9 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 18 ? THE NUCLEUS RADIOACTIVE DECAY PARTICLES AND RAYS Complete the following nuclear reactions. + X ? + ? Y + ? Z + 23892U + X ? 23993Np + 0-1e 23892U + 10n ? 23993Np + 0-1e 23993Np ? Y + 0-1e 23993Np ? 23994Pu + 0-1e 137N ? Z + 01e 137N ? 136C + 01e 51. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS Balancing the following nuclear reaction would require the use of one of the following particles 234Th --> 234Pa + [ ] 23490Th --> 23491Pa + AZ? Using algebra: 234 = 234 + A A = 0 90 = 91 + Z Z = -1 Particle = 0-1e A. He B. e C. e D. E. e 52. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS Which one of the following nuclear emissions has the largest mass number? A. alpha emission 42He B. beta emission 0-1e C. positron emission 0+1e D. gamma emission 00? E. emission from electron capture 00X-ray 1. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS When decays, the emission consists consecutively of an particle, then two - particles, and finally another particle. The resulting stable nucleus is 21484Po ----> 42He + 0-1e + 0-1e + 42He + YX? 84 = 2 -1 -1 + 2 + X X = 82 = Pb 214 = 4 + 0 + 0 + 4 + Y Y = 206 20682Pb A. Bi B. Bi C. Pb D. Pb E. Tl 20. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS When the nuclide undergoes electron capture, the product nuclide is 4422Ti + 0-1e ----> 4421Sc A. B. C. D. E. 36. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS Identify the unknown nuclide X in the following nuclear reaction: Let ?Y? be the atomic charge of X and let ?Z? be the atomic mass of X. 51 + 1 = Y + 0 Y = 52; therefore the atom = tellurium 121 + 1 = Z + 1 Z = 121 12152Te A. B. C. D. E. 45 CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS Determine the numbers of protons, neutrons, and electrons in a 235U3+ ion. Uranium has an atomic number of 92. Therefore it has 92 protons and 92 electrons. It also has (235-92=) 143 neutrons. A charge of 3+ means that it has 89 electrons. D protons neutrons electrons A. 92 143 92 B. 95 140 92 C. 95 143 92 D. 92 143 89 E. 89 146 89 40. CHEM 162-2000 FINAL EXAM CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY PARTICLES AND RAYS Which of the following statements regarding nuclear reactions is false? A. Nuclear reactions involve substantially greater energy changes than ordinary chemical reactions. B. Nuclear decay follows first order kinetics. C. Nuclear reactions do not require conservation of mass. D. The rate of nuclear decay is not temperature dependent. E. Nuclear reactions are typically the same for different isotopes of the same element. E 44 Chem 162-2004 Final Exam + Answers Chapter 18 - The nucleus Radioactive decay particles and rays Which one of the following nuclear transformations is not correct? A. Au + B. U He + Th C. K e + Ar D. Bi e + E. C e + B The sum of the atomic numbers, and the sum of the mass numbers must be equal on both sides of the equation. The sum of the atomic numbers on both sides of the equation are not equal in ?C?, with an atomic number of 19 on the left side, and a total atomic number of 20 on the right side. 33 Chem 162-2004 Final Exam + Answers Chapter 18 - The nucleus Radioactive decay particles and rays Consider the following decay series. U XYZ What is Z? A. Ra B. Th C. Pa D. Th E. U ? = 42He ? = 0-1e U 23490Th23491Pa23492U 42. Chem 162-2005 Final Exam + Answers Chaper 18 - The Nucleus Radioactive decay particles and rays When atoms of beryllium-9 are bombarded with alpha particles, neutrons are produced. What new nuclide is also formed? A. B. C. D. E. 42He + 94Be ? 10n + 126? 42He + 94Be ? 10n + 126C 24 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 18 ? THE NUCLEUS RADIOACTIVE DECAY KINETICS A sample of uranium ore is found to contain 6.77 mg 238U and 1.50 mg of 206Pb. Assuming the lead decayed from the uranium according to the reaction 238U ? 206Pb, what is the age of the ore? The half-life of the decay is 4.51×109 years. A. 2.53×109 years B. 8.36×109 years C. 6.15×108 years D. 1.48×109 years E. 4.15×108 years The initial quantity of 238U can be determined by converting the 1.50 mg of 206Pb back into 238U. However, this is not possible without knowing the atomic weight of 238U and 206Pb. But we can make an approximation that in both cases the mass numbers are equal to the atomic masses. Although this is a crude approximation, since we are doing this for both atoms most of the error should cancel out. 0.00150g Pb/206gmol-1 x 238gmol-1 = 0.001733 g 238U were converted into 206Pb Therefore the initial 238U is 6.77 mg + 1.73 mg = 8.50 mg 238U. First find k. t1/2 = 0.693/k 4.51x109 = 0.693/k k = 1.537 x 10-10 ln(Nt) = -kt + ln(No) ln(6.77) = ((-1.537 x 10-10) x (t)) + ln(8.50) Age of ore = 1.48 x 109 years 55. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY KINETICS A sample of 32P decays at a rate of 2.5 x 1010 atoms/second. The half-life (t1/2) of 32P is 14.3 days. What is the decay rate (or the number of atoms per second undergoing decay) of this sample after 30 days? 14.3 days x 24 h/d x 60 m/h x 60 s/m = 1235520 s t1/2 = 0.693/? 1235520 = 0.693/? ? = 5.609 x 10-7 ln(At/Ao) = -?t ln (At/2.5 x 1010) = -5.609 x 10-7 x 30 x 24 x 60 x 60 At = 5.8 x 109 A. 6.3 x 106 atoms/second B. 1.2 x 104 atoms/second C. 2.9 x 108 atoms/second D. 5.8 x 109 atoms/second E. 1.0 x 1010 atoms/second 18. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY KINETICS A sample of wood from an old Egyptian tomb is found to have a carbon-14 activity of 7.07 disintegrations per minute per gram of C, while a sample of wood from a newly-cut tree has a carbon-14 activity of 15.3 disintegrations per minute per gram of C. The half-life of carbon-14 is 5730 years. Calculate the age of the old wood. This appears to be a problem involving the Integrated Rate Law, ln(At/Ao) = -?t ? is not provided, but can easily be determined, having been given t1/2. t1/2 = 0.693/? 5730 = 0.693/? ? = 0.00012094 ln(At/Ao) = -?t ln(7.07/15.3) = -0.00012094 x t t = 6383 years A. 5100 years B. 6400 years C. 3600 years D. 7200 years E. 9200 years 22. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY KINETICS The half-life of 25Mg is 21 hours. What is the activity in disintegrations per second of a sample containing 1.00 x 10-5 mol of 25Mg? Understanding the terminology is not obvious. ?Disintegrations per second? is the same thing as disappearance of Mg per unit time, which is the rate of disappearance. Rate = kN When we are talking about emission counts (as in Geiger counter clicks), N is the number of atoms. N = (1.00 x 10-5 mol) x (6.022 x 1023 atoms/mol) = 6.022 x 1018 atoms t1/2 = 21 hours x (60 min/hr) x (60 sec/min) = 75600 seconds t1/2 = 0.693/k 75600 = 0.693/k k = 9.17 x 10-6 Rate = kN Rate = (9.17 x 10-6) x (6.022 x 1018) = 5.52 x 1013 dis s-1 A. 5.5 x 1013 dis s-1 B. 3.9 x 1020 dis s-1 C. 6.0 x 1018 dis s-1 D. 1.8 x 1012 dis s-1 E. 4.3 x 1015 dis s-1 35. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY KINETICS Which of the following statements concerning nuclear reactions is false? All statements are true except ?E?. Note that temperature is not a variable in any of the kinetics equations for nuclear decay (A = ?N; ln(At/Ao) = -?t; t1/2 = 0.693/?) A. Radioactive decay obeys first-order kinetics. B. A nuclear reaction often results in the formation of an element not initially present. C. Different isotopes of a given element undergo different nuclear reactions. D. The energy changes in nuclear reactions are very much larger than those in ordinary chemical reactions. E. The rates of nuclear reactions depend strongly on temperature. 36. CHEM 162-2000 FINAL EXAM CHAPTER 18 - NUCLEAR CHEMISTRY RADIOACTIVE DECAY KINETICS 90Sr, a ?--emitter found in radioactive fallout, has a half-life of 28.1 years. What percentage of a 90Sr sample will remain after 100.0 years? A. 8.5% B. 0.34% C. 63% D. 82% E. 76% A 39 Chem 162-2004 Final Exam + Answers Chapter 18 - The nucleus Radioactive decay kinetics A rock containing U and Pb was examined to determine its approximate age. Analysis showed the ratio of Pb atoms to U atoms to be 0.250. Assuming that no lead was originally present, that all the Pb formed over the years has remained in the rock, and that the number of the nuclides in intermediate stages of decay between U and Pb is negligible, calculate the age of the rock. The half-life of U is 4.5 x 109 years. A. 4.8 x 108 years B. 3.5 x 108 years C. 1.4 x 109 years D. 7.1 x 108 years E. 9.1 x 109 years The ratio of Pb atoms to U atoms is 0.250. This means that there is now a ratio of 0.250 Pb atoms to 1.000 U atoms. Since the total number of atoms do not change, i.e., 23892U atoms become 20682Pb atoms, this corresponds to meaning that originally there were 1.250 U atoms and 0 Pb atoms. So the U atoms went from 1.250 to 1.000 in a certain amount of time. Half-life of U is 4.5 x 109 yr. t1/2 = 0.693/k k = 0.693/t1/2 k = 0.693/(4.5 x 109) = 1.54 x 10-10 ln(At/Ao) = -kt ln(1.000/1.250) = -(1.54x10-10) x t t = 1.45 x 109 39. Chem 162-2005 Final Exam + Answers Chaper 18 - The Nucleus Radioactive decay kinetics The half-life of molybdenum-99 is 67.0 h. What mass remains of a 1.000 mg sample of after 335 h? A. 0.083 mg B. 0.056 mg C. 0.031 mg D. 0.062 mg E. 0.019 mg Nuclear decay follows first order kinetics. t1/2 = 0.693/k 67.0 = 0.693/k k = 0.01034 ln(Nt) = -kt + ln(No) ln(Nt) = (-0.01034 x 335) + ln(1.000) Nt = 0.0313 mg Note that the units cancel in first order reactions, so [N] and [Nt] in mg is OK. 33. Chem 162-2005 Final Exam + Answers Chaper 18 - The Nucleus Radioactive decay kinetics The remnants of an ancient fire in a cave in showed a decay rate of 3.1 counts per minute per gram of carbon. Assuming that the decay rate of in freshly cut wood is 13.6 counts per minute per gram of carbon, calculate the age of the remnants. The half life of is 5750 years. A. 10000 years B. 9000 years C. 12000 years D. 6000 years E. 16000 years 14C is an unstable isotope of carbon. 12C is a stable isotope. Carbon-14 is generated at the same rate as it decomposes. Therefore, the concentration of C14 remains constant. C14 and C12 make 14C CO2 and 12C CO2, respectively. CO2 is made from this constant ratio of 14C to 12C. Therefore, the 14CO2/12CO2 ratio is constant. Transpiration or respiration during the life of the plant or animal provides a constant ratio of 14C CO2 to 12C CO2. Therefore, there is a constant ratio of 14C to 12C in all living cells. Upon death the C in the cells is no longer replaced due to intake of fresh CO2. Since 14C is an unstable isotope, the ratio of 14C:12C decreases. Half-life of 14C = 5730 years t1/2 = 0.693/k 5730 years = 0.693/k k = 1.209 x 10-4 ln(Nt) = -kt + ln(No) ln(3.1) = (-1.209 x 10-4 x t) + ln(13.6) t = 1.22 x 104 years = 12000 years 16 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 18 ? THE NUCLEUS NUCLEAR BINDING ENERGY The nuclide 86Sr has a mass of 85.9094 amu. What is the binding energy per nucleon for this nuclide? A. 1.3610-12 J/nucleon B. 9.4210-13 J/nucleon C. 1.4910-10 J/nucleon D. 3.0810-12 J/nucleon E. 2.4410-12 J/nucleon Mass of a proton: 1.0073 amu Mass of a neutron: 1.0087 amu Strontium has 38 protons and (86-38 = ) 48 neutrons. (38 x 1.0073) + (48 x 1.0087) = 86.695g = theoretical mass of the Sr nucleus. Actual Sr nuclear mass: 85.9094g. (A nuclide is a nucleus.) 86.695 ? 85.9094 = 0.7856 g = mass defect/mole (0.7856g/mol) x (1/6.022 x 1023 atom/mol) x (1/86nucleons/atom) = 1.5169 x 10-26 g mass defect/nucleon = 1.5169 x 10-29 kg mass defect/nucleon E = mc2 E = 1.5169 x 10-29 x (3 x 108)2 = 1.365 x 10-12 J/nucleon 53. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY NUCLEAR BINDING ENERGY If the formation of an alpha particle from two protons and two neutrons resulted in a decrease in mass of 0.0305 u (atomic mass units), the nuclear binding energy for this alpha particle will be 11H + 11H + 10n + 10n ----> 42He + 0.0305 g (931.5 MeV/1 g) x 0.0305 g = 28.41 MeV A. 14.2 MeV B. 56.8 MeV C. 28.4 MeV D. 20.0 MeV E. 5.00 MeV 14. CHEM-2001 FINAL EXAM + ANSWERS CHAPTER 18 - NUCLEAR CHEMISTRY NUCLEAR BINDING ENERGY A Cu nucleus has a mass of 64.9119 u. Calculate the binding energy of a Cu nucleus. The proton mass is 1.0073 u and the neutron mass is 1.0087 u. Nucleon mass = Nuclear mass + Mass defect The lower number means that Cu has 29 protons. The upper number means that the sum of the protons and neutrons is 65. Therefore, Cu has 36 neutrons. Nucleon mass = (29 x 1.0073) + (36 x 1.0087) = 65.5249 65.5249 = 64.9119 + X X = 0.0613 g Binding energy = 931.5 MeV/g 931.5 x 0.0613 = 571.01 MeV A. 103 MeV B. 571 MeV C. 613 MeV D. 276 MeV E. 784 MeV 50. Chem 162-2005 Final Exam + Answers Chapter 18 - The Nucleus Nuclear binding energy Calculate the nuclear binding energy per nucleon for the nucleus (atomic mass = 4.0026 amu, =1.0078 amu, 01 n = 1.0087 amu) x 10-13 J = 1 MeV, 1 amu = 1.66 x 10-27 kg) A. 2.15 MeV/nucleon B. 7.13 MeV/nucleon C. 5.12 MeV/nucleon D. 14.2 MeV/nucleon E. 3.56 MeV/nucleon 42He = 2 protons + 2 neutrons Calculated nuclear mass = Actual nuclear mass + mass defect Calc?d mass of 2 protons + 2 neutrons = Actual He nuclear mass + mass defect Mass of electrons could have been provided, but is probably not provided because it cancels out. That is, the atomic mass of two protons is on the left side of the equation, and this contains two electrons which should be subtracted. The atomic mass of one helium atom is on the right side of the equation, and this contains two electrons which should be subtracted. (2 x 1.0078) + (2 x 1.0087) = 4.0026 + mass defect/mol Mass defect per mole = 0.0304 g/mol 0.0304/(6.022 x 1023) = 5.048 x 10-26 g = mass defect/atom (5.048 x 10-26)/4 = 1.262 x 10-26 g = mass defect/nucleon 1.262 x 10-26 g/nucleon x 1kg/1000 g = 1.262 x 10-29 kg = mass defect/nucleon E = mc2 E = 1.262 x 10-29 kg x (3 x 108m/s)2 = 1.136 x 10-12 J 1.136 x 10-12 J x 1MeV/(1.60 x 10-13J) = 7.10 MeV 17 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 18 ? THE NUCLEUS MISCELLANEOUS Radioactivity is dangerous both because of the ionizing ability of the radiation, but also due to the penetrating power. Which of the following correctly matches the type of radiation with its penetrating power and ability to ionize? Effects of Radiation 13 CHEM 162-2006 FINAL EXAM + ANSWERS CHAPTER 18 ? THE NUCLEUS MISCELLANEOUS Nuclear reactors produce energy by maintaining critical mass. What does critical mass mean? A. The minimum mass of a nuclide for fission to occur. B. Enough neutrons are being produced by the decay to cause one new atom to split for each decay. C. The temperature of the material is high enough to melt. D. Enough neutrons are being produced by the decay to cause more than one new atom to split for each decay. E. The energy of the ejected particles is high enough to produce electricity. ?Critical mass? is defined as the minimum mass of a particular fissionable nuclide, in a given volume, that is required to sustain a nuclear chain reaction. ?A? is close to the definition, but it doesn?t say anything about a sustained chain reaction. If the mass of the nuclide is below the critical mass then fission would begin and then fizzle out. ?B? speaks of a chain reaction, but it doesn?t mention minimal mass. ?B? would have been better written if it said, ?the minimum nuclide mass in which enough neutrons are being produced by the decay to cause one new atom to split for each decay?. ?C? is irrelevant. ?D? is a supercritical process, the type of process used to make a nuclear bomb. ?E? is nonsense. 28. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 18 - ELEMENT GROUPS 1A THROUGH 4A GROUP 1A ELEMENTS Which of the following is not an important use of alkali metals in living matter? A. Use of Li2CO3 in drugs See p. 861. Used to treat manic-depression. B. Use of sodium and potassium salts in the formulation of acidic drugs Solubilizes acidic drugs. C. Use of table salt in foods Supplies Na+ to the body (for transmission of nerve signals, and supplies Cl- to the body (for the production of stomach acid). ET note: This is a stretch. The primary purpose of addition of table salt to food is to improve taste and act as a preservative. D. Use of sodium and potassium as a major component of bone structure No. Calcium and phosphorous are the major components of bone structure. E. Use of NaOH and KOH in the manufacture of soap ET note: Whereas NaOH and KOH are involved in the manufacture of soap, it is a stretch to consider that as an important use in living matter. 34. 1A Chem 162-2004 Final Exam + Answers Chapter 19 - Groups 1A through 4A Which oxide is amphoteric? A. MgO B. CO C. BaO D. BeO An amphoteric substance can behave as an acid or a base. (A) MgO is a salt, Mg2+ + O2- The Mg atom gives up two electrons to the oxygen leaving the Mg with 1s22s22p6, having a full set of electrons. The oxygen has the same electronic configuration. Both have a stable octet of electrons. The magnesium ion doesn?t behave as an acid, although the oxygen ion behaves as a base. (B) The Lewis structure of carbon monoxide has three bonds between the C and O, and a non-bonding pair of electrons on both the C and O. Both atoms can act as a base, but neither atom is electron deficient so as to behave as an acid. (C) BaO is a salt, Ba2+ and O2-. The argument provided for MgO applies to BaO also. (D) BeO: The Lewis structure for BeO has three bonds between the Be and O and a lone pair of electrons on the oxygen. Whereas oxygen, having a lone pair of electrons, can act as a base, beryllium is deficient in electrons (it doesn?t have an octet) and can behave as an acid). (E) CaO is similar to MgO and BaO. 37 Chem 162-2004 Final Exam + Answers Chapter 19 - Groups 1A through 4A Which reaction is unknown? A. K + 2H+ 2K+ + H2 B. 2Li + H2 2LiH C. 4Li + O2 2Li2O D. 2Cs + Cl2 2CsCl E. Na + O2 NaO2 Based on information in table 19.6, p. 921 of Zumdahl and Zumdahl, ?E? is the correct answer. When I did this problem I couldn?t come up with any reasonable answer. Once I learned that the correct answer is ?E? I tried to justify it. I thought about the oxidation potential of Na being the lowest of the group 1 metals, but standard potentials are for aqueous reactions and this reaction is non-aqueous. Hence, standard oxidation-reduction potentials are not relevant to solving this problem. I thought about ionization energies, but lithium has a higher ionization energy than sodium, which would suggest that the two options containing lithium might be the correct answer, but the two options containing lithium are not consistent with the fact that ?E? is the correct answer. So although there must be some concept associated with this answer, neither I (nor Dr. Boikess) know of any. So why is ?E? the correct answer? Because it?s listed as the correct answer in Zumdahl and Zumdahl table. (This is not my idea of a good question.) 28. Chem 162-2005 Final Exam + Answers Chapter 19 - Groups IA through 4A Elements What are the expected products of the following reaction? Na(s) + H2O(l) ? (not balanced) A. Na+(aq) + (aq) + H2O(l) B. Na+(aq) + (aq) + H2 (g) C. NaH (aq) + O2(g) D. Na+(aq) + O2-(aq) + H2(g) E. Na(H2O)2+(aq) Na is a strong reducing agent, and is therefore strongly oxidized. Hence, Na ? Na+ + 1e- Since Na is oxidized, something must be reduced. The only equations which show something reduced is B, C and D. In B and D, H+ (in H2O) ? Ho (in H2). In C, although H+ was reduced to H- (hydride), an oxidation (formation of O2) also occurs, which is not likely since Na has a much greater oxidation potential than oxygen (in ). So that leaves ?B? and ?D? as the remaining choices. So what will form, a hydroxide or an oxide? Since we know from the periodic table trends that the upper right elements form acidic aqueous solutions from their oxides, and the lower left elements form basic solutions from their oxides, then it follows that oxides in water are unstable and [metal oxides] have a tendency to form hydroxides. Hence, ?B? is the most likely answer. 38. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 19 - ELEMENT GROUPS 1A THROUGH 4A GROUP 2A ELEMENTS Alkali metals such as sodium or potassium are industrially prepared by one of the following processes. A. Electrolysis of aqueous solution of their salts. B. Electrolysis of their molten salts. See p. 856 C. Reduction of their oxides with hydrogen. D. Reduction of their salts with graphite. E. Reduction of their sulfides with hydrogen. 47. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 19 - ELEMENT GROUPS 1A THROUGH 4A GROUP 2A ELEMENTS The dirty grayish film of soap (referred to as soap scum) corresponds to one of the following compositions (where R corresponds to a long chain hydrocarbon). A. none of the compositions correspond to soap scum. B. Ca(RCO2)2 Yes. A calcium salt of a soap is the insoluble soap scum. C. Na(RSO3) No. A sodium salt of a detergent is soluble. D. Na(RCO2) No. A sodium salt of a soap is soluble. E. Ca(RSO3)2 No. A calcium salt of a detergent is soluble. 35. Chem 162-2005 Final Exam + Answers Chapter 19 - Groups IA through 4A Group IIA Elements Which of the following oxides has some acidic properties A. Li2O B. BaO C. CaO D. MgO The berrylium in BeO forms a covalent bond with the oxygen due to the small size of the Be, and its corresponding closeness of its effective nuclear charge with the bonding electrons in oxygen. All of the other listed group II atoms only form ionic bonds with oxygen. Due to the strength of the BeO bond, when BeO reacts with H2O it forms Be(OH)2, which is an acid. (It also forms Be2+ + , making it amphoteric.) All of the other group II oxides only form bases upon hydrolysis. 34. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 19 - ELEMENT GROUPS 1A THROUGH 4A GROUP 3A ELEMENTS Aluminum metal is prepared industrially from one or more of the following processes: i) Reduction of Al2S3 with hydrogen No. ii) Heating of Al2O3 to high temperatures No. iii) Electrolysis of Al2O3 dissolved in molten Na3AlF6 Yes. iv) Recycling of aluminum cans Yes. A. i and ii B. ii and iii C. iii and iv D. i and iii E. i and iv 43. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 20 - ELEMENT GROUPS 5A THROUGH 8A GROUP 5A ELEMENTS The molecular formulas of two important phosphorus oxides are P4O10 and P4O6 are anhydrides of phosphoric acid and phosphorous acid, respectively. i) P4O4 ii) P4O10 + 6 H2O ----> 4 H3PO4 (phosphoric acid) iii) P4O6 + 6 H2O ----> 4 H3PO3 (phosphorous acid) iv) P4O2 v) P6O4 A. i and iii B. iii and iv C. ii and iii D. i and ii E. i and v 48 Chem 162-2004 Final Exam + Answers Chapter 20 - Groups 5A through 8A A general trend among the representative elements is that A. non metallic elements in the second period are more likely to form double bonds. B. elements are less reactive as you move down the column in all families. C. inclusion as a representative element depends on abundance. D. their reactivity depends solely on atomic radius. E. electronegativity decrease across each period. (A). True. Non metallic elements in the second period are more likely to form double bonds. Notice that carbon forms double bonds, as do nitrogen and oxygen. (B) False. In group I, francium is more active than Li. (C) False. Inclusion just depends on isolation and identification, not abundance Some synthetic elements decay so rapidly that they have 0 abundance at a given time, yet they are listed. (D) False. Reactivity depends on many factors, such as effective nuclear charge. (E) False. The convention is normally to go from left to right; as such, electronegativity increases across each period. 48. Chem 162-2005 Final Exam + Answers Groups 5A through 8A Group 6A Elements The best explanation for the fact that oxygen forms only OF2 with fluorine while sulfur forms SF2, SF4, and SF6 is that A. oxygen is more electronegative than sulfur B. oxygen has a higher first ionization energy than sulfur C. sulfur has d orbitals available for bonding D. oxygen has a higher second ionization energy than sulfur E. oxygen can form hydrogen bonds Oxygen, being in the second period, has no d orbitals and therefore cannot have more than an octet of electrons around it. Therefore, although theoretically it can form OF2 and OF4, it cannot form OF6. Sulfur, being in the third period, has d orbitals and therefore can have more than an octet of electrons around it. Therefore, it can form SF2, SF4 and SF6. 52 Chem 162-2004 Final Exam + Answers Chapter 20 - Groups 5A through 8A Which compound or ion has a trigonal bipyramidal shape? A. ICl3 B. BrF5 C. IBrCl4 D. PF5 (A) ICl3 = T-shaped (B) BrF5 = Square pyramidal (C) IBrCl4 = Square pyramidal (D) PF5 = Trigonal bipyramidal (E) SF6 = Octahedral 43. Chem 162-2005 Final Exam + Answers Groups 5A through 8A Group 7A Elements What is the molecular structure of ClF5? A. seasaw B. trigonal bipyramidal C. square pyramidal D. square planar E. octahedral . . . . : F : : F : . . . . . . : F ? Cl ? F : . . | . . : F : . . Octahedral electron group structure, but square pyramidal molecular structure. 29. CHEM-2002 FINAL EXAM + ANSWERS CHAPTER 20 - ELEMENT GROUPS 5A THROUGH 8A GROUP 8A ELEMENTS Given that 40Ar is more abundant in nature than the other inert gases; which of the following statements is/are true about 40Ar and other inert gases.? i) Noble gases are all heavier than air. No. He is less dense than N2, and therefore less dense than air. ii) 40Ar is made from the decay of 40K. Yes. 4019K ----> 4020Ar + 0-1e iii) Helium escapes from the earth's atmosphere to a greater extent than argon because its molar mass is only one tenth that of argon. True. A. iii only B. i and iii C. i and ii D. i only E. ii and iii 43 Chem 162-2004 Final Exam + Answers Chapter 20 - Groups 5A through 8A Group 8A Elements Which compound or ion is correctly associated with its shape? A. KrBr2 V-shaped B. RnO4 Tetrahedral C. XeO64- Trigonal bipyramidal D. XeO2F2 Square pyramidal Tetrahedral Do Lewis structure followed by VSEPR. (A) KrBr2 Linear (B) RnO4 Tetrahedral (C) XeO64- Octahedral (D) XeO2F2 Seesaw planar 51. Chem 162-2005 Final Exam + Answers Groups 5A through 8A Group 8A Elements Which of the following forms covalent compounds? I. He II. Ar III. Xe IV. Kr A. I, II B. II, III C. I, II, III D. II, III, IV , IV Kr, Xe and Rn are the noble gases with the lowest ionization energy. This allows them to form compounds with atoms, such as F, which are highly electronegative. Although Kr and Xe form such compounds, possible compounds with Rn are difficult to study due to the radioactivity of Rn. 1
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