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Chap 10 answers - 83 The first two facts indicate that Pu...

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Answers 1. B 2. 36.2 3. B 4. E 5. E 6. D 7. A 8. C 9. B 10. D 11. A 12. 1.9 x 10 2 13. B 14. C 15. B 16. C 17. B 18. E 19. 4.5 x 10 9 20. D 21. D 22. E 23. D 24. A 25. 5.5 26. 6.5 x 10 3 27. A 28. C 29. E 30. A 31. D 32. C 33. E 34. B 35. A 36. D 37. A 38. C 39. B 40. B 41. A
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42. C 43. E 44. E 45. C 46. E 47. C 48. C 49. D 50. D 51. C 52. D 53. E 54. B 55. B 56. D 57. A 58. C 59. C 60. C 61. D 62. D 63. 75 64. A 65. C 66. C 67. 7 x 10 5 68. 7.7 69. E 70. A 71. E 72. D 73. C 74. B 75. B 76. B 77. C 78. C 79. D 80. A 81. 1. phosphorus-32 -> beta particle + sulfur-32 2.
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2.i. lithium-8 -> beta particle + beryllium-8 2.ii. beryllium-8 -> alpha particle + alpha particle 3. potassium-40 + electron -> argon-40 4. nitrogen-13 -> positron + carbon-13 5. thorium-232 -> alpha particle + radium-228 82. 8.29
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Unformatted text preview: 83. The first two facts indicate that Pu is not a significant threat outside of the body. However, if Pu gets inside of the body (ingested or inhaled), the electrochemical data indicate that it can be easily oxidized to Pu 4+ : Pu -> Pu 4+ + 4e-, E o = 1.28 V The large positive potential indicates that oxidation of Pu is quite favorable. Now, because the chemistry of Pu 4+ is similar to Fe 3+ , Pu 4+ will tend to concentrate in tissues where Fe 3+ is found. For example, Pu 4+ could concentrate in bone marrow where red blood cells are produced (recall, hemoglobin has a central Fe 3+ ion). Once concentrated there, the alpha particles produced from its decay can do significant damage to the tissue. 84. 3270 85. B 86. C 87. D 88. B 89. E 90. D...
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