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Science Chemistry Chemistry: An Atoms-Focused Approach

The balanced nuclear equation describing the synthesis of U u o 118 294 The nuclide produced by the alpha decay process of U u o 118 294 . The nuclide produced in part b when it undergoes alpha decay process. The nuclide produced in part c when it undergoes alpha decay process. The element that occurs in nature and that has physical and chemical properties similar to U u o . Solution: C f 98 249 + C a 20 48 → U u o 118 294 + 3 n 0 1 U u h 116 290 U u q 114 286 C n 112 282 U u o 118 294 is a member of the noble gas family. The elements from the noble gas family have similar chemical and physical properties. The radon from the noble gas family is a natural occurring element with properties similar to U u o 118 294 . Explanation: U u o has atomic number 118; α particles are helium-4 nuclei. We can combine the element symbols and atomic numbers with the given mass numbers to write symbols of the nuclides involved in the reaction. To write a balanced nuclear equation, the sum of the subscripts (the atomic number) of the nuclides on the left side of the reaction equals the sum of the subscripts on the right. The sum of the superscripts (the mass numbers) is also equal to the sum of the superscripts of products. The balanced nuclear equation describing the synthesis of U u o 118 294 by bombarding C f 249 target with C a 48 nuclei is C f 98 249 + C a 20 48 → U u o 118 294 + X Z A s u m o f s u b s c r i p t o f r e a c t a n t = s u m o f s u b s c r i p t o f p r o d u c t 98 + 20 = 118 + s u b s c r i p t o f X 0 = s u b s c r i p t o f X So, the sum of subscripts or reactants is equal to the sum of subscripts of products, so the subscript of X is 0 . s u m o f s u p e r s c r i p t o f r e a c t a n t = s u m o f s u p e r s c r i p t o f p r o d u c t 249 + 48 = 294 + s u p e r s c r i p t o f X 3 = s u p e r s c r i p t o f X The superscript of X is 3 . So it will produce neutrons after nuclear reaction. C f 98 249 + C a 20 48 → U u o 118 294 + 3 n 0 1 In the alpha decay process of U u o 118 294 , the mass number will decrease by 4 and atomic number will decrease by 2 . The balanced nuclear equation describing the alpha decay process of U u o 118 294 is U u o 118 294 → U u h 116 290 + H e 2 4 In the alpha decay process of U u h 116 290 , the mass number will decreases by 4 and atomic number will decrease by 2 . The balanced nuclear equation describing the alpha decay process of U u h 116 290 is U u h 116 290 → U u q 114 286 + H e 2 4 In the alpha decay process of U u q 114 286 , the mass number will decreases by 4 and atomic number will decrease by 2 . The balanced nuclear equation describing the alpha decay process of U u q 114 286 is U u q 114 286 → C n 112 282 + H e 2 4 U u o 118 294 is a member of the noble gas family. The elements from the noble gas family have similar chemical and physical properties. The radon from the noble gas family is a natural occurring element, and it has similar properties with U u o 118 294 . Conclusion: The balanced nuclear reaction can be written by considering the Z and A on both sides of the nuclear reaction.

The balanced nuclear equation describing the synthesis of U u o 118 294 The nuclide produced by the alpha decay process of U u o 118 294 . The nuclide produced in part b when it undergoes alpha decay process. The nuclide produced in part c when it undergoes alpha decay process. The element that occurs in nature and that has physical and chemical properties similar to U u o . Solution: C f 98 249 + C a 20 48 → U u o 118 294 + 3 n 0 1 U u h 116 290 U u q 114 286 C n 112 282 U u o 118 294 is a member of the noble gas family. The elements from the noble gas family have similar chemical and physical properties. The radon from the noble gas family is a natural occurring element with properties similar to U u o 118 294 . Explanation: U u o has atomic number 118; α particles are helium-4 nuclei. We can combine the element symbols and atomic numbers with the given mass numbers to write symbols of the nuclides involved in the reaction. To write a balanced nuclear equation, the sum of the subscripts (the atomic number) of the nuclides on the left side of the reaction equals the sum of the subscripts on the right. The sum of the superscripts (the mass numbers) is also equal to the sum of the superscripts of products. The balanced nuclear equation describing the synthesis of U u o 118 294 by bombarding C f 249 target with C a 48 nuclei is C f 98 249 + C a 20 48 → U u o 118 294 + X Z A s u m o f s u b s c r i p t o f r e a c t a n t = s u m o f s u b s c r i p t o f p r o d u c t 98 + 20 = 118 + s u b s c r i p t o f X 0 = s u b s c r i p t o f X So, the sum of subscripts or reactants is equal to the sum of subscripts of products, so the subscript of X is 0 . s u m o f s u p e r s c r i p t o f r e a c t a n t = s u m o f s u p e r s c r i p t o f p r o d u c t 249 + 48 = 294 + s u p e r s c r i p t o f X 3 = s u p e r s c r i p t o f X The superscript of X is 3 . So it will produce neutrons after nuclear reaction. C f 98 249 + C a 20 48 → U u o 118 294 + 3 n 0 1 In the alpha decay process of U u o 118 294 , the mass number will decrease by 4 and atomic number will decrease by 2 . The balanced nuclear equation describing the alpha decay process of U u o 118 294 is U u o 118 294 → U u h 116 290 + H e 2 4 In the alpha decay process of U u h 116 290 , the mass number will decreases by 4 and atomic number will decrease by 2 . The balanced nuclear equation describing the alpha decay process of U u h 116 290 is U u h 116 290 → U u q 114 286 + H e 2 4 In the alpha decay process of U u q 114 286 , the mass number will decreases by 4 and atomic number will decrease by 2 . The balanced nuclear equation describing the alpha decay process of U u q 114 286 is U u q 114 286 → C n 112 282 + H e 2 4 U u o 118 294 is a member of the noble gas family. The elements from the noble gas family have similar chemical and physical properties. The radon from the noble gas family is a natural occurring element, and it has similar properties with U u o 118 294 . Conclusion: The balanced nuclear reaction can be written by considering the Z and A on both sides of the nuclear reaction.

Solution Summary: The author explains that the balanced nuclear equation describing the synthesis of _118294Og is a member of the noble gas family.

Chemistry: An Atoms-Focused Approach

Chemistry: An Atoms-Focused Approach

14th Edition

ISBN: 9780393912340

Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster

Publisher: W. W. Norton & Company

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Chemistry: An Atoms-Focused Approach

Ch. 21 - Prob. 21.1VP Ch. 21 - Prob. 21.2VP Ch. 21 - Prob. 21.4VP Ch. 21 - Prob. 21.5VP Ch. 21 - Prob. 21.6VP Ch. 21 - Prob. 21.7VP Ch. 21 - Prob. 21.8VP Ch. 21 - Prob. 21.9VP Ch. 21 - Prob. 21.10VP Ch. 21 - Prob. 21.11VP

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Plutonium was detected in trace amounts in natural uranium deposits by Glenn Seaborg and his associates in 1941. They proposed that the source of this 239Pu was the capture of neutrons by 238U nuclei. Why is this plutonium not likely to have been trapped at the time the solar system formed 4.7109 years ago?

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