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Science Chemistry CHEMISTRY ATOM FOCUSED EBK W/ A.C. >I<

The following nuclear equations describing the preparation of isotopes used for nuclear medicine: L i 6 + n 1 → H 3 + ? O 16 + H 3 → F 18 + ? F e 56 + ? → C o 57 + n 1 S b 121 + H 4 e → ? + 2 n 1 Solution: L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 26 56 + H 1 2 → C o 27 57 + n 0 1 S b 51 121 + H 2 4 e → I 53 123 + 2 n 0 1 Explanation: For the balanced nuclear reaction, the sum of charges, i.e., atomic numbers (subscript) and mass numbers (superscript) must be equal on both sides of the reaction (reactants and products side). If the atomic and mass number of the rest of the particles is known, then we can determine the missing particle in the balanced nuclear reaction. L i 6 + n 1 → ? → H 3 + ? The given reaction is L i 6 + n 1 → ? + H 3 Here, the sum of charge on the left side is 3 since the charge on Li (Lithium) is 3 and that on neutron (n) is 0 , and the total charge on right side is 1 . So, the missing particle or nuclide on the right side must have a charge equal to 2 . The sum of mass numbers on the left side is 7 , and that on right side is 3 . So the missing nuclide must have mass number 4 . The nuclide with atomic number 2 and with mass number 4 is He, and it can represented as H e 2 4 or α particle. Therefore, the complete balanced equation is L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 16 + H 3 → F 18 + ? In this equation, the atomic number of O (Oxygen) is 8 , atomic number of H (hydrogen) is 1 , and that of F (Fluorine) is 9 . The sum of charges on both sides is 9 . The sum of mass numbers on the left side is 19 , and that on the right side is 18 . So, the missing particle must have a charge 0 and mass number 1 . Hence we can use the particle with charge 0 and mass number 1, which is n 0 1 . Therefore, the equation is O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 56 + ? → C o 57 + n 1 In this equation, we need to identify the reactant of the first step, so consider the first step, and the equation is F e 56 + ? → C o 57 + n 1 The atomic number of F e is 26 , and that of Co is 27 . The sum of charges on the left side is 26 , and that on the right side is 27 . Therefore, the missing particle must have 1 charge. The sum of mass numbers on the left side is 56, and that on the right side is 58 , so the missing particle must have mass number as 2 . The element having charge 1 and mass number 2 is Deuterium, and it is represented as H 1 2 . The complete balanced equation is F e 26 56 + H 1 2 → C o 27 57 + n 0 1 d) S b 121 + H 4 e → ? + 2 n 1 The atomic number of S b (antimony) is 51 , and that of He is 2 , so the sum of charges on the left side of the reaction is 53 , and that on the right side is 0 since the charge on neutron (n) is 0 . The sum of mass numbers on the left side is 125 , and that on the right side is 2 . So the missing particle or nuclide must have atomic number 53 and mass number 123 , which is iodine, represented as I 53 123 . Therefore, the balanced nuclear equation is S b 51 121 + H 2 4 e → I 53 123 + 2 n 1 Conclusion: We determined the sum of charges and mass numbers on both sides of the reaction to identify the missing particle or nuclide that balanced the nuclear equations.

The following nuclear equations describing the preparation of isotopes used for nuclear medicine: L i 6 + n 1 → H 3 + ? O 16 + H 3 → F 18 + ? F e 56 + ? → C o 57 + n 1 S b 121 + H 4 e → ? + 2 n 1 Solution: L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 26 56 + H 1 2 → C o 27 57 + n 0 1 S b 51 121 + H 2 4 e → I 53 123 + 2 n 0 1 Explanation: For the balanced nuclear reaction, the sum of charges, i.e., atomic numbers (subscript) and mass numbers (superscript) must be equal on both sides of the reaction (reactants and products side). If the atomic and mass number of the rest of the particles is known, then we can determine the missing particle in the balanced nuclear reaction. L i 6 + n 1 → ? → H 3 + ? The given reaction is L i 6 + n 1 → ? + H 3 Here, the sum of charge on the left side is 3 since the charge on Li (Lithium) is 3 and that on neutron (n) is 0 , and the total charge on right side is 1 . So, the missing particle or nuclide on the right side must have a charge equal to 2 . The sum of mass numbers on the left side is 7 , and that on right side is 3 . So the missing nuclide must have mass number 4 . The nuclide with atomic number 2 and with mass number 4 is He, and it can represented as H e 2 4 or α particle. Therefore, the complete balanced equation is L i 3 6 + n 0 1 → α 2 4 + H 1 3 O 16 + H 3 → F 18 + ? In this equation, the atomic number of O (Oxygen) is 8 , atomic number of H (hydrogen) is 1 , and that of F (Fluorine) is 9 . The sum of charges on both sides is 9 . The sum of mass numbers on the left side is 19 , and that on the right side is 18 . So, the missing particle must have a charge 0 and mass number 1 . Hence we can use the particle with charge 0 and mass number 1, which is n 0 1 . Therefore, the equation is O 8 16 + H 1 3 → F 9 18 + n 0 1 F e 56 + ? → C o 57 + n 1 In this equation, we need to identify the reactant of the first step, so consider the first step, and the equation is F e 56 + ? → C o 57 + n 1 The atomic number of F e is 26 , and that of Co is 27 . The sum of charges on the left side is 26 , and that on the right side is 27 . Therefore, the missing particle must have 1 charge. The sum of mass numbers on the left side is 56, and that on the right side is 58 , so the missing particle must have mass number as 2 . The element having charge 1 and mass number 2 is Deuterium, and it is represented as H 1 2 . The complete balanced equation is F e 26 56 + H 1 2 → C o 27 57 + n 0 1 d) S b 121 + H 4 e → ? + 2 n 1 The atomic number of S b (antimony) is 51 , and that of He is 2 , so the sum of charges on the left side of the reaction is 53 , and that on the right side is 0 since the charge on neutron (n) is 0 . The sum of mass numbers on the left side is 125 , and that on the right side is 2 . So the missing particle or nuclide must have atomic number 53 and mass number 123 , which is iodine, represented as I 53 123 . Therefore, the balanced nuclear equation is S b 51 121 + H 2 4 e → I 53 123 + 2 n 1 Conclusion: We determined the sum of charges and mass numbers on both sides of the reaction to identify the missing particle or nuclide that balanced the nuclear equations.

CHEMISTRY ATOM FOCUSED EBK W/ A.C. >I<

CHEMISTRY ATOM FOCUSED EBK W/ A.C. >I<

2nd Edition

ISBN: 9780393657159

Author: Gilbert

Publisher: NORTON

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CHEMISTRY ATOM FOCUSED EBK W/ A.C. >I<

Ch. 21 - Prob. 21.1VP Ch. 21 - Prob. 21.2VP Ch. 21 - Prob. 21.3VP 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 Ch. 21 - Prob. 21.12VP Ch. 21 - Prob. 21.13QA Ch. 21 - Prob. 21.14QA Ch. 21 - Prob. 21.15QA Ch. 21 - Prob. 21.16QA Ch. 21 - Prob. 21.17QA Ch. 21 - Prob. 21.18QA Ch. 21 - Prob. 21.19QA Ch. 21 - Prob. 21.20QA Ch. 21 - Prob. 21.21QA Ch. 21 - Prob. 21.22QA Ch. 21 - Prob. 21.23QA Ch. 21 - Prob. 21.24QA Ch. 21 - Prob. 21.25QA Ch. 21 - Prob. 21.26QA Ch. 21 - Prob. 21.27QA Ch. 21 - Prob. 21.28QA Ch. 21 - Prob. 21.29QA Ch. 21 - Prob. 21.30QA Ch. 21 - Prob. 21.31QA Ch. 21 - Prob. 21.32QA Ch. 21 - Prob. 21.33QA Ch. 21 - Prob. 21.34QA Ch. 21 - Prob. 21.35QA Ch. 21 - Prob. 21.36QA Ch. 21 - Prob. 21.37QA Ch. 21 - Prob. 21.38QA Ch. 21 - Prob. 21.39QA Ch. 21 - Prob. 21.40QA Ch. 21 - Prob. 21.41QA Ch. 21 - Prob. 21.42QA Ch. 21 - Prob. 21.43QA Ch. 21 - Prob. 21.44QA Ch. 21 - Prob. 21.45QA Ch. 21 - Prob. 21.46QA Ch. 21 - Prob. 21.47QA Ch. 21 - Prob. 21.48QA Ch. 21 - Prob. 21.49QA Ch. 21 - Prob. 21.50QA Ch. 21 - Prob. 21.51QA Ch. 21 - Prob. 21.52QA Ch. 21 - Prob. 21.53QA Ch. 21 - Prob. 21.54QA Ch. 21 - Prob. 21.55QA Ch. 21 - Prob. 21.56QA Ch. 21 - Prob. 21.57QA Ch. 21 - Prob. 21.58QA Ch. 21 - Prob. 21.59QA Ch. 21 - Prob. 21.60QA Ch. 21 - Prob. 21.61QA Ch. 21 - Prob. 21.62QA Ch. 21 - Prob. 21.63QA Ch. 21 - Prob. 21.64QA Ch. 21 - Prob. 21.65QA Ch. 21 - Prob. 21.66QA Ch. 21 - Prob. 21.67QA Ch. 21 - Prob. 21.68QA Ch. 21 - Prob. 21.69QA Ch. 21 - Prob. 21.70QA Ch. 21 - Prob. 21.71QA Ch. 21 - Prob. 21.72QA Ch. 21 - Prob. 21.73QA Ch. 21 - Prob. 21.74QA Ch. 21 - Prob. 21.75QA Ch. 21 - Prob. 21.76QA Ch. 21 - Prob. 21.77QA Ch. 21 - Prob. 21.78QA Ch. 21 - Prob. 21.79QA Ch. 21 - Prob. 21.80QA Ch. 21 - Prob. 21.81QA Ch. 21 - Prob. 21.82QA Ch. 21 - Prob. 21.83QA Ch. 21 - Prob. 21.84QA Ch. 21 - Prob. 21.85QA Ch. 21 - Prob. 21.86QA Ch. 21 - Prob. 21.87QA Ch. 21 - Prob. 21.88QA Ch. 21 - Prob. 21.89QA Ch. 21 - Prob. 21.90QA Ch. 21 - Prob. 21.91QA Ch. 21 - Prob. 21.92QA Ch. 21 - Prob. 21.93QA Ch. 21 - Prob. 21.94QA Ch. 21 - Prob. 21.95QA Ch. 21 - Prob. 21.96QA Ch. 21 - Prob. 21.97QA Ch. 21 - Prob. 21.98QA Ch. 21 - Prob. 21.99QA Ch. 21 - Prob. 21.100QA Ch. 21 - Prob. 21.101QA Ch. 21 - Prob. 21.102QA Ch. 21 - Prob. 21.103QA Ch. 21 - Prob. 21.104QA Ch. 21 - Prob. 21.105QA Ch. 21 - Prob. 21.106QA Ch. 21 - Prob. 21.107QA Ch. 21 - Prob. 21.108QA Ch. 21 - Prob. 21.109QA Ch. 21 - Prob. 21.110QA Ch. 21 - Prob. 21.111QA Ch. 21 - Prob. 21.112QA Ch. 21 - Prob. 21.113QA Ch. 21 - Prob. 21.114QA Ch. 21 - Prob. 21.115QA Ch. 21 - Prob. 21.116QA Ch. 21 - Prob. 21.117QA Ch. 21 - Prob. 21.118QA Ch. 21 - Prob. 21.119QA Ch. 21 - Prob. 21.120QA Ch. 21 - Prob. 21.121QA Ch. 21 - Prob. 21.122QA Ch. 21 - Prob. 21.123QA Ch. 21 - Prob. 21.124QA Ch. 21 - Prob. 21.125QA Ch. 21 - Prob. 21.126QA Ch. 21 - Prob. 21.127QA Ch. 21 - Prob. 21.128QA Ch. 21 - Prob. 21.129QA Ch. 21 - Prob. 21.130QA

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