Chapter23ISM

Chapter23ISM - CHAPTER 23 NUCLEAR CHEMISTRY Problem...

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CHAPTER 23 NUCLEAR CHEMISTRY Problem Categories Biological : 23.88, 23.90, 23.95. Conceptual : 23.14, 23.15, 23.16, 23.27, 23.47, 23.49, 23.50, 23.51, 23.52, 23.59, 23.60, 23.61, 23.63, 23.64, 23.65, 23.71, 23.73, 23.76, 23.77, 23.81, 23.82, 23.84, 23.92, 23.96. Descriptive : 23.35, 23.36, 23.48, 23.57, 23.58, 23.62, 23.69, 23.75, 23.78. Difficulty Level Easy : 23.5, 23.6, 23.14, 23.15, 23.16, 23.17, 23.18, 23.23, 23.25, 23.28, 23.29, 23.33, 23.34, 23.49, 23.55, 23.57, 23.62, 23.66, 23.68, 23.79, 23.80, 23.82. Medium : 23.13, 23.19, 23.20, 23.24, 23.26, 23.27, 23.35, 23.36, 23.47, 23.50, 23.51, 23.56, 23.58, 23.59, 23.60, 23.63, 23.65, 23.67, 23.69, 23.71, 23.72, 23.75, 23.76, 23.77, 23.81, 23.83, 23.84, 23.85, 23.93, 23.96. Difficult : 23.30, 23.48, 23.52, 23.53, 23.54, 23.61, 23.64, 23.70, 23.73, 23.74, 23.78, 23.86, 23.87, 23.88, 23.89, 23.90, 23.91, 23.92, 23.94, 23.95. 23.5 (a) The atomic number sum and the mass number sum must remain the same on both sides of a nuclear equation. On the left side of this equation the atomic number sum is 13 (12 + 1) and the mass number sum is 27 (26 + 1). These sums must be the same on the right side. The atomic number of X is therefore 11 (13 2) and the mass number is 23 (27 4). X is sodium 23 ( 23 11 Na ). (b) 11 Xis Hor p (d) 56 26 Xis Fe (c) 1 0 Xis n (e) 0 1 β 23.6 Strategy: In balancing nuclear equations, note that the sum of atomic numbers and that of mass numbers must match on both sides of the equation. Solution: (a) The sum of the mass numbers must be conserved. Thus, the unknown product will have a mass number of 0. The atomic number must be conserved. Thus, the nuclear charge of the unknown product must be 1. The particle is a β particle. 135 135 53 54 IX e + ⎯⎯→ 0 1 β (b) Balancing the mass numbers first, we find that the unknown product must have a mass of 40. Balancing the nuclear charges, we find that the atomic number of the unknown must be 20. Element number 20 is calcium (Ca). 40 0 19 1 K+ ⎯⎯ →β 40 20 Ca (c) Balancing the mass numbers, we find that the unknown product must have a mass of 4. Balancing the nuclear charges, we find that the nuclear charge of the unknown must be 2. The unknown particle is an alpha ( α ) particle. 59 1 56 27 0 25 Co + n Mn + 4 2 α (d) Balancing the mass numbers, we find that the unknown products must have a combined mass of 2. Balancing the nuclear charges, we find that the combined nuclear charge of the two unknown particles must be 0. The unknown particles are neutrons. 235 1 99 135 92 0 40 52 U+ n S r+ Te+2 1 0 n
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CHAPTER 23: NUCLEAR CHEMISTRY 650 23.13 We assume the nucleus to be spherical. The mass is: 22 23 1g 235 amu 3.90 10 g 6.022 10 amu ×= × × The volume is, V = 4/3 π r 3 . 3 33 6 3 10 41 c m (7.0 10 pm) 1.4 10 cm 3 11 0p m −− ⎛⎞ × × = × ⎜⎟ × ⎝⎠ V The density is: 22 36 3 3.90 10 g 1.4 10 cm × = × 14 3 2.8 10 g/cm × 23.14 Strategy: The principal factor for determining the stability of a nucleus is the neutron-to-proton ratio ( n / p ). For stable elements of low atomic number, the n / p ratio is close to 1. As the atomic number increases, the n / p ratios of stable nuclei become greater than 1. The following rules are useful in predicting nuclear stability. 1) Nuclei that contain 2, 8, 20, 50, 82, or 126 protons or neutrons are generally more stable than nuclei that do not possess these numbers. These numbers are called magic numbers . 2) Nuclei with even numbers of both protons and neutrons are generally more stable than those with odd numbers of these particles (see Table 23.2 of the text).
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Chapter23ISM - CHAPTER 23 NUCLEAR CHEMISTRY Problem...

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