Final_review07

Final_review07 - Material for final exam (Friday 05/04/06,...

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Material for final exam (Friday 05/04/06, 8:00 a.m.) 1/3: new material (chapters 18,14) 2/3: old material (see exams 1-4 for coverage, review keys !) Total: ~11-13 problems. Some problems may combine concepts from different chapters. Kinetics Nuclear Chemistry Equilibrium Acids and Bases Buffers and Solubility Thermodynamics Electrochemistry Transition metals No material from Ch. 17 or “Special research topics” Memorize Spectrochemical Series (p. 654) Equations will be provided, except the fundamental ones Sample exam (Spring 2006) with key is posted on website
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In each of the following nuclear rxns, supply the missing particle and determine the type of rxn: (a) 60 Co -> 60 Ni + X (b) 97 Tc + X -> 97 Mo (c) 239 Pu -> 235 U + X 1. Determine Z : (a) 60 27 Co -> 60 28 Ni + X (b) 97 43 Tc + X -> 97 42 Mo (c) 239 94 Pu -> 235 92 U + X 2. Balance Z & A , and determine the particle (a) 60 27 Co -> 60 28 Ni + X 60=60 +0; A -> A=0 ; 27=28 -1 -> Z=-1 -> X= 0 -1 e (b) 97 43 Tc + X -> 97 42 Mo 97 + A = 97 -> A=0 ; 43 + Z = 42 -> Z=-1 -> X= 0 -1 e (c) 239 94 Pu -> 235 92 U + X 239 = 235 + A -> A=4 ; 94 = 92 + Z -> Z=2 -> X= 4 2 He 3. Write balanced rxns and determine the type of transformation: (a) 60 Co -> 60 Ni + 0 -1 e ( β -particle production) (b) 97 Tc + 0 -1 e -> 97 Mo (electron capture) (c) 239 Pu -> 235 U + 4 2 He ( α -particle production)
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Calculate the binding energy per nucleon for 3 1 H. The atomic mass of 3 1 H is 3.01605 u. We’ll need the following constants: m e =9.1093819x10 -31 kg, m p =1.6726216x10 -27 kg, m n =1.6749272x10 -27 kg, u = 1.66x10 -27 kg, c=2.9979x10 8 m/s 1. Calculate mass of a 3 1 H nucleus : m( 3 1 H) = atomic mass-m e =3.01605x1.66x10 -27 - 9.1093819x10 -31 = 5.00660x10 -27 – 0.00091x10 -27 = 5.00569x10 -27 kg 2. Calculate sum of masses of elementary particles that compose 3 1 H: sum =m p +2m n = 1.6726216x10 -27 + 2x1.6749272 x10 -27 = 5.02248x10 -27 kg 3. Calculate mass defect Δ m : Δ m = m( 3 1 H) – sum = 5.00569x10 -27 - 5.02248x10 -27 = - 0.01679x10 -27 kg Note: mass defect is negative -> energy is released when 3 1 H is formed from proton and two neutrons 4. Calculate total binding energy: E = Δ mxc 2 : E = 0.01679x10 -27 x(2.9979x10 8 ) 2 = 1.509x10 -12 J 5. Calculate binding energy per nucleon,
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Final_review07 - Material for final exam (Friday 05/04/06,...

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