1311ex1_f05 - k = 1.3807 x 10-23 J K-1 R = 0.082057 l atm...

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k = 1.3807 x 10 -23 J K -1 R = 0.082057 l atm mol -1 K -1 = 8.314 J mol -1 K -1 = 1.987 cal mol -1 K -1 k= R / N A N A = 6.022 x 10 23 mol -1 1 kcal = 4.184 kJ c = 2.998 x 10 8 m s -1 c = λν E = -13.6(Z/n) 2 eV 1 eV = 1.6022 x 10 -19 J 1 eV = 96.485 kJ mol -1 E = h ν χ = (IE+EA)/2 Δ E = -2.18 x 10 -8 J (Z 2 ) (1/n 2 f - 1/n 2 i ) E = -me 4 Z 2 /8 ε o n 2 h 2 E= mc 2 λ = h/mv 1 nm = 10 -9 m 1 D = 10 -8 cm =100 pm ε o = 8.85 x 10 -12 J -1 C 2 m -1 4 πε o = 1.113 x 10 -10 C 2 J -1 m -1 Molecular shapes linear bent trigonal planar trigonal pyramidal tetrahedral trigonal bipyramidal seesaw T-shaped square planar square pyramidal octahedral
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R n , R ( r ) ' 4 Z 3 ( n & R & 1)! n 4 a 3 o [( n % R )!] 3 1/2 2 Zr na o e & Zr / na o L 2 R % 1 n % R Ψ n, R ,m R (r, θ , φ ) = R n, (r) · Θ ,m R · Φ m R 1,0 (r) = 2Z 3/2 a o -3/2 e -Zr/a o R 2,0 (r)=(1/2 o 2)Z 3/2 a o -3/2 (2-Zr/a o )e -Zr/2a o R 2,1 (r)=(1/2 o 6)Z 3/2 a o -3/2 (Zr/a o )e -Zr/2a o R 3,0 (r)=(1/4 o 3)Z 3/2 a o -3/2 (27-18Zr/a o +2Z 2 r 2 /9a o 2 )e -Zr/3a o Θ ,m ( θ ) = sin m θ ·P m (cos θ ) Φ m ( φ ) = (1/ o 2 π )e ± im φ R m ΘΦ 00 1 / o 21 / o 2 π 10 ( o 3/2)cos θ 1/ o 2 π 1 ( o 3/2)sin θ 1/ o 2 π e ± i φ 20 ( o 5/8)(3cos 2 θ -1) 1/ o 2 π
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4 B r 2 R r x y Name_______________________________________ Chem 1311 First Exam 16 September 2005 In taking this examination, you are agreeing to adhere to the GT academic honor code. At a minimum this requires that you utilize only the materials supplied to you, and that you do not give help to, or accept help from, others. No electronic devices, including calculators, are allowed for this exam.
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1311ex1_f05 - k = 1.3807 x 10-23 J K-1 R = 0.082057 l atm...

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