Chem - Chem 209 Week 4 Solutions Jaime Ambrosio and Cheng Cheng 1 a = hc/E =(6.626E-34(3E8 m/s(3.61E-19 J = 5.51E-7 m = 550 nm(yellow/green b = c

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Chem 209- Week 4 Solutions Jaime Ambrosio and Cheng Cheng 1. a) λ = hc/∆E = (6.626E-34)*(3E8 m/s)/(3.61E-19 J) = 5.51E-7 m = 550 nm (yellow/green) b) λ = c/ ν = (3E8 m/s)/(93.5E6 Hz) = 3.21 m E = h ν = (6.626E-34 Js)*(93.5E6 Hz) = 6.19E-26 J 2. a) r n = (n 2 /Z)*a 0 = (4 2 /10)*.529Å = .8464 Å b) ∆p = m e ∆v ∆p∆x ≥ h/4 π ∆x = (.01)*(.8464E-10 m)= 8.646E-13 m m e ∆v*(8.646E-13 m) ≥ h/4 π ∆v ≥ 7E7 3. a) ν = c/ λ = R H Z 2 (1/n f 2 – 1/n i 2 ) (3E8 m/s)/(117.3E-9 m) = (3.2898E15 Hz)*(1/3 2 – 1/4 2 )*Z 2 Z=4, Be 3+ 4. (.372gCO2/.137gsample)*(1molCO2/44g)*(1molC/1molCO2)*(12g/1molC)= 74.11% C (.091gH2O)/.137gsample)*(1molH2O/18g)*(2molH/1molH2O)*(1g/molH)= 7.43% H (.0158gN/.183gsample)= 8.63% N 100-74-7-8= 9.8% O (74.11gC)*(1mol/12g) = 6.17mol 10mol C (7.43gH)*(1mol/1g) = 7.37mol 12mol H (8.63gN)*(1mol/14g) = .616mol 1mol N (9.828gO)*(1mol/16g) = .614mol 1mol O empirical formula:
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This note was uploaded on 06/02/2008 for the course CHEM 2090 taught by Professor Zax,d during the Fall '07 term at Cornell University (Engineering School).

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