Sol8 - #1 C O2 →CO2 mo m 2 ×16 12 = Ro c = → mc = o = 96 g × = 36 g mc 12 Ro c 2 × 16#2 4Al 3 O2 →2Al2O3(2 ×16 × 3(2 ×16 × 3 → mo =

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Unformatted text preview: #1. C + O2 →CO2 mo m 2 ×16 12 = Ro c = → mc = o = 96 g × = 36 g mc 12 Ro c 2 × 16 #2. 4Al + 3 O2 →2Al2O3 (2 ×16) × 3 (2 ×16) × 3 → mo = m Al × RO Al = 54 × = 48 g 27 × 4 27 × 4 #3. mH = 1.67 ×10 −27 kg , me = 9.1× 10 −31 kg RO Al = mH 1.67 × 10 − 27 kg N= = = 1835 me 9.1× 10 −31 kg #4. 1.6 × 10 −19 C × N = 5 × 10 −6 C N = 3.125 × 1013 #5. v = c = fλ 3 × 108 m / s f= = = 2 × 1018 Hz −10 λ 1.5 ×10 m c 1 #6. = 1.097 ×107 m −1 × ( λ λ = 6.56 ×10 −7 m #7. 1 11 −) 2 2 32 1 ) λ 32 λ = 1.03 ×10 −7 m → 103nm = 1.097 × 107 m −1 × (1 − not visible ‐> because of visible portion between 400nm and 700nm v = c = fλ #8. (a) (b) E = hf = 6.626 × 10 −34 J ⋅ s × 4.62 ×1014 Hz f= #9. (a) c λ f= = 3 ×108 m / s = 4.62 ×1014 Hz −9 650 ×10 m E 1.89eV = = 4.565 ×1014 Hz −15 h 4.14 × 10 eV ⋅ s v (b) = c = fλ λ= c 3 ×108 m / s = = 6.57 × 10 −7 m f 4.565 × 1014 m = 3.06 ×10 −19 J or = 4.14 ×10 −15 eV ⋅ s × 4.62 ×1014 Hz = 1.912eV ...
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This note was uploaded on 10/20/2011 for the course PHY 309L taught by Professor Sai during the Spring '08 term at University of Texas at Austin.

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