# hw2sol - H omework #2 Solutions 3.091 Fall Term 2004 From...

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3.091 Fall Term 2004 Homework #2 Solutions From the text: 3-7 . For = 440 Hz: Assume the speed of sound is 1116 ft/sec (3-6) 440 cycles s λ = 1116 ft s = 1116 ft s s 440 cycles = 2.54 ft cycle For 880 Hz: 880 cycles s = 1116 ft s = 1116 ft s s 880 cycles = 1.27 ft cycle As the frequency increased by a factor of 2 the wavelength is decreased by a factor of ½ . The speed at which the sound travels to your ear remains constant at 1116 ft/s. 3-25. 243.4kJ mole 1000J kJ 1 mole 6.022 × 10 23 molecules Cl 2 = 4.042 × 10 19 J molecule Cl 2 (3-25) E = hv = hc/ λ λ = hc/ E = 6.626 × 10 34 Js × 2.997925 × 10 8 m/s 4.042 × 10 19 J/molecule Cl 2 = 4.914 × 10 7 m This radiation falls in the visible light portion of the electromagnetic spectrum.

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3.091 Homework #2 Solutions page 2 3-201.
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## This note was uploaded on 11/02/2009 for the course CHEMISTRY 3.091 taught by Professor Donsadoway during the Fall '04 term at MIT.

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hw2sol - H omework #2 Solutions 3.091 Fall Term 2004 From...

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