p0114 - t = 2 . 69 10 16 molecules 10 10 molecules / min =...

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1.14. CHAPTER 1, PROBLEM 14 15 1.14 Chapter 1, Problem 14 We know from the text that 1 cm 3 of air at STP contains 2 . 69 · 10 19 molecules. Thus, since 1 mm = 0.1 cm, 1 mm 3 of air at STP contains 2 . 69 · 10 16 molecules. The time to count them at a rate of ten billion per minute is
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Unformatted text preview: t = 2 . 69 10 16 molecules 10 10 molecules / min = 2 . 69 10 6 minutes = 4 . 48 10 4 hours = 1868 days = 5 . 1 years Note that the final equality makes provision for leap years....
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This note was uploaded on 10/12/2009 for the course AME 309 at USC.

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