p42_041 - . 14 10 22 ) / (5 10 6 ) = 4 . 29 10 15...

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41. Sample Problem 42-6 gives the fraction of silicon atoms that must be replaced by phosphorus atoms. We Fnd the number the silicon atoms in 1 . 0 g, then the number that must be replaced, and Fnally the mass of the replacement phosphorus atoms. The molar mass of silicon is 28 . 086 g / mol, so the mass of one silicon atom is (28 . 086 g / mol) / (6 . 022 × 10 23 mol 1 )=4 . 66 × 10 23 g and the number of atoms in 1 . 0g is (1 . 0g) / (4 . 66 × 10 23 g) = 2 . 14 × 10 22 . According to Sample Problem 42-6 one of every 5 × 10 6 silicon atoms is replaced with a phosphorus atom. This means there will be (2
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Unformatted text preview: . 14 10 22 ) / (5 10 6 ) = 4 . 29 10 15 phosphorus atoms in 1 . 0 g of silicon. The molar mass of phosphorus is 30 . 9758 g / mol so the mass of a phosphorus atom is (30 . 9758 g / mol) / (6 . 022 10 23 mol 1 ) = 5 . 14 10 23 g. The mass of phosphorus that must be added to 1 . 0 g of silicon is (4 . 29 10 15 )(5 . 14 10 23 g) = 2 . 2 10 7 g....
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