Key H1 FA 2010

Key H1 FA 2010 - C111 FA 2010 Homework 1 Solution Guide and...

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C111 FA 2010 Homework 1 Solution Guide and Answer Key 1. The calculation to perform: ( 29 x 3 1905 . 0 3 . 1 + = ( 29 3 178 . 0 09 . 7 simplify numerator on the left side, and cube the 0.178 term on the right side: 1.4905 = 7.09 Do not round off until the end of the calculation. 3x 0.005639752 But remember the 1.4905 only has one decimal pl. which means a datum with only 2 sig figs multiply both sides by x and by 0.005639752; divide both sides by 7.09 to isolate the “x”: (0.005639752)(1.491) = x (3)(7.09) x = 3.95339 x 10 - 4 (report 2 sig figs because limit is set by the original 1.3 term with only 2 sig figs.) = 0.00040 (standard notation) or 4.0 x 10 - 4 (scientific notation ) 2. atom g 10 x 271 . 3 atoms 10 x 022 . 6 mole 1 x mole g 967 . 196 22 23 - = moles in the denominator of first term cancel moles in the numerator of second term. Multiply by 5 atoms, to get 1.635 x 10 - 21 g. (4 sig figs) (Notice 5 is an integer number of objects and does not limit sig. figs.). 3. With only 4 sig figs in the density data, the answer may have a maximum of 4 sig figs. Assume only platinum (Pt) present in the sample, and the purity of the sample is 100%, so this density can be used. (a)
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This note was uploaded on 01/18/2012 for the course CHEM 112 taught by Professor Lemaster during the Spring '08 term at Boise State.

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Key H1 FA 2010 - C111 FA 2010 Homework 1 Solution Guide and...

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