2010 Exam 1 Key

2010 Exam 1 Key - Name Section Number Chemistry 1314 Honors...

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N a m e Section Number Chemistry 1314 Honors Midterm Exam #1 September 15, 2010 This exam is closed book, notes, papers, etc. You may use a calculator. If you need a physical constant or unit conversion factor that is not provided, please ask. This exam must represent your own work. Useful information: N A = 6.022 x 10 23 particles/mol Scores: 1. a. / 6 b. / 6 c. / 4 d. / 4 2. a. / 6 b. / 4 c. / 8 d. / 2 3. a. / 10 b. / 10 4. a. / 2 b. / 10 b. / 8 5. / 20 Total: / 100
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CHEM 1314 H Midterm Exam #1, 2010 Page 2 1. In a highly unusual reaction, fluorine will combine with a particular element A to form three compounds. Compound I is 22.44% fluorine by mass, Compound II is 36.66% fluorine by mass, and Compound III is 46.47% fluorine by mass. a. By calculating mass ratios, show that these three compounds obey the law of multiple proportions. The law of multiple proportions states that if two elements combine to form multiple compounds, the different masses of one element that combine with a fixed mass of the other element can be expressed as a ratio of small whole numbers. For 100g of each compound I II III A 77.56 g 63.37 g 53.53 g F 22.44 g 36.66 g 46.47 g Taking an arbitrary fixed mass of fluorine (1.00 g), we need to determine the mass of A in the various compounds. Fixing the mass of the fluorine (or A): 2 points Ratioing the masses of A (or fluorine): 2 points Demonstrating simple whole number ratio: 2 points () ( ) ( ) 2 : 3 : 6 1 : 5 . 1 : 3 15 . 1 : 73 . 1 : 46 . 3 : : 15 . 1 47 . 46 53 . 53 00 . 1 : 73 . 1 66 . 36 33 . 63 00 . 1 : 46 . 3 44 . 22 56 . 77 00 . 1 : = = = = = = III m II m I m A g F g A g F g III A g F g A g F g II A g F g A g F g I A A A The masses of A for a fixed amount of O in the two compounds are in a simple whole number ratio, in accordance with the law of multiple proportions. b. Explain how these data indicate that element A must be composed of individual atoms of identical mass. For a fixed amount of F, the masses of A which combine are in simple integer multiples of a fixed mass. Therefore, there exists a discrete or particulate amount of fixed mass taken in integer units. This discrete amount of mass is the atom of A. Integer multiples (or simple whole numbers) mean we are “counting something”: 2 points We call that thing we are counting atoms: 2 points To be counting something by ratioing mass, all particles must have the same mass: 2 points
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CHEM 1314 H Midterm Exam #1, 2010 Page 3 c. Compound I has molecular formula AF 2 . Calculate the atomic mass of the unknown element A. Identify the element. Assume 100 g of AF
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2010 Exam 1 Key - Name Section Number Chemistry 1314 Honors...

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