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Exam 1 Solutions - General Chemistry I | CHEM 151, Exams of Chemistry

Material Type: Exam; Class: General Chemistry I; Subject: Chemistry; University: University of San Diego; Term: Unknown 1989;

Typology: Exams

Pre 2010

Uploaded on 08/18/2009

koofers-user-ump
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Download Exam 1 Solutions - General Chemistry I | CHEM 151 and more Exams Chemistry in PDF only on Docsity! Solutions 1a (suggested problems before Exam #1) Chem151 [Kua] 2.2 Molecular pictures must show the structures of individual particles (e.g., atoms, molecules, etc.) and the differences between phases. All these particles have monatomic units. A gas is mostly empty space, a liquid is tightly packed but not entirely regular, and a solid has a regular repeating structure. (a) liquid mercury (b) solid iron (c) gaseous neon 2.78 Molecular pictures must show that atoms of every element are conserved. The solid-gas transformation is correctly represented, but the number of molecules is not conserved. Here is one way to show a correct picture: 1.68 To identify each molecular picture, compare its contents with the definitions of element, compound, and mixture: (a) pure element (one type of atom); (b) heterogeneous mixture (one portion contains different elements than the other); (c) single chemical compound (two different elements, all molecules the same); and (d) homogeneous solution (two elements uniformly mixed). 2.14 Charge per proton and mass per proton can be found in Table 2-1. Because the total charge of the protons is simply the sum of the charges of the individual protons, to determine the number of protons, divide the total charge by the charge of a single proton: (a) #protons = protons10 x 52.1 C/e10 x 1.6022 C10 x 2.44 protonper charge charge total 7 19 21 == !! ! (b) To determine the mass of all of the protons, multiply the number of protons calculated above by the mass of a single proton: mprotons= (#protons)(mproton) = 1.52 x 10 7 protons !! " # $$ % & ' proton 1 kg 10 x 1.6726 27 = 2.54 x 10-20 kg Extra: We covered the Rutherford model in class. You should be able to explain the experimental setup, the observations, the conclusions and sketch the Rutherford model of the atom. 2.20 The left superscript in an isotopic symbol is the mass number, A, which is the sum of protons and neutrons; the left subscript is the atomic number, Z, which is the number of protons, and a right superscript indicates (protons – electrons), when this quantity is non- zero. (a) 30 protons, 36 neutrons, 28 electrons; (b) 7 p, 8 n, 7 e; (c) 35 p, 46 n, 34 e; (d) 35 p, 44 n, 36 e; and (e) 92 p, 146 n, 92 e. 2.22 An isotopic symbol has the elemental symbol prefaced by a superscript giving the mass number, A, which is the sum of protons and neutrons, and a subscript giving the atomic number, Z (the number of protons). Use the periodic table in your text to determine which elemental symbol corresponds to the atomic number. (a) Helium (He) has 2 protons, Z = 2; A = 2 + 1 = 3; 3 2 He; (b) Zinc (Zn) has 30 protons (see periodic table). 66 30 Zn; (c) Element 54 is xenon (Xe). A = 54 + 78 = 132. 132 54 Xe; (d) Nitrogen (N) has Z = 7. A = 7 + 7 = 14. 14 7 N. 2.26 A pie chart is a circular chart with each piece of the pie proportional in size to the percentage of the isotope that it represents. The relative mass is calculated by multiplying the mass number of each isotope with the fraction of its isotopic abundance and summing over all isotopes. Relative mass = 64(0.486) + 66(0.279) + 67(0.041) + 68(0.188) + 70(0.006) = 65.469
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