Chapter_2_ISM

# Chapter_2_ISM - Chap ter 2 Practice Exer cises 2.1 The f...

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Chapter 2 15 Practice Exercises 2.1 The first sample has a ratio of 1.25 g Cd 0.357 g S Therefore, the second sample must have the same ratio of Cd to S: 1.25 g Cd x 0.357 g S 3.50 g S = Cross-multiplication gives, (1.25 g Cd)(3.50 g S) = x(0.357 g S) (1.25g Cd)(3.50 g S) x 0.357g S = x = 12.3 g Cd 2.2 Compare the ratios of the mass of the compound before heating and the mass of the iron after heating, if they are the same, the compounds are the same. Sample Ratio A 25.36 g 16.11 g = 1.574 B 15.42 g 8.28 g = 1.86 C 7.85 g 4.22 g = 1.86 D 11.87 g 7.54 g = 1.57 Compounds A and D are the same, as are compounds B and C. 2.3 240 94 Pu The bottom number is the atomic number, found on the periodic table (number of protons). The top number is the mass number (sum of the number of protons and the number of neutrons). Since it is a neutral atom, it has 94 electrons. 2.4 35 17 Cl contains 17 protons, 17 electrons, and 18 neutrons. 2.5 We can discard the 17 since the 17 tells the number of protons which is information that the symbol "Cl" also provides. In addition, the number of protons equals the number of electrons in a neutral atom, so the symbol "Cl" also indicates the number of electrons. The 35 is necessary to state which isotope of chlorine is in question and therefore the number of neutrons in the atom. 2.6 2.24845 × 12 u = 26.9814 u 2.7 Copper is 63.546 u ÷ 12 u = 5.2955 times as heavy as carbon 2.8 (0.198 × 10.0129 u) + (0.802 × 11.0093 u) = 10.8 u 2.9 (a) 1 Ni, 2 Cl (b) 1 Fe, 1 S, 4 O (c) 3 Ca, 2 P, 8 O (d) 1 Co, 2 N, 12 O, 12 H

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Chapter 2 16 2.10 (a) 2 N nitrogen, 4 H hydrogen, 3 O oxygen (b) 1 Fe iron, 1 N nitrogen, 4 H hydrogen, 2 S sulfur, 8 O oxygen (c) 1 Mo molybdenum, 2 N nitrogen, 11 O oxygen, 10 H hydrogen (d) 6 C carbon, 4 H hydrogen, 1 Cl chlorine, 1 N nitrogen, 2 O oxygen 2.11 This is a balanced chemical equation, and the number of each atom that appears on the left is the same as that on the right: 1 Mg, 2 O, 4 H, and 2 Cl. 2.12 Mg(OH) 2 ( s ) + 2HCl( aq ) MgCl 2 ( aq ) + 2H 2 O 2.13 6 N, 42 H, 2 P, 20 O, 3 Ba, and 12 C are on both the products and the reactants sides of the equation, therefore the reaction is balanced. 2.14 The term "octa' means eight, therefore there are 8 carbon atoms in octane. The formula for an alkane is C n H 2n+2 , so octane has 8 carbons and ((2 × 8) + 2) = 18 H. The condensed formula is CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 , and the structural format is: C C C C C C C C H H H H H H H H H H H H H H H H H H 2.15 The condensed formula is CH 3 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 2 CH 3 The structural formula is: C C C C C C C C C H H H H H H H H H H H H H H H H H C H H H H H 2.16 (a) Propanol: CH 3 CH 2 CH 2 OH (b) Butanol: CH 3 CH 2 CH 2 CH 2 OH C C C O H C H H H H H H H H H 2.17 (a) Fe: 26 protons and 26 electrons (b) Fe 3+ : 26 protons and 23 electrons (c) N 3– : 7 protons and 10 electrons (d) N: 7 protons and 7 electrons 2.18 (a) O: 8 protons and 8 electrons (b)
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## This note was uploaded on 02/13/2011 for the course CHEMISTY 155 taught by Professor Bullock during the Spring '11 term at UCSB.

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Chapter_2_ISM - Chap ter 2 Practice Exer cises 2.1 The f...

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