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Chapter 2 Atoms and Elements 29 Cumulative Problems 2.95 Given: 7.83 g HCN sample 1: 0.290 g H; 4.06 g N; 3.37 g HCN sample 2 Find: g C in sample 2 Conceptual Plan: g HCN sample 1 C in HCN sample 1 ratio g C to g HCN 1 g C in HCN sample 2 gC HCN HCN HCN Solution: 7.83 g HCN 0.290 g H 4.06 N = 3.48 C 3.37 7.83 3.48 C = Check: The units of the answer (g C) are correct. The magnitude of the answer is reasonable because the sample size is about half the original sample size and the gC are about half the original g C. 2.97 Given: in CO, mass ratio O:C = 1.33:1; in compound X, mass ratio O:C = 2:1 Find: formula of X Conceptual Plan: Determine the mass ratio of 0:0 in the two compounds. Solution: For 1 gram of C 2gO 1.33 in g compound in CO X 1.5 So the ratio of to C in compound X has to be 1.5:1, and the formula is C₂O₃. Check: The answer is reasonable because it fulfills the criteria of multiple proportions and the mass ratio of O:C is 2:1. 2.99 Given: = 4.00151 amu Find: charge-to-mass ratio C/kg Conceptual Plan: Determine total charge on and then amu g kg +1.60218 1.66054 proton 1 1000 Solution: 1 atom 2 protons X +1.60218 proton C = 3.20436 atom X 10⁻¹⁹ C 4.00151 amu X 1.66054 X X 1 kg = 6.64466742 X 10²⁷ kg 1 atom 1 1 atom 3.20436 10⁻¹⁹ X 6.64466742 1 X = 4.82245 X 10⁷ C/kg Check: The units of the answer (C/kg) are correct. The magnitude of the answer is reasonable when compared to the charge-to-mass ratio of the electron. 2.101 Given: graph of mass and % abundance of Pb Find: atomic mass of Pb Conceptual Plan: % abundance relative fraction abundance mass Eu % abundance of each isotope sum of all % abundance Atomic mass = (fraction of isotope n) (mass of isotope n) 2.7 46.0 Solution: fraction = = 0.0141 fraction = 190.9 100.0 fraction = fraction = = Atomic mass = (fraction of isotope X (mass of isotope n) n = amu) + 0.2410(206 amu) + amu) + 0.5238(208 amu) = 207.24 amu = 207 amu Check: The units of the answer, (amu) are correct. The magnitude of the answer is reasonable because the fraction of the heavier isotopes is greater than that of the lighter isotopes. 2.103 A Z = number of neutrons. 236 90 = 146 neutrons. So any nucleus with 146 neutrons is an isotone of Some would be and Copyright © 2017 Pearson Education, Inc.