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chapter02v2 QRT

# chapter02v2 QRT - Chapter 2 Atoms and Elements 25...

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Chapter 2: Atoms and Elements 25 End-of-Chapter Solutions for Chapter 2 Summary Problem Answer: (a) silver, Ag (b) Z = 47, A = 109, (c) Atomic weight = 107.8 (d) Group 1B, metal (e) (i) 9.27 × 10 –3 mol Ag (ii) 5.58 × 10 21 Ag atoms (iii) 1.70 × 10 12 m Strategy and Explanation: Given the number of electrons, protons, and neutrons in the atom, determine the identity of the element, its symbol, the atomic number, the mass number, its group in the periodic table, and whether it is a metal, nonmetal or metalloid. Given the isotopic abundance and masses of two isotopes of this element, determine its atomic weight. Given the mass of the elements, determine the number of moles and the number of atoms. Given the diameter of atoms of this elements, determine how many meters long a chain of this many atoms would be. (a) The atomic number is the same as the number of protons. The number of protons is the same as the number of electrons in an uncharged atom. The element is silver, with a symbol of Ag. (b) Atomic number = Z = number of protons = 47. Mass number = A = Z + number of neutrons = 47 + 62 = 109. (c) The element is found in Group 1B, one of the groups of the transition metals of the periodic table. It is a metal. (d) Calculate the weighted average of the isotope masses. Every 10000 atoms of the element contains 5184 atoms of the 107 Ag isotope and 4816 atoms of the 109 Ag isotope. 5184 atoms 107 Ag 10000 Ag atoms " 106.905 amu 1 atom 107 Ag # \$ % % & ( ( + 4816 atoms 109 Ag 10000 Ag atoms " 108.905 amu 1 atom 109 Ag # \$ % % & ( ( = 107.8 amu/Ag atom (e) (i) Using the atomic weight calculated in (d), we can say that the molar mass is 107.8 g/mol. 1.00 g Ag " 1 mol Ag 107.8 g Ag = 9.27 × 10 –3 mol Ag (ii) Use Avogadro’s number to determine the actual number of silver atoms. 9.27 × 10 –3 mol Ag " 6.022 " 10 23 Ag atoms 1 mol Ag atoms = 5.58 × 10 21 Ag atoms (iii) The diameter of one atom is 304 pm. Multiply the number of atoms by the atomic diameter, and use metric conversions to determine the number of meters. 5.58 × 10 21 Ag atoms × (304 pm) " 1 " 10 # 12 m 1 pm = 1.70 × 10 12 m Reasonable Answer Check: The periodic table lists an atomic weight the same as the one calculated (within the given significant figures). The number of moles is smaller than the number of grams. The number of atoms is very large. The length of the atom chain is quite long, considering the small size of each atom; however, given the large number of atoms it makes sense that the chain would be long. Questions for Review and Thought Review Questions 1. The coulomb (C) is the fundamental unit of electrical charge. 2. Millikan devised an experiment to determine the charge of an electron. It consisted of a chamber with electrically charged plates on the top and bottom. (Figure 2.2). Tiny oil droplets were sprayed into the chamber. As the droplets settled slowly through the chamber, they were exposed to x-rays. This caused electrons from gas molecules in the air to be transferred to the oil droplets. Using a small telescope to observe the tiny droplets, he then adjusted the electrical charge on the plates so that the upward pull of the electrostatic charges just balanced the downward motion of the droplet due to gravity. From this, he was able to calculate the charge

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