solut02 - CHAPTER TWO ATOMS, MOLECULES, AND IONS...

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1 CHAPTER TWO ATOMS, MOLECULES, AND IONS Development of the Atomic Theory 20. 1.188 1.188 =1.000; 1.188 2.375 = 1.999; 1.188 3.563 = 2.999 The masses of fluorine are simple ratios of whole numbers to each other, 1:2:3. 21. From Avogadro&s hypothesis, volume ratios are equal to molecule ratios at constant temperature and pressure. Therefore, we can write a balanced equation using the volume data, Cl 2 + 3 F 2 2 X. Two molecules of X contain 6 atoms of F and two atoms of Cl. The formula of X is ClF 3 for a balanced reaction. 22. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (i.e., depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadro&s hypothesis implies that volume ratios are equal to molecule ratios at constant temperature and pressure. H 2 + Cl 2 2 HCl. From the balanced equation, the volume of HCl produced will be twice the volume of H 2 (or Cl 2 ) reacted. 23. To get the atomic mass of H to be 1.00, we divide the mass that reacts with 1.00 g of oxygen by 0.126, i.e., 0.126 0.126 = 1.00. To get Na, Mg, and O on the same scale, we do the same division. Na: 0.126 2.875 = 22.8; Mg: 0.126 1.500 = 11.9; O: 0.126 1.00 = 7.94 H O Na Mg Relative Value 1.00 7.94 22.8 11.9 Accepted Value 1.0079 15.999 22.99 24.31
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2 The atomic masses of O and Mg are incorrect. The atomic masses of H and Na are close. Something must be wrong about the assumed formulas of the compounds. It turns out the correct formulas are H 2 O, Na 2 O, and MgO. The smaller discrepancies result from the error in the assumed atomic mass of H. The Nature of the Atom 24. Deflection of cathode rays by magnetic and electric fields led to the conclusion that they were negatively charged. The cathode ray was produced at the negative electrode and repelled by the negative pole of the applied electric field. 25. β particles are electrons. A cathode ray is a stream of electrons ( β particles). 26. Density of hydrogen nucleus (contains one proton only): V nucleus = cm 10 _ 5 = ) cm 10 _ (5 (3.14) 3 4 = r 3 4 3 40 - 3 14 - 3 π cm g/ 10 _ 3 = cm 10 _ 5 g 10 _ 1.67 = density = d 3 15 3 40 - -24 Density of H-atom (contains one proton and one electron): V atom = cm 10 _ 4 = ) cm 10 _ (1 (3.14) 3 4 3 24 - 3 8 - d = cm g/ 0.4 = cm 10 _ 4 g 10 _ 9 + 10 _ 1.67 3 3 24 - -28 -24 Since electrons move about the nucleus at an average distance of about 1 × 10 -8 cm, then the diameter of an atom is about 2 × 10 -8 cm. Let’s set up a ratio: cm 10 _ 2 cm 10 _ 1 = model of diameter mm 1 = atom of diameter nucleus of diameter 8 - -13 , Solving: diameter of model = 2 × 10 5 mm = 200 m 27. First, divide all charges by the smallest quantity, 6.40 × 10 -13 .
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CHAPTER 2 ATOMS, MOLECULES, AND IONS 3 10 _ 6.40 10 _ 2.56 13 - -12 = 4.00; 0.640 7.68 = 12.00; 0.640 3.84 = 6.00 Since all charges are whole number multiples of 6.40 × 10 -13 zirkombs then the charge on one electron could be 6.40 × 10 -13 zirkombs. However, 6.40 × 10 -13 zirkombs could be the charge of two electrons (or three electrons, etc.). All one can conclude is that the charge of an electron is 6.40 × 10 -13 zirkombs or an integer fraction of 6.40 × 10 -13 .
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solut02 - CHAPTER TWO ATOMS, MOLECULES, AND IONS...

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