Chatper 2 Text solutions

Chatper 2 Text solutions - CHAPTER 2 ATOMS, MOLECULES, AND...

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1 CHAPTER 2 ATOMS, MOLECULES, AND IONS 18. Law of conservation of mass: mass is neither created nor destroyed. The total mass before a chemical reaction always equals the total mass after a chemical reaction. Law of definite proportion: a given compound always contains exactly the same proportion of elements by mass. For example, water is always 1 g hydrogen for every 8 g oxygen. Law of multiple proportions: When two elements form a series of compounds, the ratios of the mass of the second element that combine with 1 g of the first element always can be reduced to small whole numbers. For CO 2 and CO discussed in section 2.2, the mass ratios of oxygen that react with 1 g carbon in each compound are in a 2 : 1 ratio. 19. From Avogadro’s hypothesis (law), 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. 20. a. The composition of a substance depends on the numbers of atoms of each element making up the compound (depends on the formula of the compound) and not on the composition of the mixture from which it was formed. b. Avogadro’s hypothesis (law) 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. 21. Hydrazine: 1.44 × 1 10 " g H/g N; Ammonia: 2.16 × 1 10 " g H/g N; Hydrogen azide: 2.40 × 2 10 " g H/g N; Let's try all of the ratios: 0240 . 0 144 . 0 = 6.00; 0240 . 0 216 . 0 = 9.00; 144 . 0 216 . 0 = 1.50 = 2 3 All the masses of hydrogen in these three compounds can be expressed as simple whole- number ratios. The g H/g N in hydrazine, ammonia, and hydrogen azide are in the ratios 6 : 9 : 1.
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2 CHAPTER 2 ATOMS, MOLECULES, AND IONS 22. The law of multiple proportions does not involve looking at the ratio of the mass of one element with the total mass of the compounds. To illustrate the law of multiple proportions, we compare the mass of carbon that combines with 1.0 g of oxygen in each compound: Compound 1: 27.2 g C and 72.8 g O (100.0 - 27.2 = mass O) Compound 2: 42.9 g C and 57.1 g O (100.0 - 42.9 = mass O) The mass of carbon that combines with 1.0 g of oxygen is: Compound 1: O g 8 . 72 C g 2 . 27 = 0.374 g C/g O Compound 2: O g 1 . 57 C g 9 . 42 = 0.751 g C/g O 1 2 374 . 0 751 . 0 # ; this supports the law of multiple proportions as this carbon ratio is a whole number. 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, that is, 0.126/0.126 = 1.00. To get Na, Mg, and O on the same scale, we do the same division. Na:
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This note was uploaded on 09/21/2008 for the course CHEM 21 taught by Professor Mukundan during the Fall '08 term at Duke.

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Chatper 2 Text solutions - CHAPTER 2 ATOMS, MOLECULES, AND...

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