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1、Chapter 1 1.1 (a) Pure element: i, v (b) mixture of elements: vi (c) pure compound: iv (d) mixture of an element and a compound: ii, iii 1.5 (a) 7.5 cm, 2 significant figures (sig figs) (b) 140C, 2 sig figs 1.8 1.9 (a) Heterogeneous mixture (b) homogeneous mixture (heterogeneous if there are undisso
2、lved particles) (c) pure sub- stance (d) homogeneous mixture.1.11 (a) S (b) K (c) Cl (d) Cu (e) Si (f) N (g) Ca (h) He1.13 (a) Lithium (b) alu- minum (c) lead (d) sulfur (e) bromine (f) tin (g) chromium (h) zinc1.15 C is a compound; it contains carbon and oxygen. Ais a compound; it contains at least
3、 carbon and oxygen. B is not defined by the data given; it is probably a compound be- cause few elements exist as white solids.1.17 Physical prop- erties: silvery white; lustrous; melting point 649C; boiling point 1105C; density at 20C pounded into sheets; drawn into wires; good conductor. Chemical
4、properties: burns in air; reacts with 1.19 (a) Chemical (b) physical (c) physical (d) chemical (e) chemical1.21 (a) Add water to dissolve the sugar; filter this mixture, collecting the sand on fil- ter paper and the sugar water in a flask. Evaporate water from the flask to reproduce solid sugar. (b)
5、 Heat the mixture until sulfur melts, then decant the liquid sulfur.1.23 (a) (b)(c)(d)(e)(f) (g)(h)(i)1.25 (a) (b) 0.40 nm (c)1.27 (a)Carbon tetrachlo- ride, is more dense than water, carbon tetrachloride will sink rather than float on water. (b) 1.609 kg (c)1.29 (a) Calculated density The substance
6、 is probably toluene, density (b) 40.4 mL ethylene glycol (c)nickel 1.3146 nm1.33 (a) 17C (b) 422.1F (c) 506 K (d) 108F (e) 1644 K1.35 Exact: (c), (d), and (f)1.37 (a) 3 (b) 2 (c) 5 (d) 3 (e) 51.39 (a)(b) (c)(d)(e) 1.41 (a) 21.11 (b) 237.4 (c) 652 (d) 1.43 (a)(b) 1.45 (a) 76 mL (b) 50 nm (c)(d) (e)(
7、f)1.47 (a) (b) 88.5 m (c)(d)(e) (f)1.49 (a)(b) (c)(d)1.51 52 kg air1.53 Use the cm as a unit for comparison The order of length from short- est to longest is 1.55 (a) $0.91 (b) Since coins come in integer numbers, 7 coins are required. 1.57 Composition is the contents of a substance; structure is th
8、e arrangement of these contents.1.60 8.47 g O; the law of con- stant composition1.63 (a) Volume (b) area (c) volume (d) density (e) time (f) length (g) temperature1.66 (a) Densi- ty of peat density of topsoil It is not correct to say that peat is “lighter” than topsoil. Volumes must be specified in
9、order to compare masses. (b) Buy 7 bags of peat (6.9 are needed).1.69 (a)(b) 2.54 L4.67 gcm3 2.5 gcm3.=0.13 gcm3,= 14 in. 6 57 cm 6 1.1 m. 1.1 m = 110 cm.14 in. L 35 cm; 57 cm = 57 cm;(1 in. L 2.5 cm). 26 mL9.64 kmL 4 * 102 mg1.2 * 102 L707.9 cm3 1.420 Ls46.6 kmhr$0.499L 4.32 * 105 s6.151 * 10-3 Ls1
10、.55 gL 2.3 * 102 g6.88 * 10-4 s 1 in.3= (2.54)3 cm312 in. = 1 ft2.54 cm = 1 in., 7.66 * 10-2 -3.572 * 10-22.579 * 10-48.543 * 10-3 6.570 * 1051.025 * 102 4.6 * 10-8 m; 1.11 * 103 g 0.866 gmL.= 0.86 gmL.=50.35 cm3 1.00 gmL;1.59 gmL, 1.59 gmL.575 mm 2.55 * 10-2 g1 * 10-121 * 10-31 * 10-9 1 * 1031 * 10
11、61 * 10-61 * 10-151 * 10-2 1 * 10-1 Cl2. 1.738 gcm3;= = GivenFind 1 km 0.62 min mi/hrkm/hrkm/minkm/s 1 hr 60 min 1 hr 60 min (c)The sphere weighs 1950 kg, or 4300 lb; the thief is unlikely to be able to carry it.1.73 (a) Average speed (b) pace or 1.76 (a)or (b) 1.79 (a) 61.5% Au by mass (b) 15 carat
12、 gold1.81 The separa- tion is successful if two distinct spots are seen on the paper. To quantify the characteristics of the separation, calculate a refer- ence value for each spot: distance travelled by spot distance travelled by solvent. If the values for the two spots are fairly different, the se
13、paration is successful. Chapter 2 2.1 (a) The path of the charged particle bends because it is re- pelled by the negatively charged plate and attracted to the pos- itively charged plate. (b)(c) increase (d) decrease 2.3 The particle is an ion. 2.5 Formula: name: io- dine pentafluoride; the compound
14、is molecular.2.7 Postulate 4 of the atomic theory states that the relative number and kinds of atoms in a compound are constant, regardless of the source. Therefore, 1.0 g of pure water should always contain the same relative amounts of hydrogen and oxygen, no matter where or how the sample is obtai
15、ned.2.9 (a) (b) The numbers in part (a) obey the law of multiple proportions. Multi- ple proportions arise because atoms are the indivisible entities combining, as stated in Daltons atomic theory.2.11 (1) Elec- tric and magnetic fields deflected the rays in the same way they would deflect negatively
16、 charged particles. (2) Ametal plate exposed to cathode rays acquired a negative charge. 2.13 (a) If the positive plate were lower than the negative plate, the oil drops “coated” with negatively charged electrons would be attracted to the positively charged plate and would descend much more quickly. (b) The more times a measure- ment is repeated, the better the chance of detecting and com- pensating for experimental errors. Millikan wanted to demonstrate the validity of his result via i
