1 (21)

Prévia do material em texto

1 Matter and Measurement Solutions to Exercises -10°C: 1.4955 10³ g 0.917 g X 1000 1L 3 = 1.6309 = 1.63 L 1 cm³ (b) No. If the soft-drink bottle is completely filled with 1.50 L of water, the 1.63 L of ice cannot be contained in the bottle. The extra volume of ice will push through any opening in the bottle, or crack the bottle to create an opening. 1.71 mass of toluene = 58.58 g 32.65 g = 25.93 g volume of toluene = 25.93 g 0.864g 1 mL = 30.0116 = 30.0 mL volume of solid = 50.00 mL- 30.0116 mL = 19.9884 = 20.0 mL density of solid = 19.9884 32.65g = 1.72 (a) V = 4/3 π = 4/3 π (28.9 cm)³ = 1.0111 10⁵ = 1.01 10⁵ = 1.01 10⁵ cm³ 1.0111 X 10⁵ cm³ 19.3 cm³ g X 453.59 1 lb = 4,302 = 4.30 10³ lb (b) No. The sphere weighs 4300 pounds, more than two tons. The student is unlikely to be able to carry the it without asssistance. 1.73 1.00 gal battery acid X 1 4 gal qt X 1000 1.0567 mL qt 1.28 mL = 4,845.3 = 4.85 X 10³ g battery acid 4.8453 10³ g battery acid X g = 1846 = 1.85 10³ g sulfuric acid 38.1 sulfuric acid 100 g battery acid 1.74 (a) 14 40 20 lb peat 30 X (2.54)³ 1in³ cm³ X 453.6g = 0.13 g/cm³ peat 40 1.9 lb gal soil 1 4 gal qt 1.057 1L qt X 1 1 mL 10⁻³ L X 1cm 1 mL 3 453.6 = soil No. Volume must be specified in order to compare mass. The densities tell us that a certain volume of peat moss is "lighter" (weighs less) than the same volume of top soil. (b) 1 bag peat = 14 20 30 = 8.4 10³ in³ 15.0ft 3.0 in in² = 129,600 = 1.3 10⁵ in³ peat needed 129,600 in³ 1 bag in³ = 15.4 = 15 bags (Buy 16 bags of peat.) 1.75 8.0 OZ 1lb 453.6 lb 2.70g = 84.00 = 84 3 84 cm³ X 1 cm = 0.018 mm 15