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14 Chapter 1 Matter, Measurement, and Problem Solving V = = 1.5876 = 3.520 X 10² cm³ = 3.5 X 10² X (2.54 = 21 in³ Check: The units (in³) are correct. The magnitude of the answer seems correct considering the number of grams. Two significant figures are allowed to reflect the significant figures in 3.5 lb. Truncate the nonsignificant digits because the first nonsignificant digit is a 2. 1.119 Given: cylinder dimensions: length = 2.16 in; radius = 0.22 in; m = 41 g Find: density (g/m³) Conceptual Plan: in cm then 1, r V then m, V d 2.54 cm V = in Solution: 2.16 in X in = 5.4864 cm = 0.22 in 2.54 in cm = 0.5588 cm = r V = = (5.4864 = 5.3820798 cm³ d = V = 5.3820798 41 = 7.6178729 = cm³ Check: The units (g/cm³) are correct. The magnitude of the answer seems correct considering that the value of the density of iron (a major component in steel) is Two significant figures are allowed to reflect the significant figures in 0.22 in and 41 g. Truncate the nonsignificant digits because the first nonsignificant digit is a 1. 1.121 Given: 185 cubic yards (yd³) of Find: mass of the H₂O (pounds) Other: d(H₂O) = 1.00 at Conceptual Plan: yd³ m³ cm³ g lb lb (1.094 1.00 cm³ 453.59 Solution: 185 X (1.094 X (100 1.00 X lb = 3.114987377 X 10⁵ lb = 3.11 X 10⁵ lb Check: The units (lb) are correct. The magnitude of the answer (10⁵) makes physical sense because a pool is not a small object. Three significant figures are allowed because the conversion factor with the least precision is the density [1.00 g/cm³ - (3 significant figures) and the initial size has three significant figures. Truncate after the last digit because the first nonsignificant digit is a 4. 1.123 Given: 15 liters of gasoline Find: kilometers Other: 52 mi/gal in the city Conceptual Plan: L gal mi km 1 gal 52 mi 1 km 3.785 L 1 gal 0.6214 mi Solution: X 1 gal X 52 gal X 0.6214 = 3.316327941 X 10² km = 3.3 X 10² km Check: The units (km) are correct. The magnitude of the answer (10²) makes physical sense because the dominating conversion factor is the mileage, which increases the answer. Two significant figures are allowed because the conversion factor with the least precision is 52 mi/gal (two significant figures) and the initial volume has two significant figures. Truncate the last digit because the first nonsignificant digit is a 1. It is best to put the answer in scientific notation so that it is clear how many significant figures are expressed. 1.125 Given: radius of nucleus of the hydrogen atom = 1.0 X 10⁻¹³ cm; radius of the hydrogen atom = 52.9 pm Find: percent of volume occupied by nucleus (%) Conceptual Plan: cm m then pm m then r V then % 100 cm 10¹² pm Solution: 1.0 10⁻¹³ cm 100 = 1.0 X 10⁻¹⁵ m and 52.9 pm X = 5.29 X 10⁻¹¹ m V = Substitute into % V equation. Copyright © 2017 Pearson Education, Inc.