Chapter 3 Assignment Answers 57. Lissa was accurate (her average is 47.34) but not precise. Lamont was fairly accurate, and precise. Leigh Anne was not accurate, but was precise. 58. a. infinite b. 4 59. a. 98.5 L b. 0.000763 cg 60. a. 43 g b. 225.8 L 61. 59(a) 9.85 x 101 c. infinite d. infinite c. 57.0 m c. 92.0 kg d. 12.2 °C d. 32.4 m3 59(b) 7.63 x 10-4 59(c) 5.70 x 101 59(d) 1.22 x 101 60(a) 4.3 x 101 60(b) 2.258 x 102 60(c) 9.20 x 101 60(d) 3.24 x 101 62. Error = (experimental value β accepted value) Percent error = 63. a. second |error| x 100 accepted value b. meter c. kelvin d. kilogram 66. K = °C + 273. 962 °C + 273 = 1235 K 67. A conversion factor is a ratio of equivalent measurements. 68. The two measurements must equal each other. 69. The unit in the denominator of the conversion factor must be identical to the unit of the given or starting measurement. This is so the units cancel out. 1π 70. a. 157 ππ π₯ b. 42.7 πΏ π₯ 100 ππ = 1.57 ππ 1000 ππΏ c. 261 ππ π₯ 1πΏ 1π = 42,700 ππΏ π₯ 109 ππ 1000 ππ 1π = 2.61 π₯ 10β4 ππ d. 0.065 ππ π₯ 1000 π 1 ππ 1π e. 642 ππ π₯ 100 ππ 2 f. 8.25 π₯ 10 ππ π₯ π₯ π₯ 10 ππ 1π = 650 ππ 1 ππ β3 = 6.42 π₯ 10 ππ 1000 π 1π 100 ππ π₯ 109 ππ 1π = 8.25 π₯ 109 ππ 73. a. 283 mg b. 0.283 g c. 0.000283 kg d. 6.6 g e. 660 cg f. 0.0066 kg g. 0.28 mg h. 0.028 cg i. 2.8 x 10-7 kg 74. ππππ ππ‘π¦ = πππ π π£πππ’ππ 76. Calculate the density of the bar, and compare it to goldβs density. π= πππ π 57.3 π π = = 12 π£πππ’ππ 4.7 ππ3 ππ3 According to Table 3.6 on page 90, this does not match goldβs density of 19.3 g/cm3. 79. e, d, c, f, a, b 80. a. accurate and precise b. inaccurate and precise c. inaccurate and imprecise 82. Germaniumβs melting point is 1210 K, or 937 °C. This is lower than goldβs melting point of 1064 °C. So germanium would melt first. 86. Use density as a conversion factor: 1 ππ3 50.0 π π₯ = 4.20 π₯ 104 ππ3 β3 1.19 π₯ 10 π Or you could use the density formula and solve for V. 88. 91. 1 πππ¦ π₯ 24 βπ 1 πππ¦ 1.5 π₯ 108 ππ π₯ 93. 5.52 97. 112 π ππ3 ππ βπ π₯ π₯ π₯ 60 πππ 1 βπ 1000 π π₯ 1 ππ 1 ππ 1000 π 1000 π 1 ππ π₯ ( π₯ π₯ 0.15 s πππ π‘ 1 πππ 1π 3.0 π₯ 108 π 10 ππ 3 π₯ π₯ 60 πππ π₯ 1 πππ 60 π 60 π 1 πππ ) = 5.52 1 ππ 1 βπ 1 πππ 60 π = 3.6 ππππ’π‘ππ πππ π‘ = 8.3 ππππ’π‘ππ ππ ππ3 = 31.1 π π
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