2.1 · Measurement and CalculationPractice · 测量与计算

65 mi/h × $\dfrac{1.609\ \text{km}}{1\ \text{mi}}$ × $\dfrac{1000\ \text{m}}{1\ \text{km}}$ × $\dfrac{1\ \text{h}}{3600\ \text{s}}$ = 29.051 m/s. 1000 and 3600 are exact; 65 has 2 significant figures and 1.609 has 4, so report 2: 29 m/s.

Volume = 31.9 − 25.4 = 6.5 mL. The subtraction keeps 0.1 mL, which leaves only 2 significant figures. Density = 52.47 ÷ 6.5 = 8.072; the mass has 4 significant figures, but the volume has only 2, so report 8.1 g/mL. Keeping 4 because the mass has 4 is a common mistake.

The denominator first: 101.3 − 99.8 = 1.5. Subtraction keeps 0.1, leaving only 2 significant figures. The numerator, 8.314 × 298.2, has 4 each. The final division keeps the fewest, 2: 1652.8232 → 1.7 × 10³. Scientific notation is needed to show exactly 2 significant figures; 1650 or 1653 implies too many.

1 ft = 12 × 2.54 = 30.48 cm, and for volume the factor must be cubed: 1 ft³ = (30.48 cm)³ = 28316.84659 cm³ = 28316.84659 mL. Divide by 1000: 28.32 L. The factors are exact, so 1.00 (3 significant figures) sets the precision: 28.3 L. Multiplying by 30.48 only once is a common mistake.

152 lb × $\dfrac{453.6\ \text{g}}{1\ \text{lb}}$ × $\dfrac{1\ \text{kg}}{1000\ \text{g}}$ = 68.95 kg, and 68.95 kg × 5.0 mg/kg = 344.7 mg. 5.0 has only 2 significant figures, so report $3.4 \times 10^2$ mg.

Precision is how closely the trials agree (small range); accuracy is how close the average is to the accepted value. A: average 2.70, range 0.02; B: average 2.92, range 0.01; C: average 2.71, range 0.37; D: average 2.82, range 0.70. B’s trials are nearly identical, so precise, but the average is about 0.2 above 2.70, so not accurate; a systematic error is likely. A is precise and accurate, C averages out right but scatters, and D is neither.

Density first: 43.3 ÷ 5.00 = 8.66 g/cm³ (3 significant figures). Percent error = |8.66 − 8.96| ÷ 8.96 × 100% = 3.3%. The difference, 0.30, has only 2 significant figures, so the answer has 2. The measured density is too low.

Density = mass ÷ volume. Weighing the metal wet adds the water’s mass, so the mass and the density come out high. Air bubbles take up volume and reading from above gives too high a reading; both make the volume too large and the density too low. A balance that reads low makes the mass and density too low.

Distance: 5.00 km ÷ 1.609 km/mi = 3.1075 mi; time: 21.5 min ÷ 60 min/h = 0.3583 h. Speed = 3.1075 ÷ 0.3583 = 8.672 mi/h. 5.00 and 21.5 have 3 significant figures, 1.609 has 4, and 60 is exact, so report 3: 8.67 mi/h.

The sum in parentheses is good to 0.1: 12.0 + 0.36 = 12.36, worth 3 significant figures, but it is an intermediate result, so don’t round yet. Multiply by 1.50: 12.36 × 1.50 = 18.54, and keep 3: 18.5. Rounding the sum to 12.4 first gives 18.6, with the last digit wrong.