2.1 · Measurement and CalculationPractice · 测量与计算

Start counting at the first nonzero digit (5): 5, 2, and the final 0, so 3. Leading zeros only locate the decimal point and don’t count; a trailing zero after the decimal point shows the measurement was made to that place, so it counts.

In scientific notation, count only the digits in front: 1.030 has 4 significant figures (the zero between nonzero digits and the trailing zero after the decimal point both count). The $10^4$ only sets the size and doesn’t change the count.

Keep the first three digits, 2, 3, and 4, and look at the fourth: it is 5, so round up to 23.5.

For addition and subtraction, look at decimal places: 12.1 is known only to 0.1 g, so the sum keeps 1 decimal place. 15.556 → 15.6 g.

For multiplication and division, count significant figures: 4.0 has 2 and 2.25 has 3, so keep 2. 4.0 × 2.25 = 9, reported as 9.0 cm²; write the trailing zero.

1 km = 1000 m. Use the factor $\dfrac{1000\ \text{m}}{1\ \text{km}}$ so km cancels: 2.5 km × 1000 m/km = 2500 m (or $2.5 \times 10^3$ m, showing 2 significant figures).

1 L = 1000 mL, so 350 mL × $\dfrac{1\ \text{L}}{1000\ \text{mL}}$ = 0.35 L. Converting to a larger unit makes the number smaller.

Counted numbers and defined relationships are exact, such as 1 dozen = 12 or 1 m = 100 cm. Anything read from an instrument has an estimated last digit and limited significant figures. 1 mi ≈ 1.609 km is a rounded value with only 4 significant figures.

Percent error = |measured − accepted| ÷ accepted × 100% = |9.6 − 9.8| ÷ 9.8 × 100% = 2%. The difference, 0.2, has only 1 significant figure.

The three results agree within 0.02 g, so they are precise; but their average, about 5.21 g, is far from the true 5.50 g, so they aren’t accurate. A miscalibrated balance (a systematic error) could cause this.