Question

1. the number of significant figures in the measurement 170.040 km is
2. the number of significant figures in the measurement 210 cm is
3. to two significant figures, the measurement 0.0255 g should be reported as
4. what is the measurement 111.009 mm rounded off to four significant digits?
5. rounded to four significant figures, 1.245633×10^8 is
6. express the sum (addition) of 7.68 m and 5.0 m using the correct number of significant digits.
7. express the product (multiplication) of 2.2 mm and 5.00 mm using the correct number of significant digits.
8. the sum of 314.53 km and 32 km is correctly expressed as
9. divide 5.7 m by 2 m. the quotient is correctly reported as
10. the dimensions of a rectangular solid are measured to be 1.27 cm, 1.3 cm, and 2.5 cm. to the correct number of significant figures, the volume should be recorded as (v = lwh)
11. three samples of 0.12 g, 1.8 g, and 0.562 g are mixed together. the combined mass of all three samples, expressed to the correct number of significant figures, should be recorded as
12. the diameter of a carbon atom is 0.000000000154 m. what is this number expressed in scientific notation?
13. the expression of 5008 km in scientific notation is

problem

1. how many millimeters are in 347 cm? (1 cm = 10 mm) 10×347=(3470mm)
2. how many mg are in 3.45 kg?
3. how many grams of h₂o are in 63.4 ml of h₂o?

Explanation:

Step1: Recall significant - figure rules

Non - zero digits are always significant. Zeros between non - zero digits are significant. Trailing zeros after a decimal point are significant.
For 13. The number 170.040 km:
Non - zero digits 1, 7, 4; zero between 7 and 4; trailing zeros after decimal. So there are 6 significant figures.

Step2: For 14. The number 210 cm:

If there is no decimal, the trailing zero may or may not be significant. In the absence of other information, we assume it is not significant. So there are 2 significant figures.

Step3: For 15. Rounding 0.0255 g to two significant figures

The first two significant figures are 2 and 5. The next digit is 5, so we round up. 0.0255 rounds to 0.026 g.

Step4: For 16. Rounding 111.009 mm to four significant digits

The first four significant digits are 111.0 (since the next digit 0 is less than 5, we do not round up). So it is 111.0 mm.

Step5: For 17. Rounding 1.245633×10⁸ to four significant figures

The first four significant digits are 1.246 (since the next digit 5 causes the previous digit 5 to round up). So it is 1.246×10⁸.

Step6: For 18. Adding 7.68 m and 5.0 m

7.68 + 5.0=12.68. 5.0 has one decimal place. Rounding to one decimal place, the sum is 12.7 m.

Step7: For 19. Multiplying 2.2 mm and 5.00 mm

2.2×5.00 = 11.0. 2.2 has two significant figures, so the product is 11 mm².

Step8: For 20. Adding 314.53 km and 32 km

314.53+32 = 346.53. 32 has no decimal places. Rounding to no decimal places, the sum is 347 km.

Step9: For 21. Dividing 5.7 m by 2 m

5.7÷2 = 2.85. 2 has one significant figure. Rounding to one significant figure, the quotient is 3 m.

Step10: For 22. Calculating volume of rectangular solid

V = lwh=1.27×1.3×2.5 = 4.0775. 1.3 has two significant figures. Rounding to two significant figures, V = 4.1 cm³.

Step11: For 23. Adding 0.12 g, 1.8 g, and 0.562 g

0.12+1.8 + 0.562=2.482. 1.8 has one decimal place. Rounding to one decimal place, the sum is 2.5 g.

Step12: For 24. Expressing 0.000000000154 m in scientific notation

Move the decimal point 10 places to the right. So it is 1.54×10⁻¹⁰ m.

Step13: For 25. Expressing 5008 km in scientific notation

Move the decimal point 3 places to the left. So it is 5.008×10³ km.

Step14: For 27. Converting 3.45 kg to mg

1 kg = 1000000 mg. So 3.45×1000000 = 3450000 mg = 3.45×10⁶ mg.

Step15: For 28. Assuming density of H₂O is 1 g/mL

If density d = 1 g/mL and V = 63.4 mL, then m = dV=1×63.4 = 63.4 g.

Answer:

1. 6
2. 2
3. 0.026 g
4. 111.0 mm
5. 1.246×10⁸
6. 12.7 m
7. 11 mm²
8. 347 km
9. 3 m
10. 4.1 cm³
11. 2.5 g
12. 1.54×10⁻¹⁰ m
13. 5.008×10³ km
14. 3.45×10⁶ mg
15. 63.4 g