Significant Figures ReQuiz How many significant figures are in each of the following numbers? (7 pts each) 1) 2.400 ____ 5) 12.00300 ____ 2) 2.30 ____ 6) 0.00120 ____ 3) 0.6010 ____ 7) 0.0102 ____ 4) 26,000 ____ 8) 370.000 ____ 9) Is there an instance when an object 5.0 cm wide will not pass through an opening that is 5 cm wide? Explain. (5 pts) 10) How do significant figures represent the accuracy of a measuring device? (5 pts) 11) What is the volume of the graduated cylinder? (2pts) For chemistry help, visit www.chemfiesta.com © 2002 Cavalcade Publishing – All Rights Reserved 12) How long is the screw? (2pts) 13) What is the length of the stick? (2pts) (7 pts each) 14) Solve: 2.12 L + 0.34 L = ____ 15) Solve: 0.4430 kg – 0.20 kg = ____ 16) Solve: 2.11 cm * 0.021 cm = ____ 17) Solve: 8.80 m / 2.2 m = ____ For chemistry help, visit www.chemfiesta.com © 2002 Cavalcade Publishing – All Rights Reserved Significant Figures Worksheet - Answers How many significant figures are in each of the following numbers? 1) 2.400 4 5) 12.00300 7 2) 2.30 3 6) 0.00120 3 3) 0.6010 4 7) 0.0102 3 4) 26,000 2 8) 370.000 6 9) Is there an instance when an object 5.0 cm wide will not pass through an opening that is 5 cm wide? Explain. (5 pts) YES. If the opening is 5 cm wide, it means the opening could range from 4.5 cm to 5.4 cm. If the object is 5.0 cm wide, it could have an actual size of 4.96 cm to 5.04 cm. If the opening is actually between 4.5 and 4.96, the object will not be able to pass through it. 10) How do significant figures represent the accuracy of a measuring device? (5 pts) The more digits to the right of the decimal point indicate the greater precision of the measuring device given the same units of measure. 11) What is the volume of the graduated cylinder? (2pts) 37 mL. The acceptable range is 36 mL to 38 mL. 12) How long is the screw? (2pts) 5.11 cm. The acceptable range is 5.10 cm to 5.12 cm. 13) What is the length of the stick? (2 pts) 4.4 cm. The acceptable range is 4.3 cm to 4.5 cm. (7 pts each) 14) Solve: 2.12 L + 0.34 L = 2.46 L 15) Solve: 0.4430 kg – 0.20 kg = 0.24 kg 16) Solve: 2.11 cm * 0.021 cm = 0.044 cm2 17) Solve: 8.80 m / 2.2 m = 4.0 For chemistry help, visit www.chemfiesta.com © 2002 Cavalcade Publishing – All Rights Reserved
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