# A 1 14275 023 14598 1459810 2 b 2 79500 250 318 31810

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Chapter 2 / Exercise 54
Introductory Chemistry
Zumdahl/Decoste Expert Verified
(12 points) a. 145.75 + (2.3 × 10 –1 ) 142.75 + 0.23 = 145.98 = 1.4598*10* 2 b. 79,500 ÷ (2.5 × 10 2 ) 79,500 / 250 = 318 = 3.18*10 2 c. (7.0 × 10 –3 ) – (8.0 × 10 –4 ) 0.007 - 0.0008 = 0.0062 = 6.2*10 -3 d. (1.0 × 10 4 ) × (9.9 × 10 6 ) 10,000 * 9,900,000 = 99,000,000,000 = 9.9*10 10 (Reference: Chang 1.31) 1 Copyright © 2017 by Thomas Edison State University . All rights reserved.
##### We have textbook solutions for you!
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Chapter 2 / Exercise 54
Introductory Chemistry
Zumdahl/Decoste Expert Verified
4. What is the number of significant figures in each of the following measurements? (8 points) a. 4867 mi 4 significant digits b. 56 mL 2 significant digits c. 60,104 ton 5 significant digits d. 2900 g 4 significant digits e. 40.2 g/cm 3 3 significant digits f. 0.0000003 cm 1 significant digit g. 0.7 min 1 significant digit h. 4.6 × 10 19 atoms 2 significant digits (Reference: Chang 1.33) 5. Express the results in the proper number of significant figures. (12 points) a. 5.6792 m + 0.6 m + 4.33 m = 10.6092 = 10.6 b. 3.70 g – 2.9133 g = 0.7867 = 0.79 c. 4.51 cm × 3.6666 cm = 16.536366 = 16.536 d. (3 × 10 4 g + 6.827g)/(0.043 cm 3 – 0.021 cm 3 ) = 1363946.682 = 1364000 (Reference: Chang 1.35) 6. Carry out the following conversions: (16 points) a. 22.6 m to dm = 1 m=10 dm => 22.6=226 dm b. 25.4 mg to kilograms 1000 mg=1 g = 1000 g=1 kg => 2.54*10 -5 kg c. 556 mL to liters 1000 mL=1 L => 0.556 L => 5.56*10 -1 L d. 10.6 kg/m 3 to g/cm 3 = 10.6 kg/m 3 * 1000 g/1 kg * 1 m 3 /1000000 cm 3 = 0.106 g/cm 3 => 1.06 * 10 -1 g/cm 3 (Reference: Chang 1.39) 7. The average speed of helium at 25°C is 1255 m/s. Convert this speed to miles per hour (mph). (4 points) 1255 m/s * 1 mile / 1609.35m * 60 s/1 min * 60 min/1 hr = 2807.34458 mph => 2807.34 mph (Reference: Chang 1.41) 2 Copyright © 2017 by Thomas Edison State University . All rights reserved.
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