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

This preview shows page 1 - 3 out of 3 pages.

We have textbook solutions for you!
The document you are viewing contains questions related to this textbook.
Introductory Chemistry
The document you are viewing contains questions related to this textbook.
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!
The document you are viewing contains questions related to this textbook.
Introductory Chemistry
The document you are viewing contains questions related to this textbook.
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.

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture

  • Left Quote Icon

    Student Picture