Chapter_1_ISM

# Chapter_1_ISM - Chapter 1 1 Practice Exercises 1.1 V = 3 4...

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Unformatted text preview: Chapter 1 1 Practice Exercises 1.1 V = 3 4 r 3 ! , the SI unit for radius, r, is meters, the numbers 4 3 and π do not have units. Therefore, the SI unit for volume is meter 3 or m 3 . 1.2 Force equals mass × acceleration ( F = ma ), and acceleration equals change in velocity divided by change in time ( a = change in v change in t ), and velocity equals distance divided by time ( v = d t ). Put the equations together: F = m change in v change in t ! " # \$ % & F= m d change in t change in t ! " # \$ # \$ # \$ # \$ % & = m 2 change in d change in t ! " # \$ # \$ % & The unit for mass is kilogram (kg); the unit for distance is meter (m) and the unit for time is second (s). Substitute the units into the equation above: Unit for force in SI base units = kg 2 m s ! " # \$ % & or kg m s –2 1.3 ( ) F C 9 F 9 F t = t 32 F = 86 C 32 F = 187 F 5 C 5 C ! " ! " + + # \$ # \$ # \$ # \$ % & % & ! ! ! ! ! ! ! ! The tool that relates the two temperature scales is the Celsius - Fahrenheit conversion tool. 1.4 ( ) ( ) C F 5 C 5 C t = t 32 F = 50 F 32 F = 10 C 9 F 9 F ! " ! " # # \$ % \$ % \$ % \$ % & ’ & ’ ! ! ! ! ! ! ! ! To convert from °F to K we first convert to °C. In the ( ) ( ) C F 5 C 5 C t = t 32 F = 68 F 32 F = 20 C 9 F 9 F ! " ! " # # \$ % \$ % \$ % \$ % & ’ & ’ ! ! ! ! ! ! ! ! T K = (273 °C + t C ) 1 K 1 C ! " # \$ ° % & = (273 °C + 20 °C) 1 K 1 C ! " # \$ ° % & = 293 K 1.5 (a) 21.0233 g + 21.0 g = 42.0233 g: rounded correctly to 42.0 g (b) 10.0324 g / 11.7 mL = 0.8574 g / mL: rounded correctly to 0.857 g / mL (c) ( ) 14.24 cm 12.334 cm 2.223 cm 1.04 cm ! " = 148.57 cm: rounded correctly to 149 cm 1.6 (a) 32.02 mL – 2.0 mL = 30. mL (b) 54.183 g – 0.0278 g = 54.155 g (c) 10.0 g + 1.03 g + 0.243 g = 11.3 g (d) 43.4 in × 1 ft 12 in. ! " # \$ % & = 3.62 ft (1 and 12 are exact numbers) (e) 1.03 m 2.074 m 3.9 m 12.46 m 4.778 m ! ! ! + = 0.48 m 2 Chapter 1 2 1.7 ( ) 2 2 2 2 2 30.48 cm 1 m m = 124 ft = 11.5 m 1 ft 100 cm ! " ! " # \$ # \$ % & % & 1.8 (a) ( ) 3 ft 12 in. in. = 3.00 yd = 108 in. 1 yd 1 ft ! " ! " # \$# \$ % & % & (b) ( ) 5 1000 m 100 cm cm = 1.25 km = 1.25 10 cm 1 km 1 m ! "! " # \$ %\$ % & ’& ’ (c) ( ) 1 m 100 cm 1 in. 1 ft ft = 3.27 mm = 0.0107 ft 1000 mm 1 m 2.54 cm 12 in. ! "! "! "! " # \$# \$# \$# \$ % &% &% &% & (d) km 20.2 mile 1.609 km 1 gal km = = 8.59 L L 1 gal 1 mile 3.785 L ! " ! "! " # \$# \$# \$ % &% & % & 1.9 Density = mass volume Density of the object = 3 365 g 22.12 cm = 16.5 g/cm 3 The object is not composed of pure gold since the density of gold is 19.3 g/cm 3 . 1.10 The density of the alloy is 12.6 g/cm 3 . To determine the mass of the 0.822 ft 3 sample of the alloy, first convert the density from g/cm 3 to lb/ft 3 , then find the weight....
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## This note was uploaded on 02/13/2011 for the course CHEM 155 taught by Professor Bullock during the Spring '11 term at UCSB.

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Chapter_1_ISM - Chapter 1 1 Practice Exercises 1.1 V = 3 4...

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