Chapter_1_ISM

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This note was uploaded on 02/13/2011 for the course CHEM 155 taught by Professor Bullock during the Spring '11 term at UCSB.

Page1 / 14

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online