Chapter 5, Part 3 Measurements and Calculations

5.6 Dimensional Analysis 1 in. = 2.54 cm is an equivalence statement and can be turned into a conversion factor: 1 in./2.54 cm or2.54 cm/1 in. Any equivalence statement can be used in this manner.

Example: 62 cm = ? In. 62 cm x 1 in./2.54 cm = 24 in. (note sig figs) Example: 26.2 mi. = ? Km mi. X 1760 yd/1 mi. X 1 m/1.094 yd x 1 km/10 3 m = 42.1 km Embedded problem: (convert the speed of light from 3.00 x 10 8 meters per second to miles per hour.)

5.7 Temperature Conversions FahrenheitCelsiusKelvin b.p. water f.p. water Absolute zero (note: no degree symbol “ o ” is used in the Kelvin scale. It is an absolute scale.) Conversions: Celsius to Kelvin: add 273 to Celsius

Figure 5.6: The three major temperature scales.

Figure 5.7: Converting 70 degrees Celsius to Kelvin units.

Figure 5.8: Comparison of the Celsius and Fahrenheit scales.

Example: 70. o C = ? K 70. o C = 343 K Example: 77 K = ? o C 77 K = -196 o C Problem: Which temperature is colder, 172 K or -75 o C? (172 K)(Why?)

5.8 Density- amount of matter present in a given volume of a substance. Density = mass/volume Mass = density x volume Volume = Mass/Density Specific gravity- ratio of the density of a given liquid to the density of 4 o C (has no unit)

Example: mL of a certain liquid has a mass of g. Calculate density. D = M/V = g/23.50 mL = g/mL Problem: 35.8 mL of a cleaner has a mass of 28.1 g. Which of the four substances in the table on p. 144 is the major component? D = M/V = 28.1 g/35.8 mL = g/mL Therefore, isopropyl alcohol. Example: V = M/D = 225 g/13.6 g/mL = 16.5 mL