Conversion Factors : I

Transcription

1 Conversion Factors : I Equivalence factors can be turned into conversion factors by dividing one side into the other. 1 mi 5280 ft 1 mile = 5280 ft 1 = = 5280 ft 1 mi 1 in = 2.54 cm 1 in 2.54 cm 1 = = 2.54 cm 1 in In converting one set of units for another, the one desired is on top in the conversion factor, and the old one is on the bottom so the old units are canceled out. EP 10: Convert 29,141 ft into miles. 1 mi 29,141 ft = mi 5280 ft

2 EP 9: An experiment shows that the mass of a gold atom is 16.4 times the mass of a 12 C atom of mass 12 amu. Express the mass of the gold atom in amu and in g. 1 amu = x g m Au = 16.4 m C 16.4 x 12 amu = 197. amu g m( g ) = m( amu) amu g amu = g amu

3 Conversion Factors - II Conversion factors can be combined. EP 11: Convert 3.56 lbs/hr into units of milligrams/sec: lb = 1 kg 1000g = 1 kg 1000 mg = 1 g 1 hr = 3600 s lb kg 1000g 1000mg hr mg 3.56 =448 hr lb kg g 3600 s s

4 m 3 Liter = L = dm 3 cm 3 = ml

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6 Conversion Factors - IV metric volume to liters EP 12: The volume of the world s oceans is 1.35 x 10 9 km 3. How many liters of water are in the oceans? conversion factors: 1 km = 1000 m 1 L = 1 dm 3 = 10-3 m 3 or 1000 L = 1 m x 10 9 km 3 x (10 3 m/1 km ) 3 x ( 10 3 L/m 3 ) = 1.35 x L

7 Calculations involving density (2) EP 14: A small rectangular slab of lithium has the dimensions 20.9 mm by 11.1 mm by 11.9 mm. Its mass is 1.49 x 10 3 mg. What is the density of lithium in g/cm 3? Step 1: task: Calculate the density of lithium in g/cm 3. Step 2: given information: The mass is 1.49 x 10 3 mg and the dimensions are 20.9 by 11.1 by 11.9 mm.

8 Step 3: strategy: Density = mass/volume, so the volume needs to be calculated. For a rectangular slab, volume = length x width x height. Need to convert units. Step 4: set up the problem: density = mass volume mg d = 20.9mm 11.1mm 11.9mm Are we ready to calculate? No! The units aren t right.

9 density = mass volume mg 1g 10mm d = 20.9mm 11.1mm 11.9mm 1000 mg 1cm mg 1g 10 mm = 20.9mm 11.1mm 11.9mm 1000 mg 1cm Step 5: right units? mg 1g 10 mm d = 20.9mm 11.1mm 11.9mm 1000 mg 1cm

10 Step 6: calculate, sig figs mg 1g 10 mm g = 3 3 d = mm 11.1mm 11.9mm 1000 mg 1cm cm Step 7: check result Lithium is less dense than water, which you might not have expected. The units are right.

11 Volume by Displacement EP 15: An irregularly shaped metal object has a mass of g. When it is placed into a 2.00 liter graduated cylinder containing ml of water, the final volume of the water in the cylinder is ml Calculate the density of the metal in g/cm 3.1 ml= 1cm 3. [Solve using Archimedes Principle] Step 1: task: Calculate the density of the metal in g/cm 3. Step 2: given information: The mass is g. The initial and final volumes are ml and ml.

12 Step 3: strategy: The metal displaces water, which leads to the volume increase. The volume of the metal is the difference in the two volumes of water. Use definition of density = mass/volume to calculate density. Step 4: set up the problem: mass density = = volume Step 5: Units right? mass difference in volumes of grad cylinder = g g cm cm = cm 3 3 3

13 Step 6: calculate, sig figs g g density = = cm cm 3 3 Step 7: check result The metal is more dense than water. The units are right.

14 Computer Chips EP 16: Future computers might use memory bits which require an area of a square with μm sides. (a) How many bits could be put on a 1.00 in x 1.00 in computer chip? (b) If each bit required that 25.0 % of its area to be coated with a gold film 10.0 nm thick, what mass of gold would be needed to make one chip? The density of gold is 19.3 g/cm 3. Step 1: task: (a) Calculate number of bits in given area (b) Calculate mass of gold in volume to be calculated Step 2: given information: (a) The bit and chip have areas (0.250μm) 2 and (1.00 in) 2. (b)the Au coating of known density is 10.0 nm thick, and occupies 25% of the (1.00 in) 2 area of the chip.

15 Step 3: strategy: (a) The number of bits is the area of the chip divided by the area of a bit. (b)the volume of gold is equal to the volume of a rectangular solid of height 10.0 nm and area x 1.00 in 2. The mass is density x volume. Step 4: set up problem (a): number of bits = area of chip area of bit in 2.54 cm 1 m = 2 ( m) Step 6: Calculate, sig figs (a) in 100 cm 10 number of bits =

16 Step 4: set up problem (a): mass = density volume -9 g 2 10 m 100 cm 2.54 cm 3 ( ) cm nm m 1in = in 10.0 nm Step 6: Calculate, sig figs (a) 5 mass = g 2

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18 Temperature Scales and Interconversions Kelvin ( K ) - The Absolute temperature scale begins at absolute zero and only has positive values. Water freezes at K Celsius ( o C) -Commonly used scale around the world and in laboratories. Water freezes at 0 o C, and boils at 100 o C. Fahrenheit ( o F) -Commonly used scale in America for our weather reports. Water freezes at 32 o F, and boils at 212 o F. ( ) ( T K = T C) ( ) 9 ( T F = T C) T( C) = T( K) ( ) 5 ( T C = T F) 32 9

19 Temperature Conversions EP 17: The boiling point of Liquid Nitrogen is o C, what is the temperature in Kelvin and degrees Fahrenheit? T (K) = T ( o C) T (K) = = K T ( o F) = 9/5 T ( o C) + 32 T ( o F) = 9/5 ( o C) +32 = o F EP 18: The normal body temperature is 98.6 o F, what is it in Kelvin and degrees Celsius? T ( o C) = 5/9 x [ T ( o F) - 32] T ( o C) = 5/9 x [ 98.6 o F - 32] = 37.0 o C T (in K) = T (in o C) T (in K) = 37.0 o C = K

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21 Answers to example problems EP 1 12 ft 2 EP 2 50 mi/hr EP 3a two EP 3b four EP 3c four EP 3d six EP 3e four EP 3f three EP 3g one EP 3h five EP 3i seven EP 3j three EP 3k three EP 4: 12.7 EP 5: 13.9 EP EP EP EP g EP mi EP mg/s EP x L EP x 10 5 g EP g/cm 3 EP g/cm 3 EP 16 (a) x bits (b) 3.11 x 10-5 g EP ºF EP K