Solutions to: Units and Calculations Homework Problem Set Chemistry 145, Chapter 1

Transcription

1 to: Units and Calculations Homework Problem Set Chemistry 145, Chapter 1 1. Give the name and abbreviation of the SI Unit for: a. Length meter m b. Mass kilogram kg c. Time second s d. Electric Current amp A e. Temperature kelvin K. Give the abbreviation and describe what the following units are used to measure: a. meter m length b. liter L volume c. cubic centimeter cm 3 volume d. milliliter ml volume e. degree Celcius o C temperature f. kelvin K temperature (absolute) 3. Give the name and the abbreviation (without looking in the book) of the SI or metric prefix for: a pico p b mega M c nano n d centi c e milli m f giga G g kilo k h micro m 4. Express the following numbers in scientific notation with the appropriate number of significant figures: a = x b = x 10 5 c = x 10-5 d x 10 7 = x 10 9

2 Perform the following calculations and give the answer in scientific notation with the correct number of significant figures. (Note: The software used to perform these calculations does not round significant figures. The first answer given is the result of the calculation. The second number is appropriately rounded) a = = ( ) ( ) j = k. 70. = l = Perform the following mathematical operations and give the answer with the correct number of significant figures: a. b = c = d = e = f = g = h = i. ( ) ( ) ( ) = b ( ) = c = d = (note: first add the numerator to get 1634, but only has 3 SF.)

3 7. Perform the following unit conversions. Express your answers in scientific notation with the appropriate number of significant figures. (NOTE: the answers given below do not have the correct number of significant figures, but you should be able to determine this by now. Remember that most of these conversion factors are exact numbers, so the answer is limited by the number of significant figures as is given in the problem.) a. Convert inches into: feet, meters, centimeters, millimeters and kilometers in = ft 1ft = in ( 78.01in) 1ft 1in = ft 78.01in = m 1m = in ( 78.01in) 1m in = m 78.01in = cm 1in = cm ( 78.01in).54cm 1in = cm 78.01in = mm 1in = mm 5.4mm ( 78.01in) = 1in mm 78.01in = km 1km = m 1m ( 1km) = in m in 1km = 1m in ( 78.01in) 1km = km in

4 b. Convert feet into miles, kilometers and meters ft = mi 1mi = ft ( 14511ft) 1mi 580ft = mi 14511ft = km 1km = ft ( 14511ft) 1km ft = km 14511ft = m 1m = ft 1m ( 14511ft) = ft m c. Convert 15.4 meters into kilometers, centimeters, millimeters, micrometers, and nanometers. µm 10 6 m 15.4m = km 1km = m nm 10 9 m ( 15.4m) 1km 1000m 15.4m = cm 1cm = m 15.4m = mm 1mm = m = km 1cm ( 15.4m) 10 = cm m 1mm ( 15.4m) 10 3 = mm m 15.4m = µm 1 µm = m 1 µm ( 15.4m) 10 6 = µm m 15.4m = nm 1nm = m 1nm ( 15.4m) 10 9 = nm m

5 d. Convert 98.6 o F into o C and K. Convert o F to o C: ( ) 5 9 = Convert o C to K: = e. Convert 75 miles per hour into: km per hour and m s mi = km 1km 75 mi = mi 1km mi = km 75 mi = msec -1 1mi = m 1 = sec 75 mi m 1 1mi = msec -1 sec f. Convert 3.15 m into ft, in, and cm 3.15m = ft 3.15m = in 3.15m = cm 1m = ft 3.15m 3.808ft = ft 1m 1m = in 3.15m 39.37in = in 1m 1m = cm 3.15m 100cm = cm 1m

6 g. Convert 500 cm 3 into in 3, m 3, L and ml 500cm 3 = in 3 500cm 3 = m 3 500cm 3 500cm 3 = liter = ml 1in = cm 500cm 3 3 1in = in 3.54cm 1m = cm 500cm 3 3 1m = m 3 100cm 1liter = cm 3 500cm 3 1mL = cm 3 500cm 3 1liter 1000cm 3 1mL 1cm 3 = liter = ml

7 8. In the movie Goldfinger, James Bond foils a plot to break into Fort Knox. 007 does some quick mental calculations to determine the feasibility of removing the gold. If the price of gold is $14.00 per troy ounce ( grams), what is the mass (in kg) of 1 million dollars of gold? What is the volume of 1 million dollars of gold in L. How much would this gold be worth today if the price of gold is $ per kilogram? dollar := 1 First how much does 1 million dollars of gold weigh at $14.00 per troy ounce. troy_ounce := troy_ounce := troy_ounce Note, this is: 14dollar ( ) 1 troy_ounce dollar = 14dollar troy_ounce gm gm 1troy_ounce.10 3 kg = lb.10 3 kg = ton 1kg = kg 10 3 gm The volume of this gold may be determined from the density of gold (19.3 g/cm 3 ). You will have to look this up somewhere. (Your textbook is a good place to start, another good reference is the CRC Handbook of Chemistry and Physics. You should know where this is in the Library) density = mass volume A few notes on Mathcad, the program used to prepare these answers. This will help you learn how to read these documents. density Ag := 19.3gm cm 3 mass Ag :=.10 3 kg mass Ag volume Ag := density Ag volume Ag = liter The equals sign. The bold ( = ) is for "symbolic" math, it just shows an equation like a book would. The equals with a colon (:=) is for "defining" a variable. The regular equals (=) is for calculating a value. Units. Mathcad automatically calculates units and displays answers using SI units, unless told otherwise. I will sometimes show each step in a unit conversion, but after the first chapter, you should be comfortable with this and I will just display the results. How much would this gold be worth today if the price of gold is $ per kilogram? price Ag := dollar kg Value := mass Ag price Ag Mathcad does not round. Since properly rounding each answer is a lot of work, you are responsible for checking this. Value = dollar

8 9. The following experiment is performed with an unknown liquid. The liquid is added to a graduated cylinder with a mass of grams. After 0.00 ml of the liquid is added to the cylinder, the mass is grams. Is the liquid water? How do you know? If it is not water, what could it be? Information given in the problem: mass cylinder := gm volume liquid := 0.00mL mass total := gm From the total mass and the mass of the cylinder you can determine the mass of the liquid: mass liquid := mass total mass cylinder mass liquid = gm Now that the mass of the liquid and the volume of the liquid are known, the density may be determined. This is useful because the density of a compound is constant and may be compared with tables of known values. density = mass volume density liquid := mass liquid volume liquid density liquid = gm ml 1 The density of water at 5 o C, from the CRC Handbook of Chemistry and Physics, is g/ml. Since this is very close to the density of the unknown liquid, the unknown COULD be water (It does not have to be since some other liquid could have the same density).

9 Now repeat the above calculations for the second unknown. The experiment is repeated with a second liquid, after 1.3 ml of this liquid is added the mass of the cylinder is grams. Is the second liquid ethanol? How do you know? If it is not ethanol, what could it be? Information given in the problem: mass cylinder := gm volume liquid := 1.3mL mass total := gm From the total mass and the mass of the cylinder you can determine the mass of the liquid: mass liquid := mass total mass cylinder mass liquid = gm Now that the mass of the liquid and the volume of the liquid are known, the density may be determined. This is useful because the density of a compound is constant and may be compared with tables of known values. density = mass volume density liquid := mass liquid volume liquid density liquid = gm ml 1 The density of ethanol at 0 o C, from the CRC Handbook of Chemistry and Physics, is gm/ ml. This is not consistant with the value calculated in this problem. To find possible compounds, glance tough a table of densities and find something with a density close to this experimental value. This problem set and the solutions were prepared by: Scott Van Bramer Department of Chemistry Widener University Chester, PA svanbram@science.widener.edu