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. 10-1 pico p b. 10 6 mega M c. 10-9 nano n d. 10 - centi c e. 10-3 milli m f. 10 9 giga G g. 10 3 kilo k h. 10-6 micro m 4. Express the following numbers in scientific notation with the appropriate number of significant figures: a. 10980000000 = 1.098 x 10 10 b. 414100= 4.141 x 10 5 c. 0.00009516= 9.516 x 10-5 d. 746.5 x 10 7 = 7.465 x 10 9
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. 30.84 + 9.74 = 40.58000000 4.05810 1 1363.0 = 408.00000000 4.110 ( ) ( 8.09 10 4 ) j. 7.601 10 7 = 6.1490900 10 4 6.14910 4 k. 70. = 4.9804000 10 3 4.9310 3 1 l. 94773 = 307.85378 3.078510 6. Perform the following mathematical operations and give the answer with the correct number of significant figures: a. b. 30.84 + 9.74487 = 40.58487000 4.05810 1 c. 30.845 + 9.74 = 40.58500000 4.05810 1 d. 30.845 + 9.75 = 40.59500000 4.06010 1 e. 145 + 1.5410 6 = 145.00000154 1.4510 f. 40.79 1.1843 = 39.60568000 3.96010 1 g. 1.430.848 = 1.164000 1.1 h. 1.430.8488 = 1.1304040 1.1 i. ( 9.00810 4 ) ( 6.5710 7 ) ( ) 6.5310 4 = 8.9979980 10 15 9.0010 15 b ( 1.46010 3 ) 53.109 = 78.48970178 78.5 c. 57.49 1 = 7.5781940 7.578 d. 7.110 + 94 7.50810 4 = 0.0176345.1810 (note: first add the numerator to get 1634, but only has 3 SF.)
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 78.01 inches into: feet, meters, centimeters, millimeters and kilometers. 78.01in = 6.50083333 ft 1ft = 1.00000000 in ( 78.01in) 1ft 1in = 6.50083333 ft 78.01in = 1.98145400 m 1m = 39.37007874 in ( 78.01in) 1m 39.37007874in = 1.98145400 m 78.01in = 198.14540000 cm 1in =.54000000 cm ( 78.01in).54cm 1in = 198.14540000 cm 78.01in = 1.98145400 10 3 mm 1in = 5.40000000 mm 5.4mm ( 78.01in) = 1in 1.98145400 10 3 mm 78.01in = 1.98145400 10 3 km 1km = 1.00000000 10 3 m 1m ( 1km) = 39.37007874 in 110 3 m 39.37007874in 1km = 1m 3.93700787 10 4 in ( 78.01in) 1km 3.9370078710 4 = 1.98145400 10 3 km in
b. Convert 14511 feet into miles, kilometers and meters. 14511ft =.7489545 mi 1mi = 580.00000000 ft ( 14511ft) 1mi 580ft =.7489545 mi 14511ft = 4.49580 km 1km = 380.83989501 ft ( 14511ft) 1km 380.83989501ft = 4.49580 km 14511ft = 4.49580 10 3 m 1m = 3.8083990 ft 1m ( 14511ft) = 3.8083990ft 4.49579 10 3 m c. Convert 15.4 meters into kilometers, centimeters, millimeters, micrometers, and nanometers. µm 10 6 m 15.4m = 0.0154000 km 1km = 1000.00000000 m nm 10 9 m ( 15.4m) 1km 1000m 15.4m = 1.5400000 10 3 cm 1cm = 0.01000000 m 15.4m = 1.5400000 10 4 mm 1mm = 0.00100000 m = 0.0154000 km 1cm ( 15.4m) 10 = 1.5400000 10 3 cm m 1mm ( 15.4m) 10 3 = 1.5400000 10 4 mm m 15.4m = 1.5400000 10 7 µm 1 µm = 1.00000000 10 6 m 1 µm ( 15.4m) 10 6 = 1.5400000 10 7 µm m 15.4m = 1.5400000 10 10 nm 1nm = 1.00000000 10 9 m 1nm ( 15.4m) 10 9 = 1.5400000 10 10 nm m
d. Convert 98.6 o F into o C and K. Convert o F to o C: ( 98.6 3) 5 9 = 37.00000000 Convert o C to K: 37 + 73.15 = 310.15000000 e. Convert 75 miles per hour into: km per hour and m s -1. 75 mi = 10.70080000 km 1km 75 mi = 0.6137119 mi 1km 0.6137119mi = 10.70080043 km 75 mi = 33.5800000 msec -1 1mi = 1.60934400 10 3 m 1 = 3.60000000 10 3 sec 75 mi 1.60910 3 m 1 1mi 3.610 3 = 33.51 msec -1 sec f. Convert 3.15 m into ft, in, and cm 3.15m = 49.1845615 ft 3.15m = 3.5885718 10 4 in 3.15m =.31500000 10 5 cm 1m = 3.8083990 ft 3.15m 3.808ft = 49.1784660 ft 1m 1m = 39.37007874 in 3.15m 39.37in = 3.58848 10 4 in 1m 1m = 100.00000000 cm 3.15m 100cm =.31500000 10 5 cm 1m
g. Convert 500 cm 3 into in 3, m 3, L and ml 500cm 3 = 30.5118705 in 3 500cm 3 = 5.00000000 10 4 m 3 500cm 3 500cm 3 = 0.50000000 liter = 500.00000000 ml 1in =.54000000 cm 500cm 3 3 1in = 30.5118705 in 3.54cm 1m = 100.00000000 cm 500cm 3 3 1m = 5.00000000 10 4 m 3 100cm 1liter = 1.00000000 10 3 cm 3 500cm 3 1mL = 1.00000000 cm 3 500cm 3 1liter 1000cm 3 1mL 1cm 3 = 0.50000000 liter = 500.00000000 ml
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 (31.1035 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 $9068.77 per kilogram? dollar := 1 First how much does 1 million dollars of gold weigh at $14.00 per troy ounce. troy_ounce := troy_ounce := 7.148571410 4 troy_ounce Note, this is: 14dollar ( ) 1 troy_ounce 110 6 dollar = 14dollar 7.1485714 10 4 troy_ounce 31.1035gm 31.1035gm 1troy_ounce.10 3 kg = 4.8946 10 3 lb.10 3 kg =.44713111 ton 1kg =.167857 10 3 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 = 114.9068330 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 $9068.77 per kilogram? price Ag := 9068.77 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 =.0136694 10 7 dollar
9. The following experiment is performed with an unknown liquid. The liquid is added to a graduated cylinder with a mass of 54.6789 grams. After 0.00 ml of the liquid is added to the cylinder, the mass is 74.615 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 := 54.6789gm volume liquid := 0.00mL mass total := 74.615gm 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 = 19.9460000 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 = 0.99713000 gm ml 1 The density of water at 5 o C, from the CRC Handbook of Chemistry and Physics, is 0.99707 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).
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 7.7364 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 := 54.6789gm volume liquid := 1.3mL mass total := 7.7364gm 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 = 18.05750000 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 = 0.84776995 gm ml 1 The density of ethanol at 0 o C, from the CRC Handbook of Chemistry and Physics, is 0.7893 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 19013 svanbram@science.widener.edu http://science.widener.edu/~svanbram