Problem #1: How many palladium atoms would it take to encircle the Earth at its equator? (The atomic diameter of Pd is 140 pm; the equatorial diameter of the Earth is 40,075.02 km.)
Problem #2: In America, a car's gasoline efficiency is measured in miles/gallon. In Europe, it is measured in km/L. If your car's gas mileage is 40.0 mi/gal, how many liters of gasoline would you need to buy to complete a 142 km trip? Use the following conversions: 1 km = 0.6214 mi and 1 gal = 3.7854 L
Problem #3: Convert 2.1 Å to nanometers
Solution:
1) Convert Å to meter:
1 Å = 10-8 cm (by definition)10-8 cm x (1 m / 100 cm) = 10-10 m
1 Å = 10-10 m
2) Convert 2.1 x 10-10 m to nm:
2.1 x 10-10 m times (109 nm / 1 m) = 0.21 nm
Problem #4: Convert 0.0030 mm to nm
Solution:
3.0 x 10-3 mm x (106 nm / 1 mm) = 3.0 x 103 nm
Problem #5: Convert 6.25 x 10¯4 s to ms
The answer is 0.625 ms
Problem #6: Suppose a football, diameter 22 cm, is scaled up so that it becomes as big as the earth, diameter 13000 km. Calculate whether an atom of diameter 0.32 nm will become as big as:
a) a pin head, diameter 1 mm
b) a coin, diameter 1.9 cm
c) a football, diameter 22 cm
d) a weather balloon, diameter 1.8 m
Solution:
1) Convert all measurements to the same value. I used meters:
football: 0.22 m
earth: 13000000 m
atom: 3.2 x 10¯10 m
2) Solve via a ratio and proportion technique:
football diameter : earth diameter = atom diameter : an unknown value0.22 m ÷ 13000000 m = 3.2 x 10¯10 ÷ x
Solving for x gives 0.0189 m
3) Answer b (1.9 cm) is the correct answer, since 0.0189 m = 1.9 cm.
Problem #7: A material will float on the surface of a liquid if the material has a density less than that of a liquid. Given that the density of water is approximately 1.0 g/mL, will a block of material having a volume of 1.2 x 104 in3 and weighing 350 lb float or sink when replaced in a reservoir of water?
Solution:
1) Convert cubic inches to cubic centimeters, then to mL:
1 in3 = 1 in x 1 in x 1 insince 1 inch = 2.54 cm:
1 in3 = 2.54 cm x 2.54 cm x 2.54 cm = 16.387 cm3since 1 mL = 1 cm3, we know this:
16.387 cm3 = 16.387 mL
2) Convert the cubic inches of the block of material to mL:
1.2 x 104 in3 x 16.387 mL/in3 = 196644 mL
3) Convert pounds to grams:
350 lb x 453.6 g lb¯1 = 158760 g
4) Calculate the density of the material and compare it to the density of water:
158760 g / 196644 ml = 0.807 g/mLSince this is less than the density of water, we conclude that the block of material will float.
Problem #8: Convert 10.6 kg/m3 to mg/μL
Solution:
1) Convert kg to mg:
10.6 kg/m3 times (106 mg/kg) = 1.06 x 107 mg/m3
2) Convert m3 to L:
I have made a video on this topic: How to Convert between Cubic Meters and Liters.1 m3 = 1000 dm3 = 1000 L
1.06 x 107 mg/m3 times (1 m3 / 1000 L) = 1.06 x 104 mg/L
3) Convert L to μL:
1.06 x 104 mg/L x (1 L / 106 μL) = 1.06 x 10-2 mg/μL = 0.0106 mg/μL
Suppose the problem had been to convert kg/m3 to mg/mL? Then, the last part of the solution would look like this:
1.06 x 104 mg/L x (1 L / 103 mL) = 1.06 x 101 mg/mL = 10.6 mg/mL10.6 kg/m3 = 10.6 mg/mL
Problem #9: Light travels at a speed of 3.00 x 1010 cm/s. What is the speed of light in km/h?
Solution:
1) Convert cm/s to km/s:
answer = 3.00 x 105 km/s
2) Convert seconds to hours:
answer = 1.08 x 109 km/hr
Problem #10: Convert 22.43 gal/min to L/seconds.
Solution:
1) Convert gallons to liters:
22.43 gal/min x (3.7854 L/gal) = 84.9067863 L/minAll I did when I typed up this solution was go to Google and type "convert 22.43 gal to L" and it gave me the answer I pasted in above.
2) Convert minutes to seconds:
84.9067863 L/min x (1 min / 60 sec) = 1.415 L/sec
Problem #11: Convert 0.02515 ft3 to cm3
Solution:
1) Convert 1 foot to inches, then centimeters:
1 foot = 12 inches (by definition)12 inches x (2.54 cm / 1 inch) = 30.48 cm
2) Calculate what one cubic foot would be in cubic centimeters:
30.48 cm x 30.48 cm x 30.48 cm = 28317 cm3
3) Determine answer to problem:
0.02515 ft3 x (28317 cm3 / one ft3) = 712.2 cm3
Now, go to Google and type this in the search box:
convert 0.02515 cubic foot to cubic centimeters
Pretty amazing!
Problem #12: Gasoline has a density of 0.76 g/mL. Find the mass of 4.25 gallons in pounds.
Solution:
1) Convert g/mL to pounds per mL:
0.76 g/mL x (one lb / 453.59 g) = 0.0016755 lb/mL
Convert lb/mL to lb/gal:
0.0016755 lb/mL x (3 785.4 mL/1 gal) = 6.34244 lb/galTry this in Google: convert 0.76 g/mL to pounds/gallon
3) Find the mass of 4.25 gal:
6.34244 lb/gal x 4.25 gal = 27.0 pounds
Problem #13: How many slugs are there in 54.29 kg?
Hint: use Google to convert slugs to kilograms
A slug is a very old school unit of mass. The ChemTeam has heard of it (as well as dyne and erg), but has never used it in a computation, either as a student or as a teacher.
The ChemTeam would shout "Erg!" when putting the shot in high school and college. He thought this was quite funny, knowing that the erg is a unit of energy.
The ChemTeam is easily amused!
Problem #14: If a container measures 3.2 gallons of volume, how many liters does it occupy?
Hie thee off to the great Googlizer!
Problem #15: The three dimensions of a box are measured to be 1.3 in, 3.4 in, and 5.9 in. What is the volume of the box in liters?
Solution: A video of the solution to problem #15
Problem #16: The density of water is 1.00 g/cm3. What is the density of water expressed in lb/ft3?
Solution:
1) Convert 1.00 ft3 to cm3:
one foot = 12 inches (by definition)12 inches times (2.54 cm/inch) = 30.48 cm
30.48 cm x 30.48 cm x 30.48 cm = 28316.8466 cm3
2) Convert 28316.8466 grams to pounds:
28316.8466 grams divided by 453.59237 grams/pound = 62.43 poundsThere are 62.43 lb/ft3
I used 28316.8466 g because water has a density of 1.00 g/cm3.
Problem #17: (a) Convert 38.0 centiliters to kiloliters; (b) How many Mg are there in 100,000,000 mg? (c) Convert 0.050 kg to mg.
Solution:
(a) 38.0 cL x (1 kL / 105 cL) = 3.80 x 10¯4 kL
(b) 1 x 108 mg x (1 Mg / 109 mg) = 0.1 Mg
0.050 kg x (106 mg / 1 kg) = 5.0 x 104 mg
Note that, in each case, a '1' is associated with the larger of the two units in the conversion factor. Those units are, in turn, kL, Mg, and kg.
Problem #18: Suppose your dorm room is 11.8 ft wide by 12 ft long and 8.5 ft high and has an air conditioner that exhanges air at a rate of 1200 L/min. How long would it take the air conditioner to exhange the air in your room once?
Solution:
1) Convert feet to decimeters:
from Google, one foot = 3.048 dmtherefore:
11.8 ft times 3.048 dm/ft = 35. 9664 dm
12 ft times 3.048 dm/ft = 36.576 dm
8.5 ft times 3.048 dm/ft = 25.908 dm
2) Calculate the volume in cubic decimeters:
35. 9664 dm times 36.576 dm times 25.908 dm = 34082.156 dm3
3) Compute time required:
34082.156 dm3 / 1200 L/min = 28.40 minRemember, L = dm3, so those two units cancel exactly.
Problem #19: The concentration of carbon monoxide in an urban apartment is 41 μg/m3. What mass of carbon monoxide in grams is present in a room measuring 1.00 foot by 12.0 feet by 22.0 feet?