Gen Chem Challenge Problems: Chapter 1: Vining
1.1
A rib-eye steak in Canada costs 17.83 Canadian dollars per kilogram. Given an exchange rate of
1.12 CAD per US dollar, what is the cost of the steak, in US dollars and cents per pound?
________ USD/lb
17.83 CAD
1 USD
1 kg
454 g



= 7.23USD/lb
1 kg
1.12 CAD
1000 g
1 lb
1.2
Gasoline in Canada costs 1.24 Canadian dollars per liter. A can of soft drink says it contains 12
oz, also reported as 345 mL, of soda. A half gallon of milk says that it contains 64 oz. Given an
exchange rate of 1.12 CAD/US dollar, what is the cost of Canadian gasoline, in US dollars per
gallon?
________ UDS/gallon
1.24 CAD
1 USD
1L
345 mL
64 oz




= 4.07 USD/gallon
1L
1.12 CAD
1000 mL
12 oz
0.5 gallons
1.3
A roll of aluminum foil has a mass of 16 oz and is said to contain 75 ft2 of foil. The density of Al
is 2.70 g/cm3.
What is the thickness of the foil, in cm?
_________ cm
Step 1.
The volume of foil is found through the mass and density.
16 oz 
1 lb
454 g
1 cm3


= 168 cm3
16 oz
1 lb
2.70 g
Step 2.
 12 in 
area = 75 ft  

 1 ft 
2
2
2
 2.54 cm 
4
2
 
 = 6.97 x 10 cm
1
in


volume = area  thickness
thickness =
1.4
volume
168 cm3
=
= 0.0024 cm
area
6.97 x 104 cm2
Find the inner diameter of the opening of a very thin cylindrical glass tube, given the following
experimental information:
The tube has a length of 24.53 mm. The mass of the glass tube is 3.258 g. It is then filled with
mercury and the mass remeasured. It is found to be 3.446 g.
The density of mercury is 13.546 g/cm3.
__________ cm
The volume of the tube is found by the mass of mercury it contains.
Mass mercury = 3.446 g – 3.258 g = 0.188 g Hg
Volume = 0.188 g mercury 
1 cm3
= 0.0139 cm3
13.546 g
The area of the opening is found using the volume and length.
area =
volume
=
length
0.0139 cm3
= 0.00566 cm 2
1 cm
24.53 mm 
10 mm
The area of a circle is given by:
area =  (radius)2
Therefore,
radius =
radius =
area

area

=
0.00566 cm 2
= 0.0424 cm
3.1415
Finally, the diameter is equal to twice the radius, 2 x 0.0424 cm = 0.0848 cm.
1.5
Sea water is collected at the beach and found to contain several grains of pure, quartz sand. The
mixture is filtered and the solid sand placed in a vial labeled A. Half the remaining liquid is
placed in vial B. The other half of the liquid is heated till it boils. The gas that escapes is collected
and cooled. The liquid that reforms is placed in vial C. The solid that remains after the boiling is
placed in vial D.
Categorize each of the following:
1. the original sample
4. sample in vial C
choices:
pure element
2. sample in vial A
5. sample in vial D
pure compound
3. sample in vial B
homogeneous mixture
heterogeneous mixture
Answers:
1. heterogeneous mixture
2. pure compound (quartz, SiO2)
3. homogenous mixture
4. pure compound (water)
5. heterogeneous mixture (a mixture of salts from the sea water)