chem

2. General Chemistry
Principles and Modern
Applications, 10th Ed.,
Petucci, Pearson Press
If you are major in chemical
technology, it would be great
to buy one
If not, borrow one from library
or other books which cover
similar materials

3. Tutorial Questions
1, 3, 6, 10, 12, 16, 22, 24,
26, 29, 39, 43, 53, 57,
65, 84, 89, 100, 111

4. SI Units: Temperature
Temperature conversion between the Celsius and Fahrenheits scales
Kelvin from Celsius T(K) = t (oC) + 273.15
Fahrenheit from Celsius T(oF) = 9/5 t (oC) + 32
Celsius from Fahrenheit t(oC) = 5/9 [t (oF) -32]

5. Rules for Counting Significant Figures
All nonzero digits are significants.
Any zeros preceding the decimal point are not significant
Any zeros following the decimal point and preceding the first nonzero
digit are not significant

6. Question 1:
What are the principal reasons that one theory might be adopted over a
conflicting one?
One theory is preferred over another if it can correctly predict a wider range of
phenomena and if it has fewer assumptions.
Question 3:
A common belief among scientists is that there exists an underlying
order to nature, Einstein described this belief in the words “God is
subtle, but He is not malicious.” What do you think Einstein meant by
this remark?
For a given set of conditions, a cause, is expected to produce a certain result or
effect. Although these cause-and-effect relationships may be difficult to unravel at
times (“God is subtle”), they nevertheless do exist (“He is not malicious”).

7. Question 6:
Describe the necessary characteristics of a scientific theory
For a theory to be considered as plausible, it must, first and
foremost, agree with and/or predict the results from
controlled experiments. It should also involve the fewest
number of assumptions (i.e., follow Occam’s Razor). The best
theories predict new phenomena that are subsequently
observed after the appropriate experiments have been
performed.

8. Question 10:
Indicate whether each sample of matter listed is a substance or a
mixture; if it is a mixture, indicate whether it is homogeneous or
heterogenerous.
(a) a wooden beam
Heterogeneous mixture: We can clearly see air pockets within the solid matrix. On
close examination, we can distinguish different kinds of solids by their colors.
(b) red ink
Homogeneous mixture: Modern inks are solutions of dyes in water. Older inks
often were heterogeneous mixtures: suspensions of particles of carbon black
(soot) in water.
(c) distilled water
Substance: This is assuming that no gases or organic chemicals are dissolved in
the water.
(d) freshly squeezed orange juice
Heterogeneous mixture: The pieces of orange pulp can be seen through a
microscope. Most “cloudy” liquids are heterogeneous mixtures; the small
particles impede the transmission of light.

9. Question 12:
What type of change-physical or chemical-is necessary to separate the
following?
(a) Sugar from a sand/sugar mixture
Physical: This is simply a mixture of sand and sugar (i.e., not chemically
bonded).
(b) Iron from iron oxide (rust)
Chemical: Oxygen needs to be removed from the iron oxide.
(c) Pure water from seawater
Physical: Seawater is a solution of various substances dissolved in water.
(d) Water from a slurry of sand in water
Physical: The water-sand slurry is simply a heterogeneous mixture.

10. Question 16:
Express each value in exponential form. Where appropriate, include
units in your answer.
(a) Solar radiation received by Earth: 173 thousand trillion watts
173 thousand trillion watts = 173,000,000,000,000,000 W= 1.73 x 1017 W
(b) Average human cell diameter: 1 ten-millionth of a meter
one ten millionth of a meter = 1 ÷ 10,000,000 m =1 x 10-7 m
(c) The distance between the centers of the atoms in silver metal: 142
trillionths of a meter
(trillionth = 1 x 10-12) hence, 142 x 10-12 m or 1.42 x 10-10 m
(d)
(5.07×104)×(1.8 × 10−3)2
0.065+(3.3 × 10−2)
=
0.164
0.098
= 1.7

11. Question 22:
Perform the following calculations; express each number and the answer in exponential
form and with the appropriate number of significant figures.
a) (320 x 24.9)/0.080= 99600
1.0 x 105
b) (432.7 x 6.5 x 0.002300)/(62 x 0.103) = 1.012976
1.0 x 100
c) (32.44 + 4.9 – 0.304)/82.94 = 37.03 / 82.94 = 0.4465
(3 sig. fig.) (4 sig. fig.)
4.47 x 10-1
d) (8.002 + 0.3040)/(13.4 – 0.066 + 1.02) = 8.3060 / 14.35 = 0.5786
(4 sig. fig) (3 sig. fig.)
5.79 x 10-1
A. The result of multiplication or division may contain only as many significant
figures as the least precisely known quantity in the calculation.
B. The result of addition or subtraction must be expressed with the same number
of digits beyond the decimal point as the quantity carrying the smallest number
of such digits.

12. Question 24:
Express the result of each of the following calculations in exponential
form and with the appropriate number of significant figures.
a) (4.65 ×104
)×(2.95×10−2
) × (6.663 × 10−3
) × 8.2 = 7.49477 x 101
7.5 x 101
b)
1912×(0.0077×104)×(3.12×10−3)
(4.18×10−4)3 = 6.289 x 1012 6.3 x 1012
c) (3.46× 103
)× 0.087 × 15.26 × 1.0023=4.604 x 103 4.6 x 103
d)
(4.505×10−2)2×1.080×1545.9
0.03203×103 = 1057883 x 10-1 1.058 x 105

13. Question 1-24:
Express the result of each of the following calculations in exponential
form and with the appropriate number of significant figures.
e.
−3.61×10−4 + (3.61×10−4)2+4(1.00)(1.9×10−5)
2×(1.00)
=
−3.61×10−4 + 13.03×10−8+7.60×10−5
2×(1.00)
=
−3.61×10−4 + 0.01303×10−5+7.60×10−5
2×(1.00)
=
−3.61×10−4 + 7.61×10−5
2×(1.00)
=
−0.361×10−3 +8.72×10−3
2×(1.00)
=
8.36×10−3
2×(1.00)
= 4.2 x 10-3
Quadratic formula
3.612 = 13.0321

14. Question 26:
Use the concept of significant figures to criticize the way in
which the following information was presented. “The
estimated proved reserve of natural gas as of January 1,
1982, was 2,911,346 trillion cubic feet.”
If the proved reserve truly was an estimate, rather than an
actual measurement, it would have been difficult to estimate
it to the nearest trillion cubic feet. A statement such as
2,911,000 trillion cubic feet (or even 3 x 1018 ft3 ) would have
more realistically reflected the precision with which the
proved reserve was known.

15. Question 29:
Perform the following conversions from non-SI to SI units. (Use
information from the inside back cover, as needed.)
a) 68.4 in. = 174 cm
68.4 x 2.54 = 174 cm
b) 94 ft = 29 m
94 ft ÷ 3.2808 ft/1 m = 29 m
c) 1.42 lb = 644 g
1.42 lb x 453.6g/lb = 644 g
d) 248 lb = 112 kg
248 lb x 453.6 g/lb ÷1000 = 112 kg
e) 1.85 gal = 7.00 dm3 (1L = 1 dm3)
1.85 gal x 3.785412 L/1 gal = 7.00 dm3
f) 3,72 qt = 3.52 x103 mL
3.72 qt ÷ 1.056688 qt/1L x 1000 mL/1L = 3.52 x 103 mL
1 in. = 2.54 cm
l b = 453.6 g
1 gal = 3.785412 L
1L = 1.056688 qt
1 m = 3.2808 ft

16. Question 39:
In a user’s manual accompanying an American-made automobile, a
typical gauge pressure for optimal performance of automobile tires is
32 lb/in2. What is this pressure in grams per square centimetre and
kilograms per square meter? g/cm2
kg/m2
Pressure = 32 lb/in2 x 453.6 g/lb x (1/2.54 cm)2
= 2.2 x 103 g/cm2
Pressure =2.2 x 103 g/cm2 x (1/1000)kg/(1/100)2m2
= 2.2 x 104 kg/m2
lb = 453.6 g 1 in. = 2.54 cm

17. Question 43:
The absolute zero of temperature is -273.15 oC. Should it be
possible to achieve a temperature of - 465 oF? Explain.
Let us determine the Fahrenheit equivalent of absolute
zero
oF =
9𝑜𝑜
F
5 𝑜𝑜
𝐶𝐶
(oC)+32 oF =
9𝑜𝑜
F
5 𝑜𝑜
𝐶𝐶
(-273.15oC)+32 oF
=-459.7oF
So, a temperature of -465 oF cannot be achieved because
it is below absolute zero.

18. Question 53:
A fertilizer contains 21% nitrogen by mass. What mass of the fertilizer,
in kilograms, is required for an application requiring 225 g of nitrogen?
1.1 kg
∵ In x amount of fertilizer, 21% of which is 225 g of nitrogen
∴ x = 225 g ÷ 0.21
= 1071 g
= 1.1 kg

19. Question 57:
The following densities are given at 20 oC: water, 0.998 g/cm3; iron, 7.86 g/cm3;
aluminium, 2.70 g/cm3. Arrange the following items in terms of increasing mass.
a) A rectangular bar of iron, 81.5 cm x 2.1 cm x 1.6 cm
b) A sheet of aluminium foil, 12.12 m x 3.62m x 0.003 cm
c) 4.051 L of water
m = d x V
We start by determining the mass of each item.
(a) mass of iron bar = (81.5 cm x 2.1 cm x 1.6 cm) x 7.86 g/cm3
= 2.2 x 103 g iron
(b) mass of Al foil = (12.12 m x 3.62 m x 0.003 x 10-2 m) x 2.70 gAl/cm3
= 4 x 103 g Al
(c) mass of water = 4.051 L x 1000 cm / L x 0.998 g/cm3
= 4.04 x 103 g water
In order of increasing mass, the items are: iron bar < aluminum foil < water.
Please bear in mind, however, that, strictly speaking, the rules for significant
figures do not allow us to distinguish between the masses of aluminum and
water.

20. Question 65:
A solution of sucrose in water is 28.0% sucrose by mass and has a density
of 1.118 g/mL. What mass of sucrose, in grams, is contained in 3.50L of
this solution?
m = d x V
Use the percent composition as a conversion factor
Stepwise approach:
3.50 L ×
1000 𝑚𝑚𝑚𝑚
1𝐿𝐿
= 3.50 x 103 mL (unit conversion)
m = D x V = 3.50 x 103 mL x
1.118 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
1 𝑚𝑚𝑚𝑚
= 3.91 x 103 g soln (mass of solution of
sucrose)
3.91 x 103 g soln x
28.0 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
100 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
= 1.1 0 x 103 g sucrose (mass of pure sucrose)
Conversion pathway approach:
Mass of source = 3.50L ×
1000 𝑚𝑚𝑚𝑚
1𝐿𝐿
×
1.118 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
1 𝑚𝑚𝑚𝑚
×
28.0 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
100 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
= 1.1 0 x 103 g sucrose

21. Question 84:
A piece of high-density Styrofoam measuring 24.0 cm by 36.0 cm by 5.0
cm floats when placed in a tub of water. When a 1.5 kg book is placed
on top of the Styrofoam, the Styrofoam partially sinks, as illustrated in
the diagram below. Assuming that the density of water is 1.00 g/mL,
what is the density of Styrofoam?
First, calculate the volume of the piece of Styrofoam:
V= 36.0 cm x 24.0 cm x 5.0 cm = 4.32 x 103 cm3
Archimede’s Law: F = G (weight of displaced fluid = weight of object in fluid)
ρH2Ogv = mg
Since the object floats, it means that the water is exerting a force
equivalent to the mass of Styrofoam/book times the acceleration due to
gravity (g). We can factor out g, and are left with masses of Styrofoam and
water: ρH2Ov = m
mass of water = mass of book + mass of Styrofoam
24.0 cm
5.0 cm
36.0 cm
2.0cm

22. Question 84 (continued):
Calculate the volume of water displaced (using dimensions in the figure):
V= 36.0 cm x 24.0 cm x 3.0 cm = 2.592 x 103 cm3
The mass of displaced water is given as:
m = D×V = 1.00 g/mL × 2.592 x 103 cm3 = 2.592 x 103 g
mass of book + mass of Styrofoam = mass of water
1.5 × 103 g + D × 103 cm3 = 2.592 × 103 g
Solving for D, we obtain:
D = 0.25 g/cm3

23. Question 89:
Blood alcohol content (BAC) is sometimes reported in weight-volume
percent and, when it is, a BAC of 0.10% corresponds to 0.10 g ethyl
alcohol per 100 mL of blood. In many jurisdictions, a person is
considered legally intoxicated if his or her BAC is 0.10%. Suppose that
a 68 kg person has a total blood volume of 5.4 L and breaks down ethyl
alcohol at a rate of 10.0 grams per hour. How many 145 mL glasses of
wine, consumed over three hours, will produce a BAC of 0.1% in this
68 kg person? Assume the wine has density of 1.01 g/mL and is 11.55
ethyl alcohol by mass.
First, determine the amount of alcohol that will cause a BAC of 0.10%:
Mass of ethanol =
0.100𝑔𝑔 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒
100 𝑚𝑚𝑚𝑚 𝑜𝑜𝑜𝑜 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏
× 5400 𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 = 5.4 𝑔𝑔 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒

24. This person’s body metabolizes alcohol at a rate of 10.0 g/h. Therefore, in
3 hours, this person metabolizes 30.0 g of alcohol. For this individual to
have a BAC of 0.10% after 3 hours, he must consume 30.0 + 5.4 = 35.4 g of
ethanol.
Now, calculate how many glasses of wine are needed for a total intake of
35.4 g of ethanol:
Volume of wine = 35.4 g ethanol ÷ 11.55 % ÷ 1.01 g/mL = 303.46 mL
Number glasses of wine = 303.46 ÷ 145 = 2.1 glasses of wine

25. Question 100:
A good example of a homogeneous mixture is
a) A cola drink in a tightly capped bottle
b) Distilled water leaving a distillation apparatus
c) Oxygen gas in a cylinder used in welding
d) The material produced in a kitchen blender
The answer is (a), because the gas is fully dissolved in the
liquid and remains there until the cap is removed. (b) and (c)
are pure substances and therefore not mixtures, and material
in a kitchen blender is heterogeneous in appearance.

26. Question 111:
Water, a compound , is a substance. Is there any circumstance under
which a sample of pure water can exist as a heterogeneous mixture?
Explain.
In short, no, because a pure substance by definition is
homogeneous. However, if there are other phases of the same
pure substance present (such as pure ice in pure water), we
have a heterogeneous mixture from a physical standpoint.