1

Assignment 1. Questions from chapters 1 and 2 of McMurry and Fay
Question numbers are from the fourth edition.
Chapter 1. Chemistry: Matter and Measurement

1.1

(a) Cd (b) Sb

1.2

(a) silver

1.3

(a) Ti, metal
(d) Sc, metal

1.4

The three Acoinage metals@ are copper (Cu), silver (Ag), and gold (Au).

1.5

(a) The decimal point must be shifted ten places to the right so the exponent is S10. The
S
result is 3.72 x 10 10 m.
(b) The decimal point must be shifted eleven places to the left so the exponent is 11. The
result is 1.5 x 1011 m.

1.6

(a) microgram (b) decimeter
(c) picosecond
(d) kiloampere
(e) millimole

1.7

(c) Am

(b) rhodium

(c) rhenium

(d) cesium

(b) Te, semimetal
(e) At, semimetal

(e) argon

(f) arsenic

(c) Se, nonmetal
(f) Ar, nonmetal

5 o
5
x ( F _ 32) = x (98.6 _ 32) = 37.0 o C

9
9
K = o C + 273.15 = 37.0 + 273.15 = 310.2 K
o

C=

1.8

(a) K = oC + 273.15 = S78 + 273.15 = 195.15 K = 195 K
9
9
(b) o F = ( x o C) + 32 = ( x 158) + 32 = 316.4o F = 316 o F
5
5
(c) oC = K S 273.15 = 375 S 273.15 = 101.85oC = 102oC
9
9
o
F = ( x o C) + 32 = ( x 101.85) + 32 = 215.33o F = 215o F
5
5

1.9

d=

1.10

volume = 9.37 g x


1.11

The actual mass of the bottle and the acetone = 38.0015 g + 0.7791 g = 38.7806 g. The
measured values are 38.7798 g, 38.7795 g, and 38.7801 g. These values are both close to
each other and close to the actual mass. Therefore the results are both precise and
accurate.

m
27.43 g
=
= 2.212 g/ cm 3
V 12.40 cm 3
1 mL
= 6.32 mL
1.483 g

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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
1.12

(a) 76.600 kg has 5 significant figures because zeros at the end of a number and after the
decimal point are always significant.
(b) 4.502 00 x 103 g has 6 significant figures because zeros in the middle of a number are
significant and zeros at the end of a number and after the decimal point are always

significant.
(c) 3000 nm has 1, 2, 3, or 4 significant figures because zeros at the end of a number and
before the decimal point may or may not be significant.
(d) 0.003 00 mL has 3 significant figures because zeros at the beginning of a number are
not significant and zeros at the end of a number and after the decimal point are always
significant.
(e) 18 students has an infinite number of significant figures since this is an exact number.
S
(f) 3 x 10 5 g has 1 significant figure.
(g) 47.60 mL has 4 significant figures because a zero at the end of a number and after the
decimal point is always significant.
(h) 2070 mi has 3 or 4 significant figures because a zero in the middle of a number is
significant and a zero at the end of a number and before the decimal point may or may not
be significant.

1.13

(a) Since the digit to be dropped (the second 4) is less than 5, round down. The result is
3.774 L.
(b) Since the digit to be dropped (0) is less than 5, round down. The result is 255 K.
(c) Since the digit to be dropped is equal to 5 with nothing following, round down. The
result is 55.26 kg.
24.567

1.14

(a)

g


+ 0.044 78 g This result should be expressed with 3 decimal places. Since the
24.611 78 g

digit to be dropped (7) is greater than 5, round up. The
result is 24.612 g (5 significant figures).
(b)
4.6742 g / 0.003 71 L = 1259.89 g/L
0.003 71 has only 3 significant figures so the result of the division should have only 3
significant figures. Since the digit to be dropped (first 9) is greater than 5, round up. The
result is 1260 g/L (3 significant figures), or 1.26 x 103 g/L.
0.378 mL

(c)

+ 42.3 mL
_ 1.5833 mL

This result should be expressed with 1 decimal place. Since the

41.0947 mL

digit to be dropped (9) is greater than 5, round up. The
result is 41.1 mL (3 significant figures).

1.15

The level of the liquid in the thermometer is just past halfway between the 32oC and 33oC

2



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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
marks on the thermometer. The temperature is 32.6oC (3 significant figures).
1.16

9
9
x o C) + 32 = ( x 1064) + 32 = 1947 o F
5
5
Ballpark estimate: oF . 2 x oC if oC is large. The melting point of gold . 2000oF.

(a) Calculation:

o

F=(

(b) r = d/2 = 3 x 10 6 m = 3 x 10 4 cm; h = 2 x 10 6 m = 2 x 10 4 cm
S
S
S
Calculation: volume = πr2h = (3.1416)(3 x 10 4 cm)2(2 x 10 4 cm) = 6 x 10 11 cm3
Ballpark estimate: volume = πr2h . 3r2h . 3(3 x 10S4 cm)2(2 x 10S4 cm) . 5 x 10S11 cm3
S

1.17


S

S

S

1 carat = 200 mg = 200 x 10S3 g = 0.200 g
Mass of Hope Diamond in grams = 44.4 carats x

0.200 g
= 8.88 g
1 carat

1 ounce = 28.35 g
Mass of Hope Diamond in ounces = 8.88 g x

1 ounce
= 0.313 ounces
28.35 g

1.18

An LD50 value is the amount of a substance per kilogram of body weight that is a lethal
dose for 50% of the test animals.

1.19

mass of salt = 155 lb x


453.6 g
1 kg
4 g
= 281.2 g or 300 g
x
x
1 lb
1000 g
1 kg

Understanding Key Concepts

1.20

1.21

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Chapter 1 S Chemistry: Matter and Measurement
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1.22

red B gas; blue B 42; green B sodium

1.23

The element is americium (Am) with atomic number = 95. It is in the actinide series.


1.24

(a) Darts are clustered together (good precision) but are away from the bullseye (poor
accuracy).
(b) Darts are clustered together (good precision) and hit the bullseye (good accuracy).
(c) Darts are scattered (poor precision) and are away from the bullseye (poor accuracy).

1.25

(a) 34.2 mL (3 significant figures)

(b) 2.68 cm (3 significant figures)

1.26

The 5 mL graduated cylinder is marked every 0.2
mL and can be read to ∀ 0.02 mL. The 50 mL
graduated cylinder is marked every 2 mL and can
only be read to ∀ 0.2 mL. The 5 mL graduated
cylinder will give more accurate measurements.

1.27

A liquid that is less dense than another will float on top of it. The most dense liquid is
mercury, and it is at the bottom of the cylinder. Because water is less dense than mercury
but more dense than vegetable oil, it is the middle liquid in the cylinder. Vegetable oil is
the least dense of the three liquids and is the top liquid in the cylinder.

Additional Problems

Elements and the Periodic Table
1.28

114 elements are presently known. About 90 elements occur naturally.

1.29

The rows are called periods, and the columns are called groups.
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Chapter 1 S Chemistry: Matter and Measurement
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1.30

There are 18 groups in the periodic table. They are labeled as follows:
1A, 2A, 3B, 4B, 5B, 6B, 7B, 8B (3 groups), 1B, 2B, 3A, 4A, 5A, 6A, 7A, 8A

1.31

Elements within a group have similar chemical properties.

1.32

1.33

1.34
A semimetal is an element with

properties that fall between those of
metals and nonmetals.

1.35

(a) The alkali metals are shiny, soft, low-melting metals that react rapidly with water to
form products that are alkaline.
(b) The noble gases are gases of very low reactivity.
(c) The halogens are nonmetallic and corrosive. They are found in nature only in
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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
combination with other elements.
1.36

Li, Na, K, Rb, and Cs

1.37

Be, Mg, Ca, Sr, and Ba

1.38

F, Cl, Br, and I

1.39


He, Ne, Ar, Kr, Xe, and Rn

1.40

(a) gadolinium, Gd

(b) germanium, Ge

(c) technetium, Tc

(d) arsenic, As

1.41

(a) cadmium, Cd

(b) iridium, Ir

(c) beryllium, Be

(d) tungsten, W

1.42

(a) Te, tellurium
(d) Ar, argon

(b) Re, rhenium
(e) Pu, plutonium


(c) Be, beryllium

1.43

(a) B, boron
(d) Os, osmium

(b) Rh, rhodium
(e) Ga, gallium

(c) Cf, californium

1.44

(a) Tin is Sn: Ti is titanium.
(c) Potassium is K: Po is polonium.

(b) Manganese is Mn: Mg is magnesium.
(d) The symbol for helium is He. The
second letter is lowercase.

1.45

(a) The symbol for carbon is C.
(c) The symbol for nitrogen is N.

(b) The symbol for sodium is Na.
(d) The symbol for chlorine is Cl.


Units and Significant Figures
1.46

Mass measures the amount of matter in an object, whereas weight measures the pull of
gravity on an object by the earth or other celestial body.

1.47

There are only seven fundamental (base) SI units for scientific measurement. A derived
SI unit is some combination of two or more base SI units.
Base SI unit: Mass, kg;
Derived SI unit: Density, kg/m3

1.48

(a) kilogram, kg

1.49

(a) kilo, k

1.50

A Celsius degree is larger than a Fahrenheit degree by a factor of

1.51

A kelvin and Celsius degree are the same size.

(b) meter, m


(b) micro, Φ

(d) cubic meter, m3

(c) kelvin, K

(c) giga, G

6

(d) pico, p

(e) centi, c

9
.
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Chapter 1 S Chemistry: Matter and Measurement
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1.52

The volume of a cubic decimeter (dm3) and a liter (L) are the same.

1.53


The volume of a cubic centimeter (cm3) and a milliliter (mL) are the same.

1.54

Only (a) is exact because it is obtained by counting. (b) and (c) are not exact because
they result from measurements.
4.8673 g

1.55

The result should contain only 1 decimal place. Since the digit to

_ 4.8 g
0.0673 g

be dropped (6) is greater than 5, round up. The result is 0.1 g.

1.56

cL is centiliter (10-2 L)

1.57

(a) deciliter (10-1 L)
(c) micrometer (10-6 m)

1.58

1 mg = 1 x 10-3 g and 1 pg = 1 x 10-12 g
1 x 10 _ 3 g

1 pg
x
= 1 x 109 pg/mg
1 mg
1 x 10 _12 g

(b) decimeter (10-1 m)
(d) nanoliter (10-9 L)

35 ng = 35 x 10-9 g
35 x 10 _ 9 g
1 pg
x
= 3.5 x 10 4 pg / 35 ng
_12
35 ng
1 x 10 g
1.59

1 µL = 10-6 L

20 mL = 20 x 10-3 L
1.60

(a)

1 µL
= 106 µ L/L
_6
L

10
20 x 10 _ 3 L 1 µ L
x _ 6 = 2 x 10 4 µ L/ 20 mL
20 mL
10 L

5 pm = 5 x 10-12 m
100 cm
5 x 10-12 m x
= 5 x 10-10 cm
1m
1 nm
5 x 10-12 m x
= 5 x 10-3 nm
_9
1 x 10 m
3

(b)

 1m 
-6
3
8.5 cm x 
 = 8.5 x 10 m
100
cm


3


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
3

 10 mm 
3
3
8.5 cm x 
 = 8.5 x 10 mm
 1 cm 
3

(c)

1 x 10 _ 3 g
= 0.0652 g
1 mg
1 pg
1 x 10 _ 3 g
65.2 mg x
x
= 6.52 x 1010 pg
_12
1 x 10 g

1 mg
65.2 mg x

1.61

(a)
(b)
(c)
(d)

A liter is just slightly larger than a quart.
A mile is about twice as long as a kilometer.
An ounce is about 30 times larger than a gram.
An inch is about 2.5 times larger than a centimeter.

1.62

(a) 35.0445 g has 6 significant figures because zeros in the middle of a number are
significant.
(b) 59.0001 cm has 6 significant figures because zeros in the middle of a number are
significant.
(c) 0.030 03 kg has 4 significant figures because zeros at the beginning of a number are
not significant and zeros in the middle of a number are significant.
(d) 0.004 50 m has 3 significant figures because zeros at the beginning of a number are
not significant and zeros at the end of a number and after the decimal point are always
significant.
(e) 67,000 m2 has 2, 3, 4, or 5 significant figures because zeros at the end of a number
and before the decimal point may or may not be significant.
(f) 3.8200 x 103 L has 5 significant figures because zeros at the end of a number and after
the decimal point are always significant.


1.63

(a) $130.95 is an exact number and has an infinite number of significant figures.
(b) 2000.003 has 7 significant figures because zeros in the middle of a number are
significant.
(c) The measured quantity, 5 ft 3 in., has 2 significant figures. The 5 ft is certain and the
3 in. is an estimate.

1.64

To convert 3,666,500 m3 to scientific notation, move the decimal point 6 places to the left
and include an exponent of 106. The result is 3.6665 x 106 m3.

1.65

Since the digit to be dropped (3) is less than 5, round down. The result to 4 significant
figures is 7926 mi or 7.926 x 103 mi.
Since the digit to be dropped (2) is less than 5, round down. The result to 2 significant
figures is 7900 mi or 7.9 x 103 mi.

1.66

(a) To convert 453.32 mg to scientific notation, move the decimal point 2 places to the
left and include an exponent of 102. The result is 4.5332 x 102 mg.
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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
(b) To convert 0.000 042 1 mL to scientific notation, move the decimal point 5 places to
the right and include an exponent of 10S5. The result is 4.21 x 10S5 mL.
(c) To convert 667,000 g to scientific notation, move the decimal point 5 places to the
left and include an exponent of 105. The result is 6.67 x 105 g.
1.67

(a) Since the exponent is a negative 3, move the decimal point 3 places to the left to get
0.003 221 mm.
(b) Since the exponent is a positive 5, move the decimal point 5 places to the right to get
894,000 m.
(c) Since the exponent is a negative 12, move the decimal point 12 places to the left to
get 0.000 000 000 001 350 82 m3.
(d) Since the exponent is a positive 2, move the decimal point 2 places to the right to get
641.00 km.

1.68

(a) Since the digit to be dropped (0) is less than 5, round down. The result is 3.567 x 104
or 35,670 m (4 significant figures).
Since the digit to be dropped (the second 6) is greater than 5, round up. The result is
35,670.1 m (6 significant figures).
(b) Since the digit to be dropped is 5 with nonzero digits following, round up. The result
is 69 g (2 significant figures).
Since the digit to be dropped (0) is less than 5, round down. The result is
68.5 g (3 significant figures).
(c) Since the digit to be dropped is 5 with nothing following, round down. The result is
4.99 x 103 cm (3 significant figures).
(d) Since the digit to be dropped is 5 with nothing following, round down. The result is

S
2.3098 x 10 4 kg (5 significant figures).

1.69

(a) Since the digit to be dropped (1) is less than 5, round down. The result is 7.000 kg.
(b) Since the digit to be dropped is 5 with nothing following, round down. The result is
1.60 km.
(c) Since the digit to be dropped (1) is less than 5, round down. The result is 13.2 g/cm3.
(d) Since the digit to be dropped (1) is less than 5, round down. The result is 2,300,000.
or 2.300 000 x 106.

1.70

(a) 4.884 x 2.05 = 10.012
The result should contain only 3 significant figures because 2.05 contains 3 significant
figures (the smaller number of significant figures of the two). Since the digit to be
dropped (1) is less than 5, round down. The result is 10.0.
(b) 94.61 / 3.7 = 25.57
The result should contain only 2 significant figures because 3.7 contains 2 significant
figures (the smaller number of significant figures of the two). Since the digit to be
dropped (second 5) is 5 with nonzero digits following, round up. The result is 26.
(c) 3.7 / 94.61 = 0.0391
The result should contain only 2 significant figures because 3.7 contains 2 significant
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Chapter 1 S Chemistry: Matter and Measurement

_____________________________________________________________________________
figures (the smaller number of significant figures of the two). Since the digit to be
dropped (1) is less than 5, round down. The result is 0.039.

(d)

5502.3
24
+ 0.01
5526.31

This result should be expressed with no decimal places. Since the

digit to be dropped (3) is less than 5, round down. The result is
5526.

(e)

86.3
+ 1.42
_ 0.09

This result should be expressed with only 1 decimal place. Since

87.63
the digit to be dropped (3) is less than 5, round down. The result is
87.6.

(f) 5.7 x 2.31 = 13.167
The result should contain only 2 significant figures because 5.7 contains 2 significant

figures (the smaller number of significant figures of the two). Since the digit to be
dropped (second 1) is less than 5, round down. The result is 13.

1.71

3.41 _ 0.23
3.18
x 0.205 =
x 0.205 = 0.12457 = 0.125
5.233
5.233
Complete the subtraction first. The result has 2 decimal places and 3 significant figures.
The result of the multiplication and division must have 3 significant figures. Since the
digit to be dropped is 5 with nonzero digits following, round up.

(a)

5.556 x 2.3 5.556 x 2.3
=
= 3.08 = 3.1
4.223 _ 0.08
4.143
Complete the subtraction first. The result of the subtraction should have 2 decimal places
and 3 significant figures (an extra digit is being carried until the calculation is completed).
The result of the multiplication and division must have 2 significant figures. Since the
digit to be dropped (8) is greater than 5, round up.

(b)

Unit Conversions

1.72

(a) 0.25 lb x

453.59 g
= 113.4 g = 110 g
1 lb

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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
(b) 1454 ft x

12 in 2.54 cm
1m
= 443.2 m
x
x
1 ft
1 in
100 cm
2

2

 1.6093 km   1000 m 

12
2
(c) 2,941,526 mi2 x 
 x
 = 7.6181 x 10 m
1
mi
1
km

 


1.73

2.54 cm
1m
x
= 0.14 m
1 in
100 cm
1 kg
= 30.08 kg
(b) 66.31 lb x
2.2046 lb
3.7854 L 1 x 10 _ 3 m 3
(c) 0.5521 gal x
x
= 2.090 x 10 _ 3 m 3
1 gal

1L
mi 1.6093 km 1000 m
1h
1 min
m
(d) 65
x
x
x
x
= 29
h
1 mi
1 km
60 min 60 s
s

(a) 5.4 in x

3

 1m 
3
(e) 978.3 yd x 
 = 748.0 m
1.0936
yd


3


2

2

 1.6093 km   1000 m 
6
2
(f) 2.380 mi x 
 x
 = 6.164 x 10 m
1
mi
1
km

 

2

2

1.74

(a) 1 acre-ft x

1 mi 2
 5280 ft 
3
x

 = 43,560 ft
640 acres  1 mi 
3

 5280 ft  1 acref- ft
(b) 116 mi 3 x 
= 3.92 x 108 acre- ft
 x
3
 1 mi  43,560 ft

1.75

(a) 18.6 hands x

1/ 3 ft 12 in 2.54 cm
x
x
= 189 cm
1 hand 1 ft
1 in
3

3

3

3

 1/ 3 ft   12 in   2.54 cm   1 m 

3
(b) (6 x 2.5 x15) hands x 
 x
 x
 x
 = 0.2 m
 1 hand   1 ft   1 in   100 cm 
3

1.76

(a)
(b)

(c)
(d)
(e)

200 mg 1000 mL
x
= 2000 mg/L
100 mL
1L
200 mg 1 x 10 _ 3 g
1 µg
x
x
= 2000 µ g/mL
100 mL
1 mg

1 x 10 _ 6 g
200 mg 1 x 10 _ 3 g 1000 mL
x
x
= 2 g/L
100 mL
1 mg
1L
200 mg 1 x 10 _ 3 g 1000 mL
1 ng
1 x 10 _ 6 L
x
x
x
x
= 2000 ng/ µ L
100 mL
1 mg
1L
1 x 10 _ 9 g
1 µL
2 g/L x 5 L = 10 g

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Chapter 1 S Chemistry: Matter and Measurement
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1.77

8.65 stones x

1.78

55

1.79

160 lb x

14 lb
= 121 lb
1 stone

mi 5280 ft 12 in 2.54 cm
1h
2.5 x 10 _ 4 s
cm
x
x
x
x
x
= 0.61
h
1 mi
1 ft

1 in
3600 s
1 shake
shake

1 kg
= 72.6 kg
2.2046 lb
20 µ g
1 mg
72.6 kg x
x
= 1.452 mg = 1.5 mg
1 kg
1 x 103 µ g

Temperature

1.80

1.81

1.82

1.83

1.84

9
F = ( x oC) + 32

5
9
o
F = ( x 39.9 oC) + 32 = 103.8o F
5
9
o
F = ( x 22.2oC) + 32 = 72.0o F
5
o

(goat)
(Australian spiny anteater)

9

For Hg: mp is  x (_ 38.87)  + 32 = _ 37.97 o F
5

9

For Br2: mp is  x (_ 7.2)  + 32 = 19.0 o F
5

9

For Cs: mp is  x (28.40)  + 32 = 83.12o F
5

9



For Ga: mp is  x (29.78)  + 32 = 85.60o F
5


5 o
5
x ( F _ 32) = x (6192 _ 32) = 3422o C
9
9
K = oC + 273.15 = 3422 + 273.15 = 3695.15 K or 3695 K
o

o

C=

9
9
F = ( x o C) + 32 = ( x 175) + 32 = 347 o F
5
5

Ethanol boiling point 78.5oC
173.3oF
200oE
o
o
S179.1 F

0oE
Ethanol melting point S117.3 C
o
o
200 E
200 E
(a)
=
= 1.021 o E/ o C
o
o
o
[78.5 C _ (_117.3 C)] 195.8 C

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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
o

o

200 E
200 E
=
= 0.5675 o E/ o F
o

o
[173.3 F _ (_179.1 F)] 352.4 o F
200
(c) o E =
x ( o C + 117.3)
195.8
200
H2O melting point = 0oC; o E =
x (0 + 117.3) = 119.8o E
195.8
200
H2O boiling point = 100oC; o E =
x (100 + 117.3) = 222.0o E
195.8
200
200
x ( o F + 179.1) =
x (98.6 + 179.1) = 157.6o E
(d) o E =
352.4
352.4
352.4
352.4 



o
(e) o F =  o E x
 _ 179.1 = 130 x
 _ 179.1 = 50.0 F

200 
200 


Since the outside temperature is 50.0oF, I would wear a sweater or light jacket.

(b)

1.85

S33.4oC
S28.1oF
100oA
NH3 boiling point
NH3 melting point
S77.7oC
S107.9oF
0oA
o
o
100 A
100 A
(a)
=
= 2.26 o A / o C
[_ 33.4 _ (_ 77.7 o C)] 44.3o C
o
o
100 A
100 A

(b)
=
= 1.25 o A / o F
[_ 28.1 _ (_107.9o F)] 79.8o F
100
(c) o A =
x ( o C + 77.7)
44.3
100
x (0 + 77.7) = 175o A
H2O melting point = 0oC; o A =
44.3
100
H2O boiling point = 100oC; o A =
x (100 + 77.7) = 401o A
44.3
100
100
(d) o A =
x ( o F + 107.9) =
x (98.6 + 107.9) = 259o A
79.8
79.8

Density

1.86

250 mg x


500 lb x

1.87

1 x 10 _ 3 g
= 0.25 g;
1 mg

V = 0.25 g x

1 cm 3
= 0.18 cm 3
1.40 g

453.59 g
1 cm 3
= 226, 795 g; V = 226, 795 g x
= 161,996 cm 3 = 162, 000 cm 3
1 lb
1.40 g

1L
= 11.2 L
0.0899 g
1L
For Cl2: V = 35.45 g x
= 11.03 L
3.214 g

For H2: V = 1.0078 g x


13


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________

m
220.9 g
g
g
=
= 11.4
= 11
3
3
3
V (0.50 x 1.55 x 25.00) cm
cm
cm

1.88

d=

1.89

d = 2.40 mm = 0.240 cm

r = d/2 = 0.120 cm and V = πr2h
m
0.3624 g
d= =
= 0.534 g/ cm 3
2
V ( 3.1416 ) ( 0.120 cm ) ( 15.0 cm )

1.90

d=

1.91

The explosion was caused by a chemical property. Na reacts violently with H2O.

m
8.763 g
8.763 g
g
g
=
=
= 2.331
= 2.33
3
3
V (28.76 _ 25.00) mL 3.76 mL
cm
cm


General Problems
1.92

(a) selenium, Se

1.93

(a) Element 117 is a halogen because it would be found directly below At in group 7A.
(b) Element 119
(c) Element 115 would be found directly below Bi and would be a metal. Element 117
might have the properties of a semimetal.
(d) Element 119, at the bottom of group 1A, would likely be a soft, shiny, very reactive
metal forming a +1 cation.

1.94

(b) rhenium, Re

(c) cobalt, Co (d) rhodium, Rh

NaCl melting point = 1074 K
C = K S 273.15 = 1074 S 273.15 = 800.85oC = 801oC
9
9
o
F = ( x o C) + 32 = ( x 800.85) + 32 = 1473.53o F = 1474 o F
5
5
NaCl boiling point = 1686 K

o
C = K S 273.15 = 1686 S 273.15 = 1412.85oC = 1413oC
9
9
o
F = ( x o C) + 32 = ( x 1412.85) + 32 = 2575.13o F = 2575o F
5
5
9

9

o
F =  x o C  + 32 =  x (_ 38.9)  + 32 = _ 38.0 o F
5

5


o

1.95

1.96

V = 112.5 g x

1.97

15.28


1 mL
= 75.85 mL
1.4832 g

lb 453.59 g
1 gal
1L
x
x
x
= 1.831 g / mL
gal
1 lb
3.7854 L 1000 mL

14


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
1.98

1.99

V = 8.728 x 1010 lb x

0.22 in x


453.59 g 1 mL
1L
x
x
= 2.162 x 1010 L
1 lb
1.831 g 1000 mL

2.54 cm 10 mm
x
= 5.6 mm
1 in
1 cm

1 lb 8 pints
1 gal
453.59 g
1L
= 0.95861 g/mL
x
x
x
x
1 pint 1 gal
3.7854 L
1 lb
1000 mL
(b) area in m2 =


1.100 (a) density =

2

2

2

2

1 mi 2
 5280 ft   12 in   2.54 cm   1 m 
2
1 acre x
x
 x
 x
 x
 = 4047 m
640 acres  1 mi   1 ft   1 in   100 cm 
(c) mass of wood =
3

3

128 ft 3  12 in   2.54 cm  0.40 g
1 kg
= 1450 kg = 1400 kg
1 cord x
x

x
 x
 x
3
1 cord  1 ft   1 in  1 cm
1000 g
(d) mass of oil =
42 gal 3.7854 L 1000 mL 0.85 g
1 kg
= 135.1 kg = 140 kg
1 barrel x
x
x
x
x
1 barrel
1 gal
1L
1 mL 1000 g
(e) fat Calories =
32 servings 165 Calories 30.0 Cal from fat
0.5 gal x
x
x
= 792 Cal from fat
1 gal
1 serving
100 Cal total
1.101 amount of chocolate =
105 mg caffeine 1.0 ounce chocolate

= 14 ounces of chocolate
2.0 cups coffee x
x
1 cup coffee
15 mg caffeine
14 ounces of chocolate is just under 1 pound.
1.102 (a) number of Hershey=s Kisses =
453.59 g 1 serving 9 kisses
= 199 kisses = 200 kisses
2.0 lb x
x
x
1 lb
41 g
1 serving
41 g
1 serving 1 mL
(b) Hershey=s Kiss volume =
= 3.254 mL = 3.3 mL
x
x
1 serving 9 kisses 1.4 g
230 Cal 1 serving
(c) Calories/Hershey=s Kiss =
= 25.55 Cal/kiss = 26 Cal/kiss
x
1 serving 9 kisses
(d) % fat Calories =
13 g fat 9 Cal from fat
1 serving

x
x
x 100% = 51% Calories from fat
1 serving
1 g fat
230 Cal total
1.103 Let Y equal volume of vinegar and (422.8 cm3 S Y) equal the volume of oil.
Mass = volume x density
397.8 g = (Y x 1.006 g/cm3) + [(422.8 cm3 S Y) x 0.918 g/cm3]
397.8 g = (1.006 g/cm3)Y + 388.1 g S (0.918 g/cm3)Y

15


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
397.8 g S 388.1 g = (1.006 g/cm3)Y S (0.918 g/cm3)Y
9.7 g = (0.088 g/cm3)Y
9.7 g
Y = vinegar volume =
= 110 cm3
3
0.088 g/ cm
3
oil volume = (422.8 cm S Y) = (422.8 cm3 S 110 cm3) = 313 cm3

1.104


5 o
5
x ( F _ 32) ; Set o C = o F : o C = x ( o C _ 32)
9
9
9
Solve for o C : o C x = o C _ 32
5
9
o
( C x ) _ o C = _ 32
5
4
o
C x = _ 32
5
5
o
C = (_ 32) = _ 40o C
4
The Celsius and Fahrenheit scales Across@ at _ 40o C (_ 40o F).
o

C=

volume = 1.30 cm x 5.50 cm x 3.00 cm = 21.45 cm3
0.235 g
mass = 21.45 cm 3 x
= 5.041 g
1 cm 3

Lead:
volume = (1.15 cm)3 = 1.521 cm3
11.35 g
mass = 1.521 cm 3 x
= 17.26 g
1 cm 3
total mass = 5.041 g + 17.26 g = 22.30 g
total volume = 21.45 cm3 + 1.521 cm3 = 22.97 cm3
22.30 g
average density =
= 0.971 g/ cm 3 so the cork and lead will float.
22.97 cm 3

1.105 Cork:

60 s
+ 25 s = 505 s
1 min
Convert 293.2 K to oF 293.2 S 273.15 = 20.05oC
9
o
F = ( x 20.05) + 32 = 68.09o F
5
3. o F
Final temperature = 68.09oF + 505 s x 0 = 93.34 o F
60 s
5
o
C = x (93.34 _ 32) = 34.1o C
9


1.106 Convert 8 min, 25 s to s.

8 min x

19.7325 g
= 0.7893 g/mL
25.00 mL
total mass = metal mass + ethyl alcohol mass = 38.4704 g

1.107 Ethyl alcohol density =

16


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
ethyl alcohol mass = total mass S metal mass = 38.4704 g S 25.0920 g = 13.3784 g
1 mL
= 16.95 mL
ethyl alcohol volume = 13.3784 g x
0.7893 g
metal volume = total volume S ethyl alcohol volume = 25.00 mL S 16.95 mL = 8.05 mL
25.0920 g
metal density =
= 3.12 g/mL
8.05 mL


1.108 Average brass density = (0.670)(8.92 g/cm3) + (0.330)(7.14 g/cm3) = 8.333 g/cm3
2.54 cm
length = 1.62 in x
= 4.115 cm
1 in
2.54 cm
diameter = 0.514 in x
= 1.306 cm
1 in
volume = πr2h = (3.1416)[(1.306 cm)/2]2(4.115 cm) = 5.512 cm3
8.333 g
mass = 5.512 cm3 x
= 45.9 g
1 cm 3
1.109 35 sv = 35 x 109 m
s

3

3

3

  100 cm   1 mL   60 s 
18
(a) gulf stream flow =  35 x 109 m  
 
 = 2.1 x 10 mL/min
3 
s   1 m   1 cm   1 min 



mL   60 min 

 1.025 g 
21
18
(b) mass of H2O =  2.1 x 1018

 ( 24 h ) 
 = 3.1 x 10 g = 3.1 x 10
min   1 h 

 1 mL 

kg
1 min

 1000 mL  
(c) time = (1.0 x 1015 L ) 

 = 0.48 min
18
 1 L   2.1 x 10 mL 
1.110 (a) Gallium is a metal.
(b) Indium, which is right under gallium in the periodic table, should have similar
chemical properties.
0.2133 lb
453.59 g
1 in .3

= 5.904 g/cm3
(c) Ga density =
x
x
3
3
1 in .
1 lb
(2.54 cm )

(d) Ga boiling point
Ga melting point

2204oC
29.78oC

1000oG
0oG

o
o
o
1000 G _ 0 G
1000 G
=
= 0.4599 oG/oC
o
o
o
2174.22 C

2204 C _ 29.78 C

o

G = 0.4599 x (oC S 29.78)
G = 0.4599 x (801 S 29.78) = 355oG

o

17


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
The melting point of sodium chloride (NaCl) on the gallium scale is 355oG.

18


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________
Chapter 2. Atoms, Molecules and Ions

2

2.1


First, find the S:O ratio in each compound.
Substance A: S:O mass ratio = (6.00 g S) / (5.99 g O) = 1.00
Substance B: S:O mass ratio = (8.60 g S) / (12.88 g O) = 0.668
S : O mass ratio in substance A 1.00
3
=
= 1.50 =
S : O mass ratio in substance B 0.668
2

2.2

0.0002 in x

2.3

1 x 1019 C atoms x

2.4

75
34

2.5

35
17

2.54 cm 1 Au atom

x
= 2 x 10 4 Au atoms
•8
1 in
2.9 x 10 cm

1.5 x 10 _10 m
1 km
1 time
x
x
= 37.4 times . 40 times
C atom
1000 m 40, 075 km

Se has 34 protons, 34 electrons, and (75 S 34) = 41 neutrons.

Cl has (35 S 17) = 18 neutrons.

37
17

Cl has (37 S 17) = 20 neutrons.

2.6

The element with 47 protons is Ag. The mass number is the sum of the protons and the
neutrons, 47 + 62 = 109. The isotope symbol is 109
47 Ag .


2.7

atomic mass = (0.6917 x 62.94 amu) + (0.3083 x 64.93 amu) = 63.55 amu

2.8

2.15 g x

2.9

H
H
|
|
H _ C _
N
_ H
|
H phantomC

2.10

Figure (b) represents a collection of hydrogen peroxide (H2O2) molecules.

2.11

adrenaline, C9H13NO3

2.12


(a) LiBr is composed of a metal (Li) and nonmetal (Br) and is ionic.

1 amu
1 Cu
x
= 2.04 x 10 22 Cu atoms
_ 24
1.6605 x 10 g 63.55 amu

19


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Chapter 1 S Chemistry: Matter and Measurement
_____________________________________________________________________________

2.13

(b) SiCl4 is composed of only nonmetals and is molecular.
(c) BF3 is composed of only nonmetals and is molecular.
(d) CaO is composed of a metal (Ca) and nonmetal (O) and is ionic.
Figure (a) most likely represents an ionic compound because there are no discrete
molecules, only a regular array of two different chemical species (ions). Figure (b) most
likely represents a molecular compound because discrete molecules are present.
HF is an acid. In water, HF dissociates to produce H+(aq).
Ca(OH)2 is a base. In water, Ca(OH)2 dissociates to produce OH!(aq).
LiOH is a base. In water, LiOH dissociates to produce OH!(aq).
HCN is an acid. In water, HCN dissociates to produce H+(aq).


2.14

(a)
(b)
(c)
(d)

2.15

(a) CsF, cesium fluoride

2.16

(a) vanadium(III) chloride, VCl3
(c) copper(II) sulfide, CuS

2.17

red B potassium sulfide, K2S; green B strontium iodide, SrI2; blue B gallium oxide, Ga2O3

2.18

(a) NCl3, nitrogen trichloride
(c) S2F2, disulfur difluoride

(b) P4O6, tetraphosphorus hexoxide
(d) SeO2, selenium dioxide

2.19


(a) disulfur dichloride, S2Cl2
(c) nitrogen triiodide, NI3

(b) iodine monochloride, ICl

2.20

(a)
(b)
(c)
(e)

2.21

(a) lithium phosphate, Li3PO4
(c) manganese(II) nitrate, Mn(NO3)2

2.22

Drawing 1 represents ionic compounds with one cation and two anions. Only (c) CaCl2 is
consistent with drawing 1.
Drawing 2 represents ionic compounds with one cation and one anion. Both (a) LiBr and
(b) NaNO2 are consistent with drawing 2.

2.23

(a) HIO4, periodic acid

2.24


A normal visual image results when light from the sun or other source reflects off an
object, strikes the retina in our eye, and is converted into electrical signals that are
processed by the brain. The image obtained with a scanning tunneling microscope, by
contrast, is a three-dimensional, computer-generated data plot that uses tunneling current
to mimic depth perception. The nature of the computer-generated image depends on the
identity of the molecules or atoms on the surface, on the precision with which the probe

(b) K2O, potassium oxide

(c) CuO, copper(II) oxide

(d) BaS, barium

(b) manganese(IV) oxide, MnO2
(d) aluminum oxide, Al2O3

Ca(ClO)2, calcium hypochlorite
Ag2S2O3, silver(I) thiosulfate or silver thiosulfate
(d) Sn(NO3)2, tin(II) nitrate
NaH2PO4, sodium dihydrogen phosphate
Pb(CH3CO2)4, lead(IV) acetate
(f) (NH4)2SO4, ammonium sulfate
(b) magnesium hydrogen sulfate, Mg(HSO4)2
(d) chromium(III) sulfate, Cr2(SO4)3

(b) HBrO2, bromous acid

20

(c) H2CrO4, chromic acid



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Chapter 2 S Atoms, Molecules, and Ions
______________________________________________________________________________
tip is made, on how the data are manipulated, and on other experimental variables.
Understanding Key Concepts
2.25

Drawing (a) represents a collection of SO2 molecules. Drawing (d) represents a mixture
of S atoms and O2 molecules.

2.26

To obey the law of mass conservation, the correct drawing must have the same number of
red and yellow spheres as in drawing (a). The correct drawing is (d).

2.27

Figures (b) and (d) illustrate the law of multiple proportions. The
2.

2.28.

(a) alanine, C3H7NO2

2.29

A Na atom has 11 protons and 11 electrons [drawing (b)].

A Ca2+ ion has 20 protons and 18 electrons [drawing (c)].
A FS ion has 9 protons and 10 electrons [drawing (a)].

(b) ethylene glycol, C2H6O2

mass ratio is

(c) acetic acid, C2H4O2

2.30
2.31

(a) MgSO4

(b) Li2CO3

(c) FeCl2

(d) Ca3(PO4)2

Additional Problems
Atomic Theory
2.32

The law of mass conservation in terms of Dalton=s atomic theory states that chemical
reactions only rearrange the way that atoms are combined; the atoms themselves are not
changed.
The law of definite proportions in terms of Dalton=s atomic theory states that the
chemical combination of elements to make different substances occurs when atoms join
together in small, whole-number ratios.


2.33

The law of multiple proportions states that if two elements combine in different ways to
form different substances, the mass ratios are small, whole-number multiples of each
other. This is very similar to Dalton=s statement that the chemical combination of
elements to make different substances occurs when atoms join together in small, wholenumber ratios.

21


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Chapter 2 S Atoms, Molecules, and Ions
______________________________________________________________________________
2.34

First, find the C:H ratio in each compound.
Benzene:
C:H mass ratio = (4.61 g C) / (0.39 g H) = 12
Ethane:
C:H mass ratio (4.00 g C) / (1.00 g H) = 4.00
Ethylene:
C:H mass ratio = (4.29 g C) / (0.71 g H) = 6.0
C : H mass ratio in benzene 12
3
=
=
C : H mass ratio in ethane 4.00 1
C : H mass ratio in benzene 12 2

=
=
C : H mass ratio in ethylene 6.0 1
C : H mass ratio in ethylene 6.0 3
=
=
C : H mass ratio in ethane
4.00 2

2.35

First, find the C:O ratio in each compound.
Carbon suboxide: C:O mass ratio = (1.32 g C) / (1.18 g O) = 1.12
Carbon dioxide: C:O mass ratio = (12.00 g C) / (32.00 g O) = 0.375
C : O mass ratio in carbon suboxide 1.12 3
=
=
C : O mass ratio in carbon dioxide 0.375 1

2.36

(a) For benzene:
1 amu
1 C atom
4.61 g x
x
= 2.31 x 10 23 C atoms
_ 24
1.6605 x 10 g 12.011 amu
1 amu

1 H atom
0.39 g x
x
= 2.3 x 10 23 H atoms
_ 24
1.6605 x 10 g 1.008 amu
C 2.31 x 10 23 C atoms 1 C
=
=
H 2.3 x 10 23 H atoms 1 H
A possible formula for benzene is CH.
For ethane:

1 amu
1 C atom
x
= 2.01 x 10 23 C atoms
_ 24
1.6605 x 10 g 12.011 amu
1 amu
1 H atom
1.00 g x
x
= 5.97 x 10 23 H atoms
_ 24
1.6605 x 10 g 1.008 amu
C 2.01 x 10 23 C atoms 1 C
=
=
H 5.97 x 10 23 H atoms 3 H

A possible formula for ethane is CH3.
4.00 g x

For ethylene:
1 amu
1 C atom
x
= 2.15 x 10 23 C atoms
_ 24
1.6605 x 10 g 12.011 amu
1 amu
1 H atom
0.71 g x
x
= 4.2 x 10 23 H atoms
_ 24
1.6605 x 10 g 1.008 amu
C 2.15 x 10 23 C atoms 1 C
=
=
H 4.2 x 10 23 H atoms 2 H

4.29 g x

22


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Chapter 2 S Atoms, Molecules, and Ions

______________________________________________________________________________
A possible formula for ethylene is CH2.
(b) The results in part (a) give the smallest whole-number ratio of C to H for benzene,
ethane, and ethylene, and these ratios are consistent with their modern formulas.
2.37

2.38

1 amu
1 C atom
x
= 6.62 x 10 22 C atoms
_ 24
1.6605 x 10 g
12.011 amu
1 amu
1 O atom
1.18 g x
x
= 4.44 x 10 22 O atoms
_ 24
1.6605 x 10 g
15.9994 amu
C 6.62 x 10 22 C atoms 1.5 C
=
=
;
O 4.44 x 10 22 O atoms 1 O
therefore the formula for carbon suboxide is C1.5O, or C3O2.


1.32 g x

g
)(6.02 x 10 23 H atoms) = 1.01 g
H atom
This result is numerically equal to the atomic mass of H in grams.
g
)(6.02 x 10 23 O atoms) = 16.0 g
(b) (26.558 x 10 _ 24
O atom
This result is numerically equal to the atomic mass of O in grams.
(a) (1.67 x 10 _ 24

2.39

The mass of 6.02 x 1023 atoms is its atomic mass expressed in grams.
(a) If the atomic mass of an element is X, then 6.02 x 1023 atoms of this element weighs
X grams.
(b) If the mass of 6.02 x 1023 atoms of element Y is 83.80 g, then the atomic mass of Y is
83.80. Y is Kr.

2.40

Assume a 1.00 g sample of the binary compound of zinc and sulfur.
0.671 x 1.00 g = 0.671 g Zn;
0.329 x 1.00 g = 0.329 g S
1 amu
1 Zn atom
0.671 g x
x

= 6.18 x 10 21 Zn atoms
1.6605 x 10 _ 24 g
65.39 amu
1 amu
1 S atom
0.329 g x
x
= 6.18 x 10 21 S atoms
_ 24
1.6605 x 10 g
32.066 amu
Zn 6.18 x 10 21 Zn atoms 1 Zn
=
=
; therefore the formula is ZnS.
S
6.18 x 10 21 S atoms
1S

2.41

Assume a 1.000 g sample of one of the binary compounds.
0.3104 x 1.000 g = 0.3104 g Ti;
0.6896 x 1.000 g = 0.6896 g Cl
1 amu
1 Ti atom
0.3104 g x
x
= 3.90 x 10 21 Ti atoms
_ 24

1.6605 x 10 g
47.88 amu
1 amu
1 Cl atom
0.6896 g x
x
= 1.17 x 10 22 Cl atoms
_ 24
1.6605 x 10 g
35.453 amu

23


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Chapter 2 S Atoms, Molecules, and Ions
______________________________________________________________________________
Cl 1.17 x 10 22 3
=
=
Ti 3.90 x 10 21 1
Assume a 1.000 g sample of the other binary compound.
0.2524 x 1.000 g = 0.2524 g Ti;
0.7476 x 1.000 g = 0.7476 g Cl
1 amu
1 Ti atom
0.2524 g x
x
= 3.17 x 10 21 Ti atoms

_ 24
1.6605 x 10 g
47.88 amu
1 amu
1 Cl atom
0.7476 g x
x
= 1.27 x 10 22 Cl atoms
_ 24
1.6605 x 10 g
35.453 amu
Cl 1.27 x 10 22 4
=
=
Ti 3.17 x 10 21 1

Elements and Atoms

2.42

The atomic number is equal to the number of protons.
The mass number is equal to the sum of the number of protons and the number of neutrons.

2.43

The atomic number is equal to the number of protons.
The atomic mass is the weighted average mass (in amu) of the various isotopes for a
particular element.

2.44


Atoms of the same element that have different numbers of neutrons are called isotopes.

2.45

The mass number is equal to the sum of the number of protons and the number of
neutrons for a particular isotope.
For 14
6 C , mass number = 6 protons + 8 neutrons = 14.
14
For 7 N , mass number = 7 protons + 7 neutrons = 14.

2.46

The subscript giving the atomic number of an atom is often left off of an isotope symbol
because one can readily look up the atomic number in the periodic table.

2.47

Te has isotopes with more neutrons than the isotopes of I.

2.48

(a) carbon, C (b) argon, Ar (c) vanadium, V

2.49

137
55


Cs

2.50

(a)

220
86

Rn

(b)

210
84

Po

2.51

(a)

140
58

Ce

(b)

60

27

Co

2.52

(a)

15
7

(c)

197
79

Au

N , 7 protons, 7 electrons, (15 S 7) = 8 neutrons

24


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Chapter 2 S Atoms, Molecules, and Ions
______________________________________________________________________________
(b)

60

27

(c)

131
53

I , 53 protons, 53 electrons, (131 S 53) = 78 neutrons

(d)

142
58

Ce, 58 protons, 58 electrons, (142 S 58) = 84 neutrons

(a)
(b)
(c)
(d)

27

Al , 13 protons and (27 S 13) = 14 neutrons
S , 16 protons and (32 S 16) = 16 neutrons
64
Zn , 30 protons and (64 S 30) = 34 neutrons
207
Pb , 82 protons and (207 S 82) = 125 neutrons


2.54

(a)
(c)

24
12
104
46

(b)
(d)

58
28
183
74

2.55

(a)
(c)

202
80
184
76

(b)
(d)


195
78
209
83

2.56

(0.199 x 10.0129 amu) + (0.801 x 11.009 31 amu) = 10.8 amu for B

2.57

(0.5184 x 106.9051 amu) + (0.4816 x 108.9048 amu) = 107.9 amu for Ag

2.58

24.305 amu = (0.7899 x 23.985 amu) + (0.1000 x 24.986 amu) + (0.1101 x Z)
Solve for Z. Z = 25.982 amu for 26Mg.

2.59

The total abundance of all three isotopes must be 100.00%. The natural abundance of
29
Si is 4.67%. The natural abundance of 28Si and 30Si together must be 100.00% S 4.67%
= 95.33%. Let Y be the natural abundance of 28Si and [95.33 S Y] the natural abundance
of 30Si.
28.0855 amu = (0.0467 x 28.9765 amu) + (Y x 27.9769 amu)
+ ([0.9533 S Y] x 29.9738 amu)
_1.842
Solve for Y. Y =

= 0.922
_1.997
28
30
Si natural abundance = 92.2%
Si natural abundance = 95.33 S 92.2 = 3.1%

2.53

Co, 27 protons, 27 electrons, (60 S 27) = 33 neutrons

32

Mg , magnesium
Pd, palladium

Hg, mercury
Os, osmium

Ni, nickel
W, tungsten
Pt , platinum
Bi, bismuth

Compounds and Mixtures, Molecules and Ions

2.60

(a)
(b)

(c)
(d)

muddy water, heterogeneous mixture
concrete, heterogeneous mixture
house paint, homogeneous mixture
a soft drink, homogeneous mixture (heterogeneous mixture if it contains CO2 bubbles)

2.61

(a) 18 karat gold, (b) window glass, and (d) liquefied air are homogeneous mixtures.
(c) Tomato juice is a heterogeneous mixture because the liquid contains solid pulp.
25