SCI20UnitAreview Answers

(1)

S

CIENCE

20

U

NIT

A: C

HEMICAL

C

HANGE

(2)

You may find the following Internet sites useful:

• Alberta Education, http://www.education.gov.ab.ca

• Learning Technologies Branch, http://www.education.gov.ab.ca/ltb

• Learning Resources Centre, http://www.lrc.education.gov.ab.ca Science 20

Unit A: Chemical Change

Chapter and Unit Review Suggested Answers

Exploring the electronic information superhighway can be educational and entertaining. However, be aware that these computer networks are not censored. Students may unintentionally or purposely find articles on the Internet that may be offensive or inappropriate. As well, the sources of information are not always cited and the content may not be accurate. Therefore, students may wish to confirm facts with a second source.

This document is intended for

Students ✓

Teachers ✓

Administrators Home Instructors General Public Other

Copyright © 2006, Alberta Education. This resource is owned by the Crown in Right of Alberta, as represented by the Minister of Education, Alberta Education, 10155 – 102 Street, Edmonton, Alberta, Canada T5J 4L5. All rights reserved.

This resource was developed by or for Alberta Education. It may be reproduced in any form, including photocopying, without written permission from Alberta Education provided that associated copyright notices are also reproduced.

(3)

Chapter 1 Review Questions

Knowledge

1. a. Because the atomic number is 4, the atom is beryllium (Be).

b. atomic number = 4 number of neutrons atomic mass atomic number=

-=

-= =

9 01 4 5 01 5

. .

¨ rounded to the nearest whole number

atomic mass = 9.01 number of neutrons = ?

The atom has 5 neutrons.

c. e e 4p 5n 2 2

-d.

Be

e. Beryllium has two valence electrons (as shown in 1.c. and 1.d.).

f. Be2 +

This ion has a net charge of 2 + because it has two valence electrons. Since metals have a weaker attraction for electrons, it is much easier to lose two electrons than it is to gain six; so, beryllium loses two electrons and acquires a charge of 2 +.

2. a. Because the atom has an atomic number of 16, the atom must be sulfur (S).

b. atomic number = 16 number of neutrons atomic mass atomic number=

-=

32 06 16 16 0

. . 66 16

= ¨ rounded to the nearest whole number

atomic mass = 32.06 number of neutrons = ?

The atom has 16 neutrons.

c. 6e 8e

e 16p 16n

-2

d.

S

e. Sulfur has six valence electrons (as shown in 2.c. and 2.d.)

(4)

-3. a. Since ionic compounds contain a metal and a non-metal, potassium iodide, KI(s), and magnesium oxide, MgO(s), are both examples of ionic compounds. Since molecular compounds contain two non-metals, carbon tetrachloride, CCl₄(l), and hydrogen fluoride, HF(g), are molecular compounds.

b.

C

Cl

H

F

K

_O

2–

_Mg

2+ 4. The ionic bond in the calcium carbonate is a strong bond that holds the ions firmly in place. As a result,

a great deal of heat can be absorbed before the ions have enough energy to move. The amount of energy absorbed from the fire protects the contents of the safe from damage. Metals are good conductors of heat, whereas ionic compounds are poor conductors of heat and act as better insulators.

5. a. NaF(s) is an ionic compound. This compound will dissolve in water due to dissociation and be classified as an electrolyte.

b. CH₄(g) is a molecular compound. This compound will not dissolve in water and be classified as a non-electrolyte.

c. Li₂O(s) is an ionic compound. This compound will dissolve in water due to dissociation and be classified as an electrolyte.

d. AlCl₃(s) is an ionic compound. This compound will dissolve in water due to dissociation and be classified as an electrolyte.

e. C₅H₁₂(l) is a molecular compound. This compound will not dissolve in water and be classified as a non-electrolyte.

C

Cl

F

H

(5)

6. Dissociation is the separation of an ionic compound into individual ions in a solution. The following diagram shows sodium and chloride ions forming associations with charged regions of water molecules.

+

–

+

d+

d+ d+

2d+

7. When an ionic compound dissociates, the solute is separated into its constituent ions. The atoms that form molecules of a molecular compound do not separate when they dissolve. The individual molecules remain intact within the solution.

8. A concentrated solution of sulfuric acid contains more particles of solute in a given volume of solvent. Therefore, a concentrated solution may react quicker and for a longer period of time until all the solute particles have reacted.

9. Cold water on the skin dilutes the acid by increasing the volume of the solvent. This helps reduce the corrosive effects of the acid because there are fewer solute particles in a given volume.

Applying Concepts

10. a. m= 280 g n m

M

=

= =

280

0 817 900 333 0 818

g 342.34 g/mol

mol mol

. .

=

(

)

+

12 22 11

12 12 01 22 1 01

of C of H of O

g/mol g/m

. . ool g/mol

g/mol

(

)

+

(

)

=

11 16 00 342 34

. .

n= ?

The sample contains 0.818 mol of sugar.

b. n= 0.817 900 333 mol C n

V

=

=0 817 900 333. mol

2.00 L

V= 2.00 L

(6)

11. a. m= 16.0 g n m M = = = = ¥ -16 0

0 091 816 825 4 9 18 10 2

. . . g 174.26 g/mol mol mol

=

(

)

+

(

)

+

2 1 4

2 39 10 32 06

of K of S of O

g/mol g/mol

. . 44 16 00

174 26 . . g/mol g/mol

(

)

=

n= ?

The sample contains 9.18 ¥ 10- 2_{mol of potassium sulfate.}

b. n= 0.091 816 825 4 C n

V

=

= =

0 091 816 825 4

0 184 . . mol 0.500 L mol/L

V = ¥

=

500 1

0 500

mL L

1000 mL L

.

C= ?

The molar concentration of the solution is 0.184 mol/L.

12. parts per million = 100 ppm parts per million ppm

parts per solute solution solute = ¥ = m m m 106 million ppm ppm

ppm g

0.0500 g solution 10 100 10 500 5 6 6 ¥ = ¥ = = m

.000 10_¥ -2_g

m_solution 0.500 kg g 1 kg g = ¥ = 1000 500

m_solute= ?

The solution contains 5.00 ¥ 10- 2_{g of lead.}

13. a. m_solvent= 250 g parts per million ppm g g solute solvent = ¥ = ¥ - ¥ m m 10

9 5 10

250 10 6 6 . 66 0 038 ppm ppm = .

m_solute= 9.5 ¥ 10- 6 _g

parts per million = ?

The concentration of arsenic in this water is 0.038 ppm.

b. The concentration of this water is 0.038 ppm, which is above the maximum level of 0.025 ppm allowed by Health Canada, but far below the lethal dose of 60 ppm for a typical adult.

(7)

d. Napoleon suspected that he was being poisoned while he was held captive on the island of St. Helena. A few days before his death, Napoleon requested that his doctors make a full examination of his body, particularly of his stomach. After his death, the doctors determined that the cause of death was a perforated stomach ulcer that had turned cancerous. Although this evidence is consistent with the possible effects of long-term arsenic poisoning, it is not conclusive. Fortunately, Napoleon’s staff kept locks of his hair, which were then passed down within these families through generations. Eventually, modern forensic techniques were applied to the hair samples and traces of arsenic were found.

It is still unclear whether the source of the arsenic was deliberate poisoning or accidental exposure. One line of research suggested that the pigment used in the wallpaper in his room may have had trace amounts of arsenic. The arsenic in these pigments may have been converted into a toxic vapour that could have entered the air in the damp environment of St. Helena.

14. a. V = ¥

=

250 1

0 250

mL L

1000 mL L . C n V n CV = = =

(

)(

)

=

0 500 0 250

0 125

. .

.

mol/L L

mol

C= 0.500 mol/L

n= ?

The solution contains 0.125 mol of solute.

b. V= 5.00 L C n

V n CV = = =

(

)(

)

=

0 0250 5 00

0 125

. .

.

mol/L L

mol

C= 0.0250 mol/L

n= ?

The solution contains 0.125 mol of solute.

15. a. First, determine the number of moles of solute.

V = ¥

=

250 1

0 250

mL L

1000 mL L . C n V n CV = = =

(

)(

)

=

0 146 0 250

0 0365

. .

.

mol/L L

mol

C= 0.146 mol/L

n= ?

The solution contains 0.0365 mol of sugar.

Next, determine the mass of solute.

n= 0.0365 mol n m

M m nM = = =

(

)(

)

=

0 0365 342 34 12 5

. .

.

mol g/mol

g

=

(

)

+

12 22 11

12 12 01 22 1 01

of C of H of O

g/mol g/m

. . ool g/mol

g/mol

(

)

+

(

)

=

11 16 00 342 34

. .

(8)

b. First, determine the number of moles of solute.

V= 4.00 L C n

V n CV

= =

=

(

)(

)

=

0 150 4 00

0 600

. .

.

mol/L L

mol

C= 0.150 mol/L

n= ?

The solution contains 0.600 mol of potassium sulfate.

Next, determine the mass of solute.

n= 0.600 mol n m

M m nM

= =

=

(

)(

)

=

0 600 174 26 105

. mol . g/mol

g

=

(

)

+

(

)

+

2 1 4

2 39 10 32 06

of K of S of O

g/mol g/mol

. . 44 16 00

174 26

. .

g/mol g/mol

(

)

=

m= ?

The solution contains 105 g of potassium sulfate.

16. a. First, determine the number of moles of sugar.

m= 3.89 g n m

M

=

3 89 342 34

0 011 362 972 5 .

. .

g g/mol

mol

=

(

)

+

12 22 11

12 12 01 22 1 01

of C of H of O

g/mol g/m

. . ool g/mol

g/mol

(

)

+

(

)

=

11 16 00 342 34

. .

n= ?

Next, determine the volume of solution.

n= 0.011 362 972 5 mol C n V

V n

C

=

= =

0 011 362 972 5 0 0675 0 168

. . .

mol mol/L L

C= 0.0675 mol/L

V= ?

(9)

b. First, determine the number of moles of potassium sulfate.

m= 6.26 g n m

M = = = 6 26 174 26 0 035 923 333

. . . g g/mol mol

=

(

)

+

(

)

+

2 1 4

2 39 10 32 06

of K of S of O

g/mol g/mol

. . 44 16 00

174 26 . . g/mol g/mol

(

)

=

n= ?

Next, determine the volume of solution required.

n= 0.035 923 333 mol C n

V V n C = = = =

0 035 923 333 0 0250 1 44 . . . mol mol/L L

C= 0.0250 mol/L

V= ?

The volume of solution required is 1.44 L.

17. a. C= 0.783 mol/L C n V n CV = = =

(

)(

)

= =

0 783 0 500

0 3915 0 392

. .

mol/L L

mol mol

V = ¥

=

500 1

0 500

mL L

1000 mL L

.

n= ?

The solution will require 0.392 mol of sugar.

b. n= 0.3915 mol n m

M m nM = = =

(

)(

)

=

0 3915 342 34 134

. mol . g/mol

g

=

(

)

+

12 22 11

12 12 01 22 1 01

of C of H of O

g/mol g/m

. . ool g/mol

g/mol

(

)

+

(

)

=

11 16 00 342 34

. .

m= ?

The solution will require 134 g of sugar.

18. a. C= 0.750 mol/L C n V n CV = = =

(

)(

)

= =

0 750 1 25

0 9375 0 938

. .

mol/L L

mol mol

V= 1.25 L

n= ?

(10)

b. n= 0.9375 mol n m M m nM

= =

=

(

)(

)

=

0 9375 58 44 54 8

. .

.

mol g/mol

g

M =

(

M

)

+

(

M

)

=

(

)

+

(

)

=

of Na of Cl

g/mol g/mol

g/mo

22 99 35 45

58 44

. .

. ll

m= ?

This solution requires 54.8 g of sodium chloride.

19. V_i= 95.0 mL C V C V

C C V

V

i i f f f

i i f

mol/L mL

mL mol/L

=

(

)(

)

=

2 30 95 0

80 0 2 73

. .

C_i= 2.30 mol/L

V_f mL mL

mL

=

-=

95 0 15 0

80 0

. .

.

C_f= ?

The concentration increased to 2.73 mol/L.

20. V_i= 150 mL C V C V

C C V

V

i i f f f

i i f

mol/L mL

mL mol/L

=

(

)(

)

=

3 40 150

900 0 567

.

C_i= 3.40 mol/L

V_f mL mL

mL

= +

=

150 750

900

C_f= ?

The new concentration of the solution is 0.567 mol/L.

21. Water is the solvent in the shampoo because it acts to dissolve the other ingredients.

22. Any of the ingredients besides water could be classified as a solute.

23. The head of an anionic surfactant molecule carries a negative charge. Although a water molecule is neutral, the end of the molecule containing the oxygen atom has a partial negative charge and the end of the molecule containing the two hydrogen atoms has a partial positive charge. As shown in the diagram, the end of the water molecule with the partial positive charge would be attracted to the negatively charged head of the anionic surfactant molecule.

anionic surfactant

water molecule

H H O

(11)

24. a. V_solute= 109.0 mL % %

. _%

. % /

mL 350.0 mL

solute solution

V V V

V

(

)

= ¥

=

100

109 0 ₁₀₀

31 14

V_solution= 350.0 mL

(% V/V) = ?

The percent by volume concentration of the ammonium lauryl sulfate solution is 31.14%.

b. V_solute= 106.5 mL % %

. _%

. % /

mL 350.0 mL

solute solution

V V V

V

(

)

= ¥

=

100

106 5 ₁₀₀

30 43

(% V/V) = ?

The percent by volume concentration of the ammonium laureth sulfate solution is 30.43%.

c. V_solute= 12.0 mL % %

. _%

. % /

mL 350.0 mL

solute solution

V V V

V

(

)

= ¥

=

100

12 0 ₁₀₀

3 43

(% V/V) = ?

The percent by volume concentration of the ammonium dodecylbenzene sulfonate solution is 3.43%.

d. V_solute= 2.0 mL % %

. _%

. % /

mL 350.0 mL

solute solution

V V V

V

(

)

= ¥

=

100

2 0 ₁₀₀

0 57

(% V/V) = ?

The percent by volume concentration of the dimethicone solution is 0.57%.

25. a. Because like charges repel each other, the slightly negative charge on the surface of the hair repels the negative charge on the head of the anionic surfactant. Since the oily layer of sebum and other debris is attached to the long tail of hydrocarbons on the anionic surfactant, this material is washed away as the shampoo is rinsed from the hair.

(12)

26. Water molecules have a partial negative charge on the end of the molecule occupied by the oxygen atom and a partial positive charge on the end occupied by the hydrogen atoms. The result is that water molecules can readily dissolve ionic compounds because the positive and negative ions can be pulled apart and surrounded by the oppositely charged end of water molecules.

Oily substances are molecular compounds that consist of molecules that do not have positively or negatively charged areas. In this case, water molecules are not attracted to these substances. Instead, water molecules attract each other. Since the oil and the water remain separated (do not mix), water is, therefore, a poor solvent for oil.

27. Given the large number of manufacturers in the hair-care industry, the answers to questions 27.a. and 27.b. will vary. Sample answers are given.

a. The first three ingredients listed are water, ammonium laureth sulfate, and ammonium lauryl sulfate.

b. There are 20 ingredients in total listed on the bottle of shampoo.

c. All of the ingredients must be safe to put on hair and skin. So, each ingredient would have to be tested to ensure that it does not irritate the skin and will not damage hair growth at the root and just below the skin. Since shampoo can sometimes get in your eyes, ears, and mouth, it should not be excessively irritating to these areas. Given all the body parts that could be exposed to shampoo in the shower, the product should also not promote the growth of bacteria or other harmful microbes.

Once the shampoo is rinsed off and goes down the drain and into the sewer system, it should be easily treatable so it can be released into the environment. Each ingredient should be compatible with sewage treatment processes and should not have harmful effects on the environment.

Chapter 2 Review Questions

Knowledge

1. a. reduction, gained 2 electrons, Sn4+_(aq)₊_2e-_Æ_Sn2+_(aq)

b. reduction, gained 1 electron, Na+(aq) + 1e- Æ Na(s)

c. reduction, gained 2 electrons, S(s) + 2e- Æ S2-_(aq)

d. oxidation, lost 2 electrons, 2 F-(aq) Æ F₂(g) + 2e

-e. reduction, gained 3 electrons, As(s) + 3e- Æ As3-_(aq)

f. oxidation, lost 3 electrons, Au(s) Æ Au3+_(aq)₊_3e

-g. oxidation, lost 6 electrons, 2 N3-_(aq)_Æ_N

2(g) + 6e

2. a. Li(s) Æ Li+_(aq)_{+ 1e}

-b. Cu(s) Æ Cu2+_(aq)_{+ 2e}

3. a. The substances being oxidized are Mg(s), Ni(s), and Ca(s).

b. The substances being reduced are Cl₂(g), O₂(g), and Ag+(aq).

c. The only spectator ion in these reactions is S2-_(aq).

4. a. The chlorine molecule, Cl₂(g), gains two electrons; the oxygen molecule, O₂ (g), gains four electrons; and the silver ion, Ag+1_{(aq), gains one electron.}

(13)

5. a. A lithium atom could only be oxidized since it can only lose an electron to form a Li+(aq) ion.

b. A hydride ion could only be oxidized since it can only lose an electron to form a hydrogen atom.

6. a. The hydrogen ions, H+(aq), gain electrons to form hydrogen gas.

b. The aluminium metal atoms, Al(s), lose electrons to form aluminium ions, Al3+_(aq).

c. The chlorine ions, Cl-(aq), neither gain nor lose electrons in this reaction because the charge remains the same on both sides of the equation.

d. The aluminium metal atoms, Al(s), are being oxidized because they are losing electrons.

e. n_H

2 =300 mol

n n

coefficient coefficient n

n

n n

Al H

Al H Al

H

Al H

2 2

2

2 =

=

= ¥

2 3 2 3 2 3 3000 200

mol mol

=

n_Al= ?

For this reaction, 200 mol of aluminium metal is required.

f. n_H

2 =300 mol

n n

n

n n

HCl H

HCl H HCl

H

HCl H

2 2

2

2 =

=

= ¥

=

6 3 6 3 6 33 300 600

¥ =

mol mol

n_HCl= ?

For this reaction, 600 mol of hydrochloric acid is required.

7. a. 2 HCl(aq) + Zn(s) Æ ZnCl₂(aq) + H₂(g)

b. Pb(NO₃)₂(aq) + Cu(s) Æ Cu(NO₃)₂(aq) + Pb(s)

c. 2 AgNO₃(aq) + Cu(s) Æ Cu(NO₃)₂(aq) + 2 Ag(s)

(14)

8. a. n_{C H O}

4 6 3 =3 56. mol

n n coefficient coefficient n n

C H O C H O

C H O C H O C H O

C 9 8 4

4 6 3

9 8 4

4 6 3

9 8 4 =

44 6 3

9 8 4 4 6 3 H O

C H O C H O

mol mol = = ¥ = ¥ = 2 1 2 1 2 1 3 56 7 12

n n

. .

n_{C H O}

9 8 4 =?

You could produce 7.12 mol of aspirin.

b. n_{C H O}

4 6 3 =3 56. mol

n n coefficient coefficient n n n H O C H O

H O C H O H O

C H O 2

4 6 3

2

4 6 3

2

4 6 3 =

=1

1

H

H O2 =3 56. mol

n_{H O}

2 =?

Since the mole ratio is equal to 1, this reaction will produce 3.56 mol of water.

c. n_{C H O}

4 6 3 =3 56. mol

n n coefficient coefficient n n

C H O C H O

C H O C H O C H O

C 7 6 3

4 6 3

7 6 3

4 6 3

7 6 3 =

44 6 3

7 6 3 4 6 3 H O

C H O C H O

mol mol = = ¥ = ¥ = 2 1 2 1 2 1 3 56 7 12

n n

. .

n_{C H O}

7 6 3 =?

You will need 7.12 mol of salicylic acid.

9. a. n_CO

2 =7 13 10¥ mol 5 . n n coefficient coefficient n n n CaO CO CaO CO CaO CO CaO 2 2 2 = = = ¥ 1 1

7 13. 1105_mol

n_CaO=?

(15)

b. n_CO

2 =7 13 10¥ mol 5 . n n coefficient coefficient n n n CaCO CO CaCO CO CaCO CO CaC 3 2 3 2 3 2 = =1 1 O

O3 =7 13 10¥ mol 5

.

n_CaCO

3 =?

Since the mole ratio is 1:1, the factory is using 7.13 ¥ 105_{mol of limestone every day.}

c. The mole ratios are 1:1 for all the substances in the reaction; so, the number of moles of carbon dioxide, quicklime, and limestone are all equal.

10. The ionic compounds in these reactions are molten copper(I) sulfide, Cu₂S(l), and molten copper(I) oxide, Cu₂O(l). The molecular compounds are oxygen, O₂(g), and sulfur dioxide SO₂(g). The pure metal is the molten copper, Cu(l).

11. a. m_ore= 1000 kg percentage of copper

kg kg Cu ore = ¥ = ¥ = m m 100 18 1000 100 1 8 % % . %

m_Cu= 18 kg

percentage of copper = ?

The ore contains 1.8% copper.

b. m_ore= 1000 kg parts per million ppm

kg kg ppm 58 p Ag ore = ¥ = ¥ = m m 10 0 058 1000 10 6 6 . ppm

m_Ag g 1 kg

1000 g kg

= ¥

=

58

0 058.

parts per million = ?

The concentration of silver in the ore is 58 ppm.

c. The calculations in 11.a. and 11.b. verify this statement because they are very similar. In both cases, a ratio is multiplied by a fixed number to state the ratio in a new way. In the case of parts per million, the ratio is multiplied by 1 000 000 and the unit is stated as ppm (short for parts per million). For percentage, the ratio is multiplied by 100 and the unit is stated as %, which could be translated to be parts per hundred.

12. a. n_O

2 =34 4. mol

n n

Cu S O

Cu S O 2 2 2 2 mol mol = = ¥ = ¥ = 2 3 2 3 2 3 34 4 22 9

. .

n_{Cu S}

2 =?

(16)

b. n_O

2 =34 4. mol

n n

n

n n

Cu O O

Cu O O Cu O

O Cu O

2

2 =

=

= ¥

2 3 2 3 SSO2

mol mol

= ¥ =

2 3 34 4 22 9

. .

n_{Cu O}

2 =?

Every second, 22.9 mol of copper(I) oxide is produced.

13. Sulfur dioxide is known to be a moderately toxic gas that smells like a match that was just struck. Once sulfur dioxide is released into the atmosphere, the environment hundreds to thousands of kilometres away can be subjected to the effects of acidic deposition.

In the case of dry deposition, the sulfur dioxide is directly deposited on surfaces, such as the leaves of plants in the environment. In high concentrations, dry deposition can kill the trees in close proximity to the pollution source. In locations farther from the source, the effects include reduced growth.

In the case of wet deposition, the sulfur dioxide reacts with water in the atmosphere to form sulfuric acid. It then falls to the ground in the form of rain, snow, sleet, or hail. This is often referred to as acid precipitation.

Acid deposition is particularly harmful to the natural environment in regions where the natural alkalinity, such as a lime base, of the soils and rock formations are unable to neutralize the effects of the acid. This is a major problem in eastern Canada where the natural rock consists of the granite formations of the Canadian Shield. The effects on the environment have been devastating, especially on freshwater ecosystems. Fish are very sensitive to the acid concentration in lakes, rivers, and streams. The increased acidity of the water also causes harmful metals, such as aluminium and mercury, to be leached from the bedrock. These metals can reach toxic levels for many life forms.

Although the soils in much of western Canada have a natural alkalinity that neutralizes much of the effects of acid deposition, there are lakes surrounded by granite rock formations in northern Alberta, Nunavut, and parts of the Northwest Territories that could be susceptible to the harmful effects of acidic deposition if pollution levels increase.

14. a. D= 19.4 g/mL m= ¥ =

¨

250 19 4

4

mL g

mL using density as a conversion factor .

.. .

. 85 10

4 85 10 1

4 85

3 3 ¥

= ¥ ¥

=

g

kg 1000 g kg

g V= 250 mL

m= ?

(17)

b. Answers to this question will vary depending upon the current price of gold. The following calculations are based on the market price in December of 2005.

m= 4.85 kg value=

(

kg $17 560/kg

)

(

)

=

4 85 85 166 00

.

$ .

price of gold = $17 560/kg value = ?

c. When the bits of sand, gold, and other minerals are swirled around in a shallow pan of water, the particles with the greatest mass will tend to sink to the bottom and remain closer to the middle of the pan.

Meanwhile, the less massive particles will have a greater tendency to rise to the top and be swirled to the edges of the pan.

15. a. The least reactive metal is listed at the top, and the most reactive metal is listed at the bottom. The metals increase in reactivity as you go down the right side of the list.

b. The most reactive metal ion is listed at the top, and the least reactive metal ion is listed at the bottom. The ions decrease in reactivity as you go down the left side of the list.

c. For a reaction to occur, there must be an oxidation half-reaction (electrons are donated) and a reduction reaction (electrons are accepted). If the reduction reaction appears above the oxidation half-reaction on the activity series, the half-reaction will be spontaneous.

As a rule of thumb, if the metal ion appears above the metal in the activity series, a spontaneous reaction should occur.

16.

Metal

Metal Ion

Al3+_(aq) _Zn2+_(aq) _Sn2+_(aq) Al(s) non-spontaneous spontaneous spontaneous

Zn(s) non-spontaneous non-spontaneous spontaneous

Sn(s) non-spontaneous non-spontaneous non-spontaneous

17. Metals corrode when they are exposed to other substances they can react with. Usually, this involves an oxidation reaction in which the atoms of the metal lose electrons to form metal ions that bond with a non-metal ion to form an ionic compound. This ionic compound on the surface of the metal is what people call corrosion. For example, iron is oxidized as it reacts with oxygen and water to form the type of corrosion called rust.

(18)

18.

NO3– K+

NO3– NO3–

Mg2+

Mg2+ Ni2+

2e–

e– _e–

2e–

nickel (cathode)

KNO3(aq)

voltmeter

salt bridge magnesium

(anode) – +

Mg(NO3)2(aq)

electrolytes: Mg2+_{and Ni}2+

Ni(NO3)2(aq) Ni

19.

NO3– K+

NO3– NO3–

Al3+

Al Fe2+

3e–

e– _e–

iron (cathode)

KNO3(aq)

voltmeter

salt bridge aluminium

(anode) – +

Al2(NO3)3(aq) Fe(NO₃)₂(aq)

2e–

Fe

20. The voltaic cell with the nickel and magnesium electrodes would produce the greater output on the voltmeter. This is because these two metals are farther apart in the activity series than the metals in the other cell. Since iron and aluminium are closer together in the activity series, these metals do not have a great difference in their ability to donate electrons, resulting in a lower output on the voltmeter.

21. a. Reaction 1 is an oxidation reaction because electrons are donated in this reaction.

b. Since the powdered zinc in contact with the steel top is being oxidized, this end of the cell is the anode.

c. Reaction 2 is a reduction reaction because this reaction involves gaining electrons.

d. Since the powdered HgO(s) is being reduced, this end of the cell is the cathode.

e. If this cell were connected to an external circuit, the electrons would leave the top of the cell and return through the bottom.

(19)

23. a. The cell is electrolytic because electrical energy is being used to force a chemical change.

b. An electrolytic cell involves non-spontaneous reactions. The power source forces the exchange of electrons, making it an essential part of this reaction. Without the power source, this reaction will not occur.

c. This cell is designed to plate zinc on the surface of the iron object. Electrons leave the negative end of the power source and flow toward the electrode attached to the iron object. This allows the zinc ions to combine with the electrons to form a zinc coating on the surface of the iron object.

d. Reduction occurs at the cathode of the cell. In this reaction, zinc ions are reduced to form zinc metal atoms on the surface of the iron object. The half-reaction that describes this process is

Zn2+_(aq)₊_2e-_Æ_Zn(s)

Chapter 3 Review Questions

Knowledge

1. The structural diagrams may vary slightly from those shown in this table.

Name of

Compound Chemical Formula Condensed StructuralDiagram

CH

3

alkane

1-pentyne _C_H_(l)

(20)

3

alkene

1-heptene (and other forms

of heptene)

C₇H₁₄(l)

3 alkene

2. Both continuous-chain alkanes and branched alkanes have only single bonds between carbon atoms and behave like saturated hydrocarbons. Branched hydrocarbons will have numbers included in their names to indicate the position and length of the attached groups on the molecule. Branched alkanes do not pack as tightly together as continuous-chain alkanes, so they melt and boil at lower temperatures.

3. a. and b.

c. 3-methyl-2-pentene would have the higher boiling point and melting point since it is a larger molecule. For all members of the alkene family, morecarbon atoms results in a higher boiling point and melting point.

C

H

H C

H

C

H

2-methyl-2-butene

double bond after carbon2 4-carbon chain

methy

l group on carbon 2

C C

C

H

C

H

H C

H

C

H

3-methyl-2-pentene

double_bon_d_af_te_{r ca}_rbon 2 5-carbon chain

methyl

(21)

4.

furnace

gaseous hydrocarbons

small-chained hydrocarbons and fractions with lower boiling points

long-chained hydrocarbons and fractions with higher boiling points

residue

5. The following steps describe what happens to the molecules of petroleum that undergo the process of fractional distillation:

step 1: The petroleum is vaporized in a hot furnace.

step 2: The petroleum vapour is placed into a tall column.

step 3: The hot vapours rise inside the column. As the vapour moves away from the heat source, it cools.

step 4: As the vapour cools, the molecules condense to form liquids at different places in the tower. By condensing at different locations in the tower, the fractions can be collected separately.

(22)

6. Note that the hydrogen atoms have been omitted so that the carbon atoms can be numbered clearly.

a.

C C

C

C C

1

2 3 4 5

pentane

b.

C C

C

C C

2

C

1

3 4 5 6

3-methylhexane

c.

4-ethyl-4-methyloctane

C C C

C

C C C

1 2 3

4 5 6 7 8

d.

4-ethyl-2-methyloctane

C C C

C

C C

1 2 3 4

5

3 4 5 6 7 8 9

7. First, draw the parent chain (1-propyne).

C C C H

H

Next, add the methyl branch to carbon 2.

C C C H

H

It is impossible to add a methyl branch to carbon 2 because this carbon already has four bonds.

(23)

8. a. Draw 2-pentyne.

C

C C

H

C

H

Write the balanced chemical equation.

C₅H₈(g) + 7 O₂(g) Æ 5 CO₂(g) + 4 H₂O(g)

b. Draw 5,5-diethyl-2,2-dimethyloctane.

CH3

CH2

C

CH3

CH2

CH3

C

CH2

CH3

C₁₄H₃₀(l) + 21.5 O₂(g) Æ 14 CO₂(g) + 15 H₂O(g)

2 C₁₄H₃₀(g) + 43 O₂(g) Æ 28 CO₂(g) + 30 H₂O(g)

c. Draw 2-butene.

C C C

H

C

H

Draw the saturated version of this hydrocarbon chain.

H C C

H H

C

H

C

H

(24)

d. Draw 2-methyl-3-hexyne.

C

C C

H

C

H

C

H

C C

H

H H

H

C

H

C

H

C

H

C

H

C

H

C₇H₁₂(g) + 2 H₂(g) Æ C₇H₁₆(g)

e. Answers will vary. A sample answer is given. C₈H₁₈(g) Æ C₂H₄(g) + C₃H₈(g) + C₃H₆(g)

f. Answers will vary. A sample answer is given. C₆H₁₂(g) Æ C₃H₈(g) + C₃H₄(g)

9. a. Draw ethene, the starter compound.

C

H

C

Draw the unit that repeats throughout the polymer.

C C

H H

Draw a section of the polymer chain.

C C

H H

H

C

H

C

H

H H

C

H

C

H

C

H

(25)

b. Draw propene, the starter compound.

H

C

H

C C

H CH

3

H H

C C

H

C

H

C

H

C

H

CH

3

CH

3

CH

3

10. The granola bar is the healthiest choice if you want to reduce the risks associated with heart disease. To reduce the risk of heart disease, you should choose a food that is low in saturated fat, trans fat, and dietary cholesterol. The granola bar best matches this description.

11. a. Industrially produced trans fat is produced during the partial hydrogenation of vegetable oil. Many of the unsaturated fatty acids in vegetable oil become saturated fatty acids during this process. To become saturated, the double bond must be transformed into a single bond and more hydrogen atoms have to be added to the molecule.

However, some of the unsaturated fatty acids do not have their double bond replaced by a single bond. Instead, they have the hydrogen atoms on either side of the double bond rearrange themselves. Rather than being on the same side of the double bond, one of the hydrogen switches its position so that there is a hydrogen on either side of the double bond. The word trans is Latin for “across” and indicates this new arrangement of the hydrogen atoms. That’s why this form of the fatty acid is called a trans fatty acid. If this fatty acid joins with two others to form a fat molecule, the result is called a trans fat.

b. Both saturated fatty acids and trans fatty acids are hydrocarbons consisting of long, straight chains. Because the chains are straight, these molecules can pack together tightly, allowing for stronger forces of attraction to build between the molecules than if they had a bent shape. The result is that these compounds have higher melting and boiling points than bent molecules with the same number of carbon atoms. This is why the products that contain saturated fat and trans fat tend to be solids at room temperature while the unsaturated, bent fatty acids found in plant oils are liquids at room temperature.

(26)

12. a.

step 1:

fractional distillation

step 3:

fractional distillation naptha

fraction ethene

step 2:

cracking

The Production of Ethene

C5 to C10

petroleum C1 to C4

b.

C C

H

The chemical formula for ethene is C₂H₄(g).

13. a. The parison or preform looks like a plastic test tube with a screw top neck of a pop bottle attached.

The parison tube is then reheated and placed into another mould that is shaped like a 2-L pop bottle. A steel tube is placed inside the parison. High-pressure air is released from inside the steel tube, pressing the parison against the inside of the mould to give it the shape of the bottle. The mould is then rapidly cooled and the bottle is removed.

b. Parisons reduce the cost of shipping because they are a much more compact form of the bottle. If a preform takes up one-fifth the space of a finished 2-L bottle, you can fit five times more parisons in a given shipping container. This reduces costs.

14. a. and b.

Plastic

Recycling Code Full Name of Plastic Properties ApplicationsPackaging

Products Made from Recycling

This Plastic

1 PETE

polyethylene terephthalate

• clear and tough • forms a good

barrier to gas and moisture

bottles for holding soft drinks, water, mouthwash, and salad dressing

used for spinning fibre, known as polyester (carpets, fibrefill insulation for sleeping bags and winter coats, as well as “polar fleece” garments)

(27)

2 HDPE

high-density polyethylene

• stiff and strong • shows good

chemical resistance • not a good

barrier to gases

toys, and

containers for milk, yogurt, margarine, juice, cosmetics, shampoo, and trash

bottles for laundry detergent and motor oil, plastic buckets, flower pots, and edging for gardens

3 V

polyvinyl chloride or PVC or vinyl

• versatile

• can be made into rigid or flexible materials

food wrap, wire and cable insulation, pipes, carpet backing, and window frames

rain gutters for houses, mud flaps for trucks, and garden hose

4 LDPE

low-density polyethylene

• transparent, tough film • good barrier to

moisture

grocery store bags, bread and frozen food bags, and “squeezable” food dispensing bottles for products (e.g., ketchup)

compost bins, trash cans, and furniture

5 PP

polypropylene

• strong, tough • resistant to

heat, chemicals, grease, and oil

bottle tops, frozen food containers, ketchup bottles, and yogurt and margarine containers

brooms, brushes, ice scrapers, and plastic cafeteria trays

6 PS

polystyrene

• versatile

• can be rigid or in a foam

compact disc containers, egg cartons, and plastic cups and cutlery

light switch plates, rulers, and foam plates and cups

15. a. It is important for recycled plastics to be made into items that will have a relatively long useful life because these items cannot be recycled again, since most plastics only have a single reuse capability.

(28)

Unit A Review Questions

1. a. solute b. solvent c. solution

2. a. The two electrolyte solutions are II and III.

b. The two non-electrolyte solutions are I and IV.

3. Although answers may vary, each response should make reference to the associations that form between the ions in the solute and water. Due to the partial charge on each end of a water molecule, both the positive and negative ions in solution are surrounded by water molecules. An electrolytic solution contains dissociated (free) ions that can move throughout the solution.

4. Answers will vary but key aspects of diagrams are indicated.

a. In this diagram, both positive and negative ions are surrounded by water molecules.

water

water +

–

O

H

2d

–

d+

b. In this diagram, the solute has no charge. So, the water molecules are attracted to one another instead.

a molecule of CO2(g)

c. In this diagram, many particles of the solute are present in a given volume of solution.

(29)

5.

Reaction

Half-Reaction for the First Element

Mentioned Oxidation or Reduction

Number of Electrons Gained or Lost A Cu(s) Æ Cu2+_(aq)₊_2e- _oxidation _{2 lost} B 2H+_(aq)₊_2e-_Æ_H

2(g) reduction 2 gained

C Fe(s) Æ Fe2+_(aq)₊_2e- _oxidation _{2 lost} D Ni2+_(aq)₊_2e- _Æ_Ni(s) _reduction _{2 gained}

6. The activity series for metals and metal ions ranks metals and their ions relative to one another with respect to their reactivity. These rankings can predict whether the combination of a metal and a metal ion will result in a spontaneous or non-spontaneous reaction. This information is useful when determining whether a metal will react in a particular solution and when designing a voltaic cell.

7. a. Magnesium metal is being oxidized.

b. Oxygen is being reduced.

c. There are no spectator ions in this reaction.

d. The oxygen is being reduced, so it gains electrons. Each oxygen atom gains two electrons, producing a net gain of four electrons.

e. The magnesium is being oxidized, so it loses electrons. Each magnesium atom loses two electrons, resulting in a net loss of four electrons.

8. a. Silver metal is being oxidized.

b. Hydrogen is being reduced.

c. The sulfate ion is a spectator in this reaction.

d. The hydrogen is being reduced, so it gains electrons. Each hydrogen atom gains one electron, producing a net gain of two electrons.

e. The silver is being oxidized, so it loses electrons. Each silver atom loses one electron, resulting in a net loss of two electrons.

9. a. The hydrogen ion has already lost its only electron. Its only possible change is to gain an electron and be reduced.

b. The gold ion, Au+(aq), can become a more positive ion by losing electrons to become Au3+_{(aq). Therefore,} the gold ion can be oxidized. The gold ion, Au+(aq), can also become a neutral atom by gaining an

electron. Therefore, the gold ion can be reduced as well.

(30)

10.

Name of

Compound Chemical Formula Condensed StructuralDiagram

Alkane, Alkene, or Alkyne

Saturated or Unsaturated

2,2-dimethylpentane C7H16(l)

CH

3

CH

3

CH

2

CH

2

C

CH

3

CH

3 alkane saturated

vary.)

11.

parent chain with six carbon atoms and a triple bond

triple bond located after carbon 2

two ethyl branches

both ethyl branches bonded to carbon 4

4,4-diethyl-2-hexyne

CH

3

C

CH

2

CH

3

CH

3

CH

2

(31)

12. a. Determine the number of moles of AgNO₃(aq).

m =25.4 g n m

M = = = = 25 4 169 88

0 149 517 306 3 0 150 . . . . g g/mol mol mol

=

(

)

+

(

)

+

of Ag of N of O

g/mol g/mol

3

107 87. 14 01. 3 116 00

169 88 . . g/mol g/mol

(

)

=

n= ?

The solution contains 0.150 mol of silver nitrate.

Determine the molar concentration of the solution.

n= 0.149 517 306 3 mol C n

V

=

= =

0 149 517 306 3 0 150 0 997 . mol . . mol L /L

V = ¥

=

150 1

0 150

mL L

1000 mL L

.

C= ?

The molar concentration of this solution is 0.997 mol/L.

b. C= 0.392 mol/L C n

V n CV = = =

(

)(

)

= ¥

-0 392 0 050

0 020 10 2

. .

.

mol/L L

mol or 2.0 mol

V =0.050 L

n= ?

This solution contains 2.0 ¥ 10-2_{mol of solute.}

c. parts per million ppm ppm

=

(

)

=

2 0 010 0 020

. .

parts per million ppm

parts per solute solution solute = ¥ = m m m 106 million ppm ppm

ppm kg

solution

10 0 020

10 4 00

8 0 10

6 6 8 ¥ = ¥ = ¥ -m . . . kkg

m_solution= 4.00 kg

m_solute= ?

The mass of lead dissolved in this water sample is 8.0 ¥ 10- 8_kg.

d. parts per million = 58 ppm parts per million ppm

pa solute solvent solvent solute = ¥ = m m m m 106

rrts per million ppm 1.00 kg ppm ppm kg ¥ = ¥ = ¥ 10 58 10

1 72 10

6 6

4

.

m_solute= 1.00 kg

(32)

13. a. n_N

2 =20 0. mol

n n

n

n n

NH N

NH N NH

N

NH N 3

2

3

2

3

2

3 2

=

= ¥

=

2 1 2 1 2 11 20 0 40 0

¥ =

. .

mol mol

n_NH

3 =?

You could make 40.0 mol of ammonia.

b. n_N

2 =20 0. mol

n n

n

n n

H N

H N H

N

H N

2

2 2

=

= ¥

3 1 3 1 3 1 20..

. 0 60 0

mol mol

=

n_H

2 =?

You will require 60.0 mol of hydrogen gas.

14. The ingredient that could be classified as the solvent is water.

15. Any of the other ingredients could be classified as a solute.

16. Water molecules are neutral objects with a net charge of zero. However, there is a partial positive charge on the end of the molecule occupied by the hydrogen atoms, which is balanced by a partial negative charge on the end occupied by the oxygen atom. Since the head of the cationic surfactant has a positive charge, the surrounding water molecules arrange themselves so that their end with the partial negative charge is closest.

water molecule

O

2d

-d+

d+

+

H H

17. a. A molecule of the cationic surfactant carries a positive charge on its head, so it would tend to be attracted to the slight negative charge of the outer surface of the hairs. The oil-based ingredients designed to give hair its shine found on the tail of the cationic surfactant are carried to the shafts of hair as well.

(33)

18. a. V_solute= 7.0 mL % % . _% . % / mL mL solute solution

V V V

V

(

)

= ¥ = ¥ = 100 7 0 500 100 1 4

V_solution= 500 mL

(% V/V ) = ?

The percent by volume concentration of the distearyldiammonium chloride is 1.4%.

b. V_solute= 6.0 mL % %

. % . % / mL mL solute solution

V V V

V

(

)

= ¥ =

₍

(

)

₎

¥ = 100 6 0 500 100 1 2

V_solution= 500 mL

(% V/V ) = ?

The percent by volume concentration of the cetyl alcohol solution is 1.2%.

c. V_solute= 2.0 mL % %

. % . % / mL mL solute solution

V V V

V

(

)

= ¥ =

₍

(

)

₎

¥ = 100 2 0 500 100 0 40

V_solution= 500 mL

(% V/V ) = ?

The percent by volume concentration of polysorbate-85 is 0.40%.

19. a. 3 SnCl₂(aq) + 2 Al(s) Æ 2 AlCl₃(aq) + 3 Sn(s)

b. Au(NO₃)₃(aq) + 3 Ag(s) Æ 3 AgNO₃(aq) + Au(s)

c. 2 HNO₃(aq) + Cu(s) Æ Cu(NO₃)₂(aq) + H₂(g)

d. Zn(s) + H₂SO₄(aq) Æ H₂(g)+ ZnSO₄(aq)

20. a. Sodium is being oxidized because each sodium atom loses one electron to become a sodium ion, Na+(aq).

b. Chlorine is being reduced because each chlorine atom gains one electron to become a chloride ion, Cl-(aq).

c. A total of two electrons is transferred in this reaction.

d. n_Na= 75.0 mol n

n coefficient coefficient n n n n Cl Na Cl Na Cl Na Cl Na 2 2 2 2 = = = ¥ = 1 2 1 2 1 22 75 0 37 5 ¥ = . . mol mol n_Cl

2 =?

(34)

21. Since the platinum did not react with the silver ions or iron ions, its half-reaction must appear above those of silver and iron in the activity series for metals and metal ions. The most unreactive metals appear near the top of the chart.

22. Reactions are spontaneous when the reduction half-reaction appears above the oxidation half-reaction in the activity series.

a. The reduction of the chromium ion appears below the oxidation of copper metal. Therefore, this reaction is non-spontaneous.

b. The reduction of the hydrogen ion appears above the oxidation of zinc metal. Therefore, this reaction is spontaneous.

c. The reduction of the hydrogen ion appears below the oxidation of silver metal. Therefore, this reaction is non-spontaneous.

d. The reduction of the silver ion appears above the oxidation of nickel metal, so this reaction is spontaneous.

23. The methods used to prevent the corrosion of metal surfaces include covering the surface with paint, oil, and a thin metal coating (electroplating). All of these coatings act as barriers to prevent contact by compounds that would oxidize the metal.

24. Plastics do not corrode like metals do, so the strategy might work. In terms of advantages, a plastic body on a car would make the car lighter and improve fuel economy. One important disadvantage is that in the event of a collision, a plastic body might not provide the same level of protection as a metal body. Another disadvantage is that once the car reaches the end of its lifetime, the car’s body cannot be recycled.

25.

NO3– K+

NO3– NO3–

Mg2+

Ag+

2e–

e– _e–

e–

Mg

silver (cathode) KNO3(aq)

anions _{salt bridge} cations magnesium

(anode)

Mg(NO3)2(aq) AgNO3(aq)

+ –

Ag

26. a. The zinc electrode is gaining mass because the copper ions are coming out of the solution and are being reduced by the zinc metal being oxidized. This would also account for the colour change in the copper nitrate solution. As the copper ions come out of the solution, the solution becomes a fainter colour.

(35)

27. a. Clay used by a ceramic artist is a mixture of alumina, silica, and water.

b. The compounds used to give colour to a glaze include iron oxide, chromium oxide, magnesium dioxide, and copper carbonate.

c. All of the compounds listed in 27.b. are classified as ionic compounds because they consist of both a metal and a non-metal.

d. The compounds used to make glazes are mixed in careful proportions with water to form aqueous solutions. These solutions are mixed to set concentrations expressed in terms of percent by volume.

28. a. Reaction 1 is a reduction reaction because it requires the reactants to gain electrons. In this case, the silver is reduced.

b. Since the powdered silver oxide, Ag₂O(s), is being reduced, this must be the cathode. Reduction always occurs at the cathode.

c. Reaction 2 is an oxidation reaction because it is releasing (or losing) electrons. In this case, the zinc is oxidized.

d. Since the powdered zinc is being oxidized, this must be the anode. The anode is always the place where oxidation occurs.

e. Electrons always leave the anode and travel through the external circuit to the cathode. Therefore, the electrons should leave the anode at the top of the cell and return to the cathode at the bottom of the cell.

29. The device shown in the illustration is called a battery.

30. a. The zinc electrode is being oxidized, making the zinc electrode the anode. Oxidation always occurs at the anode.

b. Electrons always leave the anode and travel to the external circuit. Therefore, the anode is the negative electrode. The diagram confirms this because the negative terminal contacts the negative zinc electrode at the bottom of the battery.

31. a. Both manganese(IV) and ammonium are reduced in this reaction because they both gain electrons. The manganese begins as a Mn4+_{(aq) ion on the reactant side of the equation and ends as a Mn}3+_(aq) ion on the product side. This indicates that the manganese has gained an electron. The ammonium ion, NH₄+(aq), begins as a positive ion on the reactant side and ends as neutral ammonia on the reactant side. The ammonium has gained an electron.

(36)

32. e–

electrode (anode)

electrolyte bath containing Ni2+_(aq)

power source

cathode

drill bit Fe(s)

33.

parent chain with five carbons and a triple bond

triple bond located after carbon 2

one methyl branch

one ethyl branch

3-ethyl-3-methyl-2-pentyne

ethyl group bonded to carbon 3

methyl branch bonded to carbon 3

CH

3

C

b. The products of this cracking reaction are butene, C₄H₈(g), propene, C₃H₆(g), and propane, C₃H₈(g).

35. a. C₄H₁₀(g) + 6.5 O₂(g) Æ 4 CO₂(g) + 5 H₂O(g)

2 C₄H₂(g) + 13 O₂(g) Æ 8 CO₂(g) + 10 H₂O(g)

(37)

c. Draw 3-hexene.

C

C C

H

C

H

C

H

C

C C

H

C

H

C

H

H H

H

C₆H₁₂(g) + H₂(g) Æ C₆H₁₄

36. Draw the starter compound.

C C

H

C

H

H C

H

C

CH3

H

C

CH3

H

C

CH3

H

C

CH3

H

C

CH3

H

C

CH3

H

C

CH3

H

C

CH3

H

37. a. Saturated fats have hydrocarbon chains that have only single carbon-carbon bonds. Unsaturated fats may have a double or a triple carbon-carbon bond within the chain.

b. Indicators that this is a healthy food choice is that it contains omega-6 fatty acids and omega-3 fatty acids. Other indicators that this is a positive food choice is the presence of protein and some food energy.