Science 20 Unit A: Chapter and Unit Review Suggested Answers · PDF filescience 20 unit a: chemical change chapter and unit review suggested answers

SCIENCE 20UNIT A: CHEMICAL CHANGE

CHAPTER AND UNIT REVIEW SUGGESTED ANSWERS

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Science 20Unit A: Chemical ChangeChapter and Unit Review Suggested Answers

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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

4p5n

22

-

-

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

e16p16n

-

-

-2

d. S

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

f. S2 -

This ion has a net charge of 2 - because it has six valence electrons. Since non-metals have a stronger attraction for electrons, it is much easier to gain two electrons than it is to lose six; so, sulfur gains two electrons and acquires a charge of 2 -.

Science 20: Unit A 3 Chapter and Unit Review Suggested Answers

Science 20 © 2006 Alberta Education (www.education.gov.ab.ca)

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, CCl

4(l), and hydrogen fluoride, HF(g), are molecular compounds.

b. C Cl H F K I Mg O

c. A carbon tetrachloride molecule, CCl4(l), uses A hydrogen fluoride molecule, HF(g), uses covalent

covalent bonds to hold the particles together. bonds to hold the particles together.

K+ Cl–

Cl– K+

K+ Cl–

Cl– K+

K+ Cl– K+ Cl–

Cl– K+ Cl– K+

Mg2+ O2–

O2– Mg2+

Mg2+ O2–

O2– Mg2+

Mg2+ O2– Mg2+ O2–

O2– Mg2+ O2– Mg2+

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. C5H

12(l) is a molecular compound. This compound will not dissolve in water and be classified as a

non-electrolyte.

CCl

ClCl Cl

FH

A potassium iodide crystal uses ionic A magnesium oxide crystal uses ionic bonds bonds between potassium ions, K+(aq), between magnesium ions, Mg2 +(aq), and oxide

and chloride ions, Cl-(aq). ions, O2 -(aq).



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+

d+

2d+

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 mM

=

=

==

280

0 817 900 333

0 818

g342.34 g/mol

mol

mol

.

.

M M M M= ( ) + ( ) + ( )

= ( ) +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 nV

=

=

=

0 817 900 333

0 409

.

.

mol2.00 L

mol/L

V = 2.00 L

C = ?

The molar concentration of the solution is 0.409 mol/L.



11. a. m = 16.0 g n mM

=

=

== ¥ -

16 0

0 091 816 825 4

9 18 10 2

.

.

.

g174.26 g/mol

mol

mol

M M M M= ( ) + ( ) + ( )

= ( ) + ( ) +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 nV

=

=

=

0 091 816 825 4

0 184

.

.

mol0.500 L

mol/L

V = ¥

=

500 1

0 500

mL L1000 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

millionppm

ppm ppm

g

0.0500 g

solution10

10010

500

5

6

6

¥

= ¥

==

m

.000 10 2¥ - g

m

solution0.500 kg

g

1 kg

g

= ¥

=

1000

500

msolute

= ?

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

13. a. msolvent

= 250 g parts per million ppm

g g

solute

solvent

= ¥

= ¥ ¥-

m

m10

9 5 10250

10

6

6. 66

0 038

ppm

ppm= .

msolute

= 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.

c. The people of Bangladesh have significant problems with arsenic contamination in their drinking water. Because about 18% of the water wells are contaminated with arsenic, at least 40 million people are thought to be at risk. For thousands of years, a natural source of the arsenic contamination came from run-off from the Himalayan mountains, which picks up the arsenic from rock formations and carries it to the fertile plains where the people live. A more recent source of contamination has been the use of fertilizers and modern agricultural techniques. These have dramatically intensified the problems.



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 L1000 mL

L.

C nV

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 nV

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 L1000 mL

L.

C nV

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 mM

m nM

=

== ( )( )=

0 0365 342 34

12 5

. .

.

mol g/mol

g

M M M M= ( ) + ( ) + ( )

= ( ) +12 22 11

12 12 01 22 1 01

of C of H of O


g/mol

( ) + ( )=

11 16 00

342 34

.

.

m = ?

The solution contains 12.5 g of sugar.



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

V = 4.00 L C nV

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 mM

m nM

=

== ( )( )=

0 600 174 26

105

. . mol g/mol

g

M M M M= ( ) + ( ) + ( )

= ( ) + ( ) +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 mM

=

=

=

3 89342 34

0 011 362 972 5

..

.

g g/mol

mol

M M M M= ( ) + ( ) + ( )

= ( ) +12 22 11

12 12 01 22 1 01

of C of H of O


g/mol

( ) + ( )=

11 16 00

342 34

.

.

n = ?

Next, determine the volume of solution.

n = 0.011 362 972 5 mol C nV

V nC

=

=

=

=

0 011 362 972 50 0675

0 168

..

.

mol mol/L

L

C = 0.0675 mol/L

V = ?

The volume of solution required is 0.168 L.



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

m = 6.26 g n mM

=

=

=

6 26174 26

0 035 923 333

..

.

g g/mol

mol

M M M M= ( ) + ( ) + ( )

= ( ) + ( ) +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 nV

V nC

=

=

=

=

0 035 923 3330 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 nV

n CV

=

== ( )( )==

0 783 0 500

0 3915

0 392

. .

.

.

mol/L L

mol

mol

V = ¥

=

500 1

0 500

mL L1000 mL

L.

n = ?

The solution will require 0.392 mol of sugar.

b. n = 0.3915 mol n mM

m nM

=

== ( )( )=

0 3915 342 34

134

. . mol g/mol

g

M M M M= ( ) + ( ) + ( )

= ( ) +12 22 11

12 12 01 22 1 01

of C of H of O


g/mol

( ) + ( )=

11 16 00

342 34

.

.

m = ?

The solution will require 134 g of sugar.

18. a. C = 0.750 mol/L C nV

n CV

=

== ( )( )==

0 750 1 25

0 9375

0 938

. .

.

.

mol/L L

mol

mol

V = 1.25 L

n = ?

The solution requires 0.938 mol of sodium chloride.



b. n = 0.9375 mol n mM

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. Vi = 95.0 mL C V C V

CC V

V

i i f f

fi i

f

mol/L mL mL

mol/L

=

=

= ( )( )

=

2 30 95 080 0

2 73

. ..

.

Ci = 2.30 mol/L

Vf

mL mL

mL

= -=

95 0 15 0

80 0

. .

.

Cf = ?

The concentration increased to 2.73 mol/L.

20. Vi = 150 mL C V C V

CC V

V

i i f f

fi i

f

mol/L mL mL

mol/L

=

=

= ( )( )

=

3 40 150900

0 567

.

.

Ci = 3.40 mol/L

Vf

mL mL

mL

= +=

150 750

900

Cf = ?

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.

anionicsurfactant

watermolecule

H

H O

–



24. a. Vsolute

= 109.0 mL % %

.%

. %

/

mL

350.0 mL

solute

solution

V VV

V( ) = ¥

= ¥

=

100

109 0100

31 14

Vsolution

= 350.0 mL

(% V/V) = ?

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

b. Vsolute

= 106.5 mL % %

.%

. %

/

mL

350.0 mL

solute

solution

V VV

V( ) = ¥

= ¥

=

100

106 5100

30 43

Vsolution

= 350.0 mL

(% V/V) = ?

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

c. Vsolute

= 12.0 mL % %

.%

. %

/

mL

350.0 mL

solute

solution

V VV

V( ) = ¥

= ¥

=

100

12 0100

3 43

Vsolution

= 350.0 mL

(% V/V) = ?

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

d. Vsolute

= 2.0 mL % %

.%

. %

/

mL

350.0 mL

solute

solution

V VV

V( ) = ¥

= ¥

=

100

2 0100

0 57

Vsolution

= 350.0 mL

(% 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.

b. While combing hair, static charge builds up between the comb and the hair. If the comb has one charge and each hair has the opposite charge, the hair would be attracted to the comb. Since they have the same charge, adjacent hairs would repel one another and would tend to separate. The overall effect of a large number of hairs repelling and separating would be that the hairs would tend to stand up and “fly away.”



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.


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 F

2(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 N2(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 Cl

2(g), O

2(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, O

2 (g), gains four electrons;

and the silver ion, Ag+1(aq), gains one electron.

b. The magnesium metal atom, the nickel metal atom, and the calcium metal atom each lose two electrons.



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.

Applying Concepts

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. nH2

mol= 300 n

n

coefficient

coefficient

n

n

n n

Al

H

Al

H

Al

H

Al H

2 2

2

2

=

=

= ¥

= ¥

23

2323

3000

200

mol

mol=

nAl

= ?

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

f. nH2

mol= 300 n

n

coefficient

coefficient

n

n

n n

HCl

H

HCl

H

HCl

H

HCl H

2 2

2

2

=

=

= ¥

=

63

63633

300

600

¥

=

mol

mol

nHCl

= ?

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

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

2(g)

b. Pb(NO3)

2(aq) + Cu(s) Æ Cu(NO

3)

2(aq) + Pb(s)

c. 2 AgNO3(aq) + Cu(s) Æ Cu(NO

3)

2(aq) + 2 Ag(s)

d. 2 AuCl3(aq) + 3 Zn(s) Æ 3 ZnCl

2(aq) + 2 Au(s)



8. a. nC H O4 6 3

mol= 3 56. 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

=

= ¥

= ¥

=

21

2121

3 56

7 12

n n

.

.

nC H O9 8 4

= ?

You could produce 7.12 mol of aspirin.

b. nC H O4 6 3

mol= 3 56. 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

=

= 11

HH O2 mol= 3 56.

nH O2

= ?

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

c. nC H O4 6 3

mol= 3 56. 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

=

= ¥

= ¥

=

21

2121

3 56

7 12

n n

.

.

nC H O7 6 3

= ?

You will need 7.12 mol of salicylic acid.

9. a. nCO2

mol= ¥7 13 105. n

n

coefficient

coefficient

n

n

n

CaO

CO

CaO

CO

CaO

CO

CaO

2 2

2

=

=

= ¥

11

7 13. 1105 mol

nCaO

= ?

Since the mole ratio is 1:1, the factory is producing 7.13 ¥ 105 mol of quicklime every day.



b. nCO2

mol= ¥7 13 105. n

n

coefficient

coefficient

n

n

n

CaCO

CO

CaCO

CO

CaCO

CO

CaC

3

2

3

2

3

2

=

= 11

OO3mol= ¥7 13 105.

nCaCO3

= ?

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, O

2(g), and sulfur dioxide SO

2(g). The pure metal is the

molten copper, Cu(l).

11. a. more

= 1000 kg percentage of copper

kg

kg

Cu

ore

= ¥

= ¥

=

m

m100

18

1000100

1 8

%

%

. %

mCu

= 18 kg

percentage of copper = ?

The ore contains 1.8% copper.

b. more

= 1000 kg parts per million ppm

kg

kgppm

58 p

Ag

ore

= ¥

= ¥

=

m

m10

0 058

100010

6

6.

ppm

mAg

g1 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. nO2

mol= 34 4. n

n

n n

Cu S

O

Cu S O

2

2

2 2

mol

mol

=

= ¥

= ¥

=

23

2323

34 4

22 9

.

.

nCu S2

= ?

Every second, 22.9 mol of copper(I) sulfide is required.



b. nO2

mol= 34 4. n

n

coefficient

coefficient

n

n

n n

Cu O

O

Cu O

O

Cu O

O

Cu O

2

2

2

2

2

2

2

=

=

= ¥

23

23 SSO2

mol

mol

= ¥

=

23

34 4

22 9

.

.

nCu O2

= ?

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 = ¥

=

¨25019 4

4

mL g

mL using density as a conversion factor

.

..

.

.

85 10

4 85 101

4 85

3

3

¥

= ¥ ¥

=

g

kg1000 g

kg

g

V = 250 mL

m = ?

The mass of the fine grains of gold that would fill a small coffee cup is 4.85 kg.



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 half-reaction (electrons are accepted). If the reduction half-reaction appears above the oxidation half-reaction on the activity series, the 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.

The best way to prevent metals from participating in the oxidation reaction that produces corrosion is to put a barrier between the atoms of the pure metal and the non-metal ions that tend to promote corrosion. One kind of barrier is to coat the exterior of the metal with another metal that is less likely to corrode. Chrome is plated over steel on automobile parts, and gold can be plated over nickel and other metals in jewellery. Another approach is to use paint as a barrier. This is done with the hulls of ships, car-body panels, and garden tools.



18.

NO3– K+

NO3– NO3

–Mg2+

Mg2+ Ni2+

2e–

e– e–

2e–

nickel(cathode)

KNO3(aq)

voltmeter

salt bridgemagnesium(anode)

+–

Mg(NO3)2(aq)

electrolytes: Mg2+ and Ni2+

Ni(NO3)2(aq)Ni

19.

NO3– K+

NO3– NO3

–Al3+

Al Fe2+

3e–

e– e–

iron(cathode)

KNO3(aq)

voltmeter

salt bridgealuminium(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.

22. Zinc is a more reactive metal than iron, and it reacts first. The oxide coating produced by the reaction of zinc and oxygen is more stable than pure zinc and is, therefore, harder to break down. The result is that iron remains protected from exposure to air and water by the layer of zinc and by the coating of zinc oxide that eventually covers the exterior of the zinc.



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)


Knowledge

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

Name ofCompound

Chemical Formula

Condensed StructuralDiagram

Alkane, Alkene, or

Alkyne

3,4-diethylhexane C10H22(l) CH2 CHCH3 CH

CH2

CH2 CH3

CH2

CH3

CH3

alkane

3-methyl-4-propyloctane C12H26(l) CH3 CH

CH2

CH2 CH2 CH2 CH3

CH3

CH

CH2

CH2

CH3

alkane

2,2-dimethylbutane C6H14(l) CH3

CH3

CH2 CH3C

CH3

alkane

1-pentyne(and other forms

of pentyne)

C5H8(l)(unbranched)

CH3 CH2 CH2 CHC alkyne

continued on next page



3-ethyl-1-pentene C7H14(l)

CH2 CH CH2 CH3CH

CH2

CH3

alkene

3-hexene C6H12(l) CH2 CHCH3 CH2 CH3CH alkene

3-octyne C8H14(l) CH2 C CCH3 CH2 CH2 CH2 CH3 alkyne

2-ethyl-4-methyl-1-pentene

C8H16(l)

CH2 C CH2

CH2 CH3

CH3

CH CH3

alkene

1-heptene(and other forms

of heptene)

C7H14(l)(unbranched)

CH2 CH CH2 CH2 CH2 CH2 CH3 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, more carbon atoms results in a higher boiling point and melting point.

C

C

C

H

H C

H

H

HH H

C

H

H

H2-methyl-2-butene

double bond after carbon 2

4-carbon chain

methyl group on carbon 2

C C

C

H

C

H

HH C

H

H

HH H

C

H

HH3-methyl-2-pentene

double bond after carbon 2

5-carbon chain

methyl group on carbon 3



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.

Fractions with high boiling points (the largest molecules in the mixture) condense first at the bottom of the column, and fractions with low boiling points (the smallest molecules in the mixture) condense higher in the column. Those fractions that are gaseous at normal temperatures are collected at the top of the column as gases.



Applying Concepts

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 C

2C

1

3 4 5 6

3-methylhexane

c.

4-ethyl-4-methyloctane

C C C

C

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

C

C

CC C1 2 3 4

5

6

7

8

e.

3-methylhexane

C C C

C

C

C

C

6

5

4

3 2 1

f.

4 ethyl-5 methyl-1-octyne

C C C

C

C

C

C

C

C C C

8

7

6

5 4 3 2 1

g.

3,4,5-trimethyl-4-nonene

C C C

C

C

C

C

C C C C C

1

2

3 4 5 6 7 8 9

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

C C C HH

H

H

Next, add the methyl branch to carbon 2.

C C C HH

H

H

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

≠ Start numbering here!



8. a. Draw 2-pentyne.

CC C C

H

H

H

HH HC

H

H

Write the balanced chemical equation.

C5H

8(g) + 7 O

2(g) Æ 5 CO

2(g) + 4 H

2O(g)

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

CH3

CH3

CH2 CH2C

CH3

CH2

CH2 CH2 CH3

CH3

C

CH2

CH3


C14

H30

(l) + 21.5 O2(g) Æ 14 CO

2(g) + 15 H

2O(g)

2 C14

H30

(g) + 43 O2(g) Æ 28 CO

2(g) + 30 H

2O(g)

c. Draw 2-butene.

C C C

H

H

H

HHC

H

HH

Draw the saturated version of this hydrocarbon chain.

H C C

H H

H HC

H

HC

H

HH


C4H

8(g) + H

2(g) Æ C

4H

10(g)



d. Draw 2-methyl-3-hexyne.

CC C C

H

H

H

H

HC

C

H

HH H

H C

H

H


C CH

H H

HC

H

HC

H

HC

H

HC

H

HH

H

CH H


C7H

12(g) + 2 H

2(g) Æ C

7H

16(g)

e. Answers will vary. A sample answer is given.

C8H

18(g) Æ C

2H

4(g) + C

3H

8(g) + C

3H

6(g)

f. Answers will vary. A sample answer is given.

C6H

12(g) Æ C

3H

8(g) + C

3H

4(g)

9. a. Draw ethene, the starter compound.

CH

H

H

HC

Draw the unit that repeats throughout the polymer.

C C

H H

H H

Draw a section of the polymer chain.

C CH H

HCH

CH

HH HCH

HCH

HCH

HCH

H



b. Draw propene, the starter compound.

H

H

H

H

CCCH

H


C C

H CH3

H H


C C

H

H

C

H

H

CC

H

H H H

C

H

CH3 CH3 CH3

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.

c. Nutritionists recommend that you should strictly limit the amount of trans fatty acids in your diet to the minimum levels possible. Studies indicate that these compounds dramatically increase the risks associated with heart disease. In addition, nutritionists suspect that trans fats may interfere with other processes within the body that may lead to other health problems.


12. a.

step 1:fractionaldistillation

step 3:fractionaldistillation

napthafraction ethene

step 2:cracking

The Production of Ethene

C5 to C10

petroleum C1 to C4

b. C C

H

H

H

H

The chemical formula for ethene is C2H

4(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

Packaging Applications

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)

continued on next page


2

HDPE

high-densitypolyethylene

• 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-densitypolyethylene

• 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.

b. Reducing consumption is the most effective way to help the environment. Since very little plastic can be reused or recycled, it ends up in landfills. If society does not reduce its consumption of plastics, more petroleum must be pumped out of the ground and then refined to make all these products that will eventually get thrown away. Since the supply of petroleum is limited and the space available for landfill is limited, the model of ever-increasing consumer consumption is not sustainable.


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

+

–

OH

H2d–

d+

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.

d. In this diagram, there are fewer particles of the solute in a given volume of solution.


5.

Reaction

Half-Reaction for the First Element

MentionedOxidation or Reduction

Number of Electrons

Gained or Lost

A Cu(s) Æ Cu2+(aq) + 2e- oxidation 2 lost

B 2H+(aq) + 2e- Æ H2(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.

c. A silver ion has lost a valence electron. If it were to gain that electron to become silver metal, it could be reduced.


10.

Name of Compound

Chemical Formula

Condensed StructuralDiagram

Alkane, Alkene,

or Alkyne

Saturated or

Unsaturated

2,2-dimethylpentane

C7H16(l) CH3

CH3

CH2 CH2C

CH3

CH3 alkane saturated

4-propyloctane C11H24(l)

CH2CH3 CH2

CH2

CH CH2 CH2 CH2 CH3

CH2

CH3

alkane saturated

3-octene C8H16(l) CH2CH3 CH CH CH2 CH2 CH2 CH3 alkene unsaturated

3-heptyne(Answers may

vary.)C7H12(l) CH2CH3 C C CH2 CH2 CH3 alkyne unsaturated

2-ethyl-1-hexene C8H16(l)

CH3 C CH2

CH2

CH3

CH2 CH2 CH3

alkene unsaturated

methane CH4(g) CH4 alkane saturated

2,3-dimethyl- 2-pentene

C7H14(l) CH3 C CH2 CH3C

CH3

CH3

alkene unsaturated

2-methylpropane C4H10(g)CH3 CH3CH

CH3

alkane saturated

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

CH3 C C

CH2

CH3

CH3

CH2

CH2 CH3


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

m = 25.4 g n mM

=

=

==

25 4169 88

0 149 517 306 3

0 150

..

.

.

g g/mol

mol

mol

M M M M= ( ) + ( ) + ( )= ( ) + ( ) +

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 nV

=

=

=

0 149 517 306 30 150

0 997

. mol.

. mol L

/L

V = ¥

=

150 1

0 150

mL L1000 mL

L.

C = ?

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

b. C = 0.392 mol/L C nV

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

millionppm

ppmppm

kg

solution10

0 02010

4 00

8 0 10

6

6

8

¥

= ¥

= ¥ -

m

..

. kkg

msolution

= 4.00 kg

msolute

= ?

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

solventsolute

= ¥

=

m

m

mm

106

rrts per millionppm

1.00 kg

ppmppm

kg

¥

= ¥

= ¥

10

5810

1 72 10

6

6

4.

msolute

= 1.00 kg

msolvent

= ?

It would take 1.72 ¥ 104 kg (or 17.2 t) of ore to produce 1.00 kg of silver metal.


13. a. nN2

mol= 20 0. n

n

coefficient

coefficient

n

n

n n

NH

N

NH

N

NH

N

NH N

3

2

3

2

3

2

3 2

=

=

= ¥

=

21

21211

20 0

40 0

¥

=

.

.

mol

mol

n

NH3= ?

You could make 40.0 mol of ammonia.

b. nN2

mol= 20 0. n

n

coefficient

coefficient

n

n

n n

H

N

H

N

H

N

H N

2

2

2

2

2

2

2 2

=

=

= ¥

= ¥

31

3131

20..

.

0

60 0

mol

mol=

n

H2= ?

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.

watermolecule

O2d-

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.

b. Conditioned hair tends to reduce the tendency of the hair to stand on end and “fly away” because the tendency of the hairs to have a slight negative charge and repel one another has been reduced by the positive charge added by the cationic surfactants.


18. a. Vsolute

= 7.0 mL % %

. %

. %

/

mL mL

solute

solution

V VV

V( ) = ¥

= ¥

=

100

7 0500

100

1 4

Vsolution

= 500 mL

(% V/V ) = ?

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

b. Vsolute

= 6.0 mL % %

.%

. %

/

mL mL

solute

solution

V VV

V( ) = ¥

= ( )( ) ¥

=

100

6 0500

100

1 2

Vsolution

= 500 mL

(% V/V ) = ?

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

c. Vsolute

= 2.0 mL % %

.%

. %

/

mL mL

solute

solution

V VV

V( ) = ¥

= ( )( ) ¥

=

100

2 0500

100

0 40

Vsolution

= 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 AlCl

3(aq) + 3 Sn(s)

b. Au(NO3)

3(aq) + 3 Ag(s) Æ 3 AgNO

3(aq) + Au(s)

c. 2 HNO3(aq) + Cu(s) Æ Cu(NO

3)

2 (aq) + H

2(g)

d. Zn(s) + H2SO

4(aq) Æ H

2(g)

+ ZnSO

4(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. nNa

= 75.0 mol n

n

coefficient

coefficient

n

n

n n

Cl

Na

Cl

Na

Cl

Na

Cl Na

2 2

2

2

=

=

= ¥

=

12

12122

75 0

37 5

¥

=

.

.

mol

mol

n

Cl2= ?

This reaction will require 37.5 mol of chlorine gas.


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 cationssalt bridgemagnesium

(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.

b. The voltage reading is zero because the electrons do not have to travel through the wire as the copper ions and the zinc metal are already exchanging electrons in the beaker on the right. The mistake made in setting up this cell is that the copper electrode should be placed in the copper solution and the zinc electrode placed in the zinc solution.


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 Mn3+(aq) ion on the product side. This indicates that the manganese has gained an electron. The ammonium ion, NH

4+(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.

b. This device is designed to produce electricity for an external circuit by means of a spontaneous chemical reaction. Since the reaction is spontaneous, the reduction half-reaction should appear above the oxidation half-reaction. Therefore, the reduction half-reaction for manganese(IV) oxide should appear above the half-reaction for zinc.


32. e–

electrode (anode)

electrolyte bathcontaining Ni2+(aq)

powersource

cathode

drill bitFe(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

CH3 C C

CH2

CH3

CH3

CH2 CH3

This is impossible. Thiscarbon has 6 bonds.Carbons can only have amaximum number of 4 bonds.

34. a. C10

H22

(g) Æ C4H

8(g) + C

3H

6(g) + C

3H

8(g)

b. The products of this cracking reaction are butene, C4H

8(g), propene, C

3H

6(g), and propane, C

3H

8(g).

35. a. C4H

10(g) + 6.5 O

2(g) Æ 4 CO

2(g) + 5 H

2O(g)

2 C4H

2(g) + 13 O

2(g) Æ 8 CO

2(g) + 10 H

2O(g)

b. Draw 2,2-dimethylhexane.

CH3 C CH2

CH3

CH3

CH2 CH2 CH3


C8H

18 + 12.5 O

2(g) Æ 8 CO

2(g) + 9 H

2O(g)

2 C8H

18 + 25 O

2(g) Æ 16 CO

2(g) + 18 H

2O(g)


c. Draw 3-hexene.

CC C C

H

H

HH

HC

H

HH

HC

H

HH


CC C C

HHHH

HC

H

HH

HC

H

HH HH


C6H

12(g) + H

2(g) Æ C

6H

14

36. Draw the starter compound.

C CH

CH

HH CH

H

H

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.

c. Ingredients listed in the nutrition facts that are less desirable in terms of good health are the presence of saturated fat and cholesterol. Both of these substances are associated with a higher risk of developing heart disease.


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Science 20 Unit A: Chapter and Unit Review Suggested Answers · PDF filescience 20 unit a: chemical change chapter and unit review suggested answers