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.
5.
b.
In this diagram, the solute has no charge. So, the water molecules are attracted to one
another instead.
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.
Reaction
Half-Reaction for the First Element
Mentioned
Oxidation or
Reduction
Number of Electrons Gained or
Lost
oxidation
2 lost
H2(g)
reduction
2 gained
Fe2+(aq) + 2e–
oxidation
2 lost
reduction
2 gained
A
Cu(s)
B
2H+(aq) + 2e–
C
Fe(s)
D
Ni2+(aq) + 2e–
Cu2+(aq) + 2e–
Ni(s)
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.
Alkane,
Alkene,
or Alkyne
Saturated
or
Unsaturated
C7H16(l)
alkane
saturated
4-propyloctane
C11H24(l)
alkane
saturated
3-octene
C8H16(l)
alkene
unsaturated
3-heptyne
(Answers may vary.)
C7H12(l)
alkyne
unsaturated
2-ethyl-1-hexene
C8H16(l)
alkene
unsaturated
methane
CH4(g)
alkane
saturated
2,3-dimethyl2-pentene
C7H14(l)
alkene
unsaturated
Name of
Compound
Chemical
Formula
2,2dimethylpentane
Condensed Structural
Diagram
2-methylpropane
C4H10(g)
alkane
11.
12. a. Determine the number of moles of AgNO3(aq).
m = 25.4 g
M = (M of Ag) + (M of N) + 3(M of O)
= (107.87 g/mol) + (14.01 g/mol) + 3(16.00 g/mol)
= 169.88 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=?
The molar concentration of this solution is 0.997 mol/L.
b.
C = 0.392 mol/L
V = 0.050 L
n=?
c.
This solution contains 2.0 × 10–2 mol of solute.
graphic
msolution = 4.00 kg
msolute = ?
The mass of lead dissolved in this water sample is 8.0 × 10–8 kg.
saturated
d.
parts per million = 58
ppm
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.
You could make 40.0 mol of ammonia.
b.
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.
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
b.
18. a.
the tail of the cationic surfactant are carried to the shafts of hair as well.
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.
Vsolute = 7.0 mL
Vsolution = 500 mL
(% V/V ) = ?
b.
The percent by volume concentration of the distearyldiammonium chloride is
1.4%.
Vsolute = 6.0 mL
Vsolution = 500 mL
(% V/V ) = ?
c.
The percent by volume concentration of the cetyl alcohol solution is 1.2%.
Vsolute = 2.0 mL
Vsolution = 500 mL
(% V/V ) = ?
The percent by volume concentration of polysorbate-85 is 0.40%.
19. a.
3 SnCl2(aq) + 2 Al(s)
2 AlCl3(aq) + 3 Sn(s)
b.
Au(NO3)3(aq) + 3 Ag(s)
c.
2 HNO3(aq) + Cu(s)
d.
Zn(s) + H2SO4(aq)
H2(g) + ZnSO4(aq)
Sodium is being oxidized because each sodium atom loses one electron to
become a sodium ion, Na+(aq).
Chlorine is being reduced because each chlorine atom gains one electron to
become a chloride ion, Cl–(aq).
A total of two electrons is transferred in this reaction.
20. a.
b.
c.
d.
3 AgNO3(aq) + Au(s)
Cu(NO3)2 (aq) + H2(g)
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.
26. a.
27.
28.
29.
30.
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.
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.
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.
The device shown in the illustration is called a battery.
a. The zinc electrode is being oxidized, making the zinc electrode the anode. Oxidation
b.
31. a.
b.
always occurs at the anode.
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.
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, 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.
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 halfreaction should appear above the oxidation half-reaction. Therefore, the reduction halfreaction for manganese(IV) oxide should appear above the half-reaction for zinc.
32.
33.
34. a.
b.
35. a.
b.
C10H22(g)
C4H8(g) + C3H6(g) + C3H8(g)
The products of this cracking reaction are butene, C4H8(g), propene, C3H6(g), and
propane, C3H8(g).
C4H10(g) + 6.5 O2(g)
4 CO2(g) + 5 H2O(g)
2 C4H2(g) + 13 O2(g)
8 CO2(g) + 10 H2O(g)
Draw 2,2-dimethylhexane.
Write the balanced chemical equation.
C8H18 + 12.5 O2(g)
8 CO2(g) + 9 H2O(g)
c.
2 C8H18 + 25 O2(g)
Draw 3-hexene.
16 CO2(g) + 18 H2O(g)
Draw the saturated version of this hydrocarbon chain.
Write the balanced chemical equation.
C6H12(g) + H2(g)
C6H14
Draw
the
starter
compound.
36.
Draw the unit that repeats throughout the polymer.
Draw a section of the polymer chain.
37. a.
b.
c.
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.
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.
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.
Science 20 © 2006, Alberta Education