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Grade 11 Chemistry Review
The following questions highlight the main knowledge and skills from grade 11 chemistry. A good understanding
of the concepts covered in grade 11 chemistry is essential for success in grade 12 chemistry and you may need to
do some independent review of some material if you do not have a clear understanding of the concepts.
1. Use arrows to show the periodic trends for the following properties.
Increasing
Atomic Radius
Increasing
Metal Reactivity
Increasing 1st
Ionization Energy
Increasing
Electronegativity
Increasing
Non-Metal Reactivity
Increasing
Metallic Properties
Increasing
Electron Affinity
2. Rank the following elements in order of increasing 1st Ionization Energy. Explain your choice.
N, Ba, Al, Mg
Ba, Mg, Al, N
- Ba is lowest and furthest to the left. The effective nuclear charge is the least and it has the most
energy levels resulting in a weak force of attraction making the valence electrons easier to remove.
- Mg is higher up than Ba therefore fewer energy levels meaning the valence electrons are closer and
held more tightly making them harder to remove. It is further left than Al therefore it has a lower
effective nuclear charge making the electrons easier to remove than Al.
- Al is further right than Mg giving it a higher effective nuclear charge making its electrons harder to
remove than Mg. It is lower and further left than N making its electrons easier to remove.
- N is highest and furthest to the right. The effective nuclear charge is the greatest and it has the fewest
energy levels resulting in a strong force of attraction making the valence electrons harder to remove.
3. Rank the following elements in order of decreasing 1st Atomic Radius. Explain your choice.
O, F, S, Br
Br, S, O, F
- Br has the greatest number of energy levels and therefore will be the largest.
- S has the next highest number of energy levels
- O has the same number of energy levels as F but has a lower effective nuclear charge therefore its
electrons are held in as tightly making it larger than F.
- F is highest up so it has the fewest energy levels and is furthest to the right resulting in the highest
effective nuclear charge which means the electrons are held most tightly making it the smallest atom.
4. Why is the ionic radius of a negative ion larger than the atomic radius of the corresponding neutral atom?
Adding more electrons to the valence shell results in a greater repulsion causing everything to spread out
more.
5. Why is the ionic radius of a positive ion smaller than the radius of the corresponding neutral atom?
Removing an electron results in an overall positive charge which increases the attraction for the electrons
pulling them in more closely resulting in a smaller radius.
6. Draw the full electron configuration for the following elements.
A) F 1s22s22p5
2
B) Ti 1s22s22p63s23p64s23d2
C) Sn 1s22s22p63s23p64s23d104p65s24d105p2
7. Distinguish between an ionic bond, polar covalent bond, pure covalent bond and coordinate covalent
bond.
Ionic : electrostatic attraction between a positive and negative ion
Polar Covalent: Unequal sharing of an electron pair between 2 atoms
Pure Covalent: Equal sharing of an electron pair between 2 atoms
Coordinate Covalent: Covalent bond where both electrons in the shared pair come from the same atom
8. Draw Lewis Dot diagrams for the following compounds.
A) KI
B) CH4
C) CO32-
D) XeF6
**bonds should be dots also
9. Determine which of the following molecules contain polar covalent bonds and if the molecules are polar.
A) Br2 non-polar bonds and molecule
C) PCl3 polar bonds, non-polar molecule
B) SrCl2 ionic
D) HF polar bond and molecule
10. How is it possible for a non-polar molecule to have polar covalent bonds?
If a molecule has polar bonds but is symmetrical the molecule will be non-polar because there is an equal
distribution of charges.
11. Distinguish between intramolecular and intermolecular forces. Give examples of each.
Intra: Within a molecule ex. Covalent bonds
Inter: Between molecules ex. London dispersion, dipole-dipole and H bonds
12. Explain why polar molecules have higher boiling and melting points than non polar molecules.
Polar molecules will have both London Dispersion forces and Dipole-Dipole forces where non polar
molecules only have London Dipersion forces. This results in the polar molecules being held more tightly
together taking more energy to disperse the molecules and therefore higher melting and boiling points.
13. Why does chloromethane have a higher solubility in water than methane? Draw a diagram to aid in your
explanation.
Methane is a symmetrical molecule therefore the only intermolecular forces present are London
dispersion. Chloromethane is asymmetrical due to the Cl which means it has London Dispersion and
dipole-dipole forces. Since chloromethane has greater intermolecular forces, it will then have a higher
melting and boiling point.
14. What is the charge on the stable ion formed by the following elements?
A) Na = Na+
C) Ba = Ba2+
B) S = S2-
D) I = I-
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15. Complete the following word equations, identify the type of reaction and write balanced chemical
equations. (BEWARE NOT ALL WILL RESULT IN A REACTION)
A) iron + copper (II) sulphate  copper + iron (II) sulphate
Fe + CuSO4  Cu + FeSO4
Single displacement
B) potassium bromide + iodine  NO REACTION (Br is more reactive than I so I will not displace Br)
C) sodium chlorate  sodium chloride + oxygen
2 NaClO3 2 NaCl + 3 O2
Decomposition
D) ethane (dicarbon hexahydride) + oxygen  carbon dioxide + water
2 C2H6 + 7 O2  4 CO2 + 6 H2O
Complete combustion
E) magnesium oxide + carbon dioxide  magnesium carbonate
MgO + CO2  MgCO3
synthesis
F) calcium hydroxide calcium oxide + water
Ca(OH)2  CaO + H2O
decomposition
G) sulphur trioxide + water sulfuric acid
SO3 + H2O  H2SO4
synthesis
H) lithium hydroxide + phosphoric acid lithium phosphate + water
3 LiOH + H3PO4  Li3PO4 + 3 H2O
neutralization
16. Determine the molar mass for the following compounds.
A) Mg(SCN)2
B) SrCl2∙ 4H2O
140.49 g/mol
230.52 g/mol
C) CsHS2O2
230.06 g/mol
17. Determine the number of molecules in 45.2 g of calcium carbonate.
MM = 100 g/mol
# of molecules = nNA
m = 45.2 g
= (0.452 mol)(6.02 x 1023)
n = m/MM
= 2.72 x 1023
= (45.2 g)/100 g/mol
Therefore there are 2.72 x 1023 molecules.
= 0.452 mol
18. Determine the number of moles of HCl in 36 mL of a 0.15 M solution.
C= 0.15 M
V= 0.036 L
n = cV n = (0.15M)(0.036 L)
n = 0.0054 mol
19. Determine the percent composition of the following compounds.
A) SO3
B) CuSO3
MM = 80.07 g/mol
MM = 139.00 g/mol
% S= 32.07 g/mol/80.07 g/mol x 100%
% Cu= 63.55 g/mol/139.00 g/mol x 100%
= 40.05 %
= 45.72 %
% O= 48.00 g/mol/80.07 g/mol x 100%
% S= 32.07 g/mol/139.00 g/mol x 100%
= 59.95 %
= 23.07 %
% O= 48.00 g/mol/139.00 g/mol x 100%
= 34.53 %
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20. A sample of copper (II) sulphate pentahydrate weighs 6.8 g. What mass of the sample is due to the
water?
CuSO4∙5 H2O
MM= 249.62 g/mol
mwater = % water X mass of hydrate
= 90 g/mol /249.62 g/mol x 6.8 g
= 2.5 g
21. A compound contains 85.6 % C and 14.4 % H.
A) Determine the empirical formula of the compound.
Species %/MM of element
÷ by smallest
Whole # Ratio
C
85.6/12.01 = 7.127
7.127/7.127 = 1
1
H
14.4/1.01 = 14.257
14.257/7.127 = 2
2
Therefore the empirical formula is CH2
B) If the molar mass of the compound is 42.1 g/mol, what is the molecular formula of the compound.
MMEF = 14.03 g/mol
MMMF = 42.1 g/mol
# of EF = MMMF/ MMSF
= 42.1 g/mol/14.03 g/mol
=3
Therefore the molecular formula is C3H6
22. What is the concentration in % w/v and mol/L of the solution prepared by dissolving 13.6 g of sodium
hydroxide in 450 mL of water?
MM = 40 g/mol
c = n/V
n = 13.6 g/40 g/mol
= 0.34/0.45 L
= 0.34 mol
= 0.76 mol/L
% w/v = 13.6 g/450g x 100%
= 3.0 % w/v
23. Write an ionic equation for the dissociation of calcium chloride in water.
CaCl2(s)  Ca2+(aq) + 2 Cl-(aq)
24. Complete the following table by indicating the products of the double displacement reaction and by
circling any insoluble products.
Reactants
NaCl
MgSO4
Na2CO3
CaCl2
PbNO3
Li3PO4
KC2H2O2
BaS
NaNO3
PbCl2
LiCl
Na3PO4
KCl
NaC2H2O2
BaCl2
Na2S
PbSO4
Mg(NO3)2
Li2SO4
Mg3(PO4)2
K2SO4
Mg(C2H2O2)2
BaSO4
MgS
25. Sketch a typical solubility curve for a solid and a gas.
Solid
NaNO3
PbCO3
Li2CO3
Na3PO4
K2CO3
NaC2H2O2
BaCO3
Na2S
Gas
Solubility
Solubility
Temperature
Temperature
PbCl2
Ca(NO3)2
LiCl
Ca3(PO4)2
KNO3
Ca(C2H2O2)2
BaCl2
CaS
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26. Using an example to help you, distinguish between dissociation and ionization.
Dissociation is the splitting of ions when a substance is put into water such as when sodium chloride is
dissolved in water, the sodium and chloride ions are separated so they are free in solution. Ionization is
the formation of ions. For example when hydrogen chloride gas is dissolved in water, it ionizes to form
hydronium and chloride ions.
27. Lead (II) nitrate reacts with potassium iodide to produce a bright yellow solid. Write a balanced chemical
equation, an ionic equation and a net ionic equation for the reaction. MAKE SURE TO INCLUDE STATES.
Balanced Chemical Equation: Pb(NO3)2(aq) + 2 KI(aq)  PbI2(s) + 2 KNO3(aq)
Full Ionic Equqtion:
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq)  PbI2(s) + 2 K+(aq) + 2 NO3-(aq)
Spectator Ions Crossed Out:
Pb2+(aq) + 2 NO3-(aq) + 2 K+(aq) + 2 I-(aq)  PbI2(s) + 2 K+(aq) + 2 NO3-(aq)
Net Ionic Equation:
Pb2+(aq) + 2 I-(aq)  PbI2(s)
28. Distinguish between an Arrhenius acid and a Bronsted-Lowry acid?
An Arrhenius acid is a solution with acidic properties such as a pH below 7 and a Bronsted-Lowry acid is a
species that donates a proton in a given reaction.
29. Give an example of a conjugate acid-base pair.
H2O and OH-, H2CO3 and HCO330. Which of the following could be an amphiprotic substance.
A) H2O
B) SO42C) HCO3-
D) OH-
31. Determine the pH of solutions with the following hydronium concentrations.
A) [H3O+ (aq)] = 2.5 x 10-3
B) [H3O+ (aq)] = 6.76 x 10-5
+
pH = -log [H3O ]
pH = -log [H3O+]
= -log (2.5 x 10-3 mol/L)
= -log (6.76 x 10-5 mol/L)
= 2.60
= 4.170
32. Determine the hydronium concentrations for solutions with the following pHs.
A) pH= 5.60
B) pH= 2.31
+
-pH
[H3O ] = 10
[H3O+] = 10-pH
-6
= 2.5 x 10
= 4.9 x 10-3
C) pH= 12.80
[H3O+] = 10-pH
= 1.6 x 10-13
33. Magnesium metal reacts with hydrochloric acid to produce magnesium chloride and hydrogen gas. How
many hydrogen molecules are produced when 60.0 g of hydrochloric acid are reacted with excess
magnesium.
Mg(s) + HCl(aq)  MgCl2(aq) + H2(g)
nhydrogen = 1.74 mol
Molar Ratio
1
1
1
1
NA = 6.02 x 1023 entities/mol
m (g)
60.0
molecules of H2= 1.74 mol x 6.02 entities/mol
MM (mol/g)
24.3
36.46
= 1.05 x 1024 molecules
n (mol)
1.74
1.74
Therefore 1.05 x 1024 molecules of hydrogen gas are produced.
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34. Determine the mass of silver chloride (AgCl) produced when 75.0 mL of a 4.50 M solution of silver nitrate
(AgNO3) reacts with 15.0 g of sodium chloride (NaCl).
NaCl is the limiting factor
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3 (aq)
NAgCl = 0.257 mol
Molar Ratio
1
1
1
1
MMFe = 143.32 g/mol
C (mol/L)
4.50
mFe = 0.257 mol x 143.32 g/mol
V (L)
0.0750
= 36.8 kg
m (g)
15.0
36.8
MM (g/mol)
58.44
143.32
n (mol)
0.338
0.257
0.257
Therefore 36.8 g of silver chloride can be produced.
35. When solid iron (II) oxide is heated in the presence of carbon monoxide gas, iron metal and carbon
dioxide gas are produced.
A) Write a balanced chemical equation for the reaction.
FeO(s) + CO(g)  Fe(s) + CO2(g)
B) If 74.20 kg of iron (II) oxide and 40.30 kg of carbon monoxide are reacted, what is the mass of iron
that can be produced.
FeO is the limiting factor
FeO(s) + CO(g)  Fe(s) +
CO2(g)
nFe = 1033 mol
Molar Ratio
1
1
1
1
MMFe = 71.85 g/mol
m (g)
74 200
40 300
57 690
mFe = 1033 mol x 55.85 g/mol
MM (mol/g)
71.85
28.01
55.85
= 57.69 kg
n (mol)
1033
1439
1033
Therefore 57.69 kg of iron can be produced.
C) If 48.90 kg of iron metal is actually produced what is the % yield of the reaction?
% yield = Actual/Theoretical x 100 %
= 48.90 kg/57.69 kg x 100%
= 84.95 %
Therefore the percentage yield of the reaction is 84.95 %.
36. 20.0 mL of 0.20 M NaOH reacts with 25.0 mL of 0.25 M HCl (aq). What mass of sodium chloride would be
produced and what would the pH of the solution be after the reaction was complete?
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
NaOH is the limiting factor
nNaCl = 0.0040 mol
Molar Ratio
1
1
1
1
MMNaCl = 58.44 g/mol
C (mol/L)
0.20
0.25
mNaCl = 0.0040 mol x 58.44 g/mol
V (L)
0.0200
0.0250
= 0.23 g
n (mol)
0.0040
0.0063
0.0040
Reacted HCl = 0.0040 mol
Excess HCl = 0.0063 mol- 0.0040 mol
= 0.0023 mol
Therefore the mass of sodium chloride produced is 0.23 g
moles of hydronium = 0.0023 mol
and the pH of the final solution is 1.29.
[H3O+] = 0.0023 mol/0.045 L
= 0.051 mol/L
pH = -log [H3O+]
= -log (0.051 mol/L)
= 1.29