appendix d - prairie spirit school...

October 2007

Grade 12 Prototype Examination Chemistry Course Code 8212

Barcode Number Month Day

Date of Birth

Appendix D For more information, see the Table of Specifications.

- i - (Chemistry, Prototype Exam)

(October 2007)

Chemistry Time: Two and One-Half Hours

Chemistry is an open-book examination. Any number of authorized textbooks may be used. Students’ notebooks may be allowed into the examination room. The laboratory manual may be included as part of the notebook. However, examinations/quizzes/prototype examinations are NOT considered to be part of a student’s notebook and, therefore, are NOT allowed into the examination room.

Calculators may be used. Only silent hand-held calculators designed for mathematical computations such as logarithmic, trigonometric, and graphing functions are permissible. Computers, calculators with QWERTY keyboards, calculators capable of symbolic manipulation, and electronic writing pads are not allowed. Calculators that have built-in notes (definitions or explanations in alpha notation) that cannot be cleared are not permitted. All calculators must be cleared of programs.

You are allowed to use a print dictionary. No other forms of a dictionary (i.e., electronic) or translation dictionaries are permitted.

Do not spend too much time on any question. Read the questions carefully. All items are multiple-choice questions which will be machine scored. Record your answers on the Student Examination Form which is provided. Each question has four suggested answers, one of which is better than the others. Select the best answer and record it on the Student Examination Form as shown in the example below:

Example: Answers: 1. In which of the following substances does

sulfur have the highest oxidation number? A. 2H S

B. 2 4H SOC. 2SOD. 2 2 3Na S O

Student Examination Form: 1. A B C D E Use an ordinary HB pencil to mark your answers on the Student Examination Form. If you change your mind about an answer, be sure to erase the first mark completely. There should be only one answer marked for each question. Be sure there are no stray pencil marks on your answer sheet. If you need space for rough work, use the space in the examination booklet beside each question.

Do not fold either the Student Examination Form or the examination booklet. Be sure to complete the blue identification section of the Student Examination Form. Upon completion of the examination, place your Student Examination Form behind the examination booklet and insert both in the same envelope. Be sure to seal the envelope and complete the information requested on the face of the envelope. (Some of the questions on this exam reproduced in whole or in part with the permission of the Minister of Education, Province of Alberta, Canada, 1997.)

- ii - (Chemistry Prototype Exam)

(October 2007)

Chemistry 30 The following tables are provided with this examination.

• Solubility of Common Compounds in Water

• Relative Strengths of Acids in Aqueous Solution at Room Temperature, 25°C

• Standard Electrode Potentials for Half-Reactions

• Periodic Table

• pH Ranges of Common Indicators

• Formula Sheet

Solubility of Common Compounds in Water

Rule Negative Ions Positive Ions Solubility 1 essentially all Li+ , Na+ , K+ , Rb+ , Cs+ , Fr+ soluble 2 essentially all H+ soluble 3 essentially all

4NH + soluble

4 nitrate, 3NO − essentially all soluble

Ag+ low solubility 5 acetate, 3CH COO−

all others soluble Ag+ , 2Pb + , 2

2Hg + , Cu+ , Tl+ low solubility 6 bromide, Br−

chloride, Cl−

iodide, I−

all others soluble

2Ca + , 2Sr + , 2Ba + , 2Ra + , 2Pb + , Ag+ , 22Hg + low solubility 7

sulfate, 24SO −

all others soluble Li+ , Na+ , K+ , Rb+ , Cs+ , Fr+ , H+ , , 4NH +

2Be + , 2Mg + , 2Ca + , 2Sr + , 2Ba + , 2Ra +

soluble 8 sulfide, 2S −

all others low solubility Li+ , Na+ , K+ , Rb+ , Cs+ , Fr+ , H+ , ,

, 4NH +

2Sr + 2Ba + , 2Ra + , Tl+

soluble 9 hydroxide, OH−

all others low solubility Li+ , Na+ , K+ , Rb+ , Cs+ , Fr+ , H+ , 4NH + soluble 10 carbonate, 2

3CO −

phosphate, 34PO −

sulfite, 23SO −

all others low solubility

Substances are considered soluble if they dissolve enough to give ion concentrations above 0.1 moles per litre at room temperature.

(Adapted from Chemistry: Experimental Foundations, by Parry, R. W.; Steiner, L. E.; Tellefsen, R. L.; Dietz, P. M. © 1981 by Prentice-Hall, Inc. Used by permission of Pearson Education, Inc., 1987.)

- iii - (Chemistry Prototype Exam)

(October 2007)

RELATIVE STRENGTHS OF ACIDS IN AQUEOUS SOLUTION AT ROOM TEMPERATURE, 25°C

Acid Reaction Ka

perchloric acid 4 4HClO (aq) H (aq) ClO (aq)+ −→ + very large

hydriodic acid HI(aq) H (aq) I (aq)+ −→ + 3.2 × 109

hydrobromic acid HBr(aq) H (aq) Br (aq)+ −→ + 1.0 × 109

hydrochloric acid HCl(aq) H (aq) Cl (aq)+ −→ + 1.3 × 106

sulfuric acid 2 4 4H SO (aq) H (aq) HSO (aq)+ −→ + 1.0 × 103

nitric acid 3 3HNO (aq) H (aq) NO (aq)+ −→ + 2.4 × 101

oxalic acid HOOCCOOH(aq) H (aq) HOOCCOO (aq)+⇔ + − 5.4 × 10–2

sulfurous acid 2 2(SO H O)+ 2 3 3H SO (aq) H (aq) HSO (aq)+ −⇔ + 1.7 × 10–2

hydrogen sulfate ion 24 4HSO (aq) H (aq) SO (aq)− + −⇔ + 1.3 × 10–2

phosphoric acid 3 4 2 4H PO (aq) H (aq) H PO (aq)+ −⇔ + 7.1 × 10–3

hydrogen telluride 2H Te(aq) H (aq) HTe (aq)+ −⇔ + 2.3 × 10–3

hydrofluoric acid HF(aq) H (aq) F (aq)+ −⇔ + 6.7 × 10–4

nitrous acid 2 2HNO (aq) H (aq) NO (aq)+ −⇔ + 5.1 × 10–4

hydrogen selenide 2H Se(aq) H (aq) HSe (aq)+ −⇔ + 1.7 × 10–4

benzoic acid 6 5 6 5C H COOH(aq) H (aq) C H COO (aq)+⇔ + − 6.6 × 10–5

acetic acid 3CH COOH(aq) H (aq) CH COO (aq)+ −⇔ + 3 1.8 × 10–5

carbonic acid 2 2(CO H O)+ 2 3 3H CO (aq) H (aq) HCO (aq)+ −⇔ + 4.4 × 10–7

hydrogen sulfide 2H S(aq) H (aq) HS (aq)+ −⇔ + 1.0 × 10–7

dihydrogen phosphate ion 22 4 4H PO (aq) H (aq) HPO (aq)− + −⇔ + 6.3 × 10–8

hydrogen sulfite ion 23 3HSO (aq) H (aq) SO (aq)− + −⇔ + 6.2 × 10–8

hypochlorous acid HClO(aq) H (aq) ClO (aq)+ −⇔ + 2.9 × 10–8

ammonium ion 4 3NH (aq) H (aq) NH (aq)+ +⇔ + 5.7 × 10–10

hydrogen carbonate ion 23 3HCO (aq) H (aq) CO (aq)− + −⇔ + 4.7 × 10–11

hydrogen telluride ion 2HTe (aq) H (aq) Te (aq)− + −⇔ + 1.0 × 10–11

hydrogen peroxide 2 2 2H O (aq) H (aq) HO (aq)+ −⇔ + 2.4 × 10–12

monohydrogen phosphate ion 2 34 4HPO (aq) H (aq) PO (aq)− + −⇔ + 4.4 × 10–13

hydrogen sulfide ion 2HS (aq) H (aq) S (aq)− + −⇔ + 1.2 × 10–15

ammonia 3 2NH (aq) H (aq) NH (aq)+ −⇔ + very small

- iv - (Chemistry Prototype Exam)

(October 2007)

.0 mol L⋅

2F (g) 2e 2F− −+ ⇔

Standard Electrode Potentials for Half-Reactions Ionic concentrations of 1 in water at 25 °C. All ions are aqueous.

Half-reaction E° (volts)

+ 2.87 2

4 2MnO 8H 5e Mn 4 H O− + − ++ + ⇔ +3

2 21

O (g) 2H 2e H O2

+ −+ + ⇔

2Br ( ) 2e 2Br− −+ ⇔l

3 2NO 4 H 3e NO(g) 2H O− + −+ + ⇔ +

Ag e Ag(s)+ −+ ⇔

3 2 2NO 2H e NO (g) H O− + −+ + ⇔ +3 2Fe e Fe+ − ++ ⇔

2I (s) 2e 2I− −+ ⇔2Cu 2e Cu(s)+ −+ ⇔

24 2 2SO 4H 2e SO (g) 2H O− + −+ + ⇔ +4 2Sn 2e Sn+ − ++ ⇔

2S(s) 2H 2e H S(g)+ −+ + ⇔

22H 2e H (g)+ −+ ⇔3Fe 3e Fe(s)+ −+ ⇔2Pb 2e Pb(s)+ −+ ⇔2Sn 2e Sn(s)+ −+ ⇔2Ni 2e Ni(s)+ −+ ⇔2Cd 2e Cd(s)+ −+ ⇔

Fe 2e+ −+ ⇔3Cr 3e Cr(s)+ −+ ⇔2Zn 2e Zn(s)+ −+ ⇔2Mn 2e Mn(+ −+ ⇔

3Al 3e Al(s)+ −+ ⇔2Mg Mg(s)

e Na(s)+ −+ ⇔2Ca 2e Ca(s)+ −+ ⇔2Ba 2e Ba(s)+ −+ ⇔

Cs e Cs(s)+ −+ ⇔

K e K(s)+ −+ ⇔

Li e Li(s)+ −+ ⇔

Au 3e Au(s)+ −+ ⇔

2Cl (g) 2e 2Cl− −+ ⇔2 3

2 7 2Cr O 14H 6e 2Cr 7H O− + − ++ + ⇔ +2

2 2MnO (s) 4 H 2e Mn 2H O+ − ++ + ⇔ +

– 1.66

2e+ −+ ⇔ – 2.37

Na – 2.71

– 2.87

– 2.90

– 2.92

– 2.92

– 3.00

+ 0.80

+ 0.78

+ 0.77

+ 0.53

+ 0.34

+ 0.17

+ 0.15

+ 0.14

0.00

– 0.04

– 0.13

– 0.14

– 0.25

– 0.40 2 Fe(s) – 0.44

– 0.74

– 0.76

s) – 1.18

+ 1.50

+ 1.36

+ 1.33

+ 1.28

+ 1.23

+ 1.06

+ 0.96

+ 1.52

- v - (Chemistry Prototype Exam)

(October 2007)

Periodic Table of Elements 1 18

1 H

Hydrogen 1.01 2 13 14 15 16 17

2 He

Helium 4.00

3 Li

Lithium 6.94

4 Be

Beryllium 9.01

11 Atomic Number

Na Atomic Symbol

Sodium Element name

22.99 Average Atomic mass

5 B

Boron 10.81

6 C

Carbon 12.01

7 N

Nitrogen 14.01

8 O

Oxygen 16.00

9 F

Fluorine 19.00

10 Ne Neon 20.18

11 Na

Sodium 22.99

12 Mg

Magnesium 24.31 3 4 5 6 7 8 9 10 11 12

Aluminum

13 Al 26.98

14 Si

Silicon 28.09

15 P

Phosphorus 30.97

16 S

Sulfur 32.07

17 Cl

Chlorine 35.45

18 Ar Argon 39.95

19 K

Potassium 39.10

20 Ca

Calcium 40.08

21 Sc

Scandium 44.96

22 Ti

Titanium 47.87

23 V

Vanadium 50.94

24 Cr

Chromium 52.00

25 Mn

Manganese 54.94

26 Fe Iron 55.85

27 Co Cobalt 58.93

28 Ni

Nickel 58.69

29 Cu

Copper 63.55

30 Zn Zinc 65.41

31 Ga

Gallium 69.72

32 Ge

Germanium 72.64

33 As

Arsenic 74.92

34 Se

Selenium 78.96

35 Br

Bromine 79.90

36 Kr

Krypton 83.80

37 Rb

Rubidium 85.47

38 Sr

Strontium 87.62

39 Y

Yttrium 88.91

40 Zr

Zirconium 91.22

41 Nb

Niobium 92.91

42 Mo

Molybdenum 95.94

43 Tc

Technetium (98)

44 Ru

Ruthenium 101.07

45 Rh

Rhodium 102.91

46 Pd

Palladium 106.42

47 Ag Silver 107.87

48 Cd

Cadmium 112.41

49 In

Indium 114.82

50 Sn Tin

118.71

51 Sb

Antimony 121.76

52 Te

Tellurium 127.60

53 I

Iodine 126.90

54 Xe Xenon 131.29

55 Cs

Cesium 132.91

56 Ba

Barium 137.33

57-70 *

71 Lu

Lutetium 174.97

72 Hf

Hafnium 178.49

73 Ta

Tantalum 180.95

74 W

Tungsten 183.84

75 Re

Rhenium 186.21

76 Os

Osmium 190.23

77 Ir

Iridium 192.22

78 Pt

Platinum 195.08

79 Au Gold

196.97

80 Hg

Mercury 200.59

81 Tl

Thallium 204.38

82 Pb Lead

207.21

83 Bi

Bismuth 208.98

84 Po

Polonium (209)

85 At

Astatine (210)

86 Rn Radon (222)

87 Fr

Francium (223)

88 Ra

Radium (226)

89-102 **

103 Lr

Lawrencium

(262)

104 Rf

Rutherfordium

(261)

105 Db

Dubnium (262)

106 Sg

Seaborgium (266)

107 Bh

Bohrium (264)

108 Hs

Hassium (269)

109 Mt

Meitnerium (268)

110 Ds

Darmstadtium

(271)

111 Rg

Roentgenium (272)

112 Uub

Ununbium (285)

113 Uut

Ununtrium (284)

114 Uuq

Ununquadium

(289)

115 Uup

Ununpentium

(288)

116 Uuh

Ununhexium (292)

117 Uus

Ununseptium

( ? )

118 Uuo

Ununoctium (293)

*§ Lanthanoid Series

57 La

Lanthanum 138.91

58 Ce

Cerium 140.12

59 Pr

Praseodymium

140.91

60 Nd

Neodymium 144.24

61 Pm

Promethium (145)

62 Sm

Samarium 150.36

63 Eu

Europium 151.96

64 Gd

Gadolinium 157.25

65 Tb

Terbium 158.93

66 Dy

Dysprosium 162.50

67 Ho

Holmium 164.93

68 Er

Erbium 167.26

69 Tm

Thulium 168.93

70 Yb

Ytterbium 173.04

**¥ Actinoid Series

89 Ac

Actinium (227)

90 Th

Thorium (232)

91 Pa

Protactinium 231.04

92 U

Uranium 238.03

93 Np

Neptunium (237)

94 Pu

Plutonium (244)

95 Am

Americium (243)

96 Cm Curium

(247)

97 Bk

Berkelium (247)

98 Cf

Californium (251)

99 Es

Einsteinium (252)

100 Fm

Fermium (257)

101 Md

Mendelevium (258)

102 No

Nobelium (259)

- vi - (Chemistry, Prototype Exam)

(October 2007)

pH Ranges of Common Indicators

Indicator

pH range

Colour at low end of

range

Colour at middle of

range

Colour at high end of range

methyl violet 0.0–1.6 yellow green blue

orange IV 1.4–2.8 red orange yellow

methyl yellow 2.9–4.0 red orange yellow

bromophenol blue 3.0–4.6 yellow green blue

methyl orange 3.2–4.4 red orange yellow

bromocresol green 3.8–5.4 yellow green blue

methyl red 4.8–6.0 red orange yellow

chlorophenol red 5.2–6.8 yellow orange red

litmus 5.5–8.0 red purple blue

bromothymol blue 6.0–7.6 yellow green blue

phenol red 6.6–8.0 yellow orange red

phenolphthalein 8.2–10.0 colourless pink red

thymolphthalein 9.4–10.6 colourless light blue blue

alizarin yellow 10.1–12.0 yellow orange red

indigo carmine 11.4–13.0 blue green yellow

(Lide, David R., ed. CRC Handbook of Chemistry and Physics: A Ready-Reference of Chemical and Physical Data. 87th ed. Boca Raton: Taylor & Francis Group, 2006.)

- vii - (Chemistry, Prototype Exam)

(October 2007)

FORMULA SHEET

Solubility:

[ ]

6

9

1

1 1 2 2 1 1 2 2

grams of soluteppm

1 10 grams of solventfor water, 1 mL 1 g

grams of soluteppb

1 10 grams of solvent

mol L or M

amount of solute (moles)Molarity (M)

volume of solution (litres)C V C V or M V M V

manumber of moles

−

⎫= ⎪× ⎪ =⎬⎪=⎪× ⎭

= ⋅

=

= =

=ss m

or nmolar mass molar mass

=

Equilibrium: [ ][ ]

ProductsK

Reactants=

Thermodynamics: H H Hfp fr∆ ° = Σ∆ ° − Σ∆ °

H bond energies of bonds broken bond energies of bonds formed(in reactants) (in products)

∆ = Σ − Σ

Q mc T= ∆ (for water, 1 1c 4.18 J g C− −= ⋅ ⋅ ° )

Acid-Base: a a b b a a b b

3

14 143

M V M V or C V C V

pH log H O or pH log H

H OH 1 10 or H O OH 1 10

pH pOH 14

+ +

+ − − + −

= =

⎡ ⎤ ⎡ ⎤= − = −⎣ ⎦ ⎣ ⎦

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤= × = ×⎣ ⎦ ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

+ =

−

Oxidation-Reduction:

2 2Zn Zn Cu Cu+ + represents an example of an electrochemical cell

Percent Error: accepted value experimental value% error 100

accepted value−

= ×

- 1 - (Chemistry, Prototype Exam)

(October 2007)

GRADE 12 DEPARTMENTAL EXAMINATION

CHEMISTRY 30, PROTOTYPE EXAM VALUE

100 (50 × 2)

Answer the following 50 questions on the computer sheet entitled “Student Examination Form.”

1. When dilute solutions of the following pairs of ions are mixed, the pair

which would result in the formation of a precipitate is

A. 2Ba (aq), OH (aq)+ −

B. 2 24Pb (aq), SO (aq)+ −

C. 2Zn (aq), Cl (aq)+ −

D. 2 2Ca (aq), S (aq)+ −

2. The addition of which of the following compounds would cause magnesium sulfate, to precipitate when added to a saturated solution of

4MgSO ,

4MgSO (aq)?

A. NaCl

B. LiOH

C. 2 4K SO

D. 4 3NH NO

3. A student combined a solution of with a solution of A precipitate formed. Which of the following represents

the net ionic equation for this reaction?

3 2Ba(NO ) (aq)

2 4Na SO (aq).

A. 2

4 22Na (aq) SO (aq) Na SO (s)+ −+ → 4

a (aq) 2NO (aq) 2Na (aq) SO (aq)

BaSO (s) 2Na (aq) 2NO (aq)

+ − + −

+ −

+ + + →

+ +

B. 3 2 2 4 4 3Ba(NO ) (aq) Na SO (aq) BaSO (s) 2NaNO (aq)+ → +

C. B 2 23 4

4 3

D. 2 24 4Ba (aq) SO (aq) BaSO (s)+ −+ →


(October 2007)

4. A student wishes to separate and 2 2Sr (aq), Pb (aq),+ + Ag (aq)+ ions one at a time from a solution, by forming three separate precipitates. According to the solubility table supplied with this examination, which of the following procedures should be used?

A. Add filter; add 3CH COO ,− I ,− filter; add 2

4SO ,− filter

B. Add Cl filter; add ,− 24SO ,− filter; add 2S ,− filter

C. Add I filter; add ,− 24SO ,− filter; add OH ,− filter

D. Add filter; add 2S ,− 34PO ,− filter; add Br ,− filter

5. The volume of distilled water which must be added to 25.0 mL of HCl(aq) in order to reduce the [HCl(aq)] to 10.60 mol L−⋅ 10.15 mol L−⋅ is

A. 0.025 L B. 0.075 L C. 0.13 L D. 0.15 L

6. Chemists make solutions of lower concentration by using a technique called serial dilution. Starting with a stock solution of 5.0 M, a chemist transfers 100.0 mL from the previous solution and dilutes the new solution to a total volume of 250.0 mL. After 3 successive dilutions, the final concentration of the third solution would be

A. 2.0 M B. 1.6 M C. 0.67 M D. 0.32 M

7. A 115 L sample of polluted water contains 22.0 g of ions. The concentration of magnesium ions in ppb is

2Mg (aq)+

A. 51.91 10 ppb×

B. 37.87 10 ppb×

C. 124.3 10 ppb×

D. 26.38 10 ppb−×


(October 2007)

2

8. Liquid water can be decomposed into its elements by passing a current through it. The reaction for this process is

2 22H O( ) 2H (g) O (g)→ +l H° 571.5 kJ∆ =

A student carried out an experiment to determine H∆ ° for the above reaction. If the experimental value for the above reaction was 539.2 kJ, the percent error of the experiment is

A. 2.83% B. 5.65% C. 5.99% D. 6.43%

9. Consider the following equation.

2 2 32Fe(s) 3CO (s) Fe O (s) 3CO(g) + → + H 26.6 kJ∆ ° = +

Which equation is equivalent to the one above?

A. 2 3 2Fe O (s) 3CO(g) 26.6 kJ 2Fe(s) 3CO (g)+ + → +

B. 2 2 34 Fe(s) 6CO (g) 2Fe O (s) 6CO(g) 53.2 kJ+ → + +

C. 2 2 32Fe(s) 3CO (g) Fe O (s) 3CO(g)+ → + H 26.6 kJ∆ = −

D. 2 2 32Fe(s) 3CO (g) 26.6 kJ Fe O (s) 3CO(g)+ + → +


(October 2007)

10. Scientists have proposed that the burning of methane involves the following four-step reaction pathway.

4 2 3

1CH (g) O (g) CH OH( )

2+ → l H 164.2 kJ∆ = −

3 2 2

1CH OH( ) O (g) CH O(g) H O(g)

2+ → +l 2 H 118.7 kJ∆ = −

2 2

1CH O(g) O (g) HCOOH(g)

2+ → H 308.8 kJ∆ = −

2 2 2

1HCOOH(g) O (g) CO (g) H O(g)

2+ → + H 210.6 kJ∆ = −

The enthalpy change for the reaction C 4 2 2 2H 2O (g) CO (g) 2H O(g)+ → +

J

3CH ,− 2

is

A. H 427.5 k∆ = +B. H 190.1 kJ∆ = +C. H 367.4 kJ∆ = −D. H 802.3 kJ∆ = −

11. Ethane, H C can be made from ethene, 3 2H C CH ,= according to the following reaction.

Bond )∆ ⋅ -1H°(kJ mol C C− 348 C H− 412 C C= 612 H H− 436

Using the bond enthalpies given in the table above, an estimated H∆ for the reaction would be

A 124 kJ−B. 412 kJ−

C. 700 kJ− D. 1810 kJ−


(October 2007)

12. A student heated a piece of aluminum in a lab and collected the following data.

specific heat capacity of aluminum 1 10.897 J g C− −⋅ ⋅ ° initial temperature 14.5 ° C final temperature 51.0 °C amount of heat added 3 831 J

Based on this data, the mass of the piece of aluminum is

A. 48.5 g B. 84.0 g C. 117 g D. 151 g

13. Information about four theoretical chemical reactions is given below.

I. 2 33B X 31.1 kJ B X+ + → 2

II. 2 2A 4 B 2 AB 42.5 kJ+ → + III. 2 2X Y 2XY+ → H 40.9 kJ∆ = − IV. 4 2 2C 5G 2C G+ → 5 H 22.6 kJ∆ = +

The endothermic reaction(s) is(are)

A. I and IV only B. II and III only C. II only D. III only


(October 2007)

Use the following information to answer questions 14, 15, 16, and 17.

Saskatchewan motorists are encouraged to carry a safety survival kit in their vehicles. Recommended items include matches, high energy content food, candles, and a thermal blanket. Chemical “hot packs” are a common item also found in the kit. These “hot packs” are known to contain iron and activated carbon which acts as a catalyst. Oxygen in the air is allowed to combine with the iron by simply removing the plastic packaging. The reaction is

2 2 34 Fe(s) 3O (g) 2Fe O (s) H 1652 kJf+ → ∆ ° = −

14. Which of the following would increase the rate of the reaction?

A. Reduce the amount of activated carbon.

B. Use powdered iron.

C. Increase the amount of 2 3Fe O .D. Lower the initial temperature of the pack.

15. What is the value of Hf∆ ° for the reaction 2 2 3

32Fe(s) O (g) Fe O (s)?

2+ →

A. 11652 kJ mol−− ⋅B. 1826 kJ mol−− ⋅C. 1184 kJ mol−− ⋅D. 1236 kJ mol−⋅


(October 2007)

16. Which diagram best represents the data given? A. B. C. D.

17. The activated carbon is not written in the overall reaction because

A. it speeds up the rate of reaction. B. it causes the reaction equation to become unbalanced. C. it causes a different reaction mechanism. D. it would be found in equal amounts before and after the reaction.

______________________________


(October 2007)

Use the following diagram to answer questions 18 and 19.

18. Adding a catalyst to the reaction would have which of the following

affects?

A. Both X and Y would become smaller. B. X would stay the same and Y would become smaller. C. X would become smaller and Y would become larger. D. Both X and Y would become larger.

19. How would increasing the temperature of the reaction affect the shape of the diagram?

A. The curve would shift to the right. B. The activation energy would be increased. C. The potential energy of the products would be decreased. D. The curve would not change.

______________________________


(October 2007)

20. Crushing a solid increases the rate at which it dissolves because

A. more molecules of solute are available to the solvent. B. another substance is created. C. particle size is increased. D. the pressure on the system is reduced.

21. Chlorine from CFC’s used in air conditioners, refrigerators, aerosol propellants, and plastic foam is thought to be instrumental in the breakdown of ozone through the following mechanism:

Step 1: 3 2Cl O ClO O+ → +Step 2: 3 2ClO O Cl 2O+ → +

The reaction intermediate is

A. Cl

B. 2O

C. ClO

D. 3O


(October 2007)

2

Use the following information to answer questions 22, 23, 24, 25, 26, 27, and 28.

Hydrogen sulfide gas, produces the sour, rotten egg odour often found around oil wells and oil battery sites. Typically, is burned off at the well site. If the reaction took place in a closed system, the following equilibrium would occur.

2H S(g),

2H S(g)

2 2 22H S(g) 3O (g) 2H O(g) 2SO (g)+ ⇔ + H 1036 kJ∆ = −

At equilibrium, the following concentrations were found:

2[H S(g)] 0.650 M,=

2 2 2[O (g)] 0.400 M, [H O(g)] 0.225 M, and [SO (g)] 0.110 M.= = = 22. The equilibrium constant expression for this reaction is

A. 2 22 3

2 2

[SO ]Keq

[H S] [O ]=

B. 2 2

2 2

[H O][SO ]Keq

[H S][O ]=

C. 2

2 2

2[SO ]Keq

2[H S] 3[O ]=

+

D. 2 2

2 22 3

2 2

[H O] [SO ]Keq

[H S] [O ]=

23. Referring to the given information, the value of would be Keq

A. 0.022 7 B. 0.047 5 C. 0.064 0 D. 0.083 5


(October 2007)

24. Another equilibrium was established using the same chemical reaction.

Initially, 12[H S] 0.350 mol L−= ⋅ and When

equilibrium is established,

12[O ] 0.400 mol L .−= ⋅

12 2[SO ] [H O] 0.0600 mol L .−= = ⋅

2H S(g) 2O (g) 2H O(g) 2SO (g) [Initial] 0.350 M 0.400 M [Change] [Equilibrium] 0.060 0 M 0.060 0 M

The equilibrium concentration of is 2O (g)

A. 0.060 0 M B. 0.090 0 M C. 0.180 M D. 0.310 M

25. The size of the equilibrium constant is valuable because it

A. indicates how long a system will remain at equilibrium. B. reflects the rate of the reaction. C. shows whether the reactants or products are favoured. D. will determine if energy will be required.

26. When a stress is applied, equilibrium may shift towards reactants or products, or equilibrium may not be affected. A student used LeChátelier’s principle to complete the following chart.

Row Stress Applied Direction of Shift Effect on 2[O ]I add a catalyst no shift no effect II increase pressure right decrease III increase temperature left increase IV decrease volume left increase

In which row of the chart did the student make an error?

A. I B. II C. III D. IV


(October 2007)

27. Which of the following will change the value of the equilibrium constant, , for this reaction? Keq

A. Increase the pressure. D. Increase the temperature. C. Decrease the concentration of steam. B. Add more hydrogen sulfide gas.

28. The graph represents the reaction once it has reached equilibrium,

The most appropriate conclusion that can be drawn from the graph is that

A. at equilibrium, products are favoured over reactants. B. a catalyst was used to obtain equilibrium. C. at equilibrium, neither reactants nor products are favoured over

each other. D. at equilibrium, reactants are favoured over products.

______________________________


(October 2007)

Use the following information to answer questions 29, 30, and 31.

Hydrochloric acid (HCl) is used in the production of vinyl. Shelley is a laboratory technician working in a factory and is completing a titration experiment to test the HCl solution being used in the production process. She is using an unlimited quantity of distilled water, 1.75 g of NaOH(s) to create a basic solution and a stock solution of 6.00 M HCl from the company’s supplier to make an acidic solution for testing.

29. The technician correctly created a 0.085 0 M NaOH(aq) solution and

would like to check the final result using an acid-base indicator. Using the pH Ranges of Common Indicators table supplied with this examination, which indicator is matched with the colour it would have in the NaOH (aq) solution?

A. Orange IV – the solution is orange B. Thymol blue – the solution is orange C. Indigo carmine – the solution is yellow D. Phenolphthalein – the solution is colourless

30. The technique used by Shelley in performing the titration involves the dilution of the original HCl solution. A weaker solution is required due to the volume of the base needed to neutralize even a small amount of highly concentrated acid. Errors can be caused when diluting a solution. Which of the following technologies would be best suited to determine the concentration of the 6.00 M HCl solution more accurately?

A. temperature probe B. pH probe C. conductivity sensor D. ammeter

31. In the titration experiment, 15.0 mL of 0.0850 M NaOH was neutralized by a 0.150 M HCl solution. The volume of HCl used was

A. 5.75 mL B. 8.50 mL C. 12.4 mL D. 18.6 mL

______________________________


(October 2007)

32. Three unknown solutions labelled I, II and III were tested and these data were obtained:

Solution Conductivity pH

I good 8.5 II good 1.4 III poor 5.4

The acidic solutions are

A. I and II only B. I, II, and III C. I and III only D. II and III only

33. In the reaction represented by the equation

2 3 2 3 3H SO H O H O HSO+ −+ ⇔ + , the Brønsted-Lowry acids are

A. 2 3 3H SO and H O+

B. 2 3H O and H O+

C. 2 3 3H SO and HSO −

D. 2 3H O and HSO −

34. The conjugate base of 2 4H PO − is

A. 3 4H PO

B. 24HPO −

C. 34PO −

D. OH −

35. Which species is NOT amphiprotic?

A. HS (aq)−

B. 2H O( )l

C. 4NH (aq)+

D. 2 3H BO (aq)−


(October 2007)

36. represents a polyprotic acid. As this acid dissociates, the value of each successive step in the dissociation series would be expected to

3H X(aq) Ka

A. increase. B. decrease. C. stay the same. D. approach 21 10−×

37. In acidic solutions, it is possible to determine an value because

[OH (aq)]−

A. acids are solutions containing 2H O( ).lB. acids contain large amounts of hydroxide ion.

C. acids can be neutralized by bases.

D. acids have a higher pH than bases.

38. A weak acid has a of Ka64.0 10 .−× The [H ]+ of a

solution of this acid is

10.010 mol L−⋅

A. 8 14.0 10 mol L− −× ⋅B. 4 12.0 10 mol L− −× ⋅C. 4 14.0 10 mol L− −× ⋅D. 3 12.0 10 mol L− −× ⋅

39. Kelly tested portions of a solution with three indicators to determine the approximate pH. The results are given in the following table.

Indicator Colour bromocresol green blue indigo carmine blue thymolphthalein blue

The approximate pH of the solution is

A. 5.0 B. 9.0 C. 10.8 D. 11.6


(October 2007)

40. The most significant difference between electrolytic and electrochemical cells is that electrochemical cells are

A. endothermic B. catalyzed C. driven D. spontaneous

41. Which of the following changes is reduction?

A. Ca(s) becomes 2+Ca (aq)B. becomes 3

3PO (aq)− 34PO (aq)−

C. becomes 2 2I (s) I (aq)−

D. NaCl(s) becomes N and a (aq)+ Cl (aq)−

42. In an experiment, a student reacted four metals and their corresponding aqueous ions in all possible combinations in order to determine an activity series. The following chart summarizes the data.

2+Fe (aq) 3+Al (aq) +Ag (aq) 2+Pb (aq)

Fe(s) — — reaction reaction Al(s) reaction — reaction reaction Ag(s) — — — — Pb(s) — — reaction —

The list of metals in order of decreasing tendency to lose electrons is

A. Fe(s), Al(s), Pb(s), Ag(s) B. Pb(s), Al(s), Fe(s), Ag(s) C. Ag(s), Fe(s), Pb(s), Al(s) D. Al(s), Fe(s), Pb(s), Ag(s)


(October 2007)

43. Balance the following equation.

2 32 7 3 2___ Cr O ___ H ___ NO ___ Cr ___ NO ___ H O− + + −+ + → + +

When the redox equation is balanced, the coefficients of the equation are

A. 1, 6, 2, 2, 2, 3 B. 1, 14, 2, 2, 3, 3 C. 1, 5, 3, 6, 2, 7 D. 1, 14, 3, 2, 3, 8

44. According to the table of standard electrode potentials, which is the strongest reducing agent?

A. K+

B. KC. Cl−

D. 2Cl

45. Which change of Z species could cause to change to 2X (aq)− X (aq)?+

A. 2Z (aq) Z (aq)− +→B. 3Z (aq) Z(s)− →C. 3Z(s) Z (aq)+→D. 3Z(s) Z (aq)−→

46. Consider the following reactions.

Ag (aq) e Ag(s)+ −+ → E 0.80° = + V

V2I (s) 2e 2I (aq)− −+ → E 0.53° = +

The potential of the cell in which the net ionic reaction is is 22 Ag (aq) 2I (aq) 2 Ag(s) I (s)+ −+ → +

A. + 0.27 V B. + 1.07 V C. + 1.33 V D. + 2.13 V


(October 2007)

47. A metallic object is electroplated with silver using a solution of silver nitrate. Which of the following is correct?

A. The positive electrode increases in mass. B. The concentration of Ag (aq)+ ions in solution decreases. C. The reaction occurring at the negative electrode is

Ag (aq) e Ag(s).+ −+ → D. Reduction occurs at the positive electrode.

Use the following information to answer questions 48, 49, and 50.

The diagram shown represents an electrochemical cell of 2 2Zn(s) Zn (aq) Cu (aq) Cu(s)+ + .

48. The electrochemical cell as drawn will not operate. What structure is

missing?

A. salt bridge B. external wire C. battery D. cathode


(October 2007)

For the next two questions, assume that the missing structure has been added.

49. The number that represents the solution is 3 2Zn(NO ) (aq)

A. 4 B. 3 C. 2 D. 1

50. The number that represents the flow of anions from the solution would be

3 2Cu(NO )

A. 8 B. 7 C. 6 D. 5

- i - (Chemistry, Prototype Exam - Answer Key)

(October 2007)

GRADE 12 DEPARTMENTAL EXAMINATION CHEMISTRY, PROTOTYPE EXAM — Answer Key

1. B 11. A 21. C 31. B 41. C 2. C 12. C 22. D 32. D 42. D 3. D 13. A 23. A 33. A 43. A 4. A 14. B 24. D 34. B 44. B 5. B 15. B 25. C 35. C 45. D 6. D 16. C 26. D 36. B 46. A 7. A 17. D 27. B 37. A 47. C 8. B 18. B 28. A 38. B 48. A 9. D 19. D 29. C 39. C 49. A 10. D 20. A 30. B 40. D 50. C

1. B. is the only combination that forms a solid 4PbSO 2. C. This is the common ion effect. Any compound containing or 2Mg + 2

4SO − will cause to precipitate from the solution. 4MgSO

3. D. Net ionic equations include reacting species only. No spectator ions are

included. 4. A. only Ag+ is insoluble with 3CH COO .− 2Pb + and remain. 2Sr +

only is insoluble with 2Pb + I − . 2Sr + remains. is insoluble with 2Sr + 2

4SO − . 5. B. 1 1 2 2C V C V=

1 12

2

(0.60 mol L )(0.025L) (0.15 mol L )(V )

V 0.100L total volume

0.100L 0.025L 0.075L added volume

− −⋅ = ⋅

=

− = 6. D. 1st dilution 1 1 2 2C V C V=

2

2

(5.0 M)(0.100 L) (C )(0.250 L)

C 2.0 M

=

=

2nd dilution 1 1 2 2C V C V=

2

2

(2.0 M)(0.100 L) (C )(0.250 L)

C 0.80 M

=

=

3rd dilution 1 1 2 2C V C V=

2

2

(0.80 M)(0.100 L) (C )(0.250 L)

C 0.32 M

=

=

- ii - (Chemistry, Prototype Exam - Answer Key)

(October 2007)

7. A. 2

5

22.0 g of Mg 22.0 g115 L of water 1.15 10 g

+

=×

(1 mL = 1g for water)

5 9

5

22.0 g x1.15 10 g 1.0 10 g

x 1.91 10 ppb

=× ×

= ×

8. B. % error A E

100A−

= ×

571.5 539.2

100571.5

5.65%

−= ×

=

9. D. is a positive value so it would be placed on the left side of the original

equation. H∆

10. D. 4 2 3

1CH (g) O (g) CH OH( )

2+ → l H 164.2 kJ∆ = −

3 2 2

12O (g) H O(g)

2+ → +lCH OH( ) CH O(g) H 118.7 kJ∆ = −

2 2

1O (g)

2+ →CH O(g) HCOOH(g) H 308.8 kJ∆ = −

2 2 2

1O (g) CO (g) H O(g)

2+ → +HCOOH(g) H 210.6 kJ∆ = −

4 2 2 2CH 2O (g) CO (g) 2H O(g)+ → + H 802.3 kJ∆ = − 11. A. bond energies of bonds broken – ∆ = ∑ ∑ bond energies of bonds formed

H [4C H C C H H] [6C H C C]H [4(412) 612 436] [6(412) 348]H 2696 2820H 124 kJ

∆ = − + + − − − + −∆ = + + − +∆ = −∆ = −

=

12. C.

1 1

1

Q mc T

3831 J (m0.897 J g C )(51.0 14.5 C)3831 J

m32.74 J g

m 117 g

− −

−

= ∆

= ⋅ ⋅ ° −

=⋅

=

° 13. A. Endothermic reactions have the energy value on the left hand side of the

equation or as a positive H∆ .

- iii - (Chemistry, Prototype Exam - Answer Key)

(October 2007)

°

14. B. A. reducing the catalyst slows down the reaction. B. powdered iron has a larger surface area so will react faster. C. more will not create more heat because it is a product. 2 3Fe O D. a lower initial temperature will slow down the reaction. 15. B. is for 1 mole of , therefore Hf∆ 2 3Fe O

11652 kJH 826 kf 2 moles

J mol−−∆ ° = = − ⋅

16. C. The reaction starts very easily once exposed to air so it must have a low

activation energy. The reaction releases heat so it must be exothermic. 17. D. The catalyst is found in the same form before and after the reaction. 18. B. A catalyst lowers the activation energy of the reaction (Y) but has no effect

on the H∆ (X). 19. D. Only a catalyst will affect the curve. 20. A. Crushing a solid increases the surface area allowing more molecules to

dissolve right away. 21. C. An intermediate is created in one step and used up in a subsequent step

in the mechanism.

22. D. a

b

[Products]Keq

[Reactants]=

23. A. 2 2

2 22 3

2 2

[H O] [SO ]Keq

[H S] [O ]=

2 2

2 3

(0.225) (0.110)Keq

(0.650) (0.400)K 0.0227eq

=

=

24. D. 22H S 23O ⇔ 22H O 22SO (g)

0.350 M 0.400 M 0 0 – 0.060 0 M – 0.090 0 M + 0.060 0 M + 0.060 0 M

0.290 M 0.310 M 0.060 0 M 0.060 0 M

a and b represent the coefficients from the balanced equation

25. C. products favoured K 1eq >

K products = reactants 1

1eq =

K reactants favoured eq <

- iv - (Chemistry, Prototype Exam - Answer Key)

(October 2007)

26. D. Decreasing volume will favour the side of the equation with the least moles of gas. The equilibrium will shift right to the four moles of gas, away from the five moles of gas on the left side. A shift away from will decrease its concentration.

2O (g)

27. B. Only temperature will affect the value. Keq 28. A. The line representing product concentration is higher on the graph,

therefore, products are favoured 29. C. 14

131.0 10[H ] 1.18 10 M

0.0850 MpH 12.9

−+ −×

= = ×

=

- indigo carmine is yellow near pH = 13 30. B. The concentration of HCl is directly related to [H ]+ which can be

measured using a pH probe. 31. B. a a b b

a

a

M V M V

(0.150 M)(V ) (0.0850 M)(15.0 mL)

V 8.50 mL

=

=

=

32. D. Acidic solutions have a pH below 7. The conductivity will be good for

strong acids (pH near 1) and poor for weak acids (pH near 6). 33. A. A Brønsted-Lowry acid loses an H .+ For this reversible reaction,

for the forward reaction and 2 3H SO

3H O+ for the reverse reaction. 34. B. must act as an acid and lose an 2 4H PO (aq)− H+ to become the conjugate

base 24HPO .−

35. C. To be amphiprotic, the substance must be able to gain and lose H .+

cannot gain an H4NH + .+ 36. B. In a dissociation series of a polyprotic acid, each successive acid formed

becomes weaker. The values will decrease. Ka 37. A. All water solutions will have 14[H ][OH ] 1.0 10 .+ − −= ×

- v - (Chemistry, Prototype Exam - Answer Key)

(October 2007)

38. B. 2

26

1

2 8

4 1

[H ]Ka

[acid]

[H ]4.0 10

0.01 mol L

[H ] 4.0 10 mol L

[H ] 2.0 10 mol L

+

+−

−

+ −

+ −

=

× =⋅

= × ⋅

= × ⋅

39. C. bromocresol green – blue pH 5.4≥ indigo carmine – blue pH 11.4≤ thymolphthalein – blue pH 10.6≥ The pH must be between 10.6 and 11.4 40. D. Electrochemical cells are spontaneous, electrolytic cells require a battery. 41. C. Reduction results in a substance decreasing in oxidation number.

decreases from 0 to – 1. 2I

42. D. The element most likely to lose electrons would be listed first and is the

element that will react with all three of the other ions. The order continues by placing the elements in order by the decreasing number of ions they will react with.

1−

−

+ −

+ + ⇔ + +

43. A.

32 7 3 2

6 2 3 5

gains 3 e 2 6 eloses 3 e 2 6 e

Cr O 6H 2NO 2Cr 2NO 3H O− −

− −

+ + −+ +

× =

× = 44. B. The substance which is the strongest reducing agent must be oxidized and

will be the lowest half-reaction in the table. 45. D. is oxidation. The half–reaction causing the change must be

reduction.

2X X− → +

46. A.

2

2

2 Ag (aq) 2e 2 Ag(s)

2I (aq) I (s) 2e2 Ag 2I 2 Ag(s) I (s)

+ −

− −

+ −

+ →

→ ++ → +

E 0.80

E 0.53E 0.27

° = +° = −° = +

V

VV

47. C. In an electrolytic cell, the negative electrode is the cathode which is the

site of the reduction half-reaction. 48. A. A salt bridge between the two solutions is needed to complete the circuit. 49. A. The zinc half-cell is the anode site which supplies electrons – the right

side of the diagram. Label 4 is for the electrolyte solution. 50. C. The copper half-cell is the left side of the diagram. The anions will flow

away from the electrode which is label 6.

appendix d - prairie spirit school...

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