Appendix D - · PDF fileOctober 2006 (Revised 2007) Grade 12 Prototype Examination . Chemistry...

October 2006 (Revised 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 2006)

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 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 may have been reproduced in whole or in part with the permission of the Minister of Education, Province of Alberta, Canada, 1997.)

- ii - (Chemistry Prototype Exam)

(October 2006)

Chemistry 30

Revised 2007

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+ , , K , R , C , Na+ + b+ s+ 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+ , , , C , T 2Pb + 2

2Hg + u+ l+ 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+

2Be

, , K , R , C , , , ,

, Mg , , , , Ra Na+

+

+

2+

b+

2Ca +

s+

2Sr +

Fr+

2Ba +

H+4NH +

2 +

soluble 8 sulfide, 2S −

all others low solubility Li+

2Sr +

, , K , R , C , , , ,

, , , T Na+

2Ba

+

+ Ra

b+

2+

s+

l+

Fr+ H+4NH + soluble 9 hydroxide, OH−

all others low solubility Li+ , , K , R , C , , , Na+ + b+ s+ Fr+ H+

4NH + soluble 10 carbonate, 23CO −

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 2006)

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

Revised 2007

Revised 2007

- iv - (Chemistry Prototype Exam)

(October 2006)

Standard Electrode Potentials for Half-Reactions Ionic concentrations of 1.0 mol in water at 25 °C. All ions are aqueous. L⋅

Half-reaction E° (volts)

2F (g) 2e 2F− −+ ⇔ + 2.87 2

4 2MnO 8H 5e Mn 4H O− ++ + ⇔ +3

− + + 1.52

Au 3e Au(s+ ⇔ )

3

+ − + 1.50

2Cl (g) 2e 2Cl− −+ ⇔ + 1.36 2

2 7 2Cr O 14H 6e 2Cr 7H O− + − ++ + ⇔ + + 1.33 2

2 2MnO (s) 4H 2e Mn 2H O+ − ++ + ⇔ + + 1.28

21

O (g) 2H 2e H O2

+ −+ + ⇔ 2 + 1.23

2Br ( ) 2e 2Br− −+ ⇔− + −

+ 1.06

3 2NO 4H 3e NO(g) 2H O+ + ⇔ + + 0.96

Ag e Ag(s+ ⇔ )

2+

2+

+ − + 0.80

3 2NO 2H e NO (g) H O− + −+ + ⇔ + + 0.78 3 2Fe e Fe+ −+ ⇔ + 0.77

2I (s) 2e 2I− −+ ⇔2+ −

+ 0.53

Cu 2e Cu(s)+ ⇔ + 0.34 2

4 2SO 4H 2e SO (g) 2H O− + −+ + ⇔ + + 0.17 4 2Sn 2e Sn+ −+ ⇔ + 0.15

2S(s) 2H 2e H S(g)+ −+ + ⇔+ −

+ 0.14

22H 2e H (g)+ ⇔3+ −

0.00

Fe 3e Fe(s)+ ⇔ – 0.04 2Pb 2e Pb(s)+ −+ ⇔ – 0.13 2Sn 2e Sn(s)+ −+ ⇔ – 0.14 2Ni 2e Ni(s)+ −+ ⇔ – 0.25 2Cd 2e Cd(s)+ −+ ⇔2+ −

– 0.40

Fe 2e Fe(s)+ ⇔3

– 0.44

Cr 3e Cr(s)+ −+ ⇔2

– 0.74

Zn 2e Zn(s)+ −+ ⇔2

– 0.76

Mn 2e Mn(s)+ −+ ⇔3

– 1.18

Al 3e Al(s+ −+ ⇔2

)

)

– 1.66

Mg 2e Mg(s)+ −+ ⇔ – 2.37

Na e Na(s)+ −+ ⇔ – 2.71 2Ca 2e Ca(s)+ −+ ⇔ – 2.87 2Ba 2e Ba(s)+ −+ ⇔ – 2.90

Cs e Cs(s)+ −+ ⇔ – 2.92

K e K(s+ −+ ⇔ – 2.92

Li e Li(s)+ −+ ⇔ – 3.00

- v - (Chemistry Prototype Exam)

(October 2006)

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

13 Al

Aluminum26.98

14 Si

Silicon 28.09

15 P

Phosphorus30.97

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

Vanadium50.94

24 Cr

Chromium52.00

25 Mn

Manganese54.94

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

Germanium72.64

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

Zirconium91.22

41 Nb

Niobium 92.91

42 Mo

Molybdenum95.94

43 Tc

Technetium(98)

44 Ru

Ruthenium101.07

45 Rh

Rhodium 102.91

46 Pd

Palladium 106.42

47 Ag Silver 107.87

48 Cd

Cadmium112.41

49 In

Indium 114.82

50 Sn Tin

118.71

51 Sb

Antimony121.76

52 Te

Tellurium127.60

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

Tantalum180.95

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)

104Rf

Rutherfordium

(261)

105Db

Dubnium (262)

106Sg

Seaborgium(266)

107Bh

Bohrium (264)

108Hs

Hassium (269)

109Mt

Meitnerium(268)

110 Ds

Darmstadtium

(271)

111Rg

Roentgenium(272)

112Uub

Ununbium(285)

113Uut

Ununtrium(284)

114Uuq

Ununquadium

(289)

115Uup

Ununpentium

(288)

116Uuh

Ununhexium(292)

117Uus

Ununseptium

( ? )

118Uuo

Ununoctium (293)

*§ Lanthanoid Series

57 La

Lanthanum138.91

58 Ce

Cerium 140.12

59 Pr

Praseodymium

140.91

60 Nd

Neodymium144.24

61 Pm

Promethium(145)

62 Sm

Samarium 150.36

63 Eu

Europium151.96

64 Gd

Gadolinium157.25

65 Tb

Terbium 158.93

66 Dy

Dysprosium162.50

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

Protactinium231.04

92 U

Uranium 238.03

93 Np

Neptunium(237)

94 Pu

Plutonium (244)

95 Am

Americium(243)

96 CmCurium

(247)

97 Bk

Berkelium(247)

98 Cf

Californium(251)

99 Es

Einsteinium(252)

100Fm

Fermium (257)

101Md

Mendelevium(258)

102No

Nobelium (259)

- vi - (Chemistry, Prototype Exam)

(October 2006)

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.)

Revised 2007

- vii - (Chemistry, Prototype Exam)

(October 2006)

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 mor n

molar mass molar mass=

Equilibrium: [ ][ ]

ProductsK

Reactants=

Thermodynamics: H H Hfp frΔ ° = ΣΔ ° − ΣΔ °

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

Δ = Σ − Σ

Q mc T= Δ (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 2006)

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. A precipitate forms in a sample of drinking water when (sulfate) ions are added, but not when (chloride) ions are added. A cation which could cause this effect is

24SO −

−Cl

A. )aq(Ca 2+

B. )aq(Ag+

C. )aq(Na+

D. )aq(Mg 2+

2. The concentration of a solution made by dissolving 11.7 g of NaCl(s) in

enough distilled water to make 250 mL of the solution is A. -1Lmol 10.0 ⋅ B. -1Lmol 40.0 ⋅ C. -1Lmol 80.0 ⋅ D. -1Lmol 0.1 ⋅


(October 2006)

Use the solubility graph shown below to answer question 3.

100

200

300

400

500

600

700

800

900

1000

1100

1200

1300

1400

1500

00 10 20 30 40 50 60 70 80 90 100

Solu

bilit

y(g

ram

s o f

so l

ute/

litre

HO

)2

Temperature (°C)

Solubility Curves for Selected Solutes

NaNO3

KNO3

NaCl

HCl

3. Using the solubility graph shown above, g of sodium nitrate will form a saturated solution in 1 L of water at a temperature of

2001

A. 57 C° B. 50 C° C. 4 5 C° D. 40 C°


(October 2006)

4. In water analysis, it is sometimes necessary to separate the ions in solution using selective precipitation.

A solution is known to contain and . Which pair of

ions listed below could be used in the order given to selectively precipitate the two cations from solution?

)aq(Be 2+ )aq(Ra 2+

A. )aq(CO,)aq(S 2

32 −−

B. )aq(SO,)aq(OH 23

−−

C. )aq(SO,)aq(Br 24

−−

D. )aq(COOCH),aq(NO 33−−

5. What volume of 12.0 is needed to make 5.00 L of a

0.500 aqueous solution? )aq(HClLmol 1−⋅

1Lmol −⋅

A. 12.0 L B. 0.208 L C. 12.0 mL D. 0.208 mL 6. When lead nitrate solution, , is added to a solution of

sodium iodide, , a yellow precipitate results. The net ionic equation representing this reaction is

)aq()NO(Pb 23

)aq(NaI

A. )aq(NaNO2)s(PbI)aq(NaI2)aq()NO(Pb 3223 +→+

B. )aq(NO2)aq(Na2)s(PbI)aq(I2)aq(Na2)aq(NO2)aq(Pb 3232 −+−+−+ ++→+++

C. 22Pb (aq) 2I (aq) PbI (s)+ −+ →

D. )aq(NaNO2)aq(NO2)aq(Na2 33 →+ −+

7. In a high school chemistry lab, a student accidentally mixed a solution of

with a solution of . A precipitate formed as a result.

)aq(CaCl 2 )aq(CONa 32

What is the overall ionic equation for the reaction described above? A. )aq(Cl2)aq(Na2)s(CaCO)aq(CO)aq(Na2)aq(Cl2)aq(Ca 3

23

2 −+−+−+ ++→+++

B. )s(CaCO)aq(CO)aq(Ca 32

32 →+ −+

C. )aq(NaCl)s(CaCO)aq(CO)aq(Na)aq(Cl)aq(Ca 32

32 +→+++ −+−+

D. )aq(NaCl2)s(CaCO)aq(CONa)aq(CaCl 3322 +→+


(October 2006)

8. Using bond enthalpies, an estimated for the reaction, °ΔH

would be HH

NNHH

HH2NN −→−+≡

A. }N–NH–N4{}H–H2NN{ +−+≡ B. }N–NH–N{}H–H2NN{ +−+≡ C. }HH2NN{}N–NHN4{ −+≡−+− D. }HH2NN{}N–NHN{ −+≡−+−

9. Hydrogen peroxide decomposes according to the following equation.

)g(O)(OH2)(OH2 2222 +→

Using the °Δ fH values listed below, calculate the heat of reaction for

this process.

Substance °fΔH (kJ) )(OH 22 8.187− )(OH2 8.285−

)g(O2 0 A. kJ0.196+ B. kJ0.98+ C. kJ0.98− D. kJ0.196− 10. Which of the following reactions is endothermic?

A. kJ 76.242H)g(OH)g(O21

)g(H 222 −=Δ→+

B. kJ 6.2049H)g(OH4)g(CO3)g(O5)g(HC 22283 −=Δ+→+

C. kJ 2.1306)s(AlF)g(F23

)s(Al 32 +→+

D. H kJ 88.131H)g(H)g(CO)s(C)g(O 22 +=Δ+→+


(October 2006)

)

11. While calculating a value experimentally in the lab, a student determined the value to be . After consulting a chemistry handbook, the student discovered the accepted value to be

HΔHΔ

.

1molkJ535 −⋅−HΔ

molkJ547 1−⋅− The percent error was A. 2.2% B. 12.0% C. 65.2% D. 97.8% 12. A student wishes to determine the energy content of a marshmallow.

The marshmallow being used is composed of sucrose A 0.83 g sample was burned in a calorimeter and the following data were collected:

12 22 11(C H O ).

mass of marshmallow burned 0.83 g mass of water 75.0 g initial water temperature 22.0 °C final water temperature 24.9 °C The heat of combustion for sucrose is ( CHΔ A. 1374 kJ mol−− ⋅ B. 1647 kJ mol−− ⋅ C. 1909 kJ mol−− ⋅ D. 11218 kJ mol−− ⋅


(October 2006)

13. The molar heat of formation for some sodium halides are given below.

2

1Na(s) I (s) NaI(s) H 287.8 kJ

2+ → Δ = −

2

1Na(s) Br ( ) NaBr(s) H 361.1 kJ

2+ → Δ = −

2

1Na(s) Cl (g) NaCl(s) H 411.2 kJ

2+ → Δ = −

2

1Na(s) F (g) NaF(s) H 570.7 kJ

2+ → Δ = −

A generalization that can be made about these compounds is that A. iodine forms stronger bonds than chlorine does. B. bromine compounds are less stable than chlorine compounds. C. fluorine forms the weakest bond of the halides listed. D. sodium halides absorb heat when formed. 14. Which of the following is an endothermic change? A. Hot coffee cooling to room temperature B. Steam condensing to water on a cold window C. Frost forming on a windshield D. Ice melting


(October 2006)

Use the diagram below to answer questions 15 and 16.

15. The activation energy for the reverse reaction is represented by A. 1 B. 2 C. 3 D. 4 16. Δ for the forward reaction is represented by H A. 1 B. 2 C. 3 D. 4

________________________________________

17. The rate of a chemical reaction depends A. only on the collision frequency. B. only on the collision efficiency. C. on both the collision frequency and the collision efficiency. D. on neither the collision frequency nor the collision efficiency.


(October 2006)

18. The decomposition of “A” proceeds in a 3-step mechanism: Step 1 CXA +→ Step 2 E2DX +→ Step 3 B2DC →+ The net or overall reaction for the decomposition of “A” is A. B2DC →+ B. CXA +→ C. E2B2A +→ D. EDBCB ++→+ 19. Assume that the rate of reaction doubles for every 10 °C rise in

temperature. A student determined that a certain reaction reached completion in 20 minutes at 0 °C. If the same reaction was carried out at 30 °C, it would be expected to reach completion in

Revised 2007

A. 20 minutes. B. 10 minutes. C. 5 minutes. D. 2.5 minutes.


(October 2006)

20. The graph that represents the reaction whose rate could be increased the most by an increase in temperature is

A. B. C. D.


(October 2006)

)

21. Which of the following factors would NOT affect the rate of a reaction? A. Adding a catalyst B. Increasing the amount of reactants C. Decreasing the surface area of a solid D. Increasing the pressure on a liquid 22. Ammonia ( is produced according to the following equation. 3NH (g)

2 2 3N (g) 3H (g) 2NH (g)+ ⇔

When the system has reached equilibrium, the value of the equilibrium

constant, can be changed by K ,eq A. increasing the pressure. B. increasing the temperature. C. adding more nitrogen. D. decreasing the concentration of ammonia. 23. When a state of equilibrium between two opposing chemical reactions is

reached, A. both reactions continue but the net change is zero. B. the reactions stop. C. 50% of the original reactants have been changed into the final products. D. the rates of the opposing reactions are no longer equal. 24. If the system ⇔ 2HCl(g) is at

equilibrium, injecting more into the system will )g(Cl)g(H 22 + HCl)mol kJ/ 5.92H( −=Δ

)g(H2

A. cause the system to cool slightly. B. lower the in the system. ]Cl[ 2

C. lower the [HCl] in the system. D. have no effect on the or [HCl]. ]Cl[ 2

25. Which of the systems below would be an example of an equilibrium? A. Dry ice forming in a beaker of warm water )g(CO2

B. An unsaturated solution of )aq(MgCl2

C. A pot of boiling water at constant temperature D. A saturated solution of sugar dissolved in water in a beaker at 25 °C


(October 2006)

]

26. Anhydrous ammonia, an important fertilizer used in Saskatchewan’s agriculture industry, is created by the following process:

)g(NH2)g(H3)g(N 322 ⇔+

An experiment was carried out with initial [ ] and initial

The [ at equilibrium was found to be 0.24 M. M96.0N 2 =

[ ] .M72.0H2 = 3NH

[ ]2N [ ]2H [ ]3NH

[Initial] 0.96 M 0.72 M - [Change] [Equilibrium] 0.24 M

Based on the above data, the eqK value is

A. 0.16 B. 1.5 C. 2.6 D. 120

27. The graph of concentration versus time over three time intervals for the

equilibrium heat)g(Z)g(Y)g(X ++⇔ is given below.

- X (g)Δ - Y (g)

- Z (g)

Con

cen

trat

ion

ΔΔΔΔΔΔ

ΔΔ

ΔΔΔΔΔΔ

ΔΔΔΔΔ

Δ

Δ

The stress applied at was 1t A. a decrease in temperature. B. a decrease in pressure. C. the addition of a catalyst. D. a decrease in [X(g)].


(October 2006)

]

28. is produced according to the following reaction. 3SO

)g(SO2)g(O)g(SO2 322 ⇔+ The equilibrium expression for this process is

A. [ ] [

[ ]

2

2 22

3

SO OKeq

SO=

B. [ ]

[ ][ ]3

2 2

SOK eq SO O

=

] C.

[ ][ ] [

2

32

2 2

SOKeq

SO O=

D. [ ]

[ ][ ]3

2

2 2

SOK eq

SO O=

29. The following graph shows the change in pH of a solution as a function of

the volume of hydrochloric acid added to 40.0 mL of 1 sodium hydroxide solution.

-15 Lmol 1028. ⋅× −

The pH of the acid used is

approximately A. 2 B. 5 C. 7 D. 12 30. Using the K table supplied with this examination, the strongest acid

listed below is a

A. H )aq(SO32

B. H )aq(CO32

C. HCl )aq( D. HNO )aq(3


(October 2006)

31. Consider the following acid-base systems: →+ −− (aq)HSO(aq)HCO 33 )aq(SO)aq(COH 2

332−+

→+ −− )aq(PO)aq(HSO 343 )aq(SO)aq(HPO 2

32

4−− +

→+ −− )aq(HS)aq(HCO3 )aq(SH)aq(CO 22

3 +−

Identify the substance that, in the systems described above, behaves one

time as an acid and another time as a base. A. )aq(HCO3

−

B. )aq(HS−

C. )aq(HSO3−

D. )aq(HPO 24

−

32. A student added 200 mL of 0.125 potassium hydroxide solution

to 100 mL of 0.100 nitric acid solution. Using the pH Ranges of Common Indicators table supplied with this examination, which indicator is matched with the colour it would have in the new solution?

1Lmol −⋅1Lmol −⋅

A. Orange IV – the solution is orange B. Thymol blue – the solution is orange C. Indigo carmine – the solution is green D. Phenolphthalein – the solution is colourless 33. Which of the following substances would be amphiprotic (amphoteric)? A. 42SOH B. KOH C. NaCl D. OH2

34. When butter becomes rancid, it has a sour taste. One would also expect

it to A. react with Zn(s) to form H2(g). B. neutralize HCl(aq). C. turn litmus blue. D. have pH > 7.


(October 2006)

)

35. A student was asked to prepare 150 mL of 0.20 M HCl solution from a stock solution with a concentration of 6.0 M. What volume of stock solution is required?

A. 5.0 mL B. 30 mL C. 180 mL D. 4 500 mL 36. HClO (aq), is a weak acid, which has a . The [ ] of a

0.85 M solution of this acid is

8109.2aK −×= +H

A. 1Lmol85.0 −⋅ B. 14 Lmol106.1 −− ⋅× C. 17 Lmol100.1 −− ⋅× D. 18 Lmol105.2 −− ⋅× 37. Polyprotic acids dissociate in successive steps by A. releasing one at a time. )aq(OH−

B. gaining one at a time. )aq(OH−

C. releasing one at a time. )aq(H+

D. gaining one at a time. )aq(H+

38. In a sample of orange juice, the . The pH of the

orange juice is

111026.1]OH[ −− ×=

A. 10.9 B. 7.94 C. 3.10 D. 1.65 39. In a titration experiment, 15.0 mL of 0.250 M HCl was neutralized by

22.5 ml of a calcium hydroxide solution of unknown concentration. The concentration of the was

( 2)OH(CaCa( )2OH

A. 0.266 1Lmol −⋅ B. 0.174 1Lmol −⋅ C. 0.133 1Lmol −⋅ D. 0.0833 1Lmol −⋅


(October 2006)

40. Which of the following (as written from left to right) are oxidation reactions? (All ions are aqueous.)

1. −− →+ Cl2e2)g(Cl2

2. −− +→ e2)s(II2 2

3. −++ +→ eFeFe 32

4. )s(Ale3Al 3 →+ −+

5. 2SeS −−− →+ A. All of the above are oxidation reactions. B. Equations (2) and (3) are oxidation reactions. C. Equations (1) and (4) are oxidation reactions. D. Equations (1), (4), and (5) are oxidation reactions. 41. If you wish to store a solution containing Sn(NO3)2 without having a

reaction between the solution and the container, the container could be made of

A. aluminum (Al). B. nickel (Ni). C. copper (Cu). D. zinc (Zn). 42. The equation fragment that would be found in a reduction half-reaction is A. )aq(X)aq(X 32 ++ → B. )g(R)aq(R2 2→−

C. )aq(Z)s(Z 3+→ D. )aq(R)aq(R 2 ++ → 43. In the equation, , the substance that is

oxidized is )s(I)aq(F2)g(F)aq(I2 22 +→+ −−

A. )aq(I−

B. )g(F2

C. )aq(F−

D. )s(I2


(October 2006)

3

44. Which method will slow the corrosion of iron?

A. Connecting iron to a piece of lead B. Connecting iron to the negative terminal of a power source C. Connecting iron to the positive terminal of a power source D. Wrapping tin around iron 45. Consider the following equation for the rusting of iron.

2 24Fe(s) 3O (g) 2Fe O (s)+ →

Which of the following statements about this reaction is NOT true? A. This equation is an example of an oxidation-reduction reaction. B. Iron is the reducing agent in this reaction. C. Oxygen is oxidized in this reaction. D. Iron changes in oxidation number from 0 to +3. 46. To electroplate a penny with nickel, the penny must be connected to the

source of direct current as the A. cathode, and must be placed in a solution containing . )aq(Ni 2+

B. anode, and must be placed in a solution containing . )aq(Ni 2+

C. cathode, and must be placed in a solution containing . )aq(Cu 2+

D. anode, and must be placed in a solution containing . )aq(Cu 2+


(October 2006)

)

)

47. A homebuilder makes the decision to install aluminum roofing material, using iron nails. Which of the statements below explains why this was not a good chemical decision?

A. The iron nails will be oxidized and turn green. B. The aluminum roof will slowly oxidize and disintegrate around the

iron nails. C. The aluminum will be reduced, and the roof will leak. D. The iron nails will slowly disintegrate due to reduction.

Use the diagram below to answer questions 48, 49, and 50.

Mg(s)

1.0 mol LMg(NO ) (aq)

⋅ -1

3 21.0 mol L

Cu(NO ) (aq)⋅ -1

3 2porous cup

Cu(s)

eV

48. The anode is made of A. )s(Cu

B. aq()NO(Cu 23

C. )s(Mg

D. aq()NO(Mg 23

49. The reduction half-reaction for the cell is

A. )s(Cue2)aq(Cu 2 →+ −+

B. −+ +→ e2)aq(Cu)s(Cu 2

C. )s(Mge2)aq(Mg 2 →+ −+

D. −+ +→ e2Mg)s(Mg 2


(October 2006)

50. The theoretical potential difference for the cell is A. volts. 712.− B. volts. 032.− C. +2.03 volts. D. +2.71 volts.

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

(October 2006)

GRADE 12 DEPARTMENTAL EXAMINATION CHEMISTRY, PROTOTYPE EXAM — Answer Key

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

1. A. A. , insoluble with , soluble with )aq(Ca 2+ 2

4SO − −Cl

B. , insoluble with , insoluble with )

)

)

aq(Ag+ 24SO − −Cl

C. , soluble with , soluble with aq(Na+ 24SO − −Cl

D. , soluble with , soluble with aq(Mg 2+ 24SO − −Cl

2. C. mol200.0g5.58

mol1NaClg7.11 =×

[ ] 1Lmol80.0L25.0mol200.0

Lmol −⋅===

3. A. 1200 g crosses the sodium nitrate solubility curve at 57°. 4. B. A. Both cations are soluble with and insoluble with . 2S− 2

3CO −

B. will precipitate will precipitate −OH 2Be .+ 23SO − 2Ra .+

C. Both cations are soluble with −Br . D. Both cations are soluble with both anions.

L208.0V

)L00.5)(L/mol500.0()V)(L/mol0.12(

VCVC

1

1

2211

==

=5. B. 6. C. Net ionic equations include reacting species only. No spectator ions are

included. 7. A. Overall ionic equations show all spectators ions as well as those involved

in the formation of the precipitate.

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

(October 2006)

]

N]-NH-[4N-H]-2HN[Nproducts in formedbonds of energies bond -

reactantsinbrokenbondsofenergies bond

++≡=Σ

ΣH =Δ8. A.

H°Δ9. D.

[ ] [kJ0.196

)8.187(20)8.285(2frHfpH

−=−−+−=

°ΣΔ−°ΣΔ= 10. D. A. exothermic because HΔ is negative B. exothermic because Δ is negative H C. exothermic because energy is located on the product side D. endothermic because Δ is positive H

11. A. % error 100A

EA= (see formula sheet) ×−

%2.2

%100547

535547

=

×−

=

( )( )( )1 1

1C

0.83 gmoles of sucrose = 0.00243 mol

342 g/molQ mc t

75.0g 4.18 J g C 24.9 C 22.0 C

Q 909 J = 0.909 kJ0.909kJ

H 374 kJ mol0.00243mol

− −

−

=

= Δ

= ⋅ ⋅ −

=

Δ = = ⋅

12. A. since heat is released 1H 374 kJ molc

−Δ = − ⋅ 13. B. The formation of NaF releases the most energy. Therefore, the Na

bond is the strongest, and NaF is the most stable halide listed. Likewise, NaCl is more stable than NaBr.

F−

14. D. Only D requires energy to take place; all the others release energy. 15. C. The activation energy is measured from the energy of the products (the

starting point of the reverse reaction) to the highest point on the graph. 16. D. = energy of products - energy of reactants Δ Η 17. C. According to collision theory, the rate of a chemical reaction depends on

BOTH collision frequency and efficiency.

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

(October 2006)

E2B2A

B2DC3StepE2DX2Step

CXA1Step

+→

→++→+→18. C. cancel substances

that are on both reactant and product sides of equations

19. D. Generally, the rate of most reactions doubles for every 10 °C increase in

temperature. 0 °C – 20 m, 10 °C – 10 m, 20 °C – 5 m, 30 °C – 2.5 m 20. D. The dashed line represents activation energy ( . The graph with the

fewest particles with activation energy is D because only molecules to the right have enough energy to react. The rate of this reaction would benefit most from an increase in temperature.

)Ea

21. D. Pressure changes affect only gases at equilibrium. 22. B. Changing concentration, pressure, or adding a catalyst has no effect on

the value for a system at equilibrium. Only a change in temperature

will change

K eqK .eq

23. A. The rate of the forward and reverse reactions are equal, so no visible

change is noticed. 24. B. An increase in [ shifts the equilibrium to the right to use up some of

the extra . This results in a decrease in [ and an increase in [HCl]. ]H2

2H ]Cl2

25. D. A. This is not a closed system because the CO2(g) can escape. B. MgCl2(s) → MgCl2(aq) No reactant remains; therefore, it is not

an equilibrium. C. This is not a closed system because the H2O(g) can escape. D. The solution is the only closed system with reversible reactions

occurring at the same rate. A sealed container is needed only for systems involving gases.

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

(October 2006)

26. B.

[ ]2N [ ]2H [ ]3NH

[Initial] 0.96 M 0.72 M -- [Change] 0.12 M 0.36 M 0.24 M [Equilibrium] 0.84 M 0.36 M 0.24 M

[ ][ ]

[ ]

[ ][ ]

[ ] M36.0M36.0M72.0equilH

M36.0xM24.0

x23

equilNHusedH

M84.0M12.0M96.0equilN

M12.0xM24.0

x21

equilNHusedN

2

3

2

2

3

2

=−=

===

=−=

===

( )

( )( )5.147.1

36.084.0

24.0eqK

3

2

===

27. D. The vertical line for X indicates a sudden decrease in concentration.

X was removed from the system.

- co-efficients in balanced equation are used as exponents in eqK expression

28. C. b

a

]act[Re]Prod[

eqK =

29. B. The last flat part of the curve is the pH of the acid. The top flat portion is

the pH of the base. 30. C. Of the four acids listed, HCl(aq) has the highest value in the table;

therefore, it is the strongest acid.

aK

31. A. The substance must lose an and gain an +H +H . Equation 1 - , gains 323 COHHCO →− +H

Equation 3 - , loses 233 COHCO −− → +H

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

(October 2006)

32. C. This is an example of an excess acid-base titration. moles of acid = CV = 0.100 mol/L × 0.100 L = 0.0100 moles of base = CV = 0.125 mol/L × 0.200 L = 0.0250 excess base = 0.0250 – 0.0100 = 0.0150 moles

[ ] = −OHL300.0

moles0150.0 = 0.0500 M

7.12 3.114 pH

3.10500.0log pOH=−=

=−=

Indigo carmine will be green. 33. D. can lose or gain to form or . OH2

+H

−OH

+OH3

Choice A. can only lose Choices B. and C. can’t gain or lose . .H+ +

H 34. A. Sour taste indicates an acid. Acids react with metals to produce .

The other three choices are all characteristics of bases. )g(H2

35. A. BBAA VMVM =

( )( ) ( )(

mL0.5L0050.V

VL/mol0.6L15.L/mol20.0

B

B

=== )

36. B. (aq)ClO(aq)HHClO(aq) −+ +⇔

]HClO[

]ClO[]H[aK

−+

= ] [ClO] [HLet x −+ ==

14828 Lmol106.1]H[x105.2x85.0

)x)(x(109.2 −−+−− ⋅×==×==×

37. C. (aq)HX(aq)HX(aq)H2

−+ +→ H 2X (aq) H (aq) X (aq)− + −→ + 38. C. ]OH[]H[K w

−+=

( ) 10.31094.7logpH

1094.7][H

]1026.1[][H1000.1

4

4

1114

=×−=×=

×=×

−

−+

−+−

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

(October 2006)

39. D. BBAA VM2VM =

( )( ) ( )( )1

3-

B

B

Lmol0833.0L0450.0mol1075.3

M

L0225.0M2L015.0mol/L250.0

−⋅=×

=

=

40. B. Equations 2 and 3 have the electron(s) on the right side of the

equation. All of the other equations have the electrons on the left side, which is reduction. Oxidation involves the loss of electrons.

)e( −

41. C. The copper half-reaction is the only one higher than Sn in the Standard

Electrode Potentials table. would react with any metal below it in the table.

2Sn+

42. D. A reduction half-reaction involves an element whose oxidation number

decreases (electrons are gained). The other choices show increases in oxidation number.

43. A. The oxidized substance must be a reactant (left side of the equation) and

must increase in oxidation number. shows an oxidation number change from –1 to 0.

2II →−

44. B. Connecting the iron to the negative terminal will cause the process of

corrosion to run in reverse like an electrolytic cell. Tin and lead will cause iron to rust because they are higher in the Electrode Potentials table and will pull electrons from iron, causing oxidation. Connecting iron to the positive terminal draws in electrons causing the Fe(s) to lose electrons and form rust.

45. C. Oxygen is reduced. Its oxidation number changes from 0 to –2. 46. A. The penny needs an electron source (cathode) and a supply of to

turn into the Ni(s) plating. Reduction occurs at the cathode of electrochemical and electrolytic cells. (Remember red-cats.)

)aq(Ni 2+

47. B. The aluminum half-reaction has a smaller E° value, making it the

oxidation half-reaction for the resulting cell. 48. C. Electrons flow in the direction of anode to cathode, thus making Mg the

anode. 49. A. Copper is the cathode and will undergo reduction. 50. D.

basein2OH−

Sign is reversed because the reaction is written as an oxidation.

+ 2.37 V + 0.34 V + 2.71 V

Appendix D - · PDF fileOctober 2006 (Revised 2007) Grade 12 Prototype Examination . Chemistry...

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Transcript of Appendix D - · PDF fileOctober 2006 (Revised 2007) Grade 12 Prototype Examination . Chemistry...