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Foundation Program
Chemistry
TRANSITION RESOURCE BOOK
Sample Exams
UNSW Foundation Studies
UNSW Global Pty Limited
UNSW
Sydney NSW 2052 Australia
Copyright  2019
All rights reserved. Except under the conditions
described in the Copyright Act 1968 of Australia
and subsequent amendments, this publication
may not be reproduced, in part or whole,
without the permission of the copyright owner


NOTES

Some changes have occurred over the years to the Chemistry
syllabus:

‘Ionisation Energy’ and the ‘Structure of the Atom’ were moved
from Term 2 to Term 1, therefore the MID Program samples do not
have these sorts of questions in them. Look in the sample FINAL
papers for typical exam style questions on these topics.

Due to the limited time in Term 1, Electrolysis will be assessed in
the Final Exam (not in the MID Program exam).


Student Number: __________________

Family Name: __________________

Other Names: __________________

Class Group: __________________


Transition Program

Chemistry
Trial Mid–Program Examination A
Time Allowed: 1.5 hour
Reading Time: 5 minutes
TrialMIDProgram (Paper A)


Question Possible Mark
1 6
2 6
3 6
4 6
5 6
6 10
7 10
8 10
TOTAL 60
F O U N D A T I O N S T U D I E S

UNSW Foundation Studies
UNSW Global Pty Limited
UNSW
Sydney NSW 2052 Australia

















Copyright  2019


















All rights reserved. Except under
the conditions described in the
Copyright Act 1968 of Australia
and subsequent amendments,
this publication may not be
reproduced, in part or whole,
without the permission of the
copyright owner.



DIRECTIONS TO CANDIDATES

1. Check that your copy of the examination paper is complete.
This examination contains ten questions:
• Questions 1 to 5 @ 6 marks each
• Questions 6 to 8 @ 10 marks each

The total marks available are 60.

2. Ensure that you have in the room:
• a black or blue pen for writing answers
• an electronic calculator
• no unauthorised material

3. Five (5) minutes reading time is allowed. During reading time do not write at all.

4. Ensure that your name and examination number are clearly printed on the front
page on the blank answer booklet.

5. Answer all questions in the answer booklet.

6. Hand in the answer booklet to the supervisor at the end of the examination.



1. There are seven elements whose names begin with the letter “b”. From this set of 7
elements write down the name of:

(a) the largest atomic radius

(b) an element which exists as diatomic molecules

(c) an element which belongs to the halogen family

(d) an element which is classified as an actinide

(e) an element which has some metallic and some non-metallic properties

(f) the most chemically reactive metallic element.

(The name of an element may be used more than once or not at all).

2. Many household cleaners contain ammonia dissolved in water. In a consumer test, 25.0 mL
of one brand of household cleaner were titrated against 0.822 M hydrochloric acid
solution. The graph below shows the variation in pH during the titration.
(a) Using a net ionic equation briefly explain why the ammonia solution has a high pH
at the beginning of the titration.

(b) Write down the name of one indicator which would be suitable for this titration
and briefly describe how its colour changes during the titration.

(c) Calculate the molar concentration of ammonia in the household cleaner.




Volume hydrochloric acid added (mL)
0 5.0 10.0 15.0 20.0 25.0 30.0 35.0
2.00
4.00
6.00
8.00
10.00
12.00
14.00
pH

3. Five different samples of the insoluble compound marshite provided the following
analyses:

Sample Number 1 2 3 4 5
Mass of marshite analysed (g) 2.40 3.30 5.10 6.33 8.43
Mass of copper obtained (g) 0.80 1.10 1.70 2.11 2.81
Mass of iodine obtained (g) 1.60 2.20 3.40 4.22 5.62

(a) Evaluate the ratio: (moles of copper atoms) ÷ (moles of iodine atoms) formed by
analysis of marshite

(b) Write down the chemical formula and systematic name of marshite

(c) Marshite is precipitated when copper (II) sulfate solution is mixed with
potassium iodide solution. Solid iodine is also produced. Write a net ionic
equation to represent this reaction.




4. (a) Carbon monoxide gas can be made by blowing carbon dioxide gas over hot
carbon. Write a thermochemical equation for this reaction.

(b) Calculate the standard enthalpy of combustion (∆Hc in kJ/mol) of carbon monoxide.

(c) In a steelworks carbon monoxide reacts with iron (III) oxide according to:
3CO (g) + Fe2O3 (s) → 2Fe (s) + 3CO2 (g) ∆H = − 22 kJ
Use these data to calculate the standard enthalpy of formation (∆Hf in kJ/mol) of
iron (III) oxide.




5. A student was given four test tubes marked A, B, C and D. Each contained one of the
liquids: silver nitrate solution, sodium hydroxide solution, sodium carbonate solution and
sucrose (table sugar) solution, though not necessarily in that order. The student added 5 mL
of 1 M hydrochloric acid solution to each test tube and obtained the following results.

Solution Change When Hydrochloric Acid Added
A Bubbles of gas produced
B No visible change but temperature increased
C White precipitate formed
D No change was detected

Identify the contents and write a net ionic equation for the reaction in:

(a) test tube A

(b) test tube B

(c) test tube C


6. The solubility of barium hydroxide (in grams per litre of solution) at various temperatures is
tabulated below.
Temperature (oC) 0 20 60 80
Solubility (g/L) 16.7 38.9 209.4 1014.0

(a) What mass of barium hydroxide would be dissolved in 250 mL of a saturated
solution at 60 °C?

(b) Calculate the molar concentration of the solution in (a)

(c) If the solution in (a) was cooled to 20 oC calculate the mass of crystals which
would be produced.

(d) Calculate the molar concentration of hydroxide ions present in the solution in (c)
at 20 oC after all the crystals were removed.

(e) Calculate the pH of the solution in (c) at 20 oC after the crystals were removed

7. The graph below shows the electrical conductivity of the resulting solution formed
when 80.0 mL of 0.100 M barium chloride solution is added slowly to 400 mL of a
0.0100 M solution of silver sulfate.

V1
Electrical
conductivity
of the mixture
Vo

Volume of barium chloride solution
0


(a) Determine the number of moles of silver sulfate required to make 400 mL of a
0.0100 M solution of silver sulfate.

(b) Write a net ionic equation for any chemical reaction that occurred when the
barium chloride solution is added to the silver sulfate solution.

(c) Briefly explain why the solution conducts electricity at volume V0 but not
significantly at volume V1. assessed in FINAL not in MID- PROGRAM EXAM

(d) What volume of barium chloride solution has been added between volume V0 and
volume V1 ?

(e) Briefly explain why the electrical conductivity increases after volume V1.
assessed in FINAL not in MID- PROGRAM EXAM






8. The density of formaldehyde vapour is 1.226 g/L at 101.3 kPa and 25.0oC.

(a) How many moles of formaldehyde are present in 1.00 L of the vapour under these
conditions?

(b) What is the mass, in grams, of one mole of formaldehyde?

(c) If formaldehyde is a compound of carbon, hydrogen and oxygen what is its
chemical formula?

(d) The toxic amount of formaldehyde in one litre of air is 6.0 µg. How many
molecules are present in 6.0 µg of formaldehyde?

(e) Formaldehyde undergoes the following reaction at 300 oC:
Formaldehyde gas + Ammonia
gas
→ urotropine gas + steam
(6 litres) (4 litres) (1 litre) (6 litres)
If all volumes were measured at the same temperature and pressure write down
the chemical formula of urotropine.




DATA SHEET


Avogadro's constant, NA 6.02 × 1023 mol–1
Universal Gas Constant, R 8.314 J K–1 mol–1
Specific heat of water 4.180 J g–1 K–1
Density of water 1.00 g mL–1



Molar volume of ideal gas:

at 101.3 kPa and 0oC 22.41 L mol–1
at 101.3 kPa and 25oC 24.46 L mol–1



Unit / Symbol Equivalences:

∆Hf = ∆fH
M = mol L–1



Standard enthalpy of formation (∆Hf) :

CO(g) –111 kJ mol–1
CO2(g) –393 kJ mol–1









TrialMIDProgramTransition A (Solutions)

1 (a) barium
(b) bromine
(c) bromine
(d) berkelium
(e) boron
(f) barium

2 (a) Ammonia has a high pH at the beginning of the titration because it is a weak base and
produces hydroxide ion in solution
NH3(aq) + H2O(l) ⇔ NH4+(aq) + OH–(aq)
(b) The titration involves a strong acid (HCl) and a weak base (NH3) and the titration curve
shows that the pH at the equivalence point is approximately pH 5. The best indicator is
methyl red with a pH colour change interval of pH 4.4 to 6.2. The colour change is
from base to acid so will be yellow to orange (to red past end-point).
(c) NH3(aq) + HCl(aq)) ⇔ NH4 Cl(aq)
Moles HCl = 0.822 × 0.0125 = 0.01027(5) mole
Moles NH3 = moles HCl = 0.010275 = [NH3] × 0.02500
[NH3] = 0.411 molar

3. (a) The table shows that the mass of copper to mass of iodine is always 1.00 to 2.00 hence
any the calculation can be based on any pair of data points however the best answer will
be obtained by using the largest values (as this minimizes errors):
Moles Cu = 2.81/63.55 =0.0442(1) mole
Moles I = 5.62/126.9 = 0.0442(8)
Moles Cu/Moles I = 0.04428/0.04421 = 1.00(1)
(b) CuI, Copper(I) iodide
(c) 2Cu2+(aq) + 4I–(aq)  2CuI(s) + I2(s)

4. (a) C(s) + CO2(g)  2CO(g) ∆H = +171 kJ/mol
(b) CO(g) + ½ O2(g)  CO2(g) ∆H = ( –393) – (–111 + ½ × 0) = –282 kJ/mol
(c) 3CO(g) + Fe2O3(s)  3CO2(g) + 2Fe(s)
∆H =(3 × –393 + 2× 0) – (3 × –111 + ∆Hf(Fe2O3 (s)) = –22 kJ/mol
∆Hf(Fe2O3 (s)) = –824 kJ/mol

5. (a) Bubbles of gas form when hydrochloric acid are added to A therefore the only possibility
listed is that A must be a solution (liquid) that contains sodium carbonate:
2H+(aq) + CO32–(aq)  CO2(g) + H2O(l)
(b) The only combination that does not produce a precipitate or a gas yet does react with
hydrochloric acid is NaOH. The ionic equation for a strong acid/strong base
combination is:
H+(aq) + OH–(aq)  H2O(l)
(c) A white precipitate is produced when silver nitrate and hydrochloric acid react together:
Ag+(aq) + Cl–(aq)  AgCl(s)

6. (a) Mass of Ba(OH)2 in 250 mL at 60oC = 0.250 × 209.4 = 52.4 g
(b) Moles Ba(OH)2 (in 1 litre) = 209.4/(137.3 + 2 × 16.00 + 2 × 1.008) = 1.221 moles
[Ba(OH)2] = 1.221 mol/L
(c) Mass of Ba(OH)2 in 250 mL at 20oC = 250 × 38.9 = 9.7 g
Mass of solid crystals on cooling = 52.4 – 9.7 = 42.7 g
(d) Moles Ba(OH)2 (in 1 litre) = 38.9/(137.3 + 2 × 16.00 + 2 × 1.008) = 0.227(1)
moles
[Ba(OH)2] = 0.227 mol/L
Therefore [OH–] = 2 × 0.227 = 0.454 mol/L
(e) pOH = 0.34 ∴ pH = 13.66


7. (a) Moles Ag2SO4 = [Ag2SO4] × volume = 0.0100 × 0400 = 0.00400 moles
(b) Since the solutions are reacting the reactants must be aqueous. The electrical
conductivity drops to zero indicating that no soluble ionic compounds are present at the
equivalence point.
2Ag+(aq) + SO42–(aq) + Ba2+(aq) + 2Cl–(aq)  2AgCl(s) + BaSO4(s)
(c) The solution at Vo is a solution of silver sulfate. Silver sulfate is a strong electrolyte and
is completely dissociated into silver ions and sulfate ions and these ions are free to move
through the solution and carry electric charge (conduct electricity) through solution.
At the equivalence point, V1, all the reactants have reacted and there are no mobile ions
in solution (they have precipitated out of solution) and hence are not free to move
through solution so cannot the mixture at V1, cannot conduct electricity.
(d) Moles Ag2SO4 = 0.00400 moles
Moles BaCl2 = 0.00400 = 0.100 × Volume therefore Volume = 0.0400 L = 40.0 mL
(e) As previously stated, there are no mobile ions in solution at the equivalence point, V1,
and thus the mixture at V1, cannot conduct electricity. Adding more barium chloride
solution to this mixture increases the amount of barium chloride in excess and the
concentration of ions (barium ions and chloride ions) increases in solution (as there is no
longer any silver sulfate to react, it being the limiting reagent after V1). The barium ions
and chloride ions are free to move through the solution and thus the solution can conduct
electricity.

8. (a) Either n = PV/RT = (101.3 × 1.00)/(8.3141 × 298) = 0.0408(8) mole = 0.0409
moles
or n = volume/molar volume = 1.00/24.46 = 0.409 mole
(b) n = mass/molar mass therefore molar mass = 1.226/0.0409 = 29.99 = 30.0
The mass of one mole of formaldehyde is 30.0 gram
(c) CxHyOz Hence 12 x + y + 16 z = 30.0
The largest value for z is 1 hence formaldehyde contains only 1 oxygen. If formaldehyde
contains 1 oxygen the largest value for x is 1 and hence formaldehyde must only contain
1 carbon. This suggests that formaldehyde must contain 2 H atoms.
Therefore formaldehyde is CH2O.
(d) n = mass/molar mass = 6.0 × 10–6/30.0 = 2.0 × 10–7
Number of molecules = moles × Avogadro’s number = 2.0 × 10–7 × 6.02 × 1023
Number of molecules = 1.2 × 1017
(e) 6CH2O + 4 NH3  1 Urotropine + 6 H2O
Hence urotropine = C6H12N4




Student Number: __________________

Family Name: __________________

Other Names: __________________

Class Group: __________________




Transition Program

Chemistry
Trial Mid–Program Examination B
Time Allowed: 1.5 hours
Reading Time: 5 minutes
Question Possible Mark
1 6
2 6
3 6
4 6
5 6
6 10
7 10
8 10
TOTAL 60
F O U N D A T I O N S T U D I E S

UNSW Foundation Studies
UNSW Global Pty Limited
UNSW
Sydney NSW 2052 Australia

















Copyright  2019


















All rights reserved. Except under
the conditions described in the
Copyright Act 1968 of Australia
and subsequent amendments,
this publication may not be
reproduced, in part or whole,
without the permission of the
copyright owner.



DIRECTIONS TO CANDIDATES

1. Check that your copy of the examination paper is complete.
This examination contains ten questions:
• Questions 1 to 5 @ 6 marks each
• Questions 6 to 8 @ 10 marks each

The total marks available are 60.

2. Ensure that you have in the room:
• a black or blue pen for writing answers
• an electronic calculator
• no unauthorised material

5. Five (5) minutes reading time is allowed. During reading time do not write at
all.

6. Ensure that your name and examination number are clearly printed on the
front page on the blank answer booklet.

5. Answer all questions in the answer booklet.

6. Hand in the answer booklet to the supervisor at the end of the
examination.





1. 1.00 g samples of two metals (A and B), each with a valency of +2, were treated with
50.0 mL of an excess of hydrochloric acid; 445 mL of gas at RTP were produced by the
reaction of metal A with hydrochloric acid. There was no reaction when metal B was treated
with hydrochloric acid.

(a) Write an equation for the reaction between metal A and hydrochloric acid.






(b) Calculate the atomic mass, in u or amu, for metal A.






(c) Suggest a possible identity for metal B. Briefly justify your choice of identity for metal B.













2. A student set out to measure the concentration of a certain ammonia solution. This
was done by titrating 25.0 mL of the ammonia solution against 0.106 M sulfuric
acid solution to produce ammonium sulfate solution.

(a) The sulfuric acid was added using the instrument shown in the diagram
below.
Write down the name of the instrument and the reading (in mL) shown.






(b) Name an indicator that would be suitable for detecting the equivalence point of
this titration and briefly describe how its colour would change during the
titration.







(c) Determine the concentration of the ammonia solution used in this titration.


3. During the electrolysis of silver nitrate solution the following reaction takes place:
4AgNO3(aq) + 2 H2O(l) → 4 Ag(s) + 4 HNO3(aq) + O2(g)

(a) Water and solid silver nitrate are non-conductors of electricity. Briefly explain
how a silver nitrate solution is able to conduct an electric current whilst both
water and solid silver nitrate are not able to conduct electricity.
Electrolysis assessed in FINAL not in MID- PROGRAM EXAM







(b) At which electrode (negative or positive) does the silver form? Briefly explain
your choice.
Electrolysis assessed in FINAL not in MID- PROGRAM EXAM









(c) Briefly explain why the pH decreases during the electrolysis of silver nitrate
solution.













4. The following table shows the data obtained when several 100 mL samples of silver nitrate
solution were treated with different volumes of 0.0525 M calcium chloride solution. The
silver chloride that was produced by this reaction was filtered, dried and then weighed.

Volume of calcium chloride
solution added (mL)
Volume of silver
nitrate solution (mL)
Mass of silver
chloride precipitate (g)
20.0 100.0 0.301
40.0 100.0 0.602
60.0 100.0 0.903
80.0 100.0 1.204
100.0 100.0 1.355
120.0 100.0 1.355
140.0 100.0 1.355
160.0 100.0 1.355

(a) Write a nett ionic equation for the reaction that occurs.






(b) Calculate the minimum volume of the calcium chloride solution required to completely
react with 100 mL of the silver nitrate solution.










(c) Calculate the concentration of the silver nitrate solution.


5. Consider the representation of the periodic table shown below:

A D
E
J L M G

R
Q



Using only the letters on shown in the table above:

(a) Write the letter for the metallic element with valency 3.

(b) Write the letter for the element with the largest molar volume.

(c) Write the letters for three elements which you would expect to have similar properties

(d) Write an equation for the reaction (if any) of element “E” with water.

(e) Give the letter for the most reactive metal.

(f) Which of these elements would exist, at 298 K and 100 kPa, as diatomic molecules?



6. A known volume of ammonia gas, (44.8 mL at STP), was passed repeatedly over hot iron
until all the ammonia had decomposed. The iron was unchanged. On cooling the mixture to
the original conditions, the volume of gas was found to have doubled. This gas mixture was
then passed over heated copper(II) oxide (to remove hydrogen) and again cooled to the
original conditions.

(a) Calculate the number of moles of ammonia used in this experiment.








(b) Why did the gas volume double? Explain using an equation.










(c) Write an equation for the reaction with copper(II) oxide and the hydrogen.





(d) What mass of copper(II) oxide reacted in the reaction in (c).






(e) What volume of gas (at STP) remains after the reaction in (c).

7. The salt deposits at Strassfurt, Germany, contain a mixture of potassium chloride and
magnesium chloride. The aqueous solubility curves of these two compounds are shown
below:
(a) Calculate the molar solubility of potassium chloride at 25°C.









(b) 100 mL of a hot aqueous solution contains 50.0 gram of potassium chloride and 50.0
gram of magnesium chloride dissolved in it. Calculate the concentration of all the
chloride ions in this solution.










(c) Calculate the mass of each compound which would crystallise out of the solution if
the solution in (b) was cooled to 20°C.









Potassium chloride
Temperature (ºC)
0 20 40 60 80 100
So
lu
bi
lit
y
(g
/1
00
m
L
so
lu
tio
n)
20
40
60
80 Magnesium
chloride

(d) Draw a neat, labelled sketch of the apparatus suitable for separating the crystals from
the solution in (c).









(e) Potassium chloride dissolves in ether but magnesium chloride does not. Briefly
describe, using a flow chart if necessary, a method for obtaining a pure sample of
potassium chloride from a mixture of the two solids.








8. The following diagram represents the enthalpy relationships between hydrogen, oxygen,
water and hydrogen peroxide. Use the data to answer the following questions.

(a) Write a thermodynamic equation to represent the enthalpy of formation (∆Hf) of
liquid hydrogen peroxide.






(b) Write a thermochemical equation to represent the enthalpy of solution (∆Hsol) of
hydrogen peroxide.







E
N
T
H
A
L
P
kJ/mol
H2(g) + O2(g)
H2O2(g)
H2O2(l)
H2O(g)
H2O2(aq)
)
H2O(l)
H2O(s)
0
-133
-188
-191
-242
-285
-291

(c) What is the maximum amount of heat energy released when 1.50 mole of H2O2(l) is
dissolved in water?






(d) Calculate the enthalpy change (∆H) for the reaction:
2H2O2(aq) → O2(g) + 2H2O(l)






(e) A 0.100 M hydrogen peroxide solution was prepared. The hydrogen peroxide
decomposed according to the equation in part (c). What temperature change would be
observed when 0.500 L of this 0.100 M hydrogen peroxide solution decomposes?
Assume the density of the solution is 1.00 g/mL and the heat capacity of the solution is
4.18 J g–1 K–1.







UNSW Foundation Studies
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DATA SHEET

Avogadro's constant, NA 6.02 × 10
23 mol–1
Universal Gas Constant, R 8.314 J K–1 mol–1
Specific heat of water 4.180 J g–1 K–1
Density of water 1.00 g mL–1



Molar volume of ideal gas:
at 101.3 kPa and 0oC 22.41 L mol–1
at 101.3 kPa and 25oC 24.46 L mol–1



Unit / Symbol Equivalences:

∆Hf = ∆fH
M = mol L–1









TRIAL MIDPROGRAM CHEMISTRY B EXAM: ANSWERS:

1 (a) A + 2HCl  ACl2 + H2
(b) moles H2 = 0.445/24.46 = 0.01882 mole
moles A = moles H2 = 0.01882 mole
Atomic mass A = 1.00/0.01882 = 55.0 amu
(c) Metal B does not react with hydrochloric acid so it is possibly copper

2 (a) Burette: 29.60 mL (the second decimal place must be estimated for full marks)
(b) Methyl orange red: yellow → orange → red. (the intermediate colour must be
indicated for full marks)
(c) 2NH3(aq) + H2SO4(aq) → (NH4)2SO4(aq)
[NH3] × 0.0250 = 2 × 0.106× 0.02960 ∴ [NH3] = 0.251 Molar.


3 (a) Pure water does not contain ions (at a sufficient concentration) to allow electrical
conductivity to occur. Pure (solid) silver nitrate contains ions but these ions
(Ag+(s) and NO3–(s)) are trapped in their lattice positions and not freely moving (they
only vibrate) so are not free to move to carry electrical charge.
Silver nitrate is a strong electrolyte and dissociates completely when dissolved in
water (the ionic lattice is destroyed) and the aquated ions (Ag+(aq) and NO3–(aq))are
free to move between the electrodes and to conduct electricity through the solution
(NOTE: Some answers really do require a high level of detail to obtain full marks - most
students do not perform well on this sort of extended discussion question as they
leave out too much detail).
(b) Ag+(aq) migrate to the cathode which is the negative electrode in electrolysis cells
because opposite electrical charges attract (positive attracted to negative)
(c) pH = –log10[H+]
Nitric acid is produced during this reaction and nitric acid is a strong acid:
HNO3(aq)  H+(aq) + NO3–(aq).
The pH decreases because the concentration of H+ increases in solution and since pH
(NOTE: For full marks to be awarded students need to define pH and to identify why it
decreases by identifying the compound that causes hydrogen ion concentration to
increase.)

4. (a) 2AgNO3(aq) + CaCl2(aq)  2AgCl(s) + Ca(NO3)2(aq)
Ag+(aq) + Cl–(aq)  AgCl(s)
(b) Mole mass AgCl = 107.9 + 35.45 = 143.35
Moles AgCl = 1.355/143.35 = 0.009452 mole
Moles CaCl2 = ½ moles AgNO3 = 0.004726 mole
volume CaCl2 = 0.04726/0.0525 = 89.99 mL = 90.0 mL
(c) Moles AgNO3 = Moles AgCl 0.009452 mole
[AgNO3] = 0.009452/0.100 = 0.0945 molar

5. (a) M
(b) D
(c) A, J, R
(d) 2E + 2H2O  2EOH + H2
(e) Q
(f) G


6. (a) mole mass NH3 = 0.0448/22.41 = 0.00200 mole
(b) 2NH3(g)  N2(g) + 3H2(g)
2 mole of gas decomposes to give 4 moles of gas therefore 2 litres of gas would
decompose to give 4 litre of gas
(Avogadro’s law: Volume gas ∝ moles gas)
(c) CuO(s) + H2(g)  Cu(s) + H2O(l)
(d) moles CuO = Moles of H2.
Moles of H2 = 3/2 × moles NH3 = 3/2 × 0.00200 = 0.00300 mole
Moles CuO = 0.00300
therefore mass CuO = 0.00300 × (63.55 + 16.00) = 0.239 gram
(e) At STP, water is not a gas therefore the only gas remaining (after the hydrogen is
removed) is nitrogen. The moles of nitrogen is half the moles of ammonia hence the
volume of nitrogen is half the volume of ammonia:
Vol nitrogen = ½ × 44.8 = 22.4. mL

7. (a) mole mass KCl = 39.10 + 35.45 = 74.55
Mass KCl (dissolved at 25°C) = 36 gram
Moles KCl = 36/74.55 = 0.48 mole
[KCl] = moles/volume = 0.48/0.100 = 4.8 molar
(b) [KCl] = (50/74.55) ÷ 0.100 = 6.7 molar
[MgCl2] = (50/95.21) ÷ 0.100 = 5.25 = 5.3 molar
[chloride ion] = 6.7 + 2 × 5.2 = 17.2 molar
(c) At 20°C the solubility of KCl is approximately 33 g (from graph) so 17 g of KCl
precipitates from a solution that contained 50 g of KCl.
At 20°C the solubility of MgCl2 is approximately 54 g (from graph) so no MgCl2 will
precipitate from a solution that only contains 50 g.
(d) Draw a filter funnel and a beaker (don’t forget the filter paper)

(e) Add ether – KCl dissolves. Filter the mixture (the liquid is a mixture (solution of
KCl and ether) the solid is pure MgCl2. Allow the liquid to evaporate as this will
leave pure KCl as a white solid.

8. (a) O2(g) + H2(g)  H2O2(l) ∆Hf = –188 kJ/mol
(b) H2O2(l)  H2O2(aq) ∆Hsol = –191 – (–188) = –3 kJ/mol
(c) heat = 3 × 1.50 = 4.5 kJ
(d) 2H2O2(aq)  O2(g)  2H2O(l)
∆H = {0 + 2 × (–285)} – {2 × –191} = – 188 kJ/mol
(e) Moles H2O2 = 0.500 × 0.100 = 0.0500 mole
Heat released = ½ × 0.050 × 188 = 4.70 kJ releases
Heat = m×SH×∆T ∴ ∆T = 4700/(500 × 4.18) = 2.25 oC




Student Number: __________________

Family Name: __________________

Other Names: __________________

Class Group: __________________




Transition Program

Chemistry
Trial Mid–Program Examination C
Time Allowed: 1.5 hours
Reading Time: 5 minutes
TrialMIDProgram (Paper C)
Question Possible Mark
1 6
2 6
3 6
4 6
5 6
6 10
7 10
8 10
TOTAL 60
F O U N D A T I O N S T U D I E S

UNSW Foundation Studies
UNSW Global Pty Limited
UNSW
Sydney NSW 2052 Australia

















Copyright  2019


















All rights reserved. Except under
the conditions described in the
Copyright Act 1968 of Australia
and subsequent amendments,
this publication may not be
reproduced, in part or whole,
without the permission of the
copyright owner.


DIRECTIONS TO CANDIDATES

1. Check that your copy of the examination paper is complete.
This examination contains ten questions:
• Questions 1 to 5 @ 6 marks each
• Questions 6 to 8 @ 10 marks each

The total marks available are 60.

2. Ensure that you have in the room:
• a black or blue pen for writing answers
• an electronic calculator
• no unauthorised material

7. Five (5) minutes reading time is allowed. During reading time do not
write at all.

8. Ensure that your name and examination number are clearly printed on
the front page on the blank answer booklet.

5. Answer all questions in the answer booklet.

6. Hand in the answer booklet to the supervisor at the end of the
examination.










BLANK PAGE

1. Sodium hydrogencarbonate is manufactured by the Solvay Process.
This consists of four steps:
(i) Calcium carbonate is heated until it changes into
calcium oxide and carbon dioxide.
(ii) Calcium oxide reacts with ammonium chloride forming
calcium chloride, ammonia and water.
(iii) Ammonia, water and carbon dioxide combine to
produce ammonium hydrogencarbonate.
(iv) Ammonium hydrogencarbonate and sodium chloride
react together to form sodium hydrogencarbonate and
ammonium chloride.
(a) (i) Write a balanced chemical equation for the reaction described
in step (i)

………………………………………………………………………………………….

(ii) Write a balanced chemical equation for the reaction described
in step (ii)

………………………………………………………………………………………….

(b) Determine the limiting reagent when 50.0kg of calcium oxide and
50.0 kg of ammonium chloride are reacted as described in step (ii).

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

(c) Determine the maximum mass of ammonia that can be produced if
10.0 moles of calcium oxide is reacted with 10.0 moles of ammonium
chloride as described in reaction (ii)

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….




2. Two glass flasks have volumes and temperatures as shown.


Flask X Flask Y

Flask X contains 1.50 moles of carbon dioxide gas. Flask Y contains 22.0
grams of dinitrogen monoxide.
Evaluate the following ratios:

(a) (mass of gas in X) ÷ (mass of gas in Y)

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….



(b) (number of gas molecules in X) ÷ (number of gas molecules in Y)

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….



(c) (pressure of gas in X) ÷ (pressure of gas in Y)

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….



6.0 litres
50 oC
2.0 litres
50 oC

3. When heated, aluminium and carbon react together producing aluminium
carbide. Several samples of this compound were analysed and the results are
shown on the following graph.

















(a) Calculate the greatest mass of aluminium carbide which could be
obtained by heating 15.0g aluminium with 15.0g carbon.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

(b) Using the data on the graph determine the simplest (empirical) formula
for aluminium carbide.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(c) When water is added to aluminium carbide it forms aluminium
hydroxide and methane. Write a balanced chemical equation for this
reaction.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


0 2.0 4.0 6.0 8.0 10.0
Mass of aluminium in aluminium carbide (g)
4.0
3.0
2.0
1.0
0
Mass of
carbon in
aluminium
carbide
(g)

4. The minerals nitre (sodium nitrate) and saltpetre (potassium nitrate) are
sometimes found mixed together.
(a) Sodium nitrate is soluble in alcohol but potassium nitrate is not. In a
series of numbered steps briefly explain how you would obtain a pure
solid sample of sodium nitrate from a mixture of the two compounds.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

(b) The following graph shows their solubilities in water (in grams of solid
present in 100 mL of saturated solution) at various temperatures.
Calculate the molar concentration of a saturated potassium nitrate solution at 20oC.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

(c) 100.0 mL of a hot aqueous solution contains 50.0 grams of sodium
nitrate and 50.0 grams of potassium nitrate dissolved in it. If this
solution was cooled to 20 oC calculate the mass of each compound
which would crystallise out of the solution.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

5. The following diagram shows some energy relationships (in kilojoules per
mole) involving several compounds of barium.




Use these data to calculate:
(a) the enthalpy of formation (∆Hf, in kJ/mole) for barium oxide.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

(b) the enthalpy of solution (∆Hsol, in kJ/mole) for barium hydroxide.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

(c) the enthalpy change (∆H, in kJ) for the reaction of barium hydride
with water forming solid barium hydroxide and hydrogen gas.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

6.

(1) (2)

25.0 mL HCl vs KOH 25.0mL HF vs KOH
mL KOH added mL KOH added


The titration curves were obtained when 0.100 M solutions of hydrochloric acid and
hydrofluoric acid were titrated separately with two different samples of the same
potassium hydroxide solution.

(a) Briefly explain why the initial pH of the HF solution was higher than that of
the HCl solution.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(b) Write a net ionic equation to represent the reaction which occurs in titration (1).

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….




(c) Suggest a suitable indicator for titration (2) and briefly describe its colour
change during the reaction.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(d) Calculate the molar concentration of the potassium hydroxide solution.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(e) Briefly explain why the same volume of potassium hydroxide was needed to
cause neutralisation in both titrations.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….




7. Argon was discovered in 1894. J. Rayleigh and W. Ramsay burned magnesium
in air producing a solid mixture of magnesium oxide and magnesium nitride
(Mg3N2). Almost 1% of the air was left unchanged. They concluded that this
unreactive gas was a new element which they named argon.


(a) Write a balanced equation for the reaction which
produced magnesium nitride.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(b) Briefly describe how they could prove that the unreactive gas was a
new element, and not an impurity or some strange new compound of
nitrogen and oxygen.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(c) Rayleigh and Ramsay found that argon was monatomic with a density
of 1.717 g/L at S.T.P. Using their data calculate the atomic weight of
argon in a.m.u.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….




(d) The modern value for the atomic weight of argon is 40.0 a.m.u. Using
this value calculate the mass of one argon atom in grams.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(e) Briefly explain why argon (atomic weight = 40.0 a.m.u.) is placed
before potassium (atomic weight = 39.1 a.m.u.) in the periodic table.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….




8. Addition of 1.00 M hydrochloric acid solution to magnesium produces a
colourless gas.

(a) Write a net ionic equation to represent this reaction.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(b) Calculate the smallest volume of 1.00 M hydrochloric acid solution
which would react completely with 5.00 grams of solid magnesium.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(c) Would the volume of gas produced at 25 oC and 100 kPa be smaller,
larger or the same if 5.00 grams of aluminium were used instead of
5.00 grams of magnesium?
Briefly justify your answer.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….




(d) A colourless gas is also produced if magnesium carbonate is used
instead of magnesium.
Write a net ionic equation to represent this reaction.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….


(e) Would the volume of gas produced at 25 oC and 100 kPa be smaller,
larger or the same if 5.00 grams of magnesium were reacted with
excess 1.00 M HCl or if 5.00 g of magnesium were reacted with excess
0.500 M HF?
Briefly justify your answer.

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….





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DATA SHEET

Avogadro's constant, NA 6.02 × 1023 mol–1
Universal Gas Constant, R 8.314 J K–1 mol–1
Specific heat of water 4.180 J g–1 K–1

Molar volume of ideal gas:
at 101.3 kPa and 0 oC (S.T.P.) 22.41 L mol–1
at 101.3 kPa and 25 oC (R.T.P.) 24.46 L mol–1









TRIAL MIDPROGRAM CHEMISTRY EXAM PAPER C: ANSWERS:

1. (a) (i) CaCO3 → CaO + CO2
(ii) CaO + 2NH4Cl → CaCl2 + 2NH3 + H2O
(b) Moles CaCO3 = 50,000/(40.1 + 16.0) = 891 mol
Moles NH4Cl = 50,000/(14.0 + 4.0 + 35.5) = 935 mol
Therefore CaO is in excess and ammonium chloride is limiting (as 891
moles of CaO would require 2 × 891 = 1782 mole of NH4Cl for complete
reaction and there is only 935 mole of CaO)
(c) 10.0 moles of CaO would require 20.0 moles NH4Cl for complete reaction
hence NH4Cl is limiting in this case.
Mole NH3 = Moles NH4Cl
mass NH3/17.0 = 10.0
mass NH3 = 170 g (3 significant figures)


2. (a) mass X : mass Y = (1.50 x 44.0) : 22.0 = 3.00
(b) nx : ny = (mol in X) : (mol in Y) = (1.50) : (22.0 / 44.0) = 3.00
(c) P = nRT/V So Px : Py = (1.5 RT/ 6.0) : (0.5 RT/ 2.0) = 1.0
(NOTE: To score full marks in this type of question it is necessary to
establish the relationship between moles and pressure)

3. (a) From the graph 6.0 g Al combines with 2.0 g C
So 15.0 g Al combines with 5.0 g C to form 20.0 g aluminium carbide
(b) (mol C atoms) : (mol Al atoms) = (2.0 / 12.0) : (6.0 / 27.0) = 0.75 = ¾
So empirical formula is Al4C3
(c) Al4C3 + 12H2O → 4Al(OH)3 + 3CH4

4. (a) (i) Add alcohol to the mixture to dissolve the sodium nitrate.
(ii) Filter the mixture from step (i)
(iii) Crystallise solid sodium nitrate from the filtrate of step (ii).
(b) Mass of dissolved potassium nitrate in 100 mL is approximately 31 g hence
1 litre is 310 grams (a range of values around 30 to 35 g accepted).
So molar concentration = 310 / 101.1 = 3.1 M
(c) 19 grams (approximately) of potassium nitrate; zero sodium nitrate.
5. (a) ∆Hf = –554 kJ/mol
(b) ∆Hsol = – 48 kJ/mol
(c) BaH2 (s) + 2H2O (l) → Ba(OH)2 (s) + 2H2 (g)
∆H = (sum of ∆Hf of products) – (sum of ∆Hf of reactants)
= (–945 + 0) – (–179 + 2 × –285) = –196 kJ


6. (a) Only a small percentage of HF molecules dissociate into
H+(aq) and F– (aq) ions so that the H+(aq) concentration in 0.10 M HF
solution is very much less than 0.10 M; therefore its pH is greater than 1.
However essentially all the HCl molecules dissociate so that the H+(aq)
concentration in the HCl solution is 0.10 M and its pH is 1.
(b) HCl is a strong acid and KOH is a strong base hence spectator ions of Cl–
and K+ can be cancelled: The required ionic equation is:
H+ (aq) + OH– (aq) → H2O (l)
(c) Phenolphthalein is suitable; it is colourless at the start of the titration and
changes to faint pink at the equivalence point.
(d) Moles of KOH = moles of acid = (25.0 × 10-3 × 0.10) = 2.5 × 10-3
Concentration of KOH = mol / L = (2.5 × 10-3) / (20.0 × 10-3) = 0.125 M
(e) 1 mole of KOH reacts with 1 mole of acid whether the acid is strong (HCl)
or weak (HF). Because the quantities of the two acids are equal so are the
quantities of KOH required to neutralise them.

7. (a) N2 + 3 Mg → Mg3N2
(b) The emission spectrum of the gas would be analysed and compared with
those of known elements.
(c) Molar mass = 22.41 x 1.717 = 38.53 grams. So their calculated value for
the atomic weight = 38.53 u (which was obviously inaccurate).
(d) 6.02 × 1023 atoms = 40.0 g
Therefore 1 atom = 40.0 ÷ 6.02 × 1023 = 6.64 × 10-23 g
(e) Argon is placed in the same periodic group (noble gases) as the elements it
resembles. Potassium has similar properties to the alkali metals and is
grouped with them.

8. (a) Mg(s) + 2 H+(aq) → H2(g) + Mg2+(aq)
(b) Moles Mg = 5.00/24.3 = 0.206 mol
∴ Moles H+ = 2 × 0.206 = 0.412 mol
∴ Volume 1.00 M HCl = 0.412 L = 412 mL
(c) Al(s) + 3 H+(aq) → 3/2 H2(g) + Al3+(aq)
Moles Al = 5.00/27.0 = 0.185
∴ Moles H+ = 3 × 0.185 = 0.555 mol
The calculations are needed to justify that Al will react with the greater
volume o acid (as more moles of acid will react with 5.00 g of Al than with
5.00 g of Mg)
(d) MgCO3(s) + 2 H+(aq) → CO2(g) + Mg2+(aq) + H2O(l)
(e) Magnesium is in excess in each case hence the same volume of gas
(hydrogen) is produced in each case (as yields are always based on limiting
reagents).




Transition Program
Chemistry
Sample Final Examination A
Time Allowed: 2.5 hours
Reading Time: 5 minutes

Paper A
Question Mark
1 6 marks
2 6 marks
3 6 marks
4 6 marks
5 6 marks
6 6 marks
7 6 marks
8 6 marks
9 6 marks
10 6 marks
11 10 marks
12 10 marks
13 10 marks
14 10 marks
15 10 marks
16 10 marks
Total
Student Number: _____________________
First Name: _____________________
Family Name: _____________________
Class Group _____________________

UNSW Foundation Studies
UNSW Global Pty Limited
UNSW
Sydney NSW 2052 Australia

















Copyright  2017


















All rights reserved. Except under
the conditions described in the
Copyright Act 1968 of Australia
and subsequent amendments,
this publication may not be
reproduced, in part or whole,
without the permission of the
copyright owner.


DIRECTIONS TO CANDIDATES

1. Check that your copy of the examination paper is complete.
This examination contains ten questions:
• Questions 1 to 10 @ 6 marks each
• Questions 11 to 16 @ 10 marks each

The total marks available are 120.

2. Ensure that you have in the room:
• a black or blue pen for writing answers
• an electronic calculator
• no unauthorised material

3. Five (5) minutes reading time is allowed. During reading time do not
write at all.

4. Ensure that your name and examination number are clearly printed on
the front page on the blank answer booklet.

5. Answer all questions in the answer booklet.

6. Hand in the answer booklet to the supervisor at the end of the
examination.






1. A pure sample of the blue mineral chalcanthite was analysed according to the following
scheme.

(a) Write a net ionic equation to represent the reaction in Step 3.
(b) Write a net ionic equation to represent the reaction in Step 4.
(c) Using the data above determine the chemical formula of chalcanthite.





2. The following table shows boiling points of fluorine and the hydrogen halides.
Substance Fluorine Hydrogen fluoride
Hydrogen
chloride
Hydrogen
bromide
Hydrogen
iodide
Boiling point (K) 85 292 188 206 238
With reference to the intermolecular forces involved, briefly explain:
(a) the trend in boiling points among hydrogen chloride, hydrogen bromide and
hydrogen iodide.
(b) the difference in the boiling points of hydrogen fluoride and hydrogen chloride.
(c) the difference in the boiling points of fluorine and hydrogen chloride.



3. The list below contains three pairs of isomers. Draw the structural formula of each
compound listed and briefly describe a quick, easily visible chemical test you could
perform in the laboratory to distinguish between each member of the pair. (Balanced
equations are unnecessary)
(a) butanal and butanone
(b) butanoic acid and methyl propanoate
(c) 1-butene and cyclobutane





4. The diagram below shows details of a fuel cell developed by N.A.S.A. in 1979.


(a) Write an ionic equation to represent the overall reaction occurring in this cell.
(b) Which electrode (left or right on the diagram) is the cathode and what is its polarity
(negative or positive)?
(c) Calculate the E.M.F. generated by this cell under standard conditions.

5. Flask A contains 10.0 grams of argon gas at 50 oC. Flask B contains 5.00 grams of
propyne (C3H4) gas at 50 oC. The tap between them is closed.

(a) Evaluate the ratio:
(number of gas particles in A) / (number of gas particles in B).
(b) Evaluate the ratio:
(average speed of gas particles in A)/ (average speed of gas particles in B).
(c) The tap was opened and the two gases mixed. Predict the reading on the pressure
gauge at 50 oC.

6. In 1911, H. Geiger and E. Marsden fired a beam of alpha particles from a radioactive
material at a very thin sheet of gold.
They observed that almost all the alpha particles went straight through the gold but a few
were deflected through large angles.
(a) Briefly explain why most of the alpha particles went straight through the gold atoms.
(b) Briefly explain what caused a few of the alpha particles to be scattered.
(c) In 1920, J. Chadwick repeated this experiment using an equivalent thickness of silver
in place of the gold. He obtained similar results but found that the alpha particles
were scattered less by the silver than by gold. Briefly explain this observation.

Fluorescent
screen
Beam of
alpha
particles
Source
Thin sheet of
gold foil

7. Briefly outline, with relevant net ionic equations, how you would convert:
(a) solid nickel carbonate to a solution of nickel chloride.
(b) nickel chloride solution to solid nickel hydroxide.
(c) solid nickel hydroxide to a solution of nickel nitrate.
(If convenient, you may use a flow chart to answer this question)




8. The pH of blood is maintained at 7.4 by several chemical reactions in the body. One of
these is the reversible reaction between sodium dihydrogenphosphate (NaH2PO4) and
sodium hydrogenphosphate (Na2HPO4) :
H2PO4–(aq) ⇌ H+(aq) + HPO42–(aq)
(a) Calculate the hydrogen ion concentration in a solution whose pH is 7.4.
(b) What term describes a mixture of chemicals which maintains a solution at a
particular pH ?
(c) Briefly explain how this mixture is able to absorb small numbers of H+(aq) or OH–
(aq) ions without changing the pH.



9. (a) Draw a Lewis (electron dot) formula to represent an ammonia molecule and predict
its shape using the V.S.E.P.R. theory.
(b) Draw a Lewis (electron dot) formula to represent a boron trifluoride molecule and
predict its shape using the V.S.E.P.R. theory.
(c) When mixed, ammonia and boron trifluoride undergo an addition reaction to form a
compound of formula NH3BF3. Draw a structural formula to represent this
compound and name the type of bond which joins the ammonia and boron trifluoride
molecules together.

10. Cyanic acid (HOCN) is a weak acid with dissociation constant (Ka) = 1.2 × 10–4
(a) Write a net ionic equation to represent the dissociation of cyanic acid in aqueous
solution.
(b) Write down the formula for the conjugate base of cyanic acid.
(c) Calculate the pH of a 0.10 M cyanic acid solution.
(d) Calculate the percentage of cyanic acid molecules which are dissociated in a 0.100 M
solution of the acid.
(e) Pure cyanic acid changes into its isomer:
H – O – C≡N(g)  O = C = N – H (g)
Use average bond energies to predict the enthalpy change (in kJ) for this reaction.





11. A saturated solution of magnesium hydroxide has a pH of 10.50 and is known as “milk of
magnesia”.
(a) Calculate the hydroxide ion concentration in milk of magnesia.
(b) Write down an expression for the solubility product constant (KSP) of magnesium
hydroxide and determine its value.
(c) Calculate the aqueous solubility (in grams/litre) of magnesium hydroxide.
(d) When 1M ammonia solution is added to milk of magnesia a white precipitate forms.
Briefly explain the formation of this precipitate using appropriate ionic equations.
(e) The precipitate in (d) dissolves when mixed with a 1M ammonium chloride solution.
Using Le Chatelier’s Principle briefly explain why this occurs.


12. (a) The element rubidium exists naturally as a mixture of two isotopes:
rubidium–85 and rubidium-87.
Calculate the percentage of each isotope in a sample of the metal.
(b) Rubidium reacts with bromine according to:
2Rb + Br2 ⇌ 2RbBr
(m. pt. = 39oC) (m. pt. = –7oC) (m. pt. = 699oC)
Briefly explain why the melting point of rubidium bromide is much higher than that
of rubidium or bromine.
(c) Sizes of the particles involved in the reaction in (b) are tabulated below:
Element Atomic radius (nm) Ionic radius (nm)
Rubidium 0.248 0.147
Bromine 0.114 0.195
Briefly explain why the radius of rubidium decreases and the radius of bromine
increases during this reaction.
(d) The first ionisation energy of bromine is 11400 kJ/mol. Writer a thermochemical
equation to represent the first ionisation energy of bromine.




13. The amount of sulfur dioxide in polluted air can be measured by reacting it with potassium
permanganate solution which undergoes the following half reaction:
MnO4–(aq) + 8H+(aq) + 5e–  Mn2+(aq) + 4H2O(l)
(a) Sulfur dioxide is converted into sulfuric acid. Write an ionic half equation to
represent this conversion and state whether it is oxidation or reduction.
(b) Briefly explain why no special indicator is needed to detect the end point of the
reaction between sulfur dioxide and potassium permanganate solution.
(c) In one analysis, 25.0 mL of 0.0100 M potassium permanganate solution reacted
completely with all the sulfur dioxide present in 10.0 litres of polluted air at R.T.P.
Calculate the number of moles of sulfur dioxide involved in this reaction.
(d) Calculate the concentration (in mg/L) of sulfur dioxide in the polluted air.
(e) If barium chloride solution was added to the mixture produced in (c) calculate the
mass of barium sulfate which would be precipitated.


14. Methylcyclopentane and cylcohexane form an equilibrium mixture according to the
following equation:
Methylcyclopentane ⇌ Cyclohexane
At 25oC the above reaction forms an equilibrium mixture which
contains 87.5% cyclohexane and 12.5% methylcyclopentane.
(a) Rewrite the equation showing correct structural (graphic) formulas for both
compounds.
(b) Evaluate the equilibrium constant (Kc) for this reaction at 25 oC.
(c) 4.00 moles of methylcyclopentane were added to a 1.00 litre container which was
then sealed. The above reaction occurred and was allowed to reach equilibrium at
25oC. Calculate the number of moles of methylcyclopentane which were present in
the container at equilibrium.
(d) At 75oC the value of the equilibrium constant (Kc) for this reaction is 3.0. If the
mixture in (c) was allowed to reach equilibrium at 75 oC how many moles of
methylcyclopentane would be present ?
(e) Is the reaction exothermic or endothermic ? Briefly justify your answer.


15. The diagram below shows some energy relationships (in kilojoules) among potassium,
chlorine and potassium chloride.



(a) Briefly explain the difference between “K+ (l) + Cl– (l)” and “K+ (aq) + Cl– (aq)”.
(b) Using this chart determine the value (in kJ/mole) of:
the enthalpy of atomisation (∆Ha ) of potassium
the bond energy (ΒΕ) of chlorine
the enthalpy of formation (∆Hf ) of potassium chloride
the enthalpy of fusion (∆Hm ) of potassium chloride
the enthalpy of solution (∆Hsol ) of potassium chloride
the ionisation energy (I.E.) of potassium

(c) Determine the value for the electron affinity of chlorine (marked “X” on the
diagram).

16. The flow chart below summarises several reactions involving ethanal (acetaldehyde):




(a) Draw a Lewis (electron dot) diagram to represent an ethanal molecule.
(b) Briefly explain why ethanal is very soluble in water.
(c) The toxic concentration of ethanal in 1 litre of air is 0.360 mg. Calculate the number
of molecules present in 0.360 mg of ethanal.
(d) Addition of water to ethene forms Compound X which, when heated with copper
metal, changes into ethanal and hydrogen. Write a balanced equation for the
conversion of X to ethanal and use oxidation numbers to prove that this is a redox
reaction.
(e) Ketene reacts with hydrogen to form ethanal according to :
ketene gas + hydrogen gas  ethanal gas
(1.0 L at R.T.P.) (1.0 L at R.T.P.) (1.0 L at R.T.P.)
Draw a structural formula to represent a ketene molecule and name the class to
which this organic reaction belongs.



DATA SHEET

Avogadro's constant, NA 6.02 × 1023 mol–1
Faraday's constant, F 96486 C mol–1
Universal Gas Constant, R 8.314 J K–1 mol–1
Specific heat of water 4.180 J g–1 K–1

Molar volume of ideal gas:
at 101.3 kPa and 0oC (S.T.P.) 22.41 L mol–1
at 101.3 kPa and 25oC (R.T.P.) 24.46 L mol–1

Standard enthalpies of formation (∆Hf ):
CO (g) –111 kJ mol–1
CH4 (g) – 75 kJ mol–1
CH3CHO (g) –166 kJ mol–1

Average bond energies:
O –H 463 kJ mol–1
C – O 358 kJ mol–1
C ≡ N 891 kJ mol–1
C = O 745 kJ mol–1
C = N 615 kJ mol–1
N – H 391 kJ mol–1

Standard reduction potentials:

Cr3+(aq), Cr2+(aq) – 0.41 V
Fe3+(aq), Fe2+(aq) + 0.77 V






Answers to Transition Final Exam - Sample A

1. (a) Zn (s) + Cu2+ (aq)  Cu (s) + Zn2+ (aq)
(b) Ba2+ (aq) + SO42– (aq)  BaSO4 (s)
(c) mol Cu : mol BaSO4 : mol H2O = (0.64/63.5) : (2.33/233.4) : (0.90/18.0)
= 0.010 : 0.0100 : 0.050
So formula of chalcanthite is CuSO4·5H2O
(Check: 0.0100 mole CuSO4.5H2O does weigh 2.50 g)

2. (a) HCl is the most polar of these three molecules. HBr and HI are less polar.The boiling
points for this series of compounds suggests dispersion forces are more important
than dipole-dipole forces.
The boiling points increase because the strength of the dispersion forces between the
molecules becomes progressively larger as the number of electrons in each increases.
(b) Hydrogen fluoride and hydrogen chloride are both polar molecules, however
hydrogen fluoride is much more polar than hydrogen chloride (difference in
electronegativity. Hydrogen fluoride has hydrogen bonds between its molecules
which result in a higher boiling point than in hydrogen chloride which does not form
hydrogen bonds.
(c) Fluorine is non polar covalent molecule, hence only dispersion forces act between
molecules. Hydrogen chloride is a polar covalent molecule thus both dispersion and
dipole/dipole forces act between its molecules. Both fluorine and hydrogen chloride
have the same number of electrons so their dispersion forces are of similar strength
however hydrogen chloride also has dipole forces between its molecules so has a
higher boiling point.

3. (a) Butanal (CH3–CH2–CH2–CHO) would change the colour of potassium dichromate
solution from orange to green; butanone (CH3–CH2–CO–CH3) would not.
(b) Butanoic acid (CH3–CH2–CH2–COOH) would give bubbles of carbon dioxide when
added to a solution of sodium carbonate.
methyl propanoate (CH3–O–CO–CH2–CH3) would not form bubbles when added to
a solution of sodium carbonate.
(c) 1–butene (CH2=CH-CH2-CH3) would immediately change bromine solution from
brown to colourless; cyclobutane { (CH2)4} would not.

4. (a) Fe3+(aq) + Cr2+ (aq)  Cr3+ (aq) + Fe2+ (aq)
(b) The cathode is on the left hand side and is positive.
(c) E°Cell = E°cathode - E°anode = +0.77 – (–0.41) = +1.18 V

5. (a) mol in A : mol in B = (10.0 / 40.0) : (5.00 / 40.0) = 2.0
(b) Temperature is a measure of average kinetic energy. The temperatures of each gas
are equal therefore the average molecular kinetic energies are the same; since the
masses of argon and propyne particles are both 40.0 amu the ratio of their average
speeds is 1.0.
(c) P = nRT/V = ((10.0 + 5.00)/ 40.0) × 8.314 × 323 / 10.0 = 101 kPa
6. (a) Most of the volume of a gold atom is empty space so that the alpha particles met no
obstacles in travelling through it.
(b) At the centre of an atom is a small, positively charged nucleus which repels any
alpha particles (also positively charged) which come near it.
(c) The positive charge on a silver atom’s nucleus is +47 and this is less repulsive to
alpha particles than a gold nucleus whose charge is +79.
7. (a) Add hydrochloric acid to solid nickel carbonate:
NiCO3 (s) + 2H+ (aq)  Ni2+ (aq) + H2O (l) + CO2 (g)
(b) Add sodium hydroxide to nickel chloride solution then filter:
Ni2+ (aq) + 2OH– (aq)  Ni(OH)2 (s)
(c) Add nitric acid to nickel hydroxide:
2H+(aq) + Ni(OH)2 (s)  Ni2+(aq) + 2H2O (l)
8. (a) [H+] = 10–pH = 10 –7.4 = 4.0 × 10–8 M
(b) Buffer
(c) Added H+ (aq) ions react with HPO42– (aq) ions so they are removed and cannot alter
the pH. Added OH– (aq) ions react with H2PO4– (aq) ions to form HPO42– (aq) so
they are removed and cannot alter the pH.
Added acid is removed by reaction with HPO42–:
HPO42– (aq) + H+ (aq)  H2PO4– (aq)
Added base is removed by reaction with H2PO4–:
H2PO4– (aq) + OH– (aq)  HPO42– (aq)

Alternative answer and reasoning for part (c):
The equilibrium mixture that forms with these ions is represented by the following
equation:
H2PO4– (aq) ⇌ HPO42– (aq) + H+ (aq)
Le Chatelier’s principle predicts that the system will shift to minimise any
disturbance therefore adding hydrogen ion causes the system to shift to the left to
minimise the increase in [H+] thus minimising the change in pH. Similiarly, adding
hydroxide ion causes a decrease in [H+]. Le Chatelier’s Principle predicts the system
will shift to the right to minimise the decrease in [H+] thus minimising the change in
pH.

9. (a) The Lewis diagram for ammonia: The nitrogen in
ammonia is directly attached to three H atoms
and to a lone pair giving an AB3E system
(VSEPR theory) hence ammonia is pyramidal with a HNH bond angle of
approximately 109 – 2 = 107o.
(b) The Lewis diagram for boron trifluoride:
The boron in boron trifluoride is only directly
attached to three fluorine atoms giving an AB3
system (VSEPR theory) hence boron trifluoride is
trigonal (flat triangular).
(c) The link between the two molecules is
a dative covalent bond.

10. (a) HOCN (aq) ⇌ H+ (aq) + OCN– (aq)
(b) OCN–
(c) A RICE table is not necessary but may help you to get the right answer (and help the
marker to give you marks even if your final answer is incorrect)
Reaction HOCN (aq) ⇌ H+ (aq) + OCN– (aq)
Initial 0.10 0 0
Change – x + x + x
Equilibrium 0.10 – x ≈ 0.10 + x + x
If x << 0.10 than 0.10 – x ≈ 0.10
Ka = [H+][ OCN–]/[HOCN] = x2/0.10 = 1.2 × 10–4
∴ x = [H+] = [ H+] = 3.5 × 10–3
∴ pH = –log10[1.2 × 10–4] = 2.5
(d) Dissociation = (3.5 × 10–3 / 0.10 ) × 100 = 3.5 %
(e) H – O – C≡N(g)  O = C = N – H (g)
∆H = BEreactants – BEproducts
∆H = (463 + 358 + 891) – (745 + 615 + 391)
∆H = –39 kJ/mol


11. (a) pOH = 3.5 ∴ [OH–] = 10–3.5 = 3.16 × 10–4 M
(b) [Mg2+] = ½ [OH–]
Ksp = [Mg2+] × [OH–]2 = (3.16 × 10–4/2) × (3.16 × 10–4)2 = 1.6 × 10–11
(c) Molar solubility = (3.16 × 10–4 / 2) = 1.6 × 10–4 M
Mass solubility = 1.6 × 10–4 × 58.3 = 9.3 × 10–3 g/L
(d) Ammonia is a weak base and forms hydroxide ions in solution:
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH–(aq) System 1
Milk of magnesia is a saturated solution of magnesium hydroxide.
Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH–(aq) System 2
Le Chatelier’s principle predicts an increase in [OH–] causes system 2 to shift to the
left to minimise the increase in [OH–] and a precipitate of Mg(OH)2 forms.
(e) Addition of NH4+(aq) from NH4Cl drives the system 1 in part (d) above to the left
reducing the [OH–]. Le Chatelier’s principle predicts that system 2 will shift to the
right to minimise the decrease in [OH–] hence Mg(OH)2 dissolves.
12. (a) In 100 Rb atoms there are y 85Rb atoms and (100 – y) 87Rb atoms.
Their total mass is 85y + 87(100 – y) = 8550 ∴ y = 75
Hence 75% 85Rb and 25% 87Rb if Ar = 85.5
or 76.5% 85Rb and 23.5% 87Rb if Ar = 85.47
(b) Solid rubidium bromide is held together by ionic bonds between the Rb+ and Br–
ions. These are stronger than the metallic bonds in solid rubidium or the dispersion
forces between bromine molecules so RbBr has the highest melting point.
(c) The size of a particle is determined by the distance of its outermost electron from the
nucleus. When Rb forms Rb+ it loses its single electron in the outer (5th) shell and so
shrinks in size. When Br forms Br – it gains an electron which repels all the other
electrons in the outer shell thus causing the particle to expand in size.
(d) Br(g)  Br+(g)  e–(g) ∆H = 11400 kJ/mol
13. (a) SO2 (aq) + 2H2O (l) ⇌ SO42– (aq) + 4H+ (aq) + 2e–; oxidation.
(b) During the reaction MnO4– ions (purple) change to Mn2+ ions which are colourless.
The colour change that indicates the equivalence point is the appearance of a faint
pink colour indicating an excess of MnO4– ions.
(c) 2MnO4–(aq) + 5SO2(aq) + 2H2O(l)  2 Mn2+(aq) + 5SO42–(aq) + 4H+(aq)
Mol MnO4– = 0.010 x 0.0250 = 2.5 × 10–4
Mol SO2 = 5/2 × 2.50 × 10–4 = 6.25 × 10–4
(d) Mass of SO2 = 64.1 × 6.25 × 10–4 = 0.0400 grams
[SO2] = 0.0400 × 1000 / 10.0 = 4.00 mg / L
(e) Mol BaSO4 = mol SO2 = 6.25 × 10–4 ∴ mass = 233.4 × 6.25 × 10–4 = 0.146 g

14. (a) methylcyclopentane ⇌ cyclohexane




(b) Kc = [cyclohexane] / [ methylcyclopentane ] = 87.5 / 12.5 = 7.0
(c) A RICE table is not necessary, but may be used:
Reaction Methylcyclopentane ⇌ Cyclohexane
Initial 4.0 0.00
Change – x + x
Equilibrium 4.0 – x x
Kc = x/(4.0 – x) = 7.0 ∴ x = 3.5 M
[Methylcyclopentane] = 4.00 – 3.5 = 0.50 M
∴Moles Methylpentane =0.50 mol (as volume is 1.00 litre).
(d) The same approach as used in part (c), with the Kc = 3.0:
Kc = x/(4.0 – x) = 3.0 ∴ x = 3.0 M
[Methylcyclopentane] = 4.00 – 3.0 = 1.0 M
∴Moles Methylpentane =1.0 mol (as volume is 1.00 litre).
(e) Temperature is the only variable that changes the value of the equilibrium constant.
In increasing the temperature from 25oC to 75oC the equilibrium constant decreases
from 3.5 to 3.0 indicating that Le Chatelier shift has been towards reactants which is
consistent with heat being a product and the reaction, as written, being exothermic.

15. (a) “K+(l) + Cl–(l)” refers to liquid (molten) KCl which is a pure compound (KCl) that
has been heated to a temperature above its melting point.
“K+(aq) + Cl–(aq)” refers to an aqueous solution of KCl which is a mixture
containing KCl and water.
(b) i. ∆Hatomisation (KCl) = +90 kJ / mol
ii. Bond Energy(Cl–Cl) = +242 kJ / mol
iii. ∆Hformation (KCl(s)) = – 436 kJ / mol
iv. ∆Hmelting (KCl) = +27 kJ / mol
v. ∆Hsolution (KCl(s)) = +19 kJ / mol
vi. First ionisation energy (K) = + 425 kJ / mol
(c) (X + 710) = (121 + 425 + 90 + 436) ∴ X = – 362 kJ (exothermic)


16. (a)

(b) Both water and ethanal are small polar covalent molecules. Ethanal can form strong
dipole/dipole bonds (called hydrogen bonds) with water molecules therefore ethanal
is soluble in water.

(c) Moles ethanal = 0.360 × 10–3/44.0 = 8.18 × 10–6 mol
Number of molecules = 8.18 × 10–6 × 6.02 × 1023 = 4.93 × 1018 molecules
(d) X is ethanol: C2H6O  C2H4O + H2
Oxidation numbers: “C” changes from –2 to –1; “H” changes from +1 to 0
or C2H6O + ½ O2  C2H4O + H2O)
Oxidation numbers: “C” changes from –2 to –1; “O” changes from 0 to –2
(e) 1 × Ketene + H2  C2H4O
The molecular formula for ketene must be C2H2O:
Ketene is CH2=C=O; the reaction is an addition reaction.



Transition Program
Chemistry
Sample Final Examination B
Time Allowed: 2.5 hours
Reading Time: 5 minutes

Paper B
Question Mark
1 6 marks
2 6 marks
3 6 marks
4 6 marks
5 6 marks
6 6 marks
7 6 marks
8 6 marks
9 6 marks
10 6 marks
11 10 marks
12 10 marks
13 10 marks
14 10 marks
15 10 marks
16 10 marks
Total
Student Number: _____________________
First Name: _____________________
Family Name: _____________________
Class Group _____________________

UNSW Foundation Studies
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conditions described in the Copyright Act
1968 of Australia and subsequent
amendments, this publication may not be
reproduced, in part or whole, without the
permission of the copyright owner.


DIRECTIONS TO CANDIDATES

1. Check that your copy of the examination paper is complete.
This examination contains ten questions:
• Questions 1 to 10 @ 6 marks each
• Questions 11 to 16 @ 10 marks each

The total marks available are 120.

2. Ensure that you have in the room:
• a black or blue pen for writing answers
• an electronic calculator
• no unauthorised material

3. Five (5) minutes reading time is allowed. During reading time do not write at all.

4. Ensure that your name and examination number are clearly printed on the front
page on the blank answer booklet.

5. Answer all questions in the answer booklet.

6. Hand in the answer booklet to the supervisor at the end of the examination.

1. The compound “salt of Saturn” reacts with hydrochloric acid forming acetic acid and a
precipitate of lead chloride.
(a) Write a net ionic equation to represent the precipitation reaction.
(b) The acetic acid solution required 28.6 mL of 0.700 M sodium hydroxide solution for
complete neutralisation.
Calculate the number of moles of acetic acid which were produced.
(c) Write down the chemical formula of the compound “salt of Saturn”.




2. Three organic compounds (X, Y and Z) have the following properties:
Compound Molecular weight (u.)
Effect on blue
litmus Effect on KMnO4 solution
X 60.0 Change to red No change
Y 60.0 No change Changes to colourless
Z 60.0 No change No change
(a) Compound X reacts with magnesium metal to produce a gas. Write a balanced
chemical equation to represent this reaction.
(b) When compound Y reacts with KMnO4 solution one of the products is propanone.
Draw a structural formula to represent compound Y.
(c) Compound Z has a sweet smell. Name compound Z and the class of organic
compounds to which it belongs.

3. The electrical conductivity of several different liquids was tested in the equipment shown
below:

The following results are obtained:
Liquid Conductivity
Pure acetic acid None
Acetic acid dissolved in water Poor
Pure liquid ammonia None
Ammonia dissolved in water Poor
Mixture of acetic acid and ammonia
dissolved in water Good
Using appropriate ionic equations briefly explain why:
(a) acetic acid conducted a small current when dissolved in water.
(b) ammonia conducted a small current when dissolved in water.
(c) a mixture of acetic acid and ammonia in water conducted a large current.

4. The pH of a saturated solution of silver hydroxide (AgOH) is 10.15.
(a) Calculate the molar concentration of hydroxide ions in a saturated silver hydroxide
solution.
(b) Evaluate the solubility product (Ksp) of silver hydroxide.
(c) Silver hydroxide forms when solid silver oxide (Ag2O) dissolves in water. Calculate
the maximum mass of solid silver oxide which could dissolve in 1.0 litre of water.













5. Propene and cyclopropane both react with hydrogen to form propane. When heated
propane decomposes into a mixture of ethene and methane.

(a) Using structural formulas write a balanced chemical equation for the conversion of
cyclopropane to propane.
(b) Using enthalpy of formation data write a thermochemical equation for the
conversion of propene to propane under standard conditions.
(c) Using bond energy data write a thermochemical equation for the conversion of
propane into ethene and methane under standard conditions.

Cyclopropane
PropanePropene
Ethene
and
methane
Add hydrogen gas
(using a catalyst) Heat
Add hydrogen gas
(using a catalyst)

6. The following table lists physical properties of four different substances (A-D).
Substance Melting Point (oC)
Boiling
Point (oC)
Electrical
Conductivity at R.T.P.
A -205 -190 None
B 727 1050 None
C 1490 2900 Good
D -104 8 None

The substances, not necessarily in order, are:
• cobalt (Co)
• carbon monoxide (CO)
• cobalt chloride (CoCl2)
• phosgene (COCl2).
(a) Identify cobalt chloride and briefly justify your selection.
(b) Identify phosgene and briefly justify your selection.
(c) Draw a Lewis (electron dot) formula to represent a phosgene molecule and use the
VSEPR Theory to predict its shape.

7. For a stomach X-ray a patient is fed a “barium meal” which is actually a suspension of
solid barium sulfate.
(a) Calculate the molar concentration of barium ions in a saturated aqueous solution of
barium sulfate.
(b) It is important to keep the concentration of barium ions as low as possible because
they are poisonous to humans. A “barium meal” contains a mixture of barium
sulfate and aqueous potassium sulfate. Briefly explain, using Le Chatelier’s
Principle or otherwise, why the concentration of barium ions in this solution is less
than that in (a).
(c) The solubility of barium carbonate is similar to that of barium sulfate. However
barium carbonate is never used in a barium meal because the stomach fluids are very
acidic. Briefly explain, using an appropriate ionic equation, why barium carbonate
cannot be used as a barium meal.


8. Consider the following flowchart:


(a) Addition of acidified potassium permanganate solution to 2-butene forms a single
organic product (Compound X). Name Compound X and draw its structural formula.
(b) Addition of hydrogen chloride gas to 2-butene forms a single organic product
(Compound Y). Name Compound Y and draw its structural formula.
(c) Ozonolysis of 2-butene forms a single organic product (Compound Z). Name
Compound Z and draw its structural formula.
9. One mole of a hydrocarbon (compound A) weighs 56.0 grams. A reacts with water to
produce two isomeric compounds (B and C). Compound B reacts with potassium
permanganate solution forming an organic acid (D). Compound C does not react with
potassium permanganate solution.

(a) Calculate the density (in grams per litre) of gaseous A at 120 kPa and 87 oC.
(b) Write down the name of Compound D and draw a structural formula to represent its
molecule.
(c) Write down the name of Compound A and also the name of the class of
hydrocarbons to which it belongs.

KMnO4 +
H2SO4
O3 then
Zn + H2O
HCl
2-butene
Compound X Compound Y Compound Z

10. The graph below shows the boiling points (in K) of the noble gases and hydrogen halides.


Name the type of interparticle forces and identify the strongest interparticle force which is
responsible for:
(a) keeping xenon atoms clustered together in the liquid state.
(b) causing hydrogen iodide to boil at a higher temperature than xenon.
(c) causing hydrogen fluoride to boil at a higher temperature than hydrogen iodide.

2 3 4 5 PERIOD
100
200
300
Xe
Kr
Ar
Ne
HF
HCl
HBr
HI
BOILING
POINT

11. The table below lists some properties of elements whose names begin
with the letter “b”.
Element A. No. A. W. m.p.(oC) b.p.(oC) Isotopes (natural)
Barium 56 137.34 714 1640 130,132,135,136,137,138
Berkelium 97 (247) 1050 unknown
Beryllium 4 9.012 1280 2770 9
Bismuth 83 208.98 271 1560 209
Boron 5 10.81 2300 2550 10, 11
Bromine 35 79.91 -7 58 79, 81
(a) Estimate the percentage of boron atoms in nature which have a
mass number of 11.
(b) Draw a Lewis (electron dot) diagram to represent a boron bromide molecule and
briefly describe its shape.
(c) Write down the number of protons, electrons and neutrons present in a beryllium
Be2+ ion.
(d) Write down the name of the non-radioactive element from this list which has the
lowest first ionisation energy.
(e) Briefly explain why the melting and boiling points of barium are much lower than
those of beryllium.


12. Potassium formate (methanoate) can be produced by heating carbon monoxide with
sodium hydroxide: CO + KOH → KHCO2
(a) Draw a Lewis (electron dot) diagram to represent the carbon monoxide molecule.
(b) Draw a Lewis (electron dot) diagram to represent the hydroxide ion (OH–) in
potassium hydroxide.
(c) The diagram represents a formate (methanoate) ion. Use the VSEPR theory to
predict the size of the angle marked “X” and briefly justify your answer.
(d) Potassium formate (methanoate) dissolves in water producing an alkaline solution.
Write a net ionic equation to represent the hydrolysis reaction that occurs.
(e) Show that the reaction in (d) is a Bronsted-Lowry reaction and draw a circle around
each base present in the equation.


13. The diagram shows a Daniell cell in which the two electrolytes are separated by a porous
container through which ions can diffuse.
(a) Name the metal which forms the anode of this cell.
(b) Write a balanced half equation to represent the process which occurs at the cathode
of this cell.
(c) Calculate the E.M.F. generated by this cell under standard conditions.
(d) If this cell pushes 0.100 moles of electrons through an external circuit calculate the
change in mass of the copper can and indicate whether it is an increase or decrease.
(e) As the cell operates Zn2+(aq) ions diffuse out of the pot into the copper(II) sulfate
solution. Briefly explain why this occurs.

14. The diagram below shows some energy relationships among sodium, fluorine and sodium
fluoride. All enthalpies are in kilojoules.



(a) Briefly explain the difference between “Na+(l) + F–(l)” and “Na+(aq) + F– (aq)”
(b) Using this chart determine the value (in kJ/mole) of :
• the enthalpy of atomisation of sodium
• the bond energy of fluorine
• the ionisation energy of sodium
• the enthalpy of vaporisation of fluorine
• the enthalpy of formation of sodium fluoride
• the enthalpy of solution of sodium fluoride
(c) Determine the value for the electron affinity of fluorine (marked “X” on the
diagram.).


15. Lead ions react with metallic tin according to
Sn (s) + Pb2+ (aq) ⇌ Pb (s) + Sn2+ (aq) ∆H = -10 kJ
100.0 mL of 0.100 M lead (II) nitrate solution were added to 2.00 g of solid tin
at 25 oC. At equilibrium the mass of solid tin present in the container was 1.11 g.
(a) Calculate the molar concentration of Sn2+(aq) ions in the solution at equilibrium.
(b) Write down a mathematical expression relating the equilibrium constant (Kc) to the
ion concentrations and evaluate Kc at 25 oC.
The concentration of Sn2+(aq) was monitored over a period of time when three
separate disturbances (X, Y and Z) were made to the system. The results are shown
graphically below:
(c) At which point (X, Y or Z) was the mixture cooled ?
(d) At which point (X, Y or Z) was water added to the mixture ?
(e) At which point (X, Y or Z) was some NaCl solution added to the mixture ?
(Briefly justify your answers for (c), (d) and (e)).

16. (a) Flask A contains 5.00 g of hydrazine gas (N2H4) in 6.00 litres. Flask B
contains 5.00 g of oxygen in 4.00 litres. Both flasks are at 127 oC.

Evaluate the ratio:
(number of hydrazine molecules in A) ÷ (number of oxygen molecules in B).
(b) When the tap between the two flasks is opened the gases react to form nitrogen and
steam. Write a balanced chemical equation to represent this reaction.
(c) Using oxidation numbers show that the reaction between hydrazine and oxygen is a
redox reaction.
(d) Calculate the pressure gauge reading at 127 oC when the reaction is complete.
(e) Calculate the partial pressure of the nitrogen gas in the mixture in (i).


DATA SHEET

Avogadro's constant, NA 6.02 × 1023 mol–1
Universal Gas Constant, R 8.314 J K–1 mol–1
Specific heat of water 4.180 J g–1 K–1

Molar volume of ideal gas:
at 101.3 kPa and 0 oC (S.T.P.) 22.41 L mol–1
at 101.3 kPa and 25 oC (R.T.P.) 24.46 L mol–1

Standard enthalpies of formation (∆Hf):
propene (gas) + 20 kJ mol–1
propane (gas) – 104 kJ mol-1

Average bond energies:
C – C 348 kJ mol–1
C = C 614 kJ mol–1

Standard reduction potentials:
Zn2+(aq), Zn (s) – 0.76 V
Cu2+(aq), Cu (s) + 0.35 V

Solubility product constant:
BaSO4 (s) 1.0 × 10-10
















































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Answers to Final Exam - Sample B
1. (a) Pb2+(aq) + 2Cl– (aq) → PbCl2 (s)
(b) Mol of acetic acid = mol of NaOH = 0.0286 x 0.700 = 0.0200
(c) Mol PbCl2 = 2.78 / 278.1 = 0.0100 = mol Pb in the salt
∴ Pb(CH3COO)2 (Check: 0.0100 mol Pb(CH3COO)2= 3.25 g)
2. (a) X is acetic acid: 2CH3COOH + Mg → Mg(CH3COO)2 + H2
(b) Y is 2-propanol: CH3 - CH(OH) - CH3
(c) Z is methyl formate (methanoate) and is an ester.

3. (a) Acetic acid is a weak acid and dissociated to form a few ions in aqueous solution and
these ions are free to move through the solution (cations to the cathode and anions to
the anode) thus conducting electric charge (current) through the solution:
CH3COOH (aq) → CH3COO–(aq) + H+ (aq)
Alternative approach to part (a):
(a) Acetic acid is a weak acid and hydrolyses to form a few ions in aqueous solution and
these ions are free to move through the solution (cations to the cathode and anions to
the anode) thus conducting electric charge (current) through the solution:
CH3COOH (aq) + H2O(l) → CH3COO–(aq) + H3O+ (aq)
(b) Ammonia is a weak base that undergoes hydrolysis in water to produce some
(<100% hydrolysis) hydroxide ions and ammonium ions in solution. Ammonium
ions are free to move to the cathode, hydroxide ions to the anode, and the movement
of ions between the electrodes is the electric current:
NH3 (aq) + H2O → NH4+ (aq) + OH– (aq)
(c) Ammonia (a weak base) and acetic acid (a weak acid) react together forming
ammonium acetate (an ionic compound) which is completely dissociated in solution
and hence produces a large concentration of anions and cations which can freely
move between the electrodes and so can conduct a large current.
CH3COOH (aq) + NH3 (aq) → NH4+ (aq) + CH3COO– (aq)

4. (a) pOH = 3.85 ∴ [OH– (aq)] = 10–3.85 = 1.4 × 10–4 M
(b) Ksp = [Ag+ (aq)] × [OH– (aq)] = (1.4 × 10-4)2 = 2.0 × 10-8
(c) Ag2O + H2O → 2AgOH
Mol Ag2O dissolved = ½ × mol AgOH
Mass Ag2O dissolved = ½ × 1.4 × 10-4 × 231.8 = 0.016 g


5. (a)

(b) CH3CH=CH2(g) + H2(g)  CH3CH2CH3(g)
∆Hreaction = Σ(∆Hf of products) – Σ(∆Hf of reactants)
= (–104) – (+20) = –124 kJ/mol
The thermochemical equation:
C3H6 (g) + H2 (g)  C3H8 (g) ∆H = –124 kJ/mol
(c) CH3CH2CH3(g)  CH2=CH2(g) + CH4(g)
∆Hreaction = Σ(Bond energy of reactants) – Σ(Bond energy of products)
= (2 × 348 + 8 C–H) – (614 + 8 C–H) = +82 kJ/mol
The thermochemical equation:
CH3CH2CH3(g)  CH2=CH2(g) + CH4(g) ∆H = –124 kJ/mol
NOTE: Structural, or condensed structural, formulae must be used in the answers to parts
(a), (b) and (c) of this question.

6. (a) The best approach to answering this sort of question is to classify the possible
compounds as belonging to a particular lattice type (see below) and then to match the
identify substance A to D based on their physical properties..
Cobalt is a metallic substance which will conduct in the solid state therefor Cobalt is
substance C
Carbon monoxide is a polar covalent molecule. Covalent molecules tend to boil at
temperatures much less than 727 oC thus CO is likely to be either A or D. CO is a gas
at room temperatures and does not become a liquid on a cold day
hence D is not CO as its melting point is too high. CO is substance A.
Cobalt chloride is an ionic compound and ionic compounds tend to have higher melting
points than polar covalent compounds (as + to – attractions tend to be stronger than δ+
to δ– attractions).CoCl2 is substance B.
(b) Phosgene is the remaining substance. Phosgene is substance D.
(c) The Lewis diagram for phosgene is:
This shows carbon is bonded to three other atoms, giving an AB3 system that
VSEPR theory predicts will adopt a Trigonal (triangular planar) shape with bond
angles of 120o about the central carbon atom.



7. (a) BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq)
[Ba2+] = [SO42–] and [Ba2+] × [SO42–] = 1.0 × 10-10 ∴ [Ba2+] = 1.0 × 10-5 M
(b) BaSO4(s) ⇌ Ba2+(aq) + SO42–(aq)
Adding potassium sulfate causes the sulfate ion concentration to increase. Le
Chatelier’s principle predicts the system will shift to minimise this increase in sulfate
ion concentration by moving towards the left hand side (precipitating solid) as it
reaches a new equilibrium position.
Alternative approach to part (b):
BaSO4(s) ⇌Ba2+(aq) + SO42–(aq)
[Ba2+] × [SO42-] = 1.0 × 10-10 : Adding potassium sulfate increase [SO42-] (a
common ion effect). The concentration of Ba2+ decreases to ensure that equilibrium
is maintained.
(c) The stomach contains acidic substances and these acidic materials will react with
barium carbonate:
BaCO3 (s) + 2H+ (aq)  Ba2+ (aq) + H2O (l) + CO2 (g)
The soluble would Ba2+ (aq) ions are poisonous hence barium carbonate cannot be
used as a barium meal.
8. (a) ethanoic acid: CH3COOH.
(b) 2–chlorobuane: CH3CH2ClCH2CH3
(c) ethanal: CH3CHO
9. (a) Moles of A (in 1.00 litre) = PV/RT = (120 × 1.00)/(8.314 360) = 0.0400 mol
Mass of A (in 1.00 litre) = 0.0400 × 56.0 = 2.24 g so density = 2.24 g/L
(several alternative approaches can be used, all are correct as long as the logic is
correctly followed and applied)
(b) A must be C4H8; B is a primary alcohol, C is a tertiary alcohol and D is
methylpropanoic acid. CH3 - CH( CH3)-COOH
(c) A is methylpropene and is an alkene.
10. (a) Xenon atoms are non–polar therefor the only forces that operate between Xe atoms
are dispersion forces.
(b) HI is not very polar but a weak dipole/dipole force operates between its molecules as
well as the dispersion force. Xenon is non–polar therefore only a dispersion force
operates between xenon atoms.
Hydrogen iodide is the larger species (diatomic whereas Xenon is monatomic)
hydrogen iodide has the larger dispersion force, an additional (very) weak
dipole/dipole force giving it the larger total intermolecular force and higher boiling
point.
(c) Hydrogen fluoride is a small, very polar molecule. The dispersion force acting
between HF molecules is small (as it is a small molecule). The dipole/dipole forces
acting between HF molecules are very strong and are called “hydrogen bonds”. HI
is a large non-polar molecule and its dispersion force is larger than the dispersion
force in HF, but smaller than HF’s H–bond.

11. (a) In 100 B atoms there are y 10B atoms and (100 - y) 11B atoms

Their total mass = 10y + 11(100 - y) = 1081 amu ∴ y = 19
Therefore 81% of boron atoms are 11B.
(b) The Lewis diagram for Boron bromide BBr3 is:
The Lewis diagram shows that the central boron atom in boron tribromide is attached
to only three other atoms (no lone pairs) and thus is anAB3 system. VSEPR theory
predicts that this arrangement will be trigonal (flat triangular) with bond angles of
120o.
(c) 4 protons, 2 electrons and 5 neutrons.
(d) Barium
(e) Both barium and beryllium are group 2 metals. Metallic bonding is the result of the
attraction between the metal cation and their “sea of shared electrons”. Both
beryllium and barium form +2 cations and so the number of electrons in the sea of
electrons is the same. Beryllium is a smaller atom than barium and thus the strength
of attraction between a beryllium ion and the electrons in the sea of electrons is
larger and thus beryllium has the higher boiling point.

: Br :
..
..
: Br :
..
..
: Br :
..
..B
Beryllium ions are smaller hence its nucleus
is closer to its sea of electrons and is more
strongly attracted to its sea of electrons
hence beryllium’s metallic bond is stronger
and beryllium has the higher boiling point.

12. (a) :C:::O:
(b)
(c) Carbon in the formate ion is connected to three other atoms hence is surrounded by
three groups of bonding electrons which will repel each other to a trigonal planar
arrangement with a bond angle of 120o.
(d) A Brønsted/Lowry reaction is on in which a hydrogen ion is transferred from an acid
to a base (to form the corresponding conjugate acid and base.
HCO2–(aq) + H2O(l) ⇌ HCO2H(aq) + HO–(aq)
(e) A Brønsted/Lowry reaction is on in which a hydrogen ion is transferred from an acid
to a base (to form the corresponding conjugate acid and base.
HCO2–(aq) and HO–(aq) are the two bases in this reaction.

13. (a) Zinc is the more active metal and will be more easily oxidised than copper.
Oxidation occurs at the anode hence zinc is the anode.
(b) Cu2+ (aq) + 2e– → Cu (s)
(c) E°cell = E°cathode – E°anode = (+0.35) – (–0.76) = +1.11 V
(d) Moles Cu = ½ × moles of electrons = 0.100 / 2 = 0.0500 mol
∴ Mass of Cu = 0.0500 × 63.5 = 3.18 g. The copper can increases in mass.
(e) Zn2+ ions are cations that will migrate to the cathode to replace the Cu2+ ions that
are being removed from solution as solid copper plates onto the They are needed to
maintain electrical neutrality in the can where copper ions are continually changing
into copper atoms.

14. (a) “Na+(l) + F– (l)” refers to liquid (molten) NaF and is a pure compound.
“Na+(aq) + F– (aq)” refers to a solution of NaF dissolved in water and is a mixture.
(b) ∆Hatomisation Na = +109 kJ/mol
Bond Energy F–F = +160 kJ/mol
∆H1st Ionisation energy Na = +502 kJ/mol
∆Hvaporisation fluorine = +10 kJ/mol
∆Hformation NaF(s) = -574 kJ/mol
∆Hsolution NaF(s) = +10 kJ/mol
(c) X + 920 = 574 + 109 +502 +80 ∴ X = –345 kJ/mol (exothermic)



15. (a) Mass of Sn reacting = 2.00 – 1.11 = 0.89 g
Moles of Sn reacting = 0.89/118.7 of tin = 0.007497 = 0.00750 mol
[Sn2+] = 0.00750 / 0.100 = 0.0750 M
(b) A RICE table is not needed but is included for completeness.
Reaction Sn(s) + Pb2+(aq) ⇌ Pb(s) + Sn2+(aq)
Initial – 0.10 – 0
Change – –0.075 – +0.075
Equilibrium – 0.025 – +0.075
Solids are not included in calculations involving equilibrium constants and are not
shown in this RICE table
Kc = [Sn2+] / [Pb2+] = 0.075 / 0.025 = 3.0
(c) Le Chatelier’s principle predicts that system will shift to minimise the decrease in
heat that occurs when the system is cooled the system. This is an exothermic
reaction hence heat can be considered to be a product. Decreasing the heat content
favours the forward direction which will result in an increase in [Sn2+] therefore the
temperature was decreased at time Z.
(d) Adding water will cause a sudden decrease in the concentration of [Sn2+] therefore
water was added at time X.
(e) A precipitate of PbCl2 (s) forms when NaCl is added to the equilibrium mixture. Le
Chatelier’s principle predicts the system will shift to minimise the decrease in [Pb2+]
by shifting to the left thus reducing [Sn2+ (aq)] as occurs at time Y.
16. (a) Moles N2H4 = 5.00/(2 × 14.07 + 4 × 1.008) 0.1554 mole = 0.155 mol
Moles O2 = 5.00/(32.0) = 0.1563 = 0.156 mole
(Moles N2H4/Moles O2) = 0.1554/0.1563) = 0.9946 = 0.995
(b) N2H4 + O2 → N2 + 2H2O
(c) Oxidation number of “N” changes from -2 to 0; oxidation number of “O” changes
from 0 to -2
(d) Mol N2H4 = mol O2 = 5.0 / 32.0 = 0.156 so mol N2 formed = 0.156 and mol
H2O formed = 0.312. So total mol gas after reaction = 0.468
P = nRT/V = 0.468 x 8.314 x 400 / 10.0 = 156 kPa
(e) One third of the product molecules are N2 so their pressure = 156 / 3 = 52 kPa



Transition Program
Chemistry
Sample Final Examination C
Time Allowed: 2.5 hours
Reading Time: 5 minutes

Paper C

Question Mark
1 6 marks
2 6 marks
3 6 marks
4 6 marks
5 6 marks
6 6 marks
7 6 marks
8 6 marks
9 6 marks
10 6 marks
11 10 marks
12 10 marks
13 10 marks
14 10 marks
15 10 marks
16 10 marks
Total
Student Number: _____________________
First Name: _____________________
Family Name: _____________________
Class Group _____________________

UNSW Foundation Studies
UNSW Global Pty Limited
UNSW
Sydney NSW 2052 Australia

















Copyright  2017






















All rights reserved. Except under the
conditions described in the Copyright Act
1968 of Australia and subsequent
amendments, this publication may not be
reproduced, in part or whole, without the
permission of the copyright owner

DIRECTIONS TO CANDIDATES

1. Check that your copy of the examination paper is complete.
This examination contains ten questions:
• Questions 1 to 10 @ 6 marks each
• Questions 11 to 16 @ 10 marks each

The total marks available are 120.

2. Ensure that you have in the room:
• a black or blue pen for writing answers
• an electronic calculator
• no unauthorised material

3. Five (5) minutes reading time is allowed. During reading time do not write at all.

4. Ensure that your name and examination number are clearly printed on the front
page on the blank answer booklet.

5. Answer all questions in the answer booklet.

6. Hand in the answer booklet to the supervisor at the end of the examination.


1. “Goslarite” is a mineral with formula ZnSO4.yH2O. It can be analysed according to the
following scheme:


Goslarite
(1.44 g)
Barium sulfate
(1.17 g)
Dissolve
in water
Add excess barium chloride
solution then filter
Colourless
solution
Colourless
solution
Step 1. Step 2.


(a) Write a net ionic equation to represent the reaction that occurs in Step 2.
(b) Calculate the number of moles present in 1.17 grams of barium sulfate.
(c) Calculate the value of “y” in ZnSO4.yH2O.



2. For the eleven elements with atomic numbers 10 to 20 inclusive write down the name of
the element which:

(a) has the highest melting point

(b) has the lowest boiling point

(c) has the lowest first ionisation energy

(d) has the highest electronegativity

(e) has an 5 electrons in its valence electron shell

(f) has 14 neutrons and 13 protons in each of its nuclei.
(An element’s name may be used more than once).




3. Briefly explain why a potassium (K+) ion is:

(a) smaller than a potassium atom

(b) larger than a calcium (Ca2+) ion

(c) larger than a sodium (Na+) ion.




4. The electrical conductivities of hydrogen bromide and sodium bromide when solid,
liquid and in aqueous solution are listed in the following table:

Compound Solid Liquid Aqueous Solution
Hydrogen bromide None None Conducts
Sodium bromide None Conducts Conducts

(a) Briefly explain, using a diagram if necessary, why sodium bromide conducts in the
liquid state but not as a solid.
(b) Briefly explain, using a diagram if necessary, why hydrogen bromide does not
conduct when in either the liquid or solid state.
(c) Briefly explain, using an appropriate equation, why hydrogen bromide conducts in
aqueous solution.




5. When heated, solid ammonium sulfite changes into ammonia, steam and sulfur dioxide:
(NH4)2SO3  2NH3 + H2O + SO2
(a) Draw a Lewis (electron dot) diagram to represent the ammonium ion (NH4+) and use
the VSEPR theory to predict its shape.
(b) Draw a Lewis (electron dot) diagram to represent the sulfite ion (SO32−) and use the
VSEPR theory to predict its shape.
(c) The diagram represents a sulfur dioxide molecule. Use the VSEPR theory to predict
the size of the angle marked “X” and briefly justify your answer.

O = S  O


X

6. Hydrochloric acid and hydrofluoric acid are neutralised by sodium hydroxide:
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O(l) ∆H = − 57.7 kJ
HF (aq) + NaOH (aq) → NaF (aq) + H2O(l) ∆H = − 47.7 kJ

(a) Write a net ionic equation to represent the neutralisation of hydrochloric acid by
sodium hydroxide solution.
(b) Using a net ionic equation, briefly explain why the enthalpy change for the
neutralisation of hydrochloric acid is different to the enthalpy change for the
neutralisation of hydrofluoric acid.
(c) Calculate the standard enthalpy change (in kJ/mol of hydrofluoric acid) for the
reaction: HF (aq) → H+ (aq) + F− (aq)




7. The electrodes of an electric cell or battery can be identified with pole-finding paper.
This is made by soaking filter paper in sodium sulfate solution to which some
phenolphthalein indicator has been added. When the two electrodes of a cell are joined
with pole-finding paper an electrolysis reaction takes place and a deep red-violet colour
soon appears around one of them.








(a) Name the ion that is responsible for the phenolphthalein turning red-violet?
(b) Write an ion-electron half-equation to represent the formation of this ion from water
during the electrolysis.
(c) At which electrode (positive or negative) of the cell does the pole-finding paper turn
red-violet? Briefly justify your answer.





8. Mylanta© antacid tablets are used for the relief of indigestion.
Mylanta
ROLLTABS
12 ANTACID TABLETS
Active Ingredients: Calcium Carbonate 317 mg
and Magnesium Hydroxide 64 mg


(a) A human stomach contains 0.063 M hydrochloric acid solution. Calculate the pH of
0.063 M hydrochloric acid solution.
(b) Write a net ionic equation for the reaction of solid calcium carbonate with
hydrochloric acid solution.
(c) Calculate the volume of 0.063 M hydrochloric acid solution which would react
completely with 64 milligrams of solid magnesium hydroxide.





9. An old process for manufacturing sodium hydroxide involves reacting a saturated
solution of sodium carbonate with a saturated solution of calcium hydroxide.

(a) Write an ionic equation for this reaction.

(b) 1.00 litre of saturated calcium hydroxide (containing 1.85 grams of solute) was
completely converted into 1.00 litre of sodium hydroxide solution by this process.
Calculate the pH of this sodium hydroxide solution.

(c) A saturated solution of calcium carbonate contains 9.34 mg/L of CaCO3. Calculate
the Ksp for CaCO3.





10. The diagram shows two flasks (A and B), each of which has a volume of 4.00 litres and is
fitted with a pressure gauge. Flask A contains 5.00 grams of nitrogen dioxide gas. Flask B
contains 2.00 grams of methylhydrazine (CN2H6) gas. The temperature of both flasks is 127 oC.

(a) Calculate the ratio:
(number of molecules in Flask A) ÷ (number of molecules in Flask B)
(b) Calculate the ratio: (pressure in Flask A) ÷ (pressure in Flask B)
(c) When the tap is opened the gases mix and react together producing nitrogen, carbon
dioxide and steam:
10 NO2 (g) + 4 CN2H6 (g)  9 N2 (g) + 4 CO2 (g) + 12 H2O (g)
Calculate the reading on the right side pressure gauge at 127 oC when the reaction is
complete.






11. Heating propane gas converts it into a mixture of methane and ethene:
propane  methane + ethene
The standard enthalpy change for this reaction is + 81 kJ/mol of propane.
(a) Rewrite this equation using structural formulas for the three compounds.
(b) Using information on the Data Sheet, calculate the standard enthalpy of formation
(∆Hf, in kJ/mol) of ethene.
(c) Using information on the Data Sheet, calculate the bond energy (in kJ/mol) of
carbon-carbon double bonds.
(d) Ethene and oxygen can be made to react together to form a gaseous compound called
“oxirane”:
ethene + oxygen gas → oxirane
(1.0 litre) (0.50 litre) (1.0 litre)
If all gas volumes are measured at R.T.P. write a fully balanced equation
for this reaction.
(e) The toxic quantity of oxirane in one litre of air is 0.18 mg. How many molecules are
there in 0.18 mg of oxirane?







12. The solubility of copper chromate (CuCrO4) in water is 1.9 × 10-3 mol/L.

(a) Calculate the solubility of copper chromate in water in grams per litre.
(b) Using Le Chatelier’s principle or otherwise, briefly explain why the solubility of
copper chromate decreases if some solid copper chloride is added to the water.
(c) Write an expression relating the solubility product (Ksp) of copper chromate to the
concentration of its ions in a saturated aqueous solution.
(d) Calculate the solubility product (Ksp) of copper chromate in water.
(e) Predict whether a precipitate of copper chromate would form if 100 mL of
2.00 × 10-3 M Cu(NO3)2 solution was mixed with 100 mL of 4.00 × 10-3 M
Na2CrO4 solution. Briefly justify your answer.





13. Pure silicon is obtained by the following reaction:
SiCl4 (g) + 4Na (s) → Si (s) + 4NaCl (s)
Some properties of these substances are tabulated below:

Substance Melting point (oC)
Electrical conductivity
(as solid)
Electrical
conductivity (when
melted)
SiCl4 – 68 Non-conductor Non-conductor
Na + 98 Conductor Conductor
Si + 1410 Non-conductor Non-conductor
NaCl + 801 Non-conductor Conductor

(a) Briefly explain how sodium is able to conduct an electric current.
(b) Briefly describe the particles, and the type of attractive forces that exist between
them, in solid sodium chloride.
(c) Briefly explain why silicon has a much higher melting point than the other three
substances.
(d) Draw a Lewis (electron dot) diagram to represent a SiCl4 molecule and describe its
shape.
(e) Briefly explain why SiCl4 has a much lower melting point than the other three
substances.







14. A 0.100 M ammonia solution in water has a pH of 11.1
(a) Calculate the molar concentration of hydroxide ions in a solution whose pH is
equal to 11.1
(b) Write a net ionic equation to represent the reaction of ammonia with water to
produce hydroxide ions.
(c) Briefly explain whether ammonia is acting as an acid or a base in this reaction.
(d) If solutions of ammonia and magnesium sulfate are mixed together a precipitate is
formed. Write a net ionic equation to represent the formation of a precipitate when
the two solutions are mixed.
(e) If some solid ammonium sulfate is added to the mixture in (d) it is seen that the
precipitate dissolves. Using Le Chatelier’s Principle briefly explain why the
precipitate dissolved.

15. A solution of potassium permanganate can be standardised by titration with solid
oxalic (ethanedioic) acid:
KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O (unbalanced)
29.00 mL of a certain potassium permanganate solution (acidified with sulfuric acid) were
required to react completely with 0.100 g of solid oxalic acid H2C2O4.
















(a) Using oxidation numbers, show that this reaction involves oxidation and reduction.
(b) Briefly explain the meaning of the term “standardised”.
(c) Briefly explain why an indicator was not needed for this titration.
(d) The permanganate changed according to:
MnO4– (aq) + 8H+ (aq) + 5e → Mn2+ (aq) + 4H2O (l)
Write an ion-electron half equation to represent the change which occurred to the
oxalic acid during the titration.
(e) Calculate the molar concentration of the potassium permanganate solution used in
the titration.






Solid oxalic acid
Potassium
permanganate
solution

16. (a) The formula of lactic acid is:
(a) Write the correct (I.U.P.A.C.) name of lactic acid.
(b) The acid dissociation constant (Ka) for lactic acid is 1.38 × 10–4. Calculate the pH of
0.100 M lactic acid solution.
(c) Calculate the percentage of lactic acid molecules in a 0.100 M aqueous solution of
the compound which have become ionised.
(d) Potassium lactate CH3 −CH(OH) −COOK dissolves in water producing an alkaline
solution. Write a net ionic equation to represent the hydrolysisa reaction that occurs
and draw a circle around the conjugate base of lactic acid.
(e) Calculate the pH of 0.0200 M potassium lactate solution.


DATA SHEET

Avogadro's constant, NA 6.02 × 1023 mol–1
Universal Gas Constant, R 8.314 J K−1 mol–1
Specific heat of water 4.180 J g−1 K–1

Molar volume of ideal gas:
at 101.3 kPa and 0 oC 22.41 L mol−1
at 101.3 kPa and 25 oC 24.46 L mol-1

Standard enthalpies of formation (∆Hf ):
methane (g) − 75 kJ mol−1
propane (g) − 104 kJ mol−1


Average bond energies:
C — C + 346 kJ mol−1

Standard reduction potentials (Eo) in alkaline conditions:
Zn(OH)2 (s), Zn (s) – 1.22 V
NiO2 (s), Ni(OH)2 (s) + 0.49 V

Unit / Symbol Equivalences:

∆Hf = ∆ f H
Ksp = Ks
M = mol L-1

Answers to Final Exam - Sample C
1. (a) Ba2+(aq) + SO42-(aq) → BaSO4(s); (b) 5.01 × 10-3 mol; (c) y = 7.00

2. (a) silicon; (b) neon; (c) potassium; (d) chlorine; (e) phosphorus; (f) aluminium.

3. (a) A K+ ion has 18 electrons hence 2,8,8 with 3 occupied electron shells. A K atom 19
electrons and the 19th electron is in the 4th shell which is further from the nucleus thus
making the atom larger than the ion.
(b) A Ca2+ ion has the same configuration as a K+ ion but its extra proton pulls these
electrons closer to the nucleus thus making the ion smaller than K+.
(c) A Na+ ion is 10 electrons and thus only 2 occupied electron shells and it is smaller.

4. (a) Sodium bromide is made of Na+ and Br¯ ions. These can move, and so form a
current, in the liquid state but cannot move when NaBr is solid.
(b) Solid and liquid HBr consist of uncharged molecules so that a current of
moving charges is impossible.
(c) Hydrogen bromide dissolves in water forming ions:
HBr (g) → H+ (aq) + Br¯ (aq)
These ions can move and thus produce a current.


5. (a)
tetrahedral;


(b) 2- pyramidal;

(c) <120 o; VSEPR theory predicts 3 sets of electrons surround the sulfur atom will
repel to give a triangular planar arrangement
6. (a) H+(aq) + OH–(aq)  H2O(l) .
(b) HCl(aq) is a strong acid and is fully dissociated in solution. HF(aq) is a weak acid
and is mostly in the undissociated form in solution:
HF(aq) + OH–(aq)  H2O(l) + F–(aq).
(c) HF(aq)  H+(aq) + F–(aq) ∆H = –47.7 - (-57.7) = +10.0 kJ/mol

7. (a) Hydroxide ion (turns phenolphthalein red)
(b) 2H2O (l) + 2e– → H2 (g) + 2OH¯ (aq)
(c) Mg(OH)2 + 2 HCl  MgCl2 + 2H2O
Moles Mg(OH)2 0.064/(24.3 + 2 ×17.0) = 0.00110 mole
Moles HCl = 0.00220 mole
Vol HCl = 0.00200/0.63 =0.0348 L = 35 mL (to 2 significant figures)
8. (a) pH = –log10(0.063) = 1.20,
(b) CaCO3 (s) + 2H+ (aq) → Ca2+ (aq) + H2O (l) + CO2 (g)
(c) Moles HCl = 2×moles Mg(OH)2 = 0.064/(24.3 + 34.0) = 0.00220 mole
Mole HCl = 0.00220/0.063 = 0.0348 L = 35 mL (to 2 significant figures)


+
H

9. (a) CO32- (aq) + Ca2+ (aq) → CaCO3(s);
(b) 12.70
(c) Moles CaCO3 = 0.00934/(40.1 + 12.0 + 48.0) = 9.33 × 10–5 ,
Ksp = 8.71 × 10–9

10. (a) 2.50;
(b) 2.50 ((PV = nRT, hence ratio of pressures is the same as ratio of moles if T and V
are constant);
(c) 113 kPa

11. (a) CH3 – CH2 – CH3 → CH4 + CH2 = CH2
(b) + 52 kJ/mol
(c) + 611 kJ/mol
(d) 2C2H4 + O2 → 2C2H4O
(e) 2.5 × 1018

12. (a) 0.34 g/L
(b) Addition of extra Cu2+ (aq) ions from dissolved CuCl2 causes the system to
adjust so as to remove some of these i.e. more solid CuCrO4 forms
i.e. the solubility decreases
(c) Ksp = [Cu2+] × [CrO42-]
(d) Ksp = 3.6 × 10-6
(e) ion product = 2.00 × 10-6 so no precipitation occurs.

13. (a) Sodium atoms each have one loosely-held valence electron which can
move under an applied voltage and thus form a current.
(b) Sodium and chloride ions attract each other with ionic bonds (the electrostatic
force of attraction between cations and anions.
(c) Silicon atoms attract each other with covalent bonds which are stronger
than the other types of attractive force.
(d) tetrahedral shape





(e) SiCl4 is non-polar so has only dispersion forces acting between its
molecules. These are much weaker than the other types of force so the
molecules need less heat energy to separate them.
14. (a) Concentration of OH¯(aq) = 10-2.9 = 1.26 × 10-3 M
(b) NH3 (aq) + H2O (l) → NH4+ (aq) + OH¯ (aq)
(c) NH3 is acting as a base because its aqueous solution contains more
OH¯ (aq) ions than does water therefore [OH–] > [H+]
(d) Mg2+ (aq) + 2 OH¯ (aq) → Mg(OH)2 (s)
(e) Le Chatelier’s principle predicts addition of NH4+ (aq) ions forces the reaction in
(b) to the left thus reducing the concentration of OH¯(aq). This then forces the
reaction in (d) to left causing Mg(OH)2 (s) to dissolve.
Cl x Si
x
x
Cl
Cl
Cl
xx x
x
x
x
x
x
x
x
x
x
x
x
x
x x
x x
x x
x x
x x


15. (a) KMnO4 + H2C2O4 + H2SO4 → MnSO4 + K2SO4 + CO2 + H2O
(“Mn” is reduced from +VII to +II; “C” is oxidised from +III to +IV)
(b) “standardised” = “accurately measured concentration”.
(c) KMnO4 changes from violet to colourless when it is oxidised to Mn2+ At the
endpoint, the colourless solution in the conical flask turns pale pink from a slight
excess of MnO4–
(d) H2C2O4 (aq) → 2CO2 (g) + 2H+ (aq) + 2e
(e) Conctn. = 0.0153 M

16. (a) 2-hydroxypropanoic acid
(b) [H+] = 3.71 × 10-3 M so pH = 2.43
(c) 3.71 %
(d) C3H5O3– (aq) + H2O (l) ⇌ C3H6O3 (aq) + OH– (aq)
(e) pH = 8.08




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