cs chem hkdse mock ans ee

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HKDSE-CS (CHEM) © Times Publishing (Hong Kong) Ltd

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HKDSE-CS (CHEM) © Times Publishing (Hong Kong) Ltd.2

Sample Mock Paper

Section A

1. D 2. C 3. C 4. B 5. C

6. D 7. B 8. C 9. C 10. C

11. D 12. B 13. D 14. C 15. A

16. C 17. C 18. D 19. B 20. D

21. B 22. A 23. D 24. D

1. Complete combustion means that burning is with

sufficient oxygen so that carbon dioxide and water

are the only combustion products.

2. The first step is to identify the repeating unit of the

polymer which is C

H

C2H5

C

H

H

. Once the repeating

unit is identified, change the carbon-carbon bond

in the chain to a carbon-carbon double bond and

draw the structure of an alkene. The alkene is the

monomer of the polymer. In the case, the monomer

is

H

H

CCH

H

H

C

H

H

C

H

. The name of the

monomer is but-1-ene.

3. The equation for the reaction is:

AgNO3(aq) + NaBr(aq) AgBr(s) + NaNO3(aq)

Both silver nitrate and sodium bromide are salts.

When the solutions of these salts are mixed

together, the cations and anions in the solution will

interact with each other. In this case, silver ions

interact with bromide ions to form a solid which

is not soluble in water and hence a precipitate is

formed. This type of reaction is often called double

replacement reaction or precipitation reaction.

Displacement reaction is a reaction between a metal

with an aqueous solution of another metal which is

less reactive.

Neutralization reaction is a reaction between an acid

and a base. Silver nitrate and sodium bromide areneither acids nor alkalis, they are salts.

There is no change of oxidation numbers involved

in the reaction. It is not a redox reaction.

4. The equation for the reaction is:

2Cu(s) + O2(g) 2CuO(s)

The mole ratio of CuO to Cu = 1 : 1

Number of moles of CuO produced

= Number of moles of Cu used

=63.5 g

63.5 g mol –1 = 1 mol

Mass of 1 mole of CuO= (63.5 + 16.0) g mol

–1 × 1 mol = 79.5 g

5. Ethanoic acid is a weak acid and sodium hydroxide

is a strong alkali. The salt formed is not neutral

that the equivalence point of the titration is around

pH 9 to 10. The indicator used for the tit rat ion

should have a colour change around the pH range

of the equivalence point so that the end point of the

titration matches with the equivalence point of the

reaction. The indicator that has a pH range around

9 to 10 is thymol blue.

6. The pH value of pure water is 7. Look at the tableagain.

Indicator pH range

Colour

in acidic

medium

Colour in

alkaline

medium

Congo red 3.1 – 4.9 Blue Red

Bromothymol

blue6.0 – 7.0 Yellow Blue

Thymol blue 8.0 – 9.6 Yellow Blue

Methyl blue 10.6 – 13.4 Blue Pale violet

Congo red will show the colour in alkaline medium

(red), bromothymol blue will show the colour in

alkaline medium (blue), thymol blue will show the

colour in acidic medium (yellow), and methyl blue

will show the colour in acidic medium (blue).

7. The oxidation numbers of the elements in the

compounds are shown in the following table:

NaNO2 Na = +1, O = –2, N = +3

Ca(NO3)2 Ca = +2, O = –2, N = +5

(NH4)2SO4 H = +1, S = +6, O = –2, N = –3

(NH2)2CO H = +1, C = +4, O = –2, N = –3



HKDSE-CS (CHEM) © Times Publishing (Hong Kong) Ltd3

8. In the Zn/Cu chemical cell, Zn dissolves in the

solution and releases electrons which flow along the

external circuit to t he copper electrode. Hence, Zn

is the negative terminal of the cell.

As Zn Zn2+

+ 2e –

is an oxidation reaction, zinc

electrode is the anode.

9. The resulting solution is a mixture of Cu2+

(aq)

(blue) and Cr 3+(aq) (green). Hence, its colour is

bluish-green.

The oxidation number of H has not changed in the

reaction (+1 in H+(aq) and +1 in H2O(l )). It is not

the oxidizing agent.

The oxidation number of Cu has changed from +1

in Cu+(aq) to +2 in Cu

2+(aq). Hence, Cu

+(aq) is

oxidized.

The oxidation number of Cr has decreased from +6

in Cr 2O72–

(aq) to +3 in Cr 3+

(aq).

10. Standard enthalpy change of formation of carbon

monoxide ( x ) refers to the enthalpy change of thefollowing equation:

C(graphite) +12

O2(g)

CO(g)……(1)∆H

o = x

This equation can be linked to two other equations:

C(graphite) + O2(g) CO2(g)……(2) ∆H o

= y

CO(g) +12

O2(g) CO2(g)……(3) ∆H o

= z

Equation (1) is the same as equation (2) – equation

(3), hence x = y – z .

11. The atomic number of nitrogen is 7. A nitrogen

atom has 7 electrons. It has to gain 3 electrons to

form nitride ion (N3–

) to achieve the electronic

structure of neon. Hence, the nitride ion has 10

electrons.

12. Answer A shows a one-stage exothermic reaction.

Answer B shows a two-stage exothermic reaction.

Answer C shows a one-stage endothermic reaction.

Answer D shows a two-stage endothermic reaction.

13. Acidified potassium permanganate solution is an

oxidizing agent. When it reacts with a reducing

agent, it will change to Mn2+

(aq) (colourless) and

its colour will fade. Propene reacts with acidified potassium permanganate solution to form propane-

1,2-diol. Acidified potassium permanganate

oxidizes both propanol and propanal to form

propanoic acid.

14. The two particles have different numbers of protons

They are particles of different elements. As thei

numbers of protons are not the same as thei

numbers of electrons, they are ions. A is F –

and B i

Na+.

15. To obtain an accurate titration results, both the

pipette and burette should be rinsed with water and

then the solution they are going to hold.If the pipette is not rinsed with sodium hydroxide

solution before drawing the solution, the sodium

hydroxide solution delivered to the conical flask

would be diluted (as there is water in the pipette)

The amount of hydrochloric acid required to

neutralize it would be less than the “correct”

amount.

If the burette is rinsed with water only but not with

hydrochloric acid afterwards, the hydrochloric acid

in the burette would be diluted. Hence, more “dilute”

hydrochloric acid would be required to bring the

titration to the end point.

16. This experiment shows that potassium manganate(VII

is an ionic compound. Colourless K +, which carrie

positive charge, will move to the negative electrode

and purple MnO4

– will move to the positive

electrode. This experiment cannot show tha

potassium manganate(VII) is an oxidizing agent.

17. Lithium is a metal and is located above sodium in

the Period Table. Metals react with water to form

metal hydroxide and hydrogen gas.

18. Increase in temperature (∆

T ) = 34.8°

C – 19.6°

C= 15.2°C = 15.2 K

Amount of heat given out = mc ∆T

= (100 + 1.0) g × 4.2 J g –1

K –1

× 15.2 K

= 6 448 J

Number of moles of Li used =1.0 g

6.9 g mol –1

= 0.145 mol

Heat given out per mole of Li =6 448 J

0.145 mol –1

= 44 469 J mol –1

= 44.5 kJ mol

–1

∴ The enthalpy change of reaction per mole

of lithium is –44.5 kJ mol –1

.

19. Carbon monoxide is colourless and odourless. It i

a toxic gas because it combines with haemoglobin

in our blood readily and prevents haemoglobin from

carrying oxygen to various parts of our body.




20. The products of electrolysis of sea water are

hydrogen, chlorine and sodium hydroxide

solution. Hydrogen and chlorine are gases at room

temperature and pressure and they react to form

hydrochloric acid.

21. Quartz is a mineral found in granite. Quartz is a

compound of silicon and oxygen.

22. Fraction Y has a higher boiling range than fraction

X, therefore, fraction Y is more viscous (i.e. less

runny), darker in colour (has higher molecular

masses) and more difficult to burn (less volatile).

23. As the atoms of Group II metals are smaller than

that of Group I metals in the same period, their

outermost electrons are held more tightly by the

nucleus of the atom. As a result, group II metal

atoms are less reactive and have a higher melting

point and density. All metals are basically shiny.

24. The equations suggest that 1 mole of hydrogen will

use up 0.5 mole of oxygen but 1 mole of methane willuse up 2 moles of oxygen. Hence, answer A is wrong.

The equations and the enthalpy changes of

combustion also suggest that 1 mole of hydrogen

(2.0 g) gives out 285.8 kJ of heat while 1 mole of

methane (16.0 g) gives out 890.4 kJ of heat. Hence,

for the same number of moles, hydrogen gives out

less heat.

For gases, the volume is proportional to the number

of moles. Hence, for the same volume of gases,

hydrogen gives out less heat.

Since 2.0 g of hydrogen give out 285.8 kJ of heat,16.0 of hydrogen will give out 2 286.4 kJ of heat.

Hence, for the same masses of gases, hydrogen

gives out more heat.

Section B

1. (a) Isotopes are atoms of the same element having

different number of neutrons / mass number. 1

(b) Let Y be the relative abundance of 20 Ne. 1

20 y + 21(0.002 7) + 22(1 – 0.002 7 – y ) = 20.19

21.997 3 – 2 y = 20.19

y = 0.904

(c) Neon has a stable octet electronic structure. 1

(d) As a gas to fill fluorescent (neon) tube. 1

2. (a) Add aluminium to copper(II) nitrate solution.

The blue (green) solution turns pale blue

(green) / colourless; and a reddish brown

precipitate appears. 1

4

Aluminium is more reactive than copper and

can displace copper from its salt solution.

2Al(s) + 3Cu2+

(aq) 2Al3+

(aq) + 3Cu(s) ½

Add copper to silver nitrate solution. The

colourless solution turns pale blue (green)

and a greyish precipitate appears. 1

Copper is more reactive than silver and can

displace silver from its salt solution.Cu(s) + 2Ag

+(aq) Cu

2+(aq) + 2Ag(s) ½

Hence, the relative reactivities of the three

metals are in the order of Zn > Cu > Ag. 1

(b) silver oxide: by heating alone 1

copper(II) oxide: by heating with carbon 1

aluminium oxide: by electrolysis of the

molten oxide 1

3. (a) As both lead( II) hydro xide and zinc

hydroxide are amphoteric, both hydroxides

are soluble in excess sodium hydroxidesolution. / Both Pb

2+and Zn

2+are soluble

in excess sodium hydroxide solution.

Hence, sodium hydroxide is not appropriate. 1

Cl –

forms precipitate with Pb2+

but not

Zn2+

. However, PbCl2 is fairly soluble in

hot water and that the molar mass of PbCl2

is lower than that of PbSO4. Error involved

in gravimetric analysis is greater. 1

Hence, sodium sulphate was chosen.

(b) Filter the precipitate. Wash the precipitate

with a minimum amount of cold water.

Dry the precipitate by pressing it betweendry filter papers / heating it in an oven

/ by leaving it in a desiccator overnight.

Measure the mass of the precipitate by

using an electronic balance. 2

(c) Formula mass of lead(II) sulphate

= 207.2 + 32.1 + 16.0 × 4 = 303.3

Formula mass of lead(II) nitrate

= 207.0 + 2(14.0 + 16.0 × 3) = 331.2 ½

Mass of lead(II) nitrate in the sample

=3 25 331 2

303 3

. .

.

×= 3.549 g ½

Hence, the percentage by mass of lead(II)

nitrate in the sample

=3 549

5 00

.

. × 100% = 71.0% 1

7

6



HKDSE-CS (CHEM) © Times Publishing (Hong Kong) Ltd5

4. (a) A dative covalent bond is a covalent bond

in which one of the two atoms takes the two

electrons for the sharing. 1+

OH

H

H

dative covalent

bond

½ + ½

(b) Dilute sulphuric acid:

(1) Dilute sulphuric acid reacts with

magnesium to give hydrogen and

magnesium sulphate. ½

Mg(s) + H2SO4(aq)

MgSO4(aq) + H2(g) 1

(2) Dilute sulphuric acid acts as an acid. ½

Concentrated sulphuric acid:(1) Hot and concentrated sulphuric

acid reacts with magnesium to

give sulphur dioxide, magnesium

sulphate and water. 1

Mg(s) + 2H2SO4(l )

MgSO4(s) + 2H2O(g) + SO2(g) 1

(2) Concentrated sulphuric acid acts

as an oxidizing agent. 1

(c) C12H22O11(s)

12C(s) + 11H2O(g) 1

5. (a) A and D ½ + ½

HOOC COOH + n HOCH2CH2OH(s)

OC COOCH2CH

2O

n

+ (2n – 1) H2O 1

(b) (i) E ½

NH2 O

nCH3CH2CHCOOH ( n –1)H2O+HNCHC

CH2

CH3

n

1

(ii) C ½

CH3

CH3

n CH2 CH2CCCOOH

n

O OHC

1

8

5

6. (a) H+(aq) + OH

– (aq) H2O(l ) 1

(b) Standard enthalpy change of neutralization 1

(c) All the three values ∆H 1, ∆H 2 and ∆H 3 are

negative.

Increasing order of magnitude:

∆H 2, ∆H 1, ∆H 3 (∆H 3 is the most negative.) 1

All the three reactions use 1 mole of acid

and 1 mole of alkali.

In reaction (3), the dissolving of solid

sodium hydroxide in water is a highly

exothermic process. Hence ∆H 3 is the most

negative. 1

In reaction (2), CH3COOH is a weak acid.

Most CH3COOH exists in molecular form.

CH3COOH(aq)

CH3COO – (aq) + H

+(aq

Energy is required to ionize CH3COOH

into ions during neutralization. Hence, ∆H 2

is the least negative. 1

7. (a) Anode (oxidation):

Zn(s) + 2OH – (aq)

ZnO(s) + H2O(l ) + 2e –

1

Cathode (reduction):

Ag2O(s) + H2O(l ) + 2e –

2Ag(s) + 2OH – (aq) 1

(b) Silver oxide cell and zinc-carbon cell are

non-rechargeable (primary cells) while

lithium ion cell is rechargeable (secondary

cell). 1

(c) (i) A TV remote control does not require

a large current and is used only

intermittently.

Silver oxide cell is not used because it

does not have the right size. ½

Lithium ion cell is not used because it

is very expensive. ½

Zinc-carbon cell is used because it has

the right size and is inexpensive. 1

(ii) A mobile phone requires a large

and steady current and may be used

continuously for a period of time.

Silver oxide cell is not used because its

current is small. ½

5

conc. H2SO4

heat




Zinc-carbon cell is not used because

it does not provide a steady current,

especially after being used continuously

for some time. ½

Lithium ion cell is used because it can

provide a large and steady current for a

long time. 1

8. (a) Perform a flame test. 1

The colour of the flame is lilac (crimson

through cobalt glass) if K +

is present. 1

(b) Anion Y is SO32–

(aq). ½

Ba2+

(aq) forms white precipitate of

BaSO3(s) with SO32–

(aq). However, the

white precipitate redissolves to give SO2(g)

on the addition of hydrochloric acid.

Ba2+

(aq) + SO32–

(aq) BaSO3(s) ½

BaSO3(s) + 2H+

(aq)

Ba2+

(aq) + H2O(l ) + SO2(g) ½

However, when SO32–

(aq) is first treated

with Cl2(aq), it is oxidized to SO42–

(aq).

Ba2+

(aq) still forms white precipitate of

BaSO4(s) with SO42–

(aq). But BaSO4 does

not redissolve in hydrochloric acid.

Cl2(aq) + H2O(l ) + SO32–

(aq)

2H+(aq) + 2Cl

– (aq) + SO4

2– (aq) 1

Ba2+

(aq) + SO42–

(aq) BaSO4(s) ½

9. Air pollutants due to incomplete combustion:

(Max: 4)

Carbon particulates, unburnt hydrocarbon,

carbon monoxide 1

They are formed due to incomplete combustion

of gasoline in the combustion engine. 1

Harmful effects:

Carbon particulates: cause disease in respiratory

system, lung cancer ½

Unburnt hydrocarbon: carcinogenic ½

Carbon monoxide: forms carboxyhaemoglobin

with haemoglobin in blood. Blocks the transfer

of oxygen to different parts of body. People

may die because of lack of oxygen. 1

7

5

Air pollutants due to high temperature of

the combustion engine: (Max: 2)

Nitrogen oxides, NxOy ½

At high temperature in the combustion engine ,

N2 in air reacts with O2 to give NO, which is

then further converted to NO2.

N2(g) + O

2(g) 2NO(g)

2NO(g) + O2(g) 2NO2(g) ½

(Equations are optional)

Harmful effects:

Generate ph otoch emi cal sm og which is

harmful to respiratory system.

NO2 dissolves in rain water to form acid rain

which corrodes building structure, damages

plants and kills aquatic lives. 1

Effective communication (3 marks)The ability to present ideas in a precise

manner, including the proper use of chemical

terms (this mark should not be awarded to

answers which contain a lot of incorrect /

superfluous material); 1

The ability to present ideas in a systematic

manner (i.e. the answer is easy to follow); 1

The ability to present the answer in paragraph

form and to express ideas using full sentences. 1

6

3

cs chem hkdse mock ans ee

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