hkdse is 2021 panel 3

HKDSE IS 2021

PANEL 3PAPER ONE

ANS.: condensation / polymerisation

ER:

Fair.

Many candidates were unable to recognise from the given chemical equation that the type of reaction involved

was ‘condensation’ or ‘polymerisation’.

A few candidates wrongly proposed ‘hydrolysis’, in which water acts as one of the reactants. However, water is

one of the products formed in the given reaction.

Some wrongly proposed ‘irreversible reaction’, which is an over-generalised answer.

ANS.: (C6H10O5)n + 6n O2 → 6n CO2 + 5n H2O

ER:

The performance was poor. Most candidates were unable to give the correctly balanced chemical

equation as follows:

(C6H10O5)n + 6n O2 → 6n CO2 + 5n H2O

Many of them missed out n in the reaction coefficients on the product side. This suggests that

candidates were weak in balancing chemical equations, and might not have realised that matters

are not created nor destroyed in chemical reactions. That is, the number of each kind of atoms

should be the same on the reactant side and on the product side.

ANS.:

There are hydrogen bonds among water molecules. (1)

Lots of energy is absorbed from the hot object to break these strong bonds to

raise the temperature of water. (1)

Satisfactory performance.

Most candidates correctly stated that the attraction between water molecules is hydrogen

bond, which is a relatively strong type of intermolecular attraction. Some candidates did

not read the question carefully. They gave vague answers, such as the intermolecular

attraction in water is strong dipole-dipole attraction.

Some did not state the type of intermolecular attraction involved but instead gave an

explanation which was erroneously based on the high specific heat capacity of water.

5 × 4200 × (100 − 25) + 1 × 2.26 × 106 (1 M)

= 1575000 + 2260000

= 3835000 J (1 A)

Satisfactory performance.

While most candidates realised that the calculation consisted of two parts, i.e. (1) heating up 5 kg of

water from 25oC to 100oC, and (2) vaporising 1 kg of water at 100oC; many of them erroneously

perceived that only 4 kg of water was heated up in (1).

The unit of energy should be J or kJ. Some candidates gave the wrong unit, J kg−1.

ANS.: The white powder may be magnesium oxide / MgO (1)

Fair performance. The burning of magnesium in air gives magnesium oxide as the major product.

However, some candidates wrongly stated that magnesium salt, such as magnesium chloride or magnesium

sulphate, is formed.

In addition, a few candidates simply wrote ‘oxide’ as the answer. Candidates who failed to state that

magnesium oxide is formed were unable to give the correct electron diagram.

The burning magnesium reacts with the heated water / steam to give out lots

of hydrogen gas, (1)

which would cause explosion / which is also a combustible. (1)

The performance was very poor.

Many candidates did not read the question carefully. They appeared to have overlooked the words ‘with reference to

the reaction’ and did not mention in their answers that at high temperatures, magnesium reacts with steam to form

hydrogen, which is combustible/explosive. Some candidates gave vague answers such as ‘magnesium would react

more vigorously with water at high temperatures’.

The crust formed by the dry powder cuts off oxygen supply from air, which is required for supporting

combustion. (1)

OR

The crust formed by the dry powder dissipates the heat of the burning magnesium to lower its

temperature to below that required to sustain the fire. (1)

ER:

Well answered. Many candidates knew that the crust formed can cut off oxygen supply, which is

required for the burning to be sustained.

As shown in the table, the relative yield of barley decreases with pH, i.e. the

yield decreases with increasing acidity.

Satisfactory performance. However, some candidates simply stated that the relative yield of

barley is zero at pH 4.7, without mentioning that the relative yield of barley decreases with pH.

oat

Excellent performance.

Very good performance. Most candidates were able to recognise that the

reaction involved is a neutralisation.

Lime neutralises / reacts with the acidic substances in the soil.

(6.8 − 5.7) × 3 (1 M)

= 3.3 tons per hectare (1 A)

Satisfactory performance.

Most candidates knew how to do the calculation. Candidates were expected to give their answers

in ‘tons per hectare’. Some candidates gave their answers in ‘tons’ and were unable to get the

mark for accuracy.

Nitrogen-fixing bacteria residing in the root nodules (1) convert atmospheric nitrogen into

ammonium compounds to support the growth of soyabean. (1)

With a smaller number of root nodules in soyabean plants growing at soil pH 4.7, nitrogen fixation

activity is lower than that growing at pH 6.8. This results in a lower yield than that grown at pH 6.8.

(1)

Fair performance. This question covered the following three key points: (1)

the given information showed that the relative yield of soyabean is higher at

pH 6.8 than at pH 4.7; (2) this difference is due to the nitrogen-fixing

bacteria residing in the root nodules of soyabean; (3) a description of how

ammonium compounds are provided by the nitrogen fixation activity of these

bacteria help soyabean growth. Most candidates were able to mention only

one to two of these key points.

Chlorine atom has 7 electrons in its outermost shell. (1)

The sharing of an electron (i.e. forming a covalent bond) with another Cl atom

allows both atoms to achieve a stable octet electron arrangement. (1)

Satisfactory performance.

Most candidates demonstrated an understanding that a chlorine atom has 7 electrons in its outermost

shell and is unstable. They then proceeded to say that chlorine atoms will form diatomic Cl2molecules to attain stability.

This was a very generalised saying as atoms with 5 outermost electrons (e.g. N) and those with 6

outermost electrons (e.g. O) can also form diatomic molecules. Candidates should be more specific in

their answers to show an understanding that both Cl atoms could achieve a stable octet by sharing one

electron with each other.

ANS:

Let x be the relative abundance of Cl-35.

35.5 = 35x + 37(1− x) (1 M)

x = (37 −35.5) ÷ 2

= 0.75 (or 75% ) (1A)

Fair. Some candidates did not know how to calculate the relative abundance of Cl-35. For those

who did, some wrongly gave the answer as 75. They failed to realise that the relative

abundance of an isotope is always smaller than 1. It this case, it should be 75% or 0.75.

Beta decay / decay (1)

Fair. Most candidates knew that decay was involved, but very few of them were able to

write the equation for the decay correctly. They overlooked that in writing nuclear equations,

the mass number and the atomic number on the two sides of the equation must be balanced.

In decays, an electron is emitted from the atomic nucleus, i.e. −10𝑒 should appear on the

right-hand side of the equation. Some candidates wrongly put −10𝑒 on the left-hand side,

which suggests that they did not have a good understanding of the concepts of radioactive

decay.

For geological dating (1)

Its long half-life makes it a possible agent for geological dating. (1)

Satisfactory performance. Candidates who demonstrated knowledge that Cl-36 has a long

half-life were also able to correctly point out that it can be used for geological dating.

HKDSE IS 2021

PANEL 3PAPER TWO

X is chlorine / Cl2 (1)

NaClO(aq) + 2HCl(aq) → Cl2(g) + NaCl(aq) + H2O(l) (1)

Fair. Very few candidates were able to write the balanced

equation correctly. The weaker candidates failed to state that

chlorine gas is produced.

a broken line curve showing:

a steeper slope at the initial stage (1)

a final volume at 2 times that of the original one (1)

With a higher concentration of HCl(aq), there are more particles / ions per unit volume. This allows a higher collision frequency and hence a higher reaction rate will be achieved. (1)

With a doubled concentration of HCl(aq), there are 2 times of particles / ions in the same volume of HCl(aq) used. Hence, a doubled volume of the product (Cl2) is formed when the reaction is completed. (1)

Volume ofX collected

Time

Fair. The correct sketch should show (1) an increase in the initial rate of the reaction,

and (2) the total volume of Cl2(g) liberated in the repeated experiment should be double

that in the original experiment. While most candidates were able to show (1) in their

sketch, some failed to show (2). The explanation part should be of book work and the

concepts involved should have been tested from time to time. Yet, quite a number of

candidates were unable to produce a well-reasoned explanation.

As a catalyst, the palladium provides an alternative reaction pathway (1)

with a lower activation energy (1) to speed up the reaction.

Fair. Catalysts work by providing an alternative reaction pathway with a lower activation energy. Candidates’

performance demonstrated a common misconception – a catalyst works by lowering the activation energy of

the reaction. Some candidates wrote that palladium can shorten the time for the reaction to attain equilibrium.

They did not realise that reaction (I), which involves a palladium catalyst, is an irreversible reaction.

As catalyst is regenerated / is not consumed in the reaction and can be

reused, (1)

only a small amount of catalyst is required for the reaction. (1)

Fair. This question tests students’ understanding of the importance of catalysts in industrial processes.

Candidates should realise that a catalyst, besides being able to speed up a reaction, is regenerated at the end of

the reaction. Thus, a small amount of catalyst can be used for producing a large amount of product. This

explains the cost-effectiveness of using catalysts in chemical industrial processes. Yet, many candidates failed

to demonstrate such an understanding.

addition (polymerisation) (1)

The performance was poor.

This question tests candidates’ ability to apply their understanding of polymerisation to an unfamiliar

situation. Although polyvinyl acetate (PVAc) is not covered in the curriculum, it was expected that

candidates would be able to deduce, from the given structure, that PVAc is an addition polymer and the

structure of its monomer. Unfortunately, most candidates were unable to do so.

The added plasticiser will embed between the polymer chains. (1)

This will weaken the attraction between the polymer chains. (1)

The polymer chains can slide over each other more easily and thus will make the

PVAc emulsion less viscous.

The performance was poor. Very few candidates were able to demonstrate an understanding that the addition of a

plasticiser into a polymer gives a semi-solid, and that the presence of plasticiser molecules among the polymeric

molecules leads to a weakening of the intermolecular attraction of the latter.

The performance was poor. The unsatisfactory performance of candidates showed that they were weak in

balancing a chemical equation and did not know what type of chemical reaction is involved. Candidates who

realised the reaction is a hydrolysis were able to deduce that Y is ethanoic acid.

When borax solution is added, B(OH)4– ions bind the PVA molecules together /

bind to PVA molecules by hydrogen bonds, (1)

the PVA molecules are less free to move and form a semi-solid / a gel that

cannot flow as free as the PVA solution. (1)

Fair. Quite a number of candidates realised that hydrogen bonds are formed between PVA

molecules and B(OH)4− ions. However, many of them were unable to correlate this change at

the molecular level to the formation of a semi-solid, the latter being a macroscopic property.

The acid reacts with OH– ions and so the concentration of OH– ions in the

borax solution decreases. (1)

The equilibrium will shift to the right and thus the amount of B(OH)4– ions

decreases. (1)

As the amount of B(OH)4– ions decreases, there are fewer B(OH)4

– ions to bind

the PVA molecules together. (1)

Thus, the slime will become softer.

The performance was poor. Two chemical equilibria are involved in this scenario, one of them being B(OH)4−⇌ B(OH)3

+ OH− and the other being shown in Figure 3. Many candidates knew that adding an acid to slime would lead to a

decrease in concentration of OH− ions. But their explanation in terms of chemical equilibrium was poorly presented.

Many simply wrote the equilibrium position shifted to the right or to the left, without specifying which chemical

equilibrium they were referring to. Overall, most candidates failed to demonstrate a good grasp of the concepts of

chemical equilibrium.

THANK YOU!!