How many NESA Chemistry questions cover Polymers?

AusGrader has 73 NESA Chemistry questions on Polymers, all with instant AI grading and detailed marking feedback.

Describe the structure of amylose and amylopectin by completing the table. | | Amylose | Amylopectin | | :--- | :--- | :--- | | **Monomer** | | | | **Glycosidic linkage** | | | | **Chain structure** | | | | **Shape** | | |

| | Amylose | Amylopectin | |---|---|---| | **Monomer** | $\alpha$-glucose | $\alpha$-glucose | | **Glycosidic bonds** | 1,4-glycosidic bonds | both 1,4 and 1,6-glycosidic bonds | | **Chain structure** | Linear | Branched | | **Shape** | Helix | Spherical |

NESA Chemistry Polymers

15 sample questions with marking guides and sample answers · Avg. score: 51.1%

2023

QCAA

Paper 2

2 marks

Polylactic acid (PLA) and low-density polyethylene (LDPE) are both used to produce plastic wrapping film.

Plastic	Composition	Density (g/cm $^3$ )	Tensile stress (MPa)	Elongation (%)	Degradation rate
PLA	plant-based	1.24	60	6	slow
LDPE	petrochemical-based	0.92	12	148	none

Analyse the data to discuss one advantage and one disadvantage of using PLA rather than LDPE to produce plastic wrapping film.

Advantage:

Disadvantage:

Reveal Answer

Advantage: An advantage is that PLA is plant based, therefore it uses (renewable) natural resources while LDPE is produced from non-renewable fossil fuels.
Disadvantage: PLA has less % elongation than LDPE, therefore PLA would stretch less.

Marking Criteria

Descriptor	Marks
identifies an advantage of using PLA using data	1
identifies a disadvantage of using PLA using data	1

2021

SCSA

1 mark

Which of the following characteristics influence how a particular polymer might be used?

(i) The amount of cross-linking between the hydrogen atoms in the polymer.
(ii) The length of the carbon chains in the polymer.
(iii) The functional groups present in the monomer used to synthesise the polymer.
(iv) The melting point of the polymer.

ii, iii and iv only

i and ii only

ii and iii only

i, ii and iv only

Reveal Answer

ii, iii and iv only

Correct Answer

This is correct because chain length, functional groups, and melting point all determine a polymer's physical and chemical properties, dictating its applications. Statement (i) is incorrect because cross-linking involves covalent bonds between polymer chains, not between hydrogen atoms (which can only form one bond).

i and ii only

This is incorrect because statement (i) is false; hydrogen atoms cannot form cross-links as they only form one bond. Additionally, it omits valid characteristics (iii) and (iv).

ii and iii only

This is incorrect because it omits statement (iv). The melting point is a crucial characteristic that determines the temperature range in which a polymer can be used without losing its structural integrity.

i, ii and iv only

This is incorrect because statement (i) is false. Cross-linking occurs between polymer chains via covalent bonds involving carbon or other elements, not hydrogen atoms.

Q23

2024

SCSA

1 mark

Q23

1 mark

Which of the following is not a use of polytetrafluoroethene?

non-stick frypans

electrical wires

coatings for tanks

lubricants

Reveal Answer

non-stick frypans

Incorrect. Polytetrafluoroethene (PTFE), commonly known as Teflon, is widely used to make non-stick frypans due to its high heat resistance and low coefficient of friction.

electrical wires

Correct Answer

Correct. General electrical wires are typically insulated with polyvinyl chloride (PVC) or polyethylene, making this the correct answer for what is not a primary use of PTFE.

coatings for tanks

Incorrect. PTFE is highly unreactive and resistant to corrosive chemicals, making it an ideal protective coating for industrial chemical storage tanks.

lubricants

Incorrect. Because of its extremely low coefficient of friction, PTFE is frequently used as a dry lubricant to reduce wear between moving surfaces.

Q22

2024

QCAA

Paper 1

3 marks

Q22a

2 marks

Describe the type of reaction that occurs when amino acid monomers are joined to form polypeptides.

Reveal Answer

A condensation reaction occurs when amino acid monomers are joined to form a peptide bond and water is removed.

Marking Criteria

Descriptor	Marks
Identifies condensation reaction	1
Describes that water is removed when the amino acids are joined to form polypeptides	1

Q22b

1 mark

Identify the bond formed when amino acid monomers join to form a dipeptide.

Reveal Answer

A peptide bond is formed.

Marking Criteria

Descriptor	Marks
Identifies peptide bond	1

Q22

2023

QCAA

Paper 1

4 marks

Q22a

2 marks

Write a balanced chemical equation to describe how polytetrafluorethene (PTFE) is produced from its monomer.

Reveal Answer

$nF_2C=CF_2 \rightarrow \left[ F_2C-CF_2 \right]_n$

Marking Criteria

Descriptor	Marks
describes formulas for tetrafluorethene monomer and polytetrafluorethene polymer	1
provides a balanced equation	1

Q22b

2 marks

Determine whether polytetrafluorethene is an addition or condensation polymer. Explain your reasoning.

Reveal Answer

Addition polymer
The double bond in tetrafluorethene is broken to allow the monomers to join.

Marking Criteria

Descriptor	Marks
identifies addition polymer	1
explains double bond in monomer is broken to allow the formation of polymer	1

Q16

2025

NESA

1 mark

Q16

1 mark

A single straight strand of polyester was produced through a condensation reaction of 1000 molecules of 3-hydroxypropanoic acid, HOCH $_2$ CH $_2$ COOH.

What is the approximate molar mass of the strand (in g mol $^{-1}$ )?

72 062

72 080

90 060

90 078

Reveal Answer

72 062

This incorrectly assumes 1000 water molecules are lost. A straight chain of 1000 monomers forms 999 bonds, so only 999 water molecules are lost during the condensation reaction.

72 080

Correct Answer

The molar mass is calculated by taking the mass of 1000 monomers ( $1000 \times 90.078$ g mol $^{-1}$ ) and subtracting the mass of the 999 water molecules lost during condensation ( $999 \times 18.016$ g mol $^{-1}$ ), yielding approximately 72 080 g mol $^{-1}$ .

90 060

This subtracts the mass of only a single water molecule from the total mass of 1000 monomers, rather than subtracting one water molecule per ester bond formed.

90 078

This is simply the mass of 1000 unreacted monomers ( $1000 \times 90.078$ g mol $^{-1}$ ) and fails to account for the mass of water lost during the condensation reaction.

2024

QCAA

Paper 1

1 mark

Polytetrafluorethene (PTFE) has a higher melting point than polypropene (PP) due to the

C–F bonds being non-reactive.

fluorine atoms forming stable C–F covalent bonds.

dispersion forces between closely packed fluorocarbon chains.

dipole–dipole interaction between fluorine and carbon atoms within the fluorocarbon chain.

Reveal Answer

C–F bonds being non-reactive.

Chemical reactivity refers to the stability of the molecule against chemical changes, whereas the melting point is a physical property determined by the strength of intermolecular forces.

fluorine atoms forming stable C–F covalent bonds.

The strength of covalent bonds determines the polymer's thermal stability (resistance to decomposition), but the melting point is governed by the intermolecular forces holding separate chains together.

dispersion forces between closely packed fluorocarbon chains.

Correct Answer

PTFE chains are linear and rigid, allowing them to pack very closely. This dense packing results in strong cumulative London dispersion forces between the chains, requiring significant energy to overcome.

dipole–dipole interaction between fluorine and carbon atoms within the fluorocarbon chain.

Melting involves overcoming forces between different polymer chains (intermolecular), whereas interactions between atoms within the same chain (intramolecular) determine the chain's structure and stiffness.

Q22

2024

SCSA

1 mark

Q22

1 mark

Which of the following does not contribute to the specific properties and uses of polymers in plastics? The

length of the polymer chains.

amount of cross-linking between the chains.

types of intermolecular forces that exist between the chains.

process used to produce the polymer.

Reveal Answer

length of the polymer chains.

The length of polymer chains directly affects the strength and melting point of the plastic due to increased intermolecular interactions as the chains get longer.

amount of cross-linking between the chains.

The amount of cross-linking between chains significantly alters a polymer's rigidity and heat resistance, which determines whether a plastic is thermoplastic or thermosetting.

types of intermolecular forces that exist between the chains.

Intermolecular forces, such as hydrogen bonding or van der Waals forces, dictate the tensile strength, flexibility, and melting point of the polymer.

process used to produce the polymer.

Correct Answer

While the production process can influence the final structure, it is the resulting structural features (like chain length, cross-linking, and intermolecular forces), not the manufacturing process itself, that directly determine the polymer's properties.

2022

QCAA

Paper 1

1 mark

Identify the type of reaction that occurs when ethene undergoes polymerisation to form polyethene.

addition

elimination

substitution

condensation

Reveal Answer

addition

Correct Answer

This is an addition reaction because the carbon-carbon double bonds ( $C=C$ ) in the ethene monomers break to form single bonds, allowing the molecules to add together into a long chain without any by-products.

elimination

Elimination reactions involve removing atoms to form a double or triple bond, which is the opposite of this process where double bonds are consumed to form single bonds.

substitution

Substitution reactions involve replacing one atom or group with another, whereas this reaction involves the joining of monomers through the rearrangement of bonds.

condensation

Condensation polymerization involves the joining of monomers with the release of a small molecule (like water), but the polymerization of ethene produces no by-products.

2022

VCAA

1 mark

In a protein, hydrogen bonding takes place during the formation of the

secondary, tertiary and quaternary structures only.

primary, secondary and tertiary structures only.

tertiary and quaternary structures only.

primary and tertiary structures only.

Reveal Answer

secondary, tertiary and quaternary structures only.

Correct Answer

Hydrogen bonds are essential for stabilizing the secondary structure (alpha helices and beta sheets), tertiary structure (side-chain interactions), and quaternary structure (interactions between multiple polypeptide subunits).

primary, secondary and tertiary structures only.

The primary structure is formed exclusively by covalent peptide bonds linking amino acids together, not by hydrogen bonds.

tertiary and quaternary structures only.

While hydrogen bonds do occur in tertiary and quaternary structures, this option incorrectly excludes the secondary structure, which is primarily defined and stabilized by hydrogen bonds.

primary and tertiary structures only.

The primary structure does not involve hydrogen bonding, and this option fails to include the secondary and quaternary structures where hydrogen bonding is a key stabilizing force.

Q29

2024

VCAA

1 mark

Q29

1 mark

A reaction between sulfur atoms in proteins

contributes to the quaternary structure through ionic bonding.

contributes to the secondary structure through hydrogen bonding.

leads to the formation of covalent bonds, and contributes to the tertiary structure.

leads to the formation of disulfide linkages, and contributes to the primary structure.

Reveal Answer

contributes to the quaternary structure through ionic bonding.

The reaction between sulfur atoms forms covalent disulfide bonds, not ionic bonds.

contributes to the secondary structure through hydrogen bonding.

Secondary structure is primarily stabilized by hydrogen bonds between the amino acid backbone, not by reactions between sulfur atoms.

leads to the formation of covalent bonds, and contributes to the tertiary structure.

Correct Answer

Sulfur atoms in cysteine residues react to form covalent disulfide bonds, which play a key role in stabilizing the 3D folding, or tertiary structure, of proteins.

leads to the formation of disulfide linkages, and contributes to the primary structure.

While the reaction does lead to disulfide linkages, these contribute to the tertiary or quaternary structure, not the primary structure (which is simply the linear sequence of amino acids).

2024

QCAA

Paper 1

1 mark

Identify which polymer contains a carbonyl (C=O) group.

polyester

polyethene

polypropene

polytetrafluorethene

Reveal Answer

polyester

Correct Answer

Polyesters are formed through condensation polymerization and contain ester linkages ( $-COO-$ ), which inherently include a carbonyl group ( $C=O$ ).

polyethene

Polyethene is a hydrocarbon polymer consisting only of carbon and hydrogen chains ( $-CH_2-CH_2-$ ), so it lacks oxygen atoms and carbonyl groups.

polypropene

Polypropene is a hydrocarbon polymer made solely of carbon and hydrogen atoms, meaning it does not contain the oxygen required for a carbonyl group.

polytetrafluorethene

Polytetrafluoroethene (PTFE) consists of a carbon backbone with fluorine atoms ( $-CF_2-CF_2-$ ) and contains no oxygen or carbonyl groups.

Q29

2020

VCAA

1 mark

Q29

1 mark

Which of the following combinations of bonds can be broken during the breakdown of a protein that a person has eaten?

covalent bonds in the secondary structure and hydrogen bonds in the primary structure

covalent bonds in the tertiary structure and hydrogen bonds in the secondary structure

covalent bonds in the secondary structure and hydrogen bonds in the tertiary structure

covalent bonds in the quaternary structure and hydrogen bonds in the primary structure

Reveal Answer

covalent bonds in the secondary structure and hydrogen bonds in the primary structure

This is incorrect because secondary structure is stabilized by hydrogen bonds, not covalent bonds, and primary structure is stabilized by covalent peptide bonds, not hydrogen bonds.

covalent bonds in the tertiary structure and hydrogen bonds in the secondary structure

Correct Answer

This is correct because tertiary structure can be stabilized by covalent disulfide bonds, and secondary structure is stabilized by hydrogen bonds. Both of these bonds are broken down during protein digestion and denaturation.

covalent bonds in the secondary structure and hydrogen bonds in the tertiary structure

This is incorrect because secondary structure is formed and stabilized by hydrogen bonds between the amino acid backbone, not covalent bonds.

covalent bonds in the quaternary structure and hydrogen bonds in the primary structure

This is incorrect because the primary structure of a protein consists of a sequence of amino acids linked by covalent peptide bonds, not hydrogen bonds.

2022

QCAA

Paper 2

8 marks

Bioethanol is a renewable energy source made from biomasses such as starch and cellulosic materials. The two-step process for the conversion of starch and cellulose to bioethanol is shown.

Process	Step 1	Step 2	Conversion to glucose	Production process
Starch	Enzymatic hydrolysis ( $\alpha$ -amylase) of starch biomass to form glucose	Fermentation of glucose to form bioethanol (yeast)	Easier	Faster
Cellulose	Acid hydrolysis ( $H_2SO_4(aq)$ at 320 $^{\circ}$ C and 25 MPa) of cellulose biomass to form glucose	Fermentation of glucose to form bioethanol (yeast)	Harder	Slower

Q4a

2 marks

Identify why it is important to control the temperature during the fermentation process to produce bioethanol.

Reveal Answer

The fermentation process requires yeast as a catalyst.
Yeast is temperature sensitive.

Marking Criteria

Descriptor	Marks
Identifies fermentation requires yeast as a catalyst	1
Identifies that yeast is temperature-sensitive	1

Q4b

3 marks

Explain why cellulose is harder to convert to glucose than starch.

Reveal Answer

Cellulose is a linear polymer. The $\beta$ -glucose monomers in cellulose can pack closely together. This increases hydrogen bonding between adjacent chains, which reduces interactions with water (solvents) and makes hydrolysis of cellulose more difficult than starch.

Marking Criteria

Descriptor	Marks
Identifies cellulose is a linear polymer	1
Identifies monomers can pack closely together	1
Explains increased H-bonding between adjacent chains makes hydrolysis of cellulose more difficult	1

Q4c

3 marks

After 48 hours of fermentation, a 15% w/v glucose solution produces $37.5\ \text{g L}^{-1}$ of ethanol. Calculate the percentage yield of ethanol. Show your working. (to one decimal place)

Reveal Answer

Moles $C_6H_{12}O_6 = \frac{150}{180.18} = 0.833$

Moles $CH_3CH_2OH = 0.83 \times 2 = 1.67$

Mass $CH_3CH_2OH = 1.67 \times 46.08 = 76.72$ g

Ethanol yield = $\frac{37.5}{\text{theoretical yield}} = \frac{37.5}{76.72} = 48.9\%$

Marking Criteria

Descriptor	Marks
Determines 150 g glucose can produce 1.67 M of ethanol	1
Calculates theoretical mass of ethanol as 76.72 g	1
Calculates % yield of ethanol is 48.9%	1

2024

QCAA

Paper 2

11 marks

Bioethanol can be synthesised from plants rich in starch. Amylose and amylopectin in the starch are converted to glucose, which then undergoes fermentation to produce ethanol.

$(\text{C}_6\text{H}_{10}\text{O}_5)_x \xrightarrow{\text{enzymes}} \text{C}_6\text{H}_{12}\text{O}_6 \xrightarrow{\text{yeast}} 2(\text{C}_2\text{H}_5\text{OH}) + 2\text{CO}_2$
(starch) $\hspace{2cm}$ (glucose) $\hspace{2cm}$ (bioethanol)

Q4a

2 marks

Explain the role of enzymes in converting the amylose and amylopectin in starch to glucose.

Reveal Answer

Enzymes convert starch into glucose via hydrolysis that breaks the glycosidic bond.

Marking Criteria

Descriptor	Marks
Identifies that enzymes break glycosidic bonds in starch to form glucose	1
Explains that enzymes convert starch to glucose via hydrolysis	1

Q4b

4 marks

Describe the structure of amylose and amylopectin by completing the table.

	Amylose	Amylopectin
Monomer
Glycosidic linkage
Chain structure
Shape

Reveal Answer

	Amylose	Amylopectin
Monomer	$\alpha$ -glucose	$\alpha$ -glucose
Glycosidic bonds	1,4-glycosidic bonds	both 1,4 and 1,6-glycosidic bonds
Chain structure	Linear	Branched
Shape	Helix	Spherical

Marking Criteria

Descriptor	Marks
Describes that the monomer for amylose and amylopectin is α-glucose	1
Describes that amylose contains 1,4-glycosidic bonds and amylopectin contains 1,4 and 1,6-glycosidic bonds	1
Describes that amylose has a straight chain structure and that amylopectin has a branched chain structure	1
Describes that amylose is helical in shape and that amylopectin forms spheres	1

Q4c

3 marks

Determine whether the fermentation of glucose to bioethanol is a redox reaction. Explain your reasoning.

Reveal Answer

$0 +1 -2 \quad -2 +1 -2 +1 +4-2$
$C_6H_{12}O_6 \rightarrow 2(C_2H_5OH) + 2CO_2$

Carbon in glucose is oxidised to $CO_2$ . Oxidation number increases from 0 to +4.

Carbon in glucose is reduced to $C_2H_5OH$ . Oxidation number decreases from 0 to -2.

Marking Criteria

Descriptor	Marks
Determines fermentation is a redox reaction	1
Explains carbon in glucose is oxidised to CO₂; oxidation number increases from 0 to +4	1
Explains carbon in glucose is reduced to C₂H₅OH; oxidation number decreases from 0 to –2	1

Q4d

2 marks

Calculate the atom economy for the fermentation of glucose to bioethanol. Show your working.

Reveal Answer

Molar mass glucose = 180
Molar mass ethanol = 2(46)
(C = 12, H = 1, O = 16) = 92

atom economy $= \frac{92}{180} \times \frac{100}{1} = 51\%$

Marking Criteria

Descriptor	Marks
Determines the molar mass of glucose is 180 and molar mass of ethanol is 92	1
Calculates atom economy	1

NESA Chemistry Polymers

Frequently Asked Questions

Ready to practise NESA Chemistry?

NESA Chemistry Polymers

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Sample Answer

Frequently Asked Questions

Ready to practise NESA Chemistry?