How many VCAA Chemistry questions cover How are organic compounds analysed and used??

AusGrader has 40 VCAA Chemistry questions on How are organic compounds analysed and used?, all with instant AI grading and detailed marking feedback.

VCAA (VCE) Chemistry — How are organic compounds analysed and used? Questions & Answers | AusGrader

Q11

2021

QCAA

Paper 1

1 mark

Q11

1 mark

Enzymes can act as biological catalysts because they

A

can be denatured.

B

lower the activation energy.

C

are sensitive to pH and temperature changes.

D

increase the equilibrium constant for the reaction.

Q5

2024

QCAA

Paper 1

1 mark

Q5

1 mark

Titration is a volumetric analysis method used to

A

measure the pH of an analyte.

B

determine the volume of a titrant.

C

calculate the concentration of an identified solution.

D

prepare a standard solution of known concentration and volume.

Q2

2023

QCAA

Paper 2

3 marks

Q2

3 marks

Compare the structure of $\alpha$ -helix and $\beta$ -pleated sheets in the secondary structure of proteins.

Similarity:

Difference:

Significance:

Similarity: α-helix structure and β-pleated sheets both form H-bonds between two peptide bonds on polypeptide chains.
Difference: α-helix structure contains intra-chain H-bonds while β-pleated sheets contain inter-chain H-bonds.
Significance: α-helix structure produces a regular coiled configuration, whereas β-pleated structure produces a sheet.

Marking Criteria

Descriptor	Marks
identifies both contain H-bonds between peptide bonds in the polypeptide chain	1
identifies α-helix structure contains intra-chain H-bonds while β-pleated sheets contain inter-chain H-bonds	1
explains α-helix is coiled and β-pleated forms sheets	1

Q4

2022

QCAA

Paper 2

8 marks

Q4

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.

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.

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)

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

Q4

2024

QCAA

Paper 2

11 marks

Q4

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.

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

	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.

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

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.

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

VCAA Chemistry How are organic compounds analysed and used?

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