| Date | November 2017 | Marks available | 2 | Reference code | 17N.3.sl.TZ0.7 |
| Level | SL | Paper | 3 | Time zone | TZ0 |
| Command term | Describe | Question number | 7 | Adapted from | N/A |
Question
The development of materials with unique properties is critical to advances in industry.
Low density polyethene (LDPE) and high density polyethene (HDPE) are both addition polymers.
Outline two properties a substance should have to be used as liquid-crystal in a liquid-crystal display.
Describe how the structures of LDPE and HDPE affect one mechanical property of the plastics.
One of the two infrared (IR) spectra is that of polyethene and the other of polytetrafluoroethene (PTFE).
Deduce, with a reason, which spectrum is that of PTFE. Infrared data is given in section 26 of the data booklet.
Many plastics used to be incinerated. Deduce an equation for the complete combustion of two repeating units of PVC, (–C2H3Cl–)2.
Markscheme
Any two of:
ability to form a LC phase
chemically stable
«LC phase that is» stable over suitable temperature range
polar
OR
being able to change orientation with applied electric field
rapid switching speed «responds to changes of voltage quickly»
Accept “ability of molecules to transmit light under certain conditions” OR “rodshaped molecules” OR “stable to light/not light sensitive”.
[Max 2 Marks]
branching in LDPE prevents close packing «of chains»
LDPE is more flexible/less rigid
OR
LDPE has lower «tensile» strength
Do not accept “difference in density”.
Award [1 max] for stating “branching in LDPE AND little/no branching in HDPE”.
B AND absence «of absorption of» C–H at 2850–3090 «cm–1»
OR
B AND presence of «absorption of» C–F at 1000–1400 «cm–1»
(–C2H3Cl–)2 (s) + 5O2 (g) → 4CO2 (g) + 2H2O (l) + 2HCl (g)
correct species in reactants and products
balanced
Accept “(–C2H3Cl–)2 (s) + 5.5O2 (g) → 4CO2 (g) + 3H2O (l) + Cl2 (g)”.
Award M2 only if M1 correct.
Examiners report
Syllabus sections
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18M.1.sl.TZ2.12:
The compounds shown below have similar relative molecular masses. What is the correct order of increasing boiling point?
A. CH3COOH < (CH3)2CO < (CH3)2CHOH
B. CH3COOH < (CH3)2CHOH < (CH3)2CO
C. (CH3)2CO < CH3COOH < (CH3)2CHOH
D. (CH3)2CO < (CH3)2CHOH < CH3COOH
- 22M.2.hl.TZ1.5a(ii): Compound A and Compound B are both liquids at room temperature and pressure. Identify the...
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17N.3.hl.TZ0.7c:
Identify the type of intermolecular bonding that is responsible for Kevlar®’s strength.
- 22M.2.sl.TZ1.3c(ii): Identify the strongest force between the molecules of Compound B.
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22M.1.sl.TZ2.9:
Which statement best describes the intramolecular bonding in HCN (l)?
A. Electrostatic attractions between H+ and CN− ions
B. Hydrogen bonding
C. Van der Waals forces and hydrogen bonding
D. Electrostatic attractions between pairs of electrons and positively charged nuclei
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22M.1.sl.TZ2.12:
What is the main interaction between liquid CH4 molecules?
A. London (dispersion) forces
B. Dipole–dipole forces
C. Hydrogen bonding
D. Covalent bonding
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18M.2.sl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
- 17M.1.sl.TZ1.9: A substance has the following properties: What is the most probable structure of this...
- 17M.1.sl.TZ1.24: What is the order of increasing boiling point? A. C4H10 < CH3COOH < CH3CH2CHO <...
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17M.1.sl.TZ2.10:
Which bonds cause the boiling point of water to be significantly greater than that of hydrogen sulfide?
A. London (dispersion)
B. Covalent
C. Ionic
D. Hydrogen
- 19N.3.hl.TZ0.15c: Explain why maltose, C12H22O11, is soluble in water.
- 17N.1.sl.TZ0.11: Which of the following series shows increasing hydrogen bonding with water? A. Propane <...
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17M.2.sl.TZ2.4a.ii:
State why hydrazine has a higher boiling point than dinitrogen tetraoxide.
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17M.3.sl.TZ1.10b.ii:
Explain why the difference in their structures affects their melting points.
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18M.2.hl.TZ1.1e.i:
Suggest one reason why urea is a solid and ammonia a gas at room temperature.
- 21N.1.sl.TZ0.13: Which alcohol is least soluble in water? A. CH3OH B. CH3CH2OH C. CH3CH2CH2OH D. ...
- 18N.1.sl.TZ0.11: Which compound has the highest boiling point? A. CH3CHO B. CH3CH2F C. ...
- 18N.2.hl.TZ0.6d: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
- 17M.1.sl.TZ1.12: Which correctly states the strongest intermolecular forces in the compounds below?
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19M.2.hl.TZ1.2h:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
- 19N.3.sl.TZ0.10c: Explain why maltose, C12H22O11, is soluble in water.
- 18N.2.sl.TZ0.6b: Explain why butanoic acid is a liquid at room temperature while ethylamine is a gas at room...
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19N.2.sl.TZ0.3d(ii):
Explain why the compound C2H6O, produced in (b), has a higher boiling point than compound C2H4O, produced in d(i).
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19M.2.hl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
- 19M.1.sl.TZ1.11: Which describes an ionic compound?
- 19M.1.sl.TZ2.12: Which compound has hydrogen bonds between its molecules? A. CH4 B. CH4O C. CH3Cl D. CH2O
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19M.1.hl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
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19M.2.hl.TZ2.1d(iii):
Explain why product B is water soluble.
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18M.2.hl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
- 16N.3.sl.TZ0.18c: Justify the conclusion that recrystallization increased the purity of the product, by...
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17M.3.sl.TZ1.13d:
The solubility of a vitamin depends on its structure.
Identify the vitamin given in section 35 of the data booklet that is the most soluble in water.
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17M.3.sl.TZ2.16a:
Explain why diamorphine passes more readily than morphine through the blood-brain barrier.
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16N.3.hl.TZ0.22c:
(i) Uranium hexafluoride, UF6, is used in the uranium enrichment process that produces fuel for nuclear reactors.
State the molecular shape of uranium hexafluoride.
(ii) Explain why uranium dioxide, UO2, has a very high melting point whereas uranium hexafluoride vapourises easily into gas.
- 16N.3.sl.TZ0.18b: Suggest why isolation of the crude product involved the addition of ice-cold water.
- 16N.1.sl.TZ0.11: Between which pair of molecules can hydrogen bonding occur? A. CH4 and H2OB. CH3OCH3 and...
- 21N.1.sl.TZ0.12: The following compounds have similar relative molecular masses. What is the order of...
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17N.2.sl.TZ0.2b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group.
- 19N.3.sl.TZ0.9c: Explain why stearic acid has a higher melting point than oleic acid.
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18M.2.sl.TZ1.1e.ii:
Sketch two different hydrogen bonding interactions between ammonia and water.
- 19N.3.hl.TZ0.13b: Explain how the double-helical structure of DNA is stabilized once formed.
- 19N.2.sl.TZ0.6a(iv): Suggest why water vapour deviates significantly from ideal behaviour when the gases are...
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19N.2.hl.TZ0.3d(ii):
Explain why the compound C3H8O, produced in (a)(iv), has a higher boiling point than compound C3H6O, produced in d(i).
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16N.2.sl.TZ0.1c:
Explain why the boiling point of ethane-1,2-diol is significantly greater than that of ethene.
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19M.1.sl.TZ1.9:
What is the order of increasing boiling point?
A. CH3CH2CH2CH3 < CH3CH(OH)CH3 < CH3COCH3 < CH3CO2H
B. CH3CH2CH2CH3 < CH3COCH3 < CH3CH(OH)CH3 < CH3CO2H
C. CH3CO2H < CH3COCH3 < CH3CH(OH)CH3 < CH3CH2CH2CH3
D. CH3CH2CH2CH3 < CH3COCH3 < CH3CO2H < CH3CH(OH)CH3
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18M.1.sl.TZ1.12:
What are the strongest intermolecular forces between molecules of propanone, CH3COCH3, in the liquid phase?
A. London (dispersion) forces
B. Covalent bonding
C. Hydrogen bonding
D. Dipole–dipole forces
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17N.2.hl.TZ0.3b:
Explain why the melting points of the group 1 metals (Li → Cs) decrease down the group whereas the melting points of the group 17 elements (F → I) increase down the group.
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17M.3.sl.TZ2.11:
Suggest, in terms of its structure, why vitamin D is fat-soluble using section 35 of the data booklet.
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17N.3.sl.TZ0.10a:
Explain, at the molecular level, why vitamin D is soluble in fats. Use section 35 of the data booklet.
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17M.3.hl.TZ1.25a:
Suggest why aspirin is slightly soluble in water. Refer to section 37 of the data booklet.
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17M.2.sl.TZ1.2e.ii:
A chloride of titanium, TiCl4, melts at 248 K. Suggest why the melting point is so much lower than that of KCl.
- 19N.3.sl.TZ0.4b: Predict, with a reason, whether isotactic or atactic polypropene has the higher melting point.
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19M.2.sl.TZ2.1b(iii):
Identify the type of interaction that must be overcome when liquid ethyne vaporizes.
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18M.2.hl.TZ2.7e:
Carbon and silicon are elements in group 14.
Explain why CO2 is a gas but SiO2 is a solid at room temperature.
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18M.3.sl.TZ1.1a.i:
Part of this molecule is hydrophilic (bonds readily to water) and part hydrophobic (does not bond readily to water). Draw a circle around all of the hydrophilic part of the molecule.
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19M.2.sl.TZ1.2c:
Suggest how benzoic acid, Mr = 122.13, forms an apparent dimer, Mr = 244.26, when dissolved in a non-polar solvent such as hexane.
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20N.3.sl.TZ0.6b(ii):
State the most significant intermolecular forces in the phospholipid in b(i).
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19M.2.sl.TZ2.1d(iii):
Explain why product B is water soluble.
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22M.2.hl.TZ2.7c:
Suggest why hydrogen chloride, HCl, has a lower boiling point than hydrogen cyanide, HCN.
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17M.3.sl.TZ1.11b.i:
The drain pipe of a kitchen sink can become clogged by fatty acids, such as linoleic acid, C18H32O2, but not by the trisaccharide, raffinose, C18H32O16, containing the same number of carbon atoms.
Explain why raffinose is far more water soluble than linoleic acid.
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16N.3.sl.TZ0.8b:
The table below shows average figures for the percentage fatty acid composition of some common fats and oils.
(i) Deduce, with a reason, which fat or oil from the table above has the lowest iodine number.
(ii) Deduce, with a reason, which fat or oil from the table above is most likely to become rancid when exposed to the air.
(iii) The P/S index of a fat or oil is the ratio of polyunsaturated fat to saturated fat present. It is sometimes used to compare the relative health benefits of different lipids in the diet. Calculate the P/S index of beef fat and soybean oil.
(iv) Suggest why a P/S index of greater than 1 is considered beneficial to health.
(v) Cotton seed oil and corn oil have similar iodine numbers but the melting point of cotton seed oil is higher than that of corn oil. Suggest an explanation in terms of the structure and bonding in these two oils.
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21N.2.hl.TZ0.6c(v):
Suggest a reason that the Winkler Method used to measure biochemical oxygen demand (BOD) must be done at constant temperature.
- 20N.3.hl.TZ0.15b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest...
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17N.3.sl.TZ0.9c:
Explain how the inclusion of carbohydrates in plastics makes them biodegradable.
- 21M.1.sl.TZ2.9: Which compound has the greatest volatility under the same conditions? A. SO2 B. SiO2 C. ...
- 21M.1.sl.TZ2.12: Which is the correct order based on increasing strength? A. covalent bonds < hydrogen...
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20N.3.hl.TZ0.6a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
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17M.3.sl.TZ1.19a:
Suggest why aspirin is slightly soluble in water. Refer to section 37 of the data booklet.
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17M.3.sl.TZ2.8a:
Explain which one of these fatty acids has the highest boiling point.
- 20N.2.sl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
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18M.3.sl.TZ1.1a.ii:
When a small amount of palmitic acid is placed in water it disperses to form a layer on the surface that is only one molecule thick. Explain, in terms of intermolecular forces, why this occurs.
- 20N.3.sl.TZ0.11b: Aspirin crystals are rinsed with water after recrystallization to remove impurities.Suggest...
- 20N.2.hl.TZ0.4c: Discuss, referring to intermolecular forces present, the relative volatility of propanone and...
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20N.3.sl.TZ0.5a(ii):
Proteins are polymers of amino acids.
The mixture is composed of glycine, , and isoleucine, . Their structures can be found in section 33 of the data booklet.
Deduce, referring to relative affinities and , the identity of A1.
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20N.3.sl.TZ0.1a:
Suggest why a non-polar solvent was needed.
- 20N.3.sl.TZ0.1c: Non-polar solvents can be toxic. Suggest a modification to the experiment which allows the...
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22M.2.hl.TZ2.8a(ii):
Explain why C60 and diamond sublime at different temperatures and pressures.
- 21M.2.sl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.2.hl.TZ2.2d: Suggest, giving reasons, the relative volatilities of SCl2 and H2O.
- 21M.1.sl.TZ1.24: Which series is in order of increasing boiling point? A. CH2CH2CH3OH CH3COCH3 ...
