CAPE Chemistry June 2021 Unit 1 Paper 2 - Past Paper Solutions

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Notes relevant to each question follow the solutions.

Question 1 Notes:

1(e)    It could be that CXC has obtained permission to use past exam questions from         

          different examining bodies. Here is the source of this         

          one: https://capechemistry.blogspot.com/2021/12/source-of-cape-chemistry-june-2021-u1.html

Source Of CAPE Chemistry June 2021 U1 P2 Q1(e)

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The question below was seen before, in a 2005 paper from another examining body.

Its mark scheme follows. The only significant difference is "NaoH" in part (iv) of the CAPE exam vs. "NaOH" in the corresponding part of the original question.

CAPE Chemistry June 2009 U1 P1 - Answers And Explanations

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1. B

2. D

3. A

4. D

5. C

6. D

7. B

8. C

9. C

10. B

11. C

12. D

13. C

14. B

15. D

16. A

17. B

18. C

19. B

20. A

21. D

22. B

23. A

24. C

25. B

26. A

27. D

28. C

29. B

30. B - Detailed solution below.

31. A

32. B

33. D

34. B

35. C

36. A

37. D

38. C

39. D

40. A

41. D

42. A

43. D

44. C

45. B

Question # 30 - Detailed Solution

Hydrogen Emission Spectrum

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Theory for this syllabus section, U1 M1 SS 1.7 is broken up into 6 small, and easily manageable parts. After going through them, you will be shown how to use each part to answer the only three CAPE past paper questions on the topic. They are:

1. 2019 U1 P2 Q1(a) - Link to paper
2. 2009 U1 P2 Q1(a) - Link to paper
3. 1999 U1 P2 Q4(a)

____________________________________________________________________________

Part 1

Each element has a characteristic 'fingerprint' line emission spectrum due to its unique arrangement of electrons/electronic configuration. The simplest of these is produced by hydrogen.

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Part 2

• When sufficient energy is supplied to the atoms of hydrogen, an electron is promoted (excited) from a lower energy level to a higher energy level where it becomes unstable.
• The unstable electron will emit the excess energy as radiation and drop back to the lower energy level. Each line in the emission spectrum occurs due to the transitions of electrons moving from a higher energy level to a lower energy level.
• The energy difference between the higher energy level and lower energy level is fixed. The fixed frequencies of each line provide evidence for discrete energy levels.
• Each line represents emitted radiation of a specific wavelength and frequency. Electronic transitions from higher levels to the level, n = 2 give rise to a series of lines known as the Balmer series.

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Part 3

In summary,

• Electrical or thermal energy is passed through and absorbed by a sample of the element.
• Radiation is emitted at certain wavelengths and frequencies.
• The emission spectrum consists of separate lines at the particular frequencies.

____________________________________________________________________________

Part 4

Electron transitions occurring between energy levels associated with the hydrogen emission spectrum are shown in the diagram below.

These transitions produce the hydrogen spectrum which follows.

____________________________________________________________________________

Part 5

The three main regions of the hydrogen emission spectrum, as related to series of lines are:

Lyman series < 400 nm - ultraviolet region.

Balmer series 400 nm to 700 nm - visible region.

Paschen series > 700 nm - infrared region.

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Part 6

To calculate the energy, E, of a quantum of radiation with a frequency (ν) of 4.57 × 10¹⁴ Hz where h = 4 ×10^-13 kJ s mol^-1, we use the formula E = hν as illustrated next.

E = h × ν

   = 4 ×10^-13 kJ s mol^-1 × 4.57 × 10¹⁴ Hz

   = 182.8 kJ mol^-1

____________________________________________________________________________

How To Answer The Past Paper Questions

2019 U1 P2 Q1(a)(i) - Each point in Part 3 is one mark each.

2019 U1 P2 Q1(a)(ii) - Part 5 says it is the Balmer series.

2019 U1 P2 Q1(a)(iii) - First three points and first sentence in 4th point of Part 2 are one mark each.

2019 U1 P2 Q1(a)(iv) - Lines in Balmer series, only, of Part 4 for full marks.

2009 U1 P2 Q1(a)(i) - Answer is found in Part 1.

2009 U1 P2 Q1(a)(ii) - Draw just Balmer part of diagram in Part 5 and label as required.

2009 U1 P2 Q1(a)(iii) - First, second and fourth points in Part 2 will get you full marks.

2009 U1 P2 Q1(a)(iv) - Part 5 says it is in the visible region.

2009 U1 P2 Q1(a)(v) - Part 6 is the solution.

1999 U1 P2 Q4(a)(i) - Same as 2009 U1 P2 Q1(a)(ii)

1999 U1 P2 Q4(a)(ii) - Same as 2019 U1 P2 Q1(a)(ii)

1999 U1 P2 Q4(a)(iii) - All four points in Part 2 for full marks.

Transition Elements

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Exam Tip For 2021

Vanadium is most likely to come in 2021 (last seen 11 years ago) and copper is second most likely (last seen 12 years ago).

Colours and formulae of all species required in CAPE past papers for at least the last 15 years are shown below:

Learn these and you'll be able to handle the following questions:

1. 2019 U1 P2 Q3(d)(iii)
2. 2014 U1 P2 Q3(d)(i)
3. 2010 U1 P2 Q3(a)
4. 2009 U1 P2 Q3(d)
5. 2008 U1 P2 - Resit Q3(b)(i)
6. 2004 U2 P2 - Q2(b)(i)

Past papers from 2005 to 2016 can be found here: Past Paper Booklet

Definitions To Learn

Transition element - A d-block element whose atom has an incomplete d sub-shell or forms at least one stable ion with a partially filled d sub-shell.

Ligand - A molecule, atom or ion that bonds with a central metal atom/ion by donating an electron pair.

These are worth a maximum of 2 marks each and are the answers to:

1. 2019 U1 P2 Q3(d)(i)
2. 2014 U1 P2 Q3(a)
3. 2007 U1 P2 Q6(a)(i)

Difference Between Reactivity Series And Electrochemical Series

Reactivity And Electrochemical Series - Metals

The reactivity series can be used to compare the relative reactivities of different metals. It lists metals in order of general chemical reactivity. Metals generally react by losing electrons to form positive ions. The more readily a metal loses electrons, the more reactive it is – and the greater its strength as a reductant. Metals higher up in the series can reduce the ions of those lower down.

The standard electrode potential of a metal also indicates its strength as a reductant. The more negative the value of the standard electrode potential of a metal, the greater is its strength as a reductant. Hence, you might expect the metal reactivity series and standard electrode potentials to list metals in the same order. However, you must remember that the metal reactivity series is based on observing a range of reactions, such as displacement reactions between solid metals and solid metal oxides. Standard electrode potentials refer specifically to reactions taking place in aqueous solution.

Below is a comparison of the reactivity series with the electrochemical series, which ranks metals according to their standard electrode potentials.

The obvious discrepancy is the relative positions of sodium and calcium. Calcium is a stronger reductant than sodium according to Eo (standard electrode potential) values, but the metal reactivity series suggests that calcium is less reactive than sodium. This discrepancy arises because calcium reacts at a much slower rate, in displacement reactions for example, which in turn happens because two electrons must be removed, not one as for sodium.

Note also that aluminium reacts readily with oxygen in the air, forming a layer of stable aluminium oxide on its surface. This impervious oxide coat often causes aluminium to exhibit lower reactivity than its position in the metal reactivity series indicates.

Group IV Elements - CAPE Chemistry Unit 1

Elements - Structure And Bonding

Main Points To Be Used In Answering A Past Paper Question:

• Down the group there is a change in structure from giant molecular to giant metallic and a change in bonding from covalent to metallic.
• From C to Ge elements exhibit a giant molecular structure.
• Sn and Pb exhibit a giant metallic lattice structure.
• From C to Ge bond length between group IV atoms increases and bond energy/strength decreases.

Elements - Electrical Conductivity

Trend: Going down the group there is a general increase in electrical conductivity.

Reason: Down the group there is a gradual increase in metallic character due to an increase in delocalisation of electrons throughout the structure.

Note: There is an increase in electrical conductivity from C(diamond) to Sn. There is a decrease from Sn to Pb which has never been addressed in the past papers. Hence we use the phrase "general increase" when talking about the trend in electrical conductivity down the group. 

CAPE Chemistry Past Paper Answer - 2016 Unit 2 Paper 2 Question 5

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5(a)(i)

As applied to completely miscible binary systems, this states that the partial pressure of a constituent of a binary mixture at any given temperature is equal to the normal vapor pressure of that constituent at the stated temperature multiplied by the mole fraction of the constituent in the mixture.

5(a)(ii)

1. The interactions between components are similar to those in pure components.

2. There is no volume change on mixing the components.

5(b)(i)

An azeotropic mixture is one which boils or distils without change in composition, and in general it has a boiling point higher or lower than that of any of its pure constituents.

5(b)(ii)

An azeotrope is not a compound because its composition varies with pressure.

5(b)(iii)

Mixture X boils at a temperature T1 and at equilibrium gives off a vapor of composition Y richer in the more volatile component A. Continued distillation produces the azeotrope as the distillate while pure B is the residue.

5(c)

At equilibrium, mass of compound in water = (5-x) g and mass of compound in solvent = x g. 

Partition coefficient = concentration in water/concentration in ether, therefore  ((5-x)/100)/(x/25) = 0.2 i.e. x = 2.8 g.

Unit 1 Module 1 SS 2 Notes

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2.1

State the various forces of attraction between particles.

Ionic bonds, covalent bonds, metallic bonds, van der Waals' forces.

Both bonding (intramolecular) forces and intermolecular forces arise from electrostatic attractions between opposite charges. Bonding forces are due to the attraction between cations and anions (ionic bonding), nuclei and electron pairs (covalent bonding), or metal cations and delocalized valence electrons (metallic bonding). Intermolecular forces, on the other hand, are due to the attraction between molecules as a result of partial charges, or the attraction between ions and molecules. The two types of forces differ in magnitude, and Coulomb's law explains why:

• Bonding forces are relatively strong because they involve larger charges that are closer together.
• Intermolecular forces are relatively weak because they typically involve smaller charges that are farther apart.

Ion-Dipole Forces

When an ion and a nearby polar molecule (dipole) attract each other, an ion-dipole force results. The most important example takes place when an ionic compound dissolves in water. The ions become separated because the attractions between the ions and the oppositely charged poles of the H2O molecules overcome the attractions between the ions themselves.

Dipole-Dipole Forces

When polar molecules lie near one another, as in liquids and solids, their partial charges act as tiny electric fields that orient them and give rise to dipole-dipole forces: the positive pole of one molecule attracts the negative pole of another (diagram below).

Polar molecules and dipole-dipole forces. In a solid or a liquid, the polar molecules are close enough for the partially positive pole of one molecule to attract the partially negative pole of a nearby molecule. The orientation is more orderly in the solid (left) than in the liquid (right) because, at the lower temperatures required for freezing, the average kinetic energy of the particles is lower. (Interparticle spaces are increased for clarity.)

For molecular compounds of approximately the same size and molar mass, the greater the dipole moment, the greater the dipole-dipole forces between the molecules are, and so the more energy it takes to separate them. Consider the boiling points of the compounds in the next diagram. Methyl chloride, for instance, has a smaller dipole moment than acetaldehyde, so less energy is needed to overcome the dipole-dipole forces between its molecules and it boils at a lower temperature.

Dipole moment and boiling point. For compounds of similar molar mass, the boiling point increases with increasing dipole moment. (Note the increasing color intensities in the electron density models.) The greater dipole moment creates stronger dipole-dipole forces, which require higher temperatures to overcome.

The Hydrogen Bond

A special type of dipole-dipole force arises between molecules that have an H atom bonded to a small, highly electronegative atom with lone electron pairs. The most important atoms that fit this description are N,0, and F. The H-N, H-O,
and H- F bonds are very polar, so electron density is withdrawn from H. As a result, the partially positive H of one molecule is attracted to the partially negative lone pair on the N, 0, or F of another molecule, and a hydrogen bond (H bond) forms. Thus, the atom sequence that allows an H bond (dotted line) to form is -B:····H-A-, where both A and B are N, O, or F. Three examples are

The small sizes of N, O, and F are essential to H bonding for two reasons:

1. It makes these atoms so electronegative that their covalently bonded H is
    highly positive.


2. It allows the lone pair on the other N, O, or F to come close to the H.

The Significance of Hydrogen Bonding

Hydrogen bonding has a profound impact in many systems. Here we'll examine one major effect on physical properties and preview its enormous importance in biological systems.

How To Use Red Mud As A Catalyst

Can chemists come up with better uses of mineral resources to make catalysts that are more sustainable? For a growing number of researchers, the answer is yes, and the key is taking advantage of materials that are already out of the ground. Red mud, the noxious by-product of the Bayer process for extracting aluminum from bauxite ore, makes a good case study.

The majority of material processed in mining operations ultimately goes to waste. For every ton of alumina extracted from bauxite, more than a ton of red mud is produced; aluminum mining leaves behind some 120 million metric tons per year of the salty, highly alkaline, heavy-metal-laden material, according to the International Aluminum Institute. Some 4 billion metric tons of the  material is lying about globally, much of it held in retention ponds.

Mining companies have long tried to find ways to recycle the environmentally problematic red mud. It is a classic problem in search of a solution. One approach is neutralizing red mud with seawater or treating it with CO2 or sulfur compounds. The modified materials have been tried as fill for mining and construction, as pigment and filler for bricks and cement, and as a sorbent for water treatment. Others have looked at extracting more aluminum from red mud, or obtaining other useful metals such as sodium, copper, and nickel. But so far there have been few safe and economical large-scale applications.

On a new front, some chemists are trying to go catalytic, focusing on iron oxide, the chief component of red mud. But given the purity and properties of red mud, researchers have found it typically is not an active enough catalyst to compete against existing commercial catalysts. That’s because the mineral composition, particle size, and surface properties are important in developing heterogeneous catalysts. With red mud, finding the right combination is a work in progress.

One early sign of success comes from Foster A. Agblevor of Utah State University’s USTA Bioenergy Center and coworkers in conjunction with Pacific Northwest National Laboratory researchers. They have been testing red mud as a bulk catalyst to replace zeolites in a fluidized-bed reactor to pyrolyze biomass to make crude oil.

The team processes the biocrude oil using a traditional catalytic hydrotreating process to make a gasoline- type fuel and has tested it on a lawn mower or lawn trimmer. “We are able to run an engine on the fuel without difficulty,” Agblevor says. The Utah State researchers have applied for a patent for their process. They are working with catalyst company Nexceris to scale up catalyst production and with Wildland Forestry & Environmental to harvest wood from pinyon-juniper range lands in the western U.S. to scale up biofuel production.

The team is also expanding the scope of using red mud beyond biomass pyrolysis, Agblevor says. The researchers have applied the catalyst to coal gasification, he notes, as well as to a process for catalytic pyrolysis of waste tires for fuel production. Despite raw red mud’s ultimate utility as a catalyst, its story points to other possibilities for recovering metals that have already been extracted and used. For example, industrial processing, the use of consumer goods and medicines, and even the wearing away of jewelry leads to measurable amounts of catalyst metals such as gold, silver, and platinum accumulating at wastewater treatment plants.

Weak Base – Strong Acid Titration Curves

The titration of a weak base (NH3) with a strong acid (HCl) is shown below. Note that the curve has the same shape as the weak acid-strong base curve, but it is inverted. Thus, the regions of the curve have the same features, but the pH decreases throughout the process:

Curve for a weak base-strong acid titration. Titrating 40.00 mL of 0.1000 M NH3 with a solution of 0.1000 M HCl leads to a curve whose shape is the same as that of the weak acid-strong base curve,

but inverted. The midpoint of the buffer region occurs when [NH3] = [NH4+].

Methyl red is a suitable indicator here.

1.  The initial solution is that of a weak base, so the pH starts out above 7.00.

2.  The pH decreases gradually in the buffer region, where significant amounts of base (NH3) and conjugate acid (NH4+) are present. At the midpoint of the buffer region, the pH equals the pKa of the ammonium ion.

3.  After the buffer region, the curve drops vertically to the equivalence point, at which all the NH3 has reacted and the solution contains only NH4+ and Cl-. Note that the pH at the equivalence point is below 7.00 because Cl- does not react with water and NH4+ is acidic:

                                 NH4+(aq) + H2O(l) ⇌  NH3(aq) + H3O+(aq)

4. Beyond the equivalence point, the pH decreases slowly as excess H3O+ is added.

For this titration also, we must be more careful in choosing the indicator than for a strong acid-strong base titration. Phenolphthalein changes colour too soon and too slowly to indicate the equivalence point; but methyl red lies on the steep portion of the curve and straddles the equivalence point, so it is a perfect choice.