Danho
ZIMSEC A Level · J2023

Chemistry Paper 2 June 2023

Questions
43
Total marks
60
Time allowed
75 min

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Questions
43
Pass mark
26
Sit this paper

Answer every question in the printed order, get marked at the end, then see the answers.

The questions

Question 101

[1 marks]kinetic theory and chemical equilibrium
An assumption of the kinetic theory of gases is that
  1. Athe particles occupy most of the volume of the container
  2. Ball the particles move with the same speed
  3. Cthe particles attract one another strongly
  4. Dthe collisions between particles are perfectly elastic

Question 102

[1 marks]kinetic theory and chemical equilibrium
0.16 g of a liquid Y gave 46.00 cm346.00\ cm^3 of vapour at 100 °C and 1.02×1051.02 \times 10^5 Pa. The relative molecular mass of Y is
  1. A53
  2. B88
  3. C106
  4. D212

Question 103

[1 marks]kinetic theory and chemical equilibrium
For 2HBr(g)⇌H2(g)+Br2(g)2HBr(g) \rightleftharpoons H_2(g) + Br_2(g) the equilibrium mole fractions are HBr 0.40, H2H_2 0.03 and Br2Br_2 0.03. The value of KpK_p is
  1. A0.0023
  2. B0.0056
  3. C0.075
  4. D0.150

Question 104

[2 marks]kinetic theory and chemical equilibrium
In liquid Y, the particles are arranged
  1. Aclose together but in a disordered, loosely-packed arrangement, held by intermolecular forces that allow them to move past one another.
  2. Bfar apart in a completely random arrangement, with negligible forces acting between them at all.
  3. Cclose together in a fixed, highly ordered lattice, held rigidly in place by strong intermolecular forces that prevent any movement at all.
  4. Din a regular, repeating crystalline pattern, identical in every way to the solid state.

Question 105

[2 marks]kinetic theory and chemical equilibrium
The KpK_p expression for 2HBr(g)⇌H2(g)+Br2(g)2HBr(g) \rightleftharpoons H_2(g) + Br_2(g) is

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Question 106

[2 marks]kinetic theory and chemical equilibrium
Since the decomposition of HBr is endothermic, increasing the temperature shifts the equilibrium position

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Question 107

[1 marks]kinetic theory and chemical equilibrium
State a second assumption of the kinetic theory of gases, other than that collisions between particles are perfectly elastic.

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Question 201

[1 marks]electron configuration, redox and electrochemistry
The electronic configuration of the S2−S^{2-} ion is
  1. A1s22s22p63s21s^22s^22p^63s^2
  2. B1s22s22p63s23p41s^22s^22p^63s^23p^4
  3. C1s22s22p63s23p61s^22s^22p^63s^23p^6
  4. D1s22s22p63s23p63d21s^22s^22p^63s^23p^63d^2

Question 202

[1 marks]electron configuration, redox and electrochemistry
In the reaction Sn+4HNO3→SnO2+4NO2+2H2OSn + 4HNO_3 \rightarrow SnO_2 + 4NO_2 + 2H_2O, nitrogen
  1. Ais oxidised from +4+4 to +5+5
  2. Bis reduced from +5+5 to +4+4
  3. Cis reduced from +5+5 to +2+2
  4. Ddoes not change its oxidation state

Question 203

[1 marks]electron configuration, redox and electrochemistry
Given Eθ(H2O2/H2O)=+1.77 VE^\theta(H_2O_2/H_2O) = +1.77\ V and Eθ(I2/I−)=+0.54 VE^\theta(I_2/I^-) = +0.54\ V, the cell e.m.f. and the feasibility of the reaction between hydrogen peroxide and iodide ions are
  1. A+1.23 V+1.23\ V; the reaction is not feasible
  2. B−1.23 V-1.23\ V; the reaction is feasible
  3. C+2.31 V+2.31\ V; the reaction is feasible
  4. D+1.23 V+1.23\ V; the reaction is feasible

Question 204

[2 marks]electron configuration, redox and electrochemistry
The shapes of the s and p atomic orbitals are, respectively,
  1. Acubic, and spherical.
  2. Bspherical, and dumbbell-shaped (two lobes).
  3. Cdumbbell-shaped (two lobes), and spherical.
  4. Dboth spherical, differing from each other only in radius.

Question 205

[1 marks]electron configuration, redox and electrochemistry
The type of electrode appropriate for the standard cell formed from hydrogen peroxide and sodium iodide is

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Question 206

[2 marks]electron configuration, redox and electrochemistry
The half-equation for the reduction of hydrogen peroxide (Eθ=+1.77E^\theta = +1.77 V) is

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Question 207

[2 marks]electron configuration, redox and electrochemistry
The half-equation for the I2/I−I_2/I^- electrode (Eθ=+0.54E^\theta = +0.54 V), written as a reduction, is

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Question 301

[1 marks]bonding, periodicity and contact process
Sodium chloride is a solid at room temperature whereas SCl2SCl_2 is a liquid because sodium chloride
  1. Ahas stronger covalent bonds within its molecules
  2. Bhas a much higher relative formula mass
  3. Cis a giant ionic lattice with strong electrostatic forces
  4. Dcontains hydrogen bonds

Question 302

[1 marks]bonding, periodicity and contact process
Silicon dioxide is insoluble in water because it
  1. Areacts with water to form a gas
  2. Bhas a giant covalent structure with strong bonds throughout
  3. Cis denser than water
  4. Dis a non-polar simple molecule

Question 303

[1 marks]bonding, periodicity and contact process
One of the steps in the Contact process is represented by the equation
  1. A2SO2+O2⇌2SO32SO_2 + O_2 \rightleftharpoons 2SO_3
  2. BSO3+2H2O→H2SO3+H2O2SO_3 + 2H_2O \rightarrow H_2SO_3 + H_2O_2
  3. CS+O2→SO3S + O_2 \rightarrow SO_3
  4. DSO2+H2O→H2SO4SO_2 + H_2O \rightarrow H_2SO_4

Question 304

[3 marks]bonding, periodicity and contact process
The first ionisation energy of magnesium is higher than that of aluminium mainly because in aluminium the outermost electron is removed from
  1. Aa filled, unusually stable 3s orbital that requires noticeably more energy to break into than any other subshell available in the atom.
  2. Ba 3d orbital, which is always higher in energy than any p or s orbital in the same shell of any atom.
  3. Cthe nucleus directly, since aluminium has one more proton in its nucleus than magnesium.
  4. Da 3p orbital, which is higher in energy and further from the nucleus (better shielded) than the full 3s orbital in magnesium.

Question 305

[1 marks]bonding, periodicity and contact process
The equation for the first step of the Contact process (burning sulfur) is

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Question 306

[2 marks]bonding, periodicity and contact process
The equation for the third step of the Contact process (absorbing SO3SO_3 into concentrated sulfuric acid) is

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Question 307

[1 marks]bonding, periodicity and contact process
The equation for the fourth step of the Contact process (diluting oleum with water) is

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Question 401

[1 marks]organic chemistry - arenes and carbonyls
Methylbenzene is converted into chloromethylbenzene, C6H5CH2ClC_6H_5CH_2Cl, using
  1. Achlorine with an AlCl3AlCl_3 catalyst in the dark
  2. Bconcentrated hydrochloric acid under reflux
  3. CPCl5PCl_5 at room temperature
  4. Dchlorine in ultraviolet light

Question 402

[1 marks]organic chemistry - arenes and carbonyls
Methylbenzene is refluxed with acidified potassium manganate(VII). The organic product A is
  1. Acyclohexane
  2. Bbenzaldehyde
  3. Cbenzoic acid
  4. Dphenol

Question 403

[1 marks]organic chemistry - arenes and carbonyls
Butan-2-one is formed from butan-2-ol by
  1. Adehydration with concentrated H2SO4H_2SO_4
  2. Breduction with NaBH4NaBH_4
  3. Coxidation with acidified K2Cr2O7K_2Cr_2O_7 under reflux
  4. Dhydrolysis with aqueous NaOH

Question 404

[1 marks]organic chemistry - arenes and carbonyls
When methylbenzene reacts with bromine in the presence of FeBr3FeBr_3, the observation made is that the bromine

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Question 405

[2 marks]organic chemistry - arenes and carbonyls
The organic product formed when methylbenzene reacts with bromine in the presence of FeBr3FeBr_3 is

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Question 406

[2 marks]organic chemistry - arenes and carbonyls
The chemical test that confirms the presence of a carbonyl group in butan-2-one is: adding
  1. A2,4-dinitrophenylhydrazine (2,4-DNPH), giving an orange/yellow precipitate.
  2. BTollens' reagent, which would give a silver mirror forming on the inner walls of the test tube upon gentle warming.
  3. CFehling's solution, giving a brick-red precipitate on heating.
  4. Dbromine water, which is decolourised immediately at room temperature.

Question 407

[2 marks]organic chemistry - arenes and carbonyls
The type of organic product formed when a carbonyl compound reacts with 2,4-DNPH is called a

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Question 501

[1 marks]organic chemistry - alkanes and functional groups
Butane does not react with aqueous bromine because it
  1. Ahas a very low boiling point
  2. Bis saturated and has no reactive functional group
  3. Cis insoluble in water
  4. Dis a gas at room temperature

Question 502

[1 marks]organic chemistry - alkanes and functional groups
Butane has a lower melting point than hexane because butane molecules
  1. Aare smaller, so the van der Waals forces between them are weaker
  2. Bform fewer hydrogen bonds
  3. Ccontain stronger covalent bonds
  4. Dare more polar

Question 503

[1 marks]organic chemistry - alkanes and functional groups
X is HO−C6H4−CH2CH2OHHO-C_6H_4-CH_2CH_2OH and Y is HO−C6H4−COCH3HO-C_6H_4-COCH_3. The reagent which distinguishes them is
  1. A2,4-dinitrophenylhydrazine, which gives an orange precipitate with Y only
  2. Biron(III) chloride, which gives a violet colour with X only
  3. Csodium metal, which gives a gas with X only
  4. Daqueous sodium carbonate, which fizzes with Y only

Question 504

[2 marks]organic chemistry - alkanes and functional groups
Butane is insoluble in water because butane molecules are
  1. Apolar, but with a dipole oriented in the wrong direction to attract water.
  2. Bnon-polar and cannot form hydrogen bonds with the polar water molecules.
  3. Cdenser than water, so they sink without ever mixing into solution.
  4. Dtoo large to fit between the water molecules in the liquid.

Question 505

[1 marks]organic chemistry - alkanes and functional groups
One environmental effect of burning alkanes as fuels is that complete combustion produces

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Question 506

[1 marks]organic chemistry - alkanes and functional groups
Besides carbon dioxide, incomplete combustion of alkane fuels can also produce the toxic gas

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Question 507

[2 marks]organic chemistry - alkanes and functional groups
When compound X (HO−C6H4−CH2CH2OHHO-C_6H_4-CH_2CH_2OH) reacts with excess sodium metal, the observable change is
  1. Aeffervescence (bubbles of hydrogen gas) as both -OH groups react with the sodium.
  2. Bthe sodium dissolves completely with no visible gas evolved at all.
  3. Ca colour change from colourless to a deep violet as the reaction proceeds.
  4. Da white precipitate forms immediately because the resulting sodium salt is completely insoluble in the reaction mixture.

Question 508

[1 marks]organic chemistry - alkanes and functional groups
The type of compound formed when the phenolic -OH group of X or Y reacts with sodium is a sodium

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Question 601

[1 marks]nanoparticles and amino acids
A nanoparticle is a particle whose dimensions lie between
  1. A1 and 100 mm
  2. B1 and 100 pm
  3. C1 and 100 nm
  4. D1 and 100 μ\mum

Question 602

[1 marks]nanoparticles and amino acids
A mixture of glycine, aspartic acid and lysine is separated by electrophoresis buffered at pH 7. The amino acid which moves towards the negative electrode is
  1. Anone of them, because all are neutral at pH 7
  2. Bglycine, because it exists as a zwitterion
  3. Caspartic acid, because it has a second carboxyl group
  4. Dlysine, because its extra amino group is protonated

Question 603

[1 marks]nanoparticles and amino acids
The overall charge on a zwitterion is
  1. A−1-1
  2. Bzero
  3. C+1+1
  4. D+2+2

Question 604

[2 marks]nanoparticles and amino acids
One application of nanoparticles is in

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Question 605

[1 marks]nanoparticles and amino acids
One example of a nanoparticle used in industry is

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Question 606

[1 marks]nanoparticles and amino acids
The type of reaction that joins glycine and aspartic acid to form a peptide bond is called a

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Question 607

[3 marks]nanoparticles and amino acids
The formula of the dipeptide formed when glycine (H2NCH2COOHH_2NCH_2COOH) reacts with aspartic acid, with loss of water, is

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The answers, and why they are the answers

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