Danho
ZIMSEC A Level · N2003

Chemistry Paper 2 November 2003

Questions
32
Total marks
49
Time allowed
75 min

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Questions
32
Pass mark
20
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]mass spectrometry and bonding
2.000 g of lead was converted completely into lead(II) chloride, giving 2.682 g of the chloride. Taking Ar(Cl)=35.45A_r(Cl) = 35.45, the relative atomic mass of the lead is
  1. A103.9
  2. B207.9
  3. C311.8
  4. D415.8

Question 102

[1 marks]mass spectrometry and bonding
Lead melts at 327 °C but lead(IV) chloride melts at −15-15 °C. This is because lead has
  1. Aa giant ionic lattice while PbCl4PbCl_4 is a giant covalent solid
  2. Bstronger van der Waals forces than PbCl4PbCl_4
  3. Ca higher relative atomic mass than chlorine
  4. Da giant metallic lattice while PbCl4PbCl_4 is a simple covalent molecule

Question 103

[1 marks]mass spectrometry and bonding
In the mass spectrum of lead shown, the peaks at m/em/e = 204, 206, 207 and 208 are due to
  1. Afragments of the lead atom
  2. Bdoubly charged lead ions
  3. Cisotopes of lead
  4. Dimpurities in the sample

Question 104

[2 marks]mass spectrometry and bonding
Write the formulae (mass number nuclide notation) of the four lead nuclides present in this sample, whose peaks appear at m/e = 204, 206, 207 and 208.

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Question 105

[2 marks]mass spectrometry and bonding
The mass spectrum of this lead sample shows relative abundances of about 2% at m/e 204, 25% at m/e 206, 22% at m/e 207 and 51% at m/e 208. The average relative atomic mass of this sample is

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Question 106

[2 marks]mass spectrometry and bonding
The first lead sample (from the mass spectrum) has an average Ar of 207.2, while a second lead sample (from a lead(II) chloride experiment) has an Ar of 207.9. The most reasonable conclusion is that the two samples
  1. Amay have slightly different isotopic compositions, though the difference is small and could partly reflect experimental error
  2. Bmust be different elements entirely, since their Ar values are not identical
  3. Ccannot both be lead, since Ar should always be a whole number
  4. Dhave exactly identical isotopic compositions, and the small difference is impossible

Question 107

[1 marks]mass spectrometry and bonding
Suggest one way in which lead is released into the environment.

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Question 108

[2 marks]mass spectrometry and bonding
The average relative atomic mass of lead, 207.2, is calculated as a weighted mean of the isotope mass numbers (weighted by relative abundance) rather than a simple average of 204, 206, 207 and 208 because
  1. Athe least abundant isotope should be given the greatest weight in the calculation
  2. Ba simple average is mathematically impossible to calculate for more than two numbers
  3. Cthe isotopes are not present in the sample in equal proportions, so each mass number must count according to how abundant it actually is
  4. Dmass numbers must always be rounded to the nearest whole number before averaging

Question 201

[1 marks]energetics / Hess's law
The standard enthalpy changes of combustion of ethanol and ethanal are −1370.7-1370.7 and −1167.3 kJ mol−1-1167.3\ kJ\,mol^{-1} respectively. The enthalpy change for C2H5OH(l)+12O2(g)→CH3CHO(l)+H2O(l)C_2H_5OH(l) + \frac{1}{2}O_2(g) \rightarrow CH_3CHO(l) + H_2O(l), in kJ mol−1kJ\,mol^{-1}, is
  1. A−2538.0-2538.0
  2. B−203.4-203.4
  3. C+203.4+203.4
  4. D+2538.0+2538.0

Question 202

[1 marks]energetics / Hess's law
Ethanol is oxidised to ethanoic acid according to C2H5OH+O2→CH3COOH+H2OC_2H_5OH + O_2 \rightarrow CH_3COOH + H_2O. The mass of ethanol needed to give 5.0 g of ethanoic acid is
  1. A5.00 g
  2. B6.52 g
  3. C7.67 g
  4. D3.83 g

Question 203

[1 marks]energetics / Hess's law
The relative molecular mass of ethanoic acid measured in benzene is 120. This is because in benzene ethanoic acid
  1. Areacts with the solvent
  2. Bloses a molecule of water
  3. Cforms hydrogen bonded dimers
  4. Dionises completely

Question 204

[2 marks]energetics / Hess's law
Which statement below correctly gives Hess's Law?
  1. Aa system at equilibrium always shifts to oppose any change imposed on it
  2. Bthe enthalpy change of a reaction is always negative if the reaction releases a gas
  3. Cenergy can be neither created nor destroyed, only converted from one form into another
  4. Dthe total enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same

Question 205

[2 marks]energetics / Hess's law
Adding equations I and II together gives the overall equation for the bacterial oxidation of ethanol all the way to ethanoic acid. Write this overall equation.

Answer this when you sit the paper.

Question 206

[1 marks]energetics / Hess's law
The intermolecular force responsible for ethanoic acid molecules pairing up into dimers in benzene solution is

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Question 301

[1 marks]chemical equilibrium
4.0 mol of NO2NO_2 were placed in a 1 dm31\ dm^3 container and heated. At equilibrium, 2NO2(g)⇌2NO(g)+O2(g)2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g), the mixture contained 0.8 mol of oxygen. KcK_c, in mol dm−3mol\,dm^{-3}, is
  1. A0.09
  2. B0.36
  3. C0.85
  4. D1.07

Question 302

[1 marks]chemical equilibrium
An equilibrium mixture of colourless N2O4N_2O_4 and brown NO2NO_2 is immersed in ice-cold water. The mixture becomes
  1. Adarker brown, as the pressure falls
  2. Bpaler, as more N2O4N_2O_4 is formed
  3. Ccolourless, as the reaction stops
  4. Ddarker brown, as more NO2NO_2 is formed

Question 303

[1 marks]chemical equilibrium
The dimerisation 2NO2(g)→N2O4(g)2NO_2(g) \rightarrow N_2O_4(g) is exothermic because
  1. ANO2NO_2 is a brown gas
  2. BN2O4N_2O_4 has a lower relative molecular mass
  3. Ca new N−NN-N bond is formed
  4. Dthe number of gaseous molecules decreases

Question 304

[2 marks]chemical equilibrium
When the pressure on the equilibrium mixture of N2O4N_2O_4 and NO2NO_2 is increased, the mixture becomes
  1. Apaler, as the equilibrium shifts toward N2O4N_2O_4, the side with fewer gas molecules
  2. Bdarker brown, as the equilibrium shifts toward NO2NO_2
  3. Ccolourless, as the equilibrium mixture liquefies under pressure
  4. Dunchanged in colour, since pressure does not affect a gas-phase equilibrium

Question 305

[2 marks]chemical equilibrium
The gas-phase structure of N2O4N_2O_4 is O2N−NO2O_2N-NO_2, a single N-N bond joining two planar NO2NO_2 units. The O-N-O bond angle within each NO2NO_2 unit is closest to
  1. A90 degrees, since each nitrogen is octahedral
  2. B120 degrees, since each nitrogen is trigonal planar with no lone pair
  3. C109 degrees, since each nitrogen is tetrahedral
  4. D180 degrees, since each nitrogen is linear

Question 306

[2 marks]chemical equilibrium
For the equilibrium 2NO2(g)⇌2NO(g)+O2(g)2NO_2(g) \rightleftharpoons 2NO(g) + O_2(g), the correct expression for KcK_c is
  1. AKc=[NO]2[O2][NO2]2K_c = \dfrac{[NO]^2[O_2]}{[NO_2]^2}
  2. BKc=[NO][O2][NO2]K_c = \dfrac{[NO][O_2]}{[NO_2]}
  3. CKc=[NO]2[O2][NO2]2K_c = [NO]^2[O_2][NO_2]^2
  4. DKc=[NO2]2[NO]2[O2]K_c = \dfrac{[NO_2]^2}{[NO]^2[O_2]}

Question 401

[1 marks]organic chemistry / polymers
A compound X of empirical formula CH3OCH_3O was vapourised at 400 K; 0.20 g of X occupied 107.2 cm3107.2\ cm^3 at 100 kPa. The molecular formula of X is
  1. AC2H6O2C_2H_6O_2
  2. BC3H9O3C_3H_9O_3
  3. CC4H12O4C_4H_{12}O_4
  4. DCH3OCH_3O

Question 402

[1 marks]organic chemistry / polymers
Compound X, C2H6O2C_2H_6O_2, is oxidised mildly to a dicarboxylic acid. The structure of X is
  1. ACH3CH2OOHCH_3CH_2OOH
  2. BCH3CH(OH)2CH_3CH(OH)_2
  3. CCH3OCH2OHCH_3OCH_2OH
  4. DHOCH2CH2OHHOCH_2CH_2OH

Question 403

[1 marks]organic chemistry / polymers
Ethane-1,2-diol reacts with benzene-1,4-dicarboxylic acid to form a fibre used in clothing. The fibre is
  1. Aa polyalkene
  2. Ba polyamide
  3. Ca polypeptide
  4. Da polyester

Question 404

[2 marks]organic chemistry / polymers
Write a balanced equation for the reaction of X (ethane-1,2-diol, HOCH2CH2OHHOCH_2CH_2OH) with excess sodium.

Answer this when you sit the paper.

Question 405

[1 marks]organic chemistry / polymers
When compound X (ethane-1,2-diol) reacts with excess sodium, the gas evolved is
  1. Ahydrogen
  2. Bchlorine
  3. Coxygen
  4. Dcarbon dioxide

Question 501

[1 marks]organic chemistry / functional group tests
An orange precipitate is formed when 2,4-dinitrophenylhydrazine is added to
  1. AC6H5CONH2C_6H_5CONH_2
  2. BC6H5CH2CHOC_6H_5CH_2CHO
  3. CC6H5CH(OH)CH3C_6H_5CH(OH)CH_3
  4. DC6H5CH2CO2HC_6H_5CH_2CO_2H

Question 502

[1 marks]organic chemistry / functional group tests
A pale yellow precipitate of triiodomethane is obtained on warming alkaline I2I_2 with
  1. AC6H5CONH2C_6H_5CONH_2
  2. BC6H5CH2CHOC_6H_5CH_2CHO
  3. CC6H5CH(OH)CH3C_6H_5CH(OH)CH_3
  4. DC6H5CH2CO2HC_6H_5CH_2CO_2H

Question 503

[1 marks]organic chemistry / functional group tests
Ammonia, which turns damp red litmus paper blue, is evolved when aqueous sodium hydroxide is heated with
  1. AC6H5CH2CO2HC_6H_5CH_2CO_2H
  2. BC6H5CONH2C_6H_5CONH_2
  3. CC6H5CH2CHOC_6H_5CH_2CHO
  4. DC6H5CH(OH)CH3C_6H_5CH(OH)CH_3

Question 504

[2 marks]organic chemistry / functional group tests
Warming with ethanol (in the presence of a little acid catalyst) gives a sweet-smelling ester with
  1. AC6H5CH2CHOC_6H_5CH_2CHO
  2. BC6H5CH2CO2HC_6H_5CH_2CO_2H
  3. CC6H5CH(OH)CH3C_6H_5CH(OH)CH_3
  4. DC6H5CONH2C_6H_5CONH_2

Question 505

[2 marks]organic chemistry / functional group tests
The ester formed when compound B, C6H5CH2CO2HC_6H_5CH_2CO_2H, is warmed with ethanol is

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Question 506

[1 marks]organic chemistry / functional group tests
Heating compound C, C6H5CONH2C_6H_5CONH_2, with aqueous NaOH gives ammonia gas and the sodium salt of benzoic acid, named

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Question 507

[1 marks]organic chemistry / functional group tests
The type of organic product formed when 2,4-dinitrophenylhydrazine reacts with the carbonyl group of compound D is a

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The answers, and why they are the answers

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