Q1. (a) Calcium chloride/ calcium sulfate
(b) –Rinse a pipette with deionised water and then the hard water using a pipette filler
-Pipette 50 cm3 of the water sample into a conical flask using a pipette filler
-Make sure the bottom of the meniscus is at the mark at eye level
–Do not squeeze out the last drop
–Tap off any excess water droplets on the outside of the pipette before placing it into the conical flask
(c)(i) Eriochrome black T
(ii) Wine red
(iii) Navy blue
(d)(i) pH buffer 10
(ii) To ensure that the pH remained constant so that full complexing occurred between the edta and the calcium and magnesium ions
(e)(i) 9.3 cm3 of 0.01M edta was used
0.01 moles in 1000 cm3
0.01 ÷ 1000 = 0.00001
0.00001 × 9.3 = 0.000093 moles
(ii) Ratio of edta : hard water = 1:1
Moles of hard water = 0.000093 moles
(iii) Mass = moles × molecular mass
Molecular mass of CaCO3: 40 + 12 + 16 + 16 + 16 = 100
Mass of CaCO3 in 50 cm3 = 0.000093 × 100 = 0.0093g
0.0093 ÷ 50 = 0.000186
Mass of CaCO3 in g/L = 0.000186 × 1000 = 0.186 g
(iv) Hardness in ppm = 0.186 × 1000 = 186 ppm
(f) -Temporary hardness is caused by Ca(HCO3)2
-To test for calcium hydrogencarbonate, add hydrochloric acid
–Fizzing occurs
-The gas produced turns limewater milky (CO2)
–Add MgSO4 to calcium hydrogencarbonate
–No precipitate is formed
–A white precipitate is formed upon heating
Q2. (a)(i) X: Liebig condenser
(ii) To allow enough time for the reaction to reach completion at the correct temperature without the flask boiling dry
(iii) Reactants: -Sunflower oil/ lard
-Sodium hydroxide
Solvent: -Ethanol
(b)(i) A mixture of two immiscible liquids
(ii) Cyclohexane
(iii) When shaking the separating funnel, open the tap periodically to relieve pressure
(iv) -Add anhydrous magnesium sulfate to remove any remaining water
-Filter through filter paper into a buchner flask
-Evaporate off the cyclohexane on a hotplate in a fume cupboard
(v) 0.15 × 1.05 = 0.1575 g
(0.1575 ÷ 5) × 100 = 3.15 %
(c)(i) –Hot filtration only gets rid of insoluble impurities, such as charcoal.
-There were no insoluble impurities present, as only sodium chloride, which is a soluble impurity, was present.
(ii) Suction filtration is faster
(iii) -Determine the melting point range of both the original and recrystallised samples using a hotplate and a melting point block.
–The melting point range of the purer sample will be higher and narrower.
Q3. (a) Exothermic
Justify: –Bond formation results in exothermic reactions, where heat is given out.
–Bonds are formed between the hydrochloric acid and the sodium hydroxide.
(b)(i) So that the temperature of the mixture is the same throughout the whole mixture
(ii) Heat was given out to the surroundings
(iii) 6K
(c)(i) -When using concentrated solutions, there is a greater heat change, therefore the heat change is easier to record
-This makes calculating the heat of the reaction more accurate
(ii)
(d)(i) 150 cm3 of 1M HCl
1M = 1 mole in 1000 cm3
1 ÷ 1000 = 0.001
0.001 × 150 = 0.15 moles
(ii) Q = mc∆t
m = 310 g = 0.310 kg
c = 4.2 kJ / kg K
∆t = 6.4 K
Q = 0.31 × 4.2 × 6.4 = 8.3328 kJ
(iii) 8.3328 for 0.15 moles
For 1 mole = 8.3328 ÷ 0.15 = −55.552 kJ (exothermic)
(e) State: Carry out the reaction in a polystyrene cup
Explain: Polystyrene is a better insulator than cardboard
Q4. (a) To measure the charge on the electron
(b) State: decrease in atomic radius
Reason: increasing nuclear charge, which means the electrons are more attracted to the nucleus
(c) Fe26: 1s2, 2s2, 2p6, 3s2, 3p6, 4s2, 3d6
(d) -Collisions are not perfectly elastic
-There are intermolecular forces between the molecules of gases
(e) Acid: HNO3
Conjugate base: NO3−
(f) Oxidation number of chlorine in NaOCl: +1 (O is −2, Na is +1)
Oxidation number of chlorine in NaCl: −1 (Na is +1)
Since the oxidation number decreased, chlorine was reduced
(g) Cl2 + 2NaBr → 2NaCl + Br2
(h) State: There is an increased rate of reaction of the formation of chloromethane
Explain: -This reaction is a free radical substitution reaction
-Tetraethyllead is a source of free radicals and therefore speeds up the reaction
(i) In 100 g, there would be 54.5 g carbon, 9.1 g hydrogen and 36.4 g oxygen
Moles = mass ÷ molecular mass
Moles of carbon = 54.5 ÷ 12 = 4.54 moles
Moles of hydrogen = 9.1 ÷ 1 = 9.1 moles
Moles of oxygen = 36.4 ÷ 16 = 2.275 moles
4.54 ÷ 2.275 = 2
9.1 ÷ 2.275 = 4
Ratio of carbon : hydrogen : oxygen = 2:4:1
Therefore, the empirical formula is C2H4O
(j) The amount of oxygen that is consumed in a sample of water when it is left at 20°C in a dark room for 5 days
(k) A: Which: water vapour
Reason: -The more water vapour there is in the atmosphere, the more IR radiation from the Earth can be absorbed.
-This further heats up the atmosphere.
B: (i) water
(ii) graphite
Q5. (a)(i) Number of protons and neutrons
(ii) The average mass of all isotopes of an element, taking their abundances into account and measured against the mass of a Carbon-12 atom, which has a mass of 12
(b) -Ionisation
-Acceleration
-Separation
(c) (24 ÷ 100) × 79 = 18.96
(25 ÷ 100) × 10 = 2.5
(26 ÷ 100) × 11 = 2.86
18.96 + 2.5 + 2.86 = 24.32
(d)(i) The spontaneous breaking up of unstable nuclei with the emission of one or more different types of radiation
(ii) A neutron changes into a proton
(iii) 6C14 → −1e0 + 7N14
(iv) Explain: -The half-life of an element is the time taken for half of the sample to disintegrate
-Since Carbon-14 has a half-life of 5730 years, double the sample should have been present 5730 years ago
1.5 × 1012 × 2 = 3 × 1012 atoms
What mass: 3 × 1012 atoms were present
(3 × 1012) ÷ (6 × 1023) = 5 × 10−12 moles
Mass = moles × molecular mass
Mass = 5 × 10−12 × 14 = 7 × 10−11 g
Q6. (a) The splitting up of crude oil into various hydrocarbons based on their boiling points
(b)(i) Propane and butane
(ii) It is used to add odour to the gases, so that gas leaks can be detected
(iii) C + 2H2 + S → CH3SH ∆H = −22.8 kJ/mol
C + O2 → CO2 ∆H = −393.5 kJ/mol
H2 + ½ O2 → H2O ∆H = −285.8 kJ/mol
S + O2 → SO2 ∆H = −296.8 kJ/mol
Flip first equation: ∆H = 22.8 kJ/mol
Multiply ∆H of H2O by 2: ∆H = −571.6 kJ/mol
Add all heats together: 22.8 – 393.5 − 571.6 − 296.8 = −1239.1 kJ
(c)(i) Made by process A: 2,2,4 – trimethylpentane
Made by process B: But-2-ene and Butane
Made by process C: ethylbenzene
(ii) B: catalytic cracking
C: dehydrocyclisation
(iii) To increase octane number, so that the fuels are more likely to resist knocking and therefore are more effective fuels
(d) The boiling point of petrol is lower than the boiling point of diesel
Q7. (a)(i) The rate of change of concentration of any one product or reactant
(ii) The minimum energy required for effective collisions to occur
(b)
(c) Instantaneous rate = slope
(0.009 – 0.004) ÷ 200 = 0.000025 M/s
(d)(i) Not all collisions are effective, i.e. not all collisions reach the activation energy
(ii) Increase the temperature in the reaction vessel
(iii)
Q8. (a)(i) B:
C:
(ii) Ethanoic acid
(b)(i) Structural isomers: Compounds that have the same molecular formula but different structural formulas
Primary alcohol: An alcohol in which the carbon bonded to the -OH (hydroxyl) group is only bonded to one other carbon atom
(ii)
(iii) Propanone
(c)(i) –Propanal is capable of forming hydrogen bonds, whereas butane can only form dipole-dipole bonds.
–Hydrogen bonds are a stronger intermolecular force than dipole-dipole bonds.
(ii) Ethanoic acid is capable of forming two hydrogen bonds, whereas propan-1-ol can only form one hydrogen bond.
(d) Propanoic acid
(e) C3H6OH + Na → C3H6ONa + ½ H2
Q9. (a) A state in which the rate of the forward reaction equals the rate of the reverse reaction
(b)
(c)(i) State: the mixture turns a paler green (more colourless)
Explain: When pressure is increased, the equilibrium shifts to the side with less molecules, which is the right hand side
(ii) State: Less yield of phosgene
Explain: -High temperatures favour endothermic reactions, which is the reverse reaction.
-The equilibrium shifts left to oppose the stress.
(iii) State: no effect
Explain: -Kc value is constant at constant temperature.
-Only temperature affects Kc value.
-The charcoal catalyst will alter the rate of the forward and reverse reactions equally.
(d) 85% of the chlorine had reacted at equilibrium
100 – 85 = 15% chlorine left
(0.2 ÷ 100) × 15 = 0.03 moles chlorine present at equilibrium
Kc = 0.0142 ÷ (0.0025 × 0.0025) = 2272
(e) -Low temperatures are uneconomical because they slow down the rate of the reaction.
-There is a risk of explosions happening with high pressures.
Q10. (a) Ethene: -Add bromine water to a sample of ethene gas in a test tube.
-Shake the test tube
-Bromine water changes from yellow to colourless (decolourises)
Benzene: –No, benzene does not readily undergo an addition reaction with bromine water
Explain: -Benzene has 6 bonds that are intermediate between single and double bonds.
-This gives the molecule extra stability.
-Bromine water readily reacts with molecules that have double bonds.
Delocalised: electrons that are not associated with a single covalent bond, but instead move around a number of covalent bonds
(i) 6
(ii) 12
Health hazard: Benzene is carcinogenic
(b)(i) Ground state
(ii) Excited state
(iii) It receives energy, e.g. heat
(iv) They are unstable and temporary
(v) n=2
Modern: Electrons move as a wave (Louie de Broglie’s wave-particle duality)
Orbital: A region of space where an electron is most likely to be found
(c)(i) Moles = mass ÷ molecular mass
Molecular mass of FePO4: 56 + 31 + 16 + 16 + 16 + 16 = 151
Moles of FePO4 = 4.53 ÷ 151 = 0.03 moles
(ii) Mass = moles × molecular mass
Molecular mass of Fe2O3: 56 + 56 + 16 + 16 + 16 = 160
Ratio of FePO4:Fe2O3 = 2:1
0.03 ÷ 2 = 0.015
Mass = 0.015 × 160 = 2.4 g
(iii) Ratio of Fe2O3:H3PO4 = 1:2
Moles of H3PO4 needed = 0.015 × 2 = 0.03
6 moles in 1000 cm3
Volume needed for 1 mole = 1000 ÷ 6 = 166.67 cm3
Volume needed for 0.03 moles = 166.67 × 0.03 = 5 cm3
(iv) Ratio of Fe2O3:H2O = 1:3
0.015 × 3 = 0.045 moles
Mass = moles × molecular mass
Molecular mass of water: 1 + 1 + 16 = 18
Mass = 0.045 × 18 = 0.81 g
Volume = mass ÷ density
0.81 ÷ 1 = 0.81 cm3
Q11. (a)(i) The relative attraction of an atom to a shared pair of electrons in a covalent bond
(ii) Electronegativity of carbon: 2.55
Electronegativity of chlorine: 3.16
3.16 – 2.55 = 0.61
CCl4 is polar covalent
(iii) Valency of 4
(iv) State: Tetrahedral shape
Account: Using VSEPR = 4 bond pairs, 0 lone pairs
(v)
(b) base: a substance that dissociates in water to form OH− ions
-NH3 is a proton donor
(i) pH = −log10[H+]
pH = −log10[0.5] = 0.3
(ii) pH = −log10√(Ka× Ma)
pH = −log10√(1.8 × 10−5 × 0.5) = 2.52
Which: Ethanoic acid
Evidence: -The graph starts at a pH of approximately 2.52, which is the pH of ethanoic acid
-The vertical part of the curve takes place from approximately a pH of 7 to a pH of 11, indicating a weak acid/ strong base reaction
Property: The colour change of the indicator must occur within the vertical part of the curve
(c) A: (i) MgCl2 + Ca(OH)2 → CaCl2 + Mg(OH)2
Mg(OH)2 → MgO +H2O
Use: Used as a heat resistant material to line furnaces
Identify: Suspended solids
Measure: Particles of dust in the chimney stacks are removed by filtration
(ii) Catalyst: Platinum and Rhodium catalyst
Economic significance: This value of ∆H means that the catalyst can be kept red hot
2NO + O2 → 2NO2
3NO2 + H2O → 2HNO3 + NO
Negative impact: Ammonium nitrate is very explosive
(iii) Conditions: -High temperature (450°C)
-Low pressure (200 atmospheres)
Catalyst: Iron catalyst
CH4 + H2O ⇌ CO + 3H2
CO + H2O ⇌ CO2 + H2
Nitrogen: -Unreacted methane is burned in air
-This causes the oxygen in the air to be removed, leaving pure nitrogen
CO2 + 2NH3 → NH2CONH2 + H2O
B: (i) 2NaAlO2 + 4H2O → Al2O3.3H2O + 2NaOH
Al2O3.3H2O → Al2O3 + 3H2O
(ii) -Aluminium metal is extracted from aluminium oxide by electrolysis in steel containers lined with graphite
-The anode consists of graphite
-The cathode consists of the graphite lining of the cell
-Aluminium oxide is dissolved into cryolite
-This takes place at a temperature of 1000°C
-Reduction happens at the cathode and oxidation happens at the anode
(iii) -Aluminium smelters are usually built in countries that have cheap hydroelectric power
-Also, the recycling of aluminium is encouraged
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