Chemistry Year 12 School Closure Pack A timetable has been created for you to ensure that your work is manageable. Each day you are expected to complete 1 hr of chemistry work. For each topic: 1. Revise using the online resources (approx. 20 min) 2. Answer and mark all exam questions (approx. 30 min) 3. Add up your total, take a picture of your work and upload to Edmodo by 3.30pm You will also have an online quiz every Tuesday and Thursday. Resources: 1. chemrevise.org/revision-guides/ 2. www.youtube.com/user/MrERintoul/playlists 3. chemguide.co.uk 4. physicsandmathstutor.co.uk Ms Baig is reachable on Edmodo for feedback, tutorials and to answer any of your questions, during the school day. Name………………………………………………………………………………………………………
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Chemistry Year 12 School Closure Pack
A timetable has been created for you to ensure that your
work is manageable. Each day you are expected to complete
1 hr of chemistry work. For each topic:
1. Revise using the online resources (approx. 20 min)
2. Answer and mark all exam questions (approx. 30 min)
3. Add up your total, take a picture of your work and
upload to Edmodo by 3.30pm
You will also have an online quiz every Tuesday and
Thursday.
Resources:
1. chemrevise.org/revision-guides/
2. www.youtube.com/user/MrERintoul/playlists
3. chemguide.co.uk
4. physicsandmathstutor.co.uk
Ms Baig is reachable on Edmodo for feedback, tutorials and to
answer any of your questions, during the school day.
Name………………………………………………………………………………………………………
Year 12 Chemistry
Date Topic Title Work to be completed (Tick)
Resource provided
Outcome Online Support
1 Ch6: Equilibria
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
2 Ch11: Introduction to Organic Chemistry
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
End of day 2 o Complete Edmodo Quiz (on previous 2 chapters atleast)
Edmodo o At least 70% in biweekly chapter quiz
o Edmodo Quiz
3 Ch13: Haloalkanes
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
4 Ch2: Amount of substance
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
End of day 4 o Complete Edmodo Quiz (on previous 2 chapters atleast) by 11pm
Edmodo o At least 70% in biweekly chapter quiz
o Edmodo Quiz
5 Ch12: Alkanes
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
6 Ch4: Energetics
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
o At least 70% in
biweekly chapter quiz
End of day 6 o Complete Edmodo Quiz (on previous 2 chapters atleast) by 11pm
Edmodo o At least 70% in biweekly chapter quiz
o Edmodo Quiz
7 Ch14: Alkenes
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
8 Ch3: Bonding o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
End of day 8 o Complete Edmodo Quiz (on previous 2 chapters atleast) by 11pm
Edmodo o At least 70% in biweekly chapter quiz
o Edmodo Quiz
9 Ch8: Periodicity
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
10 Ch9: Group 2
o Revise using the online support
o Answer and mark all exam questions
o Add up your total, take a picture and upload to edmodo
-Exam questions and mark scheme
o Detailed understanding of notes
o Marked exam questions
o At least 70% in biweekly chapter quiz
o Chemrevise notes o Edmodo o Eliot Rintoul
YouTube channel o Chemguide o Maths and Physics
Tutor
End of day 10 o Complete Edmodo Quiz (on previous 2 chapters atleast) by 11pm
Edmodo o At least 70% in biweekly chapter quiz
o Edmodo Quiz
Resources:
1. https://chemrevise.org/revision-guides/
2. www.youtube.com/user/MrERintoul/playlists
3. chemguide.co.uk
4. physicsandmathstutor.co.uk
1. (a) The diagram below shows the effect of temperature and pressure on the equilibrium yield of the product in a gaseous equilibrium.
Yield/%
Temperature
10 MPa
30 MPa
50 MPa
(i) Use the diagram to deduce whether the forward reaction involves an increase or a decrease in the number of moles of gas. Explain your answer.
Change in number of moles .............................................................................
2. Acid X reacts with methanol to form ester Y according to the following equation.
A mixture of 0.25 mol of X and 0.34 mol of methanol was left to reach equilibrium in the presence of a small amount of concentrated sulphuric acid. The equilibrium mixture thus formed contained 0.13 mol of Y in a total volume of V dm3.
(a) Using X to represent the acid and Y to represent the ester, write an expression for the equilibrium constant, Kc, for this reaction.
(1)
(b) Calculate the number of moles of X, the number of moles of methanol and the number of moles of water in the equilibrium mixture.
Moles of X ..............................................................................................................
Moles of methanol ...................................................................................................
Moles of water ......................................................................................................... (3)
(c) State why the volume V need not be known in calculating the value of Kc for the reaction.
C H C O O H C H C O O C H + 2 C H O H + 2 H O ∆ H = – 1 5 k J m o l
2 2 3 3 2 – 1 2 2 3
a c i d X e s t e r Y
4
3. As a first step in the manufacture of nitric acid it has been suggested that nitrogen monoxide, NO, can be formed from nitrogen and oxygen in a reversible reaction.
(a) Write an equation for this reaction and deduce an expression for the equilibrium constant, Kc
(c) The value of Kc for this reaction is 1 × 10–5 at 1500 K.
Explain the significance of this value for an industrial chemist interested in manufacturing nitrogen monoxide by the direct combination of the elements.
Discuss, with reasons, the effects of increasing separately the temperature and the pressure on the yield of the products and on the rate of this reaction.
(6) (Total 6 marks)
5. The reaction between hydrogen and iodine can be represented by the following equation:
H2(g) + I2(g) 2HI(g) ∆H = +52 kJ mol–1 (a) Write a Kc expression for the decomposition of hydrogen iodide. At a given temperature, the
value of Kc for this reaction is 20. What will be the value of Kc for the reaction between hydrogen and iodine at this temperature?
(2)
(b) The pressure of an equilibrium mixture of hydrogen iodide, hydrogen and iodine was increased. State what, if anything, would happen to:
(i) the rates of both forward and reverse reactions; (2)
(ii) the position of equilibrium; (1)
(iii) the value of the equilibrium constant. (1)
(Total 6 marks)
7
4.2 ANSWERS TO EXAM QUESTIONS
1. (a) (i) Increase (if wrong no further marks in part (i) 1 higher P gives lower yield or moves to left 1 Eqm shifts to reduce P or eqm favours side with fewer moles 1
(ii) Endothermic if wrong no further marks in part (ii) 1 increase T increases yield or moves to right 1 Eqm shifts to reduce T or eqm favours endothermic direction 1
(b) (i) Moles of iodine = 0.023 1 Moles of HI = 0.172 1 If × 2 missed, max 1 in part (iv)
If wrong no marks in (i)
(ii) Kc = 222
HI][]I][H[ must be square brackets (penalise once in paper) 1
– if round, penalise but mark on in (iv) if Kc wrong, no marks in (iv) either but mark on from a minor slip in formula
(iii) V cancels in Kc expression or no moles same on top and bottom of expression or total moles reactants = moles products, i.e. total no of moles does not change 1
(iv) Kc = 2
2
)172.0()023.0(
1
= 0.0179 or 1.79 × 10–2
Conseq on (i) 1 Allow 0.018 or 1.8 × 10–2
(v) Kc = 55.9 or 56 1 Conseq i.e. (answer to (iv))–1
[13]
2. (a) Kc = 23
22
OH][CH [X]O][H [Y]
(1)
if Kc expression wrong lose units mark in (e) also must be [ ]
1
(b) Moles of X: 0.25 - 0.13 = 0.12 (1) Moles of methanol: 0.34 - 0.26 = 0.08 (1) Moles of water: 0.26 (1) 3
(c) Equal no. of moles on each side of equation (1) OR V cancels out (provided not incorrectly qualified)
1
1
(d) Calculation: Kc = 2
2
V0.08
V0.12
V0.26
V0.13
(1)
= 11(.4) (1) Can score all 3 conseq on (b) and (c) If different values from (c) used allow units only (conseq on correct Kc)
Units of Kc: none (1) but lose this mark if Kc is wrong even if none given
= (0.091)2/0.109 × 0.019 = 3.7 to 4.9 (1) 7 Ignore units NB Allow the answer 4 one mark as correct knowledge
(ii) Similar (types) of bond broken and made (1) Same number of the bonds broken and made (1) 2
any number if equal NB If a list given then the total number of each type of bond
broken and made must be the same
[9]
3
7. (a) Homogeneous; All reactants in the same phase or state (1)
Dynamic; Continuous or 'on-going' (1)
Equilibrium: Concentrations of reactants and products constant or rates of forward and backward reactions equal (1)
Equation; 2NH3 N2 + 3H2 (Must be decomposition) (1)
Kc; [N2][H2]3/[NH3]2 (1) 5
(b) Conditions: decomposition favoured by high temp (1)
since the reaction endothermic or logical statement with application of Le Chatelier's principle (1)
decomposition favoured by low pressure (1)
2 mole gas giving 4 moles gas or more gas moles on right (1)
4
(c) In practise low pressure means low production (1) low pressure means low rate (1)
high temperature means high rate (1) high temperature expensive (1)
Catalyst equilibrium yield unaffected (1) rates of forward and backwards reactions increased by an equal amount (1) more hydrogen produced in a given time (1)
Max 6 [15]
4
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Q1.The carboxylic acid 3-methylbutanoic acid is used to make esters for perfumes. The following scheme shows some of the reactions in the manufacture of this carboxylic acid.
(a) One of the steps in the mechanism for Reaction 1 involves the replacement of the functional group by bromine.
(i) Use your knowledge of organic reaction mechanisms to complete the mechanism for this step by drawing two curly arrows on the following equation.
(2)
(ii) Deduce the name of the mechanism in part (i).
(b) Reaction 3 is an acid-catalysed reaction in which water is used to break chemical bonds when the CN functional group is converted into the COOH functional group. Infrared spectroscopy can be used to distinguish between the compounds in this reaction.
PhysicsAndMathsTutor.com
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Deduce the name of the type of reaction that occurs in Reaction 3.
Identify one bond in (CH3)2CHCH2CN and a different bond in (CH3)2CHCH2COOH that can be used with infrared spectroscopy to distinguish between each compound. For each of these bonds, give the range of wavenumbers at which the bond absorbs. Use Table A on the Data Sheet when answering this question.
(c) When 3-methylbutanoic acid reacts with ethanol in the presence of an acid catalyst, an equilibrium is established. The organic product is a pleasant-smelling ester.
(CH3)2CHCH2COOH + CH3CH2OH
(CH3)2CHCH2COOCH2CH3
an ester + H2O
The carboxylic acid is very expensive and ethanol is inexpensive. In the manufacture of this ester, the mole ratio of carboxylic acid to ethanol used is 1 to 10 rather than 1 to 1.
(i) Use Le Chatelier’s principle to explain why a 1 to 10 mole ratio is used. In your explanation, you should not refer to cost.
M1.(a) (i) M1 double-headed curly arrow from the lone pair of the bromide ion to the C atom of the CH2
Penalise additional arrows.
M2 double-headed arrow from the bond to the O atom
As follows
2
(ii) M1 nucleophilic substitution M1 both words needed (allow phonetic spelling).
M2 1-bromo(-2-)methylpropane M2 Require correct spelling in the name but ignore any hyphens or commas.
2
(b) M1 hydrolysis For M1 give credit for ‘hydration’ on this occasion only.
M2 C≡N with absorption range 2220–2260 (cm−1) Credit 1 mark from M2 and M3 for identifying C≡N and either O–H(acids) or C=O or C–O without reference to wavenumbers or with incorrect wavenumbers.
M3 O–H(acids) with absorption range 2500–3000 (cm−1)
OR
C=O with absorption range 1680–1750 (cm−1)
OR
C–O with absorption range 1000–1300 (cm−1) Apply the list principle to M3
3
(c) (i) M1 Yield / product OR ester increases / goes up / gets more
M2 (By Le Chatelierߣs principle) the position of equilibrium is driven / shifts / moves to the right / L to R / in the forward direction / to the
PhysicsAndMathsTutor.com
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product(s)
M3 – requires a correct statement in M2
(The position of equilibrium moves)
to oppose the increased concentration of ethanol
to oppose the increased moles of ethanol
to lower the concentration of ethanol
to oppose the change and decrease the ethanol If no reference to M1, marks M2 and M3 can still score BUT if M1 is incorrect CE=0 If there is reference to ‘pressure’ award M1 ONLY.
3
(ii) M1
Catalysts provide an alternative route / pathway / mechanism
OR
surface adsorption / surface reaction occurs For M1, not simply ‘provides a surface’ as the only statement. M1 may be scored by reference to a specific example.
M2
that has a lower / reduced activation energy
OR
lowers / reduces the activation energy Penalise M2 for reference to an increase in the energy of the molecules. For M2, the student may use a definition of activation energy without referring to the term. Reference to an increase in successful collisions in unit time alone is not sufficient for M2 since it does not explain why this has occurred.
2
[12]
M2.(a) (i) (Compounds with the) same molecular formula Allow same number and type of atom for M1
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Ignore same general formula. 1
But different structural formula / different displayed formula / different structures / different skeletal formula
M2 dependent on M1 Not different positions of atoms / bonds in space.
1
(ii) But-2-ene Allow but-2-ene. Allow but 2 ene. Ignore punctuation.
1
(iii) (2)-methylprop-(1)-ene Do not allow 2-methyleprop-1-ene.
1
(iv)
Do not allow skeletal formulae. Penalise missing H and missing C
1
(b) (i) C4H8 + 2O2 → 4C + 4H2O Accept multiples.
1
PhysicsAndMathsTutor.com
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(ii) Exacerbates asthma / breathing problems / damages lungs / smog / smoke / global dimming
Ignore toxic / pollutant / soot / carcinogen. Do not allow greenhouse effect / global warming / acid rain / ozone.
Q3.Which one of the following types of reaction mechanism is not involved in the above sequence?
CH3CH2CH3 (CH3)2CHCl (CH3)2CHCN
(CH3)2CHCH2NHCOCH3 (CH3)2CHCH2NH2
A free-radical substitution
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B nucleophilic substitution
C elimination
D nucleophilic addition-elimination (Total 1 mark)
Q4. The conversion of compound A into compound B can be achieved in two steps as shown below.
The intermediate compound, X, has an absorption at 1650 cm–1 in its infra-red spectrum.
(a) Identify compound X. Explain your answer. (2)
(b) For each step in this conversion, give the reagents and essential conditions required and outline a mechanism.
(11)
(c) Show how the number of peaks in their proton n.m.r. spectra would enable you to distinguish between compounds A and B.
(2) (Total 15 marks)
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M1. (a) (i)
(ii) restricted rotation OR no rotation OR cannot rotate (1) 3
(b) (i) Mechanism:
M1 and M2 independent Curly arrows must be from a bond or a lone pair Do not penalise sticks
Penalise M1 if precedes (penalise this once) Penalise incorrect δ+ δ– for M2 Penalise + on C atom for M2 Only allow M1 for incorrect haloalkane
Role of the hydroxide ion: nucleophile (1) electron pair donor lone pair donor
NOT nucleophilic substitution
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(ii) Mechanism:
Only allow M1 and M2 for incorrect haloalkane unless RE on (i) + charge on H on molecule, penalise M1 M3 independent M2 must be to correct C–C M1 must be correct H atom Credit M1 and M2 via carbocation mechanism No marks after any attack of C by OH–
Role of the hydroxide ion: base (1) proton acceptor accepts H+
7
[10]
M2. (a) (i)
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If wrong carbocation, lose structure mark If wrong alkene, lose structure mark Can still score ¾ i.e. penalise M3 Penalise M2 if polarity included incorrectly no bond between H and Br bond is shown as or
(ii) credit secondary carbocation here if primary carbocation has been used in (i)
Ignore attack on this carbocation by 5
(b) (i) Structure:
No credit for propan-1-ol even when named correctly Credit propane-2-ol
Name: propan-2-ol (1)
Not 2-hydroxypropane
(ii) Name of mechanism: nucleophilic substitution (1) (both words) (NOT SN1 orSN2)
Mechanism:
PhysicsAndMathsTutor.com
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penalise incorrect polarity on C‑ Br (M1) Credit the arrows even if incorrect haloalkane If SN1, both marks possible
Allow one for “A has more peaks than B” when no number of peaks is given
2
[15]
PhysicsAndMathsTutor.com
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Q1.Some airbags in cars contain sodium azide (NaN3).
(a) Sodium azide is made by reacting dinitrogen monoxide gas with sodium amide (NaNH2) as shown by the equation.
2NaNH2 + N2O NaN3 + NaOH + NH3
Calculate the mass of sodium amide needed to obtain 550 g of sodium azide, assuming there is a 95.0% yield of sodium azide. Give your answer to 3 significant figures.
(iii) Which is the correct formula of magnesium azide?
Tick (✓) one box.
Mg3N
MgN
MgN6
Mg3N2
(1)
(Total 21 marks)
Q2.A sample of pure Mg(NO3)2 was decomposed by heating as shown in the equation below.
PhysicsAndMathsTutor.com
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2Mg(NO3)2(s) 2MgO(s) + 4NO2(g) + O2(g)
(a) A 3.74 × 10−2 g sample of Mg(NO3)2 was completely decomposed by heating.
Calculate the total volume, in cm3, of gas produced at 60.0 °C and 100 kPa. Give your answer to the appropriate number of significant figures. The gas constant R = 8.31 J K−1 mol−1.
Total volume of gas = ...................... cm3
(5)
(b) The mass of MgO obtained in this experiment is slightly less than that expected from the mass of Mg(NO3)2 used. Suggest one practical reason for this.
Q3.Calamine lotion can contain a mixture of zinc carbonate and zinc oxide in suspension in water. A manufacturer of calamine lotion claims that a sample contains 15.00 g of zinc carbonate and 5.00 g of zinc oxide made up to 100 cm3 with distilled water.
(a) A chemist wanted to check the manufacturer’s claim. The chemist took a 20.0 cm3 sample of the calamine lotion and added it to an excess of sulfuric acid. The volume of carbon dioxide evolved was measured over time. The chemist’s results are shown in the table.
Total moles of gas produced = 5/2 × moles of Mg(NO3)2
= 5/2 × 2.522 × 10–4 = 6.305 × 10–4
If ratio in stage 2 is incorrect, maximum marks for stage 3 is 2
1
Stage 3
PV = nRT so volume of gas V = nRT / P 1
V = = 1.745 × 10–5 m3
1
V = 1.745 × 10–5 × 1 × 106 = 17.45 cm3 = 17.5 (cm3) Answer must be to 3 significant figures (answer could be 17.4 cm3 dependent on intermediate values)
1
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(b) Some of the solid is lost in weighing product / solid is blown away with the gas 1
[6]
M3.(a) (i) Uses sensible scales. Lose this mark if the plotted points do not cover half of the paper.
Lose this mark if the graph plot goes off the squared paper
Lose this mark if volume is plotted on the x-axis 1
All points plotted correctly Allow ± one small square.
1
Smooth curve from 0 seconds to at least 135 seconds − the line must pass through or close to all points (± one small square).
Make some allowance for the difficulties of drawing a curve but do not allow very thick or doubled lines.
1
(ii) Any value in the range 91 to 105 s Allow a range of times within this but not if 90 quoted.
1
(b) (i) Using pV = nRT This mark can be gained in a correctly substituted equation.
1
100 000 × 570 × 10−6 = n × 8.31 × 293 Correct answer with no working scores one mark only.
1
n = 0.0234 mol Do not penalise precision of answer but must have a minimum of 2 significant figures.
1
(ii) Mol of ZnCO3 = 0.0234 Mark consequentially on Q6
M1 1
Mass of ZnCO3 = M1 × 125.4 = 2.9(3) or 2.9(4) g If 0.0225 used then mass = 2.8(2) g
PhysicsAndMathsTutor.com
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Q1.(a) The hydrocarbon but-1-ene (C4H8) is a member of the homologous series of alkenes. But-1-ene has structural isomers.
(i) State the meaning of the term structural isomers.
Which is stronger than IMF (VDW / dipole-dipole forces) in ethanethiol / (H bonding) is the strongest IMF
Only award M2 if M1 given, but allow IMF in ethanol are stronger than in ethanethiol for maximum 1 mark
1
(g) (i) (2,2-)dimethylpropane Ignore punctuation
1
(ii) Because molecule is smaller / less polarisable / has less surface (area) / is more spherical / molecules can’t get as close to one another (to feel the vdW forces)
Allow converse answers referring to straight chain isomers CE = 0 / 2 if breaking bonds
1
vdW intermolecular forces or vdW force between molecules are weaker or fewer
Need vdW rather than just IMF 1
(iii) 1 or one 1
(h) (i) C9H20
H20C9
1
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(ii) Thermal (cracking) If not thermal cracking CE = 0 / 2
1
High pressure AND high temperature If blank mark on Allow high P and T
1
OR
Pressure of ≥ 10 atm, ≥ 1 MPa ≥ 1000 kPa
AND temp of 400 °C ≤ T ≤ 1000 °C or 650 K ≤ T≤ 1300 K Do not allow high heat If no units for T, then range must be 650 − 1000
1
[17]
M3.(a) (i) Crude oil / oil / petroleum Do not allow ‘petrol’
1
(ii) Fractional distillation / fractionation / fractionating Not distillation alone
1
(b) (i) 5 Allow five / V
1
(ii) Chain (isomerism) Allow branched chain / chain branched / side chain (isomerism) Ignore position (isomerism) Do not allow straight chain / geometric / branched / function
1
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(c) (i) C12H26 / H26C12
Only 1
(ii) Thermal cracking If not thermal cracking, CE = 0/2 If blank mark on
1
High temperature Allow ‘high heat’ for ‘high temperature’
(400°C < T < 900°C) or (650 K < T < 1200 K) Not ‘heat’ alone If no T, units must be 650 – 900
and
High pressure (> 10 atm, > 1 MPa, >1000 kPa) 1
(iii) To produce substances which are (more) in demand / produce products with a high value / products worth more
Ignore ‘to make more useful substances’ 1
(d) (i) Corrosive or diagram to show this hazard symbol Ignore irritant, acidic, toxic, harmful
1
(ii) ( 120.5 × 100)(86 + 71 )
=76.75(%) or 76.8(%) Allow answers > 3 sig figs
1
(e) 2,2-dichloro-3–methylpentane Ignore punctuation Any order
1
PhysicsAndMathsTutor.com
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Q1.Vanadium is an important metal. Ferrovanadium, an alloy of iron and vanadium, is used to make a strong type of vanadium-steel. Pure vanadium is used in nuclear reactors.
(a) The table shows some standard enthalpy of formation data.
V2O5(s) CaO(s)
ΔHfθ / kJ mol−1 −1560 −635
In the oldest method of extraction of vanadium, V2O5 is reacted with calcium at a high temperature.
5Ca(s) + V2O5(s) 2V(s) + 5CaO(s)
Use data from the table and the equation to calculate the standard enthalpy change for this reaction.
State the type of reaction that V2O5 has undergone.
Suggest one major reason why this method of extracting vanadium is expensive, other than the cost of heating the reaction mixture.
(b) In the first stage of the extraction of antimony from a high-grade ore, antimony sulfide is roasted in air to convert it into antimony(III) oxide (Sb2O3) and sulfur dioxide.
(c) In the second stage of the extraction of antimony from a high-grade ore, antimony(III) oxide is reacted with carbon monoxide at high temperature.
(i) Use the standard enthalpies of formation in the table and the equation given below the table to calculate a value for the standard enthalpy change for this reaction.
(d) Deduce one reason why the method of extraction of antimony from a low-grade ore, described in part (a), is a low-cost process. Do not include the cost of the ore.
Q4.The alcohol 2-methylpropan-2-ol, (CH3)3COH, reacts to form esters that are used as flavourings by the food industry. The alcohol can be oxidised to produce carbon dioxide and water.
A student carried out an experiment on a pure sample of 2-methylpropan-2-ol to determine its enthalpy of combustion. A sample of the alcohol was placed into a spirit burner and positioned under a beaker containing 50 cm3 of water. The spirit burner was ignited and allowed to burn for several minutes before it was extinguished.
The results for the experiment are shown in Table 1.
Table 1
Initial temperature of the water / °C 18.1
Final temperature of the water / °C 45.4
Initial mass of spirit burner and alcohol / g 208.80
Final mass of spirit burner and alcohol / g 208.58
(a) Use the results from Table 1 to calculate a value for the heat energy released from the combustion of this sample of 2-methylpropan-2-ol. The specific heat capacity of water is 4.18 J K–1 g–1. Show your working.
M1.(a) M1 (could be scored by a correct mathematical expression) M1 ΔH = ΣΔHf (products) − ΣΔHf (reactants) OR a correct cycle of balanced equations
M2 = 5(−635) − (−1560)
= − 3175 + 1560
(This also scores M1)
M3 = − 1615 (kJ mol−1) Award 1 mark ONLY for (+) 1615
Correct answer to the calculation gains all of M1, M2 and M3
Credit 1 mark for(+) 1615 (kJ mol−1) For other incorrect or incomplete answers, proceed as follows • check for an arithmetic error (AE), which is either a transposition error or an incorrect multiplication; this would score 2 marks (M1 and M2) • If no AE, check for a correct method; this requires either a correct cycle with V2O5 and 5CaO OR a clear statement of M1 which could be in words and scores only M1
M4 Type of reaction is • reduction • redox • (or accept) V2O5 / it / V(V) has been reduced
In M4 not “vanadium / V is reduced”
M5 Major reason for expense of extraction − the answer must be about calcium
Calcium is produced / extracted by electrolysis OR calcium is expensive to extract OR calcium extraction uses electricity OR calcium extraction uses large amount of energy OR calcium is a (very) reactive metal / reacts with water or air OR calcium needs to be extracted / does not occur native
QoL
Accept calcium is expensive “to produce” but not “to source, to get, to obtain, to buy” etc. In M5 it is neither enough to say that calcium is “expensive” nor that calcium “must be purified”
5
(b) M1 2Al + Fe2O3 2Fe + Al2O3
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Ignore state symbols Credit multiples of the equation
M2 (Change in oxidation state) 0 to (+)3 OR (changed by) +3
In M2 if an explanation is given it must be correct and unambiguous
• Explosion risk with hydrogen (gas) OR H2 is flammable For M2 / M3 there must be reference to hydrogen; it is not enough to refer simply to an explosion risk For M2 / M3 with HCl hazard, require reference to acid(ic) / corrosive / toxic only
M4 The only other product / the HCl is easily / readily removed / lost / separated because it is a gas OR will escape (or this idea strongly implied) as a gas OR vanadium / it is the only solid product (and is easily separated) OR vanadium / it is a solid and the other product / HCl is a gas
In M4 it is not enough to state simply that HCl is a gas, since this is in the question.
4
[11]
M2.(a) C(s) + 2F2(g) CF4(g) State symbols essential
1
(b) Around carbon there are 4 bonding pairs of electrons (and no lone pairs) 1
Therefore, these repel equally and spread as far apart as possible
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1
(c) ΔH = Σ ΔfH products – Σ ΔfH reactants or a correct cycle 1
The student is correct because the F–F bond energy is much less than the C–H or other covalent bonds, therefore the F–F bond is weak / easily broken
Relevant comment comparing to other bonds (Low activation energy needed to break the F–F bond)
1
[10]
M3.(a) (i) 3Fe + Sb2S3 3FeS + 2Sb
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Or multiples. Ignore state symbols.
1
(ii) Fe Fe2+ + 2e−
Ignore charge on the electron unless incorrect. Or multiples. Credit the electrons being subtracted on the LHS. Ignore state symbols.
1
(b) (i) Sb2S3 + 4.5O2 Sb2O3 + 3SO2
Or multiples. Ignore state symbols.
1
(ii) SO3 or sulfur trioxide / sulfur (VI) oxide Credit also the following ONLY. H2SO4 or sulfuric acid. OR
Gypsum / CaSO4 or plaster of Paris. 1
(c) (i) M1 (could be scored by a correct mathematical expression) Correct answer gains full marks.
M1 ∆Hr = Ʃ∆Hf(products) − Ʃ∆Hf(reactants)
OR a correct cycle of balanced equations / correct numbers of moles Credit 1 mark for +104 (kJ mol−1).
M2 = 2(+20) + 3(−394) − (−705) − 3(−111)
= 40 −1182 + 705 + 333
= −1142 − (−1038)
(This also scores M1)
M3 = −104 (kJ mol−1)
(Award 1 mark ONLY for + 104) For other incorrect or incomplete answers, proceed as follows:
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• Check for an arithmetic error (AE), which is either a transposition error or an incorrect multiplication; this would score 2 marks. • If no AE, check for a correct method; this requires either a correct cycle with 3CO, 2Sb and 3CO2 OR a clear statement of M1 which could be in words and scores only M1.
3
(ii) It / Sb is not in its standard state
OR
Standard state (for Sb) is solid / (s)
OR
(Sb) liquid is not its standard state Credit a correct definition of standard state as an alternative to the words ‘standard state’. QoL
1
(iii) Reduction OR reduced OR redox 1
(d) Low-grade ore extraction / it
• uses (cheap) scrap / waste iron / steel
• is a single-step process
uses / requires less / low(er) energy Ignore references to temperature / heat or labour or technology.
1
[10]
M4.(a) (Q = mcΔT)
= 50 × 4.18 × 27.3 If incorrect (eg mass = 0.22 or 50.22 g) CE = 0 / 2
1
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= 5706 J (accept 5700 and 5710) Accept 5.7 kJ with correct unit. Ignore sign.
1
(b) Mr of 2-methylpropan-2-ol = 74(.0) For incorrect Mr, lose M1 but mark on.
1
Moles = mass / Mr
= 0.22 / 74(.0)
= 0.00297 moles 1
ΔH = –5706 / (0.002970 × 1000)
= –1921 (kJ mol–1) If 0.22 is used in part (a), answer = –8.45 kJ mol–1 scores 3
(Allow –1920, –1919) If uses the value given (5580 J), answer = –1879 kJ mol–1 scores 3 Answer without working scores M3 only. Do not penalise precision. Lack of negative sign loses M3
1
(c) ΔH = ΣΔH products – ΣΔH reactants OR a correct cycle
Correct answer with no working scores 1 mark only. 1
(d) A chemical test can be used to distinguish between separate samples of Isomer 2 and Isomer 3. Identify a suitable reagent for the test. State what you would observe with Isomer 2 and with Isomer 3.
Test reagent ...............................................................................................
Observation with Isomer 2...........................................................................
(b) The first stage in this conversion involves the reaction of hydrogen bromide with but-1-ene.
CH3CH2CH=CH2 + HBr CH3CH2CHBrCH3
Outline a mechanism for this reaction.
(4)
(c) The second stage is to convert 2-bromobutane into but-2-ene.
CH3CH2CHBrCHCH3
+ KOH CH3CH=CHCH3 + KBr + H2O
Outline a mechanism for this reaction.
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M1.B [1]
M2. (a) Pentan-2-one ONLY but ignore absence of hyphens
1
(b) Functional group (isomerism) Both words needed
1
(c) (i)
Award credit provided it is obvious that the candidate is drawing the Z / cis isomer The group needs to be CHOHCH3 but do not penalise poor C–C bonds or absence of brackets around OH Trigonal planar structure not essential
1
(ii) Restricted rotation (about the C=C)
OR
No (free) rotation (about the C=C) 1
(d)
M1 Tollens’ (reagent)
(Credit ammoniacal silver nitrate OR a description of making Tollens’)
(Do not credit Ag+, AgNO3 or [Ag(NH3)2
+] or “the silver mirror test”
M1 Fehling’s (solution) / Benedict’s
(Penalise Cu2+(aq) or CuSO4 but mark M2 and M3)
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on their own, but mark M2 and M3)
M2 silver mirror
OR black solid or black precipitate
M2 Red solid/precipitate
(Credit orange or brown solid)
M3 (stays) colourless
OR
no (observed) change / no reaction
M3 (stays) blue
OR
no (observed) change / no reaction
If M1 is blank CE = 0, for the clip Check the partial reagents listed and if M1 has a totally incorrect reagent, CE = 0 for the clip Allow the following alternatives M1 (acidified) potassium dichromate(VI) (solution); mark on from incomplete formulae or incorrect oxidation state M2 (turns) green M3 (stays) orange / no (observed) change / no reaction OR M1 (acidified) potassium manganate(VII) (solution); mark on from incomplete formulae or incorrect oxidation state M2 (turns) colourless M3 (stays) purple / no (observed) change / no reaction In all cases for M3 Ignore “nothing (happens)” Ignore “no observation”
3
(e) (i) Spectrum is for Isomer 1
or named or correctly identified The explanation marks in (e)(ii) depend on correctly identifying Isomer 1. The identification should be unambiguous but candidates should not be penalised for an imperfect or incomplete name. They may say “the alcohol” or the “alkene” or the “E isomer”
1
(ii) If Isomer 1 is correctly identified, award any two from
• (Strong / broad) absorption / peak in the range
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3230 to 3550 cm–1 or specified value in this range or marked correctly on spectrum and (characteristic absorption / peak for) OH group /alcohol group
• No absorption / peak in range 1680 to 1750 cm–1 or absence marked correctly on spectrum and (No absorption / peak for a) C=O group / carbonyl group / carbon-oxygen double bond
• Absorption / peak in the range 1620 to 1680 cm–1
or specified value in this range or marked correctly on spectrum and
(characteristic absorption / peak for) C=C group / alkene / carbon-carbon double bond
If 6(e)(i) is incorrect or blank, CE=0 Allow the words “dip” OR “spike” OR “trough” OR “low transmittance” as alternatives for absorption. Ignore reference to other absorptions e.g. C-H, C-O
2
[10]
M3. (a) Ca(OH)2 OR Mg(OH)2
Ignore name Could be ionic
1
(b) NaF or sodium fluoride
OR
NaCl or sodium chloride Either formula or name can score Do not penalise the spelling “fluoride” When both formula and name are written, • penalise contradictions • if the attempt at the correct formula is incorrect, ignore it and credit correct name for the mark unless contradictory
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• if the attempt at the correct name is incorrect, ignore it and credit correct formula for the mark unless contradictory
1
(c) NaClO OR NaOCl Ignore name (even when incorrect) The correct formula must be clearly identified if an equation is written
1
(d) Br2 (ONLY) Only the correct formula scores; penalise lower case “b”, penalise upper case “R”, penalise superscript Ignore name The correct formula must be clearly identified if an equation is written
1
(e) M1 S OR S8 OR S2
M2 I2 (ONLY) Ignore names penalise lower case “i” for iodine, penalise superscripted numbers Mark independently The correct formula must be clearly identified in each case if an equation is written
2
(f) (i) CH3CH2CH=CH2
Structure of but-1-ene. Ignore name Credit “sticks” for C-H bonds
1
(ii) CH3CH2CH2CH2OH Structure of butan-1-ol. Ignore name Credit “sticks” for C-H bonds
1
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(iii) CH3CH2CH3
Structure of propane. Ignore name Ignore calculations and molecular formula Credit “sticks” for C-H bonds Ignore the molecular ion
1
(iv) CH3CH2Br OR C2H5Br
Structure of bromoethane. Ignore name and structure of nitrile Credit “sticks” for C-H bonds
1
[10]
M4.C [1]
M5.(a) Position(al) (isomerism) 1
(b) Penalise one mark from their total if half-headed arrows are used
M1 must show an arrow from the double bond towards the H atom of the H–Br molecule
M1 Ignore partial negative charge on the double bond.
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M2 must show the breaking of the H–Br bond. M2 Penalise partial charges on H–Br bond if wrong way and penalise formal charges
M3 is for the structure of the secondary carbocation. Penalise M3 if there is a bond drawn to the positive charge
M4 must show an arrow from the lone pair of electrons on the negatively charged bromide ion towards the positively charged carbon atom of either a primary or secondary carbocation.
Penalise once only in any part of the mechanism for a line and two dots to show a bond Maximum any 3 of 4 marks for wrong reactant or primary carbocation. If Br2 is used, maximum 2 marks for their mechanism Do not penalise the use of “sticks”
NB The arrows here are double-headed 4
(c)
Penalise one mark from their total if half-headed arrows are used
M1 must show an arrow from the lone pair on oxygen of a negatively charged hydroxide ion to a correct H atom
Penalise M1 if covalent KOH
M2 must show an arrow from a C–H bond adjacent to the C–Br bond towards the appropriate C–C bond. Only award if an arrow is shown attacking the H atom of an adjacent C–H (in M1)
M3 is independent provided it is from their original molecule. Penalise M3 for formal charge on C of the C–Br or incorrect partial charges on C–Br Penalise M3 if an extra arrow is drawn from the Br of the C–Br bond to, for example, K+
Ignore other partial charges Penalise once only in any part of the mechanism for a line and two dots to show a bond. Maximum any 2 of 3 marks for wrong reactant or wrong product(if shown) or a mechanism that leads to but-1-ene
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Q1.This question is about the elements in Period 3 of the Periodic Table.
(a) State the element in Period 3 that has the highest melting point. Explain your answer.
Element ..........................................................................................................
(ii) Draw a diagram to show how one molecule of hydrogen peroxide interacts with one molecule of water. Include all lone pairs and partial charges in your diagram.
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(3)
(c) Explain, in terms of electronegativity, why the boiling point of H2S2 is lower than H2O2.
M1.(a) Silicon / Si If not silicon then CE = 0 / 3
1
covalent (bonds) M3 dependent on correct M2
1
Strong or many of the (covalent) bonds need to be broken / needs a lot of energy to break the (covalent) bonds
Ignore hard to break 1
(b) Argon / Ar If not argon then CE = 0 / 3. But if Kr chosen, lose M1 and allow M2+M3
1
Large(st) number of protons / large(st) nuclear charge Ignore smallest atomic radius
1
Same amount of shielding / same number of shells / same number of energy levels
Allow similar shielding 1
(c) Chlorine / Cl Not Cl2, Not CL, Not Cl2
1
(d) (i)
Or any structure with 3 bonds and 2 lone pairs Ignore any angles shown
1
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Or a structure with 2 bonds and 1 lone pair
1
(ii) Bent / v shape Ignore non-linear, angular and triangular Apply list principle
1
(iii) Cl2 + F2 ClF3
No multiples Ignore state symbols
1 [11]
M2.(a) Al + 1.5Cl 2 → AlCl3
Accept multiples. Also 2Al + 3Cl2 → Al2Cl6
Ignore state symbols. 1
(b) Coordinate / dative (covalent) If wrong CE=0/2 if covalent mark on.
1
Electron pair on Cl − donated to Al(Cl 3) QoL Lone pair from Cl − not just Cl Penalise wrong species.
1
(c) Al2Cl6 or AlBr3
Allow Br3Al or Cl6Al2
Upper and lower case letters must be as shown. Not 2AlCl3
1
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(d) SiCl4 / silicon tetrachloride Accept silicon(4) chloride or silicon(IV) chloride. Upper and lower case letters must be as shown. Not silicon chloride.
1
(e)
Accept shape containing 5 bonds and no lone pairs from Tl to each of 5 Br atoms. Ignore charge.
1
Trigonal bipyramid(al) 1
(f) (i) Cl — Tl — Cl
Accept this linear structure only with no lone pair on Tl 1
(ii) (Two) bonds (pairs of electrons) repel equally / (electrons in) the bonds repel to be as far apart as possible
Dependent on linear structure in (f)(i). Do not allow electrons / electron pairs repel alone.
1
(g) Second 1
[10]
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M3.(a)
Mark is for correct number of bonds and lone pair in each case. Ignore charges if shown.
2
Pyramidal / trigonal pyramid Allow tetrahedral.
1
107° Allow 107 to 107.5°.
1
(b) M1 Ionic CE = 0 / 3 if not ionic.
1
M2 Oppositely charged ions / Tl+ and Br− ions If molecules / intermolecular forces / metallic bonding, CE=0.
1
M3 Strong attraction between ions M3 dependent on M2. Allow ‘needs a lot of energy to break / overcome’ instead of ‘strong’.
1
(c) Tl + TlBr Allow multiples.
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Ignore state symbols even if incorrect. 1
[8]
M4.(a) 94−105.5° 1
(b) (i) Hydrogen bond(ing) / H bonding / H bonds Not just hydrogen
1
(ii)
OR
1 mark for all lone pairs 1 mark for partial charges on the O and the H that are involved in H bonding 1 mark for the H-bond, from Hδ+ on one molecule to lone pair on O of other molecule
3
(c) Electronegativity of S lower than O or electronegativity difference between H and S is lower
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Mark independently 1
No hydrogen bonding between H2S2 molecules
Or only van der Waals / only dipole-dipole forces between H2S2 molecules If breaking covalent bonds CE = 0
1 [7]
M5.(a) C(s) + 2F2(g) CF4(g) State symbols essential
1
(b) Around carbon there are 4 bonding pairs of electrons (and no lone pairs) 1
Therefore, these repel equally and spread as far apart as possible 1
(c) ΔH = Σ ΔfH products – Σ ΔfH reactants or a correct cycle 1
(b) State the type of structure shown by crystals of sulfur and phosphorus. Explain why the melting point of sulfur is higher than the melting point of phosphorus.
(c) Draw a diagram to show how the particles are arranged in aluminium and explain why aluminium is malleable. (You should show a minimum of six aluminium particles arranged in two dimensions.)
(e) Identify the element in Period 2 that has the highest first ionisation energy and give its electron configuration.
Element .......................................................................................................
Electron configuration .................................................................................. (2)
(f) State the trend in first ionisation energies in Group 2 from beryllium to barium. Explain your answer in terms of a suitable model of atomic structure.
(e) A general trend exists in the first ionisation energies of the Period 2 elements lithium to fluorine. Identify one element which deviates from this general trend.
M1. (a) Macromolecular/giant covalent/giant molecular/giant atomic If IMF/H-bonds/Ionic/metallic CE = 0/3 covalent bond between molecules CE = 0/3 If giant unqualified M1 = 0 but mark on
1
Many/strong covalent bonds M2 and M3 can only be scored if covalent mentioned in answer Ignore metalloid and carbon Ignore bp
1
Bonds must be broken/overcome Ignore numbers of bonds and references to energy
1
(b) (Simple) molecular QoL Do not allow simple covalent for M1 Giant covalent/ionic/metallic, CE = 0 If breaking covalent bonds CE= 0/3
1
S bigger molecule (than P) or S8 and P4 references QoL Allow more electrons in sulfur molecule or S8
Do not allow S is bigger then P Allow S molecule has a bigger Mr
Do not allow contradictions 1
So more/stronger van der Waals’ forces (to be broken or overcome) Not just more energy to break
1
(c) Regular arrangement of minimum of 6 particles in minimum of 2 rows
Ignore e– Do not allow ring arrangements OR structures bonded with electrons
1
+ charge in each one (of 6)
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Allow +, (1+, 2+ or 3+) in ions/or in words 1
Rows/planes/sheets/layers (of atoms/ions) can slide (owtte) over one another
M3 independent If ionic bonding/molecules/IMF/vdw/covalent, penalise M3 Ignore layers of electrons sliding
1
(d) Bigger charge (3+ compared to 1+) CE = 0 if molecules, ionic, covalent, IMF (Allow Al2+)
OR smaller atom/ion in Al/more protons/bigger nuclear charge 1
More free/delocalised electrons (in Al)/bigger sea of electrons in Al Accept 2 or 3 delocalised electrons compared to 1 in Na
1
Stronger metallic bonding/stronger (electrostatic) attraction between the (+) ions or nuclei and the (delocalised) electrons (or implied)
Must be implied that the electrons are the delocalised ones not the electrons in the shells. Accept converse arguments
1
[12]
M2. (a) Cross between the Na cross and the Mg cross 1
One mark for state symbols consequential on getting equation correct. Electron does not have to have the – sign on it Ignore (g) if put as state symbol with e– but penalise state symbol mark if other state symbols on e–
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2
(c) 2nd/second/2/II Only
1
(d) Paired electrons in (3)p orbital Penalise wrong number If paired electrons repel allow M2
1
repel 1
(e) Neon/Ne No consequential marking from wrong element
1
1s22s22p6/[He}2s22p6
Allow capital s and p Allow subscript numbers
1
(f) Decreases CE if wrong
1
Atomic radius increases/electron removed further from nucleus or nuclear charge/electron in higher energy level/Atoms get larger/more shells
Accept more repulsion between more electrons for M2 Mark is for distance from nucleus Must be comparative answers from M2 and M3 CE M2 and M3 if mention molecules Not more sub-shells
1
As group is descended more shielding 1
[11]
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M3. (a) 2s22p63s1
1s2 can be rewritten Allow 2s22px
22py22pz
23s1
Allow subscripts and capitals 1
(b) (i) Energy/enthalpy (needed) to remove one mole of electrons from one mole of atoms/compounds/molecules/elements
1
OR
Energy to form one mole of positive ions from one mole of atoms
OR
Energy/enthalpy to remove one electron from one atom
In the gaseous state (to form 1 mol of gaseous ions) Energy given out loses M1 M2 is dependent on a reasonable attempt at M1 Energy needed for this change X(g) → X+(g) + e(–) = 2 marks This equation alone scores one mark
1
(ii) Mg+(g) → Mg2+(g) + e(–)
Mg+(g) + e(–) → Mg2+(g) + 2e(–)
Mg+(g) – e(–) → Mg2+(g) Do not penalise MG Not equation with X
1
(iii) Electron being removed from a positive ion (therefore need more energy)/electron being removed is closer to the nucleus/Mg+
smaller (than Mg)/Mg+ more positive than Mg Allow from a + particle/species Not electron from a higher energy level/or higher sub-level More protons = 0
1
(iv) Range from 5000 to 9000 kJ mol–1
1
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(c) Increase If decrease CE = 0/3 If blank mark on
1
Bigger nuclear charge (from Na to Cl)/more protons QWC
1
electron (taken) from same (sub)shell/similar or same shielding/ electron closer to the nucleus/smaller atomic radius
If no shielding = 0 Smaller ionic radius = 0
1
(d) Lower If not lower CE = 0/3 If blank mark on Allow does not increase
1
Two/pair of electrons in (3)p orbital or implied Not 2p
1
repel (each other) M3 dependent upon a reasonable attempt at M2
1
(e) Boron/B or oxygen/O/O2
1
[13]
M4. (a) 2s2 2p6; If ignored the 1s2 given and written 1s22s22p6 mark as correct Allow capitals and subscripts
1
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(b) (i) Na+(g) → Na2+ (g) + e(–); One mark for equation and one mark for state symbols
Na+(g) + e(–) → Na2+ (g) + 2e(–); M2 dependent on M1 Allow Na+(g) – e(–) → Na(g) Allow X+(g) → X2+ (g) + e = 1 mark
2
(ii) Na(2+) requires loss of e– from a 2(p) orbital or 2nd energy level or 2nd shell and Mg(2+) requires loss of e– from a 3(s) orbital or 3rd
energy level or 3rd shell / Na(2+) loses e from a lower (energy) orbital/ or vice versa;
Not from 3p 1
Less shielding (in Na); Or vice versa for Mg
1
e(–) closer to nucleus/ more attraction (of electron to nucleus) (in Na); M3 needs to be comparative
1
(iii) Aluminium /Al; 1
(c) Decreases; If not decreases CE = 0 If blank, mark on
1
Increasing nuclear charge/ increasing number of protons; 1
Electrons in same shell or level/ same shielding/ similar shielding; 1
(d) Answer refers to Na; Allow converse answers relating to Mg.
Na fewer protons/smaller nuclear charge/ fewer delocalised electrons; Allow Mg is 2+ and Na is +. If vdw CE = 0.
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Q1.The table below shows observations of changes from some test-tube reactions of aqueous solutions of compounds Q, R and S with five different aqueous reagents. The initial colours of the solutions are not given.
BaCl2 +
HCl AgNO3 +
HNO3
NaOH Na2CO3 HCl (conc)
Q no change observed
pale cream precipitate
white precipitate
white precipitate
no change observed
R no change observd
white precipitate
white precipitate, dissolves in excess of
NaOH
white precipitate,
bubbles of a gas
no change observed
S white
precipitate no change observed
brown precipitate
brown precipitate,
bubbles of a gas
yellow solution
(a) Identify each of compounds Q, R and S. You are not required to explain your answers.
Identity of Q ...................................................................................................
(e) Magnesium burns with a bright white light and is used in flares and fireworks.
Use your knowledge of the reactions of Group 2 metals with water to explain why water should not be used to put out a fire in which magnesium metal is burning.
Q3.A student investigated how the initial rate of reaction between sulfuric acid and magnesium at 20 °C is affected by the concentration of the acid.
The equation for the reaction is
H2SO4(aq) + Mg(s) MgSO4(aq) + H2(g)
(a) The student made measurements every 20 seconds for 5 minutes. The student then repeated the experiment using double the concentration of sulfuric acid.
State a measurement that the student should make every 20 seconds. Identify the apparatus that the student could use to make this measurement.
(c) When the student had finished the investigation, an excess of sodium hydroxide solution was added to the reaction mixture. This was to neutralise any unreacted sulfuric acid. The student found that a further reaction took place, producing magnesium hydroxide.
(i) Draw a diagram to show how the student could separate the magnesium hydroxide from the reaction mixture.
(2)
(ii) Suggest one method the student could use for removing soluble impurities from the sample of magnesium hydroxide that has been separated.
Q4.(a) Anhydrous strontium chloride is not used in toothpaste because it absorbs water from the atmosphere. The hexahydrate, SrCl2.6H2O, is preferred.
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A chemist was asked to determine the purity of a sample of strontium chloride hexahydrate. The chemist weighed out 2.25 g of the sample and added it to 100 cm3 of water. The mixture was warmed and stirred for several minutes to dissolve all of the strontium chloride in the sample. The mixture was then filtered into a conical flask. An excess of silver nitrate solution was added to the flask and the contents swirled for 1 minute to make sure that the precipitation was complete.
The silver chloride precipitate was separated from the mixture by filtration. The precipitate was washed several times with deionised water and dried carefully. The chemist weighed the dry precipitate and recorded a mass of 1.55 g.
(i) Calculate the amount, in moles, of AgCl in 1.55 g of silver chloride (Mr = 143.4).
(iv) Use your answers from parts (a)(ii) and (a)(iii) to calculate the percentage by mass of strontium chloride hexahydrate in the sample. Show your working. Give your answer to the appropriate precision.
(v) Several steps in the practical procedure were designed to ensure an accurate value for the percentage by mass of strontium chloride hexahydrate in the sample.
1 Explain why the solution of strontium chloride was filtered to remove insoluble impurities before the addition of silver nitrate.
(b) Magnesium hydroxide and magnesium carbonate are used to reduce acidity in the stomach. Magnesium hydroxide can be prepared by the reaction of solutions of magnesium chloride and sodium hydroxide.
(i) Write the simplest ionic equation for the reaction that occurs between magnesium chloride and sodium hydroxide. Include state symbols in your equation.
(c) Calcium ethanoate, (CH3COO)2Ca, is used in the treatment of kidney disease. Thermal decomposition of calcium ethanoate under certain conditions gives propanone and one other product.
M2.(a) M1 Used in a barium meal / barium swallow / barium enema
OR (used to absorb) X-rays Credit a correct reference to M1 written in the explanation in M2 unless contradictory.
M2 BaSO4 / barium sulfate / it is insoluble For M2 penalise obvious reference to barium or to barium ions being insoluble.
2
(b) Mg(OH)2 + 2HCl MgCl2 + 2H2O Or multiples. Ignore state symbols.
1
(c) It / magnesium hydroxide is insoluble / insufficiently soluble / sparingly soluble / less soluble than barium hydroxide / forms low concentration solutions
Weak alkali alone is insufficient. Formation of a precipitate needs explanation.
1
(d) TiCl4 + 2Mg 2MgCl2 + Ti Or multiples. Ignore state symbols.
1
(e) M1 Hydrogen / H2 produced
OR an equation to produce hydrogen / H2
( eg Mg + 2H2O Mg(OH)2 + H2)
( eg Mg + H2O MgO + H2) For M1
Do not penalise an incorrect equation; the mark is for H2 or hydrogen. Award one mark only for ‘exothermic reaction with steam /
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H2O’ for a student who has not scored M1
M2 requires correct M1
risk of explosion
OR forms explosive mixture (with air)
OR (highly) flammable Ignore ‘violent’ reaction.
2
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M3.(a) (Measure the) volume of gas / mass of the container + contents 1
Suitable named piece of equipment Gas syringe (or inverted burette or measuring cylinder, as long as student has referred to the cylinder being filled with water) / balance. Equipment must be correct for the measurement stated.
1
(b) Any one of:
• Mass of magnesium Allow amount of magnesium.
• Surface area of magnesium 1
(c) (i) Gravity: Conical flask or beaker and funnel /
Vacuum: Sealed container with a side arm and Buchner or Hirsch funnel Must be either gravity filtration (with a V-shaped funnel) or vacuum filtration (with a side-arm conical flask) appropriately drawn.
1
Filter paper Must show filter paper as at least two sides of a triangle
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(V-shaped) for gravity filtration or horizontal filter paper for vacuum filtration.
1
(ii) Wash with / add (a small amount of cold) water Ignore filtering.
1
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M4.(a) (i) 1.08 × 10–2
Do not penalise precision but must be to at least 2 significant figures. Do not accept 1 × 10–2
1
(ii) 5.4(0) × 10–3
Allow (i) / 2 Do not penalise precision but must be to at least 2 significant figures.
1
(iii) 266.6 Lose this mark if answer not given to 1 decimal place.
1
(iv) mass = 5.4(0) × 10–3 × 266.6 = 1.44 g M1
Allow (ii) × (iii). 1
percentage = 1.44 × 100 / 2.25 = 64.0 M2
Allow consequential answer from M1
Lose this mark if answer not given to 3 significant figures. Correct answer with no working scores M2 only.
1
(v) 1 Would give an incorrect / too large mass (of silver chloride)
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Do not allow ‘to get an accurate result’ without qualification. 1
2 To remove soluble impurities / excess silver nitrate (solution) / strontium nitrate (solution) Do not allow ‘to remove impurities’. Do not allow ‘to remove excess strontium chloride solution’.
1
(b) (i) Mg2+(aq) + 2OH-(aq) → Mg(OH)2(s) Allow Mg2+(aq) + 2OH-(aq) → Mg2+(OH-)2(s) Allow multiples, including fractions. Lose mark if state symbols are missing or incorrect. Lose mark if incorrect charge on an ion.
1
(ii) Does not produce CO2 / gas which distends stomach / does not produce wind / does not increase pressure in stomach
Allow ‘prevents flatulence’ and ‘prevents burping’. Do not allow ‘gas’ without qualification.