skema markah solaf1 kertas2 2011
DESCRIPTION
Solaf Chemistry Paper 2TRANSCRIPT
Question Rubric Mark1. (a) (i) +1 1
(ii) The number of protons is more than the number of electrons by one unit.
1
(b) (i) P // T 1(ii) Two shells filled with electrons//Two occupied shells 1
(c ) Proton number = 6Nucleon number = 12 or 13
1
(d) Because Q and R have the same number of valence electron that is 1.
1
(e ) (i) QS 1(ii) Ionic bond 1(iii) High melting and boling point//
Soluble in water//Insoluble in organic solvent//Can conduct electricity in molten state or aqueous solutionAny one
1
TOTAL 9MARKING SCHEME
PART A
2. (a) Electrical energy to chemical energy 1(b) (i) Cu2+, SO4
2- , H+, OH- // Copper(II)ion, Sulphate ion, Hydrogen ion, Hydroxide ion
1
(ii) Cu2+(aq) + 2e- → Cu(s) 1
(c ) (i) From magnesium plate to copper plate through external circuit
1
(ii) Mg(s) → Mg2+(aq) + 2e- 1
(iii) The cell voltage will increase 1(d) (i)
Functional diagram – 1Labelling are correct - 1
2
(ii) Use a low electric current// The iron spoon is rotated slowly throughout the process
1
Iron spoon
Silver plate
Silver nitrate solution
TOTAL 9
3. (a) blue 1(b) CuO + H2SO4 → CuSO4 + H2O
Formula of reactants and products are correct
2
(c ) Acid completely reacts with copper(II)oxide // Acid is completely neutralized by copper(II)oxide
1
(d) Mass of CuSO4 = Number of mole of CuSO4 x molar mass of CuSO4
= 0.1 mol x [64 + 32 + 64]g mol-1 - 1 = (0.1 x 160) g = 16.0 g - 1
2
(e) 1. NaOH/NH3 solution is added drop by drop until in excess
2. Blue precipitate is insoluble/blue precipitate is soluble in excess NaOH/NH3 solution.
2
(f) (i) Carbon dioxide r: formula 1(ii) CuCO3 1
TOTAL 10
4. (a) pH scale is a scale of 0 to 14 that shows the degree of acidity or alkalinity of a solution.
1
(b) 1. Hydrochloric acid is a strong acid that ionizes completely in water to produce a high concentration of hydrogen ions.
2. Ethanoic acid is a weak acid that ionizes partially in water to produce a low concentration of hydrogen ions. (Any one)
2
(c ) Mg + 2HCl → MgCl2 + H2 1(d) (i) Number of mol of NaHCO3 = 2.1 / 84
= 0.025 mol1
(ii) Molarity of NaHCO3 = 0.025 x 1000/250 = 0.10 mol dm-3
1
(e ) Sodium hydrogen carbonate is basic 1(f) M1V1 = M2V2
0.10 x 250 = 0.05 x (250 + V) - 1 V = 250 cm3 - 1
2
(g) (i) Effervescence occurs 1(ii) NaHCO3 + HCl → NaCl + H2O + CO2 1
TOTAL 11
5. (a) Zinc / zink 1
(b) atom 1(c ) (i) 1
(ii)
[ drawing shows zinc atom in the lattice of copperLukisan menunjukkan atom zink dalam kekisi
kuprum ] 1
[ Label zinc and copper correct.Label zink dan kuprum betul ] 1
2
(d) copper has orderly arrangement of atom when force is applied, atoms are easily to slide.In brass, the presence of zinc atom disrupt the orderly arrangement of copper atom when force is applied, atom difficult to slide.
kuprum mempunyai susunan atom sekata/teratur apabila dikenakan daya, atomnya mudah menggelongsorDalam loyang kehadiran atom zink mengganggu susunan teratur atom kuprum apabila dikenakan daya, atomnya sukar menggelongsor 1
1
11
(e) (i) tin // stanum // timah 1(ii) to beautify // to improve the appearance
untuk mencantikkan // membaiki rupato withstand corrosionmengelak hakisan
1
1
TOTAL 11
Question number
Answer Mark
[ particles touchClosed arrangement Condition : rows ≥ 3]zarah bersentuhan susunan tertutup syarat : barisan ≥ 3 ]
Copper atom / Atom kuprum
Zinc atom/ Atom zink
6 (a)
functional diagram label
11
(b) (i) Oxygen 1
(ii) 2H2O2 2 H2O + O2 [formula correct and balanced] [Formula correct but not balanced, 1 mark]
2
(c) (i) Experiment II 1(ii) Concentration of hydrogen peroxide in Expt II is higher than in Expt I 1
(iii)
2 (d) Lower the activation energy ( 1)
Frequency of effective collision increases (1) 2
Total 11
PART B
II
I
hydrogen peroxide solution
manganese(IV) oxide
7 (a)
Butan-1-ol
Butan-2-ol
2-methylpropan-1-ol
2-methylpropan-2-ol
Choose any two structural formula and correct name. 4
(b)Aspect P Q
Type of compound
hydrocarbon Non-hydrocarbon
Homologousseries
alkene Carboxylic acid
Type of atom present
Contains carbon atoms and
hydrogen atoms only
Contains carbon atoms, hydrogen
atoms and oxygen atom
Solubility in water
Insoluble in water
Soluble in water
Functional group
Carbon-carbon double bond
Carboxyl group
Reaction with bromine water
Decolourises brown bromine
water
Does not change the brown colour of
bromine waterReaction with
acidified potassium
manganate(VII) solution
Decolourises purple colour of
potassium manganate(VII)
solution
Does not change the purple colour of potassium
manganate(VII) solution
General formula CnH2n, n=2,3,… CnH 2n+1COOH, n=0,1,…
1
1
1
1
1
1
1
1
Max 6
(c) Apparatus : test tubes, dropperMaterials: bromine water // acidified potassiummanganate(VII) solution, hexane, hexene
Procedure:1. 2 cm3 of liquid in bottle X is poured into two
separate test tubes.2. 2 to 3 drops of bromine water are added to two
testTubes
3.The mixture is shaken. 4.Any observation is recorded. 5. Step 1 to 3 are repeated using liquid in bottle Y to replace liquid in bottle X.
Observation:
1
1
1
111
Liquid in bottle X Liquid in bottle YBrown bromine water decolourises // purple acidified potassium manganate (VII) solution decolourises
No visible change.
Liquid in bottle X is hexeneLiquid in bottle Y is hexane
1+1
11__10
T0TAL 20
No. Mark Scheme Sub Mark
Total Mark
8 (a)(i)
(ii)
(iii)
(iv)
Blue solution X = Copper (II) sulphateColurless solution Y = potassium carbonate // sodium carbonate // ammonium carbonateDouble decomposition method
CuSO4 + K2CO3 CuCO3 + K2SO4 //
CuSO4 + Na2CO3 CuCO3 + Na2SO4 //
CuSO4 + (NH4 )2CO3 CuCO3 + (NH4 )2SO4
Correct rectants and productsBalanced equations
- Add sodium hydroxide solution (until excess)- Blue precipitate formed
//- Add ammonia aqueous / ammonium
hydroxide solution (until excess)Blue precipitate soluble in excess
11
1
11
11
2
1
2
2
(b)(i) Copper(II) oxide 1 1(ii) Materials : [25 – 100] cm3 of [0.5 – 2.0] moldm-3
copper (II)sulphate solution(any suitable answer) [25 – 100] cm3 of [0.5 – 2.0] moldm-3 sodium carbonate solution (any suitable answer) Filter paperApparatus : Filter funnel, beakers, retort stand and clamp, glass rod and 100cm3 measuring cylinder.
1
1
11
Procedures :- About [25 – 100] cm3 of [0.5 – 2.0] moldm-3
copper (II) sulphate solution is measured into a beaker.
- About [25 – 100] cm3 of [0.5 – 2.0] moldm-3 sodium carbonate solution is measured and mixed with the solution in the beaker.
- The mixture is stirred with a glass rod.- The precipitate formed is removed by
filtration.- The precipitate is rinsed with distilled water.- The precipitate is dried between the filter
paper
1
1
11
11
10
(iii)2AgNO3 + MgCl2 2AgCl + Mg(NO3)2
No of moles AgNO3 = 50 x 1.0 / 1000 = 0.05 mol
2 mol AgNO3 2 mol AgCl from the reaction0.05 mol AgNO3 0.05 mol AgCl
Mass of AgCl = 0.05 x 143.5 = 7.175 g
1
1
2
TOTAL 20
PART CQuestion
No.Explanation Mark Σ Mark
9 (a)(i)
(a)(ii)
(b) (i)
(ii)
(c)
Correct apparatus set upCorrect labelling
Volume of gas / cm3
Curve labelling axes with units
CaCO3 + 2HCl CaCl2 + CO2 + H2 O
No. of moles acid = (0.1)(50) = 5 x 10-3
1000 2 mol of HCl reacted evolve 1 mol of CO2 0.005 mol of HCl reacted evolve 0.005 mol of CO2 2 = 0.0025 mol CO2 Volume of CO2 =(0.0025)(24) = 0.06 dm3
= 60 cm3
Overall average rate of reaction =Total volume of CO2
Total time = 60 cm 3 300 s = 0.2 cm3 s-1
Rate of reaction in Expt II is higher than Expt IExperiment II is at a higher temperature,the kinetic energy of the reacting particles increases and the particles move faster
11
11
1
1
1
1
1
1
11
1
4
4
2
Time/min
Frequency of collision between marble and hydrogen ions increasesFrequency of effective collision increases
Rate of reaction in Expt III is higher than Expt IPowdered marble in Expt III has greater total surface area/ bigger surface area per unit volumePowdered marble is more exposed to collision Frequency of collision between marble and hydrogen ions increasesFrequency of effective collision increase
[ -1 if students use HCl or particles in the explaination]
1 1
1 1
1
11
Total
5
5
20
10 (a) Position of ions effect the product of electrolysis process Ions presence in the solutions are Cu2+, SO4
2-, and H+ In set I
- ion move to cathode is H+
11
1
- H+ ions is discharged at cathode to form hydrogen gas - 2 H+ + 2e H2
In set II- Ions move to cathode are Cu2+ and H+
- Cu2+ is discharged due to the lower position in the ECS // it is easier to be discharged
- Cu2+ + 2e Cu
11
11
1
max 6(b) 1. Zinc is more electropositive // Negative terminal :
Zn Zn2+ + 2e
2. Copper is less electropositive // Positive terminal:Cu2+ + 2e Cu
3. The electron move (from negative terminal to the positive terminal ) / (electrode zinc to electrode copper)
4. The flow of electron, produce the electric current//The needle of the voltmeter deflected, shows the electric current is produced
1
1
1
1
4(c)
Functional diagram : [switch on, battery, shade of the solution]
Label diagram : [iron spoon at cathode, copper rod at anode, copper(II) sulphate solution]
1
1
(i) 100 cm3 of 1.0 moldm-3 of copper(II) sulphate solution is poured into a beaker
(ii) An iron spoon is connected to the negative terminal of the battery // is connected to the cathode
1
1
Copper(II) sulphate solution
Copper electrodeIron spoon
(iii) Copper rod is connected to the positive terminal of the battery // is connected to the anode
(iv) The switch is on // can infer from the diagram(v) The observations is recorded in the table after 15 minutes.
(vi)
Electrodes ObservationAnode Copper electrode become thinnerCathode Iron spoon is coated with brown solid / metal
(vi) Half equations :
Anode : Cu Cu 2+ + 2e
Cathode : Cu 2+ + 2e Cu
1
1
11
1
1
10JUMLAH 20