Chemistry Perfect Score Module 2010 Answer Scheme
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Description
SET 1 PAPER 2: STRUCTURE QUESTION 1. (a) bromine and phenol (b) liquid (c) Nickel (d)
1 1 1
1
(e) ion (f) (i) T1 (ii) Heat absorbed by the particles/molecules is used to overcome the attraction forces between the particles/molecules in solid naphthalene. (iii) Become faster
1 1 1 1
……8
2
o
(a) T C 1 (b) t2 1 (c) Heat energy released to the surroundings is balanced by the heat release as the particles attract one another to form a solid 1 (d) (i) molecules 1 (ii) P R
1+1 1 1 1
(e) (i) liquid and solid (ii) Solid (f) Sublimation
…….9
3
(a) Functional apparatus Label
1 1
(b)
1
1
(c) 1.
Label and unit of axes X and Y 2. Scale and size of graph 3. Transfer point 4. Correct and smooth curve
1 1 1 1 O
(d) (i) Melting point is shown on the graph at 80 C (ii) Melting point is the temperature at which solid change to liquid
1 1
(e) Heat absorbed by the particles/molecules is used to overcome the attraction forces between the particles/molecules in solid naphthalene.
1 ……
10 A representation of a chemical substance using letters for atoms and subscripts for each type of atoms present in the substance.
4 (a)
(b) (i)
(ii)
1
Hydrochloric acid // nitric acid // sulphuric acid Magnesium // zinc
1
Correct formula of reactant and product Balance equation
1 1
Example: Mg + HCl
→ MgCl2
+
H2
(c)
Flow the hydrogen gas into the combustion tube for a few minutes to remove air before heating metal oxide
(d) (i)
Number of mole of copper = 1.62 64
1
// 0.025 mol 1
(ii)
Number of mole of oxygen = 0.40 // 16
0.025 mol
1
(iii)
Number of mole of of copper : Number of mole of of oxygen 0.025 : 0.025 // 1 : 1
1
CuO
1
(e)
Iron oxide // Tin oxide // Lead(II) oxide // Silver oxide
1
(f)
Burn metal M in oxygen
1 ...11
5
(a)
(i)
Magnesium // zinc Hydrochloric acid // nitric acid // sulphuric acid
1 1
(ii)
Correct formula of reactant and product Balance equation
1 1
Example: Zn + 2HCl
ZnCl2 + H2
2
(b)
(i) 1 1 1
(ii)
Empirical formula is MO 2
1
Correct formula of reactant and product Balance equation
1 1
Example: MO2 + 2H2
M + 2H2O
(c)
Flow the hydrogen hydrogen gas into the combustion tube for a few minutes to remove air air before heating metal oxide // Allow flow of hydrogen gas into the combustion tube during cooling
1
(d)
No. Magnesium is more reactive than hydrogen.
1 1
…13
6
(a) The chemical formula that that shows the simplest ratio of the number of atoms of each type of elements in the compound
1
(b) (i) Mass of magnesium = (26.4-24.0)g =2.4 g Mass of oxygen = (28.0 – 26.4) g = 1.6 g
1
(ii) moles of magnesium atoms : moles of oxygen oxygen atoms 1.6 : 2.4 // 24 16 0.1 : 0.1
1
(iii) MgO.
1
(iv) Correct formula of reactant and product Balance equation
1 1
Example: 2Mg + O2 2MgO (c) To allow oxygen to enter the crucible for complete combustion to occur. (d) Functional apparatus Label X oxide
1 1 1
Dry hydrogen gas (e) Collect the gas in a test tube.Place a burning wooden splinter at the mouth of the test tube. No pop sound. 1 Heat …..10
3
7 (a)
(i) (ii)
(b)
(i) (ii) (iii)
(iv)
(c)
(i) (i)
2.6 Period : 2 Group : 16 9 10 19 P 9
1 1 1 1 1
1. Element Q is more reactive than element R. 2. The size of atom Q is smaller tha atom R 3. The attraction forces between nucleus and valence electron electron of atom Q is stronger than atom R 4. it is easier for atom Q to receive valence electron compare to atom R.
1 1 1
Ionic bond Has high melting and boiling point // Conduct electricity in aqueous solution and molten state // Soluble in water // insoluble in organic solvent
1
1
1
1
(d) 1+1
C
P
P ….14
8 (a) (b)
(i) (ii) (iii)
2.8.2 Ionic bond 2+ 1. Atom X releases 2 electrons to form X ion 2. achieve octet electron arrangement Each ion drawn correctly 2-
2+ X
(iv) (c)
(i) (ii)
1 1 1 1
Y
Number of electrons for ion X and ion Y are correct Charge of ions are correct
1 1
Has high melting and boiling point// Conduct electricity in aqueous solution and molten state // Soluble in water // insoluble in organic solvent ZY2 12 + 2(16) // 44
1 1 1 9
……
9
(a) (b)
Metal : P/Q/R/S/T Non-metal : V/U/W/X 2.8.5
1 1 1
(c)
R
1
(d)
Electronegativity increases from Q to U 1. The atomic size size decreases from Q to U 2. The nuclei attraction forces between electrons and nucleus increases from Q to U
1 1
(e)
+
Q
1 1
4
(f)
(g)
X
1
Atom X has achieved octet electron arrangement
1
Correct formula of reactant and product Balance equation
1 1
Example: 2R (h)
10
+
U2
→ 2RU
1. form coloured ions 2. has more than one oxidation number 3. as catalyst 4. form complex ions [ any one] (a)
(b)
(c)
1
…..13
(i)
seven / 7
1
(ii)
2.8.7
1
(i)
P
+
1
(ii)
Q
1
(iii)
PQ2
1
Element U is not reactive // do not cause explosion
1
(d)
Q
Q
T
Q
Q
1 st mark: - showing the sharing of electrons
and correct number of
1
2 nd mark – correct label and correct number of el electron ectron pairs being
1
electrons in each shell
shared
(e)
low melting point & boiling point // does not conduct electricity TOTAL
5
1 9
11
(a)
(i) (ii)
(b)
(i) (ii)
2.8.1 Period : 2 Group : 16 Ionic
1 1 1 1 2-
+ P
+
Q
P
Number of electrons for ion X and ion Y are correct Charge of ions and ratio of ion X to ion Y are correct
(c)
S
S
R
S
1 1
S
(d)
Correct number of atom R and S Correct number electrons and shells
1 1
1. Compound in (b) can conduct electricity at molten or aqueous state. 2. Compound in (c) cannot conduct electricity at any state 3. Compound in (b) consist of freely moving ions 4. Compound in (c) consist consist of neutral molecules / no freely moving ion ion
1 1 1 1
OR 1. 2. 3. 4.
The melting point of compound compound (b) is high The melting point of compound compound (c) is low Ions in compound (b) are attracted by strong electrostatic forces molecules in compound (c) are attracted by weak van der Waals forces ….12
6
12 (a) (b)(i) (b)(ii) (c)(i)
Q Ion solid state state : Ions are not freely moving// ions are in a fixed position. molten state : Ion can move freely R : Gas T : Liquid
1 1 1 1 1 1
(c)(ii) 1
(c)(iii)
-
Van der waal/intermolecular forces between molecules are weak Small amount of heat is required to overcome it
1 1 .......9
PAPER 2 :ESSAY QUESTION: 13 (a) (i) The number of neutrons : 18 and 20 respectively (ii) Similarities
2
Differences
1. having the same proton number// Same number of electrons
1. different in the number of neutrons // different in the nucleon number
2. having the same valence electron// having the same chemical properties
2. different in physical properties
4 (b) (i) 1. Nucleus contains 6 proton and 6 neutron 2. Electrons move around the nucleus 3. Two shells are filled with electrons 4. There are 6 valence electron// electron arrangement is 2.6
1 1 1 1
…….4 (ii) Comparison Proton number Number of valence electron Chemical properties Number of neutron//nucleon number Physical properties Standard representation of element Any four
Diagram P 6 4 similar 6//12 different different
Q 6 4 similar 7//13 different diff erent different 4
(c) - Correct curve - Mark 71 on the y axes which is at the same level with the - X and Y axis with correct title and unit
Temperature/ ◦ C 100 71 60 Time/s
7
1 1
1……3
(ii)
1 Substance X in both solid and liquid state 2. heat energy is released 3. kinetic energy of particles decreases 4. They are closer to each other // Attraction force between the particles become stronger
1 1 1 1……4 20
14
(a)
1. 2. 3. 4.
(b)
Number of mole in 16 g of oxygen = 16/32 // 0.05 mole -3 -3 Volume occupied by 16 g of oxygen = 0.05 mole x 24 dm // 12 dm Number of mole in 22 g of CO2 = 22/44 // 0.05 mole -3 -3 Volume occupied by 22g of CO2 = 0.05 moles x 24 dm // 12 dm
Element
C
H
N
O
Mass /g
0.48
0.05
0.28
0.16
0.48/12 //0.04
0.05/1 //0.05
0.28/14 //0.02
0.16/16 //0.01
0.04/0.01 // 4
0.05/0.01 // 5
0.02/0.01 // 2
0.01/0.01 // 1
Number of mole The simplest ratio
(c)
1 1 1
1…..4
1 1 1
1 1 1 1
Empirical formula = C4H5N2O [C4H5N2O ]n = 194 [ 97 ]n = 194 n = 194/97 // 2 Molecular formula = C8H10N4O2
1…..8
1 2 3 4 5 6 7 8
1 1 1 1 1 1 1 1…8
molar mass of ammonium sulphate = 132g/mol percentage of nitrogen in ammonium sulphate = 28/132 x 100% // 21.2% molar mass of urea = 60 g/mol percentage of nitrogen in urea = 28/ 60 x 100% // 46.7% molar mass of hydrazine = 32g/mol percentage of nitrogen in hydrazine = 28/132 x 100% // 87.5% Hydrazine has the richest source of nitrogen compares with other fertilizers. The farmer should choose hydrazine
20 15
(a)
(b)
1. The proton number is 11 // Number of proton is 11 2. Nucleon number is 23 // Atomic mass is 23 3. Number of neutron neutron = 23-11 = 12 4. Nucleus contains 11p and 12n 5. Position of electron circulating the nucleus 6. Correct number shell shell consists of electron 7. Symbol of sodium is Na any 6
(i)
Formula that show simplest ratio number of atoms of each element in compound
8
1 1 1 1 1 1.....6
1
(ii)
(c)
1. Relative molecular mass for n(CH2O) = 180 // 12n + 2n + 16n = 180 2. n = 6 3. C6H12O6
1 1 1 ....4
(i) 1. 2. 3.
Element Mass/g No. of moles Ratio of moles/ Simplest ratio
Fe
Cl
2.80 2.80/56 = 0.05 0.05/0.05 = 1
5.32 5.32/35.5 = 0.15 0.15/0.05 = 3
1...4
4. Empirical formula = FeCl3 (ii)
1. 2. 3.
Formula of the reactants Formula of products Balance equation
2Fe 4. 5. 6.
+
3Cl2
→
1 1 1
1 1 1
2FeCl3
No. of moles Fe = 2.80/56 = 0.05 mol No. of moles Cl2 = (0.05 x 3)/2 = 0.075 mol 3 3 Volume of Cl2 = 0.075 x 24 = 1.8 dm / 1,800 cm
1 1 1 ...6 20
16
(a)
(b)
Formula that shows the simplest simplest ratio of the number of atoms for each element in the compound.
Element
C
H
Mass (%)
92.3
7.7
Number of moles Ratio of moles
(c)
92.3 12
= 7.7 1
7.7 1
1
= 7.7
1
1
1
Empirical formula : CH RMM of (CH) n = 78 [ 12 + 1]n = 78 13 n = 78 n = 6 Molecular formula : C6H6 Procedure: 1. Clean magnesium ribbon with sand paper 2. Weigh crucible and its lid 3. Put magnesium ribbon ribbon into the crucible and weigh weigh the crucible crucible with its lid 4. Heat strongly the crucible without its lid 5. Cover the crucible when the magnesium starts to burn and lift/raise the lid a little at intervals 6. Remove the lid when the magnesium burnt completely 7. Heat strongly the crucible crucible for a few minutes 8. Cool and weigh the the crucible with its lid and the content 9. Repeat the processes of heating, cooling and weighing until a constant mass is obtained Record all the mass
9
1…1
1
1…5 1 1 1 1 1 1 1 1 1
10. Results: Mass/g x y z
Crucible + lid Crucible + lid + magnesium Crucible + lid + magnesium oxide Calculations: Empirical formula: MgaOb / MgO Element Mass (g)
Simplest ratio of moles
(d)
17
(a)
Mg
O
y-x
z-y
x
z
y
24
16
a
b
- Cannot separate copper from magnesium oxide - Cannot weigh copper
(i) (ii)
1 1 1
y
Number of moles
1
1.The electron arrangement arrangement of atom Q : 2.7 2. The electron arrangement of atom R : 2.4 1.The number of neutrons in atom Q is 10 2. Number of electron in atom Q is 9
1 Max 12
1 1 20
1+1 2 1+1 2
(b)
1.Q and R form covalent bond. 2.Atom Q has an electron arrangement of 2.7 3.Atom R has an electron arrangement of 2.4 4.To achieve the stable electron arrangement, atom R shares electrons with atom Q. 5.One atom R contributes 4 electrons. 6. Each atom R contributes one electron. 7. Atom R shares four of its valence electrons each with 4 atoms of Q 8. molecule with the formula RQ 4.
1 1 1 1 1 1 1
1 Q
Q
Q R
Q
Notes : points 4, 7 and 8 can be obtained from the diagram
10
8
For group 1 elements, 1. Going down the group ,atomic size increases 2. the valence electron becomes further away from the nucleus 3. Forces of attraction between nucleus and the valence electron becomes weaker. 4. It is easier for the atom to donate / release the valence electrons. 5. The reactivity increases down the group
(c)
1 1 1
For Group 17 elements , 6. Atomic size increases when descending the group 7. the valence valence electrons become further away from nucleus. 8. Forces of attraction between between the protons / nucleus and the valence electrons become weaker 9. It is more difficult for the atom to accept /gain/receive electrons.
1
1 1 1
8.The reactivity decreases down the Group
1
max Total 18
(a) (i)
(ii)
Atom Y : 2.8.7 Atom Z : 2.8.8.1
1 1
Group 17 Because atom Y has 7 valence electron Period 3 Because atom Y has three shells occupied with electrons
1 1 1
2Na + Y2 → 2NaY Correct formula of reactants and product correct Balance equation
1 1
→ NaY + NaOY + H 2O Y2 + 2NaOH Correct formula of reactants and products correct Balance equation (b)
1….6
1
1…4
Ionic compound/bond 1- The electron arrangement of atom P = 2, 8, 1 / 2. 8. 1 The electron arrangement of atom Q = 2, 8, 7 / 2. 8. 7
1
2- to achieve the stable electron arrangement
1 +
3- atom of P donates / gives one electron to form P // half equation
1 -
4 - atom of Q receives / accepts one electron to form Q // half equation +
1
-
5 -the P ion and Q ions are attracted by a strong electrostatic force to form ionic bond .
1
6- formula PQ / correct diagram
1
Covalent compound / bond 7-The electron arrangement of atom R = 2, 6 / 2. 6 8-to achieve the stable electron arrangement
11
1
1 8 20
9-atom R and atom Q share electrons 10 - atom R contributes 4 electrons and atom Q contributes one electron 11- one atom R and 4 atom Q share share 4 pairs of electrons 12- to form covalent compound with the formula RQ4 / diagram
1 1
1Max 10 PAPER 3 STRUCTURE Explanation 19 (a)
(b)
Score
Suggested Answer (a) 0 s = 95.0 ° C, 30 s = 85.0 ° C, 60 s = 82.0 ° C, 90 s = 80.0 ° C, 120 s = 80.0 ° C,150 s = 80.0 ° C, 180 s = 78.0 ° C, 210 s = 70.0 ° C. Suggested Answer
3
3 Time (second) (secon d) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0
Temperature Tempe rature (° C) 95.0 85.0 82.0 80.0 80.0 80.0 78.0 70.0
(c) (i) & (ii) [ Score 3 & 3 ]
Tem Tem erat eratur uree / ° C
95 90
x
x x
80.0
x
x
x
Freezing point
x x
70
60
30
60
90
120 12
150
180
210
Time / second
d(i)
d(ii)
e
f
Suggested Answer The constant temperature at which liquid becomes a solid Suggested Answer The heat released when the particles in the liquid arrange to form solid balanced by the heat loss to the surroundings. Suggested Answer The air trapped in the conical flask is a poor conductor of heat. The air helped to minimize heat loss to the surroundings. / to ensure uniform cooling. Suggested Answer Element Compound P, R Q, S
3
3
3 3
20 (a) Observation Inference (i) White fume is released (i) Magnesium oxide is formed (ii) White solid is formed/The mass of crucible (ii) Magnesium reacts with oxygen and its content increases. (b) The mass of crucible and lid = 25.35 g. The mass of crucible, lid and magnesium ribbon =27.75 g. The mass of crucible, lid and magnesium oxide when cooled = 29.35g (c) (i) The mass of Mg= (27.75 -25.35)g =2.40g (ii) The mass of O2=(29.35-27.75)=1.60 g (iii) The number of moles Mg=0.1 mole The number of moles O = 0.1 mole The ratio of Mg : O = 1 : 1 The empirical formula is MgO. (d) 0.1 mole of Mg reacts with with 0.1 mole of O/1 mole of Mg reacts with 1 mole of O 21 (a)
Able to predict the manipulated variable, the responding variable and the constant variable completely .
Able to state how to control the manipulated variables correctly
Manipulated variable : metals of Group Group 1 elements // sodium, lithium, potassium.
Repeat the experiment by using the m etals of sodium, lithium and potassium
Responding variable: the reactivity of the reaction with water // the speed of movement on the water surface
Able to state correctly the way to control contr ol the manipulated variable To observe how fast the metals move on the surface of water.
Constant variable: size/mass of metals. Volume of water
Able to use the metal granules with the same size Use the metal granules with the same size.
21 (b) Able to state the relationship between the manipulated variable and the responding variable correctly.. Suggested answer: The reactivity of Group 1 elements increases going down the group. 21(c) 21 (d)
Able to arrange correctly the reactivity series of the metals according to descending order. Answer: potassium, sodium, lithium Able to classify the ions correctly. [to name or write all the formula of the ions correctly at the cations and anions group.] Answer: positive positive ion/ cation : sodium ion/ Na+, hydrogen ion/ H+ Negative ion/anion : hydroxide ion/ OH-
Paper 3: Essay
13
22(a)
22(b)
22(c)
[Able to state the aim of experiment accurately] To compare the reactivity of lithium, sodium and potassium based on the reaction with water and describing the effect of the solution towards the red litmus paper. Hypothesis Metals of lithium, sodium and potassium show different rate of reactivity with water and the solution formed turns red litmus paper to blue. Variables a) Manipulated variable :type of metals b) Responding variable : reactivity of reaction c) Constant variable : water and temperature
3
3
3
22(d)
[Able to list the correct and complete substances and apparatus.] Substances and Apparatus Lithium, sodium and potassium metals with water, basin, knife, forceps, blue blue litmus paper and white tile.
3
22(e)
[Able to give all the procedures correctly, steps 1 - 7] 1. Lithium metal is cut into a small piece 2. The paraffin oil oil on the surface of the metal is wiped with the filter. 3. A basin is filled with water. 4. Lithium metal is put on the surface of the water with with a pair of forceps. 5. Reactivity of the reaction is observed and recorded. 6. The experiment is repeated with sodium and potassium metals.
3
22(f)
[Able to show the accurate tabulation of data with correct title.] Metals Lithium Sodium Potassium
Observations
14
2
SET 2 1
2
(a)
Electrolysis is a process whereby an electrolyte is decomposed to its constituent elements when electric current passes through it. 2+
1
-
(b)
Pb
(c )
electrical to chemical
1
(d)
In solid : no free moving ions In molten : Ions free to move
1 1
(e)
(i) (ii)
brown gas are released 2Br → Br 2 + 2e
1 1
(f)
(i) (ii)
Lead reduction
1 1 9
(a)
(i)
/ lead(II) ions , Br Br / bromide ions
Copper ion and hydrogen ion 2+
r: Cu
(b)
(c)
(d)
1
1
+
and H
(ii)
[Able to mark the positive and the negative electrode correctly]
(i)
The blue colour turns colourless/becomes paler
(ii)
The concentration/ number of Cu
(i)
Mg
(ii)
Mg → Mg2+ + 2e
2+
2+
1
1
decreases
1
/ magnesium ion
1 1
(i)
Reduction
1
(ii)
The voltmeter reading increases
1
The distances between Mg and Cu is greater than the distance between Mg and Zn in electrochemical series
1 9
3
(a)
To allow the movement of ions.
1
(b)
e
G
e
e
Electrode P
Electrode Q
Potassium iodide solution
Chlorine water
1
Dilute sulphuric acid
(c) (i) (ii)
e
Colourless change to brown Put a few drops of starch solution.
15
1 1
4
5
A blue precipitate is formed.
1
(d)
Iodide ion // potassium iodide Loss electron//increase in oxidation number
1 1
(e)
Cl2 + 2e 2Cl
(f)
Bromine water // acidified KMnO4 solution // acidified K2Cr2O7 solution
1
(g)
0 to -1
1 10
-
1
(a)
(i) (ii) (iii)
Iron(III) nitrate // Iron (III) ions Oxidising agent Add / pour NaOH /NH3 solution Brown precipitate is formed
(b )
(i) (ii) (iii)
Zn + Cu → Zn 0 to +2 Oxidation
(c)
(i)
Magnesium // Aluminium Metal M higher than zinc in the reactivity series// Metal M reactive than zinc
(a)
2+
2+
+ Cu
1 1 1 1 1 1 1 1 1
(ii)
M , Zn , N
(i)
Mg /Zn / [ any suitable metal ] metal ] MgSO4 / ZnSO4 / [ any suitable solution ] solution ]
1 10 1 1
(ii)
Positive terminal : Cu Negative terminal : X/Mg/Zn
1 1
(iii)
+ 2e → Cu Negative terminal : X → X2+ + 2e // Mg → Mg2+ + 2e
2+
Positive terminal : Cu
1
1…. 6
(b)
(i)
Z
Y
X
W
1
(ii)
X X below Z in the electrochemical series// X less electropositive than Z 0.6 V
1 1
1…. 4
(c )
Observation on the electrolyte Experiment I 2+ The concentration of Cu ions decrease Copper ions discharge to form copper atom at cathode 2+ Cu + 2e → Cu Experiment II 2+ The concentration of Cu ions unchanged 2+ The rate of Cu ions discharge at cathode is the same the rate of copper ions formed at anode.
16
1 1 1
1 1
Observation at anode Experiment I OH ions discharge to form oxygen gas 2OH ion lower than SO 4 ion in electrochemical series. → 2H2O + O2 + 4e 4OH
1 1 1
Experiment II 2+
Copper ionise to form Cu Cu→ Cu2+ + 2e
ions
1
1…10 20
6
(a)
Reaction II Oxidation number of magnesium changes from 0 to +2, Oxidation number of zinc changes from +2 to 0 No change in oxidation number for each elements in reaction I
1 1 1
1…4 (b)
Test tube P: The solution changes colour from pale green to yellow 2+ 3+ 2Fe + Cl2 2Fe +2Cl Correct formulae of reactants and products Balance equation Test tube Q: The solution changes colour from fr om colourless to yellow/brown. 2I + Cl2 I2 + 2Cl Correct formulae of reactants and products Balance equation
1 1 1
1
1
1…6 (c )
Experiment I Reaction between carbon and oxide of metal P occurs
1
Carbon is more reactive than metal P
1
Experiment II Reaction between carbon and oxide of metal Q does not occur
1
Metal Q is more reactive than carbon
1
Experiment III Reaction between carbon and oxide of metal R occurs. Carbon is more reactive than metal R
1 1
Reaction between carbon and oxide of metal P produces flame whereas reaction between carbon and oxide of metal R produces glow. Metal P is less reactive than metal m etal R. Reactivity of metals in descending order is Q, carbon, R, P
1 1 1
Q is Aluminium // Magnesium
1..10 Total marks
17
[20]
7
(a)
(b)
+
2H + 2e H2 Correct formulae of reactants and product Balanced Properties Cell A 1. Type of cell Voltaic cell 2. Energy change Chemical electrical 3. Electrodes Positive terminal: Copper Negative terminal: Magnesium 2+ 2+ 4. Ions in electrolyte Cu , SO4 , H and OH ions 5. Half equation Positive terminal: 2+ Cu + 2e Cu Negative terminal 2+ Mg Mg + 2e 6. Observation Positive terminal: Copper plate becomes thicker Magnesium becomes thinner/dissolve
(c)
1 1 Cell B Electrolytic cell Electrical chemical Anode: Copper Cathode: Copper 2+
2-
+
2
1 1 1 -
Cu , SO4 , H and OH ions Anode: 2+ Cu Cu + 2e Cathode: 2+ Cu + 2e Cu Anode: Copper dissolves//become thinner Cathode: Copper becomes thicker
1 1 1 1
1
8
Chemicals silver plate , Silver nitrate nitrate solution 1 Procedure: 1. Iron ring is connected to the negative plate on the battery while while the silver plate is connected to the positive terminal of the battery//Iron ring is made as cathode while silver plate is made as anode 2. Both plates are immersed into the silver silver nitrate solution. 3. The circuit is completed Diagram
Functional apparatus set-up Label correctly:
1 1 1
1 1
Chemical Equation + Cathode: Ag + e Ag + Anode : Ag Ag + e
1 1
Observation: Cathode :Grey /silvery solid is deposited Anode/silver become thinner//dissolve
1 1 Total
18
10 20
8
(a) (i)
Sampel answer. 2+ → Mg Mg + 2e // any suitable equation Metal / magnesium atom lose electron electron // metal / magnesium is oxidized
1
1…2 (ii)
Experiment I 2+ Iron nail is oxidized to form Fe ions Metal P speeds up the process of rusting Because iron is more electropositive than P 2+ Dark blue precipitate indicates the presence of Fe ions
1 1 1 1
Experiment II Metal Q is oxidized to form Q ions Because metal Q is more electropositive than iron Water and oxygen accept electron to become OH ions // 4OH → 2H2O + OH + 4e Pink colour of solution indicate the presence of OH ions Arrangement : metal Q , iron , metal P.
1 1 1 1 1 Any 8….8
(b) Sample answer Zinc as a reducing agent Add zinc to iron(III) chloride solution Heat the solution
1 1 1
Filter the solution / mixture 2+ Add sodium hydroxide solution to the solution produced/ Fe Green precipitate is formed
1 1 1
Chlorine as an oxidising agent Add chlorine water to iron(II) nitrate solution Stir/ shake the solution Add sodium hydroxide solution Brown precipitate
1 1 1 1 1
Max
…10 20
Questio n 9 (a)
Rubric
Score
[Able to state the relationship correctly between the manipulated variable and the responding variable ] 3
Example: When a more/less electropositive metal in contact with iron, the metal inhibits/speeds up rusting. 9 (b)
[Able to state three variables correctly] Example: Manipulated variable: metals in contact with iron//magnesium,copper,zinc Responding variable: The intensity of blue colouration // rusting of ion Constant variable: Iron nails//temperature of solution.
19
3
[Able to state state the inference based on the observation correctly]
9(c)
Example:
9(d)
Test tube
W
X
Y
Z
Inferences
The iron nail does not rust
The iron nail does not rust
The iron nail rust quickly
The iron nail rust a little
3
[Able to write the balanced half equations correctly ] Example: 2+ Oxidation: Fe Fe + 2e Reduction: O2 + 2H2O + 4e
3 4OH
-
[Able to state the operational definition for rusting correctly ] 9(e)
Example:
9(f)
Rusting of iron is the formation of blue colouration when iron is in contact with less electropositive metals or without contact with other metals. [Able to classify all the three metals correctly]
3
Metals that can provide sacrificial protection Magnesium Zinc
Question 10(a)
Rubric Able to state the relationship between the manipulated variable and the responding variable correctly and with direction Sample answer: The higher the metal in reactivity series, the brighter the flame / glow produced // The higher the metal in reactivity series, the reactivity of the metal increases.
10(b)
3
Metals that cannot provide sacrificial protection Copper
Score
3
Able to state three variables and the way to control them correctly: Sample answer: (i) Manipulated variable Type of metals (ii) Responding variable Reactivity of metals / Brightness of flame or glow
3
(iii) Controlled variable Mass of metal powder / quantity of potassium manganate(VII) 10(c)
Able to give the operational definition accurately Sample answer: Metal that burns brightly when reacts with oxygen is the most reactive metal // Metal that glows faintly when reacts with oxygen is the least reactive metal
20
3
10(d)
Able to state one observation accurately Sample answer: Brown solid when hot, yellow solid when cold
10(e)
3
Able to state an accurate inference for this experiment: Sample answer: Magnesium is the most reactive metal 3
10(f)
Able to arrange the metals in ascending order of reactivity series of metals towards oxygen accurately Sample answer: Copper, lead, zinc, magnesium
Question number 10(g)
10(h)
10(i)
10(j)(i)
10(j) (ii)
3
Rubric
Score
Able to predict the position of iron iro n in the reactivity series of metals accurately Answer: Between zinc and lead Able to explain the relationship between the time to light up and the reactivity of metal accurately Sample answer: Magnesium is more reactive than zinc // Zinc is less reactive than magnesium Able to make the classification of more reactive metals and less reactive metals when reacts with oxygen accurately Sample answer: More reactive metals : Magnesium, zinc Less reactive metals : Lead, copper Able to record the masses accurately in two decimal places with unit Answer: 14.63g 17.03g 18.63g Able to construct a table that contains: 1. Crucible + lid, crucible + lid + magnesium, crucible + lid + magnesium oxide and mass with correct unit. 2. Transfer all the readings from (j)(i) correctly . Answer: Description Crucible + lid Crucible + lid + magnesium Crucible + lid + magnesium oxide
21
Mass (g) 14.63 17.03 18.63
3
3
3
3
3
11. (a) - Statement of the problem Score
Rubric [ Able to give the statement of problem correctly ] 3 Example : How is the effect on on rusting of iron iron when iron is in contact with another metals? (b) - variables Score Rubric [ Able to state All variables correctly ] Suggested answer : 3 Manipulated variable : Different types of metals// Different metals Responding variable : Rate of rusting // Rusting of iron Constant variable : Iron nails/temperature (c) - hypothesis Score Rubric 3 [Able to give the hypothesis accurately] Suggested answer : When a more electropositive metal m etal is in contact with iron, the metal m etal inhibits rusting // When a less electropositive metal is in contact with iron, the metal speed up u p rusting // Iron rusts faster when in contact co ntact with metal less electropositive (d) - Apparatus and materials Score Rubric [ Able to give the the list of the apparatus and substances correctly and completely] Suggested answer : Apparatus : Five test tubes, test tube rack Materials : sand paper, five iron nails, magnesium strip, zinc zinc strip, tin strip, copper 3 strip, hot agar-agar/jelly solution mixed with potassium hexacyanoferrate(III) solution and phenolphthalein indicator (e) - Procedure of the experiment Score Rubric [ Able to state all procedures correctly ] Suggested answer : 1. Clean all the metal strips with sand paper 2. Coil the metal strip around the iron nails and then put in the each test tube 3. Pour the same volume of hot agar-agar/jelly solution has been mixed with 3 potassium hexacyanoferrate(III) and phenolphthalein indicator 4. Leave the test tubes aside for one day 5. Compare the intensity of the blue and pink colour in each test tube and recorded
3 (f) - Tabulation of data Score Rubric [ Able to exhibit the tabulation of data correctly ] Tabulation of data has 6 columns and and 3 rows
2
Example : Test tube Intensity of blue colour Intensity of pink colour
A
B
C
22
D
E
Question No. 12(a)
Rubric
Skor
[Able to write the statement of the problem accurately.] Sample answer: 3 Does the difference in the position of two metals in the Electrochemical Series causes the difference in the voltage?
12(b)
[Able to state the three variables correctly.] Sample answer:
3
Manipulated variable : Type of metal Responding variable : The voltage of the cell. Constant variable : The type and concentration of electrolyte, copper electrod/ positive terminal. 12(c)
Able to give the hypothesis accurately. Sample answer: If the distance between two metals is further away in the electrochemical series, the voltage produced will be higher/bigger.
12(d)
[Able to state complete materials and apparatus.] Sample answer:
3
3
Materials: Zinc, Magnesium, Copper, Iron and Copper(II) sulphate solution/ any other electrolytes. Apparatus: Voltmeter, connecting wires, beaker.
12(e)
12(f)
Able to state a complete procedure. Sample answer: 1. Fill a beaker with copper(II) sulphate solution until it is two-thirds full. 2. Dip the zinc strip and copper strip into copper(II) sulphate solution. 3. Connect the two metals to a voltmeter(by using connecting wires) 4. Record the potential difference produced between the metals. Repeat steps 1 to 4 using iron strip, magnesium strip to replace zinc strip. Able to construct a labelled tabulation of data with suitable unit. Sample answer: Experiment Pairs of metal. Reading of voltmeter / V
3
3 I II III
23
Set 3 1
a.
Certain volume of acid completely neutralises a given volume of alkali// Certain volume of acid at which change of colour of indicator when a drop of acid is added to alkali Pink solution to colourless solution Experiment 1 = 22.40 2=22.20, 3=22.00 H2SO4 + 2NaOH Na2SO4 + 2H2O 3 (i) Average volume = (22.40 + 22.30 + 22.00)÷3 = 22.30 cm (ii) Mol of sulphuric acid, H 2SO4 = (22.30 X 1 )/1000 = 0.0223 mol (i) Functional set-up: Conical flask, Burette (ii) Label: (H2SO4, KOH and Phenolphthalein) (i) Add drops of acid a little at a time - towards the end point (ii) Conical flask with content - shaken during experiment
b. c. d. e. f. g.
2
a.
b. c d. e. f.
(i) Red (ii) Yellow (iii) Orange 3 15.00 cm (i) H2SO4 + 2KOH K2SO4 + 2H2O + (ii) H + OH H2O -3 0.1 x 20 = 2 Ma = 0.067 mol dm Ma x 15.00 1 (i) Yellow (ii) Red 3 30 cm
1 1 1 1 1 1 1 1 1 1 10 1 1 1 1 1 1 1 1 1 1 10
3 a. Ba(OH)2 + H2SO4 2H2O + BaSO4 reactants & product correct – 1, balanced -1 b. Neutralisation / double decomposition c. Methyl orange indicator changes from yellow to orange - White precipitate formed d. (i) The titration titration has achieved the end-point / there is no more free moving ions present in the beaker (ii) Ba(OH)2 + H2SO4 2H2O + BaSO4 Mb= 0.2M Ma=1.0M 3 3 Vb= 50 cm Va= x cm 1.0M(x) = 1 MaVa = 1 0.2M(50) 1 MbVb 1
x = 10 cm e. (i)
2 1 1 1 1
1
3
1
Ammeter reading/ A
0
5
Volume of sulphuric acid added/
Correct shape - 1 Correct volume indicated and ammeter reading not reaching zero (ii) - Ammeter does not show a zero reading + 2because at end-point there are free moving Na and SO4 ions OR
24
1 1 1
2
- The volume of sulphuric acid required is halved because sodium hydroxide is monobasic alkali whereas barium hydroxide is dibasic alkali / the concentration of OH ions in NaOH is half of Ba(OH) 2 4
1
1
(a) Neutralisation (b) To ensure all nitric acid is completely reacted (c) ZnO + 2HNO3 Zn(NO3)2 + H2O (d)
2
1 1 1
zinc oxide
[functional apparatus] [label] (e) Number mole of nitric acid = 2(50) = 0.1 mol 1000 Mass of salt = 0.1 x 189 = 9.45 g 1 (f) Zinc carbonate, Zinc
5
1 1 1 1 1 1 10
(a) (b) (c) (d)
Silver chloride White 3 3.0 cm (i) 0.5(6.0) = 0.003 mol 1000 (ii) Number of mole of sodium chloride = 3.0(1) = 0.003 mol 1000 0.003 mol of sodium chloride reacts with 0.003 mol of silver nitrate Therefore 1 mol of sodium chloride reacts with 1 mol of silver nitrate. + (e) Ag + Cl AgCl (f) 1.5 cm
1 1 1 1 1 1 1 1 1
Height of precipitate (cm)
1.5__________
Ι Ι Ι Ι 1.5
3
Volume of sodium chloride (cm )
[labeled axis + correct shape of graph -1] [height of precipitate and volume of NaCl required – correct & indicated – 1]
6
(a) (i) Copper(II) sulphate (ii) Heat the solution until saturated then cool Filter and then rinse the salt with distilled water Dry the salt using filter paper (b) (i) Blue precipitate Cannot dissolve in excess NaOH (ii) Cu(OH)2
25
1 1 10 1 1 1 1 1 1 1
(c) (d) (i) (ii) (iii)
7(a)
1 1 1 1 10
1. x-axis and y-axis labelled with units 2. plots transferred correctly 3. smooth curve 1. Rate of reaction of Experiment 4 is higher than Experiment 2 2. Temperature in Experiment 4 is higher. 3. The kinetic energy of thiosulphate ions is higher + 24. The frequency of collisions between the H ions and the S2O3 ions increases 5. The frequency of effective eff ective collisions increases.
(b)
Na2S2O3 + H2SO4 → Na2SO4 + S + H2O + SO2 Sulphur Concentration of sodium thiosulphate solution
(c) (d) (e)
8(a) (b) (c) (d)
(e)
1. Functional apparatus set-up 2. Complete labels 25 = 3 -1 0.78 cm s 32 + 2+ Zn + 2H → Zn + H2 1. Rate of reaction in Experiment II is higher than in Experiment I 2. In Experiment II copper(II) sulphate is a catalyst. 3. Catalyst will lower the activation energy. energ y. + 4. The frequency of effective collisions between zinc atoms and the H ions increases. 1. Rate of reaction of Experiment III is higher than in Experiment I 2. Sulphuric acid in Experiment III is a diprotic acid (hydrchloric acid is a monoprotic acid) + 3. Concentration of H ions per unit volume in Experiment III is 2 times higher than in Experiment I. + 4. The frequency of effective collisions between zinc atoms and the H ions increases
(f)
9
Ammonia White 2+ 2Pb + SO4 PbSO4 Filter the mixture
(i) Hydrogen (ii) Bring a lighted wooden splinter to the mouth of the test tube Pop sound
(a) (b) (c)
(d)
Heat released when 1 mole of silver is is displaced from its salt solution solution by copper metal. Silvery solid formed// Colourless solution of silver nitrate becomes blue// Amount of copper powder decreases + (i) No. of moles of Ag reacted = No. of moles of AgNO3 used = mv/1000 = 0.5(50)/1000 = 0.025 mol (ii) Heat released = No. of moles of Ag Ag x ΔH = 0.025 x 105 kJ = 2.625 kJ = 2625 J o (iii) Heat change = mc 2625 J = 50 (4.2) = 12.5 C Assumptions - no heat loss to surrounding - specific heat capacity of solution = specific heat capacity of water - density of solution = density of water ( any two)
26
1 1
1……3 1 1 1 1
1…..5 1 1 1 13
1
1…..2 1 1 1 1 1
1……4 1 1 1
1…..4 1 1 1….3 13
1 1 1 1 1 1 1 1
(e) Energy
+
2 Ag + Cu ΔH =
-105kJ 2+
2 Ag + Cu
1.
The position and name/formulae for the reactants and products are correct Label for the energy axis and arrow for the two levels are shown.
1 1
(f) Lower/smaller The total surface area exposed to the air is larger Heat is lost to the environment (g) To ensure all the silver nitrate solution reacted completely (h) Bigger / Higher because magnesium is more electropositive than copper.
1 1 1 1 1
2
15 10
(a) Strong acid – nitric acid/ hydrochloric acid / sulphuric acid (b) (c)
(d)
(e)
(f)
Strong alkali – sodium hydroxide / potassium hydroxide Heat released when 1 mole of water water is formed from the the reaction between an acid and an alkali - It is an exothermic reaction // heat energy is released to the surrounding - The total energy of reactants is higher than the products - 57 kJ of heat energy is released when 1 mole of water is formed ( any 2) (i) No. of moles alkali used = mv/1000 = 1(50)/1000 = 0.05 mol (ii) Heat change = mc = (50+50)x4.2x 6.5 = 2730 J Heat of neutralization = - 2730/0.05 = - 54600 J/ mol = - 54.6kJ / mol (i) The heat of neutralization for Experiment I is higher than Experiment II (ii) Ethanoic acid – weak acid, dissociates partially in water Part of heat released in Experiment t II during neutralization is absorbed to dissociate further the molecules of ethanoic acid The number of moles of water produced doubled, hence amount of heat energy released is doubled but the total volume of solution used also doubled, therefore the temperature increase remain the same
1 1 1 1 1 1 1 1 1 1 1 1
1 13
ESSAY SECTION B Question Number 10(a) 10(a)
(b)
(c) (i)
Explanation
Marks
Acid that ionises completely in water to produce high concentration of hydrogen ions
1 1..2 1..2
Number of sulphuric acid = 0.5(50)/1000 = 0.025 mol Number of hydrochloric acid = 1.0(50)/1000 = 0.05 mol Both are strong acids ionizes completely com pletely in water + To produce the same concentration of H
1 1 1 1…..4
Test
Aqueous HCl solution
Universal indicator Add zinc powder Add copper(II) oxide powder
Green to red Bubbles of colourless gas formed Blue solution formed
27
Solution of HCl in methylbenzene No changes No changes No changes
1+1 1+1
1+1….6
Question Number
(ii)
(iii)
Question Number 11 (a)
11(b)
11(c)(i)
Explanation
Aqueous hydrogen chloride solution In presence of water hydrogen chloride ionizes + to produce hydrogen ions/H which gives rise to acidic properties Aqueous HCl HCl in methylbenzene Observation Ammeter needle Ammeter needle does deflected not deflect Explanation Consists of free Consists of molecules // moving ions no free moving ions
Marks
1 1 1
1…..4 1 1 1 1…..4 20
Explanation -
3
-3
25 cm of 1.0 mol dm sodium hydroxide solution is measured and poured into a conical flask. Two drops of phenolphthalein are added to the solution. A burette is filled with nitric acid and the nitric acid is added slowly dropwise into the potassium hydroxide solution and shaken until the pink solution turned coloruless. The volume of acid added is calculated. The experiment is repeatedby using the same volume of sodium hydroxide and nitric acid but without the phenolphthalein. The salt solution formed is heated until saturated. The saturated solution is allowed to cool. The salt crystals are filtered and rinsed with distilled water. The salt crystals are dried by pressing them between sheets of filter papers. NaOH + HNO3 NaNO3 + H2O 3 50 cm of 1.0M copper(II) sulphate solution is measured and 3 added into 50 cm of 1.0M sodium carbonate solution in a beaker. The mixture is stirred and filtered to obtain green precipitate, copper(II) carbonate. The copper(II) carbonate is rinsed with water. The copper(II) carbonate is added a little at a time into into heated 3 50 cm of 1.0M dilute hydrochloric acid until some copper(II) carbonate solid no longer dissolved anymore. The mixture is filtered to remove excess copper(II) carbonate. The filtrate is copper(II) chloride solution. Lead(II) oxide Brown when hot, yellow when cold Carbon dioxide Bubble the gas through lime water and lime water turns chalky
28
Marks 1 1
1 1 1 1 1 1 1
1…..10 1 1 1 1 1
1…..6 1 1 1 1….4 20
ESSAY SECTION C 12
(a) (i)
(a)(ii)
(b)
+
Acid that will produce two moles of hydrogen ion, H from one mole of the acid in water. H2SO4 Acid that dissociates dissociates completely in water to produce high concentration concentration of + hydrogen ion, H HCl sodium hydroxide is a strong akali that undergoes complete dissociation in aqueous solution Ammonia is weak alkali that undergoes partial dissociation dissociation only The concentration of hydroxide ion in sodium hydroxide is higher than in ammonia Hence, the pH of sodium hydroxide is higher than that of ammonia.
1 1....2 1.... 2 1 2 1....2 1.... 1 1 1 1....4 1.... 4
(c)
[calculation ] 1. Molar mass of KOH = 39+16+1 = 56 2. Mol KOH = 250 x 1.0/1000 = 0.25 3. Mass = mol x molar mass = 0.25 x 56 = 14.0 gram
1 1
-3
[ preparation of 1.0 mol dm KOH ] KOH ] 4. Weigh exactly 14.0 g of KOH accurately in a weighing bottle. 5. Dissolve 14.0 g of KOH in in a little water in a beaker 3 6. transfer the contents into a 250 cm volumetric flask 7. Rinse the beaker with distilled water water and transfer all the contents into the volumetric flask 8. Distilled water is added to the volumetric flask until the calibration calibration mark.
1 1 1 1 1
-3
[ preparation of 0.1 mol dm KOH ] KOH ] [calculation ] Volume of KOH is added 9. M1 x V1 = M2 x V2 V1 = M2 x V2 / M1 3 10. = 0.1 x 250 / 1 = 25 cm 3
-3
1 1 3
11. 25.0 cm of 1.0 mol dm KOH is transfer to 250 cm 3 using 25.0 cm pipette. 12. Distilled water is added to the volumetric flask until the calibration mark.
13 (a)
(b)
[Name any insoluble salt] [Name any two suitable solution] [Write correct ionic equation]
1 1 1....12 1.... 12 20
1 1+1
1…. 4
Example: Lead(II) sulphate Lead(II) nitrate solution and sodium sulphate solution 2+ 2 Pb + SO4 PbSO4 2+ Test for Fe ion Procedure I: 3 A few drops of sodium hydroxide solution are added into 2 cm of salt solution of X in a test tube until in excess. excess. Observation : Observation : Dirty green precipitate cannot dissolve in excess sodium hydroxide solution Procedure II: 3 A few drops of ammonia solution are added into 2 cm of salt solution excess. of X in a test tube until in excess.
29
1
1
Observation : Observation : Dirty green precipitate cannot dissolve in excess ammonia solution.
Inference: 2+ Fe ion is present 2+ Test for SO4 ion 3 -3 5 cm of 1 mole dm hydrochloric acid is added into the salt solution of 3 -3 X follow by 2 cm of 1 mole dm barium chloride solution. Observation : White precipitate is formed. formed. Inference: 2SO4 ion is present.
1 1
1
1….6 (c)
Chemicals : sulphuric acid and magnesium oxide / magnesium carbonate / magnesium 3 -3 50 cm of 1.0 mol dm sulphuric acid is poured into a beaker and heated gently carefully Magnesium oxide powder is added a little at a time into the acid using spatula. The mixture is stirred well with a glass rod. Magnesium oxide powder is added continuously until some of it no longer dissolves. The mixture is filtered to remove the excess magnesium oxide. The filtrate is poured into an evaporating dish and heated gently to produce a saturated solution / heated until the filtrate is evaporated to about 1/3 of its original volume. The saturated solution is then allowed to cool to room temperature for crystalisation to occur. The magnesium sulphate crystals are filtered and dried by pressing them between a few pieces of filter paper. H2SO4 + MgO
MgSO4 + H2O
1 1 1 1 1 1
1 1
1
1….10
14 a. Minimum energy that the particles must have in order to collide to produce a chemical reaction. 1 b. i. For a given chemical reaction to occur, the particles must collide 1 with energy same or greater than the activation energy 1 and with correct orientation. 1
ii. A positive catalyst will provide an alternative reaction pathway with a lower activation energy. Hence the frequency of effective collisions between the particles increases and subsequently the rate of reaction increases.
c.
Diagram – labeled and functional 1+1
1 1 1 1 7 2
Procedure: 1. A burette is filled with water and inverted over a basin containing water. The burette is clamped to the retort stand. The water level in the burette is adjusted and the initial 1 burette reading is recorded. 3 -3 2. 50cm of 0.2 mol dm hydrochloric acid is poured into a small conical flask. 1 3. 5.0g of marble granules are added into the conical flask and start the stopwatch stopwatch simultaneously.1 4. Close the conical flask immediately with a stopper which is joined to the delivery tube and shake the conical flask steadily throughout the activity 1
30
5. The burette readings are recorded at 30 second intervals for 5 minutes 6. Step 1 to 6 are repeated using 5.0g of marble chips
1 1 Max 5
Results : Marble granules / marble chips Time(s) Burette reading/cm3
0
30
60
90
120
150
180
210
240
…….
300
Total volume of gas/cm3 3 1 Volume of gas released / cm
Marble chips
Marble granules Time taken / s 1 The gradient of graph involving marble chips is steeper than that of marble granules. This shows that the smaller the size of solid reactant, the higher the rate of reaction 1 d. When the concentration of solution increases, - the number of particles per unit volume increases - the frequency of collision between particles increases - the frequency of effective collision between particles increases - hence rate of reaction increases
15
(a) (i)
1 1 1
energy
Zn + CuSO4 ∆H =
-152 kJmol
-1
ZnSO4 + Cu
Y-axes : energy and two different level of energy The position of reactants and products correct 1
1….2 (ii) (b) (i)
- reactants have more energy // products have less energy - energy is released during the experiment // this is exothermic reaction - HCl is strong acid // CH 3COOH is weak acid - strong acid / HCl ionized completely // the degree of ionization of HCl / strong acid is 100% in water to produce + higher concentration of H // - CH3COOH / weak acid ionized partially // the degree of ionization of CH3COOH / weak acid is less than 100% in + water to produce low concentration H - when neutralization occurs, some of the heat released are absorbed by ethanoic acid / CH 3COOH to break the O-H bonds in the molecules/ ionizes the31 ethanoic molecules
1
1.…2 1
1
1
1 Max 3
(ii)
- H2SO4 is diprotic acid // HCl is monoprotic m onoprotic acid + - H2SO4 / diprotic acid produced two moles of hydrogen ion / H when one mole of the acid ionized in water // + - HCl / HCl / monoprotic acid produced one hydrogen ion / H when one mole of the acid ionized in water - In expt I, one mole of of water is formed whereas in expt III, 2 moles of water is formed, hence the heat released in expt III is twice the value of expt I
(c)
- apparatus & materials :2m - procedures :5m - table :1m - calculations : 2 m Sample answer : Apparatus : thermometer, measuring cylinder Materials : calcium nitrate soln, sodium carbonate soln, plastic/ polystyrene cup Procedures : 3 3 - measure 50 cm of 1.0 mol/ dm Ca(NO3)2 solution and poured into a plastic cup 3 3 - measure 50 cm of 1.0 mol/ dm Na2CO3 solution and poured into another plastic cup - measure and record the initial temperature of both solutions after 5 minutes - pour quickly and carefully Ca(NO3)2 solution into the plastic cup that contains Na2CO3 solution and stir continuously - measure and record the lowest temperature reached
1 1 1 1 Max 3
1
1…2 1 1 1 1
1…5
Tabulation of data : o
Initial temperature of Ca(NO3)2 ( C) o Initial temperature of Na2CO3 ( C) o Average initial temperature ( C) o Lowest temperature of the mixture ( C) o Change in temperature ( C)
Ө1 Ө2 (Ө1 + Ө2)/2 = Ө3 Ө4 Ө3- Ө4
1
Calculation : No. of moles of CaCO 3 = No. of moles of Ca(NO3)2 = mv/1000 = 1.0(50)/1000 = 0.05
heat change = mc(Ө4 - Ө3) = x kJ
1 -1
heat of reaction = + x kJmol 0.05 = + y kJmol
1….3 20
-1
16 (a) The heat of combustion of propanol is the heat energy released when 1 mole of propanol is burnt completely in the excess oxygen (b) (i) 1. Functional set-up of apparatus 2. Labelled propanol, copper can, water 3 3. Measure (100 – 250) cm of water and pour into the copper can and initial temperature is recorded 4. Weigh the spirit lamp and its content 5. Light the spirit lamp to heat the water in the can and stir 6. Extinguish the spirit lamp when the temperature increase reaches
30˚C, record the maximum temperature of water reached 7. Weigh the spirit lamp with its remain.
32
1 1 1 1 1 1 1 1
Result : 8. The initial mass of the spirit lamp + propanol = a g The final mass of of the spirit lamp + propanol = b g 9. The mass of propanol burnt = (a-b) g 10. The initial temperature of water = t1˚C The maximum temperature of water = t2˚C 11. Increase in temperature of the water = (t2 – t1) = t˚C
1 1 1 1
Calculation : RMM of propanol C3H7OH = 60
1 1
b 12. The no. of mol of propanol burnt = a60 = y mol
13. The released heat = mc = 100 x 4.2 x t = xJ 14. The heat of combustion of propanol = - Z kJ mol
1
x y
-1
J mol or
1….14
-1
(b) (ii) -
use wind shields Spirit lamp and its content is weighed immediately after the lamp is extinguished Ensure the flame touches the bottom of the copper can Stir continuously (either two of the answers) (c) The heat of combustion of propanol is higher than the methanol because propanol contain more no. of carbon and hidrogen atoms per molecule. -
16(a) (b)
(c)
(d)
(e)(i)
(e)(ii)
(f)
Burette reading 24.10, 18.00, 13.00
3
Experiment I 32.00, 37.00, 40.50, 42.00, 42.00 Experiment II 28.00, 36.50, 41.00, 42.00, 42.00, 42.00 The graph consist of: 1. Both axis are labeled and with unit 3 - y axis, volume of gas / cm - x axis, time/ s 2. All points are transferred correctly 3. Uniform scale 4. Best fit curve Experiment II Because the curve in Experiment II is steepest//the gradient is higher Manipulated variable Size of calcium carbonate /Total surface area of calcium carbonate Responding variable Rate of reaction Controlled variable Concentration and volume of hydrochloric acid Hypothesis When the total surface area of calcium carbonate increases, the rate of reaction increases 0.00 cm
3
3
3
3
3
3
3
33
1+1 1 1
(g)
17
Fast reaction
Slow reaction
Combustion Neutralization Precipitation
Photosynthesis Rusting Fermentation
3
a. Brown solid formed Blue solution of coper(II) sulphate becomes colourless Amount of zinc powder decreases o b. Initial temperature : 30.0 C o Highest temperature : 38.0 C o Temperature change : 8.0 C
c.
Heat change = mc Ө = 50 (4.2) 8.0 = 1680 J
d.
e. f. 18
It is is an exothermic reaction // heat energy energy is released to the surrounding Zinc is more electropositive than copper // zinc is higher than copper in Electrochemical Series The total energy of reactants is higher than that of product o
The temperature increase will be 16.0 C 2+
Zn + Cu
Cu + Zn
2+
(a) Aim : To compare the heat of combustion of octane and heptane (b) Hypothesis : Heat of combustion of octane is higher than heptane (c) All variables : Manipulated : Type of fuel / alkane Responding : Heat of combustion Controlled : volume of water, size of copper can (d) Material : Octane, heptane, water Apparatus : spirit lamp, thermometer, copper copper can (any suitable suitable container), measuring cylinder, weighing balance, wind shield (e) Procedure : 3 1. Measure [100-200 cm ] of water and pour into the copper can and record the temperature 2. A spirit spirit lamp is filled with heptanes and its mass is recorded 3. Adjust the height of the lamp so that the flame flame touches the bottom of the can. Light up the lamp o 4. Extinguish the lamp when the temperature increase reaches 30 C. Record highest temperature reached. 5. Weigh the the spirit lamp with heptanes and record its mass 6. Repeat the experiment by using octane to replace heptanes. (f) Tabulation of data Type of fuel / alkane o Initial temperature of water/ C o Highest temperature of water / C o Increase in temperature/ C Initial mass of spirit lamp and content / g Final mass of spirit lamp and content/ g Mass of fuel( /alkane) used/ g
34
Heptane
Octane
Set 4 No. 1(a)
Explanation Ethanoic acid o
Mark 1
1(b)
180 C Nickle / platinum
1 1
1(c)
C2H5OH
1
C2H4 + H2O
1(d)
1+1 [Any two correct structure] 1(e)(i)
Gas Q:. Brown solution turns colourless / decolourise Gas R: No change/ Brown solution remains rem ains unchanged
1(e)(ii)
Ethene has carbon-carbon double bond/ is an unsaturated hydrocarbon but ethane has carbon-carbon single bonds/ is a saturated hydrocarbon. or
1 1
1+1
Ethene can undergoes addition reaction but ethane cannot. 2
(a) (b)
Hydrogenation// addition (of hydrogen) (i) Phosphoric acid // H3PO4
35
1 1
(ii) (iii) (i) (ii)
1 1 1 1
(d)
C2H4 + H2O C2H5OH Fermentation Ethanoic acid Acidified potassium dichromate(VI) solution // Acidified potassium manganate(VII) solution C2H5OH C2H4 + H2O
(e)
(i)
Esterification
1
(ii)
- has a sweet smell// fruity smell - a neutral compound - colourless liquid - slightly soluble in water - readily soluble in organic compounds [ Choose any one ]
1
(c)
(iii)
1
1 H
O
H
H
H – C – C – O – C – C – H H
H
H Total
. 3 (a)
(i) ester (ii) –COO- // -OOC-
10
1 1
H
H ― C ― OH H ― C ― OH 1
H ― C ― OH H HOOC(CH2)7CHCH(CH2)7CH3
1
(b) unsaturated; the molecule of olive oil contains C=C double bonds. o
(c) (i) Hydrogen, nickel, 180 C (ii) hydrogenation (addition reaction) 4
1 1 1+1 + 1 1 1
a
A compound that contain element of carbon
b (i)
C2H4
1
(ii)
ethene
1
(iii)
double bond between carbon atoms
1
(a: C=C )
36
c (i)
Bromine water
1
(ii)
Halogenation// Addition of bromine
1
(iii)
Brown bromine water decolourised
1
d (i)
C2H4 + H2O
C2H5OH
1
(ii)
Alcohol
(iii) (iii)
Temperature 300 C // Pressure 60 atmosphere // concentrated
o
phosphoric acid
1 1
Total
10
5(a)
Composite material
1
5(b)(i)
A mixture of two or more m ore metals with a fixed composition //a mixture of metal and non-metal with a fixed composition
1
5(b)(ii)
Tin
1
5(c) 1
5(d)
High melting point // Resistant to thermal shock // Resistant to chemical attack // Low thermal expansion coefficient
1
5(e)(i)
Haber process
1
5(e)(ii)
N2 + 3H2
5(e)(iii)
Iron
1
5(e)(iv)
Ammonium sulphate //ammonium nitrate // ammonium phosphate
1
2NH3
[correct reactants and product, balance]
37
1+1
66
6 (a) Saponification (b) (i) Ester (ii) COO
1 1 1 1 1 1 1 1
(c) Concentrated potassium hydroxide (d) (i) Hydrophobic (ii) • Hydrophilic 'head' dissolves in water. • Hydrophobic 'tail' dissolves in grease. • Detergent ions reduce the surface tension of water. (iii)
1
2
(a) (b)
Hydrogenation// addition (of hydrogen) (i) Phosphoric acid // H3PO4
1 1
(ii) (iii) (i) (ii)
1 1 1 1
(d)
C2H4 + H2O C2H5OH Fermentation Ethanoic acid Acidified potassium dichromate(VI) solution // Acidified potassium manganate(VII) solution C2H5OH C2H4 + H2O
(e)
(i)
Esterification
1
(ii)
- has a sweet smell// fruity smell - a neutral compound - colourless liquid - slightly soluble in water - readily soluble in organic compounds [ Choose any one ]
1
(c)
(iii)
1
1 H
O
H
H
H – C – C – O – C – C – H H
H
H Total
7(a)
(i) (ii) (iii) (iv)
Saponification To precipitate the soap. To remove the glycerol and excess sodium hydroxide solution. Concentrated potassium hydroxide solution
38
10
(v)
(b)
(vi) It helps to suspend the grease particles. (i) Hydrophobic part: CH3(CH2)n CH 2 Hydrophilic part:
(ii) The formula mass of CH3(CH2)n CH2OSO3Na = 330 12 + 3 + (12 + 2)n 2)n + + 12 + 2 + 16 + 32 + 3(16) + 23 = 330 15 + 14n 14n + + 133 = 330 14n 14n = = 182 n = n = 13 (iii) The cleansing action of detergents is more effective than soaps in hard water. No 8
Marking scheme
Mark 1 1 1
(a)
i) ii) iii)
Stimulant Antidepressant Steroid
(b)
To stimulate positive emotion from the patience like self-confidence, more active and energetic
1
(c)
Usage of psychotherapeutic drugs can cause many side effects like addiction, fear, aggressiveness or death in a person.
1
(d)
Arthritis Asthma
1 1
Essay Section B No 9
(a)
Marking scheme A group of organic compound that has certain characteristics: Members of a homologous series can be represented by the same general formula. Members of a homologous series can be prepared by the same method. Members of a homologous series have similar chemical properties Successive members differ from each other by –CH2 unit. Members show a gradual change in their physical properties.
Mark 1 1 1 1 1 1 Max: 5
(b)
Percentage of carbon in pentane, C5H12
=
5(12) 5(12) + 12(1)
x 100%
1 1
= 83.33%
Percentage of carbon in pentene, C5H10 =
5(12) 5(12) + 10(1) x 100% = 85.71%
39
1
1
(c) (i)
(ii)
Percentage of carbon by mass in pentene is higher than that in pentane, hence pentene burns with a more sooty flame than pentane
1 1 Max 5
W: propanoic acid; Z: ethyl methanoate W and Z have the same molecular formulae but different structural formulae. W has the carboxyl group as the functional group while Z has the carboxylate group as the functional group.
1+1 1
W dissolves readily in water whereas Z does not. W has a sour smell. Z has a fragrant fruity smell.
1 1 1
(iii)
H
O
H
1
H―C― C―O ― C―H
H
1
1 1
H Used as food flavouring / perfume /
fragrance (iv) CH3 COOH + CH3OH CH3 COOCH3 + H2O Catalyst: Concentrated sulphuric sulphuric acid TOTAL 10
(a)
.
Element % No. of moles Ratio of moles Simplest ratio
C 52.2 52.2/12 4.35 2
H 13.0 13/1 13 6
O 34.8 34.8/16 2.175 1
Assume that the molecular formula is C2H6O. Given that the relative molecular mass is 46, n[ C2H6O] = 46 46n = 46 n=1 Therefore the molecular formula of compound X is C2H5OH.
(b)
3
CH3 COOH
(c)
3
50 cm ethanol and 25 cm of ethanoic acid are added into a round-bottomed flask. 3 5 cm of concentrated sulphuric acid is added. A Liebig condenser is fixed to the round-bottomed flask. The mixture is heated under reflux for 30 minutes. The ester, ethyl ethanoate is distilled out from the mixture at its boiling point.
+
C2H5OH
Dehydration Alumina / unglazed porcelain chips, heat
CH3 COOC2H5 + H2O
10
1 1 1
1 1..5
1 1 1 1 1
1..6
1 1+1 ..3
40
(d)
(e)
+
A long chain molecule that is formed by the joining together of smaller molecules called monomers.
H n
(b)(i) (b)(ii) (b)(iii)
1 Max: 4 1
H
--- C ― C ----
H
H
H
n
1..2 20
Mark
SO2 + H2O H2SO3 Corrodes buildings Corrodes metal structures pH of the soil decreases Lakes and rivers become acidic [Able to state any three items correctly]
1
(c)(ii)
H
1 1 1
Explanation
(c)(i)
H
C == C H
Question Number 11 (a)(i) (a)(ii)
1
When ethanoic acid is added to latex, the H ions in the acid will neutralize the negative charges on the protein membrane of the rubber particles. As a result the rubber particles parti cles will collide with each other and break the protein membrane setting free the rubber polymer molecules which then coagulate. Coagulation can be prevented by adding add ing an alkali.
Oleum 2SO2 + O2 2SO3 Moles of sulphur = 48 / 32 =1.5 Moles of SO2 = moles of sulphur = 1.5 3 Volume of SO2 = 1.5 24 dm 3 = 36 dm Pure metal are made up of same type of atoms and are of the same size. The atoms are arranged in an orderly manner. The layer of atoms can slide over each other. Thus, pure copper are ductile. There are empty spaces in between the atoms. When a pure copper is knocked, atoms slide. Thus, pure copper are malleable. Zinc. Zinc atoms are of different size, The presence of zinc atoms disturbs the orderly arrangement of copper atoms. This reduce the layer of atoms from sliding. Zinc atom
Copper atom
41
3
4
1 1 1 1 1 1
1 1 1
1 1 1 1 Max:5 1 1 1 1
6
Question Number
Explanation
Mark
1 Arrangement of atoms – 1; Label - 1 1 Max: 5 Total
No. 12
(a)
(b)
(c )
(d) (i)
(ii)
(iii)
13 (a)
Marking Criteria Part X – hydrophobic/hydrocarbon Part Y – hydrophilic/ionic Parx X – dissolves in grease Part Y – dissolves in water
Mark 1 1 1 1
1.The cloth in experiment II is clean whereas the cloth in Experiment I is still dirty. 2.In hard water,soap react with magnesium ion 3.to form scum 4.Detergent are more effective in hard water water 5.Detergent does not form scum 6.Detergent are better cleansing agen then soap to remove oily stain. Patient X : Analgeis/anpirin Patient Y: Antibiotic/penicillin/streptomycin Patient Z ; Psychotherapeutic / antidepressant Precaution: 1.Take after food. 2. Swallowed with plenty of water Explain: 1. Acidic and can cause irritation of the stomach. 2. To avoid internal bleeding/ulceratiion [precaution – 1m] [explain – 1m ] 1.To make sure all the bacteria are killed / becomes ill again – 1m 2. bacteria become more resistant. – 1m 3.Need stronger antibiotic to fight the same infection – 1m 1.Drowsiness – 1m 2. poor coordination/light-headedness – 1m TOTAL
Sources of sulphur dioxide: Volcanic eruptions Burning of fossil fuels From industries manufacturing sulphur based products [any two] Health hazards: Irritates the nose and eyes Causes bronchitis and asthma Formation of acid rain: Sulphur dioxide reacts with oxygen to form sulphur trioxide 2SO2 + O2 → 2SO3 Both oxides of sulphur dissolve in rain water to form sulphurous and sulphuric acids respectively SO2 + H2O → H2SO3 SO3 + H2O → H2SO4
42
20
Total
4
1 1 1 1 1 1 6 1 1 1
3
1 1
2
3 2
7 20
1 1 1
1 1 1 1 1 1 1 1
Effects of acid rain: Corrodes buildings and bridges Damages vegetation
1 1..max 10
(b)
Tape a steel ball bearing to the brass block Hang a weight of 1 kg at a specified height of 50 cm above the ball bearing Drop the weight and allow it to hit the steel ball bearing Use a caliper or ruler to measure the diameter of the dent made on the brass block Repeat the experiment to obtain another two readings so that an average value can be calculated The whole experiment is repeated using copper block to replace brass block Observation: Type of block
1
(a)(ii)
Number of moles of ethene = 5.6/28 = 0.2 mol Number of moles of O2 = 3 x 0.2 = 0.6 mol 3 3 Volume of O2 = 0.6 x 24 dm = 14.4 dm
1 1
P: Ethanol Q: 2-methylbut-1-ene
1 1
(c)
Mark 1+1
Colour of flame Blue flame Yellow
1
1 1..10 20
Explanation C2H4 + 3O2 2CO2 + 2H2O [Correct reactants and products, balance]
Compound A B
1
1 1
No. 14(a)(i)
(b)(ii)
1
1
Diameter of dents (cm) 2 Average
Brass Copper Results and explanation: Diameter made by copper is larger than brass Brass is harder than copper The foreign atoms (zinc atoms) in brass prevent the layers of copper atoms from sliding past each other .
(b)(i)
1 1 1
Σ Mark
4
Solubility in water Soluble Insoluble
Material: Ethanol, propanoic acid, concentrated sulphuric acid
1+1 1+1
1 1 1
Water out Liebig condenser
43 Water in Water bath
6
1+1
Procedure: 1. 2. 3. 4.
3
3
50 cm of ethanol and 25 cm of propaoic acid is put into a round-bottomed flask. 3 5 cm of concentrated sulphuric acid is added. Liebig condenser is fixed to the round-bottomed flask. The mixture is heated under reflux for 30 minutes.
Equation : C2H5OH
+ C2H5COOH
C2H5COOC2H5 + H2O
1 1 1 1
10
1
Question 15 (a)
Rubric [Able to record all the six readings correctly.] Vulcanised rubber: 2, 4, 6 Unvulcanised rubber: 4, 8, 12
15(b)
[Able to relate between the manipulated variable and the responding variable.] Vulcanised rubber is more elastic than the unvulcanised rubber// Unvulcanised rubber is less elastic than the vulcanised rubber
15(c) Variable (i) Manupilated variable Vulcanized and unvulcanized rubber// Mass of weight (ii) Responding variable Increase in length of rubber strip//elasticity (iii) Fixed variable Initial length of rubber strip
Action to be taken (i) The way to manupilate variable Repeat by replacing vulcanized rubber with unvulcanized rubber//Use weights with different masses (ii) What to observe in the responding variable To measure length of rubber strip (iii) The way to maintain the controlled variable Use the same length of vulcanized and unvulcanized rubber strips
15(d) 15(d)
[Able to make the correct inference] (i) Vulcanized rubber (ii) Presence of the sulphur cross links between the chain of rubber polymers in vulcanized rubber makes the small increase in length and can return to its original length after stretching.
15(e) 15(e)
[Able to make an operational definition correctly:] Rubber that can stretch a bit and returns to its original length when not stretched.
No
Marking scheme
44
Mark
16
(a) (b)
3 1 1 1
(c)
Hypothesis: Vulcanized rubber is more elastic than unvulcanized rubber.
3
(d)
The elasticity of the rubber strip is shown by its ability to return to its original length after it is stretched.
3
(e)
No 17
How does the elasticity of vulcanized rubber differ from that of unvulcanized rubber? (a) Variable: Manipulated Variable: Types of rubber Responding variable: Length of rubber strip Fixed variable: Size of rubber strip, str ip, Mass of weight
(a)
(b) Unvulcanized rubber: the minimum weight is 40g Vulcanized rubber could return to its original length even after the 50g weight was removed
3
Marking scheme To compare the cleansing power of soap and detergent in hard water.
Mark 3
(b)
The cleansing cleansing power of soap is weaker in sea water than the cleansing power of detergent.
3
(c)
(i) Soap and detergent (ii) Cleanliness of the clothes or amount of the greasy spots removed. (iii)Mass of soap and detergent dissolved in sea water.
1 1 1
(d)
Beaker Observation
A Only some greasy spots are removed
B Most of the greasy spots are removed.
3
(e) The cleansing power of soap is weaker in sea water compare to detergent The cleansing power of soap is weaker in sea water compare to detergent (f)
No 18
Sea water contains magnesium and calcium ions. Soap particles form insoluble calcium and magnesium salt (called scum) with these ions. 2+ 2+ Detergent particles are not precipitated out by Ca and Mg ions present in sea water and will remain in the solution to do the cleansing job.
Marking scheme (i)
Problem statement: Iron rusts more easily than steel.
3
Mark 3
(ii)
Hypothesis: Iron rusts faster than steel.
(iii)
Material: Iron nail, steel nail, agar-agar solution, potassium hexacyanoferrate(III) solution.
3
Apparatus: Test tubes. (iv)
3
Procedure: 1. Iron nail nail and steel nail are cleaned using using sandpaper. 2. Iron nail is placed into test tube A and steel nail is placed into test tube B. 3. Prepare 5 % of agar-agar solution and add several drops of
45
3
4. 5. 6.
(v)
potassium hexacyanoferrate(III) solution to the agar-agar solution. Pour the agar-agar solution into test tubes A and B until it covers the nails. Leave for 1 day. Both test tubes are observed to determine whether there there is any blue spots formed or if there are any changes on the nails.
Tabulation of data: Test tube A B
3
The intensity of blue spots 3
OR Based on the hardness of iron and steel. Problem statement: Iron is softer than steel. Hypothesis: The diameter of the dent of the steel is smaller than the diameter of the dent of iron. Material: Steel block, iron block. Apparatus: Ruler, 1 kg weight, retort stand and clamp, thread, steel ball, cellophane tape. Procedure: 1. A steel ball is attached on the surface of the the iron block using a cellophane tape. 2. The 1 kg weight is held 1 metre from from the surface of the iron iron block. 3. The weight is then released. 4. The diameter of dent formed on the iiron ron block is measured using a ruler. 5. Steps 2 to 4 are repeated on different surfaces of the iron block and the average diameter of dents is obtained. 6. The experiment is repeated by replacing the iron block with steel block. Tabulation of data: Diameter of the dent (cm) Material Reading 1 Reading 2 Iron block Steel block No 19
(i)
Marking scheme Aim: To differentiate and identify hexan-1-ol, hex-1-ene and hexane through chemical tests.
Mark
3
(ii)
(iii)
Variables Manipulated Variable: Types of reagents Responding variable: Change in colour Fixed variable: hexan-1-ol, hex-1-ene and hexane
3
Apparatus: Test tubes, test tube holder, dropper 3
46
-3
Materials: bromine water, 0.5 mol dm potassium dichromate(VI) -3 solution, 1 mol dm sulphuric acid, Liquids X, Y and Z (iv)
(v)
Procedure: 3 1. About 2 cm of each liquid X, Y and Z are poured into three separate test tubes. 3 2. 1 cm of potassium dichromate(VI) solution is added into each 3 -3 test tube followed by 1 cm of 1 mol dm sulphuric acid and heat. 3. The mixture in each test tube is then shaken well. 4. The changes in each test tube are observed and recorded. 3 5. Steps 1 to 4 are repeated using 2 cm of bromine water to replace the acidified potassium dichromate(VI) solution. Tabulation of Data: Reagent Liqiud
Acidified potassium dichromate(VI) solution.
Bromine water
3
Inference
X Y 3 Z
47
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