
Transcript
@Hak cipta BPSBPSK/SBP/2013 1 Perfect Score & X A Plus
Module/mark scheme 2013 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH
DAN SEKOLAH KLUSTER JAWAPAN MODUL PERFECT SCORE & X A-PLUS 2013
CHEMISTRY Set 1 Set 2 Set 3 Set 4 Set 5
http://cikguadura.wordpress.com/CHEMISTRY @Hak cipta
BPSBPSK/SBP/2013 2 Perfect Score & X A Plus Module/mark scheme
2013 MODULE PERFECT SCORE & X A-PLUS 2013 SET 1 :THE STRUCTURE
OF ATOM, PERIODIC TABLE OF ELEMENTS AND CHEMICAL BONDS Question No
Mark schemes Mark 1 (a) (i) Melting 1 (ii) Molecule 1 (b) The heat
energy absorbed by the particles is used to overcome the forces of
attraction between the naphthalene molecules / particles. 1 (c) The
particles move faster 1 (d) (i) X : electron Y : nucleus 1 (ii)
Electron 1 (e) (i) W and X 1 (ii) W and X atom have different
number of neutrons but same number of protons Atom// Element W and
X has different nucleon number but same proton number 1+1 10
Question No Mark schemes Mark 2 (a) No of electrons = 18, No of
neutrons = 22 1+1 (b) (i) The total number of protons and neutrons
in the nucleus of an atom 1 (ii) 40 (c) (i) 2.1 1 (ii) X (d) (i) W
and Y 1 (ii) Atom W and Y have the same number of valence electrons
1 (iii) To estimate the age of fossils /artefacts. 1 10 X X e 3p 4n
X e e X http://cikguadura.wordpress.com/ @Hak cipta
BPSBPSK/SBP/2013 3 Perfect Score & X A Plus Module/mark scheme
2013 Question No. Mark Scheme Marks 3 (a) (i) Total number of
protons and neutrons in the nucleus of an atom 1 (ii) 35 18 = 17 1
(iii) shows nucleus and three shells occupied with electron Label
12 proton, 12 neutron 1 +1 (iv) Number of electrons = 2 1 ...5 (b)
(i) Liquid 1 (ii) Q R 1+1 ...3 (c) 1st mark - Label X and Y axis
with correct unit 2 nd mark - Correct shape of curve 1+1 10
Temperature/o C Time/s 67 90 @Hak cipta BPSBPSK/SBP/2013 4 Perfect
Score & X A Plus Module/mark scheme 2013 4 a) (i) F 1 (ii) Atom
F has achieve stable/octet electron arrangement // has 8 valence
electron 1 b) (i) 2D + 2H2O 2DOH + H2 Correct reactant &
correct product Balance equation 1 1 (ii) The nuclei attraction
towards the valence electrons is weaker in atom G. More easier for
atom G to lose / release an electron to form a positively charged
ion. 1+1 c) (i) Covalent bond 1 (ii) 1 1 (iii) Cannot conduct
electricity at any state/ low melting and boiling point/.... 1 (d)
Show coloured ion//formed complex ion//has various oxidation
number//act as catalyst 1 11 5 (a) Increasing of proton number. 1
(b) (i) Na/sodium, Mg/magnesium .... 1 (ii) Atomic size decreases
across the period // Period 3. 1 (iii) 1. Number of protons in atom
increases when across the period. 2. Force of attraction between
nucleus and electrons in the shell is stronger. 1+1 ..4 (c)
Chlorine more reactive than bromine Size of chlorine atom is
smaller than bromine atom Chlorine atom is easier to receive one
electron 1+1 (d) Al3+ 1 (e) (i) Ionic compound 1 (ii) 1+1 11 Y Yx X
x x x x x E EY @Hak cipta BPSBPSK/SBP/2013 5 Perfect Score & X
A Plus Module/mark scheme 2013 6 (a) P : liquid Q : solid R : gas 1
+1+1 (b) (i) 1. P can be change to Q through freezing process. 2.
When the liquid cooled, the particles in liquid lose energy and
move slower. 3. As temperature drops, the liquid particles attract
tone another and change into solid 1 1 1 (ii) 1. P can change to R
through boiling. 2. When liquid is heated, the particles of the
liquid gain kinetic energy and move faster as the temperature
increase 3. The particles have enough energy to overcome the forces
between them and gas is formed 1 1 1 (iii) 1. R can be change to P
through condensation process. 2. When the gas cooled, the particles
in gas lose energy and move slower. 3. Particles attract one
another and change into liquid 1 1 1 (c) (i) 1. Uniform scale for
X-axis and Y-axis and labelled/size of graph plotted of graph
paper. 2. Tranfer of point 3. Smooth curve 1 1 1 (ii) 1. Dotted
line on the graph from the horizontal line to Y-axis at 80o C. 2.
Arrow mark freezing point at 80o C 1 1 (iii) 1. Heat released to
sorrounding 2. Is balanced when particles comes together to form a
solid 1 1 (iv) Supercooling 1 20 @Hak cipta BPSBPSK/SBP/2013 6
Perfect Score & X A Plus Module/mark scheme 2013 Question No.
Mark Scheme Mark 7 (a) (i) Atom R is located in Group 17, Period 3
1 + 1 (ii) Electron arrangement of atom R is 2.8.7. Group 17
because it has seven valence electron. Period 3 because it has
three shells filled with electron 1 1 1 (b) (i) Atoms P and R form
covalent bond. To achieve the stable electron arrangement, atom P
needs 4 electrons while atom R needs one electron. Thus, atom P
shares 4 pairs of electrons with 4 atoms of R, forming a molecule
with the formula PR4 // diagram 1 1 1 1 1 (ii) Atom Q and atom R
form ionic bond. Electron arrangement for atom Q is 2.8.1 and
electron arrangement for atom R is 2.8.7// Atom Q has 1 valence
electron while atow R has 7 valence electron To achieve a stable
(octet ) electron arrangement, atom Q donates 1 electron to form a
positive ion// equation Q Q+ + e Atom R receives an electron to
form ion R- //equation and achieve a stable octet electron
arrangement. R + e R- Ion Q+ and ion R- are attracted together by
the strong electrostatic forces to form a compound with the formula
QR// diagram 1 1 1 1 1 1 R R R R P Q R + - - @Hak cipta
BPSBPSK/SBP/2013 7 Perfect Score & X A Plus Module/mark scheme
2013 Question No Mark scheme Mark 8 (a) 12 represents the nucleon
number. 6 represents the proton number. 1 1 (b) Able to draw the
structure of an atom elements X. The diagram should be able to show
the following informations: 1. correct number and position of
proton in the nucleus/ at the centre of the atom. 2. correct number
and position of neutron in the nucleus/ at the centre of the atom.
3. correct number and position of electron circulating the nucleus
4. correct number of valence electrons Sample answer: or 1 1 1 1 e-
e- e- e- e- e- e- e- e- e- e- 11p 12n 1 2 3 4 e- e- e- e- e- e- e-
e- e- e- e- 11p + 12n @Hak cipta BPSBPSK/SBP/2013 8 Perfect Score
& X A Plus Module/mark scheme 2013 (c) (i) Atoms W and Y form
covalent bond. To achieve the stable electron arrangement, atom W
contributes 4 electrons while atom Y contributes one electron for
sharing. Thus, atom W shares 4 pairs of electrons with 4 atoms of
Y, forming a molecule with the formula WY4 // diagram 1 1 1 1 1
(ii) Atom X and atom Y form ionic bond. Electron arrangement for
atom X is 2.8.1 and electron arrangement for atom Y is 2.8.7 To
achieve a stable (octet )electron arrangement, atom X donates 1
electron to form a positive ion // equation X X+ + e Atom Y
receives an electron to form ion Y- //equation and achieve a stable
octet electron arrangement. Y + e Y- Ion X+ and ion Y- are
attracted together by the strong electrostatic forces to form a
compound with the formula XY// diagram 1 1 1 1 1 1 (d) The melting
point of the ionic compound/ (b)(ii) is higher than that of the
covalent compound/ (b)(i) . This is because in ionic compounds
oppositely ions are held by strong electrostatic forces. High
energy is needed to overcome these forces. In covalent compounds,
molecules are held by weak intermolecular forces. Only a little
energy is required to overcome the attractive forces. OR The ionic
compound/(b)(ii) conducts electricity in the molten or aqueous
state whereas the covalent compound/(b)(i) does not conduct
electricity. This is because in the molten or aqueous state, ionic
compounds consist of freely moving ions carry electrical charges.
Covalent compounds are made up of molecules only 1 1 1 1 1 or 1 1 1
1 1 20 X Y + - - Y Y Y WY @Hak cipta BPSBPSK/SBP/2013 9 Perfect
Score & X A Plus Module/mark scheme 2013 9 (a) (i) 1. Correct
number of shells and valence electrons 2. Black dot or label Q at
the center of the atom 1 1 (ii) 1. Group 14 2. There are 4 valence
electrons 3. Period 2 4. Atom consists of 2 shells occupied with
electrons 1 1 1 1 (b) (i) 1. Floats and moves fast on the water 2.
Hiss sound occurs 3. Gas liberates / bubble [any two] 1 1 (ii) 2Q +
2H2O 2QOH + H2 1. Correct reactant and product 2. Balanced equation
1 1 (c) (i) Compound X Sharing electron between atom B and A 1 1
(ii) Choose any one ionic compound and any one covalent compound.
Melting/boiling point Ionic compound Covalent compound 1. 2. 3.
High force of attraction between oppositely charged ions are
strong. more heat energy needs to overcome the forces. low force of
attraction between molecules are weak. less heat energy needs to
overcome the forces. 1 1 1 1 Electrical conductivity Ionic compound
Covalent compound 4. 5. Conduct in molten state or aqueous
solution. The free moving ions are able to carry electrical
charges. Not conduct electricity. Neutral molecules are not able to
carry electrical charges. 1 1 1 1 Solubility Ionic compound
Covalent compound 6 7 Soluble in water. Water molecule is polar
solvent. soluble in benzene/ toluene / any organic solvents. The
attraction forces between molecules in solute and solvent are the
same. 20 Q @Hak cipta BPSBPSK/SBP/2013 10 Perfect Score & X A
Plus Module/mark scheme 2013 10 (i) Compound formed between X and Y
Molecule formed between Z and Y Types of chemical bonds Ionic bond
is formed because X atom donates electrons and Y atom receives
electrons to achieve stable octet electron arrangement/involve
transfer electron Covalent bond is formed because Z and Y atoms
share the electrons to achieve stable electron arrangement //
Inovelve sharing of electron Boiling point and melting point High
because a lot of heat energy needed to overcome the strong
electrostatic forces between ions Low because less heat energy is
needed to overcome the weak forces of attraction between molecules
2 2 (b) 1.Correct electron arrangement of 2 ions 2.Correct charges
and nuclei are shown 3. X atom with an electron arrangement of
2.8.2 donates 2 valence electrons to achieve the stable octet
electron arrangement, 2.8. X2+ ion is formed // X X2+ + 2e- 4. Y
atom with an electron arrangement of 2.6 accept 2 electrons to
achieve the stable octet electron arrangement, 2.8. Y2- ion is
formed // Y + 2e- Y2- 5. The oppositely-charged ions, X2+ and Y2-
are attracted to each other by a strong electrostatic force. 6. An
ionic compound XY is formed 1 1 1 1 1 1 1 X X X X X X X X X X XX X
X X X X X X 2+ 2- X2+ Y2- X X X X X X X X X X XX X X X X X X X @Hak
cipta BPSBPSK/SBP/2013 11 Perfect Score & X A Plus Module/mark
scheme 2013 (c) 1. A crucible is filled with solid P until it is
half full. 2. Two carbon electrodes are dipped in the solid P and
connected to the batteries using connecting wire. 3. Switch is
turned on and observation is recorded. 4. The solid P is then
heated until it melts completely. 5. The switch is turned on again
and observation is recorded. 6. Steps 1 to 5 are repeated using
solid Q to replace solid P. 7. Observations: P does not light up
the bulb in both solid and molten states. Q lights up the bulb in
molten state only. P: naphthalene // any suitable answer Q:
lead(II) bromide // any suitable answer 1 1 1 1 1 1 1 1 1 1 1 1 1
20 11 (a) (i) Z : 2.8.7 X : 2.4 1 1 ..2 (ii) Z atom has 7 valence
electrons needs one electron X atom has 4 valence electrons ,hence
it needs 4 more electron each atom achieves stable octet electron
arrangement share electrons between them four Z atoms , each
contributes 1 electron // [ diagram one X atom contributes 4
electrons //[diagram] - four single covalent bonds are formed - the
molecular formula is XZ4 - diagram [ no. of electrons in all the
occupied shells in the X and Z atoms - correct] [ sharing of 4
pairs of single covalent bonds between 1 X atom and 4 Z atoms ] 1 1
1 1 1 1 1 1 1 1 ..10 (iii) Colourless liquid 1 b) [Procedures of
the experiment] eg. 1. Add a quarter of spatula of YZ solid and add
into a test tube. 2. Pour 2-5 cm3 of distilled water into the test
tube containing theYZ2 3. Stopper the test tube and shake well. 4.
Repeat Steps 1 to 3 using [ named organic solvent eg ether ] 5.
Observe the changes and record them in a table 1 1 1 1 1 .
[Results] Eg Solvent Observation Distilled water Colourless
solution obtained [named organic solvent] e.g ether Solid crystals
insoluble in liquid [Conclusion] eg ZY is insoluble in organic
solvent/[named organic solvent] but soluble in water. 1 1 ..7 @Hak
cipta BPSBPSK/SBP/2013 12 Perfect Score & X A Plus Module/mark
scheme 2013 No Explanation Sub Total 12 (a)(i) Y more reactive 1 5
Atomic size of Y bigger than X // The number of shell occupied with
electron atom Y more than X. 1 The single valence electron becomes
further away from the nucleus. 1 the valence electron becomes
weakly pulled by the nucleus. 1 The valence electron can be
released more easily. 1 (ii) Name : Sodium 4Na + O2 2Na2O Chemical
formulae Balance equation 1 1 1 3 (b) Put group1 metal into bottle
that contain paraffin oil Group 1 metal readily reacts with
air/moisture in atmosphere/ water 1 1 2 (c) Name : Sodium/any group
1 element Material : group 1 elements, water, Apparatus : forceps ,
knife, filter paper, basin, litmus paper. 1 1 [procedure] 3. Pour
some water into the basin 4. Group 1 metal is take out from
paraffin oil using forceps 5. A small piece of group 1 element is
cut using a small knife 6. Oil on group 1 element is dried using a
filter paper 7. The group 1 element is placed in the basin contain
water. 8. Dip a red litmus paper into water 1 1 1 1 1 1 Max 5
[observation] 9. Color of red litmus paper turn to blue 1 [chemical
equation ] Sample answer 2 Na + 2 H2O 2NaOH + H2 Chemical formulae
Balance equation 1 1 Total 20 No Explanation ` Total 13. (a)
Glucose // naphthalene // any solid covalent compound covalent
Intermolecular forces are weak Small amount of heat energy needed
to overcomes the forces 1 1 1 1 4 (b) X = 2.1 X = 2.2 Y = 2.7 // Y
= 2.6 // 1. Suitable electron aranggement 2. Ionic bond 3. to
achieve octet electron arrangement 4. One atom of X donates 1
electron to form ion X+ 5. One atom of Y receives an electron to
form ion Y- 6. Ion X+ and ion Y- are attracted together by the
strong electrostatic forces 1 1 1 1 1 1 1 7 (c) material and
apparatus; compound XY, Carbon electrode, cell, wire, crucible,
bulb/ammeter/galvanometer 1 @Hak cipta BPSBPSK/SBP/2013 13 Perfect
Score & X A Plus Module/mark scheme 2013 Procedure A crucible
is half fill with solid XY powder Dipped two carbon electrode
Connect the electrodes with connecting wire to the battery and bulb
Observed whether bulb glow Heated the solid XY in the crucible
Observed whether bulb glow Observation Solid XY - bulb does not
glow Molten XY - bulb glow Diagram Functional diagram Labeled 1 1 1
1 1 1 1 1 9 TOTAL 20 SET 1:CHEMICAL FORMULAE AND EQUATIONS Question
No Mark scheme Mark 1 (a) Molar mass is the mass of a substance
that contains one mole of the substance. Example : Molar mass of
one mole of magnesium is 24gmol-1 . 1 (b) Substance Molar mass /
gmol-1 N2 14x2 = 28 CO2 12+2(16) = 44 H2S 2(1)+ 32 = 34 H2O 2(1)+16
= 18 4 (c) Mole of water = 0.9/ 18 = 0.05 Number of molecules =
0.05 x 6.02 x 1023 = 0.3 x 1023 // 3 x 1022 Mole of carbon dioxide
= 2.2 / 44 = 0.05 1 1 1 1 @Hak cipta BPSBPSK/SBP/2013 14 Perfect
Score & X A Plus Module/mark scheme 2013 Number of molecules =
0.05 x 6.02 x 1023 = 0.3 x 1023 // 3 x 1022 Number of molecule is
simmilar 1 2 (a) (i) Volume CO2 = 0.1 mol x 24dm3 mol-1 = 2.4 dm3 1
(ii) Mass of CO2 = 0.1 mol x 44 gmol-1 = 4.4 g 1 (iii) Number of
molecules = 0.1 mol x 6.02 x 1023 1 (iv) Number of atoms = 6.02 x
1022 x 3 = 1.806 x 1023 1+1 (b) (i) Heating, cooling and weighing
processes are repeated a few times until a constant mass is
obtained. (ii) Compound Anhydrous CoCl2 H2O Mass/g (34.10-31.50)g =
2.60 g (36.26-34.10)g = 2.16 g Number of moles 2.60/130 = 0.02
2.16/18 = 0.12 Ratio of moles 0.02/0.02 = 1 0.12/0.02 = 6 Simplest
ratio of moles 1 6 1 mole of CoCl2 combines with 6 moles of H2O
Therefore, the molecular formula of hydrated cobalt(II) chloride
crystal is CoCl2.6H2O. Hence, the value of x in CoCl2.xH2O is 6. 1
1 1 (iii) Percentage of water = 6(18)2(35.5)59 )18(6 x 100% = 108 x
100% = 45.4% 238 1 1 Total 10 3 (a) (i) concentrated sulphuric acid
1 (ii) zink and hydrochloric acid[ any suitable metal and acid ] 1
(iii) Zn + 2HCl ZnCl2 + H2 (b) (i) Mole of oxygen = 46.35 - 45.15
16 = 1.2 = 0.075 16 1 @Hak cipta BPSBPSK/SBP/2013 15 Perfect Score
& X A Plus Module/mark scheme 2013 (ii) Mole of copper = 45.15
- 40.35 64 = 4.8 = 0.075 64 1 1 (iii) Empirical formula = CuO 1 (c)
(i) Collect the hydrogen gas in a test tube Put a burning wooden
splinter at the mouth of the test tube No pop sound produced. 1 1 1
(ii) To avoid the hot copper react with oxygen/air 1 (iii) Repeat
heating, cooling and weighing processes until a constant mass
obtained. 1 Total 11 4 (a) (i) Pb(NO3)2 1 (ii) AgCl 1 (b) (i) Pb2+
+ 2 Cl- PbCl2 Correct formula for reactants and product Balance
ionic equation 1+1 (ii) Qualitative aspect : Lead(II) nitrate and
sodium chloride are the reactants and lead (II) chloride and sodium
nitrate are the products // Lead(II) nitrate solution reacts with
sodium chloride solution to form lead(II) chloride precipitate and
sodium nitrate solution. Quantitative aspect : One mole of lead(II)
nitrate reacts with 2 mole sodium chloride to produce 1 mole of
lead(II) chloride and 2 mole of sodium nitrate. 1 1 (c) (i) 2
Pb(NO3)2 2 PbO + 4NO2 + O2 1 Compound Colour of the residue when
hot Colour of the residue when cold PbO Brown Yellow Gases Colour
of the gas released NO2 Brown O2 Colourless 1 1 1 Total 10 @Hak
cipta BPSBPSK/SBP/2013 16 Perfect Score & X A Plus Module/mark
scheme 2013 No Explanation Mark 5 (a) (i) Al3+ , Pb4+ 1+ 1 (ii)
Aluminium oxide Lead(IV) oxide 1 + 1 (b) (i) (CH2O)n = 60 12n + 2n
+ 16n = 60 n = 2 Molecular formula = C2H4O2//CH3COOH 1 1 1 (ii)
CaCO3 + 2CH3COOH (CH3COO)2Ca + H2O + CO2 2 (c) (i) 1.Green solid
turn Black 2. Lime water becomes cloudy 1 1 (ii) CuCO3 CuO + CO2 1
+ 1 (iii) 1. 1 mol of copper(II) carbonate decomposed into 1 mol of
copper(II) oxide and 1 mol of carbon dioxide 2. copper(II)
carbonate is in solid state, copper(II) oxide is in solid state and
carbon dioxide is in gaseous state 1 1 (iv) 1. No. of mole for
CuCO3 = 12.4 / 124 = 0.1 mol 2. 1 mol of CuCO3 produces 1 mol of
CuO Therefor No. of mole for CuO = 0.1 mol 3. Mass of CuO = 0.1 mol
X 80 g mol-1 = 8 g 1 1 1 (v) Mass of oxygen is 0.8g Simplest mol
ratio : Cu : O = 3.2/64 : 0.8/16 = 1 : 1 1 1 20 Mark 6 (a) (i)
Empirical formula of a compound is a formula that shows the
simplest whole number ratio of each atoms of each element in a
compound. 1 (ii) (ii) Substance Empirical formula C10H8 C5H4 H2SO4
H2SO4 1 1 @Hak cipta BPSBPSK/SBP/2013 17 Perfect Score & X A
Plus Module/mark scheme 2013 (b) Element Carbon Hydrogen Oxygen
Percentage (%) 62.07 10.34 27.59 Mass/ g 62.07 10.34 27.59 Mole
62.07/12 = 5.17 10.34/1 = 10.34 27.59/16 = 1.72 Simplest mole ratio
5.17/1.72 = 3 10.34/1.72 = 6 1.72/1.72 =1 Empirical formula = C3H6O
n [C3H6O ] = 116 [ 3(12) + 6(1) + 16 ] n = 116 58 n = 116 n = 2
Molecular formula = C6H12O2 1 1 1 1 1 (c) Procedure : 1. Clean
magnesium ribbon with sand paper. 2.Weigh crucible and its lid. 3.
Put magnesium ribbon into the crucible and weigh the 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 for a few minutes.
8.Cool and weigh the crucible with its lid and the content. 9.
Repeat the processes of heating, cooling and weighing until a
constant mass is obtained. 10.Record all the mass. Tabulation of
result : Description Mass/ g Crucible + lid a Crucible + lid +
magnesium b Crucible + lid + magnesium oxide c 10 1 Element
Magnesium Oxygen Mass / g b-a c-b Mole b-a/ 24 c-b / 16 Simplest
ratio of mole x y Empirical formula = MgxOy 1 1 1 Max 11 Total 20
@Hak cipta BPSBPSK/SBP/2013 18 Perfect Score & X A Plus
Module/mark scheme 2013 No Sub T 7. (a) 1. Empirical formula is the
chemical formula that shows the simplest ratio of atoms of each
element in the compound. 2. Molecular formula is the formula that
shows the actual number of atoms of each element in the compound.
3. Example : empirical formula of ethene is CH2 and the molecular
formula is C2H4 1 1 1 3 (b)(i) (ii) Element Carbon Hydrogen Oxygen
Percentage 40.00 6.66 53.33 Number of moles 40 12 3.33 6.66 1 6.66
53.33 16 3.33 Ratio of moles 1 2 1 Empirical formula is CH2O
n(CH2O) = 180 12n + 2n + 16n = 180 30n = 180 n=6 molecular formula
= C6H12O6 1 1 1 1 1 5 (c)(i) Magnesium is more reactive than
hydrogen//Position of magnesium is above hydrogen in the reactivity
series 1 (ii) Lead(II) oxide / Stanum oxide / iron oxide /
copper(II) oxide (iii) 1. Clean [5 15] cm magnesium ribbon with
sandpaper and coil it. 2. Weigh an empty crucible with its lid. 3.
Place the magnesium in the crucible and weigh again. 4. Record the
reading. 5. Heat the crucible very strongly. 6. Open and close the
lid very quickly. 7. When burning is complete stop the heating 8.
Let the crucible cool and then weigh it again 9. The heating,
cooling and weighing process is repeated until a constant mass is
recorded. 10. Description Mass(g) Crucible + lid Crucible + lid +
Mg / Zn / Al Crucible + lid + MgO / ZnO / Al2O3 10 Total 20 @Hak
cipta BPSBPSK/SBP/2013 19 Perfect Score & X A Plus Module/mark
scheme 2013 SET 2 :ELECTROCHEMISTRY Question No Mark scheme Mark
1(a) Electrical to chemical energy / Tenaga elektrik kepada tenaga
kimia 1 (b) Pure copper / Kuprum tulen 1 (c) Cu2+ and H+ 1 (d)(i)
(ii) Become thicker / brown solid formed Bertambah tebal / pepejal
perang terbentuk Cu2+ + 2e Cu 1 1 (e) Blue solution remain
unchanged // the intensity of blue solution is the same. Larutan
biru tidak berubah // keamatan warna biru larutan adalah sama. (i)
the concentration of Cu2+ ions remains the same. kepekatan ion
kuprum(II) tidak berubah (ii) the rate of ionized copper at the
anode same as the rate of discharged copper(II) ion at the cathode
. kadar pengionan kuprum di anode sama dengan kadar ion kuprum(II)
dinyahcaskan di katod 1 1 1 (f) Oxidation / pengoksidaan Copper
atom released electron to form copper(II) ion. Atom kuprum
menderMarkan / membebaskan elektron menghasilkan ion kuprum(II). 1
1 (g) Electroplating of metal // extraction of metal Penyaduran
logam // pengekstrakan logam 1 Total 11 2(a)(i) Chloride ion / Cl-
, hydroxide ion / OH- , sodium ion / Na+ and hydrogen ion / H+ Ion
klorida / Cl- , ion hidroksida /OH- , ion natrium , Na+ dan ion
hidrogen / H+ 1 (ii) Cl- . The concentration of chloride ion is
higher than hydroxide ion. Cl- . Kepekatan ion klorida lebih tinggi
daripada ion hidroksida 1 + 1 (iii) 2Cl- Cl2 + 2e 1 (b)(i)
Functional 1 Label - 1 1 1 (ii) - place lighted splinter at the
mouth of the test tube containing hydrogen gas - pop sound produced
- Letakkan kayu uji menyala ke dalam tabung uji berisi gas hydrogen
- Bunyi pop terhasil 1 1 (iii) - Sodium ion and hydrogen ions move
to the cathode, hydrogen ion is selectively discharged - hydrogen
ion is lower than sodium ion in the Electrochemical Series. - Ion
natrium dan ion hydrogen bergerak / tertarik ke katod, ion hidrogen
terpilih untuk nyahcas / discas - Ion hidrogen terletak di bawah
ion natrium dalam Siri Elektrokimia 1 1 Total 11 Carbon electrodes
Elektrod karbon Sodium sulphate solution Larutan natrium sulfat A
Oxygen gas Gas oksigen Hydrogen gas Gas hidrogen
http://cikguadura.wordpress.com/ @Hak cipta BPSBPSK/SBP/2013 20
Perfect Score & X A Plus Module/mark scheme 2013 Question No
Mark scheme Mark 3(a) Cu2+ , H+ 1 (b) Carbon electrode which
connect to copper electrode in cell A. Because oxidation takes
place Elektrod karbon yang disambung kepada elektrod kuprum dalam
sell A Kerana proses pengoksidaan berlaku 1 1 (c)(i) X silver
electrode / elektrod argentum Y impure silver electrode / elektrod
argentum tak tulen 1 1 (ii) Ag+ + e Ag 1 (d)(i) - The electrode
become thinner - Silver atom ionized / silver atom oxidized to form
silver ion - elektrod seMarkin nipis - atom argentum mengion / atom
argentum dioksidakan membentuk argentum ion. 1 1 (ii) Y : Ag Ag+ +
e Z : Ag+ + e Ag 1 1 (e) The waste chemicals emitted contain
poisonous heavy metal ions and cyanide ions / alter the pH of
water. Bahan buangan kimia dibebaskan mengandungi logam berat yang
beracun dan sianid / mengubah nilai pH air 1 11 Question No Mark
scheme Mark 4(a)(i) Lead(II) ion// Pb2+ , bromide ion// Br- Ion
plumbum(II)// Pb2+ , ion bromida// Br- 1 (ii) Sodium ion // Na+ ,
hydrogen ion// H+ , sulphate ion// SO4 2- , hydroxide ion//OH- ion
natrium // Na+ , ion hidrogen// H+ , ion sulfat // SO4 2- , ion
hidroksida //OH- 1 (b)(i) Lead / Plumbum 1 (ii) Pb2+ + 2e Pb 1
(iii) Brown gas / Gas berwarna perang 1 (c)(i) hydroxide ion / ion
hidroksida 1 (ii) Anode : Oxygen gas anod : Gas oksigen Cathode :
hydrogen gas Katod : gas hidrogen 1 1 (iii) Sodium nitrate solution
// sulphuric acid Larutan natrium nitrat // asid sulfurik (Any
suitable electrolyte) 1 9 @Hak cipta BPSBPSK/SBP/2013 21 Perfect
Score & X A Plus Module/mark scheme 2013 Rubric Mark 5(a) (i)
Q, R, S , Cu 1 . 1 (ii) positive terminal : Cu Potential difference
: 0.7 V S is higher than Cu in the Electrochemical Series 1 1 1
..... 3 (b) (i) positive terminal : copper / Cu Negative terminal :
Metal P (ii) metal P : Zinc / Zn // Magnesium/Mg (any suitable
metal) Solution Q : Zinc sulphate // magnesium sulphate (any
suitable electrolyte) 1 1 1 1 ..... 4 (c) (i) anode : greenish
yellow gas cathode : colourless gas (bubbles) 1 1 .. 2 (ii) gas X :
hydrogen gas Y : chlorine 1 1 .. 2 (iii) Anode Cathode Ions move to
/ ion attracted to Hydroxide ion/OH- Chloride ion/Cl- Hydrogen
ion/H+ , Potassium ion/K+ Ions selectively discharged Cl- H+ Reason
Concentration Cl- higher than OH- Position of hydrogen ion/H+ is
lower than potassium ion/K+ in the Electrochemical Series. Half
equation 2Cl- Cl2 + 2e 2H+ + 2e H2 1+1 1+1 1+1 1+1 . 8 Total 20
Question No Mark scheme Mark 6(a) (i) Substance R : Glucose /
ethanol (any suitable covalent compound) Substance S : Sodium
chloride solution ( any salt solution / acid / alkali) 1 1 .. 2
(ii) 1. S conducts electricity but R does not 2. S has free moving
ions // ions free to move 3. R consists of molecules / no free
moving ions 1 1 1 .. 3 (b) (i) negative terminal : zinc positive
terminal : copper 1 1 .. 2 (ii) 1. zinc electrode become thinner 2.
Zn Zn2+ + 2e 1 1 .. 2 (iii) 1. the potential difference decreases
2. iron is lower than zinc in the Electrochemical Series // iron is
less electropositive than zinc // distance between iron and 1 1
@Hak cipta BPSBPSK/SBP/2013 22 Perfect Score & X A Plus
Module/mark scheme 2013 copper is shorter than distance between
zinc and copper in the Electrochemical Series .. 2 (c) (i) Sample
answer Lead(II) bromide / lead(II) iodide /sodium chloride/sodium
iodide (any suitable ionic compound) r : substance that decompose
when heated. Example : lead(II) nitrate, lead(II) carbonate 1 (ii)
Diagram: Functional Label Observation: Anode : brown gas Cathode:
grey solid Half equation: Anode : 2Br- Br2 + 2e Cathode : Pb2+ + 2e
Pb Product: Anode : lead Cathode : bromine gas 1 1 1 1 1 1 1 1 .. 8
Total 20 Question No Mark scheme Mark 7(a) Sample answer Silver
nitrate solution Functional 1 Label - 1 Anode : Ag Ag+ + e Cathode
: Ag+ + e Ag 1 1 1 1 1 .. 5 (b) 1. metal X is more electropositive
than copper // X is higher than copper in the Electrochemical
Series 1 Carbon electrodes Elektrod karbon PbI2 // PbBr2 // NaCl
Heat Panaskan Note : Observations and half-equations are based on
the substance suggested. Silver nitrate solutionIron spoon Silver
@Hak cipta BPSBPSK/SBP/2013 23 Perfect Score & X A Plus
Module/mark scheme 2013 2. atom X oxidises to X ion // atom X
releases electron 3. copper(II) ion accepts electron to form copper
4. the concentration of copper(II) ion decreases 5. metal Y is less
electropositive than copper // Y is lower than copper in the
Electrochemical Series 1 1 1 1 .. 5 (c) Material 0.5 mol dm-3 of P
nitrate, Q nitrate, R nitrate, S nitrate solutions, metal P, Q, R
and S Apparatus Test tube, test tube rack, sand paper Procedure 1.
Clean the metal strips with sand paper 2. Pour 5 cm3 of P nitrate
solution , R nitrate solution , S nitrate solution into different
test tubes. 3. Place a strip of metal P into each test tube 4.
Record the observation after 5 minutes 5. Repeat steps 2 to 4 using
strip of metal Q, R and S to replace metal P. Observation Metal
Metal ion P Metal ion Q Metal ion R Metal ion S P X X X Q / X X R /
/ X S / / / 1 1 1 1 1 1 1 1 1 Conclusion The electropositivity of
metals increases in the order of P,Q,R,S 1 ..10 TOTAL 20 SET 2
:OXIDATION AND REDUCTION Question No Mark scheme Mark 1 (a) To
allow the flow / movement / transfer of ions through it 1 (b)
chemical energy to electrical energy 1 ( c) mark at electrodes 1
Cell 1 Cell 2 Positive electrode Negative electrode Positive
electrode Negative electrode Q P R S (d)(i) magnesium more
electropositive than copper // above copper in the Electrochemical
Series (ii) blue becomes paler / colourless 1 Concentration /
number of Cu2+ ion decreases 1 (iii) Mg Mg2+ + 2e 1 (iv) Oxidation
1 (e)(i) copper become thicker // brown solid deposited 1 (ii) zinc
1 (iii) zinc undergoes oxidation // zinc atom release electron to
form zinc ion 1 11 @Hak cipta BPSBPSK/SBP/2013 24 Perfect Score
& X A Plus Module/mark scheme 2013 Question No Mark scheme Mark
2(a) A reaction which involves oxidation and reduction occur at the
same time 1 (b) (i) green to yellow/brown 1 (ii) oxidation 1 (iii)
Fe2+ Fe3+ + e 1 (iv) 0 1 (c) (i) magnesium 1 (ii) Mg +Fe2+ Mg2+ +
Fe 1 (iii) +2 to 0 1 (d) 1. label for iron, water and oxygen 2.
ionization of iron in the water droplet (at anode) 3. flow of
electron in the iron to the edge of water droplet Water droplet O2
Iron 1 1 1 11 3 (a) Reaction A : not a redox reaction Reaction B :
a redox reaction 1 1 Reaction A: No change in oxidation number
Reaction B: Oxidation number of magnesium changes/increases from 0
to +2 // Oxidation number of zinc changes/decreases from +2 to 0 1
1.....4 (b) (i) Oxidation number of copper in compound P is + 2
Oxidation number of copper in compound Q is + 1 1 1.....2 (ii)
Compound P : Copper(II) oxide Compound Q : Copper(I) oxide
Oxidation number of copper in compound P is +2 Oxidation number of
copper in compound P is +1 1 1 1 1.....4 (iii) Substance that is
oxidised : H2 Substance that is reduced : CuO Oxidizing agent : CuO
Reducing agent : H2 1 1 1 1.....4 (c) (i) X, Z, Y 1 Y : Copper Z :
Lead X : Magnesium 1 1 1.....3 Fe Fe2+ +2e ee @Hak cipta
BPSBPSK/SBP/2013 25 Perfect Score & X A Plus Module/mark scheme
2013 2Mg + O2 2MgO // 2X + O2 2XO [Correct formulae of reactants
and product] [Balanced equation] 1 1.....2 TOTAL 20 4 (a) (i)
Iron(II) ion releases / loses one electron and is oxidised to
iron(III) ion// Oxidation number of iron in iron(II) ion increases
from +2 to +3. Iron(II) ion undergoes oxidation, Iron(II) ion acts
as a reducing agent 1 1 (ii) Iron(II) ion receives/ gain one
electron and is reduced to iron.// Oxidization number of iron in
iron(II) iron decreases from +2 to 0. iron(II) ion undergoes
reduction, Iron(II) ion acts as an oxidising agent 1 1 (b) eMgMg 22
Oxidation number of magnesium increases from 0 to +2 magnesium
undergoes oxidation CueCu 22 oxidation number of copper in
copper(II) ion decreases from +2 to 0 copper(II) ion undergoes
reduction 1 1 1 1 1 1 (c) At the negative terminal: Iron(II) ion
release / lose one electron and is oxidised to iron(III) ion. Fe2+
Fe3+ + e The green coloured solution of iron(II) sulphate turns
brown. Fe2+ act as a reducing agent. At the positive terminal:
Bromine molecules accepts electrons and is reduced to bromide ions,
Br- Br2 + 2e 2Br- The brown colour of bromine water turns
colourless. Bromine acts as an oxidising agent 1 1 1 1 1 1 1 1 1 1
20 Question No Mark scheme Mark 5 (a) 1. Mg/Al/Fe/Pb/Zn 2.
Magnesium undergoes oxidation as oxidation number of magnesium
increases from 0 to +2 and 3. Copper (II) oxide undergoes reduction
as oxidation number of copper in copper(II) oxide decreases from +2
to 0 4. Oxidation and reduction occur at the same time. 1 1 1 1 (b)
Experiment I 1. Fe2+ ion present 2. Metal X lower than iron in the
Electrochemical Series // Metal X is less electropositive than iron
3. Iron atoms releases electrons to form iron(II) ions 1 1 1 @Hak
cipta BPSBPSK/SBP/2013 26 Perfect Score & X A Plus Module/mark
scheme 2013 Experiment II 1. OH- ion present 2. Metal Y higher than
iron in the Electrochemical Series // Metal Y is more
electropositive than iron 3. Atom Y releases electrons to form Yn+
ions 4. Water and oxygen gain electron to form OH- ion // 2H2O + O2
+ 4e 4OH- 1 1 1 1 Max 3 (c) Procedure 1. One spatula of
copper(II)oxide powder and one spatula of carbon powder is placed
into a crucible 2. The crucible and its content are heated strongly
3. The reaction and the changes that occur are observed 4. Steps 1
to 3 are repeated by replacing copper(II)oxide powder with zinc
oxide powder and magnesium oxide powder. Observation Mixture
Observation Carbon and copper(II)oxide The mixture burns brightly.
The black powder turns brown Carbon and zinc oxide The mixture
glows dimly. The white powder turns grey. Carbon and magnesium
oxide No Changes 1 1 1 1 1+1 Explanation Carbon can react with
copper(II)oxide and zinc oxide Carbon more reactive than copper and
zinc / carbon is above copper and zinc in the Reactivity Series
Carbon cannot react with magnesium oxide Carbon less reactive than
magnesium / carbon is below magnesium in the Reactivity Series 1 1
1 1 20 6 Sample answer (a) Magnesium/Aluminium/zinc/iron/lead 1
Magnesium dissolve//The blue colour of copper(II)sulphate solution
become paler // brown solid deposited 1 MgMg2+ + 2e 1 Cu2+ + 2e Cu
1 Oxidising agent- Cu2+ ion / copper(II) sulphate 1 Reducing agent-
Mg 1..6 (b) sample answer Pb(NO3)2 + 2KI Pbl2 + 2KNO3 1 Oxidation
number: +2 +5 -2 +1 -1 +2 -1 +1 +5 -2 1 @Hak cipta BPSBPSK/SBP/2013
27 Perfect Score & X A Plus Module/mark scheme 2013 no changes
of oxidation number of all elements in the compounds of reactants
and products. 1 Neutralization 1...4 (c ) sample answer [Material :
Any suitable oxidizing agent (example : acidified potassium
manganate(VII) solution, acidified potassium dichromate(VI)
solution, chlorine water, bromine water), any suitable reducing
agent (example : potassium iodide solution, iron(II) sulphate
solution) and any suitable electrolyte] 1 [ Apparatus : U-tube ,
carbon electrodes , connecting wires and galvanometer] 1 Diagram
Functional 1 Labelled 1 Procedure 1 Sulphuric acid is added into a
U-tube until 1/3 full 1 2 Bromine water is added into one end of
the U-tube while potassium iodide solution is added into the other
end of the U-tube 1 3 carefully 1 4 Two carbon electrodes connected
by connecting wires to a galvanometer are dipped into the two
solution at the two ends of the U-tube. 1 Observation The colour of
bromine water change from brown to colourless// The colour of
potassium iodide solution change from colourless to yellow/brown//
The needle of the galvanometer is deflected 1 Oxidation reaction :
Br2 + 2e 2Br- 1 Reduction reaction: 2I- I2 + 2e 1 Max : 10 20 @Hak
cipta BPSBPSK/SBP/2013 28 Perfect Score & X A Plus Module/mark
scheme 2013 SET 3 :ACIDS, BASES AND SALTS Question No Mark scheme
Mark 1 (a)(i) Propanone / Methylbenzene / [any suitable organic
solvent] 1 (ii) Water 1 (b)(i) Molecule 1 (ii) Ion 1 (c) 1. Beaker
A : No observable change Beaker B : Gas bubbles released 2. H+ ion
does not present in beaker A but H+ ion present in beaker B //
Hydrogen chloride in beaker A does not show acidic properties but
hydrogen chloride in beaker B shows acidic properties 1 1 (d)(i) 1.
Correct formula of reactants and products 2. Balanced equation Mg +
2HCl MgCl2 + H2 1 1 (ii) 1. Mole of HCl 2. Mole ratio 3. Answer
with correct unit Mole HCl = // 0.005 2 mol HCl reacts with 1 mol
Mg 0.005 moles HCl reacts with 0.0025 moles Mg Mass Mg = 0.0025 x
24 // 0.06 g 1 1 1 TOTAL 10 Question No Mark scheme Mark 2 (a)(i)
Substance that ionize / dissociate in water to produce H+ ion 1
(ii) 3 1 (iii) 1. Concentration of acid / H+ ion in Set II is lower
than Set I 2. The lower the concentration of H+ ion the higher the
pH value 1 1 (iv) 1. Ethanoic acid is weak acid while hydrochloric
acid is strong acid 2. Ethanoic acid ionises partially in water to
produce low concentration of H+ ion while 3. hydrochloric acid
ionises completely in water to produce high concentration of H+ ion
1 1 1 (b)(i) 1. The pH value of sodium hydroxide in volumetric
flask B is lower than A 2. Concentration of sodium hydroxide / OH-
ion in volumetric flask B is lower than A 1 1
http://cikguadura.wordpress.com/ @Hak cipta BPSBPSK/SBP/2013 29
Perfect Score & X A Plus Module/mark scheme 2013 (ii) 1. Mole
of NaOH 2. Mass of NaOH with correct unit Mole NaOH = // 0.005 Mass
NaOH = 0.005 x 40 g // 0.2 g 1 1 (iii) 0.01 x V = 0.002 x 100 // 20
cm3 TOTAL 10 Question No Mark scheme Mark 3 (a) Pink to colourless
1 (b) Potassium nitrate 1 (c)(i) HNO3 + KOH KNO3 + H2O 1 (ii) 1.
Mole of HNO3 // Substitution 2. Mole ratio 3. Concentration of KOH
with Mole HNO3 = // 0.01 0.01mole HNO3 reacts with 0.01 mole KOH
Molarity KOH = mol dm-3 // 0.4 mol dm-3 1 1 1 (d)(i) 10 cm3 1 (ii)
1. Sulphuric acid is diprotic acid but nitric acid is monoprotic
acid // 1 mole of sulphuric acid produce 2 moles of H+ ion but 1
mole of nitric acid produce 1 mole of H+ ion 2. Concentration of H+
ion in sulphuric acid is double compare to nitric acid 3. Volume of
sulphuric acid needed is half 1 1 1 TOTAL 10 Question No Mark
scheme Mark 4 (a) Ionic compound formed when H+ ion from an acid is
replaced by a metal ion or ammonium ion 1 (b) Pb(NO3)2 1 (c) To
ensure all the nitric acid reacts completely 1 (d)(i) 1. Correct
formula of reactants and products 2. Balanced equation 2H+ + PbO
Pb2+ + H2O 1 1 @Hak cipta BPSBPSK/SBP/2013 30 Perfect Score & X
A Plus Module/mark scheme 2013 (ii) 1. Mole of acid 2. Mole ratio
3. Answer with correct unit Mole HNO3 = // 0.05 0.05 moles HNO3
produce 0.025 moles salt G Mass of salt G = 0.025 x 331 g // 8.275
g 1 1 1 (e) 1. Add 2 cm3 dilute sulphuric acid followed by 2 cm3 of
Iron(II) sulphate solution Slowly add concentrated sulphuric acid
by slanted the test tube. Then turn it upright. 2. Brown ring is
formed. 1 1 TOTAL Question No Mark scheme Mark 5 (a)(i) Salt W :
Copper(II) carbonate Solid X : Copper(II) oxide 1 1 (ii) 1. Flow
gas into lime water 2. Lime water turns cloudy / chalky 3. 1 1
(iii) Neutralisation (iv) 1. Correct formula of reactants and
products 2. Balanced equation CuO + 2HCl CuCl2 + H2O 1 1 (b) Cation
: Cu2+ ion // copper(II) ion Anion : Cl- ion // chloride ion 1 1
(c)(i) Ag+ + Cl- AgCl 1 (ii) Double decomposition reaction 1 TOTAL
Question No Mark scheme Mark 6 (a)(i) Green 1 (ii) Double
decomposition reaction 1 (b)(i) Carbon dioxide 1 (ii) CuCO3 CuO +
CO2 1 (iii) 1. Functional apparatus 2. Label 1 1 (c)(i) Sulphuric
acid // H2SO4 1 Heat Copper(II) carbonate Lime water @Hak cipta
BPSBPSK/SBP/2013 31 Perfect Score & X A Plus Module/mark scheme
2013 (ii) 1. Mole of CuCO3 2. Mole ratio 3. Answer with correct
unit Mole CuCO3 = // 0.1 0.1 moles CuCO3 produces 0.1 mole CuO Mass
CuO = 0.1 x 80 g // 8 g 1 1 1 TOTAL 7 (a) 1. Vinegar 2. Wasp sting
is alkali 3. Vinegar can neutralize wasp sting 1 1 1 (b) 1. Water
is present in test tube X but in test tube Y there is no water. 2.
Water helps ammonia to ionise // ammonia ionise in water 3. OH- ion
present 4. OH- ion causes ammonia to show its alkaline properties
5. Without water ammonia exist as molecule // without water OH- ion
does not present 6. When OH- ion does not present, ammonia cannot
show its alkaline properties 1 1 1 1 1 1 (c) 1. Sulphuric acid is a
diprotic acid but nitric acid is a monoprotic acid 2. 1 mole of
sulphuric acid ionize in water to produce two moles of H+ ion but 1
mole of nitric acid ionize in water to produce one mole of H+ ion
3. The concentration of H+ ion in sulphuric acid is double / higher
4. The higher the concentration of H+ ion the lower the pH value 1
1 1 1 (d)(i) 1. Mole of KOH 2. Molarity of KOH and correct unit
Mole KOH = // 0.25 Molarity = mol dm-3 // 1 mol dm-3 1 1 (ii) 1.
Correct formula of reactants 2. Correct formula of products 3. Mole
of KOH // Substitution 4. Mole ratio 5. Answer with correct unit
HCl + KOH KCl + H2O Mole KOH = // 0.025 0.025 mole KOH produce
0.025 mole KCl Mass KCl = 0.025 x 74.5 g // 1.86 g 1 1 1 1 1 TOTAL
20 @Hak cipta BPSBPSK/SBP/2013 32 Perfect Score & X A Plus
Module/mark scheme 2013 Question No Mark scheme Mark 8 (a)(i) 1.
PbCl2 2. Double decomposition reaction 1 1 (ii) Copper (II)
chloride : Copper(II) oxide / copper(II) carbonate , Hydrochloric
acid Lead (II) chloride : Lead (II) nitrate solution , sodium
chloride solution ( any solution that contains Cl- ion) 1 + 1 1 + 1
(b)(i) 1. S = zinc nitrate 2. T = zinc oxide 3. U = nitrogen
dioxide 4. W = oxygen 1 1 1 1 (ii) 2Zn(NO3)2 2ZnO + 4NO2 + O2 1+1
(c)(i) 1. Both axes are label and have correct unit 2. Scale and
size of graph is more than half of graph paper 3. All points are
transferred correctly 1 1 1 (ii) 1 (iii) Mole Ba2+ ion = // 0.0025
Mole SO4 2- ion = // 0.0025 Ba2+ ion : SO4 2- ion 0.0025 : 0.0025
// 1 : 1 1 1 1 (iv) Ba2+ + SO4 2- BaSO4 1 TOTAL 20 Question No Mark
scheme Mark 9 (a) 1. HCl // HNO3 2. 1 mole acid ionises in water to
produce 1 mole of H+ ion 3. H2SO4 4. 1 mole acid ionises in water
to produce 2 moles of H+ ion 1 1 1 1 (b) 1. Sodium hydroxide is a
strong alkali 2. Ammonia is a weak alkali 3. Sodium hydroxide
ionises completely in water to produce high concentration of OH-
ion 4. Ammonia ionises partially in water to produce low
concentration of OH- ion 5. Concentration of OH- ion in sodium
hydroxide is higher than in ammonia 6. The higher the concentration
of OH- ion the higher the pH value 1 1 1 1 1 1 5 @Hak cipta
BPSBPSK/SBP/2013 33 Perfect Score & X A Plus Module/mark scheme
2013 (c) 1. Volumetric flask used is 250 cm3 2. Mass of potassium
hydroxide needed = 0.25 X 56 = 14 g 3. Weigh 14 g of KOH in a
beaker 4. Add water 5. Stir until all KOH dissolve 6. Pour the
solution into volumetric flask 7. Rinse beaker, glass rod and
filter funnel. 8. Add water 9. when near the graduation mark, add
water drop by drop until meniscus reaches the graduation mark 10.
stopper the volumetric flask and shake the solution 1 1 1 1 1 1 1 1
1 1 TOTAL 20 Question No Mark scheme Mark 10 (a)(i) Substance C :
Glacial ethanoic acid Solvent D : Propanone [ or any organic
solvent] 1 1 (ii) Solution E 1. Ethanoic acid ionises in water 2.
Can conduct electricity because presence of freely moving ions 3.
blue litmus paper turns to red because of H+ ions is present
Solution F 4. Ethanoic acid exist as molecules 5. Cannot conduct
electricity because no freely moving ion 6. Cannot change the
colour of blue litmus paper because no H+ ion 1 1 1 1 1 1 (b) 1.
Measure and pour [20-100 cm3 ] of [0.1-2.0 mol dm-3 ]zinc nitrate
solution into a beaker 2. Add [20-100 cm3 ] of [0.1-2.0 mol dm-3
]sodium carbonate solution 3. Stir the mixture and filter 4. Rinse
the residue with distilled water 5. Zn(NO3)2 + Na2CO3ZnCO3 + 2NaNO3
6. Measure and pour [20-100cm3 ]of [0.1-1.0mol dm-3 ]sulphuric acid
into a beaker 7. Add the residue/ zinc carbonate into the acid
until in excess 8. Stir the mixture and filter 9. Heat the filtrate
until saturated / 1/3 of original volume 10. Cool the solution and
filter 11. Dry the crystal by pressing between two filter papers
12. ZnCO3 + H2SO4 ZnSO4 + H2O + CO2 1 1 1 1 1 1 1 1 1 1 1 1 TOTAL
20 @Hak cipta BPSBPSK/SBP/2013 34 Perfect Score & X A Plus
Module/mark scheme 2013 SET 3 :RATE OF REACTION Question No Mark
scheme Mark 1(a)(i) Set II 1 (ii) Able to draw the graph with these
criterion: 1 Labelled axis with correct unit 2. Uniform scale for X
and Y axis & size of the graph is at least half of the graph
paper 3. All points are transferred correctly 4. Curve is smooth. 1
1 1 1 (b)(i) Set I : 1.Tangen shown in graph correctly 2.Rate of
reaction = 0.19 cm3 s-1 ( +- 0.05) Set II : 1.Tangen shown in graph
correctly 2.Rate of reaction = 0.23 cm3 s-1 (+- 0.05) 1 1 1 1 (ii)
Add catalyst Increase the temperature Use smaller size/ metal
powder Increases the concentration of acid// Double the
concentration of acid but half volume [Any two] 1 1 Question No
Mark scheme Mark 2 (a) 1. Correct formulae of reactants and product
2. Balanced equation CaCO3+ 2HNO3 Ca(NO3) 2+ CO2 + H2O 1 1 (b)
Functional diagram Label 1 1 (c) 1. Mole of nitric acid 2. Mole
ratio 3. Answer with correct unit Number of moles of HNO3 = 0.2 X
50 = 0.01 mol // 1000 2 mol of HNO3 produce 1 mol of CO2 0.01 mol
of HNO3 produce 0.005 mol of CO2 1 1 1 Nitric acid Calcium
carbonate Water http://cikguadura.wordpress.com/ @Hak cipta
BPSBPSK/SBP/2013 35 Perfect Score & X A Plus Module/mark scheme
2013 Maximum volume of CO2 = 0.005 x 24 = 0.12 dm3 // 120 cm3 (d)
Experiment I = 0.12 X 1000 // 0.2 cm3 s-1 // 10 X 60 //0.12 //0.012
dm3 min-1 10 Experiment II = 0.12 X 1000 // 0.4 cm3 s-1 // 5 X 60
// 0.12 // 0.024 dm3 min-1 5 1 1 (e)(i) Rate of reaction in
Experiment II is higher than I 1 (ii) - The size of calcium
carbonate in Experiment II is smaller than Experiment I // calcium
carbonate powder in Experiment II has a larger total surface area
exposed to collision than Experiment I. - The frequency of
collision between between calcium carbonate and hydrogen ion in
Experiment II is higher than Experiment I. - The frequency of
effective collision s in Experiment II is higher than Experiment I
1 1 1 Question No Mark scheme Mark 3 (a) -Total surface area of
smaller pieces wood is larger/bigger/ greater than the bigger
pieces of wood - More surface area exposed to air for burning 1 1
(b)(i) 1. Experiment II 2. Present of catalyst /manganase(IV) oxide
in Experiment I 1 1 (ii) 1.Correct formulae of reactants and
product 2.Balanced equation 2H2O2 2H2O + O2 1 1 (iii) 1. Arrow
upward with energy label ,two levels and position of reactant and
products are correct 2. Curve of Experiment I and experiment II are
correct and label 3. Activation energy of experiment I and
experiment II are shown and labelled 1 1 1 (c)(i) 1.Correct
formulae of reactants and product 2.Balanced equation Zn + 2HCl
ZnCl2 + H2 1 1 (ii) No. of mol HCl = 50 X 0.5 // 0.025 1000 1
Energy 2H2O2 Ea 2 H2O + O2 Ea @Hak cipta BPSBPSK/SBP/2013 36
Perfect Score & X A Plus Module/mark scheme 2013 2 mol HCl : 1
mol H2 0.025 mol HCl : 0.0125 mol H2 Volume of H2 = 0.0125 x 24 //
0.3dm3 // 300 cm3 1 1 (iii) 1. Add excess zinc powder with 12.5 cm3
of 1 mol dm-3 hydrochloric acid . 2. At the same temperature OR 1.
Add excess zinc powder with 25 cm3 of 0.5 mol dm-3 hydrochloric
acid 2. At the higher temperature //present of catalyst 1 1 1 1
(iv) 1. Rate of reaction using sulphuric acid is higher 2. The
concentration of H+ ion in sulphuric acid is higher 3. Maximum
volume of gas collected is double 4. The number of mole of H+ ion
in sulphuric acid is double 1 1 1 1 20 Question No Mark scheme Mark
4 (a) 1. Temperature in refrigerator is lower than in cabinet 2.
The activity of microorganisme (bacteria) in refrigerator is lower
than in refrigerator 3. The amount of toxin produced in the
refrigerator is less then in the kitchen cabinet. 1 1 1 (b)(i) 1.
Correct formula of reactants and products 2. Mol of sulphuric acid
3. Mole ratio 4. Volume and ratio Zn + H2SO4 ------- ZnSO4 + H2 No.
Of mol H2SO4 = 1 X 50/1000 // 0.05 1 mol of H2SO4 : 1 mol of H2
0.05 mol of H2SO4 : 0.05 mol of H2 Volume of H2 = 0.05 x 24 dm3
//1.2 dm3 //0.05 x 24000//1200 cm3 1 1 1 1 (ii) Experiment I = 1200
// 15 cm3 s-1 80 Experiment II = 1200 // 7.5 cm3 s-1 160 Experiment
III = 600 // 2.5 cm3 s-1 240 1 1 1 (iii) Exp I and II 1.Rate of
reaction of Expt I is higher 2.The size of zinc in Expt I is
smaller 3.Total surface area of zinc in Expt I is bigger/larger
4.The frequency of collision between zinc atom and hydrogen ion/H+
in Expt I is higher 5. The frequency of effective collision in Exp
I is higher 1 1 1 1 1 @Hak cipta BPSBPSK/SBP/2013 37 Perfect Score
& X A Plus Module/mark scheme 2013 Exp II and III 1. Rate of
reaction in Expt II is higher 2.The concentration of sulphuric
acid/ H+ ion in Exp II is higher 3. The no. of H+ per unit volume
in Expt II is higher/greater in Expt II// 4. The frequency of
collision between zinc atom and H+ in Expt II is higher 5. The
frequency of effective collision in Expt II is higher 1 1 1 1 1 20
Question No Mark scheme Mark 5.(a) (i) N2 + 3H2 ------- 2NH3 1 + 1
(ii) Temperature : 450 550 C Pressure : 200 300 atm Catalyst :
Powdered iron// Iron filling [ Any two] 1 1 (b)(i) Example of acid
Sample answer : Hydrochloric acid / HCl// Sulphuric acid // Nitric
acid Correct formula of reactant and product Balance Sample answer
2HCl + Mg MgCl2 + H2 1 1 1 (ii) 1. Experiment I : 20 cm3 / 60 s //
0.33 cm3 s-1 2. Experiment II : 20 cm3 / 50 s // 0.4 cm3 s-1 1 1
(iii) (Catalyst) Experiment 1: 1.Pour /measure (50-100) cm3 of
(0.1-2 mol dm-3 ) hydrochloric acid . 2.Add excess zinc
powder/granules 3.Add a (2-5 cm3 ) of copper(II) sulphate solution
4.At the same temperature Experiment II : 1. Pour /measure (50-100)
cm3 of (0.1-2 mol dm-3 ) hydrochloric acid . 2. Add excess zinc
powder/granule 3. At the same temperature OR (Temperature)
Experiment 1: 1. Pour /measure (50-100) cm3 of (0.1-2 mol dm-3 )
hydrochloric acid 2. Heat acid to (30-80O C) 3. Add excess zinc
powder/granule Experiment II : 1. Pour /measure (50-100) cm3 of
(0.1-2 mol dm-3 ) hydrochloric acid . 2. Without heating 3. Add
excess zinc powder/granules OR (Concentration) Experiment 1: 1.Pour
/measure (50-100) cm3 of (0.2-2 mol dm-3 ) hydrochloric acid . 2.
Add excess zinc powder/granules 3.At the same temperature 1 1 1 1 1
1 1 1 1 1 1 1 1 @Hak cipta BPSBPSK/SBP/2013 38 Perfect Score &
X A Plus Module/mark scheme 2013 Experiment II : 1. Pour /measure
(50-100) cm3 of (0.1-1 mol dm-3 ) hydrochloric acid . 2. Add excess
zinc powder/granules 3. At the same temperature OR (Size)
Experiment 1: 1.Pour /measure (50-100) cm3 of (0.1-2 mol dm-3 )
hydrochloric acid . 2. Add excess zinc powder 3.At the same
temperature Experiment II : 1. Pour /measure (50-100) cm3 of (0.1-2
mol dm-3 ) hydrochloric acid . 2. Add excess zinc granule 3. At the
same temperature 1 1 1 1 1 1 1 1 (iv) (Catalyst)
1.Catalyst/copper(II) sulphate is used in Experiment I 2.
Catalyst/(copper(II) sulphate) lower activation energy (and provide
an alternative path) 3. More colliding particles / ions are able to
achieve that lower activation energy. 4.The frequency of effective
collision between magnesium atoms and hydrogen ion increases. 5.
The rate of reaction of Experiment I is higher. (Any 4)
(Temperature) 1. Rate of reaction in Experiment I is higher. 2. The
temperature of reaction in Experiment I is higher 3. The kinetic
energy of particles increases in Experiment I // The particles move
faster 4. Frequency of collision between magnesium atom and H+ ion
in Experiment I is higher 5. Frequency of effective collision in
Experiment I is higher (Any 4) (Concentration) 1. Rate of reaction
in Experiment II is higher 2. The concentration of acid in
Experiment I is higher 3. The number of hydrogen ion per unit
volume in Experiment II is higher 4. Frequency of collision between
magnesium atom and H+ ion in Experiment I is higher 5. Frequency of
effective collision in Experiment II is higher (Any 4) (Size)
1.Rate of reaction in Experiment I is higher 2.The size of
magnesium in Experiment I is smaller 3.Total surface area of
magnesium in Experiment I is bigger/larger 4.The frequency of
collision between magnesium atoms and hydrogen ions in Experiment I
higher 5.The frequency of effective collision between in Experiment
I is higher (Any 4) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (v) The
number of mol are same // The concentration and volume of acid are
same 1 @Hak cipta BPSBPSK/SBP/2013 39 Perfect Score & X A Plus
Module/mark scheme 2013 Question No Mark scheme Mark 6.(a) (i) 1.
First minute = 24/60 =0.4 cm3 s-1 // 24 cm3 min-1 2. 2 nd minute =
34-24/60 =0.167 cm3 s-1 // 10 cm3 min-1 1 1 (ii) 3. rate in 1 st
minute higher than 2 nd minute (vice versa) 4. concentration of
sulphuric acid / mass of zinc decreases 1 1 (iii) All hydrogen ion
from acid was completely reacts 1 (iv) A catalyst lower activation
energy provide an alternative path More colliding particles /zinc
atoms and hydrogen ions are able to overcome the lower activation
energy. The frequency of effective collisions between zinc atom and
hydrogen ion in is higher. (any 2 ) 1 1 (b) - hydrogen and oxygen
molecules collide - with correct orientation -total energy of
particles higher or equal to activation /minimum energy 1 1 1
(Temperature) Materials: 0.2 mol dm-3 sodium thiosulphate, 1.0 mol
dm-3 sulphuric acid, a piece of white paper marked X at the centre.
Apparatus: 150 cm3 conical flask, stopwatch, 50 cm3 measuring
cylinder, 10 cm3 measuring cylinder, thermometer, Bunsen burner,
wire gauze. Procedure: 1.Using a measuring cylinder, 50 cm3 of 0.2
mol dm-3 sodium thiosulphate solution is measured and poured into a
conical flask. 2.The conical flask is placed on top of a piece of
white paper marked X at the centre. 3.5 cm3 of 1.0 mol dm-3
sulphuric acid is measured using another measuring cylinder. 4.The
sulphuric acid is poured immediately and carefully into the conical
flask. At the same time, the stop watch is started 5.The mixture in
a conical flask is swirled. 6.The X mark is observed vertically
from the top of the conical flask through the solution. 7.The
stopwatch is stopped once the X mark disappears from view. 8.Step 1
7 are repeated using 50 cm3 of 0.2 mol dm-3 sodium thiosulphate
solution at 40o C, 50o C, 60 o C by heating the solution before 5
cm3 of sulphuric acid is added in. (Max 7) Conclusion When the
temperature of sodium thiosulphate solution is higher , the rate of
reaction is higher 1 1 1 1 1 1 1 1 1 1 1 @Hak cipta
BPSBPSK/SBP/2013 40 Perfect Score & X A Plus Module/mark scheme
2013 (Temperature) Materials: 0.2 mol dm-3 sodium thiosulphate, 1.0
mol dm-3 sulphuric acid, water, a piece of white paper marked X at
the centre. Apparatus: 150 cm3 conical flask, stopwatch, 50 cm3
measuring cylinder, 10 cm3 measuring cylinder, wire gauze.
Procedure: 1.Using a measuring cylinder, 50 cm3 of 0.2 mol dm-3
sodium thiosulphate solution is measured and poured into a conical
flask. 2.The conical flask is placed on top of a piece of white
paper marked X at the centre. 3.5 cm3 of 1.0 mol dm-3 sulphuric
acid is measured using another measuring cylinder. 4.The sulphuric
acid is poured immediately and carefully into the conical flask. At
the same time, the stop watch is atarted 5.The mixture in a conical
flask is swirled. 6.The X mark is observed vertically from the top
of the conical flask through the solution. 7.The stopwatch is
stopped once the X mark disappears from view. 8.Step 1 7 are
repeated by adding 5 cm3 , 10 cm3 , 15 cm3 , 20 cm3 and 40 cm3 of
distilled water .(at the same time) maintaining the total volume of
solution at 50 cm3 after dilution//table of dilution (Max 7)
Conclusion When the temperature of sodium thiosulphate solution is
higher , the rate of reaction is higher 1 1 1 1 1 1 1 1 1 1 1 SET 3
:THERMOCHEMISTRY Question No Mark scheme Mark 1 (a) Heat change
/released when 1 mol copper is displaced from copper (II) sulphate
solution by zinc 1 (b) Blue to colourless 1 (c) (i) 50 X 4.2 X 6 J
// 1260 J 1 (ii) (1.0 )(50) 1000 1 (iii) 1260 0.05 1 // 0.05 J //
25200 J mol-1 @Hak cipta BPSBPSK/SBP/2013 41 Perfect Score & X
A Plus Module/mark scheme 2013 = - 25.2 kJ mol-1 1 (d) 1. Correct
reactant and product 2. Correct two energy level for exothermic
reaction 3. Correct value heat of displacement and unit Sample
answer Energy 1 1 1 (e) (i) 3C 1 (ii) Number of mole copper
displaced is half Heat released is half / 1260 2 1 1 TOTAL 12
Question No Mark scheme Mark 2 (a) Heat of precipitation is the
heat change when one mole of a precipitate is formed from its
solution. 1 (b) To reduce heat loss to the surrounding. Reject :
prevent 1 (c) Ag+ + Cl- AgCl 1 (d) (i) The heat released =(50 + 50)
x 4.2 x 3.5 =1470 J 1 (ii) Number of moles of Ag+ = (50 x 0.5) =
0.025 mol 1000 Number of moles of Cl- = (50 x 0.5) = 0.025 mol 1000
1 1 (iii) 0.025 mole of Ag+ reacts with 0.025 mole of Cl- to form
0.025 mole of AgCl Number of moles of AgCl = 0.025 mol 1 (iv) = x
1470 J =58 800 J Heat of precipitation of AgCl = -58.8 kJ mol-1 1 1
(e) (i) Ag+ + Cl- AgCl H = -58.8kJmol-1 // AgNO3 + NaCl AgCl +
NaNO3 H = -58.8kJmol-1 1 J // 630 J Zn + CuSO4 //Zn + Cu2+ H = -
25.2 kJmol-1 ZnSO4 + Cu //Zn2+ + Cu @Hak cipta BPSBPSK/SBP/2013 42
Perfect Score & X A Plus Module/mark scheme 2013 (ii) 1. Label
axes 2. Energy levels of reactants and products correct with
formula of reactants and products 3. Heat of precipitation written
1 1 1 Total Question No Mark scheme Mark 3. (a) (i) Ethanol 1 (ii)
1260 kJ of heat energy is released when one mole of ethanol is
burnt completely in excess oxygen 1 (b) (i) No of moles of alcohol
= 0.23 / 46 = 0.005 mol 1 mol of alcohol burnt released 1260 kJ
Thus, 0.005 mol of alcohol burnt released 6.3 kJ 1 1 (ii) mc = 6.3
kJ mc = 6.3 x 1000 = 6300/ 200 x 4.2 = 7.5 0 C 1 1 ( c) Heat is
lost to the surrounding // Heat is absorbed by the apparatus or
containers // Incomplete combustion of alcohol 1 (d) (i) 1. Label
axes 2. Energy levels of reactants and products correct with
formula of reactants and products 3. Heat of combustion written 1 1
1 C2 H5 O H + 3 O2 2 CO2 + H2 O H = - 1260 kJmol-1 Ag+ + Cl- H =
-58.8kJmol-1 AgCl Energy Energy @Hak cipta BPSBPSK/SBP/2013 43
Perfect Score & X A Plus Module/mark scheme 2013 (ii) 1. Label
2. Functional 1 1 (e) (i) - 2656 kJmol-1 // 2500-2700 kJmol-1 1
(ii) 1. The molecular size/number of carbon atom per molecule
propanol is bigger/higher methanol 2. Combustion of propanol
produce more carbon dioxide and water molecules 3. More heat is
released during formation of carbon dioxide and water molecules 1 1
1 Total marks Question No Mark scheme Mark 4 (a) (i) Characteristic
Diagram 4.1 Diagram 4.2 Change in temperature Increase Decrease
Type of chemical reaction Exothermic reaction Endothermic reaction
Energy content of reactants and products The total energy content
of the reactants more than the energy content of the products The
total energy content of the reactants less than the energy content
of the products Amount of heat absorbed /realeased during breaking
of bonds Amount of heat absorbed for the breaking of bond in the
reactant is less than heat released during formation of bond in the
products Amount of heat absorbed for the breaking of bond in the
reactant is more than heat released during formation of bond in the
products 1 1 1+1 1+1 (ii) Number of moles of FeSO4 = MV 1000 =
(0.2)(50) = 0.01 mol 1000 Heat change = 0.01 x 200 kJ = 2 kJ //
2000 J Heat change = mc = 2000 (50)(4.2) = 9.5 o C 1 1 1 @Hak cipta
BPSBPSK/SBP/2013 44 Perfect Score & X A Plus Module/mark scheme
2013 (b) 1. Number of mole of Ag+ ion in both experiment = 25 x 0.5
// 0.0125 mol 1000 2. Number of mole of Cl- ion in both experiment
= 25 x 0.5 // 0.0125 mol 1000 3. Number of mole of silver chloride
formed is the same 4. Na+ ionand K+ ion not involved in the
reaction // Ag+ ionand Cl- involved in the reaction 1 1 1 1 (c) (i)
Heat change = mc = (100)(4.2)(42.2 30.2) = 5040 J / 5.04 kJ Number
of moles of HCl / H + ion = (50)(2 = 0.1 mol 1000 Number of moles
of NaOH / OH - ion = (50)(2) = 0.1 mol 1000 The heat of
neutralization = 5.04 0.1 H = - 50.4 kJ mol-1 1 1 1 1 (ii)
Temperature change is 12.0 o C // same Number of moles of sodium
hydroxide reacted when hydrochloric acid or sulphuric acid is used
is the same // 0.01 mol Number of mole of water formed when
hydrochloric acid or sulphuric acid used is the same // 0.01 mol H+
ion in excess when sulphuric acid is used 1 1 1 1 Total marks 20
Question No Mark scheme Mark 5 (a) (i) Neutralisation//Exothermic
reaction 1 (ii) Total energy content of reactant is higher than
total energy content in product 1 (iii) 1. The heat of
neutralization of Experiment 1 is higher than Experiment 2 2. HCl
is strong acid while ethanoic acid is weak acid 3. HCl ionises
completely in water to produce high concentration of H+ ion 4.
CH3COOH ionizes partially in water to produce low concentration of
H+ ion and most of ethanoic acid exist as molecules 5. In Expt
2,Some of heat given out during neutralization reaction is used to
dissociate the ethanoic acid molecules completely in water//part of
heat that is released is used to break the bonds in the molecules
of ethanoic acid that has not been ionised 1 1 1 1 1 (b) (i) No of
mol acid/alkali= 50 X 1 /1000= 0.05 Q = H X no of mol = 57.3 X 0.05
= 2.865 kJ // 2865 J 1 1 1 (ii) 2865 = 100 X 4.2 X 0 = 2865 420 1 1
@Hak cipta BPSBPSK/SBP/2013 45 Perfect Score & X A Plus
Module/mark scheme 2013 = 6.8 o C ( correct unit) 1 (iii) 1. Some
of heat is lost to the sorrounding 2. Heat is absorbed by
polystyrene cup 1 1 (c ) A B The reaction is exothermic// Heat is
released to the surrounding during the reaction The reaction is
endothermic// Heat is absorbed from the surrounding during the
reaction Heat released is x kJ when 1 mol product is formed Heat
absorbed is y kJ when 1 mol product is formed. The total energy
content in reactant is higher than total energy content in product
The total energy content in reactant is lower than total energy
content in product The temperature increases during the reaction
The temperature decreases during the reaaction Heat released during
the formation of bond in product is higher than heat absorbed
during the breaking of bond in reactant Heat absorbed during the
breaking of bond in reactant is higher than heat released during
the formation of bond in product 1 1 1 1 1 TOTAL 20 6 (a) (i) 1.
Y-axes : energy 2. Two different level of energy 1 1 (ii) 1.
reactants have more energy // products have less energy 2.energy is
released during the experiment // this is exothermic reaction 1 1
(b) No. of mol of H+ ion/OH- = 1x50/1000// 0.05 Heat change = 100x
4.2 x7//2940 Joule//2.94 kJ Heat of neutralization= -2940/0.05 =
-58800 J mol -1 //-58.8 kJ mol-1 1 1 1 1 (c) 1. Heat of combustion
of propane is higher 2. The molecular size/number of carbon atom
per molecule propane is bigger/higher 3. Produce more carbon
dioxide and water molecules//released more heat energy 1 1 1 1.
Methanol/ethanol/ propanol, Diagram: 2. -labelled diagram 3.
-arrangement of apparatus is functional 1 1 1..3 energy Zn + CuSO4
ZnSO4 + Cu H = -152 kJmol-1 @Hak cipta BPSBPSK/SBP/2013 46 Perfect
Score & X A Plus Module/mark scheme 2013 1. (100-250 cm3 )of
water is measured and poured into a copper can and the copper can
is placed on a tripod stand 2. the initial temperature of the water
is measured and recorded 3. a spirit lamp with ethanol is weighed
and its mass is recorded 4. the lamp is then placed under the
copper can and the wick of the lamp is lighted up immediately 5.
the water in the can is stirred continuously until the temperature
of the water increases by about 30o C. 6. the flame is put off and
the highest temperature reached by the water is recorded 7. The
lamp and its content is weighed and the mass is recorded . 8 max 4
Data The highest temperature of water = t2 The initial temperature
of water = t1 Increase in temperature, = t2 - t1 = Mass of lamp
after burning = m2 Mass of lamp before burning = m1 Mass of lamp
ethanol burnt, m = m1 - m2 = m ..1 Calculation : Number of mole of
ethanol, C2H5OH, n = m 46 1 The heat energy given out during
combustion by ethanol = the heat energy absorbed by water= 100x x c
x J Heat of combustion of ethanol = m c KJ mol-1 n = -p kJ/mol 1
..4 ..3 Total marks 20 @Hak cipta BPSBPSK/SBP/2013 47 Perfect Score
& X A Plus Module/mark scheme 2013 Question No Mark scheme Mark
7 (a) (i) Heat change = mc = (25+25)(4.2)(33-29) = 445 J Heat of
precipitation of AgCl = - 445 / 0.0125 = -35600 J mol-1 // 35.6 kJ
mol-1 1. The position and name /formulae of reactants and products
are correct. 2. Label for the energy axis and arrow for two levels
are shown. 1 1 1 1 (b) (i) (ii) 1. HCl is a strong acid // CH3COOH
is a weak acid. 2. HCl ionised completely in water to produce
higher concentration of H+ ion. // 3. CH3COOH ionised partially in
water to produce lower concentration of H+ ion. 4. during
neutralisation reaction, some of the heat released are absorbed by
CH3COOH molecules to dissociate further in the molecules. 1. H2SO4
is a diprotic acid// HCl is a monoprotic acid. 2. H2SO4 produced
two moles of hydrogen ion/H+ when one mole of the acid ionised in
water // 3. HCl produced one mole of hydrogen ion/ H+ when one mole
of the acid ionised in water. 4. When one mole of OH- reacts with
two moles of H+ will produce one mole of water, the heat of
neutralisation is still the same as Experiment I because the
definition of heat of neutralisation is based on the formation of
one mole of water. 4Max 3 4Max 3 (c) - apparatus and material : 2
marks - procedures : 5 marks - Table : 1 mark - Calculation : 2
marks Sample answer: Apparatus : Polystyrene cup, thermometer,
measuring cylinder. Materials : Copper (II) sulphate, CuSO4
solution, zinc powder. Procedures : 1. Measure 25 cm3 of 0.2 mol
dm-3 copper (II) sulphate, CuSO4 solution and pour it into a
polystyrene cup. 2. Put the thermometer in the polystyrene cup and
record the initial temperature of the solution. 3. Add half a
spatula of zinc powder quickly and carefully into the polystyrene
cup. 4. Stir the reaction mixture with the thermometer to mix the
reactants. 5. Record the highest temperature reached. 1 1 1 1 1 1 1
Energy AgNO3 + NaCl AgCl + NaNO3 * H = -35.6 kJ mol-1 * Accept
ionic equation @Hak cipta BPSBPSK/SBP/2013 48 Perfect Score & X
A Plus Module/mark scheme 2013 SET 4 :CARBON COMPOUNDS Question No
Mark scheme Mark 1 (a) Or 1 (b) C3H7OH + 9/2O2 3CO2 + 4H2O 1 (c)
(i) Sweet/ pleasant smell /// fruity smell 1 (ii) Methanoic acid 1
(iii) 1+1 (d) (i) Oxidation 1 (ii) Orange colour of acidified
potassium dichromate (VI) solution turns green 1 (iii) C3H7OH +
2[O] C2H5COOH + H2O 1 (e) C3H7OH C3H6 + H2O (ii) 1+1 Tabulation of
data: Initial temperature of CuSO4 solution (o C) 1 Highest
temperature of the reaction mixture (o C) 2 Temperature change (o
C) 2 - 1 ....1 Calculation : Number of mole of CuSO 4 = MV/1000 =
(0.2)(25)/1000 = 0.005 mol 1 Heat change = mc(2 - 1) = x J Heat of
displacement = x / 0.005 kJ mol-1 = y kJ mol-1 .1 TOTAL 20 propene
propanol O H H H H C O C C C H H H H
http://cikguadura.wordpress.com/ @Hak cipta BPSBPSK/SBP/2013 49
Perfect Score & X A Plus Module/mark scheme 2013 Question No
Mark scheme Mark 2 (a) (i) Fermentation 1 (ii) Ethanol 1 (iii) 1
(b) C2H5OH + 3O2 2CO2 + 3H2O 1+1 (c) (i) Ethene 1 (ii) 1 (d) Purple
to colourless 1 (e) (i) Ethyl ethanoate 1 (ii) CH3COOH + C2H5OH
CH3COOC2H5 + H2O 1+1 Question No Mark scheme Mark 3 (a)
Characteristics Explanation Same general formula CnH2n + 1OH
successive member is different from each other by CH2 Relative
atomic mass is different by 14 Gradual change in physical
properties // Melting / boiling point increase Number of carbon
atom per molecules increase // size of molecule increase Similar
chemical properties // oxidation produce carboxylic acid Have same
chemical/similar functional group Can be prepared by similar method
// can be prepared by hydration of alkene Have same chemical
properties // have same functional group 1+1 1+1 1+1 1+1 1+1 (b)
(i) (CH2O)n = 60 (12 + 2 + 16)n = 60 n = 2 1 C2H4O2 1 (ii)
Carboxylic acid 1 React with carbonate to produce carbon dioxide 1
OH C H HCH H H H H C - C H H n @Hak cipta BPSBPSK/SBP/2013 50
Perfect Score & X A Plus Module/mark scheme 2013 (iii) 2
CH3COOH + CaCO3 (CH3COO)2Ca + H2O + CO2 Correct formula of
reactants and products Balanced equation 1 1 (c) Compound P Q The
number of carbon atom 2 2 The number of hydrogen atom 4 6 number of
hydrogen atom Q is higher Type of covalent bond between // carbon/
Type of hydrocarbon Double bond / / Unsaturated Single bond/ /
Saturated Type of homologous series // // Name of compound Alkene//
Ethene // Alkane // Ethane General formula// Molecular formula of
the compound CnH2n // C2H4 CnH2n+2 // C2H6 1 1 1 1 1 Max 4 20
Question No Mark scheme Mark 4 (a) (i) 14.3 % 1 (ii) Element C H
Mass/ % 85.7 14.3 No. of moles 12 7.85 = 7.14 1 3.14 = 14.3 Ratio
of moles/ Simplest ratio 14.7 14.7 = 1 14.7 3.14 = 2 Empirical
formula = CH2 RMM of (CH2)n = 56 .............1 [(12 + 1(2)]n = 56
14n = 56 n = 14 56 = 4 ..1 Molecular formula : C4H8 ..1 1 1 1 6 max
5 (iii) [any 2] 1+1 1+1 Max 4 But-1-eneBut-2-ene 2-methylpropene
@Hak cipta BPSBPSK/SBP/2013 51 Perfect Score & X A Plus
Module/mark scheme 2013 (iv) Compound M (Butene, C4H8) has a higher
percentage of carbon atom in their molecule than butane, C4H10 .1 %
of C in C4H8 = 8)12(4 )12(4 x 100% = 56 48 x 100% = 85.7% 1 % of C
in C4H10 = 10)12(4 )12(4 x 100% = 58 48 x 100% = 82.7% ..1 .....3
(b) (i) Starch Protein / natural silk 1 1 (ii) H H CH3 H I I I I C
= C C = C I I H H 2-methylbut-1,3-diene or isoprene 1 1..2 (c) (i)
Rubber that has been treated with sulphur 1 (ii) In vulcanised
rubber sulphur atoms form cross-links between the rubber molecules
These prevent rubber molecules from sliding too much when stretched
1 1 TOTAL 20 Question No Mark scheme Mark 5 (a) (i) Hydrocarbon
Type of bond Homologous series General formula A covalent alkane
CnH2n+2 B covalent alkene CnH2n 3 3 (ii) Carbon dioxide 2C4H10 +
13O2 8CO2 + 10H2O [Chemical formulae of reactants and products]
[Balanced] 1 1 1 (iii) Hydrocarbon B. Hydrocarbon B is an
unsaturated hydrocarbon which react with bromine. Hydrocarbon A is
a saturated hydrocarbon which do not react with bromine. 1 1 1 @Hak
cipta BPSBPSK/SBP/2013 52 Perfect Score & X A Plus Module/mark
scheme 2013 (iv) Hydrocarbon B more sootiness. B has higher
percentage of carbon by mass. % of carbon by mass ; Hydrocarbon A :
4(12) 100 // 82.76 % 4(12) + 10(1) Hydrocarbon B : 4(12) 100 //
85.71 % 4(12) + 8(1) 1 1 1 1 (b) Carboxylic acid X : Propanoic acid
Alcohol Y: Ethanol 1 1 1 1 TOTAL 20 Question No Mark scheme Mark 6
(a) (i) X - any acid methanoic acid Y - any alkali ammonia aqueous
solution 1 1 (ii) 1. Methanoic acid contains hydrogen ions 2.
Hydrogen ions neutralise the negative charges of protein membrane
3. Rubber particles collide, 4. Protein membrane breaks 5. Rubber
polymers combine together 1 1 1 1 1 5 max 4 (iii) Ammonia aqueous
solution contains hydroxide ions Hydroxide ions neutralise hydrogen
ions (acid) produced by activities of bacteria 1 1 (b) (i) Alcohol
1 (ii) Burns in oxygen to form carbon dioxide and water Oxidised by
oxidising agent (acidified potassium dichromate (VI) solution) to
form carboxylic acid 1 1 (iii) Procedure: 1. Place glass wool in a
boiling tube 2. Soak the glass wool with 2 cm3 of ethanol 3. Place
pieces of porous pot chips in the boiling tube 4. Heat the porous
pot chips strongly 5. Heat glass wool gently @Hak cipta
BPSBPSK/SBP/2013 53 Perfect Score & X A Plus Module/mark scheme
2013 6. Using test tube collect the gas given off Diagram:
[Functional diagram] [Labeled porcelain chips, water, named
alcohol, heat] Test: Add a few drops of bromine water Brown colour
of bromine water decolourised 6 max 5 1 1 1 1 Total 20 Question No
Mark scheme Mark 7 (a) Carbon dioxide/ CO2 and water/ H2O Any one
correct chemical equation Example 2C4H10 + 13O2 8CO2 + 10H2O
Chemical formula of reactants Balanced 1 1 1 (b) Compound B &
Compound D Same molecular formula / C4H8 Different structural
formula 1 1 1 (c) Pour compound A/B into a test tube Add bromine
water to the test tube and shake Test tube contain compound A
unchanged Test tube contain compound B brown colour turn colourless
or Pour compound A/B into a test tube Add acidified Potassium
manganate(VII) solution to the test tube and shake Test tube
contain compound A unchanged Test tube contain compound B purple
colour turn colourless 1 1 1 1 (d) (i) Any members of carboxylic
acid and correct ester Example [Methanoic acid] [Propylmethanoate]
1 1 1 1 Heat Heat Glass wool soaked with ethanol Porcelain chips
Water @Hak cipta BPSBPSK/SBP/2013 54 Perfect Score & X A Plus
Module/mark scheme 2013 (d) (ii) Pour 2 cm3 of [methanoic acid]
into a boiling tube Add 2 cm3 of propanol/compound E into the
boiling tube Slowly/carefully/drop 1 cm3 of concentrated sulphuric
acid Heat the mixture gently Pour the mixture in a beaker that
contain water Observation : Colorless liquid with fruity smell is
formed / Colorless liquid float on water surface 1 1 1 1 1 1 TOTAL
20 Question No Mark scheme Mark 8(a) But-2-ene 2-methylpropene 1+1
1+1 (b) (i) (ii) Propanoic acid Ethanol Chemical properties for
propanoic acid: 1. React with reactive metal to produce salt and
hydrogen gas 2. React with bases/alkali to produce salt and water
3. React with carbonates metal to produce salt, carbon dioxide gas
and water 4. React with alcohol to produce ester [any three]
Chemical properties for ethanol: 1. Undergo combustion to produce
carbon dioxide and water 2. Burnt in excess oxygen to produce CO2
and H2O 3. Undergo oxidation to produce carboxylic acid / ethanoic
acid 4. React with acidified K2Cr2O7 /KMnO4 to produce carboxylic
acid / ethanoic acid 5. Undergo dehydration to produce alkene /
ethene. [Any three answers] 1 1 1 1 1 1 1 1 1 1 1 1 (c) (i) P :
Hexane Q : Hexene // Hex-1-ene (ii) Reaction with bromine //
acidified potassium manganate(VII) solution Procedure: 1 1 1 1 C C
C CH H H H H H H H C C C C H H H H H H H H @Hak cipta
BPSBPSK/SBP/2013 55 Perfect Score & X A Plus Module/mark scheme
2013 1. Pour about [2 -5 cm3 ] of P into a test tube. 2. Add 4-5
drops of bromine water / acidified potassium manganate(VII)
solution and shake. 3. Observe and record any changes. 4. Repeat
steps 1 to 3 by replacing P with Q Observation: P : Brown/ Purple
colour remains unchanged. Q : Brown/ Purple colours decolourise /
turn colourless. 1 1 1 1 1 Max 6 20 SET 4 :MANUFACTURED SUBSTANCE
IN INDUSTRY Question No Mark scheme Mark 1 (a) (i) Contact process
1 (ii) Ammonia 1 (iii) Vanadium(V) oxide, 450 o C - 500o C 1 (iv)
Ammonium sulphate 1 (v) 2NH3 + H2SO4 (NH4)2SO4 1+1 (b) (i)
Composite material 1 (ii) Correct arrangement Correct label 1 1
(iii) nC2H3Cl --( C2H3Cl )n 1 (iv) It has low thermal expansion
coefficient // resistant to thermal shock 1 TOTAL 11 Question No
Mark scheme Mark 2 (a) (i) (ii) 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 3 4 Tin
atom Copper atom http://cikguadura.wordpress.com/ @Hak cipta
BPSBPSK/SBP/2013 56 Perfect Score & X A Plus Module/mark scheme
2013 (b) (i) (ii) (iii) Oleum 2SO2 + O2 2SO3 Moles of sulphur = 48
/ 32 =1.5 Moles of SO2 = moles of sulphur = 1.5 Volume of SO2 = 1.5
24 dm3 = 36 dm3 1 1 1 1 1 1 6 (c) (i) 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. 1 1 1 1 1 1 1 Max:5 (ii) Zinc. Zinc atoms are
of different size, The presence of zinc atoms distrupt the orderly
arrangement of copper atoms. This reduce the layer of atoms from
sliding. Arrangement of atoms 1; Label - 1 1 1 1 1 1 1 Max: 5 Total
20 Question No Mark scheme Mark 3 (a) Haber process Iron N2 + 3H2
2NH3 1 1 1+1 (b) Pure copper Bronze Bronze is harder than pure
copper Tin atoms are of different size The presence of tin atoms
distrupt the orderly arrangement of copper 1 1+1 1 1 Zinc atom
Copper atom Tin atom Copper atom @Hak cipta BPSBPSK/SBP/2013 57
Perfect Score & X A Plus Module/mark scheme 2013 atoms. This
reduce the layer of atoms from sliding. 1 1 MAX 6 Procedure: 1.
Iron nail and steel nail are cleaned using sandpaper. 2. Iron nail
is placed into test tube A and steel nail is placed into test tube
B. 3. Pour the agar-agar solution mixed with potassium
hexacyanoferrate(III) solution into test tubes A and B until it
covers the nails. 4. Leave for 1 day. 5. Both test tubes are
observed to determine whether there is any blue spots formed or if
there are any changes on the nails. 6. The observations are
recorded Results: Test tube The intensity of blue spots A High B
Low Conclusion: Iron rust faster than steel. 1 1+ 1 1 1 1 1 1 1
TOTAL 20 SET 4 :CHEMICALS FOR CONSUMERS Question No Mark scheme
Mark 1 (a) (i) To improve the colour of food 1 (ii) Absorbs water
/inhibits the growth of microorganisms 1 (iii) 1. Preservative 2.
Flavouring 1 1 (b) (i) Analgesic 1 (ii) To relieve pain 1 (c) (i)
Saponification // alkaline hydrolysis 1 (ii) Hydrophobic
hydrophilic 1+1 (iii) Soap form scum/insoluble salts in hard water.
1 TOTAL 10 @Hak cipta BPSBPSK/SBP/2013 58 Perfect Score & X A
Plus Module/mark scheme 2013 Question No Mark scheme Mark 2 (a)
Examples of food preservatives and their functions: Sodium nitrite
slow down the growth of microorganisms in meat Vinegar provide an
acidic condition that inhibits the growth of microorganisms in
pickled foods 1+1 1+1 (b) (i) No // cannot Because aspirin can
cause brain and liver damage if given to children with flu or
chicken pox. // It causes internal bleeding and ulceration 1 1 (ii)
Paracetamol Codeine 1 1 (iii) 1. If the child is given a overdose
of codeine, it may lead to addition. 2. If the child is given
paracetamol on a regular basis for a long time, it may cause skin
rashes/ blood disorders /acute inflammation of the pancreas. 1 1
(c) Type of food additives Examples Function Preservatives Sugar,
salt To slow down the growth of microorganisms Flavourings
Monosodium glutamate, spice, garlic To improve and enhance the
taste of food Antioxidants Ascorbic acid To prevent oxidation of
food Dyes/ Colourings Tartrazine Turmeric To add or restore the
colour in food Disadvantages of any two food additives: Sugar
eating too much can cause obesity, tooth decay and diabetes Salt
may cause high blood pressure, heart attack and stroke. Tartrazine
can worsen the condition of asthma patients - May cause children to
be hyperactive MSG can cause difficult in breathing, headaches and
vomiting. 2 2 2 2 1 1 TOTAL 20 Question No Mark scheme Mark 3 (a)
(i) Traditional medicines are derived from plants or animals.
Modern medicines are made by scientists in laboratory and based on
substances found in nature. 1 1 (ii) Type Modern medicine
Analgesics Aspirin Paracetamol Codein Antibiotics Penicillin
Psychotherapeutic Chloropromazin Caffeina 1 1 1 1 1 1 MAX 5 (iii)
Penicillin Cause allergic reaction, diarrhoea, difficulty breathing
and easily bruising 1 @Hak cipta BPSBPSK/SBP/2013 59 Perfect Score
& X A Plus Module/mark scheme 2013 Codeine Cause addiction,
drowsiness, trouble sleeping, irregular heartbeat and
hallucinations. Aspirin Cause brain and liver damage if given to
children with flu or chicken pox. Cause internal bleeding and
ulceration 1 1 (b) Hard water contains calcium ions and magnesium
ions. Example : sea water Procedure 1. 20cm3 of hard water
(magnesium sulphate solution) is poured into two separate beakers X
and Y. 2. 50 cm3 of soap and detergent solutions are added
separately in beaker X and beaker Y. 3. A small piece of cloth with
oily stains is dipped into each beaker. 4. Each cloth is washed. 5.
The cleansing action of the soap and detergent is observed. Results
Beaker Observation X The cloth is still dirty. Y The cloth becomes
clean. Conclusion The cleansing action of detergent is more
effective than soap in hard water 1 1 1 1 1 1 1 1 1 1 @Hak cipta
BPSBPSK/SBP/2013 60 Perfect Score & X A Plus Module/mark scheme
2013 SET 5 :PAPER 3 SET 1 Rubric Score 1(a)(i) Able to give correct
observation Sample answer: Colourless solution formed//Aluminium
oxide powder dissolved in nitric acid/sodium hydroxide solution. 3
Rubric Score 1(a)(ii) Able to give the correct inference. Sample
answer Aluminium oxide is soluble in nitric acid/sodium hydroxide
solution//Aluminium oxide shows basic/acidic properties 3 Rubric
Score 1(a) (iii) Able to give the correct property of aluminium
oxide. Answer: amphoteric 3 Rubric Score 1(b) Able to state the
hypothesis correctly. Sample answer: When aluminium oxide dissolves
in nitric acid, it shows basic properties, when aluminium oxide
dissolves in sodium hydroxide solution, shows acidic properties. 3
Rubric Score 1(c) Able to state all the variables correctly.
Answer: Manipulated variable: type of solutions // nitric acid and
sodium hydroxide solution Responding variable: solubility of
aluminium oxide in acid and alkali//property of aluminium oxide
Fixed variable: aluminium oxide 3 Rubric Score 1(d) Able to state
the operational definition correctly. Sample answer. When aluminium
oxide solid is added into sodium hydroxide solution, the solid
dissolved. 3 http://cikguadura.wordpress.com/ @Hak cipta
BPSBPSK/SBP/2013 61 Perfect Score & X A Plus Module/mark scheme
2013 Rubric Score 1(e)(i) Able to give the correct observations for
both experiments. Red litmus paper turns blue Blue litmus paper
turns red 3 Rubric Score 1(e)(ii) Able to classify all the oxides
correctly. Acidic oxide Basic axide Carbon dioxide Phosphorous
pentoxide Magnesium oxide Calcium oxide 3 Rubric Score 2(a) Able to
state the observation Sample Answer: 1. Iron glowed brightly 2.
Iron ignited rapidly with bright flame. 3. Iron glowed dimly 3
Rubric Score 2(b) Able to state the observation and the way on how
to control variable Sample Answer : 1. change bromine with chlorine
and iodine 2. Ignition or glowing of halogen 3. Use the same
quantity of iron wool in each experiment. 3 Rubric Score 2(c) Able
to state the correct hypothesis by relating the manipulated
variable and responding variable Sample Answer : 1. The higher the
position of halogen in group 17 the higher the reactivity towards
iron. 2. The higher the position of halogen in group 17 the greater
the ignition or glowing reaction with iron. Rubric Score 2(d) Able
to state the inference correctly. Sample answer: The solid of
Iron(lll) bromide formed//Bromine combined with iron //Iron is
oxidized by bromine//Bromine is reduced by iron 3 Rubric Score 2(e)
Able to arrange the three position of halogen based on the
reactivity toward iron in ascending order Answer : Iodine. Bromine,
Chlorine, 3 Rubric Score 3(a) Able to give the correct arrangement
of the metals Answer: Magnesium, Y, copper 3 @Hak cipta
BPSBPSK/SBP/2013 62 Perfect Score & X A Plus Module/mark scheme
2013 Rubric Score 3(b) Able to give the name of metal Y correctly.
Answer: Zinc//Iron//Lead 3 Rubric Score 3 (c) Able to give the
three observations correctly. Answer: 1. Brown solid deposited 2.
Blue solution turns light blue 3. Zinc strip becomes pale blue. 3
Rubric Score 4(a) Able to give the problem statement correctly.
Sample answer: How is the effect of other metals on the rusting of
iron when the metals are in contact with iron. 3 Rubric Score 4(b)
Able to state the three variables correctly. Answer: Manipulated
variable: Type of metals//Zinc and copper Responding variable:
Rusting of iron Fixed variable: iron nail 3 Rubric Score 4(c) Able
to state the hypothesis correctly. Sample answer: When iron is in
contact with a more electropositive metal/zinc, rusting will not
occur, when iron is in contact with less electropositive
metal/copper, rusting will occur. 3 Rubric Score 4(d) Able to list
the apparatus and materials needed for the experiment. Apparatus:
two test tubes, test-tube rack, Materials: hot agar-agar solution
added with phenolphthalein and potassium hexacyanoferrate(III)
solution, iron nails, zinc strip, copper strip, sand paper. 3
Rubric Score 4(e) Able to give the procedures correctly Sample
answer: 1. Clean 2 pieces of iron nails, zinc strip and copper
strip with sand paper. 2. Coil the iron nails with zinc strip and
copper strip each. 3. Put the iron nails into two different test
tubes 4. Pour hot agar into each test tube until the iron nail is
immersed. 5. Leave the apparatus for about 1 day and record the
observations. 3 @Hak cipta BPSBPSK/SBP/2013 63 Perfect Score &
X A Plus Module/mark scheme 2013 Rubric Score 4(f) Able to tabulate
the data correctly Answer: Experiment Observation Iron nail coiled
with zinc Iron nail coiled with copper 2 PAPER 3 SET 2 Rubric Score
1(a) Able to construct the table correctly with the following
aspects: Experiment Ammeter reading/A I 0.0 II 0.5 III 0.0 3 Rubric
Score 1(b) Able to state the inference correctly. Sample answer:
Lead(II) bromide can conduct electricity in molten
state//Naphthalene/Glucose cannot conduct electricity in molten
state 3 Rubric Score 1(c) Able to state the type of compound
correctly Answer: ionic compound 3 Rubric Score 1(d) Able to state
all the three variables correctly: Answer: Manipulated variable:
type of compound Responding variable: ammeter reading//conductivity
of electricity Fixed variable: state of compound//ammeter 3 Rubric
Score 1(e) Able to state the hypothesis correctly. Sample answer:
Molten ionic compound can conduct electricity but molten covalent
compound cannot conduct electricity. 3 Rubric Score 1(f) Able to
state the operational definition correctly. Sample answer: When
carbon electrodes are dipped into molten lead(II) bromide, ammeter
shows a reading/ammeter needle deflects 3
http://cikguadura.wordpress.com/ @Hak cipta BPSBPSK/SBP/2013 64
Perfect Score & X A Plus Module/mark scheme 2013 Rubric Score
1(g) Able to explain the difference in conductivity of electricity
in Experiment I and II. Sample answer: In Experiment II, molten
lead(II) bromide consists of free moving ions that carry the
electrical current, In Experiment I molten naphthalene consists of
neutral molecules. 3 Rubric Score 1(h) Able to class