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SECTIO
1. (a) an
(b) Zn
(c) copp
(d) To e(e) Repe
(f)
MassNo of mRatioEmpirical
2. (a)(i)
(ii) U
(iii) W
(b) (i) 2.
(ii) X+
(c)(i) Th
(ii) X, Y,
(d) (i) 4
(ii) 2X +
3. (a)(i)
(ii) Grou
(iii) Ato
(b)(i) 2.
(ii)
(iii) Q→
(c) P an
(d)(i) ion(ii)
(e)(i) Co
N A
hydrous cal
2HCl →
r(II) oxide
sure that tat the heati
X5.32
l 0.761
formula =
.1
y have the
Z
+ O2 →
2H2O →
.4
14 and pe
P has 4 val
Q2+ + 2eR. They ha
ic bond
alent bond
ium chlorid
ZnCl2 + H2
e combustig, cooling
– 4.560 = / 64 = 0.0
XO
same 3 she
2X2O
2XOH +
iod 2
lence electr
ve the sam
n tube is tnd weighin
0.768 g12
lls occupied
2
ns and has
proton nu
tally filledprocess u
O5.520 – 50.192 / 11
with electro
2 shells occ
ber but dif
ith hydrogetil a consta
.328 = 0.19 = 0.012
ns
upied with
ferent nucle
n gasnt mass is p
2 g
lectrons
on number.
roduced
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(ii)
4. (a) Al
(b)
p
s
c
T
h
(c) Maki
(d) M1 V1
(2.0)(25
M2
(e) 2Na
(f) (i)
(ii) HClNo of m
From the
No of m
0.0
M
ali which io
H value for
dium hydr
ncentratio
he concentr
igher the co
g soap
= M2 V2
= M2(75)
= 0.67 m
2H2O
NaOHl NaOH =
equation,
0.
l HCl = MV
= M(2
= 2.5
nizes compl
odium hyd
xide is stro
of hydroxi
ation of hyd
ncentration
l dm-3
2NaOH
NaCl + H V / 1000 =
mol NaOH
5 mol NaO
/1000
0) / 1000
mol dm-3
tely in wat
oxide is 14
g alkali whi
e ions
roxide ion i
of hydroxid
+ H2
O2.0(25) / 1
: 1 mol HCl
: 0.05 mol
r to form hi
and pH valu
ch ionizes c
sodium hy
ion, the hi
00 = 0.05
HCl
igh concent
e for ammo
ompletely i
droxide is hi
gher the pH
ol
ation of hy
nia is 11
water to f
igher than a
value.
roxide ions
rm higher
mmonia. T
.
e
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5. (a) Th
between
(b) (i) H
(ii) no of
No of m
(iii) 1 m
0.025 m
(iv) 0.02
1 mol Ag
Heat of
(c)
(d) Ag+
6. (a) To
(b) (i) H
(ii) no ofno of mo
0.2 mol
0.2 mol
1 mol H2
Heat of
e heat chan
sodium chl
= mcӨ = (
mol Ag+ =
l Cl¯ = MV
l Ag+ + 1 m
l Ag+ + 0.0
mol AgCl
Cl release 5
recipitation
Cl¯ → A
reduce the
at change,
mol NaOHl HCl = MV
Cl + 0.2 m
2O release
→ 5460
eutralizatio
ge(release)
ride solutio
0 + 50) (4.
V / 1000
/ 1000 = (
ol Cl¯ → 1
25 mol Cl¯
elease 147
8800 J mol-
, ∆ H = - 58.
gCl
heat loss to
H = mcӨ
= MV / 100 / 1000 = (2
ol NaOH→
10920 J of
J mol-1
, ∆ H = - 5
when 1 mol
n and silver
) ( 30.5 –
(0.5)(50) /
.5)(50) / 1
mol AgCl
→ 0.025 m
J of heat
of heat
8 kJ mol-1
the surrou
(100 + 10
= (2.0)(10.0)(100) / 1
0.2 mol H2
eat
.6 kJ mol-1
of silver ch
nitrate solu
7) = 1470
1000 = 0.0
00 = 0.025
l AgCl
ding
) (4.2) (41
0) / 1000 =00 = 0.2
loride is for
ion.
25 mol
mol
– 28) = 10
0.2 molol
ed from th
20 J
e reaction
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(c)
(d) The
hydrochl
(e)(i) les
(ii) Etha
complete
7. (a)
(b) Zn
(c) Exp I
Exp II:
eat change
ric acid an
than - 54.
oic acid is a
ly.
H2SO4 →
: Average r
verage rate
/ released
sodium hy
kJ mol-1 //
weak acid.
ZnSO4 +
te of reacti
of reaction
hen 1 mol
roxide solu
any value l
Some of he
H2
n = 32 cm3
= 20 cm3 /
of water is
tion.
wer than -
at released
/ 120 s =
120 s = 0.1
ormed fro
54.6 kJ mol
is used to i
.27 cm3s-1
7 cm3s-1
the reactio
-1
nize the et
n between
anoic acid
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(c)
9. (a) (i)
(ii) body
(iii) PVC
(b) (i) Sa
(ii) To pr
(iii) Diss
Dissolve
(c)(i) as
(ii) To tr
(iii) To c
zinc
of an aerop
does not ru
ponification
oduce soap
lves in wat
in oils and
irin // para
at diabetes
lm the pati
lane
t
precipitate
r : Part B
grease: Hy
etamol
nt.
/ To reduc
rophobic
solubility of soap
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SECTIO
10. (a) C
Tempera
Pressure
Catalyst
(b)
P
B
(c)
S
S
S
2
C
T
P
C
S
S
H
N B
onditions fo
ture : 450°
: 200 atm
: Iron
ure copper
hich are ar
hen force i
ronze consi
hich disrup
hen force i
ulphur is bu
+ O2 →
ulphur dioxi
SO2 + O2
onditions re
emperature
ressure: 1
atalyst: Van
ulphur trioxi
O3 + H2S
leum is dilu
2S2O7 + H
r Haber Pro
onsist of at
anged in or
applied, th
ts of tin ato
the orderly
applied, th
rned in oxy
SO2
e is burne
→ 2SO3
quired:
: 450°C
tm
adium(V) o
de is dissol
4 → H2S2O
ed in water
O → 2H2
ess:
oms of the
derly mann
e layer of a
ms of differ
arrangeme
e layers of
en to form
in oxygen
ide
ed in conce
7
to form sul
O4
ame size
r
oms easily
ent size
nt of pure c
ure copper
sulphur dio
o form sulp
ntrated sulp
phuric acid
lide over o
opper atom
not easily s
ide
hur trioxide
huric acid t
e another
s
lide over on
form oleu
e another
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(d)
Sulphur dioxide gas is released to the atmosphere and cause air pollution.
The air pollution can cause respiratory problems and asthma.
Sulphur dioxide released also react with rain water to form acid rain.
Acid rain can corrode buildings, cars, increase acidity of rivers and lakes and increase
acidity of soil.
(e)(i) Procedure
1. Fill a burette with 1 mol dm-3 sulphuric acid. Record the initial burette reading, V1 .
2. Pour 25 cm3 of 1 mol dm-3 ammonia / ammonium hydroxide solution into a conical flask. Add
three drops of phenolphthalein into the conical flask.
3. Add sulphuric acid solution from the burette into the conical flask slowly. Swirl the conical
flask.
4. Stop adding sulphuric acid when the solution in conical flask turn from pink to colourless.
Record the final burette reading, V2.
5. Calculate the volume of sulphuric acid used, V = V2 – V1 6. Pour 25 cm3 of ammonia / ammonium hydroxide solution into a beaker.
7. Add V cm3 of sulphuric acid into the beaker and stir the mixture.
8. Heat and evaporate the salt solution until 1/3 its original volume.
9. Cool the saturated salt solution at room temperature.
10. The ammonium sulphate crystals are filtered.
11. The ammonium sulphate crystals are pressed between two filter papers.
(ii)
Pour 2 cm3
of ammonium sulphate solution into a test tube Add 2 cm3 of dilute hydrochloric acid followed by 2 cm3 of barium chloride solution into
the test tube
Shake the mixture
White precipitate is formed
Sulphate ion is confirmed present.
11. (a) (i) Haber process
(ii) N2 + 3H2 → 2NH3
(iii) Catalyst : IronPressure: 200 atm
Temperature : 450°C
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(b) (i) ethanol and propanoic acid
(ii) Ethanol:
Ethanol undergo dehydration process when with heated porcelain chips to produce
ethene and water
Ethanol undergo oxidation process when heated with acidified potassium
manganate(VII) to produce ethanoic acid and waterPropanoic acid:
Propanoic acid react with calcium carbonate to produce calcium ethanoate, carbon
dioxide and water
Propanoic acid react with zinc to produce zinc ethanoate and hydrogen gas
(c) Reaction with bromine water:
1. Pour 2 cm3 of hexane into a test tube.
2. Add 3 drops of bromine water into the test tube
3. Shake the mixture
4. Repeat steps 1 to 3 using hexene.5. If bromine water change from brown to colourless, the solution is hexene.
6. If no change, the solution is hexane.
Reaction with acidified potassium manganate(VII) :
1. Pour 2 cm3 of hexane into a test tube.
2. Add 3 drops of acidified potassium manganate(VII) into the test tube
3. Shake the mixture
4. Repeat steps 1 to 3 using hexene.
5. If acidified potassium manganate(VII) change from purple to colourless, the solution is
hexene.6. If no change, the solution is hexane.
13. (a)
Chlorine is more reactive than bromine
Chlorine can displace bromide ion from potassium bromide solution
Bromide ion release electron to form bromine molecule. Oxidation occur.
2Br¯ → Br2 + 2e
Chlorine molecule accept electron to form chloride ion. Reduction occur.
Cl2 + 2e → 2Cl¯
Oxidising agent: Chlorine The product is bromine molecule and potassium chloride
(i) Add 2 cm3 of 1,1,1-trichloroethane into the test tube
(ii) Shake the test tube
(iii) Record the observation on top and bottom layer of the solution
(iv) The bottom layer becomes orange. Therefore, bromine is displaced from
potassium bromide solution by chlorine.
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(b)
Procedur
1. Clamp2. PourU-tube.3. Add iruntil the4. Add clayer of5. Imme
6. Conne7. Deterpointer.8. Left t
Transfer
Negative
I
I
C
Positive C
C
R
C
T
e:
a U-tube tilute sulphu
n(II) sulphlayer of thelorine watehe solutionse carbon
ct the electine the ne
e apparatu
of electrons
terminal:
on(II) sulp
on(II) ion r
xidation oc
arbon electr
erminal:hlorine wat
hlorine acce
eduction oc
arbon electr
he electron
a retort staric acid into
ate solutionsolution rear carefully ireaches thelectrodes in
odes to a gative and p
aside for 3
:
ate is reduc
lease elect
ur
ode immers
r is oxidizin
pt electron
ur
ode immers
flow from ir
nd.the U-tube
carefully inches the heto the rightheight of 3to each of t
lvanometeositive term
0 minutes a
ing agent.
on to form
ed in iron(II
g agent.
o form chlo
ed in chlori
n(II) sulph
until its lev
o the left aight of 3 c arm of thecm.e arms.
by using cinals based
nd record t
iron(III) ion
) sulphate s
ride ion. Cl2
e water be
ate solution
l are 6 cm
m of the U-.U-tube by
nnecting won the defl
e observati
. Fe2+ → F
olution bec
+ 2e → 2
omes positi
to chlorine
way from t
tube by usi
sing a drop
ires.ction of gal
n at both e
3+ + e
mes negati
Cl¯
ive terminal
water throu
he mouths
g a droppe
er until the
anometer
lectrodes.
ve terminal
gh external
f the
wire.
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14. (a)
A
C
E
Half equ
A C
Redox re
A
C
(b) Mate
Apparat
Procedur1. Clean
2. Pour 1
3. Coppe
4. Imme
5. Conne
6. Turn
7. Recor
Observat
Anode:Cathode:
Half equ
Anode:
Cathode:
bservation:
node P: Ga
athode Q: b
lectrolyte: B
tions:
node P : 4Oathode Q:
action:
node P: hy
athode Q:
ials: iron s
s: Dry cells,
e:the iron sp
50 cm3 of 0
r plate is m
se the iron
ct both elec
n the switc
the observ
ion:
opper beco Brown soli
tions:
u → Cu2+
Cu2+ + 2
bubbles ar
rown solid
lue solution
H¯ → O2 +u2+ + 2e →
roxide ion r
opper(II) io
oon, coppe
switch, am
on and cop
.5 mol dm-3
de anode a
spoon and
trode to dry
and allow
ations at el
mes thinnerdeposited
+ 2e
→ Cu
released
eposited
turns light
2H2O + 4 Cu
lease elect
n accept ele
plate, 0.5
meter, conn
er plate wi
copper(II)
nd iron spo
opper plate
cells, amm
the current
ctrodes
on iron spo
lue // inten
on to form
ctron to for
ol dm-3 co
ecting wire,
h sand pap
ulphate sol
n is made
into copper
ter and swi
to flow for
n
sity of blue
oxygen and
m copper a
per(II) sul
beaker
r
tion into a
athode
(II) sulphat
itch using c
0 minutes
solution de
water. Oxi
om. Reduct
hate soluti
beaker
e solution.
nnecting wi
reases
ation occur
ion occur.
n, sand pa
re
.
er
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(c)
Zinc is more electropositive than copper in electrochemical series
Zinc becomes negative terminal and copper becomes positive terminal
Negative terminal:
Zinc release electron to form zinc ion. Oxidation occur. Half equation: Zn → Zn2+ + 2e
Observation: Zinc becomes thinner
The electron move from zinc electrode to copper electrode through external circuit /
external wire
Positive terminal:
Copper(II) ion accept electron to form copper atom. Reduction occur.
Half equation: Cu2+ + 2e → Cu
Observation: Copper becomes thicker
15. (a)(i) Lead(II) nitrate: neutralization reaction through the reaction between lead(II) oxide and
nitric acid
Lead(II) sulphate: Double decomposition through the reaction between lead(II) nitrate
solution and sodium sulphate solution
(ii) lead(II) nitrate solution and sodium sulphate solution
(b) Preparation of lead(II) nitrate crystal:
1. Pour 100 cm3 of 1 mol dm-3 nitric acid into a beaker and the solution is heated gently.
2. Add solid lead(II) oxide little by little until in excess.
3. Stir the mixture using glass rod.4. The heating is stopped when lead(II) oxide is no longer dissolve in nitric acid.
5. Filter the mixture
6. The filtrate is heated/evaporated until 1/3 its original volume
7. The saturated lead(II) nitrate solution is cooled to room temperature.
8. The lead(II) nitrate crystals are filtered
9. The lead(II) nitrate crystals are pressed between two filter paper
10. Chemical equation: PbO + 2HNO3 → Pb(NO3)2 + H2O
(c) Test for chloride ion:
1. Pour 2 cm3 sodium chloride solution into a test tube2. Add 2 cm3 dilute hydrochloric acid followed by 2 cm3 silver nitrate solution into the test tube
3. Shake the mixture
4. White precipitate is formed
5. Chloride ion is confirmed present
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Test for iron(II) ion:
1. Pour 2 cm3 iron(II) nitrate solution into a test tube
2. Add 2 cm3 of potassium hexacyanoferrate (III) solution into the test tube
3. Shake the mixture
4. Dark blue colouration is formed
5. Iron(II) ion is confirmed present
PAPER 3
16. (a)
Pink colouration Iron does not rust Very high intensity of dark blue colouration Iron rust very quicklyLow intensity of dark blue colouration Iron rust quickly
(b) (i) Type of metal in contact with iron
(ii) Rusting of iron / Intensity of dark blue colouration
(iii) Iron nails
(c) Fe → Fe2+ + 2e
(d) The further the distance of metal from iron in electrochemical series, the iron rust very
quickly
(e)
Metal more electropositive than iron Metal less electropositive than ironZinc Copper
tin
(f) When the iron nail coiled with less electropositive metal is immersed in jelly solutioncontaining phenolphthalein and potassium hexacyanoferrate(III) solutions, dark bluecolouration is formed.
17. (a) Reading in 2 d.p with correct unit e.g 0.80 cm
(b)
Type of metal Diameter of dent (cm) Average diameter of dent(cm)1 2 3
Pure copper Alloy X(c) Alloy X is harder than pure copper // The smaller the diameter of dent, the metal is harder
(d) Bronze // Brass
18. (a)
2.8 V 0.8 V1.4 V 0.4 V
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(b)
Pairs of metals Potential difference (V)M and Cu 2.8N and Cu 0.8P and Cu 1.4
Q and Cu 0.4
(c) The further the distance between two metals in electrochemical series, the higher the
potential difference
(d) (i) electrode M becomes thinner
(ii) Copper becomes thicker
(iii) Blue solution turns pale blue / Intensity of blue solution decreases
(e) At copper electrode, copper(II) ion accept electron to form copper atom.
The concentration of copper(II) ion in the solution decreases.
(f)(i) Pairs of metals(ii) Potential difference
(iii) copper electrode
(g) Cu, Q, N, P, M
(h)
Pairs of metals Positive terminal Voltage/ VM and N N 2.8-0.8 = 2.0N and P N 1.4-0.8 = 0.6M and P P 2.8-1.4 = 1.4(i)
Electrolyte Non-electrolyteSodium chlorideZinc sulphate
Silver chlorideLead(II) sulphate
19. (a) How to identify the solubility of salt in water? // How to identify the soluble salt and
insoluble salt when dissolve in water?
(b) Manipulated : Type of salts // Salt A and salt BResponding : Solubility of salt in waterFixed : Type of solvent // distilled water // water(c) When the salt dissolves in water, it is soluble salt whereas when the salt not dissolves inwater, it is insoluble salt(d) Apparatus : Beaker, glass rod, spatula, measuring cylinder.Materials : Salt A, salt B, distilled water
(e) 1. Measure 50 cm3 of distilled water (using measuring cylinder) and pour into a beaker.2. Add one spatula of salt A into the beaker.3. Stir the mixture (using glass rod)4. Record the solubility of salt in water.5. Repeat steps 1 to 4 using salt B.
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(f)
Type of salts ObservationSalt ASalt B
20.
Aim of experiment To investigate the effect of rusting of iron when in contact with metal
X[Magnesium] and metal Y[copper]
All variablesinvolved
Manipulated variable: Types of metal in contact with iron//Magnesium and
copper
Responding variable: Rusting or iron
Controlled variable: Iron nails// jelly solution
Statement ofhypothesis
When iron in contact with magnesium, iron does not rust /
When iron in contact with copper, the iron rust
List of materialsand apparatus:
Apparatus : Test tube, test tube rack, dropper, glass rodMaterials : Iron nails, sandpaper, magnesium ribbon, copper strip, hot jelly,
potassium hexacyanoferrate (III) and phenolphthalein
Experimentalprocedure:
1. Clean three iron nails, Magnesium ribbon and Copper strip with
sandpaper.
2. Two iron nails are coiled with magnesium and copper each
3. Place the three nails into three different test tubes
4. Add 4 drops of potassium hexacyanoferrate (III) solution followed by 4drops of phenolphthalein into hot jelly solution. Stir the mixture.5. Pour hot jelly solutions into all the test tubes until it covers the entire nail
6. Left the test tube aside for one day7. Any observation are recordedTabulation of data Pairs of metal Intensity of blue
colour
Intensity of pink
colour
Iron Nail
Iron Nail +
magnesium
Iron Nail +copper
21.
Problem statement How does the concentration of ions affect the selective discharge of ions at theanode?
All variables involved Manipulated: concentration of chloride ionsResponding: Ions discharged at anode/product at anodeConstant: carbon electrodes
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Statement ofhypothesis
When the concentration of chloride ions is higher, chloride ions will be selectivelydischarged at the anode.When the concentration of chloride ions is lower, hydroxide ions will be selectivelydischarged at the anode. //When the concentration of chloride ions is higher, the product at anode is chlorinegas.
When the concentration of chloride ions is lower, the product at anode is oxygengas
List of materials andapparatus:
Materials: 0.001 mol dm-3 hydrochloric acid, 1 mol dm-3 hydrochloric acid, bluelitmus, wooden splinter Apparatus: electrolytic cell, test tubes, measuring cylinder carbon electrodes, twodry cells, connecting wire, switch
Experimentalprocedure:
1. Measure 100 cm3 of 0.001 mol dm-3 hydrochloric acid solution using measuringcylinder and pour into an electrolytic cell.2. Dip two carbon electrodes into the solution. Invert the two test tubes containinghydrochloric acid solution onto both electrodes.3. Connect the electrodes to a switch, ammeter and two dry cells by usingconnecting wires.4. Close the switch and the observation at anode is recorded.
5. Collect the gas in a test tube at anode. Place a glowing wooden splinter into themouth of the test tube. Record the observation.6. Repeat steps 1 to 4 using 1 mol dm-3 hydrochloric acid solution to replace 0.001mol dm-3 hydrochloric acid solution.7. Collect the gas in a test tube at anode. Place a moist blue litmus paper into themouth of the test tube. Record the observation.
Tabulation of data Concentration of hydrochloric acid(mol dm-3)
Observation at Anode
0.0011.0
22.
Aim of experiment To investigate the cleansing action of soap and detergent in hard water
All variables involved Manipulated variable: Soap and detergentResponding variable: The presence of greasy stain on clothFixed variable: Volume of hard water// volume of cleaning agent
Statement ofhypothesis
The cleansing action of a detergent is more effective than soap in hard water.
List of materials andapparatus:
Materials: 5% soap solution, 5% detergent solution, hard water, two small pieces
of cloth with greasy stains.
Apparatus: 100 cm3 beaker, 50 cm3 measuring cylinder, glass rod
Experimentalprocedure:
1. Measure 50 cm3 of hard water using measuring cylinder and pour into a beaker.2. Add 50 cm3 of soap solution into the beaker. Stir the mixture.
3. Immerse a small piece of cloth with greasy stains into the mixture.4. Rub the cloth gently.5. Record the presence of greasy stain on the cloth.6. Repeat steps 1 to 5 using detergent solution.
Tabulation of data Type of cleaning agent ObservationSoap + hard waterDetergent + hard water
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23.
Problem statement How to compare the electrical conductivity between molten naphthalene andmolten lead(II) bromide?
All variables involved Manipulated: naphthalene and lead(II) bromide // Types of compoundResponding: Reading of ammeter/electrical conductivityConstant: mass of compound// carbon electrodes
Statement ofhypothesis
Molten lead(II) bromide can conduct electricity whereas molten naphthalenecannot conduct electricity.
List of materials andapparatus:
Materials: naphthalene and lead(II) bromide Apparatus: crucible, carbon electrodes, two dry cells, connecting wire, ammeter,switch, Bunsen burner, electronic balance
Experimentalprocedure:
1. Measure 50 g of naphthalene and pour into a crucible.2. Dip two carbon electrodes into the naphthalene.3. Connect the electrodes to a switch, two dry cells and ammeter usingconnecting wires.4. Heat the naphthalene until its melt.5. Close the switch and the observation is recorded.6. Repeat steps 1 to 5 using lead(II) bromide to replace naphthalene.
Tabulation of data Type of compound ObservationNaphthalene moltenLead(II) bromide molten