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Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
1 21Nov2017 | cikguadura.wordpress.com
Cadangan jawapan
Kertas 3 – KIMIA SPM 2017 – Cg Adura
[SPM04-01-P3]
(a)
Cadangan jawapan Skor
Observation Inferences
Dapat menyatakan dua pemerhatian
dengan tepat:
Contoh jawapan lengkap:
1. Wasap putih terbebas / white fume
released
2. Jisim kandungan mangkuk pijar
bertambah/ mass increases
3. Pepejal/ serbuk putih terbentuk /
white solid produce/ formed
4. Nyalaan putih/ terang/ berkilau //
burn brightly
Dapat menyatakan dua inferens yang
lengkap sepadan dengan pemerhatian:
Contoh inferens lengkap:
Magnesium oksida terbentuk/ MgO
formed
Magnesium berpadu / bertindak balas
dengan oksigen / Magnesium react with
oxygen
Magnesium teroksida/ dioksidakan
Magnesium oxidized
Dapat menyatakan satu pemerhatian
yg lengkap atau dua pemerhatian yg
kurang lengkap:
Contoh jawapan kurang lengkap:
1. Wasap terbebas // asap putih
terbentuk
2. Jisim magnesium bertambah
3. Pepejal/ serbuk terbentuk
4. Magnesium menyala
Dapat menyatakan satu inferens yang
lengkap atau dua inferens yang kurang
lengkap:
Contoh inferens kurang lengkap:
Sebatian magnesium terbentuk
Magnesium berpadu / bertindak balas
dengan udara
Magnesium teroksida/ dioksidakan
Dapat menyatakan satu idea tentang
pemerhatian:
Contoh jawapan:
1. Gas dibebaskan // gas putih
2. Terdapat pertambahan berat
3. Berat magnesium bertambah
4. Nyalaan/ baraan / cahaya
Jisim mangkuk pijar bertambah
Dapat menyatakan satu idea inferens:
Contoh jawapan:
Tindak balas berlaku
|
3
b
Cadangan jawapan Skor
Dapat mencatatkan ketiga-tiga jisim dengan 2 t. perpuluhan dengan betul:
Contoh jawapan:
Jisim mangkuk pijar dan penutup: 25.35 g
Jisim mangkuk pijar, penutup dan pita magnesium: 27.75 g
Jisim mangkuk pijar, penutup dan magnesium oksida setelah disejukkan:
29.35 g
3
Dapat mencatatkan mana-mana dua jisim dengan 2 t. perpuluhan dengan
betul
2
Dapat mencatatkan mana-mana satu jisim dengan 2 t. perpuluhan dengan
betul
1
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
2 21Nov2017 | cikguadura.wordpress.com
(i) dan (ii)
Cadangan jawapan Skor
Dapat menentukan jisim magnesium, jisim oksigen dan menunjukkan 3
langkah menentukan formula empirik magnesium oksida dengan betul:
Jisim magnesium: 27.75 -25.35 = 2.40 / 2.4 g
Jisim oksigen: 29.35 – 27.75 = 1.60 / 1.6 g
Langkah menentukan formula empirik magnesium oksida:
L1: Mol magnesium:
24
4.2
= 0.1
Mol oksigen:
16
6.1
= 0.1
L2: Nisbah Mg : O = 1:1
L3: Formula empirik = MgO
Dapat menyatakan jisim magnesium, jisim oksigen dan menunjukkan 2
langkah pertama atau menyatakan formula empirik magnesium oksida dengan
betul
Dapat menyatakan jisim magnesium atau jisim oksigen atau menyatakan
formula empirik magnesium oksida dengan betul
d
Cadangan jawapan Skor
Dapat menyatakan definasi secara operasi berdasarkan formula empirik
dengan tepat:
Contoh jawapan:
1 mol magnesium react with 1 mol oxygen
Dapat menyatakan definasi secara operasi berdasarkan formula empirik
dengan kurang tepat:
Contoh jawapan:
1 mol magnesium bertindak balas dengan oksigen
Dapat menyatakan definasi secara operasi mengikut konsep:
Contoh jawapan:
Mol magnesium dan oksigen adalah sama.
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
3 21Nov2017 | cikguadura.wordpress.com
Negeri Sembilan – 2016
Explanation score
1(a) [Able to fulfil the following criteria]
1. all readings correct
2. two decimal places
3. without unit
Answer:
Methanol = 1.55 Ethanol = 1.20 Propanol = 1.10 Butanol = 1.61 3
One of criteria is not fulfilled
2
[Able to fulfil criteria 1] 1
[No response given or wrong response] 0
Explanation score
1
(b)
[Able to state all variables correctly]
Sample answer:
Manipulated variable: Types of alcohols//[Name of alcohols]
Responding variable: Heat of combustion//Temperature
increase/change
Fixed variable: Aluminium can//Water (Reject: Volume of water)
3
[Able to state any 2 variables correctly]
2
[Able to state any 1 variable correctly] 1
[No response given or wrong response] 0
Explanation score
1
(c)
[Able to state the relationship between the manipulated variable and the
responding variable with direction correctly]
Sample answer:
The higher the number of carbon atoms per molecule of alcohol, the
higher the heat of combustion.
Note : RV  MV score 2
3
[Able to state the relationship between the manipulated variable and the
responding variable]
Sample answer:
The higher the number of carbon atoms, the higher the heat of
combustion.
2
[Able to state an idea of hypothesis]
Sample answer:
Different alcohols ,different heat of combustion. 1
[No response given or wrong response] 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
4 21Nov2017 | cikguadura.wordpress.com
Explanation score
1
(d)
Able to draw a bar chart by fulfill the following criteria:
(i) Axes are labelled with correct unit
(ii)Scale used must cover at least 50% of graph paper
(iii)Uniform spacing and width for each bar
Sample answer:[Refer to page 8]
3
Able to draw a bar chart by fulfill the following criteria:
(i) Axes are labelled with correct unit
(ii)Uniform spacing and width for each bar
2
Able to draw a bar chart by fulfill the following criteria:
(i) Axes are labelled with correct unit
(ii) All bars are combined 1
[No response given or wrong response] 0
Explanation score
1(e)
[Able to predict the heat of combustion of pentanol and draw the bar for
pentanol correctly]
Criteria : 1. Value with correct unit
2.The bar drawn as in 1(d) score 3
Answer:
2260 kJ mol-1
[Draw a correct bar chart]
3
[Able to predict the heat of combustion of pentanol and draw the bar for
pentanol]
Answer:
2260 + 20 [without unit]
[Draw a correct bar chart]
2
[Able to predict the heat of combustion of pentanol or draw the bar for
pentanol]
Answer:
2260 + 20 [without unit]
or
[Draw a correct bar chart]
1
[No response given or wrong response] 0
Explanation score
1 (f)
[Able to state the operational definition for heat of combustion correctly]
Sample answer:
The heat released/produced when 1 mole of alcohol is burnt in excess
oxygen, thermometer reading increases.
3
[Able to state the operational definition for heat of combustion]
Sample answer:
Heat released/produced when alcohol is burnt in excess oxygen. 2
[Able to state an idea of operational definition for heat of combustion]
Sample answer:
Heat released when alcohol burns.
1
[No response given or wrong response] 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
5 21Nov2017 | cikguadura.wordpress.com
Melaka 2016
Mark Scheme Mark
1(a) Able to state all pH value correctly
Answer
Acid solution Colour of indicator pH value
Acid P Red 1
Acid Q Yellowish orange 5
Acid R Yellow 6
|
3
Able to state any twopH value correctly 2
Able to state any one pH value correctly 1
No response or wrong response 0
1(b) Able to state all the variables correctly
Sample answer
Manipulated variable:
Type of acid // Acid P, Q, R and S // Four different acids
Responding variable:
pH value // Colour of indicator // Strength of acid
Fixed variable:
Concentration of acid // Universal indicator // volume of acid
3
Able to state any two variables correctly 2
Able to state any one variable // idea of all variables 1
No response or wrong response 0
1(c) Able to state all correct inferences
Acid Inference
P P is a strong acid // P ionize completely in
water // P produces high concentration of
hydrogen ions.
Q Q is a weak acid // Q ionize partially in water
// Q produces low concentration of hydrogen
ions.
R R is a weak acid // R ionize partially in water
// R produces low concentration of hydrogen
ions.
|
3
Able to state any two inferences correctly. 2
Able to state any one inferences // idea of all inferences 1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
6 21Nov2017 | cikguadura.wordpress.com
1(d) Able to state the relationship between pH value with the strength of
acid correctly.
Sample answer
When pH value increases, the strength of acid decreases.
3
Able to state the relationship between pH value with the strength of
acid less correctly.
Sample answer
pH value is inversely proportional to the strength of acid.
2
Able to give an idea
Sample answer
pH value is affected by strength of acid.
1
No response or wrong response 0
1(e) Able to state the operational definition for strong acid correctly.
1. Low pH value
2. Add universal indicator
Sample answer
Acid that shows low pH value when universal indicator is added into it.
3
Able to state the operational definition for strong acid less correctly.
1. Low pH value //
2. Add universal indicator
Sample answer
i. Acid that shows low pH value. //
ii. Add universal indicator into the acid.
2
Able to state an idea of operational definition
Sample answer:
Colour of indicator change // pH less than 7 // Ionise completely
1
No response or wrong response 0
1(f) Able to record all the burette readings accurately with two decimal
place and with correct unit. 3
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
7 21Nov2017 | cikguadura.wordpress.com
Answers
Set Initial burette reading Final burette reading
I 0.30 cm3 24.30 cm3
II 24.30 cm3 47.80 cm3
III 0.50 cm3 25.00 cm3
|
Able to record all burette readings less correctly with or without unit
Sample answers
Set Initial burette reading Final burette reading
I 0.30 24.30
II 24.30 47.80
III 0.50 25.00
or
Set Initial burette reading Final burette reading
I 0.3 cm3 24.3 cm3
II 24.3 cm3 47.8 cm3
III 0.5 cm3 25 cm3
|
2
Able to record at least four burette readings less correctly
Sample answers:
Set Initial burette reading Final burette reading
I 1.7 24.3
II 24.3 48.2
III 0.5 25
|
1
No response or wrong response 0
1(g) Able to construct a table to record the initial burette readings, final
burette readings and volume of acid for Set I, Set II and Set III that
contain:
3
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
8 21Nov2017 | cikguadura.wordpress.com
1. Correct titles
2. Readings and units
Sample answer:
Set Initial burette
reading (cm3)
Final burette
reading (cm3)
Volume of
acid (cm3)
I 0.30 24.30 24.00
II 24.30 47.80 23.50
III 0.50 25.00 24.50
|
Able to construct a less accurate table that contains the following:
1. Titles
2. Readings
Sample answer:
Set Initial burette
reading
Final burette
reading
Volume of
acid
I 0.3 24.3 24.0
II 24.3 47.8 23.5
III 0.5 25.0 24.5
|
2
Able to construct a table that contains titles and readings for initial
burette readings and final burette readings only.
Sample answer:
Set Initial burette reading Final burette reading
I 0.3 24.3
II 24.3 47.8
III 0.5 25.0
|
1
No response or wrong response 0
1(h) Able to predict the volume of acid S correctly with unit.
Answer: 50.0 cm3
3
Able to predict the volume of acid S
Answer: 50
2
Able to give an idea to predict the volume of acid S
Sample answer: more than 25 // more than acid P
1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
9 21Nov2017 | cikguadura.wordpress.com
(i)(i) Able to state the correct observation for both set.
Sample answers
Set I :
Bubbles released // Calcium carbonate dissolve in ethanoic acid
solution.
Set II :
No bubbles // Calcium carbonate does not dissolve in glacial ethanoic
acid.
3
Able to state correct observation for any one set.
Sample answers:
Set I :
Bubbles released // Calcium carbonate dissolve in ethanoic acid
solution.
or
Set II :
No bubbles // Calcium carbonate does not dissolve in glacial ethanoic
acid.
2
Able to give an idea of the observation
Sample answers:
Calcium carbonate reacts with acid // Carbon dioxide gas
1
No response or wrong response 0
(i)(ii) Able to state the relationship between the manipulated variable and
the responding variable and state the direction correctly.
Sample answer:
Presence of water causes acid to show its properties.
3
Able to state the relationship between the manipulated variable and
the responding variable less correctly.
Sample answer:
Acid will shows its acidic properties when water presence.
2
Able to state the idea of hypothesis
Sample answer:
Reaction occurs when water present.
1
No response or wrong response 0
(j) Able to classify all the acids correctly
Sample answer:
Strong acid Weak acid
Nitric acid
Sulphuric acid
Citric acid
Carbonic acid
|
3
Able to classify any three acids correctly 2
Able to classify any two acids correctly 1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
10 21Nov2017 | cikguadura.wordpress.com
Perlis 2015
Rubric Score
1(a) Able to state an observation accurately
Sample answer:
Rubber strip Y is extended longer than rubber strip X
3
Able to state the observation correctly
Sample answer:
Rubber strip Y is extended/become longer//
Rubber strip X do not extended
2
Able to state idea of the observation
Sample answer:
Rubber strip extended
1
No response or wrong response 0
Rubric Score
1(b) Able to state an accurate inference for this experiment:
Sample answer:
Rubber strip X is harder than rubber strip Y
3
Able to state the inference for this experiment:
Sample answer:
Rubber strip X is hard// Rubber strip Y is soft
2
Able to state the general inference for this experiment:
Sample answer:
X is more elastic rubber
1
No response or wrong response 0
Rubric Score
1(c) Able to write all the reading of rubber strip X and Y correctly with one
decimal place.
Answer:
Rubber strip X: 5.0, 5.0, 5.0, 6.0
Rubber strip Y: 5.0, 5.5, 6.4, 8.5
3
Able to write all the reading of rubber strip X and Y correctly
2
Able to write 3 correct reading for rubber strip X and 3 correct
reading for rubber strip Y 1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
11 21Nov2017 | cikguadura.wordpress.com
Rubric Score
1(d) Able to construct a table with correct title and units and accurately
Type of
rubber Rubber strip X Rubber strip Y
Weight (g) 0 15 30 45 0 15 30 45
Length
(cm)
5.0 5.0 5.0 6.0 5.0 5.5 6.4 8.5
3
Able to construct a table without title or units and less accurately
|
Type of
rubber
Rubber strip
X
Rubber strip Y
Weight 15 30 45 15 30 45
Length 5.0 5.0 6.0 5.5 6.4 8.5
2
Able to give the idea about to construct table
Type of
rubber
Rubber strip
X
Rubber strip Y
Length 5.0 5.0 6.0 5.5 6.4 8.5
1
No response or wrong response 0
Rubric Score
1(e)
Able to state three variables correctly:
Sample answer:
(i) Manipulated variable: Rubber strip X, Rubber strip Y
(ii) Responding variable: The length of rubber strip after weight is
removed
(iii) Fixed variable: Size of rubber strip, mass of weight
3
Able to state any 2 of the above information correctly 2
Able to state any 1 of the above information correctly 1
No response or wrong response 0
Rubric Score
1(f)
Able to state the relationship between the manipulated variable and the
responding variable correctly and with direction
Sample answer:
Rubber strip X is more elastic than rubber strip Y
3
Able to state the relationship between the manipulated variable and the
responding variable correctly and without direction
Sample answer:
The elasticity of rubber strip X is high than rubber strip Y 2
Able to state an idea of the hypothesis
Sample answer:
Rubber strip x is less elastic. 1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
12 21Nov2017 | cikguadura.wordpress.com
Rubric Score
1(g) Able to give an accurately relationship between length of rubber strip and
elasticity.
Sample answer:
The length of rubber strip increases the elasticity decreases
3
Able to give less accurately relationship between length of rubber strip and
elasticity.
Sample answer:
The length increases the elasticity decreases
2
Able to give and idea of relationship between length of rubber strip and
elasticity
Sample answer:
directly proportional
1
No response or wrong response 0
Rubric Score
1(h) Able to state the operational definition correctly
Sample answer:
The longer the length of rubber strip after weight is removed the less
elastic the rubber strip.
3
Able to state the operational definition less correctly
Sample answer:
After weight is removed the rubber strip become longer 2
Able give an idea for operational definition
Sample answer:
The length of rubber strip increase 1
No response or wrong response 0
Rubric Score
1(i) Able to explain the observation by stating all the following aspect correctly
Answer:
Rubber strip Y has more double bond between carbon and carbon atom//
Rubber strip X has less double bond between carbon and carbon atom//
Rubber strip Y easily undergoes oxidation process compared to rubber
strip X//
3
Able to explain the observation by stating any two of the aspect correctly
2
Able to explain the observation by stating any one of the aspect correctly
1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
13 21Nov2017 | cikguadura.wordpress.com
Rubric Score
1(j) Able to predict the rubber that will snap first and state the type of rubber
strip X and Y correctly
Answer:
Rubber Y will snap first
Rubber strip X : Vulcanized rubber
Rubber strip y : Unvulcanized rubber
3
Able state any two of the answer correctly
2
Able state any one of the answer correctly
1
No response or wrong response 0
Rubric Score
1(k) Able to classify all the substances correctly
Sample answer:
Substance that can
coagulate latex
Substance that cannot
coagulate latex
Nitric acid
Methanoic acid
Sodium hydroxide
Ammonia
|
3
Able to classify any three the substances correctly 2
Able to classify any two the substances correctly 1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
14 21Nov2017 | cikguadura.wordpress.com
Johor Julang – Set 2017
Measuring using numbers / Mengukur menggunakan nombor
1 Rubric Score
(a) [Able to measure the diameter of dents correctly and accurately with unit]
[Berupaya mengukur diameter lekuk dengan betul dan jitu dengan unit]
Sample answer / Contoh jawapan:
Brass : 1.20 cm, 1.00 cm, 1.20 cm
Copper : 1.35 cm, 1.60 cm. 1.50 cm
3
[Able to measure the diameter of dents without 2 decimal places and unit]
[Berupaya mengukur diameter lekuk tanpa dua titik perpuluhan atau unit]
2
[Able to state four diameters of dents without 2 decimal places and unit]
[Berupaya menyatakan empat diameter lekuk dengan betul]
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Communicate / Berkomunikasi
1 Rubric Score
(b) [Able to construct the table to record the diameters and average diameter
of dents on brass and copper block with correct:
1. Titles and units
2. Reading]
[Berupaya membina jadual untuk menrekodkan diameter dan diameter
purata blok loyang dan kuprum dengan betul:
1. Tajuk dan unit
2. Bacaan]
Sample answer / Contoh jawapan:
Type of block
Jenis blok
Diameter of dents (cm)
Diameter lekuk (cm)
Average diameter of dents
(cm)
Purata diameter lekuk (cm)I II III
Brass
Loyang
1.20 1.00 1.20 1.13
Copper
Kuprum
1.35 1.50 1.60 1.48
|
3
[Able to construct a less accurate the table that contain the following :
1. Titles without unit
2. Reading]
[Berupaya membina jadual kurang tepat yang mengandungi yang berikut :
1. Tajuk tanpa unit
2. Bacaan]
2
[Able to construct the idea of tabulation of data]
[berupaya membina idea penjadualan data]
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
15 21Nov2017 | cikguadura.wordpress.com
Observe / Memerhati
1 Rubric Score
(c)(i) [Able to state the observation correctly - Must have comparison]
[Berupaya menyatakan pemerhatiaan dengan betul – dalam bentuk
perbandingan]
Sample answer / Contoh jawapan:
The average diameter of dents on brass block is 1.13 cm and the
average diameter of dents on copper block is 1.48 cm //
The size / diameter of dents on brass block is smaller than the size /
diameter of dents on copper block //
Diameter purata lekuk pada blok loyang adalah 1.13 cm dan diameter
purata lekuk pada blok kuprum adalah 1.48 cm //
Saiz / diameter lekuk pada blok loyang lebih kecil daripada saiz /
diameter pada blok kuprum //
3
[Able to state the incomplete observation]
[Berupaya menyatakan pemerhatian yang tidak lengkap]
Sample answer / Contoh jawapan:
The size / diameter of dents on copper block is bigger //
The size / diameter of dents on brass block is smaller
Saiz / diameter lekuk pada blok kuprum lebih besar //
Saiz / diameter lekuk pada blok loyang lebih kecil
2
[Able to state the idea of observation]
[Berupaya menyatakan idea bagi pemerhatian]
Sample answer / Contoh jawapan:
The size / diameter of dents on copper block is big //
The size / diameter of dents on brass block is small
Saiz / diameter lekuk pada blok kuprum besar //
Saiz / diameter lekuk pada blok loyang kecil
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Making inference / membuat inferens
1 Rubric Score
(c)(ii) [Able to state the inference correctly]
[Berupaya menyatakan inferens dengan betul]
Sample answer / Contoh jawapan:
Brass is harder than copper // Copper is less harder than brass
Loyang lebih keras daripada kuprum // Kuprum kurang keras daripada
loyang
3
[Able to state the incomplete inference ]
[Berupaya menyatakan inferens kurang lengkap]
Sample answer / Contoh jawapan:
Brass is harder // Copper is less harder.
Loyang lebih keras // Kuprum kurang keras.
2
[Able to state the idea of inference ]
[Berupaya menyatakan idea bagi inferens]
Sample answer / Contoh jawapan:
Brass is hard // Copper is soft.
Loyang keras // Kuprum lembut.
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
16 21Nov2017 | cikguadura.wordpress.com
Interpret data / Mentafsir data
1 Rubric Score
(c)(iii) [Able to explain the arrangement of particles in the materials correctly]
[Berupaya menerangkan susunan zarah dalam bahan dengan betul]
Sample answer / Contoh jawapan:
Brass / Loyang Copper / Kuprum
1. Atomic size of zinc / foreign
atoms and copper are
different.
Saiz atom zink / atom asing
dan kuprum adalah berbeza
1. Atomic size of copper atoms
are same.
Saiz atom-atom kuprum
adalah sama
2. The presence of zinc / foreign
atoms in brass disrupts the
orderly arrangement of
copper atoms.
Kehadiran atom zink / atom
asing mengganggu susunan
kemas dan rapi atom-atom
kuprum.
2. Atoms are in closely pack in
orderly manner in layers
Atom-atom disusun dengan
kemas dan rapi dalam
lapisan
3. Layers of atom more difficult
to slide each another when
force is applied.
Lapisan-lapisan atom lebih
sukar menggelongsor antara
satu sama lain apabila daya
dikenakan
3. Layer of atoms easily sliding
when force is applied
Lapisan-lapisan atom mudah
menggelongsor apabila daya
dikenakan
|
3
[Able to state at least two point completely or three points without the
name of atoms]
[Berupaya menyatakan sekurang-kurangnya dua isi lengkap atau 3 isi
tanpa nama atom]
2
[Able to state at least one point completely]
[Berupaya menyatakan sekurang-kurangnya satu isi lengkap]
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
17 21Nov2017 | cikguadura.wordpress.com
Manipulate variables / memanipulasi pembolehubah
1 Rubric Score
(d) [Able to state all the variable and action to be taken correctly]
[Berupaya menyatakan semua pembolehubah dan tindakan yang perlu
diambil dengan betul]
Sample answer / Contoh jawapan:
Variables
Pembolehubah
Action to be taken
Tindakan yang diambil
Manipulated variable :
Pembolehubah dimanipulasi :
Type of block //
Brass block and copper block
Jenis blok //
Blok loyang dan blok kuprum
The way to manipulate variable :
Cara memanipulasikan pembolehubah :
Replace brass block with copper block
Menggantikan blok loyang dengan blok
kuprum
Responding variable :
Pembolehubah bergerakbalas:
Diameter of dents //
Hardness
Diameter lekuk //
Kekerasan
What to observe in the responding
variable:
Apa yang diperhatikan pada
pembolehubah bergerakbalas:
The measurement of diameter of dents
Pengukuran diameter lekuk
Controlled variable :
Pembolehubah ditetapkan :
Height of weight //
Mass of weight //
Size of stainless steel ball
bearing
ketinggian pemberat //
Jisim pemberat //
Saiz bebola keluli nirkarat
How to maintain the controlled
variable :
Cara untuk mengekalkan
pembolehubah ditetapkan:
Using the same height of weight//
Using the same mass of weight //
Using the same size of stainless steel
ball bearing
Gunakan tinggi pemberat yang sama
//Gunakan jisim pemberat yang sama
//Gunakan saiz bebola keluli nirkarat
yang sama
|
6
[Able to state all three variables and at least two corresponding action to
be taken correctly]
[Berupaya menyatakan semua tiga pembolehubah dan sekurang-
kurangnya dua tindakan yang perlu diambil pada pembolehubah yang
berkaitan dengan betul]
5
[Able to state at least two variables and at least two corresponding action
to be taken correctly]
[Berupaya menyatakan sekurang-kurangnya dua pembolehubah dan
sekurang-kurangnya dua tindakan yang perlu diambil pada pembolehubah
yang berkaitan dengan betul]
4
[Able to state at least two variables and at least one corresponding action
to be taken correctly]
[Berupaya menyatakan sekurang-kurangnya dua pembolehubah dan
sekurang-nurangnya satu tindakan yang perlu diambil pada pembolehubah
yang berkaitan dengan betul]
3
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
18 21Nov2017 | cikguadura.wordpress.com
[Able to state at least one variables and at least one corresponding action
to be taken correctly]
[Berupaya menyatakan sekurang-kurangnya satu pembolehubah dan
sekurang-kurangnya satu tindakan yang perlu diambil pada pembolehubah
yang berkaitan dengan betul]
2
[Able to state at least one variables correctly]
[Berupaya menyatakan sekurang-kurangnya satu pembolehubah dengan
betul]
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Making hypothesis / membuat hipotesis
1 Rubric Score
(e) [Able to state the hypothesis correctly]
[Berupaya menyatakan hipotesis dengan betul]
Sample answer / Contoh jawapan:
Brass block has a smaller size / diameter of dents than copper block
Brass is harder than copper //
Blok loyang mempunyai saiz / diameter lekuk yang lebih kecil daripada
blok kuprum.
Loyang lebih keras daripada kuprum
3
[Able to state the hypothesis less correctly]
[Berupaya menyatakan hipotesis kurang tepat]
Sample answer / Contoh jawapan:
Brass is harder //
Copper is less harder
Loyang lebih keras
Kuprum kurang keras
2
[Able to state the idea of hypothesis]
[Berupaya menyatakan idea hipotesis]
Sample answer / Contoh jawapan:
Brass is hard //
Copper is soft
Loyang keras //
Kuprum lembut
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
19 21Nov2017 | cikguadura.wordpress.com
Operational definition / Mendefinisi secara operasi
1 Rubric Score
(f) [Able to state the operational definition correctly]
[Berupaya menyatakan mendefinisi secara operasi dengan betul]
What should be done: Drop 1.0 kg weight from 0.5 meter height on the
material block.
What should be observed: Dents formed on the material block.
Apa yang perlu dibuat : Jatuhkan pemberat 1.0 kg dari ketinggian 0.5 meter
di atas blok bahan.
Apa yang diperhatikan : Lekuk terbentuk di atas blok bahan
Sample answer / Contoh jawapan:
Smaller dents formed when 1.0 kg weight dropped from 0.5 meter height
on the block.
Lekuk yang lebih kecil terbentuk apabila pemberat 1.0 kg dijatuhkan dari
ketinggian 0.5 meter di atas blok bahan.
3
[Able to state the operational definition correctly]
[Berupaya menyatakan mendefinisi secara operasi dengan betul]
What should be done : Drop weight on the material block.
What should be observed : Dents formed.
Apa yang perlu dibuat : Jatuhkan pemberat di atas blok bahan.
Apa yang diperhatikan : Lekuk terbentuk
Sample answer / Contoh jawapan:
A small dents formed when weight dropped on the block.
Lekuk yang kecil terbentuk apabila pemberat dijatuhkan di atas blok
bahan.
2
[Able to state the idea of operational definition]
[Berupaya menyatakan idea bagi mendefinisi secara operasi]
Sample answer / Contoh jawapan:
The harder the material produce a smaller dent
Bahan yang lebih keras menghasilkan lekuk yang lebih kecil
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
20 21Nov2017 | cikguadura.wordpress.com
Predict / meramal
1 Rubric Score
(g) [Able to predict the diameter of dent correctly]
[Berupaya meramalkan diameter lekuk dengan betul]
Sample answer / Contoh jawapan:
Diameter / size of dents on copper more than 1.60 cm
Diameter /saiz lekuk pada blok kuprum lebih besar daripada 1.60 cm
3
[Able to predict less correctly]
[Berupaya meramal dengaa kurang tepat]
Sample answer / Contoh jawapan:
Diameter become larger / bigger / increase
Diameter semakin besar / meningkat
2
[Able to state the idea of prediction]
[Berupaya menyatakan idea meramal]
Sample answer / Contoh jawapan:
Change
Berubah
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Clasify/mengelas
1 Rubric Score
(h) [Able to classify all materials correctly]
[Berupaya mengelaskan semua bahan dengan betul]
Sample answer / Contoh jawapan:
Alloy / Aloi Pure metal / Logam tulen
Stainless steel
Keluli nirkarat
Bronze
Gangsa
Duralumin
Duralumin
Iron
Besi
Aluminium
Aluminium
Chromium
Kromium
# if reverse, score 1
Jika terbalik, skor 1
3
[Able to classify at least four materials correctly]
[Berupaya mengelaskan sekurang-kurangnya empat bahan dengan tepat]
2
[Able to classify at least two materials correctly]
[Berupaya mengelaskan sekurang-kurangnya dua bahan dengan tepat]
1
[No response or wrong response]
[Tiada jawapan atau jawapan salah]
0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
21 21Nov2017 | cikguadura.wordpress.com
Esei
Kelantan-2017
(a) [Dapat menyatakan pernyataan masalah dengan betul.]
Sampel jawapan:
Adakah kepekatan elektrolit mempengaruhi hasil elektrolisis di anod?
3
[Dapat menyatakan pernyataan masalah dengan kurang tepat]
Sampel jawapan:
Adakah kepekatan elektrolit mempengaruhi hasil elektrolisis?
2
[Dapat memberi idea bagi pernyataan masalah.] Sampel jawapan:
Kepekatan elektrolit mempengaruhi hasil elektrolisis
1
0
Rubriks
(b) [dapat menyatakan hubungan di antara pemboleh ubah dimanipulasi dan
pemboleh ubah bergerak balas dan mempunyai arah]
Sampel jawapan:
Jika larutan natrium klorida pekat digunakan hasil di anod ialah gas
klorin. Jika larutan natrium klorida cair digunakan hasil di anod ialah
gas oksigen.
3
[dapat menyatakan hubungan di antara pemboleh ubah dimanipulasi dan
pemboleh ubah bergerak balas]
Sampel jawapan:
Jika larutan natrium klorida pekat digunakan hasil di anod ialah gas
klorin // Jika larutan natrium klorida cair digunakan hasil di anod
ialah gas oksigen.
2
[Dapat menyatakan idea bagi hipotesis]
Sampel jawapan:
Kepekatan elektrolit mempengaruhi hasil di anod
1
Tiada respons atau respons salah 0
(c) [Dapat menyatakan dua pemboleh ubah dengan betul]
Sampel jawapan
Pemboleh ubah dimanipulasi: Kepekatan elektrolit
Pemboleh ubah bergerak balas: Hasil di anod
Pemboleh ubah dimalarkan : Jenis elektrod
3
[Dapat menyatakan dua pemboleh ubah dengan betul] 2
[dapat menyatakan satu pemboleh ubah dengan betul] 1
[tiada respons atau respons salah] 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
22 21Nov2017 | cikguadura.wordpress.com
Rubriks
(d) [Dapat menyenaraikan bahan dan radas dengan lengkap]
Sampel jawapan:
Bahan: Larutan natrium klorida 0.001moldm-3 , larutan natrium klorida
1.0 moldm-3
Radas
Elektrod karbon, sel elektrolisis, elektrod karbon, wayar penyambung
dan klip buaya, sel kering, tabung uji
3
[Dapat menyenaraikan bahan dan radas dengan kurang lengkap]
Sampel jawapan:
Bahan: Larutan natrium klorida
Radas
Elektrod karbon, sel elektrolisis, elektrod karbon, wayar penyambung
dan klip buaya, sel kering, tabung uji kecil
2
[Dapat menyenaraikan bahan dan radas dengan tidak lengkap]
Sampel jawapan:
Bahan: Larutan natrium klorida
Radas
elektrod karbon, bikar, wayar penyambung dan klip buaya, sel kering,
1
[Tiada respons atau respons salah] 0
Rubriks
(e) [dapat menyenaraikan semua langkah dengan betul]
Prosedur/kaedah
1. Masukkan larutan natrium klorida 0.001 moldm-3 ke dalam sel
elektrolisis sehingga separuh penuh.
2. Masukkan larutan natrium klorida 0.001 moldm-3 ke dalam 2
tabung uji kecil sehingga penuh.
3. Telangkupkan tabung uji yang berisi larutan kepada keduadua
elektrod
4. Sambungkan kedua-dua elektrod kepada sel kering dengan
wayar penyambung dan klip buaya.
5. Pemerhatian direkodkan.
6. Langkah 1 hingga 4 diulang dengan menggunakan larutan
natrium klorida 1.0 moldm-3 .
3
[dapat menyenaraikan langkah-langkah 1, 4, 5 dan 6 ] 2
[dapat menyenaraikan langkah-langkah 1, 4 dan 5] 1
[tiada respons atau respons salah] 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
23 21Nov2017 | cikguadura.wordpress.com
2(f) [dapat membina jadual dengan aspek berikut]
Kepekatan elektrolit
(moldm-3)
Pemerhatian di anod
0.001
1.0
|
2
[dapat membina jadual dengan tidak lengkap]
Kepekatan elektrolit Pemerhatian di anod
0.001
1.0
|
1
[tiada respons atau respons salah atau jadual kosong] 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
24 21Nov2017 | cikguadura.wordpress.com
Kedah 2017
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
25 21Nov2017 | cikguadura.wordpress.com
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
26 21Nov2017 | cikguadura.wordpress.com
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
27 21Nov2017 | cikguadura.wordpress.com
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
28 21Nov2017 | cikguadura.wordpress.com
JUJ-2017
Rubric
(a)
Able to give the problem statement correctly Sample answer:
How does the effectiveness of cleansing action of two different types of
cleansing agent in hard water?
// What is the difference in effectiveness of cleansing action in hard water
using soap and detergent?
3
Able to give the problem statement less correctly
Sample answer:
How does different type of cleansing agent affect the effectiveness of
cleansing action in hard water?
// To investigate the effectiveness of different type of cleansing agent in
hard water.
2
Able to give an idea of statement of the problem Sample answer:
To study the effectiveness of cleansing action in hard water 1
No response or wrong response 0
Rubric
Able to state all variables correctly
Sample answer:
(b) Manipulated variable :Types of cleansing agent//soap and detergent
Responding variable : Effectiveness of cleansing agent//ability
to remove the oily stains on cloth Constant variable :
Hard water//Cloth with oily stains//Sizeof cloth
3
Able to state any two variables correctly 2
Able to state any one variable correctly 1
No response or wrong response 0
Rubric
Able to state the relationship between the manipulated variable and the
responding variable correctly
Sample answer:
Detergent is more effective than soap in hard water.
// Detergent in hard water can remove the oily stains on cloth while soap
cannot.
3
Able to state the relationship between the manipulated variable and the
responding variable incorrectly
(c)
Sample answer:
Detergent is more effective // Soap is less effective
// Detergent in hard water can remove the oily stains on cloth
//soap cannot remove stain on cloth.
2
Able to state an idea of hypothesis 1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
29 21Nov2017 | cikguadura.wordpress.com
Rubric
Able to give the completely list of the apparatus and materials Answer:
Apparatus:
Beakers, measuring cylinder, glass rod
Material:
Soap, detergent, [cleansing agent X dan cleansing agent Y], 0.5 - 1 mol dm- magnesium
sulphate solution/[sea water/hard water], 2 pieces of cloth with oily stains.
3
(d) Able to give the list of the apparatus and materials correctly but not completely
Answer:
Apparatus: Any containers, glass rod
Material: Cleansing agent, cloth with oily stains.
2
Able to give two materials and at least one apparatus
Answer :
Material:
Soap solution / detergent, hard water [any container]
1
No response or wrong response 0
Rubric e
Able to state all procedures correctly
Sample answer :
(e)
1. Label 2 beakers as A and B respectively
2. Fill all the beakers with [50 - 100 cm3
]of hard water / magnesium sulphate solution
3. Add cleansing agent X/detergent in beaker A and cleansing agent Y/soap in bikar B.
4. Place a piece of cloth with oily stain into each beaker
5. Stir the solution in each beaker using glass rod until no change can be observed
6. Wash the oily stain on each piece of cloth using the solution in each beaker
7. Record the observation.
3
Able to state three steps of procedures correctly Steps 2, 3, 5, 6, 2
Able to state two steps of procedures correctly Steps 2, 3, 6 1
No response or wrong response 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
30 21Nov2017 | cikguadura.wordpress.com
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
31 21Nov2017 | cikguadura.wordpress.com
Johor 2015
Explanation
(a) [Able to state the problem statement correctly]:
Suggested answer:
Which metals are harder, alloys or pure copper?
3
[Able to state the problem statement less correctly]
Suggested answer:
Are alloys are harder than pure metals?
2
[Able to state the problem statement less correctly]
Suggested answer:
Which metals are harder?
1
[No response given or wrong response] 0
Explanation
(b) [Able to state thevariables correctly]:
Suggested answer:
(a) Manipulated variable : Type of metals block
(b) Responding variable : Hardness of metals
(c) Fixed variable : Size of steel ball bearing /Height of weight
3
[Able to state thevariables less correctly]
Suggested answer:
One mistake
2
[Able to state i thevariables less correctly]
Suggested answer:
Two mistakes
1
[No response given or wrong response] 0
Explanation
(c) [Able to state thehypothesis correctly]:
Suggested answer:
Alloy is harder than pure metals / Pure metals are harder than alloy
3
[Able to state thehypothesis less correctly]
Suggested answer:
Alloy is harder
2
[Able to statethehypothesis less correctly]
Suggested answer:
Alloy produce smaller dent
1
[No response given or wrong response] 0
Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura
32 21Nov2017 | cikguadura.wordpress.com
Explanation
(d) [Able to list the apparatus and materials correctly]:
Suggested answer:
Apparatus : 1 kg of weight, thread, cellophane tape, retort stand, 1 meter of
ruler, ruler
Materials : steel ball bearing, copper block, bronze block
3
[Able to list the apparatus and materials less correctly]
Suggested answer: One mistake
2
[Able to list the apparatus and materials less correctly]
Suggested answer: Two mistakes
1
[No response given or wrong response]
Suggested answer: More than three mistakes
0
Explanation
(e) [Able to state theprocedure correctly]:
Suggested answer:
1. Put a steel ball to an alloy block using cellophane tape.
2. Hang 1 kg weight at a height of 50 cm from the top of the alloy block.
3. Put a bronze block under the weight.
4. Release the weight so that it falls on the steel ball bearing.
5. Measure the diameter of the dent formed on top of the alloy block using
meterruler.
6. Record the diameter reading in the table.
7. Repeat the steps 1 to 5 using pure metal block.
3
[Able to state theprocedure less correctly]
Suggested answer: One mistake in procedure
2
[Able to statetheprocedure less correctly]
Suggested answer: Two mistakes in procedure
1
[No response given or wrong response]
Suggested answer: More than three mistakes in procedure
0
Explanation
(f) [Able to draw the table correctly]:
Suggested answer:
Type of block Diameter of dent / cm
Bronze
Pure copper
|
2
[Able to draw the table less correctly]
Suggested answer:
One mistake in table such as
1. Table not properly drawn
2. No unit
3. No type of block mention
1
[No response given or wrong response] 0

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Kertas 3-kimia-spm-2017-cg-adura-jawapan

  • 1. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 1 21Nov2017 | cikguadura.wordpress.com Cadangan jawapan Kertas 3 – KIMIA SPM 2017 – Cg Adura [SPM04-01-P3] (a) Cadangan jawapan Skor Observation Inferences Dapat menyatakan dua pemerhatian dengan tepat: Contoh jawapan lengkap: 1. Wasap putih terbebas / white fume released 2. Jisim kandungan mangkuk pijar bertambah/ mass increases 3. Pepejal/ serbuk putih terbentuk / white solid produce/ formed 4. Nyalaan putih/ terang/ berkilau // burn brightly Dapat menyatakan dua inferens yang lengkap sepadan dengan pemerhatian: Contoh inferens lengkap: Magnesium oksida terbentuk/ MgO formed Magnesium berpadu / bertindak balas dengan oksigen / Magnesium react with oxygen Magnesium teroksida/ dioksidakan Magnesium oxidized Dapat menyatakan satu pemerhatian yg lengkap atau dua pemerhatian yg kurang lengkap: Contoh jawapan kurang lengkap: 1. Wasap terbebas // asap putih terbentuk 2. Jisim magnesium bertambah 3. Pepejal/ serbuk terbentuk 4. Magnesium menyala Dapat menyatakan satu inferens yang lengkap atau dua inferens yang kurang lengkap: Contoh inferens kurang lengkap: Sebatian magnesium terbentuk Magnesium berpadu / bertindak balas dengan udara Magnesium teroksida/ dioksidakan Dapat menyatakan satu idea tentang pemerhatian: Contoh jawapan: 1. Gas dibebaskan // gas putih 2. Terdapat pertambahan berat 3. Berat magnesium bertambah 4. Nyalaan/ baraan / cahaya Jisim mangkuk pijar bertambah Dapat menyatakan satu idea inferens: Contoh jawapan: Tindak balas berlaku | 3 b Cadangan jawapan Skor Dapat mencatatkan ketiga-tiga jisim dengan 2 t. perpuluhan dengan betul: Contoh jawapan: Jisim mangkuk pijar dan penutup: 25.35 g Jisim mangkuk pijar, penutup dan pita magnesium: 27.75 g Jisim mangkuk pijar, penutup dan magnesium oksida setelah disejukkan: 29.35 g 3 Dapat mencatatkan mana-mana dua jisim dengan 2 t. perpuluhan dengan betul 2 Dapat mencatatkan mana-mana satu jisim dengan 2 t. perpuluhan dengan betul 1
  • 2. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 2 21Nov2017 | cikguadura.wordpress.com (i) dan (ii) Cadangan jawapan Skor Dapat menentukan jisim magnesium, jisim oksigen dan menunjukkan 3 langkah menentukan formula empirik magnesium oksida dengan betul: Jisim magnesium: 27.75 -25.35 = 2.40 / 2.4 g Jisim oksigen: 29.35 – 27.75 = 1.60 / 1.6 g Langkah menentukan formula empirik magnesium oksida: L1: Mol magnesium: 24 4.2 = 0.1 Mol oksigen: 16 6.1 = 0.1 L2: Nisbah Mg : O = 1:1 L3: Formula empirik = MgO Dapat menyatakan jisim magnesium, jisim oksigen dan menunjukkan 2 langkah pertama atau menyatakan formula empirik magnesium oksida dengan betul Dapat menyatakan jisim magnesium atau jisim oksigen atau menyatakan formula empirik magnesium oksida dengan betul d Cadangan jawapan Skor Dapat menyatakan definasi secara operasi berdasarkan formula empirik dengan tepat: Contoh jawapan: 1 mol magnesium react with 1 mol oxygen Dapat menyatakan definasi secara operasi berdasarkan formula empirik dengan kurang tepat: Contoh jawapan: 1 mol magnesium bertindak balas dengan oksigen Dapat menyatakan definasi secara operasi mengikut konsep: Contoh jawapan: Mol magnesium dan oksigen adalah sama.
  • 3. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 3 21Nov2017 | cikguadura.wordpress.com Negeri Sembilan – 2016 Explanation score 1(a) [Able to fulfil the following criteria] 1. all readings correct 2. two decimal places 3. without unit Answer: Methanol = 1.55 Ethanol = 1.20 Propanol = 1.10 Butanol = 1.61 3 One of criteria is not fulfilled 2 [Able to fulfil criteria 1] 1 [No response given or wrong response] 0 Explanation score 1 (b) [Able to state all variables correctly] Sample answer: Manipulated variable: Types of alcohols//[Name of alcohols] Responding variable: Heat of combustion//Temperature increase/change Fixed variable: Aluminium can//Water (Reject: Volume of water) 3 [Able to state any 2 variables correctly] 2 [Able to state any 1 variable correctly] 1 [No response given or wrong response] 0 Explanation score 1 (c) [Able to state the relationship between the manipulated variable and the responding variable with direction correctly] Sample answer: The higher the number of carbon atoms per molecule of alcohol, the higher the heat of combustion. Note : RV  MV score 2 3 [Able to state the relationship between the manipulated variable and the responding variable] Sample answer: The higher the number of carbon atoms, the higher the heat of combustion. 2 [Able to state an idea of hypothesis] Sample answer: Different alcohols ,different heat of combustion. 1 [No response given or wrong response] 0
  • 4. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 4 21Nov2017 | cikguadura.wordpress.com Explanation score 1 (d) Able to draw a bar chart by fulfill the following criteria: (i) Axes are labelled with correct unit (ii)Scale used must cover at least 50% of graph paper (iii)Uniform spacing and width for each bar Sample answer:[Refer to page 8] 3 Able to draw a bar chart by fulfill the following criteria: (i) Axes are labelled with correct unit (ii)Uniform spacing and width for each bar 2 Able to draw a bar chart by fulfill the following criteria: (i) Axes are labelled with correct unit (ii) All bars are combined 1 [No response given or wrong response] 0 Explanation score 1(e) [Able to predict the heat of combustion of pentanol and draw the bar for pentanol correctly] Criteria : 1. Value with correct unit 2.The bar drawn as in 1(d) score 3 Answer: 2260 kJ mol-1 [Draw a correct bar chart] 3 [Able to predict the heat of combustion of pentanol and draw the bar for pentanol] Answer: 2260 + 20 [without unit] [Draw a correct bar chart] 2 [Able to predict the heat of combustion of pentanol or draw the bar for pentanol] Answer: 2260 + 20 [without unit] or [Draw a correct bar chart] 1 [No response given or wrong response] 0 Explanation score 1 (f) [Able to state the operational definition for heat of combustion correctly] Sample answer: The heat released/produced when 1 mole of alcohol is burnt in excess oxygen, thermometer reading increases. 3 [Able to state the operational definition for heat of combustion] Sample answer: Heat released/produced when alcohol is burnt in excess oxygen. 2 [Able to state an idea of operational definition for heat of combustion] Sample answer: Heat released when alcohol burns. 1 [No response given or wrong response] 0
  • 5. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 5 21Nov2017 | cikguadura.wordpress.com Melaka 2016 Mark Scheme Mark 1(a) Able to state all pH value correctly Answer Acid solution Colour of indicator pH value Acid P Red 1 Acid Q Yellowish orange 5 Acid R Yellow 6 | 3 Able to state any twopH value correctly 2 Able to state any one pH value correctly 1 No response or wrong response 0 1(b) Able to state all the variables correctly Sample answer Manipulated variable: Type of acid // Acid P, Q, R and S // Four different acids Responding variable: pH value // Colour of indicator // Strength of acid Fixed variable: Concentration of acid // Universal indicator // volume of acid 3 Able to state any two variables correctly 2 Able to state any one variable // idea of all variables 1 No response or wrong response 0 1(c) Able to state all correct inferences Acid Inference P P is a strong acid // P ionize completely in water // P produces high concentration of hydrogen ions. Q Q is a weak acid // Q ionize partially in water // Q produces low concentration of hydrogen ions. R R is a weak acid // R ionize partially in water // R produces low concentration of hydrogen ions. | 3 Able to state any two inferences correctly. 2 Able to state any one inferences // idea of all inferences 1 No response or wrong response 0
  • 6. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 6 21Nov2017 | cikguadura.wordpress.com 1(d) Able to state the relationship between pH value with the strength of acid correctly. Sample answer When pH value increases, the strength of acid decreases. 3 Able to state the relationship between pH value with the strength of acid less correctly. Sample answer pH value is inversely proportional to the strength of acid. 2 Able to give an idea Sample answer pH value is affected by strength of acid. 1 No response or wrong response 0 1(e) Able to state the operational definition for strong acid correctly. 1. Low pH value 2. Add universal indicator Sample answer Acid that shows low pH value when universal indicator is added into it. 3 Able to state the operational definition for strong acid less correctly. 1. Low pH value // 2. Add universal indicator Sample answer i. Acid that shows low pH value. // ii. Add universal indicator into the acid. 2 Able to state an idea of operational definition Sample answer: Colour of indicator change // pH less than 7 // Ionise completely 1 No response or wrong response 0 1(f) Able to record all the burette readings accurately with two decimal place and with correct unit. 3
  • 7. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 7 21Nov2017 | cikguadura.wordpress.com Answers Set Initial burette reading Final burette reading I 0.30 cm3 24.30 cm3 II 24.30 cm3 47.80 cm3 III 0.50 cm3 25.00 cm3 | Able to record all burette readings less correctly with or without unit Sample answers Set Initial burette reading Final burette reading I 0.30 24.30 II 24.30 47.80 III 0.50 25.00 or Set Initial burette reading Final burette reading I 0.3 cm3 24.3 cm3 II 24.3 cm3 47.8 cm3 III 0.5 cm3 25 cm3 | 2 Able to record at least four burette readings less correctly Sample answers: Set Initial burette reading Final burette reading I 1.7 24.3 II 24.3 48.2 III 0.5 25 | 1 No response or wrong response 0 1(g) Able to construct a table to record the initial burette readings, final burette readings and volume of acid for Set I, Set II and Set III that contain: 3
  • 8. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 8 21Nov2017 | cikguadura.wordpress.com 1. Correct titles 2. Readings and units Sample answer: Set Initial burette reading (cm3) Final burette reading (cm3) Volume of acid (cm3) I 0.30 24.30 24.00 II 24.30 47.80 23.50 III 0.50 25.00 24.50 | Able to construct a less accurate table that contains the following: 1. Titles 2. Readings Sample answer: Set Initial burette reading Final burette reading Volume of acid I 0.3 24.3 24.0 II 24.3 47.8 23.5 III 0.5 25.0 24.5 | 2 Able to construct a table that contains titles and readings for initial burette readings and final burette readings only. Sample answer: Set Initial burette reading Final burette reading I 0.3 24.3 II 24.3 47.8 III 0.5 25.0 | 1 No response or wrong response 0 1(h) Able to predict the volume of acid S correctly with unit. Answer: 50.0 cm3 3 Able to predict the volume of acid S Answer: 50 2 Able to give an idea to predict the volume of acid S Sample answer: more than 25 // more than acid P 1 No response or wrong response 0
  • 9. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 9 21Nov2017 | cikguadura.wordpress.com (i)(i) Able to state the correct observation for both set. Sample answers Set I : Bubbles released // Calcium carbonate dissolve in ethanoic acid solution. Set II : No bubbles // Calcium carbonate does not dissolve in glacial ethanoic acid. 3 Able to state correct observation for any one set. Sample answers: Set I : Bubbles released // Calcium carbonate dissolve in ethanoic acid solution. or Set II : No bubbles // Calcium carbonate does not dissolve in glacial ethanoic acid. 2 Able to give an idea of the observation Sample answers: Calcium carbonate reacts with acid // Carbon dioxide gas 1 No response or wrong response 0 (i)(ii) Able to state the relationship between the manipulated variable and the responding variable and state the direction correctly. Sample answer: Presence of water causes acid to show its properties. 3 Able to state the relationship between the manipulated variable and the responding variable less correctly. Sample answer: Acid will shows its acidic properties when water presence. 2 Able to state the idea of hypothesis Sample answer: Reaction occurs when water present. 1 No response or wrong response 0 (j) Able to classify all the acids correctly Sample answer: Strong acid Weak acid Nitric acid Sulphuric acid Citric acid Carbonic acid | 3 Able to classify any three acids correctly 2 Able to classify any two acids correctly 1 No response or wrong response 0
  • 10. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 10 21Nov2017 | cikguadura.wordpress.com Perlis 2015 Rubric Score 1(a) Able to state an observation accurately Sample answer: Rubber strip Y is extended longer than rubber strip X 3 Able to state the observation correctly Sample answer: Rubber strip Y is extended/become longer// Rubber strip X do not extended 2 Able to state idea of the observation Sample answer: Rubber strip extended 1 No response or wrong response 0 Rubric Score 1(b) Able to state an accurate inference for this experiment: Sample answer: Rubber strip X is harder than rubber strip Y 3 Able to state the inference for this experiment: Sample answer: Rubber strip X is hard// Rubber strip Y is soft 2 Able to state the general inference for this experiment: Sample answer: X is more elastic rubber 1 No response or wrong response 0 Rubric Score 1(c) Able to write all the reading of rubber strip X and Y correctly with one decimal place. Answer: Rubber strip X: 5.0, 5.0, 5.0, 6.0 Rubber strip Y: 5.0, 5.5, 6.4, 8.5 3 Able to write all the reading of rubber strip X and Y correctly 2 Able to write 3 correct reading for rubber strip X and 3 correct reading for rubber strip Y 1 No response or wrong response 0
  • 11. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 11 21Nov2017 | cikguadura.wordpress.com Rubric Score 1(d) Able to construct a table with correct title and units and accurately Type of rubber Rubber strip X Rubber strip Y Weight (g) 0 15 30 45 0 15 30 45 Length (cm) 5.0 5.0 5.0 6.0 5.0 5.5 6.4 8.5 3 Able to construct a table without title or units and less accurately | Type of rubber Rubber strip X Rubber strip Y Weight 15 30 45 15 30 45 Length 5.0 5.0 6.0 5.5 6.4 8.5 2 Able to give the idea about to construct table Type of rubber Rubber strip X Rubber strip Y Length 5.0 5.0 6.0 5.5 6.4 8.5 1 No response or wrong response 0 Rubric Score 1(e) Able to state three variables correctly: Sample answer: (i) Manipulated variable: Rubber strip X, Rubber strip Y (ii) Responding variable: The length of rubber strip after weight is removed (iii) Fixed variable: Size of rubber strip, mass of weight 3 Able to state any 2 of the above information correctly 2 Able to state any 1 of the above information correctly 1 No response or wrong response 0 Rubric Score 1(f) Able to state the relationship between the manipulated variable and the responding variable correctly and with direction Sample answer: Rubber strip X is more elastic than rubber strip Y 3 Able to state the relationship between the manipulated variable and the responding variable correctly and without direction Sample answer: The elasticity of rubber strip X is high than rubber strip Y 2 Able to state an idea of the hypothesis Sample answer: Rubber strip x is less elastic. 1 No response or wrong response 0
  • 12. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 12 21Nov2017 | cikguadura.wordpress.com Rubric Score 1(g) Able to give an accurately relationship between length of rubber strip and elasticity. Sample answer: The length of rubber strip increases the elasticity decreases 3 Able to give less accurately relationship between length of rubber strip and elasticity. Sample answer: The length increases the elasticity decreases 2 Able to give and idea of relationship between length of rubber strip and elasticity Sample answer: directly proportional 1 No response or wrong response 0 Rubric Score 1(h) Able to state the operational definition correctly Sample answer: The longer the length of rubber strip after weight is removed the less elastic the rubber strip. 3 Able to state the operational definition less correctly Sample answer: After weight is removed the rubber strip become longer 2 Able give an idea for operational definition Sample answer: The length of rubber strip increase 1 No response or wrong response 0 Rubric Score 1(i) Able to explain the observation by stating all the following aspect correctly Answer: Rubber strip Y has more double bond between carbon and carbon atom// Rubber strip X has less double bond between carbon and carbon atom// Rubber strip Y easily undergoes oxidation process compared to rubber strip X// 3 Able to explain the observation by stating any two of the aspect correctly 2 Able to explain the observation by stating any one of the aspect correctly 1 No response or wrong response 0
  • 13. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 13 21Nov2017 | cikguadura.wordpress.com Rubric Score 1(j) Able to predict the rubber that will snap first and state the type of rubber strip X and Y correctly Answer: Rubber Y will snap first Rubber strip X : Vulcanized rubber Rubber strip y : Unvulcanized rubber 3 Able state any two of the answer correctly 2 Able state any one of the answer correctly 1 No response or wrong response 0 Rubric Score 1(k) Able to classify all the substances correctly Sample answer: Substance that can coagulate latex Substance that cannot coagulate latex Nitric acid Methanoic acid Sodium hydroxide Ammonia | 3 Able to classify any three the substances correctly 2 Able to classify any two the substances correctly 1 No response or wrong response 0
  • 14. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 14 21Nov2017 | cikguadura.wordpress.com Johor Julang – Set 2017 Measuring using numbers / Mengukur menggunakan nombor 1 Rubric Score (a) [Able to measure the diameter of dents correctly and accurately with unit] [Berupaya mengukur diameter lekuk dengan betul dan jitu dengan unit] Sample answer / Contoh jawapan: Brass : 1.20 cm, 1.00 cm, 1.20 cm Copper : 1.35 cm, 1.60 cm. 1.50 cm 3 [Able to measure the diameter of dents without 2 decimal places and unit] [Berupaya mengukur diameter lekuk tanpa dua titik perpuluhan atau unit] 2 [Able to state four diameters of dents without 2 decimal places and unit] [Berupaya menyatakan empat diameter lekuk dengan betul] 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 Communicate / Berkomunikasi 1 Rubric Score (b) [Able to construct the table to record the diameters and average diameter of dents on brass and copper block with correct: 1. Titles and units 2. Reading] [Berupaya membina jadual untuk menrekodkan diameter dan diameter purata blok loyang dan kuprum dengan betul: 1. Tajuk dan unit 2. Bacaan] Sample answer / Contoh jawapan: Type of block Jenis blok Diameter of dents (cm) Diameter lekuk (cm) Average diameter of dents (cm) Purata diameter lekuk (cm)I II III Brass Loyang 1.20 1.00 1.20 1.13 Copper Kuprum 1.35 1.50 1.60 1.48 | 3 [Able to construct a less accurate the table that contain the following : 1. Titles without unit 2. Reading] [Berupaya membina jadual kurang tepat yang mengandungi yang berikut : 1. Tajuk tanpa unit 2. Bacaan] 2 [Able to construct the idea of tabulation of data] [berupaya membina idea penjadualan data] 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0
  • 15. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 15 21Nov2017 | cikguadura.wordpress.com Observe / Memerhati 1 Rubric Score (c)(i) [Able to state the observation correctly - Must have comparison] [Berupaya menyatakan pemerhatiaan dengan betul – dalam bentuk perbandingan] Sample answer / Contoh jawapan: The average diameter of dents on brass block is 1.13 cm and the average diameter of dents on copper block is 1.48 cm // The size / diameter of dents on brass block is smaller than the size / diameter of dents on copper block // Diameter purata lekuk pada blok loyang adalah 1.13 cm dan diameter purata lekuk pada blok kuprum adalah 1.48 cm // Saiz / diameter lekuk pada blok loyang lebih kecil daripada saiz / diameter pada blok kuprum // 3 [Able to state the incomplete observation] [Berupaya menyatakan pemerhatian yang tidak lengkap] Sample answer / Contoh jawapan: The size / diameter of dents on copper block is bigger // The size / diameter of dents on brass block is smaller Saiz / diameter lekuk pada blok kuprum lebih besar // Saiz / diameter lekuk pada blok loyang lebih kecil 2 [Able to state the idea of observation] [Berupaya menyatakan idea bagi pemerhatian] Sample answer / Contoh jawapan: The size / diameter of dents on copper block is big // The size / diameter of dents on brass block is small Saiz / diameter lekuk pada blok kuprum besar // Saiz / diameter lekuk pada blok loyang kecil 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 Making inference / membuat inferens 1 Rubric Score (c)(ii) [Able to state the inference correctly] [Berupaya menyatakan inferens dengan betul] Sample answer / Contoh jawapan: Brass is harder than copper // Copper is less harder than brass Loyang lebih keras daripada kuprum // Kuprum kurang keras daripada loyang 3 [Able to state the incomplete inference ] [Berupaya menyatakan inferens kurang lengkap] Sample answer / Contoh jawapan: Brass is harder // Copper is less harder. Loyang lebih keras // Kuprum kurang keras. 2 [Able to state the idea of inference ] [Berupaya menyatakan idea bagi inferens] Sample answer / Contoh jawapan: Brass is hard // Copper is soft. Loyang keras // Kuprum lembut. 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0
  • 16. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 16 21Nov2017 | cikguadura.wordpress.com Interpret data / Mentafsir data 1 Rubric Score (c)(iii) [Able to explain the arrangement of particles in the materials correctly] [Berupaya menerangkan susunan zarah dalam bahan dengan betul] Sample answer / Contoh jawapan: Brass / Loyang Copper / Kuprum 1. Atomic size of zinc / foreign atoms and copper are different. Saiz atom zink / atom asing dan kuprum adalah berbeza 1. Atomic size of copper atoms are same. Saiz atom-atom kuprum adalah sama 2. The presence of zinc / foreign atoms in brass disrupts the orderly arrangement of copper atoms. Kehadiran atom zink / atom asing mengganggu susunan kemas dan rapi atom-atom kuprum. 2. Atoms are in closely pack in orderly manner in layers Atom-atom disusun dengan kemas dan rapi dalam lapisan 3. Layers of atom more difficult to slide each another when force is applied. Lapisan-lapisan atom lebih sukar menggelongsor antara satu sama lain apabila daya dikenakan 3. Layer of atoms easily sliding when force is applied Lapisan-lapisan atom mudah menggelongsor apabila daya dikenakan | 3 [Able to state at least two point completely or three points without the name of atoms] [Berupaya menyatakan sekurang-kurangnya dua isi lengkap atau 3 isi tanpa nama atom] 2 [Able to state at least one point completely] [Berupaya menyatakan sekurang-kurangnya satu isi lengkap] 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0
  • 17. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 17 21Nov2017 | cikguadura.wordpress.com Manipulate variables / memanipulasi pembolehubah 1 Rubric Score (d) [Able to state all the variable and action to be taken correctly] [Berupaya menyatakan semua pembolehubah dan tindakan yang perlu diambil dengan betul] Sample answer / Contoh jawapan: Variables Pembolehubah Action to be taken Tindakan yang diambil Manipulated variable : Pembolehubah dimanipulasi : Type of block // Brass block and copper block Jenis blok // Blok loyang dan blok kuprum The way to manipulate variable : Cara memanipulasikan pembolehubah : Replace brass block with copper block Menggantikan blok loyang dengan blok kuprum Responding variable : Pembolehubah bergerakbalas: Diameter of dents // Hardness Diameter lekuk // Kekerasan What to observe in the responding variable: Apa yang diperhatikan pada pembolehubah bergerakbalas: The measurement of diameter of dents Pengukuran diameter lekuk Controlled variable : Pembolehubah ditetapkan : Height of weight // Mass of weight // Size of stainless steel ball bearing ketinggian pemberat // Jisim pemberat // Saiz bebola keluli nirkarat How to maintain the controlled variable : Cara untuk mengekalkan pembolehubah ditetapkan: Using the same height of weight// Using the same mass of weight // Using the same size of stainless steel ball bearing Gunakan tinggi pemberat yang sama //Gunakan jisim pemberat yang sama //Gunakan saiz bebola keluli nirkarat yang sama | 6 [Able to state all three variables and at least two corresponding action to be taken correctly] [Berupaya menyatakan semua tiga pembolehubah dan sekurang- kurangnya dua tindakan yang perlu diambil pada pembolehubah yang berkaitan dengan betul] 5 [Able to state at least two variables and at least two corresponding action to be taken correctly] [Berupaya menyatakan sekurang-kurangnya dua pembolehubah dan sekurang-kurangnya dua tindakan yang perlu diambil pada pembolehubah yang berkaitan dengan betul] 4 [Able to state at least two variables and at least one corresponding action to be taken correctly] [Berupaya menyatakan sekurang-kurangnya dua pembolehubah dan sekurang-nurangnya satu tindakan yang perlu diambil pada pembolehubah yang berkaitan dengan betul] 3
  • 18. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 18 21Nov2017 | cikguadura.wordpress.com [Able to state at least one variables and at least one corresponding action to be taken correctly] [Berupaya menyatakan sekurang-kurangnya satu pembolehubah dan sekurang-kurangnya satu tindakan yang perlu diambil pada pembolehubah yang berkaitan dengan betul] 2 [Able to state at least one variables correctly] [Berupaya menyatakan sekurang-kurangnya satu pembolehubah dengan betul] 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 Making hypothesis / membuat hipotesis 1 Rubric Score (e) [Able to state the hypothesis correctly] [Berupaya menyatakan hipotesis dengan betul] Sample answer / Contoh jawapan: Brass block has a smaller size / diameter of dents than copper block Brass is harder than copper // Blok loyang mempunyai saiz / diameter lekuk yang lebih kecil daripada blok kuprum. Loyang lebih keras daripada kuprum 3 [Able to state the hypothesis less correctly] [Berupaya menyatakan hipotesis kurang tepat] Sample answer / Contoh jawapan: Brass is harder // Copper is less harder Loyang lebih keras Kuprum kurang keras 2 [Able to state the idea of hypothesis] [Berupaya menyatakan idea hipotesis] Sample answer / Contoh jawapan: Brass is hard // Copper is soft Loyang keras // Kuprum lembut 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0
  • 19. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 19 21Nov2017 | cikguadura.wordpress.com Operational definition / Mendefinisi secara operasi 1 Rubric Score (f) [Able to state the operational definition correctly] [Berupaya menyatakan mendefinisi secara operasi dengan betul] What should be done: Drop 1.0 kg weight from 0.5 meter height on the material block. What should be observed: Dents formed on the material block. Apa yang perlu dibuat : Jatuhkan pemberat 1.0 kg dari ketinggian 0.5 meter di atas blok bahan. Apa yang diperhatikan : Lekuk terbentuk di atas blok bahan Sample answer / Contoh jawapan: Smaller dents formed when 1.0 kg weight dropped from 0.5 meter height on the block. Lekuk yang lebih kecil terbentuk apabila pemberat 1.0 kg dijatuhkan dari ketinggian 0.5 meter di atas blok bahan. 3 [Able to state the operational definition correctly] [Berupaya menyatakan mendefinisi secara operasi dengan betul] What should be done : Drop weight on the material block. What should be observed : Dents formed. Apa yang perlu dibuat : Jatuhkan pemberat di atas blok bahan. Apa yang diperhatikan : Lekuk terbentuk Sample answer / Contoh jawapan: A small dents formed when weight dropped on the block. Lekuk yang kecil terbentuk apabila pemberat dijatuhkan di atas blok bahan. 2 [Able to state the idea of operational definition] [Berupaya menyatakan idea bagi mendefinisi secara operasi] Sample answer / Contoh jawapan: The harder the material produce a smaller dent Bahan yang lebih keras menghasilkan lekuk yang lebih kecil 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0
  • 20. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 20 21Nov2017 | cikguadura.wordpress.com Predict / meramal 1 Rubric Score (g) [Able to predict the diameter of dent correctly] [Berupaya meramalkan diameter lekuk dengan betul] Sample answer / Contoh jawapan: Diameter / size of dents on copper more than 1.60 cm Diameter /saiz lekuk pada blok kuprum lebih besar daripada 1.60 cm 3 [Able to predict less correctly] [Berupaya meramal dengaa kurang tepat] Sample answer / Contoh jawapan: Diameter become larger / bigger / increase Diameter semakin besar / meningkat 2 [Able to state the idea of prediction] [Berupaya menyatakan idea meramal] Sample answer / Contoh jawapan: Change Berubah 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 Clasify/mengelas 1 Rubric Score (h) [Able to classify all materials correctly] [Berupaya mengelaskan semua bahan dengan betul] Sample answer / Contoh jawapan: Alloy / Aloi Pure metal / Logam tulen Stainless steel Keluli nirkarat Bronze Gangsa Duralumin Duralumin Iron Besi Aluminium Aluminium Chromium Kromium # if reverse, score 1 Jika terbalik, skor 1 3 [Able to classify at least four materials correctly] [Berupaya mengelaskan sekurang-kurangnya empat bahan dengan tepat] 2 [Able to classify at least two materials correctly] [Berupaya mengelaskan sekurang-kurangnya dua bahan dengan tepat] 1 [No response or wrong response] [Tiada jawapan atau jawapan salah] 0
  • 21. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 21 21Nov2017 | cikguadura.wordpress.com Esei Kelantan-2017 (a) [Dapat menyatakan pernyataan masalah dengan betul.] Sampel jawapan: Adakah kepekatan elektrolit mempengaruhi hasil elektrolisis di anod? 3 [Dapat menyatakan pernyataan masalah dengan kurang tepat] Sampel jawapan: Adakah kepekatan elektrolit mempengaruhi hasil elektrolisis? 2 [Dapat memberi idea bagi pernyataan masalah.] Sampel jawapan: Kepekatan elektrolit mempengaruhi hasil elektrolisis 1 0 Rubriks (b) [dapat menyatakan hubungan di antara pemboleh ubah dimanipulasi dan pemboleh ubah bergerak balas dan mempunyai arah] Sampel jawapan: Jika larutan natrium klorida pekat digunakan hasil di anod ialah gas klorin. Jika larutan natrium klorida cair digunakan hasil di anod ialah gas oksigen. 3 [dapat menyatakan hubungan di antara pemboleh ubah dimanipulasi dan pemboleh ubah bergerak balas] Sampel jawapan: Jika larutan natrium klorida pekat digunakan hasil di anod ialah gas klorin // Jika larutan natrium klorida cair digunakan hasil di anod ialah gas oksigen. 2 [Dapat menyatakan idea bagi hipotesis] Sampel jawapan: Kepekatan elektrolit mempengaruhi hasil di anod 1 Tiada respons atau respons salah 0 (c) [Dapat menyatakan dua pemboleh ubah dengan betul] Sampel jawapan Pemboleh ubah dimanipulasi: Kepekatan elektrolit Pemboleh ubah bergerak balas: Hasil di anod Pemboleh ubah dimalarkan : Jenis elektrod 3 [Dapat menyatakan dua pemboleh ubah dengan betul] 2 [dapat menyatakan satu pemboleh ubah dengan betul] 1 [tiada respons atau respons salah] 0
  • 22. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 22 21Nov2017 | cikguadura.wordpress.com Rubriks (d) [Dapat menyenaraikan bahan dan radas dengan lengkap] Sampel jawapan: Bahan: Larutan natrium klorida 0.001moldm-3 , larutan natrium klorida 1.0 moldm-3 Radas Elektrod karbon, sel elektrolisis, elektrod karbon, wayar penyambung dan klip buaya, sel kering, tabung uji 3 [Dapat menyenaraikan bahan dan radas dengan kurang lengkap] Sampel jawapan: Bahan: Larutan natrium klorida Radas Elektrod karbon, sel elektrolisis, elektrod karbon, wayar penyambung dan klip buaya, sel kering, tabung uji kecil 2 [Dapat menyenaraikan bahan dan radas dengan tidak lengkap] Sampel jawapan: Bahan: Larutan natrium klorida Radas elektrod karbon, bikar, wayar penyambung dan klip buaya, sel kering, 1 [Tiada respons atau respons salah] 0 Rubriks (e) [dapat menyenaraikan semua langkah dengan betul] Prosedur/kaedah 1. Masukkan larutan natrium klorida 0.001 moldm-3 ke dalam sel elektrolisis sehingga separuh penuh. 2. Masukkan larutan natrium klorida 0.001 moldm-3 ke dalam 2 tabung uji kecil sehingga penuh. 3. Telangkupkan tabung uji yang berisi larutan kepada keduadua elektrod 4. Sambungkan kedua-dua elektrod kepada sel kering dengan wayar penyambung dan klip buaya. 5. Pemerhatian direkodkan. 6. Langkah 1 hingga 4 diulang dengan menggunakan larutan natrium klorida 1.0 moldm-3 . 3 [dapat menyenaraikan langkah-langkah 1, 4, 5 dan 6 ] 2 [dapat menyenaraikan langkah-langkah 1, 4 dan 5] 1 [tiada respons atau respons salah] 0
  • 23. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 23 21Nov2017 | cikguadura.wordpress.com 2(f) [dapat membina jadual dengan aspek berikut] Kepekatan elektrolit (moldm-3) Pemerhatian di anod 0.001 1.0 | 2 [dapat membina jadual dengan tidak lengkap] Kepekatan elektrolit Pemerhatian di anod 0.001 1.0 | 1 [tiada respons atau respons salah atau jadual kosong] 0
  • 24. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 24 21Nov2017 | cikguadura.wordpress.com Kedah 2017
  • 25. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 25 21Nov2017 | cikguadura.wordpress.com
  • 26. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 26 21Nov2017 | cikguadura.wordpress.com
  • 27. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 27 21Nov2017 | cikguadura.wordpress.com
  • 28. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 28 21Nov2017 | cikguadura.wordpress.com JUJ-2017 Rubric (a) Able to give the problem statement correctly Sample answer: How does the effectiveness of cleansing action of two different types of cleansing agent in hard water? // What is the difference in effectiveness of cleansing action in hard water using soap and detergent? 3 Able to give the problem statement less correctly Sample answer: How does different type of cleansing agent affect the effectiveness of cleansing action in hard water? // To investigate the effectiveness of different type of cleansing agent in hard water. 2 Able to give an idea of statement of the problem Sample answer: To study the effectiveness of cleansing action in hard water 1 No response or wrong response 0 Rubric Able to state all variables correctly Sample answer: (b) Manipulated variable :Types of cleansing agent//soap and detergent Responding variable : Effectiveness of cleansing agent//ability to remove the oily stains on cloth Constant variable : Hard water//Cloth with oily stains//Sizeof cloth 3 Able to state any two variables correctly 2 Able to state any one variable correctly 1 No response or wrong response 0 Rubric Able to state the relationship between the manipulated variable and the responding variable correctly Sample answer: Detergent is more effective than soap in hard water. // Detergent in hard water can remove the oily stains on cloth while soap cannot. 3 Able to state the relationship between the manipulated variable and the responding variable incorrectly (c) Sample answer: Detergent is more effective // Soap is less effective // Detergent in hard water can remove the oily stains on cloth //soap cannot remove stain on cloth. 2 Able to state an idea of hypothesis 1 No response or wrong response 0
  • 29. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 29 21Nov2017 | cikguadura.wordpress.com Rubric Able to give the completely list of the apparatus and materials Answer: Apparatus: Beakers, measuring cylinder, glass rod Material: Soap, detergent, [cleansing agent X dan cleansing agent Y], 0.5 - 1 mol dm- magnesium sulphate solution/[sea water/hard water], 2 pieces of cloth with oily stains. 3 (d) Able to give the list of the apparatus and materials correctly but not completely Answer: Apparatus: Any containers, glass rod Material: Cleansing agent, cloth with oily stains. 2 Able to give two materials and at least one apparatus Answer : Material: Soap solution / detergent, hard water [any container] 1 No response or wrong response 0 Rubric e Able to state all procedures correctly Sample answer : (e) 1. Label 2 beakers as A and B respectively 2. Fill all the beakers with [50 - 100 cm3 ]of hard water / magnesium sulphate solution 3. Add cleansing agent X/detergent in beaker A and cleansing agent Y/soap in bikar B. 4. Place a piece of cloth with oily stain into each beaker 5. Stir the solution in each beaker using glass rod until no change can be observed 6. Wash the oily stain on each piece of cloth using the solution in each beaker 7. Record the observation. 3 Able to state three steps of procedures correctly Steps 2, 3, 5, 6, 2 Able to state two steps of procedures correctly Steps 2, 3, 6 1 No response or wrong response 0
  • 30. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 30 21Nov2017 | cikguadura.wordpress.com
  • 31. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 31 21Nov2017 | cikguadura.wordpress.com Johor 2015 Explanation (a) [Able to state the problem statement correctly]: Suggested answer: Which metals are harder, alloys or pure copper? 3 [Able to state the problem statement less correctly] Suggested answer: Are alloys are harder than pure metals? 2 [Able to state the problem statement less correctly] Suggested answer: Which metals are harder? 1 [No response given or wrong response] 0 Explanation (b) [Able to state thevariables correctly]: Suggested answer: (a) Manipulated variable : Type of metals block (b) Responding variable : Hardness of metals (c) Fixed variable : Size of steel ball bearing /Height of weight 3 [Able to state thevariables less correctly] Suggested answer: One mistake 2 [Able to state i thevariables less correctly] Suggested answer: Two mistakes 1 [No response given or wrong response] 0 Explanation (c) [Able to state thehypothesis correctly]: Suggested answer: Alloy is harder than pure metals / Pure metals are harder than alloy 3 [Able to state thehypothesis less correctly] Suggested answer: Alloy is harder 2 [Able to statethehypothesis less correctly] Suggested answer: Alloy produce smaller dent 1 [No response given or wrong response] 0
  • 32. Bahan Kimia SPM 2017 – Kertas 3 | Cikgu Adura 32 21Nov2017 | cikguadura.wordpress.com Explanation (d) [Able to list the apparatus and materials correctly]: Suggested answer: Apparatus : 1 kg of weight, thread, cellophane tape, retort stand, 1 meter of ruler, ruler Materials : steel ball bearing, copper block, bronze block 3 [Able to list the apparatus and materials less correctly] Suggested answer: One mistake 2 [Able to list the apparatus and materials less correctly] Suggested answer: Two mistakes 1 [No response given or wrong response] Suggested answer: More than three mistakes 0 Explanation (e) [Able to state theprocedure correctly]: Suggested answer: 1. Put a steel ball to an alloy block using cellophane tape. 2. Hang 1 kg weight at a height of 50 cm from the top of the alloy block. 3. Put a bronze block under the weight. 4. Release the weight so that it falls on the steel ball bearing. 5. Measure the diameter of the dent formed on top of the alloy block using meterruler. 6. Record the diameter reading in the table. 7. Repeat the steps 1 to 5 using pure metal block. 3 [Able to state theprocedure less correctly] Suggested answer: One mistake in procedure 2 [Able to statetheprocedure less correctly] Suggested answer: Two mistakes in procedure 1 [No response given or wrong response] Suggested answer: More than three mistakes in procedure 0 Explanation (f) [Able to draw the table correctly]: Suggested answer: Type of block Diameter of dent / cm Bronze Pure copper | 2 [Able to draw the table less correctly] Suggested answer: One mistake in table such as 1. Table not properly drawn 2. No unit 3. No type of block mention 1 [No response given or wrong response] 0