Cadangan jawapan Kertas 3 KIMIA SPM 2017 Cg Adura

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1 Bahan Kimia SPM 07 Kertas Cikgu Adura Cadangan jawapan Kertas KIMIA SPM 07 Cg Adura [SPM04-0-P] (a) Cadangan jawapan Observation Dapat menyatakan dua pemerhatian dengan tepat: Contoh jawapan lengkap:. Wasap putih terbebas / white fume released. Jisim kandungan mangkuk pijar bertambah/ mass increases. Pepejal/ serbuk putih terbentuk / white solid produce/ formed 4. Nyalaan putih/ terang/ berkilau // burn brightly Dapat menyatakan satu pemerhatian yg lengkap atau dua pemerhatian yg kurang lengkap: Contoh jawapan kurang lengkap:. Wasap terbebas // asap putih terbentuk. Jisim magnesium bertambah. Pepejal/ serbuk terbentuk 4. Magnesium menyala Dapat menyatakan satu idea tentang pemerhatian: Contoh jawapan:. Gas dibebaskan // gas putih. Terdapat pertambahan berat. Berat magnesium bertambah 4. Nyalaan/ baraan / cahaya Jisim mangkuk pijar bertambah Inferences 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 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 inferens: Contoh jawapan: Tindak balas berlaku Skor b Cadangan jawapan Dapat mencatatkan ketiga-tiga jisim dengan t. perpuluhan dengan betul: Skor Contoh jawapan: Jisim mangkuk pijar dan penutup: 5.5 g Jisim mangkuk pijar, penutup dan pita magnesium: 7.75 g Jisim mangkuk pijar, penutup dan magnesium oksida setelah disejukkan: 9.5 g Dapat mencatatkan mana-mana dua jisim dengan t. perpuluhan dengan betul Dapat mencatatkan mana-mana satu jisim dengan t. perpuluhan dengan betul Nov07 cikguadura.wordpress.com

2 Bahan Kimia SPM 07 Kertas Cikgu Adura (i) dan (ii) Cadangan jawapan Dapat menentukan jisim magnesium, jisim oksigen dan menunjukkan langkah menentukan formula empirik magnesium oksida dengan betul: Skor Jisim magnesium: =.40 /.4 g Jisim oksigen: =.60 /.6 g Langkah menentukan formula empirik magnesium oksida:. 4 L: Mol magnesium: = Mol oksigen: 6 = 0. L: Nisbah Mg : O = : L: Formula empirik = MgO Dapat menyatakan jisim magnesium, jisim oksigen dan menunjukkan 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 Dapat menyatakan definasi secara operasi berdasarkan formula empirik dengan tepat: Skor Contoh jawapan: mol magnesium react with mol oxygen Dapat menyatakan definasi secara operasi berdasarkan formula empirik dengan kurang tepat: Contoh jawapan: mol magnesium bertindak balas dengan oksigen Dapat menyatakan definasi secara operasi mengikut konsep: Contoh jawapan: Mol magnesium dan oksigen adalah sama. Nov07 cikguadura.wordpress.com

3 Bahan Kimia SPM 07 Kertas Cikgu Adura Negeri Sembilan 06 Explanation score (a) [Able to fulfil the following criteria]. all readings correct. two decimal places. without unit Answer: Methanol =.55 Ethanol =.0 Propanol =.0 Butanol =.6 One of criteria is not fulfilled [Able to fulfil criteria ] [No response given or wrong response] 0 (b) Explanation [Able to state all variables correctly] 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) [Able to state any variables correctly] [Able to state any variable correctly] score [No response given or wrong response] 0 (c) Explanation [Able to state the relationship between the manipulated variable and the responding variable with direction correctly] The higher the number of carbon atoms per molecule of alcohol, the higher the heat of combustion. Note : RV MV score [Able to state the relationship between the manipulated variable and the responding variable] The higher the number of carbon atoms, the higher the heat of combustion. score [Able to state an idea of hypothesis] Different alcohols,different heat of combustion. [No response given or wrong response] 0 Nov07 cikguadura.wordpress.com

4 Bahan Kimia SPM 07 Kertas Cikgu Adura (d) Explanation score 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 [Refer to page 8] 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 Able to draw a bar chart by fulfill the following criteria: (i) Axes are labelled with correct unit (ii) All bars are combined [No response given or wrong response] 0 (e) Explanation [Able to predict the heat of combustion of pentanol and draw the bar for pentanol correctly] Criteria :. Value with correct unit.the bar drawn as in (d) score Answer: 60 kj mol - [Draw a correct bar chart] score [Able to predict the heat of combustion of pentanol and draw the bar for pentanol] Answer: [without unit] [Draw a correct bar chart] [Able to predict the heat of combustion of pentanol or draw the bar for pentanol] Answer: [without unit] or [Draw a correct bar chart] [No response given or wrong response] 0 (f) Explanation score [Able to state the operational definition for heat of combustion correctly] The heat released/produced when mole of alcohol is burnt in excess oxygen, thermometer reading increases. [Able to state the operational definition for heat of combustion] Heat released/produced when alcohol is burnt in excess oxygen. [Able to state an idea of operational definition for heat of combustion] Heat released when alcohol burns. [No response given or wrong response] 0 4 Nov07 cikguadura.wordpress.com

5 Bahan Kimia SPM 07 Kertas Cikgu Adura Melaka 06 (a) Mark Scheme Able to state all ph value correctly Answer Mark Acid solution Colour of indicator ph value Acid P Red Acid Q Yellowish orange 5 Acid R Yellow 6 Able to state any twoph value correctly Able to state any one ph value correctly (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 Able to state any two variables correctly Able to state any one variable // idea of all variables (c) Able to state all correct inferences Acid P Q R Inference P is a strong acid // P ionize completely in water // P produces high concentration of hydrogen ions. Q is a weak acid // Q ionize partially in water // Q produces low concentration of hydrogen ions. R is a weak acid // R ionize partially in water // R produces low concentration of hydrogen ions. Able to state any two inferences correctly. Able to state any one inferences // idea of all inferences 5 Nov07 cikguadura.wordpress.com

6 Bahan Kimia SPM 07 Kertas Cikgu Adura (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. 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. Able to give an idea Sample answer ph value is affected by strength of acid. (e) Able to state the operational definition for strong acid correctly.. Low ph value. Add universal indicator Sample answer Acid that shows low ph value when universal indicator is added into it. Able to state the operational definition for strong acid less correctly.. Low ph value //. Add universal indicator Sample answer i. Acid that shows low ph value. // ii. Add universal indicator into the acid. Able to state an idea of operational definition Colour of indicator change // ph less than 7 // Ionise completely (f) Able to record all the burette readings accurately with two decimal place and with correct unit. 6 Nov07 cikguadura.wordpress.com

7 Bahan Kimia SPM 07 Kertas Cikgu Adura Answers Set Initial burette reading Final burette reading I 0.0 cm 4.0 cm II 4.0 cm cm III 0.50 cm 5.00 cm Able to record all burette readings less correctly with or without unit Sample answers Set Initial burette reading Final burette reading I II III or Set Initial burette reading Final burette reading I 0. cm 4. cm II 4. cm 47.8 cm III 0.5 cm 5 cm Able to record at least four burette readings less correctly Sample answers: Set Initial burette reading Final burette reading I.7 4. II III (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: 7 Nov07 cikguadura.wordpress.com

8 Bahan Kimia SPM 07 Kertas Cikgu Adura. Correct titles. Readings and units Set Initial burette reading (cm ) Final burette reading (cm ) I II III Volume of acid (cm ) Able to construct a less accurate table that contains the following:. Titles. Readings Set Initial burette reading Final burette reading Volume of acid I II III Able to construct a table that contains titles and readings for initial burette readings and final burette readings only. Set Initial burette reading Final burette reading I II III (h) Able to predict the volume of acid S correctly with unit. Answer: 50.0 cm Able to predict the volume of acid S Answer: 50 Able to give an idea to predict the volume of acid S more than 5 // more than acid P 8 Nov07 cikguadura.wordpress.com

9 Bahan Kimia SPM 07 Kertas Cikgu Adura (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. 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. Able to give an idea of the observation Sample answers: Calcium carbonate reacts with acid // Carbon dioxide gas (i)(ii) Able to state the relationship between the manipulated variable and the responding variable and state the direction correctly. Presence of water causes acid to show its properties. Able to state the relationship between the manipulated variable and the responding variable less correctly. Acid will shows its acidic properties when water presence. Able to state the idea of hypothesis Reaction occurs when water present. (j) Able to classify all the acids correctly Strong acid Nitric acid Sulphuric acid Weak acid Citric acid Carbonic acid Able to classify any three acids correctly Able to classify any two acids correctly 9 Nov07 cikguadura.wordpress.com

10 Bahan Kimia SPM 07 Kertas Cikgu Adura Perlis 05 Rubric (a) Able to state an observation accurately Rubber strip Y is extended longer than rubber strip X Score Able to state the observation correctly Rubber strip Y is extended/become longer// Rubber strip X do not extended Able to state idea of the observation Rubber strip extended Rubric (b) Able to state an accurate inference for this experiment: Rubber strip X is harder than rubber strip Y Able to state the inference for this experiment: Rubber strip X is hard// Rubber strip Y is soft Able to state the general inference for this experiment: Score X is more elastic rubber Rubric (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 Score Able to write all the reading of rubber strip X and Y correctly Able to write correct reading for rubber strip X and correct reading for rubber strip Y 0 Nov07 cikguadura.wordpress.com

11 Bahan Kimia SPM 07 Kertas Cikgu Adura Rubric (d) Able to construct a table with correct title and units and accurately Type of rubber Rubber strip X Rubber strip Y Weight (g) Length (cm) Score Able to construct a table without title or units and less accurately Type of rubber Rubber strip X Rubber strip Y Weight Length Able to give the idea about to construct table Type of rubber Rubber strip X Rubber strip Y Length (e) Rubric Able to state three variables correctly: (i) Manipulated variable: Rubber strip X, Rubber strip Y (ii) Responding variable: The length of rubber strip after weight is removed Score (iii) Fixed variable: Size of rubber strip, mass of weight Able to state any of the above information correctly Able to state any of the above information correctly (f) Rubric Able to state the relationship between the manipulated variable and the responding variable correctly and with direction Rubber strip X is more elastic than rubber strip Y Able to state the relationship between the manipulated variable and the responding variable correctly and without direction Score The elasticity of rubber strip X is high than rubber strip Y Able to state an idea of the hypothesis Rubber strip x is less elastic. Nov07 cikguadura.wordpress.com

12 Bahan Kimia SPM 07 Kertas Cikgu Adura Rubric (g) Able to give an accurately relationship between length of rubber strip and elasticity. The length of rubber strip increases the elasticity decreases Score Able to give less accurately relationship between length of rubber strip and elasticity. The length increases the elasticity decreases Able to give and idea of relationship between length of rubber strip and elasticity directly proportional Rubric (h) Able to state the operational definition correctly The longer the length of rubber strip after weight is removed the less elastic the rubber strip. Score Able to state the operational definition less correctly After weight is removed the rubber strip become longer Able give an idea for operational definition The length of rubber strip increase (i) Rubric 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// Score Able to explain the observation by stating any two of the aspect correctly Able to explain the observation by stating any one of the aspect correctly Nov07 cikguadura.wordpress.com

13 Bahan Kimia SPM 07 Kertas Cikgu Adura (j) Rubric 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 Score Able state any two of the answer correctly Able state any one of the answer correctly Rubric (k) Able to classify all the substances correctly Score Substance that can coagulate latex Nitric acid Methanoic acid Substance that cannot coagulate latex Sodium hydroxide Ammonia Able to classify any three the substances correctly Able to classify any two the substances correctly Nov07 cikguadura.wordpress.com

14 Bahan Kimia SPM 07 Kertas Cikgu Adura Johor Julang Set 07 Measuring using numbers / Mengukur menggunakan nombor 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] Brass :.0 cm,.00 cm,.0 cm Copper :.5 cm,.60 cm..50 cm [Able to measure the diameter of dents without decimal places and unit] [Berupaya mengukur diameter lekuk tanpa dua titik perpuluhan atau unit] [Able to state four diameters of dents without decimal places and unit] [Berupaya menyatakan empat diameter lekuk dengan betul] [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 Communicate / Berkomunikasi Rubric Score (b) [Able to construct the table to record the diameters and average diameter of dents on brass and copper block with correct:. Titles and units. Reading] [Berupaya membina jadual untuk menrekodkan diameter dan diameter purata blok loyang dan kuprum dengan betul:. Tajuk dan unit. Bacaan] Type of block Jenis blok Diameter of dents (cm) Diameter lekuk (cm) Average diameter of dents (cm) I II III Purata diameter lekuk (cm) Brass Loyang Copper Kuprum [Able to construct a less accurate the table that contain the following :. Titles without unit. Reading] [Berupaya membina jadual kurang tepat yang mengandungi yang berikut :. Tajuk tanpa unit. Bacaan] [Able to construct the idea of tabulation of data] [berupaya membina idea penjadualan data] [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 4 Nov07 cikguadura.wordpress.com

15 Bahan Kimia SPM 07 Kertas Cikgu Adura Observe / Memerhati Rubric Score (c)(i) [Able to state the observation correctly - Must have comparison] [Berupaya menyatakan pemerhatiaan dengan betul dalam bentuk perbandingan] The average diameter of dents on brass block is. cm and the average diameter of dents on copper block is.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. cm dan diameter purata lekuk pada blok kuprum adalah.48 cm // Saiz / diameter lekuk pada blok loyang lebih kecil daripada saiz / diameter pada blok kuprum // [Able to state the incomplete observation] [Berupaya menyatakan pemerhatian yang tidak lengkap] 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 [Able to state the idea of observation] [Berupaya menyatakan idea bagi pemerhatian] 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 [No response or wrong response] 0 [Tiada jawapan atau jawapan salah] Making inference / membuat inferens Rubric Score (c)(ii) [Able to state the inference correctly] [Berupaya menyatakan inferens dengan betul] Brass is harder than copper // Copper is less harder than brass Loyang lebih keras daripada kuprum // Kuprum kurang keras daripada loyang [Able to state the incomplete inference ] [Berupaya menyatakan inferens kurang lengkap] Brass is harder // Copper is less harder. Loyang lebih keras // Kuprum kurang keras. [Able to state the idea of inference ] [Berupaya menyatakan idea bagi inferens] Brass is hard // Copper is soft. Loyang keras // Kuprum lembut. [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 5 Nov07 cikguadura.wordpress.com

16 Bahan Kimia SPM 07 Kertas Cikgu Adura Interpret data / Mentafsir data Rubric Score (c)(iii) [Able to explain the arrangement of particles in the materials correctly] [Berupaya menerangkan susunan zarah dalam bahan dengan betul] Brass / Loyang. Atomic size of zinc / foreign atoms and copper are different. Saiz atom zink / atom asing dan kuprum adalah berbeza. 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.. 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 Copper / Kuprum. Atomic size of copper atoms are same. Saiz atom-atom kuprum adalah sama. Atoms are in closely pack in orderly manner in layers Atom-atom disusun dengan kemas dan rapi dalam lapisan. Layer of atoms easily sliding when force is applied Lapisan-lapisan atom mudah menggelongsor apabila daya dikenakan [Able to state at least two point completely or three points without the name of atoms] [Berupaya menyatakan sekurang-kurangnya dua isi lengkap atau isi tanpa nama atom] [Able to state at least one point completely] [Berupaya menyatakan sekurang-kurangnya satu isi lengkap] [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 6 Nov07 cikguadura.wordpress.com

17 Bahan Kimia SPM 07 Kertas Cikgu Adura Manipulate variables / memanipulasi pembolehubah Rubric Score (d) [Able to state all the variable and action to be taken correctly] 6 [Berupaya menyatakan semua pembolehubah dan tindakan yang perlu diambil dengan betul] Variables Pembolehubah Manipulated variable : Pembolehubah dimanipulasi : Action to be taken Tindakan yang diambil The way to manipulate variable : Cara memanipulasikan pembolehubah : Type of block // Brass block and copper block Jenis blok // Blok loyang dan blok kuprum Responding variable : Pembolehubah bergerakbalas: Diameter of dents // Hardness Diameter lekuk // Kekerasan Controlled variable : Pembolehubah ditetapkan : Height of weight // Mass of weight // Size of stainless steel ball bearing ketinggian pemberat // Jisim pemberat // Saiz bebola keluli nirkarat Replace brass block with copper block Menggantikan blok loyang dengan blok kuprum What to observe in the responding variable: Apa yang diperhatikan pada pembolehubah bergerakbalas: The measurement of diameter of dents Pengukuran diameter lekuk 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 [Able to state all three variables and at least two corresponding action to be taken correctly] [Berupaya menyatakan semua tiga pembolehubah dan sekurangkurangnya dua tindakan yang perlu diambil pada pembolehubah yang berkaitan dengan betul] [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] [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] Nov07 cikguadura.wordpress.com

18 Bahan Kimia SPM 07 Kertas Cikgu Adura [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] [Able to state at least one variables correctly] [Berupaya menyatakan sekurang-kurangnya satu pembolehubah dengan betul] [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 Making hypothesis / membuat hipotesis Rubric Score (e) [Able to state the hypothesis correctly] [Berupaya menyatakan hipotesis dengan betul] 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 [Able to state the hypothesis less correctly] [Berupaya menyatakan hipotesis kurang tepat] Brass is harder // Copper is less harder Loyang lebih keras Kuprum kurang keras [Able to state the idea of hypothesis] [Berupaya menyatakan idea hipotesis] Brass is hard // Copper is soft Loyang keras // Kuprum lembut [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 8 Nov07 cikguadura.wordpress.com

19 Bahan Kimia SPM 07 Kertas Cikgu Adura Operational definition / Mendefinisi secara operasi Rubric Score (f) [Able to state the operational definition correctly] [Berupaya menyatakan mendefinisi secara operasi dengan betul] What should be done: Drop.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.0 kg dari ketinggian 0.5 meter di atas blok bahan. Apa yang diperhatikan : Lekuk terbentuk di atas blok bahan Smaller dents formed when.0 kg weight dropped from 0.5 meter height on the block. Lekuk yang lebih kecil terbentuk apabila pemberat.0 kg dijatuhkan dari ketinggian 0.5 meter di atas blok bahan. [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 A small dents formed when weight dropped on the block. Lekuk yang kecil terbentuk apabila pemberat dijatuhkan di atas blok bahan. [Able to state the idea of operational definition] [Berupaya menyatakan idea bagi mendefinisi secara operasi] The harder the material produce a smaller dent Bahan yang lebih keras menghasilkan lekuk yang lebih kecil [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 9 Nov07 cikguadura.wordpress.com

20 Bahan Kimia SPM 07 Kertas Cikgu Adura Predict / meramal Rubric Score (g) [Able to predict the diameter of dent correctly] [Berupaya meramalkan diameter lekuk dengan betul] Diameter / size of dents on copper more than.60 cm Diameter /saiz lekuk pada blok kuprum lebih besar daripada.60 cm [Able to predict less correctly] [Berupaya meramal dengaa kurang tepat] Diameter become larger / bigger / increase Diameter semakin besar / meningkat [Able to state the idea of prediction] [Berupaya menyatakan idea meramal] Change Berubah [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 Clasify/mengelas Rubric Score (h) [Able to classify all materials correctly] [Berupaya mengelaskan semua bahan dengan betul] Alloy / Aloi Stainless steel Keluli nirkarat Bronze Gangsa Duralumin Duralumin # if reverse, score Jika terbalik, skor Pure metal / Logam tulen Iron Besi Aluminium Aluminium Chromium Kromium [Able to classify at least four materials correctly] [Berupaya mengelaskan sekurang-kurangnya empat bahan dengan tepat] [Able to classify at least two materials correctly] [Berupaya mengelaskan sekurang-kurangnya dua bahan dengan tepat] [No response or wrong response] [Tiada jawapan atau jawapan salah] 0 0 Nov07 cikguadura.wordpress.com

21 Bahan Kimia SPM 07 Kertas Cikgu Adura Esei Kelantan-07 (a) [Dapat menyatakan pernyataan masalah dengan betul.] Sampel jawapan: Adakah kepekatan elektrolit mempengaruhi hasil elektrolisis di anod? [Dapat menyatakan pernyataan masalah dengan kurang tepat] Sampel jawapan: Adakah kepekatan elektrolit mempengaruhi hasil elektrolisis? [Dapat memberi idea bagi pernyataan masalah.] Sampel jawapan: Kepekatan elektrolit mempengaruhi hasil elektrolisis 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. [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. 0 [Dapat menyatakan idea bagi hipotesis] Sampel jawapan: Kepekatan elektrolit mempengaruhi hasil di anod 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 [Dapat menyatakan dua pemboleh ubah dengan betul] [dapat menyatakan satu pemboleh ubah dengan betul] [tiada respons atau respons salah] 0 Nov07 cikguadura.wordpress.com

22 Bahan Kimia SPM 07 Kertas Cikgu Adura Rubriks (d) [Dapat menyenaraikan bahan dan radas dengan lengkap] Sampel jawapan: Bahan: Larutan natrium klorida 0.00moldm -, larutan natrium klorida.0 moldm - Radas Elektrod karbon, sel elektrolisis, elektrod karbon, wayar penyambung dan klip buaya, sel kering, tabung uji [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 [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, [Tiada respons atau respons salah] 0 Rubriks (e) [dapat menyenaraikan semua langkah dengan betul] Prosedur/kaedah. Masukkan larutan natrium klorida 0.00 moldm - ke dalam sel elektrolisis sehingga separuh penuh.. Masukkan larutan natrium klorida 0.00 moldm - ke dalam tabung uji kecil sehingga penuh.. 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 hingga 4 diulang dengan menggunakan larutan natrium klorida.0 moldm -. [dapat menyenaraikan langkah-langkah, 4, 5 dan 6 ] [dapat menyenaraikan langkah-langkah, 4 dan 5] [tiada respons atau respons salah] 0 Nov07 cikguadura.wordpress.com

23 Bahan Kimia SPM 07 Kertas Cikgu Adura (f) [dapat membina jadual dengan aspek berikut] Kepekatan elektrolit (moldm - ) Pemerhatian di anod [dapat membina jadual dengan tidak lengkap] Kepekatan elektrolit Pemerhatian di anod [tiada respons atau respons salah atau jadual kosong] 0 Nov07 cikguadura.wordpress.com

24 Bahan Kimia SPM 07 Kertas Cikgu Adura Kedah 07 4 Nov07 cikguadura.wordpress.com

25 Bahan Kimia SPM 07 Kertas Cikgu Adura 5 Nov07 cikguadura.wordpress.com

26 Bahan Kimia SPM 07 Kertas Cikgu Adura 6 Nov07 cikguadura.wordpress.com

27 Bahan Kimia SPM 07 Kertas Cikgu Adura 7 Nov07 cikguadura.wordpress.com

28 Bahan Kimia SPM 07 Kertas Cikgu Adura JUJ-07 (a) Rubric Able to give the problem statement correctly 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? Able to give the problem statement less correctly 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. Able to give an idea of statement of the problem To study the effectiveness of cleansing action in hard water Rubric Able to state all variables correctly (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 Able to state any two variables correctly Able to state any one variable correctly Rubric Able to state the relationship between the manipulated variable and the responding variable correctly Detergent is more effective than soap in hard water. // Detergent in hard water can remove the oily stains on cloth while soap cannot. Able to state the relationship between the manipulated variable and the responding variable incorrectly (c) 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. Able to state an idea of hypothesis 8 Nov07 cikguadura.wordpress.com

29 Bahan Kimia SPM 07 Kertas Cikgu Adura 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], mol dm- magnesium sulphate solution/[sea water/hard water], pieces of cloth with oily stains. (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. Able to give two materials and at least one apparatus Answer : Material: Soap solution / detergent, hard water [any container] Able to state all procedures correctly Sample answer : Rubric e (e). Label beakers as A and B respectively. Fill all the beakers with [50-00 cm ]of hard water / magnesium sulphate solution. 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. Able to state three steps of procedures correctly Steps,, 5, 6, Able to state two steps of procedures correctly Steps,, 6 9 Nov07 cikguadura.wordpress.com

30 Bahan Kimia SPM 07 Kertas Cikgu Adura 0 Nov07 cikguadura.wordpress.com

31 Bahan Kimia SPM 07 Kertas Cikgu Adura Johor 05 (a) Explanation [Able to state the problem statement correctly]: Suggested answer: Which metals are harder, alloys or pure copper? [Able to state the problem statement less correctly] Suggested answer: Are alloys are harder than pure metals? [Able to state the problem statement less correctly] Suggested answer: Which metals are harder? [No response given or wrong response] 0 (b) Explanation [Able to state thevariables correctly]: Suggested answer: (a) Manipulated variable (b) Responding variable (c) Fixed variable : Type of metals block : Hardness of metals : Size of steel ball bearing /Height of weight [Able to state thevariables less correctly] Suggested answer: One mistake [Able to state i thevariables less correctly] Suggested answer: Two mistakes [No response given or wrong response] 0 (c) Explanation [Able to state thehypothesis correctly]: Suggested answer: Alloy is harder than pure metals / Pure metals are harder than alloy [Able to state thehypothesis less correctly] Suggested answer: Alloy is harder [Able to statethehypothesis less correctly] Suggested answer: Alloy produce smaller dent [No response given or wrong response] 0 Nov07 cikguadura.wordpress.com

32 Explanation (d) [Able to list the apparatus and materials correctly]: Bahan Kimia SPM 07 Kertas Cikgu Adura Suggested answer: Apparatus : kg of weight, thread, cellophane tape, retort stand, meter of ruler, ruler Materials : steel ball bearing, copper block, bronze block [Able to list the apparatus and materials less correctly] Suggested answer: One mistake [Able to list the apparatus and materials less correctly] Suggested answer: Two mistakes [No response given or wrong response] Suggested answer: More than three mistakes 0 Explanation (e) [Able to state theprocedure correctly]: Suggested answer:. Put a steel ball to an alloy block using cellophane tape.. Hang kg weight at a height of 50 cm from the top of the alloy block.. 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 to 5 using pure metal block. [Able to state theprocedure less correctly] Suggested answer: One mistake in procedure [Able to statetheprocedure less correctly] Suggested answer: Two mistakes in procedure [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 [Able to draw the table less correctly] Suggested answer: One mistake in table such as. Table not properly drawn. No unit No type of block mention [No response given or wrong response] 0 Nov07 cikguadura.wordpress.com

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