Ppt chem ting 5 paper 2 (2017) answer scheme

ANSWER SCHEME

1

PPT CHEM P2 TING. 5 (2017)

a: accept ; r: reject ; adp: avoid double penalty ; wcr: wrong cancels right ; ecf: error carried forward ; bod: benefit of doubt Question Answer Marks 1 (a) (i) Silica / silicon dioxide / silicon(IV) oxide / SiO2 // 1 Silika / silikon dioksida / silicon(IV) oksida a: silicon oxide / silikon oksida r: sand / pasir / SiO (ii) Borosilicate (glass) / fused (glass) 1 (Kaca) borosilikat / (kaca) terlakur r: wrong spelling eg: borosillicate (iii) Can withstand high temperature // low thermal expansion coefficient 1 Boleh tahan suhu yang tinggi // pekali pengembangan haba yang rendah a: heat resistant // high heat resistant // can withstand heat // can withstand high heat // does not crack when heated strongly // tahan panas // r: high melting and boiling point // has high specific heat capacity (b) (i) Chloroethane / vinyl chloride // 1 Kloroetana / vinil klorida r: vinil chloride (combination of BM & BI) / vinyl klorida (combination of BM & BI) / monovinyl chloride / (ii) H Cl 1 | | C=C | | H H a: condensed formula, but must show double bond eg: CH2=CHCl r: H Cl | | C=C | | H H n

(c)

(iii) Produce acidic / poisonous / toxic gas Menghasilkan gas berasid / beracun / toksik a: release hydrogen chloride gas // release harmful / dangerous gas // release chlorine gas // release carbon monoxide // gas can be replaced with smoke / asap r: produce toxic substance // causes air pollution (too general) // 1. Photochromic glass contains silver chloride / AgCl Kaca fotokromik mengandungi argentum klorida / AgCl 2. When exposed to light, silver chloride / AgCl is converted to silver / Ag and the glass darkens

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ANSWER SCHEME

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PPT CHEM P2 TING. 5 (2017)

Apabila terdedah kepada cahaya, argentum klorida / AgCl ditukarkan kepada argentum / Ag 3. When light dims, silver / Ag is converted back to silver chloride / AgCl Apabila cahaya malap, argentum / Ag ditukarkan semula kepada argentum klorida / AgCl TOTAL

2

Question (a) (b)

(c) (d) (e)

(f)

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(i)

Answer 3 r: three / tiga / III Chlorine / Cl / Klorin a: chlorine molecule / Cl2 / chlorine gas r: chlorine atom Aluminium / Al r: aluminium oxide / aluminium oksida / Al 2O3 Silicon / Si / Silikon 1. Atomic size / Radius decreases Saiz / Jejari atom berkurangan a: jejari atom semakin pendek / saiz atom semakin kecil / mengecil 2. Attraction force between nucleus and its valence electron is stronger // Nuclei charge increases // Atom easier to receive / gain electrons Daya tarikan antara nukleus dengan elektron valens semakin kuat // Cas nuklei bertambah // Atom lebih mudah menerima elektron a: Attraction force between nucleus and its valence electron increases Daya tarikan antara nukleus dengan elektron valens bertambah // Attraction force between nucleus and its electron is stronger / increases Daya tarikan antara nukleus dengan elektron semakin kuat / bertambah // Positive charge increases Cas positif bertambah High melting and boiling point // Takat lebur dan takat didih yang tinggi // Soluble in water // larut dalam air // Insoluble in organic solvent // tidak larut dalam pelarut organik // Can conduct electricity in molten and aqueous states // boleh mengkonduksikan / mengalirkan elektrik dalam keadaan leburan dan akueus

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PPT CHEM P2 TING. 5 (2017)

(ii)

-

+ Na

2

Cl

TOTAL Question 3 (a) (i)

Answer The (chemical) formula that shows the actual number of atoms of each element in a molecule / compound // Formula (kimia) yang menunjukkan bilangan atom yang sebenar bagi setiap unsur dalam satu molekul / sebatian (ii) CH2O (iii) 1. Reactants : Carbon dioxide / CO2 and water / H2O 2. Products : Glucose / C6H12O6 and oxygen / O2 3. Mole ratio : 6 mol CO2 reacts with 6 mol H2O produces 1 mol C6H12O6 and 6 mol O2 // 6 mol CO2 bertindak balas dengan 6 mol H2O menghasilkan 1 mol C6H12O6 dan 6 mol O2 Note 1. If mol is replaced with molecule, can award P3, provided, all are replaced 2. If extra info contradicts the equation, apply wcr once only Eg: P1 P2 P3 P4 P5

(b)

(i)

X:4 Y:3 Z:2 (ii) 2 (mol) Note 1. ecf from 3(b)(i) (iii) 178 r: 178 g mol-1 (RFM cannot have any unit)

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ANSWER SCHEME Question 4 (a)

(b)

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PPT CHEM P2 TING. 5 (2017) Answer

R : C5H12 a: condensed formula eg. CH3CH2CH2CH2CH3

Marks 1

S : C5H10 a: condensed formula eg. CH2CHCH2CH2CH3 R : alkane / alkana r: pentane / pentana

1

S : alkene / alkena r: pentene / pentena

1

(c)

1

4

F1: N1:

Pent-1-ene / pent-1-ena

F2: N2:

Pent-2-ene / pent-2-ena

F3:

N3:

2-methylbut-1-ene / 2-metilbut-1-ena

F4: N4:

3-methylbut-1-ene / 3-metilbut-1-ena

F5:

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PPT CHEM P2 TING. 5 (2017)

N5: 2-methylbut-2-ene / 2-metilbut-2-ena Any 2 isomers with correct names *Apply wcr once only if more than 2 isomers and names given

(d)

(i)

(ii)

Hydrogenation / Penghidrogenan /Addition of hydrogen / Penambahan hidrogen r: addition / penambahan C5H10 + H2  C5H12

1

TOTAL Question 5 (a)

(b)

(c)

(d)

(i)

(ii)

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Answer Chemical substance that ionizes / dissociates in water to produce H+ / hydrogen ion // Bahan kimia yang mengion / bercerai dalam air untuk menghasilkan H+ / ion hidrogen a: compound / chemical / substance / solution / sebatian / bahan / larutan r: dissolve / decompose / urai X : ethanoic acid / acetic acid / CH3COOH / asid etanoik / asid asetik r: vinegar / cuka

Marks 1

Y : sulphuric acid / asid sulfurik 1. Acid X is a weak acid while acid Y is a strong acid // Acid X ionizes / dissociates partially in water while acid Y ionizes / dissociates completely in water // Acid X produce lower concentration of hydrogen ion / H+ than acid Y Asid X ialah asid lemah manakala asid Y ialah asid kuat // Asid X mengion / bercerai separa dalam air manakala asid Y mengion / bercerai sepenuhnya dalam air // asid X menghasilkan kepekatan ion hidrogen / H+ yang lebih rendah berbanding asid Y 2. Concentration of hydrogen ion / H+ in (acid Y) is higher, pH value is lower // Concentration of hydrogen ion / H+ in (acid X) is lower, pH value is higher Kepekatan ion hidrogen / H+ dalam (asid Y) lebih tinggi, nilai pH lebih rendah // Kepekatan ion hidrogen / H+ dalam (asid X) lebih rendah, nilai pH lebih tinggi 1. Correct reactants and products 2. Balanced equation H2SO4 + ZnCO3  ZnSO4 + CO2 + H2O Note: 1. If acid Y is incorrect in 5(b), consider P2 only 1. Ratio of moles From equation, 1 mol ZnCO3 : 1 mol CO2 0.5 mol ZnCO3 : 0.5 mol CO2

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PPT CHEM P2 TING. 5 (2017)

2. Volume of CO2 0.5 mol x 24 dm3 mol-1 // 12 dm3 Note: Apply ecf P1 from 5(d)(i) (iii) 1. Flow / Pass / Channel / Bubble the gas into limewater // Deliver the gas through delivery tube into limewater Salurkan / Alirkan gas ke dalam air kapur // Lalukan gas ke dalam air kapur melalui salur penghantar 2. Limewater turns chalky / milky / cloudy Air kapur menjadi keruh TOTAL

6

Question (a)

(b)

(i) (ii)

(c)

(d)

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(i) (ii)

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1

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Answer Marks Size / Total surface area of calcium carbonate // Calcium carbonate 1 chips and powder // Saiz / Jumlah luas permukaan kalsium karbonat // Ketulan dan serbuk kalsium karbonat a: size / total surface area of reactant // saiz / jumlah luas permukaan bahan tindak balas r: size / saiz Carbon dioxide / CO2 / Karbon dioksida 1 Changes in the volume of carbon dioxide / CO2 can be measured 1 easily Perubahan isi padu karbon dioksida / CO2 boleh diukur dengan mudah Any 2 factors 1. Concentration of hydrochloric acid // Kepekatan asid hidroklorik 2. Temperature / Suhu 3. Mass of calcium carbonate // Jisim kalsium karbonat Max. 2 + 2+ CaCO3 + 2 H ďƒ Ca + CO2 + H2O 1 1. Total surface area of calcium carbonate powder is bigger than 1 calcium carbonate chips Jumlah luas permukaan serbuk kalsium karbonat lebih besar berbanding ketulan kalsium karbonat 2. Frequency of collision between calcium carbonate / CaCO3 and hydrogen ion / H+ is higher in Experiment II Frekuensi perlanggaran antara kalsium karbonat / CaCO3 dan ion hidrogen / H+ dalam Ekeperimen II lebih tinggi

1

3. Frequency of effective collision between calcium carbonate / CaCO3 and hydrogen ion / H+ is higher in Experiment II Frekuensi perlanggaran berkesan antara kalsium karbonat / CaCO3 dan ion hidrogen / H+ dalam Ekeperimen II lebih tinggi

1

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ANSWER SCHEME

(e)

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PPT CHEM P2 TING. 5 (2017)

Note: 1. CaCO3 and H+ must be mentioned at least once, either in P2 or P3 Total

3

1. Steeper gradient 2. Higher quantity of product

1 1

Total TOTAL Question 7 (a)

1.

2.

3.

4.

(b)

1.

2.

3.

4.

5.

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Answer Marks Factor affecting the product of electrolysis is type of electrodes 1 Faktor yang mempengaruhi produk elektrolisis ialah jenis elektrod At anode, copper ionises to produce Cu2+ ions 1 Di anod, kuprum mengion untuk menghasilkan ion Cu2+ At cathode, Cu2+ ions in the solution are selectively chosen to be 1 discharged Di katod, ion Cu2+ dalam larutan telah dipilih untuk dinyahcas Mass of anode decreases, mass of cathode increases 1 Jisim anod berkurang, jisim katod bertambah // Anode become thinner, cathode become thicker Anod menjadi nipis, katod menjadi tebal Total 4 Nickel and copper plates are cleaned with sandpaper 1 Kepingan nikel dan kuprum dibersihkan dengan kertas pasir Nickel is connected to the positive terminal of the battery 1 Nikel disambungkan dengan terminal positif bateri Copper plate is connected to the negative terminal of the battery 1 Kepingan kuprum disambung dengan terminal negatif bateri Both electrodes are dipped into nickel(II) sulphate / nitrate / 1 chloride solution Kedua-dua elektrod dicelupkan ke dalam larutan nikel(II) sulfat / nitrat / klorida Use a low concentration electrolyte 1 Gunakan elektrolit berkepekatan rendah

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ANSWER SCHEME

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PPT CHEM P2 TING. 5 (2017)

6. Use small electric current Gunakan arus elektrik yang rendah 7. Turn the copper plate Pusingkan elektrod kuprum

1 1 Max.

(c)

(i)

(ii)

Question 8 (a)

(b)

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(i)

1. Positif terminal is copper metal / Terminal positif ialah logam kuprum 2. Negative terminal is zinc metal / Terminal negatif ialah logam zink 3. Zinc electrode become thinner/ Elektrod zink menjadi nipis 4. Colourless gas bubbles released / Gelembung gas tak berwarna dibebaskan 5. Zn → Zn2+ + 2e 6. 2H+ + 2e  H2 7. Zn + 2H+ → Zn2+ + H2 8. Chemical energy → electrical energy //Tenaga kimia → tenaga elektrik Total 1. Decreases / Berkurang 2. The distance between Fe and Cu is shorter than the distance between Zn and Cu in the electrochemical series Jarak antara Fe dan Cu lebih dekat berbanding jarak antara Zn dan Cu dalam siri elektrokimia // Fe is less electropositive than Zn Fe kurang elektropositif berbanding Zn Total TOTAL

6 1 1 1 1 1 1 1 1 8 1 1

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Answer Marks Any 2 isotopes with 2 corresponding uses 1. Cobalt-60 : treatment of cancer // destroy bacteria in food 1+1 Kobalt-60 : rawatan kanser // memusnahkan bakteria dalam makanan 2. Iodine-131 : treat thyroid disease 1+1 Iodin-131 : merawat penyakit tiroid 3. Phosphorus-32 : study absorption of phosphorus in plants 1+1 Fosforus-32 : mengkaji penyerapan fosforus dalam tumbuhan 4. Sodium-24 : detect leakage of underground pipes 1+1 Natrium-24 : mengesan kebocoran paip bawah tanah 5. Carbon-14 : carbon dating // estimate age of fossils and artifacts 1+1 Karbon-14 : pentarikhan karbon // menganggar usia fosil dan artifak 5. Uranium : source of nuclear energy // sumber tenaga nuklear 1+1 Max. 4 Any 4 answers 1. Atom X has 11 protons // Atom X mempunyai 11 proton 1 2. Atom X has 12 neutrons // Atom X mempunyai 12 neutron 1

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(ii)

(c)

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PPT CHEM P2 TING. 5 (2017)

3. protons and neutrons are located in the nucleus // proton dan neutron terletak di dalam nukleus 4. Atom X has 11 electrons // Atom X mempunyai 11 elektron 5. Atom X has 3 shells occupied with electrons // Atom X mempunyai 3 petala berisi elektron 6. electron is located in the shells orbiting the nucleus // elektron terletak di dalam petala yang mengelilingi nucleus 7. nucleon number of element X is 23 // nombor nukleon unsur X ialah 23 Max. 1. Group 1 / Kumpulan 1 r: One / I / Satu 2. because it has 1 valence electron / kerana mempunyai 1 elektron valens 3. Period 3 r: Three / III / Tiga 6. because it has 3 shells filled / occupied with electrons // kerana mempunyai 3 petala berisi elektron Total Ionic compound / Sebatian ion 1. To achieve stable octet electron arrangement // Untuk mencapai susunan elektron oktet yang stabil 2. Lithium / Li atom donates / release 1 valence electron // Atom litium / Li menderma / melepaskan 1 elektron valens 3. forming lithium ion / Li+ // membentuk ion litium / Li+ // Li ďƒ Li+ + e 4. Chlorine / Cl atom receives / accepts / gain 1 electron // Atom klorin / Cl menerima 1 elektron 5. forming chloride ion / Cl- // membentuk ion klorida / Cl- // Cl + e ďƒ Cl6. lithium ion / Li+ and chloride ion / Cl- are attracted by electrostatic force // ion litium / Li+ dan ion klorida / Cl- ditarik oleh daya elektrostatik 7. Ionic compound, (LiCl) is formed // Sebatian ion, (LiCl) terbentuk Covalent compound / Sebatian kovalen 8. Carbon atom share valence electrons with chlorine atom // Atom karbon berkongsi elektron dengan atom klorin 9. Carbon atom contributes 4 electrons and each chlorine atom contributes 1 electron // Carbon atom share electrons with 4 chlorine atoms // Atom karbon menyumbang 4 elektron dan setiap atom klorin menyumbang 1 elektron // Atom karbon berkongsi 4 elektron dengan 4 atom klorin 10. 4 single covalent bonds are formed // Covalent compound, CCl 4 is formed // 4 ikatan kovalen tunggal terbentuk // Sebatian kovalen, CCl4 terbentuk Note:

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1 1 1

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10

PPT CHEM P2 TING. 5 (2017)

1. Atom must be mentioned in P2 / P4 and P8 / P9 2. P2, P3, P4, P5, P10 can be replaced with diagram Max. TOTAL Question 9 (a) (i)

Answer Reaction / Tindak balas I : 1. Oxidation / Pengoksidaan

Homologous series / Siri homolog Molecular formula / Formula molekul

P

Q

2. Alcohol / Alkohol

3. Alkene / Alkena

5. C2H5OH

6. C2H4

8.

9.

8 20 Marks 10

R 4. Carboxylic acid / Asid karboksilik 7. CH3COOH 10.

Structural formula / Formula struktur (ii)

1. Correct formula of reactants and products 2. Balanced chemical equation

1 1

C2H4 + 3O2  2CO2 + 2H2O 3. Number of moles of compound Q = 7.0 g // 0.25 mol (2 × 12) + (4 × 1) g mol-1

1

4. Ratio of moles 1 mol compound Q / C2H4 : 2 mol CO2 0.25 mol compound Q / C2H4 : 0.5 mol CO2

1

5. Volume of CO2 = 0.5 mol × 24 dm3 mol-1 // 12 dm3 // 12000 cm3

1

Note: 1. P5 must contain correct answer and correct unit Total (b)

(i)

1. A : Propene / Propena a: prop-1-ene / prop-1-ena 2. B : Propane / Propana Total

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ANSWER SCHEME (ii)

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PPT CHEM P2 TING. 5 (2017)

Using bromine water 1. Method : Add a few drops of bromine water into hydrocarbon A and B (in separate test tubes) Kaedah : Tambah beberapa titis air bromin ke dalam hidrokarbon A dan B (dalam tabung uji berlainan) 2. Observation of hydrocarbon A : no change to brown bromine water Pemerhatian hidrokarbon A : tiada perubahan pada warna perang air bromin 3. Observation of hydrocarbon B : brown bromine water decolourised / becomes colourless Pemerhatian hidrokarbon B : warna perang air bromin dinyahwarnakan / menjadi tidak berwarna OR Using acidified potassium manganate(VII) solution 1. Method : Add a few drops of acidified potassium manganate(VII) solution into hydrocarbon A and B (in separate test tubes) Kaedah : Tambah beberapa titis larutan kalium manganat(VII) berasid ke dalam hidrokarbon A dan B (dalam tabung uji berlainan) 2. Observation of hydrocarbon A : no change to purple acidified potassium manganate(VII) solution Pemerhatian hidrokarbon A : tiada perubahan pada warna ungu larutan kalium manganat(VII) berasid 3. Observation of hydrocarbon B : purple acidified potassium manganate(VII) solution decolourised / becomes colourless Pemerhatian hidrokarbon B : warna ungu larutan kalium manganat(VII) berasid dinyahwarnakan / menjadi tidak berwarna Total TOTAL

Question 10 (a) (i)

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Answer 1. CnH2n = 28 // n(12) + 2n(1) = 28 // 12n + 2n = 28 // n = 28/14 // n=2 OR CnH2n = 42 // n(12) + 2n(1) = 42 // 12n + 2n = 42 // n = 42/14 // n=3 OR CnH2n = 56 //

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1

1

OR

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3 20 Marks 1

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ANSWER SCHEME

12

PPT CHEM P2 TING. 5 (2017)

n(12) + 2n(1) = 56 // 12n + 2n = 56 // n = 56/14 // n=4 2. Molecular formula : C2H4 / C3H6 / C4H8 3. Structural formula 4. Name

1 1+1

Ethene / Etena OR

Propene / Propena a: prop-1-1ene OR

// But-1-ene / But-1-ena

// But-2-ene / But-2-ena

2-methylpropene/ 2-metilpropena Total

4

(ii) Answer is dependent on 10(a)(i) 1. Structural formula of Y

// Ethanol

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2. Name of Y

// Propanol

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1+1

// Butan-1-ol / 1-butanol

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ANSWER SCHEME

13

PPT CHEM P2 TING. 5 (2017)

Butan-2-ol / 2-butanol 3. Structural formula of Z

4. Name of Z

// Ethanoic acid

1+1

// Propanoic acid

Butanoic acid

Note: (for P1 – P4) 1. if structural formula drawn wrongly, but name is correct, award marks for names 2. if structural formula of Y and Z are interchanged, no P1 & P3, but consider P2 & P4 if names are correct 5. Correct formula of reactants and products 6. Balanced equation C2H5OH + 2 [O]  CH3COOH // C3H7OH + 2 [O]  C2H5COOH // C4H9OH + 2 [O]  C3H7COOH

1 1

2[O] r: C2H5OH  CH3COOH (0 + 0) r: C2H5OH + O2  CH3COOH (0 + 0)

(b)

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Note: 1. if structural formulae of compound Y and Z are not from 10(a)(i), consider marks for names of compounds (P2 & P4) and chemical equation (P5 & P6) Total 1. Named alcohol eg.: ethanol / etanol 2. Pour [2 – 50] cm3 ethanol into test tube / boiling tube / round bottom flask // Tuang [2 – 50] cm3 etanol ke dalam tabung uji / tabung didih / kelalang dasar bulat Note: 1. if test tube, [2 – 5] cm3 2. if boiling tube, [2 – 10] cm3 © 2017 Hak Cipta Panitia Kimia SMK Sultan Yussuf

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PPT CHEM P2 TING. 5 (2017)

3. if round bottom flask, [20 – 100] cm3 4. if no volume, no P2 3. Named carboxylic acid eg: (glacial) ethanoic acid / asid etanoik 4. Add [2 – 50] cm3 ethanoic acid to ethanol // Tambah [2 – 50] cm3 asid etanoik kepada etanol Note: 1. sequence of P2 & P4 does not matter i.e whichever comes first or later does not matter 5. Add concentrated sulphuric acid // Tambah asid sulfurik pekat 6. Heat / reflux the mixture // Panaskan / Refluks campuran a: boil (for reflux only) 7. Correct formulae of reactants and products 8. Balanced equation C2H5OH + CH3COOH  CH3COOC2H5 + H2O Max.

1 1

1 1 1 1 7

(c)

1.

2.

3.

4.

Natural rubber Vulcanised rubber Getah asli Getah tervulkan Less elastic // easily stretched More elastic // difficult to but difficult to return to its stretch but easily returns to its original shape original shape Kurang kenyal // Mudah Lebih kenyal // Sukar diregang diregang tetapi sukar kembali tetapi mudah kembali ke ke bentuk asal bentuk asal Non-heat resistant // Tidak More heat resistant // Lebih tahan haba tahan haba a: less heat resistant // kurang tahan haba Vulcanised rubber contains sulphur cross-link which prevents rubber molecule from sliding over one another // Getah tervulkan mengandungi rangkai silang sulfur yang menghalang molekul getah daripada menggelongsor antara satu sama lain Presence of sulphur cross-links makes rubber difficult to melt // Kehadiran rangkai silang sulfur menyebabkan getah sukar melebur Max. TOTAL

1

1

1

1

3 20

Disediakan oleh:

Disemak oleh:

Disahkan oleh:

.........................................................

(CIK INTAN ZAREENA ZAHEDI)

.........................................................

(PN. NURRUL ‘AINNI MD. TENI) Ketua Unit Kimia Tingkatan 6

................................................

(PN. NORLIZA MD DERUS) Guru Kanan Sains

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