chemistry perfect score 2011 module answer

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BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

JAWAPAN MODUL PERFECT SCORE 2011

CHEMISTRY[KIMIA]

Set 1 Set 2 Set 3 Set 4 Set 5

1

JAWAPAN SET 1 PAPER 2 : STRUCTURED QUESTION Section A No. 1 Answer The formula that shows the simplest whole number ratio of atoms of each element in a compound. H2SO4 + Zn ZnSO4 + H2 Heating, cooling and weighing are repeated until a constant mass is obtained. Copper 47.70 25.30 =22.40 Mole atom 22.40 64 = 0.35 Simplest ratio 1 Empirical formula = CuO (e) (f) (g) Element Mass, g Oxygen 53.30 47.70 =5.60 5.60 16 = 0.35 1 4 2 1 Mark 1 2 1

(a) (b) (c) (d)

H2 + CuO Cu + H2O To prevent the hot copper from being oxidized again.

Magnesium ribbon

Heat 2 TOTAL No. 2 Answer (a) (i) Al2CO3 (ii) Al2(CO3)3 Al2O3 + 3CO2 (iii) The number of mole of Al2 (CO3) 3 = 70.2/ 234 = 0.3 mol Based on the balanced equation; Al2 (CO3)3 : Al2O3 1 : 1 0.3 : 0.3 Mass of Ag = 0.4 x 102 = 30.6 g (iv) Based on the balanced equation Al2 (CO3)3 : CO2 1 : 3 0.3 : 0.6 Volume of CO2 = 0.9 x 24 = 21.6 dm3 = 21600 cm3 Zinc carbonate Zinc oxide and carbon dioxide 13 Mark 1 2 1

1 1

1 1 1 1 1

(b) (i) (ii)

2

(iii) ZnCO3 No. 3 (a) (b) (i) (ii)

ZnO + CO2 TOTAL Answer

1 12 Mark 1 1 1

The number of protons found in the nucleus of an atom 7

33 16

Q1 1 1 1

(c) (d) (i) (ii) (e) (i) (ii)

P and S // Q and R Q and R Have same proton number but different nucleon number // Have same number of protons but different number of neutrons Melting point : 63 OC [values & unit must be correct]

Section AB DE Solid

Physical state

Liquid and gas 1 1 1 TOTAL 10 Mark 1 1 1 1

(iii) the heat energy absorbed by the particles is used to overcome the forces of attraction between particles

No. 4

(a) (b) (c) (d) (i)

Answer Sodium and magnesium // sodium and aluminium // magnesium and aluminium Halogen 2.8.3 Sodium, magnesium, aluminium, chlorine, argonAtomic size decreases

(ii)

From left to right : The proton number // the positive charge increases from sodium to argon The forces of attraction by the nucleus on the electrons (nuclei attraction) in the first three occupied shells become stronger Sodium burnt rapidly and brightly with a yellow flame // White fumes liberated // white solid formed

1 1 1

(e)

(i) (ii)

2Na + Cl2 2NaCl [Formula of reactants and product are correct] [Balanced equation] has high melting / boiling point // conduct electricity (iii) in molten state or aqueous solution // soluble in water TOTAL

1 1 1 10

3

No. 5 (a) (i) (ii) (b) (c) (i) X

Answer

Mark 1 1 1 1 1+1W

8 valence electron // electron arrangement 2.8 // achieve octet electron arrangement Covalent

VW4 ( (b) (a) i i (ii) ) (

W

V

W

W

(iii)

has low melting / boiling point // cannot conduct electricity in molten and solid state . // insoluble in water// soluble in organic solvent. Ionic compound Atom U donate one electron to form U+ ion Atom W accept one electron to form W- ion U+ ion and W- ion attracted to each other by strong electrostatic force / ionic bond.

1

(d)

(i) (ii)

1 1 1 1

(iii) 1 1

U

W

[Number of electron each shells are correct] [Number of charge symbol are correct] TOTAL 13

4

PAPER 2: ESSAY QUESTION Section B No. Answer 6 (a) Group 17 Period 3 Has seven valence electrons. Has three shells occupied with electron (b) (i) Between Y and X 1.Atom Y has 1 valence electron and atom X has 7 valence electron 2. to achieve octet electron arrangement 3. Atom Y loses/donates/transfers 1 electron to form ion Y+ 4. Atom X gains/receives 1 electrons from atom Y to form ion X5 Y+ ion and X- ion are attracted by a strong electrostatic force / ionic bond 6. Diagram

Mark 1 1 1 1

1 1 1 1 1

+

Y

X1

(ii) Between W and X 1. Atom W has 4 valence electrons and atom X has 7 valence electrons. 2. Each atom W contributes 4 electrons whereas each atom X contributes one electron for sharing. 3. to achieve octet electron arrangement 4. Four atoms of X share a pair of electrons with one atom W to form a WX4 molecule / Diagram 1

1 1 1

X W X W

X W

W

X W

Molecules WX4

5

(c)

Compound P : ionic bond Compound Q : Covalent bond Melting Point Compound P Ions are held by strong electrostatic forces. More energy is needed to overcome these forces. Compound Q Molecules are held by weak intermolecular forces. Only a little energy is required to overcome the forces. Or Electrical conductivity Compound P In molten state or aqueous solution , there are free moving ions Ions carry charge Compound Q In molten and solid states , no free moving ions exist as molecule TOTAL

1 1

1 1

1 1 1 1 1 1 20

No. 7

Answer (a) (i) (ii) 2.8.7, Chlorine 2Fe + 3Cl2 2FeCl3 Correct formulae of reactants and product Balanced Z,Y,X Z more reactive than X Atomic size of Z bigger than atomic size X Valence electron become further away from nucleus Valence electron to be more weakly pulled by the nucleus Valence electron can be released more easily in atom Z same/similar Same valence electron X : 2.4 Y : 2.6 to achieve octet electron arrangement one X atom contributes four electron and each two Y atoms contributes two electrons for sharing Group 16 Period 2 6 valence electron 2 shells occupied with electrons TOTAL

Mark 1+1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 20

(b) (i)

(ii) (c)

6

PAPER 2: ESSAY QUESTION Section C No. 8 (a) (i) Answer Dilute acid: Hydrochloric acid / Sulphuric acid/ Nitric acid Metal N: Magnesium / zinc Anhydrous calcium chloride To dry the hydrogen gas Mark 1 1 1 1 1 1 1 1 1 1 1

(ii)

(iii) Example: Copper(II) oxide Copper ion is reduced// reduction process Because oxidation number of copper decrease from +2 to 0 Hydrogen is oxidised// oxidation process Because oxidation number of hydrogen increase from 0 to +1 Hydrogen is reducing agent Copper(II) ion// Copper(II) oxide is oxidising agent (b) (i) Relative Molecular mass of (CH2)n = 56 (12 + 2) n = 56 n=4 Molecular formula = C4H8 Unglased porcelain chips

1

(ii)

Glass wool soaked in butanol

Heat

Water2

Procedure:

1. 2. 3. 4. 5. 6.

A small amount of glass wool soaked in butanol is placed in a boiling tube. The boiling tube is clamped horizontally The unglazed porcelain chips are placed in the middle section of the boiling tube. The boiling tube is closed with a stopper fitted with a delivery tube The unglazed porcelain chips are heated strongly. Then, the glass wool is warmed gently to vaporize the propanol. The gas released is collected in a test tube. TOTAL

1 1 1 1 1 1 20

7

No. 9

(a)

(i)

Answer Formula that shows the simplest ratio of the number of atoms for each element in the compound. Copper(II)oxide // lead(II)oxide CuO + H2 Cu + H2O // PbO + H2 Pb + H2O Magnesium oxide / zinc oxide Procedure:

Mark 1

(ii) (b) (i) (ii)

1 1+1 1

1. Clean magnesium / zinc ribbon with sand paper 2. Weigh crucible and its lid 3. Put magnesium ribbon into the crucible and weigh the crucible withits lid

1 1 1 1 1 1 1 1 1 1

4. Heat strongly the crucible without its lid 5. Cover the crucible when the magnesium starts to burn and lift/raisethe lid a little at intervals Remove the lid when the magnesium burnt completely Heat strongly the crucible for a few minutes Cool and weigh the crucible with its lid and the content Repeat the processes of heating, cooling and weighing until a constant mass is obtained 10. Record all the mass

6. 7. 8. 9.

Result: Description Crucible + lid Crucible + lid + magnesium Crucible + lid + magnesium oxide Calculation: Element Mass, g Mole Mg y-x y-x 24 =0.1 1 Mass/g x y z O z-y z-y 16 =0.1 1 1

1

Simplest ratio Empirical formula: MgO (c) Element Mass (%) Number of moles Mole ratio

1 Max 10 C 84.6 84.6/12 =7.05 1 H 15.4 15.4/1 =15.4 21

1 1

Empirical formula : CH2 RMM of (CH2)n = 70 [ 12 + 2]n = 70 14 n = 70 n = 5 Molecular formula : C5H10

1

1 1 20

8

JAWAPAN SET 2 PAPER 2 : STRUCTURED QUESTION Section A No. 1 (a) Cell II (b) (i) Magnesium electrode (ii) e

Answer

Mark 1 1V

1

Magnesium electrode

Copper electrode

(iii) (c)

Copper electrode thicker // Brown solid deposited 1. Correct formulae of reactant and product 2. Balanced equation Cu2+ + 2e Cu Electrical energy to chemical energy Blue colour remain unchange 1. Concentration / Number of mole of Cu2+ ion remain unchanged 2. Rate of Cu2+ ion discharge at cathode is the same as rate of Cu atom ionize at anode TOTAL Answer Iodine r: formula/iodide/iodine gas MnO4 - + 8 H+ + 5 e Mn2+ + 4 H2O +7 +2 reduction Potassium chloride // iron(II) sulphate // [any reducing agent] Zinc 1. Correct formulae of reactant and product 2. Balanced equation 2 Zn + O2 2 ZnO a: 2 J + O2 2 JO K,J, L Predict : no changes r: no reaction Reason : L is more reactive than J/zinc r: more electropositive TOTAL

1 1 1

(d)

(i) (ii) (iii)

1 1 1 1 10 Mark 1 1 1 1 1 1 1 1

2

No. (a) (i) (ii) (iii) (iv) (i) (ii)

(b)

(iii) (iv)

1 1 1 11

9

PAPER 2 : ESSAY QUESTION Section B No. (a) Answer Propanone is a covalent compound Propanone exist as molecule // No freely moving ion in propanone Sodium chloride is an ionic compound Sodium chloride solution has freely moving ion Cell X Voltaic cell C