stpm trials 2009 chemistry answer scheme terengganu

17
Name : ………………………………………………… NRIC : …………………………  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA  JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA MARKING SCHEME This answer paper consists of 17 printed pages CHEMISTRY PAPER 1 & 2 TRIAL 962/1 JABATAN PELAJARAN NEGERI TERENGGANU TRIAL EXAMINATION 2009

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Page 1: STPM Trials 2009 Chemistry Answer Scheme Terengganu

8/3/2019 STPM Trials 2009 Chemistry Answer Scheme Terengganu

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Name : ………………………………………………… NRIC : …………………………

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

 JABATAN PELAJARAN TERENGGANUJABATANPELAJARAN TERENGGANU JABATANPELAJA

MARKING SCHEME

This answer paper consists of  17 printed pages

CHEMISTRY

PAPER 1 & 2

TRIAL962/1

JABATAN PELAJARAN

NEGERI TERENGGANU

TRIAL EXAMINATION 2009

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OTI 2 2009 STPM

 Marking scheme OTI 22009

2

STPM CHEMISTRY 962/1

OTI 2 2009

MARKING SCHEME PAPER 1

QuestionNo.

Answer Explanation

1  A These are the isotopes of iron. Different isotopes have different numbers of 

neutron

2  D Of the 4 gasses given hydrogen gas is most likely to exhibit ideal behavior.

Refer to any graph of ( pV/nRT ) against p.

3  D Carbon dioxide cannot be solidified at 1.0 atm if the temperature is greatly

reduced. At 1.0 atm CO2 exists in both solid and gaseous forms.

4  C Actual electronic arrangement is 1s2

2s2

2p6

3s2

3p1. Total number of 

electron is 13. Thus the proton number is 13

5  C Na+

is the biggest as it has the smallest number of nuclear charge. The

smaller the nuclear charge, the bigger the ionic size.6  D Only phosphorus can exist as monatomic solid with acidic properties in its

oxides.

7  C Across period 3 from left to right, conductivity increases for the first 3

elements, then starts to decrease as the elements become metalloid, and

decreases further as elements are non-metals.

8  C In the CH3CHO molecule, hydrogen atom is bonded to carbon not oxygen.

So no intermolecular hydrogen bonds.

9  B No energy gap between the two bands, so it belongs to a metal, which is

magnesium.

10  C There are only repulsive forces exist between electron orbitals. So the lone

pair electrons push each other so that the angle between them is maximized.

11  A The graph is reversible reaction. The rate of the forward reaction is equal tothe rate of the backward reaction over time. The formation of ester is a

reversible reaction.

12  C e.g.: if the original sample = 100g

100g  → 50g  → 25g  → 12.5g  → 6.25g

t 1/2 t 1/2 t 1/2 t 1/2 

Percentage remaining after the 4th

half life = 6.25 x 100 = 6.25% 

100 

13 B

Since the forward reaction is endothermic, an increase in temperature shifts

the equilibrium to the right according to Le Chatelier’s principle, by

absorbing the excess heat. In the case, more gas particles are produced andthere is an increase in volume. Furthermore, heating will also cause the gases

to expand.

14 A

AgBr (s) Ag + (aq) + Br-(aq)

Ksp = [ Ag+

] [Br-]

= 5 x 10-13

mol2dm

-3 

But [ Ag+

] = [Br-]

[ Ag+ ] = ( 5 x 10-13

)1/2

= 7.1 x 10-7

moldm-3

 

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OTI 2 2009 STPM

 Marking scheme OTI 22009

3

Question

No.

Answer Explanation

15 B

mB = MB x Po

mW MW x Po

mB = 156.6 x (101 x 85)

100 18 x 85

= 163.66

% bromobenzene = 163.66 x 100

(1663.66 + 100)

= 62%

16 

BThe very negative Eo value of Mg shows that Mg2+ is very difficult to bereduced to Mg. Therefore, normal chemical reduction is not suitable. A

feasible way would be the electrolysis of molten Mg2+

, in this case, MgCl2

(MgCl2 has a lower m.p. than MgO).

17 A

Quantity of electricity used = I x t  

= 8 x (100 x 60)

= 48000 C

Amount of electrons passed = 48000C96000 Cmol-1

 

= 0.5 mol

Amount of O2 evolved = 0.5

4

= 0.125 mol

Volume of O2 evolved = 0.125 x 22.4

= 2.8 dm3 

18 A The enthalpy change of formation of CO is the enthalpy change when 1

mol of gaseous CO is formed from its constituent elements, that is C and O2,

at their standard states.

19 A

H = -52 + (-85) = - 137 kJmol-1

 

20 A

For cis-trans isomers, 2 different groups must be bonded to the same C at the

double bond.

B : 1 and 3 are identical

C : 1 and 4 are structural isomersD : 2 and 4 are structural isomers

2O2-

(l) O2(g) + 4e

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OTI 2 2009 STPM

 Marking scheme OTI 22009

4

Question

No.

Answer Explanation

21 C Upon reacting with Br2 , but-1- ene gives 1,2- dibromobutane and

but -2-ene gives 2,3- dibromobutane.22 

BBeing an alkene, hex-1-ene undergoes electrophilic addition readily with Br2 

in CCl4 and decolourised it. Methylbenzene is unable to do so.

CH3(CH2)3CH=CH2 + Br2 CH3(CH3)3CHBr—CH2Br

23  A

24  ARemember: Ethanoyl chloride (acyl chloride) will react with alcohols but

not with carboxylic acids.

25  B CH3CH2COOH All carbonyl compounds, which are aldehyde and ketone

react with 2,4-dinitrophenyihy- drazine to form orange precipitate that has a

high melting point.

26  A Ethanoic acid is a weak acid and it partially ionises in water to produce

hydroxonium ion, H+.

CH3COOH + H20 CH3COO-+ H

If ethanoic acid is dissolved in a base like ammonia that is able to extract the

proton (H+

) from ethanoic acid more effectively, more ethanoic acidmolecules will ionise. Thus H

+ions are produced are the strength of ethanoic

acid will increase. Thus ethanoic is a stronger acid in liquid ammonia than in

water.

27  D The reaction equation is as follows.

Remember: This molecule contains phenyihydrazine carbonyl group.

28  C Hydrogen bonds are formed between ethanol and water molecules.

Thus ethanol is soluble in water.

Whereas ethyl ethanoate (ester) that has a long carbon chain does not formhydrogen bonds with water, thus it is less soluble in water

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OTI 2 2009 STPM

 Marking scheme OTI 22009

5

Question

No.

Answer Explanation

29  C

30  D chloride

An aromatic amine with an amino group at its side chain will have similar

chemical properties as an aliphatic amine.

31 D

At pH 2, an acidic solution and 2-amino- propanoic acid (alanine)will exist

as positive ions.

32  B

The reaction results in a nucleophilic addition to the double bond and theanionic polymerisation process takes place.

33  ANaH – ion, SiH4 - covalent, H2S – covalent, HCl – covalent.

The sodium atom wiyh the biggest size among the four elements has

the lowest ionization energy and can lose electron readily to form

ionic compound.

34  AAluminium is a amphoteric. It si soluble in aqueous sodium

hydroxide.Al2O3 (p) + 2OH

-(aq) + 3H2O(l) → 2[(OH)4]

-(aq)

35  DXCl4 decomposes at room temperature showing that the X-Cl bond is

very weak. Hence, X must have the biggest size in Group 14.

PbCl4(l) →  PbCl2 (s) + Cl2 (g)

36  DThe inert pair effect increases going down the Group 14. Hence, the

+2 state becomes progressively more stable while the + 4 state

becomes less stable.

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OTI 2 2009 STPM

 Marking scheme OTI 22009

6

Question

No.

Answer Explanation

37  C The Lewis structure of N2O4 is as follows :

It is a symmetrical molecule. Hence is non-polar.

There are 3 bond pairs with no lone pairs surrounding the nitrogen

atom. Hence, its shape is triginal planar and not ‘V’ shape.

As can be seen from the Lewis diagram all the electrons in N2O4 are

paired. N2O

4is formed from the dimerisation of NO

2.2NO2 N2O4

38  C The lone pair electrons in NH3 pushes the three bonding pairs closer to

one another. Hence, its bond angle is slightly less than that of a perfect

tetrahedron, 109.5o

.

39  D KI reacts with H2SO4 to produce iodine, which is a purple fume.

KI + H2SO4  → KHSO4 + HI(g)

2HI(g) + H2SO4  → I2(g) + SO2(g) + 2H2O

KI reacts with silver nitrate to produce silver iodide, which is yellow.

KI(aq) + AgNO3 (aq) → KNO3 (aq) + AgI(s)

AgI is insoluble in either dilute or concentrated ammonia.

40  B Heterogeneous catalyst uses their empty d orbitals to form temporary bonds

with the reacting molecules. This process is known as adsorption.

41  B 1 Elevation of boiling point of solvent

2 Depression of freezing point of solvent

42 B(1,2)

According to Le Chatliers Principle,

When OH-ions are neutralized by the acid, position of eq will shift to the

right, making the solid Ca5(PO4)3OH dissolved.

When the PO43-

ions accepts H+

, the eq will shift to the right, making the

solid Ca5(PO4)3OH dissolved.

43 A( 1 )

Halogen is an donating group which decreases the acid strength.( Ka is

greater)

44 D(1,2,3)

Partition law can only be used under the following condition- the solutions must be dilute

- the solute cannot undergo dissociation or association in one solvent and

not the other

- the temperature must be fixed

45  D The Grignard reagent has the general formula, RMgX. With ketones, a

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OTI 2 2009 STPM

 Marking scheme OTI 22009

7

Question

No.

Answer Explanation

tertiary alcohol is produced.

Remember: The C—Mg and C—Li bonds are very polar. When either the

C—Mg or the C—Cl bonds undergoes fission, the carbon acts

as a nucleophile

46  A To form a diazonium salt, nitric(III) acid must react with a primary aromatic

amine solution at a temperature of 0 — 5°C. 2-methylphenylamine, is a

primary aromatic amine.

47  A Going down group 2, the size of the cation increases causing the polarizing

power to decrease. Statement 2 only indicates that the oxides are more stable

than the carbonates. It does not explain about the stability trend of the

carbonates.

48  C Zn(OH)2 and Cu(OH)2 are soluble in excess ammonia.

49  B

50  B Mol ratio of Ni : N : H = 13.98/59 : 13.27/14 : 2.79/1

= 0.236 : 0.947 : 3.79

= 1 : 4 : 16

:. Ratio of H2 : N = 8 : 4 = 2 : 1

The complex does not contain polydentate ligands. Hence, it cannot exhibits

optical isomerism.

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8

STPM CHEMISTRY

OTI 2 2009/TRIAL EXAM

MARKING SCHEME

PAPER 2

SECTION A ( Structured Question )

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 1(a)(i)  The element X contains isotopes.  1 

1(a)(ii)  Lets the percentage abundance of 79

X = x% 

Percentage abundance of 81

X = (100-x)% 

79.99 =

X = 50.5 

Percentage abundance of isotopes79

X = 50.5% Percentage abundance of isotopes

81X = 49.5% 

1 or 1 

Any 2 

1(a)(iii) 

Axis labeled

and unit 1 

Draw a correct

peak  1 

1(b)(i) 

At low pressure, the molecules are far apart , so repulsive forces among the molecules of CO2 

1 1 

1(b)(ii)  Positive deviation  1 

1(b)(iii) 

Forces of repulsion between the H2 gas molecules

which causes the speed of H2 molecule collision with the vessel wall to

increase. 

So the pressure of H2 is higher, causing > 1 (positive deviation)

1 1 

Total 10

79 80 81 m/e 

Abundance(%) 

50.5

49.5

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9

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 2 (a) (i) Zn(s) + 2Ag

+ (aq) → Zn

2+(aq) + 2Ag(s) 1

2 (a) (ii)   Half cell reaction E 0 /V 

Zn2+

(aq) + 2e→ Zn(s) -0.762Ag

+(aq) + 2e → 2Ag(s) +0.8

Zn(s) + 2Ag+

(aq) → Zn2+

(aq) + 2Ag(s)

E.m.f. of cell = 0.80 + 0.76 = 1.56 V

11

1

2 (a) (iii)  E cell — E ceii° - 0.059 log [Zn2+

]

2 [Ag]2 

1

2 (b) (i)

2 (b) (ii) -487 = 178 + 279 + (590 + 1150) + 337 + Lattice energy

Lattice energy = -3021 kJ mol-1

 

5

For every

single level

energyshown,one

mark will be

given

1+1

Total 10

3 (a) (i) CH3COCl + CH3CH2CH2OH → CH3COO CH2CH2CH3 +  HCl

Propyl ethanoate1

1

3 (a) (ii) CH3COCl + C6H5NH2  →  CH3CONH C6H5 + HCl

N-phenyl ethanamide 

1

1

3 (b) Heat with dilute sulphuric acid.

CH3CONH C6H5 + H2O → CH3COOH + C6H5NH2

1

3 (c) Structural formulae of the products :

HOCH2CHOH and NaOOCCH2COONa

│ CH3 

1+1

3 (d) (i) Heat 2-hydroxybenzoic acid with ethanoyl chloride. 1

1

3 (d) (ii) X is aspirin. It is an analgesic. 1

Total 10

Latice energy

CaS(s)

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10

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 4 (a) (i) Dinitrogen oxide, N2O 1

4 (a) (ii) NH4NO3(s) → N2O(g) + 2H2O(l) 1 

4 (b)

Explanation:

Na2O and MgO, ionic with strong electrostatic forces;Al2O3, ionic

with

Covalent character; SiO2 macromolecular with high melting

point;

Oxides of P and S,simples molecules

1

1

1

1

4 (c) (i) 1

4 (c) (ii)  3d electrons absorb energy in the visible wavelengths except the

green wavelength which it reflects and jump to the higher set of vacant

3d orbitals. Therefore, Cr3+

ions appear green.

1+1

1

Total 10

Mg

Melting point of oxides

P SSiAlNa

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11

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 5(a)  The hydrogen emission spectrum consist of discrete lines 

whereas the spectrum produced by the tungsten filament bulb is a

continuos spectrum. 

1 1 

5(b)  1s,2s,2px,2py The difference between 1s and 2s is that the size of 2s is bigger than 1s. or The difference between 2s and 2p is that 2s is spherical but that 2p is dumb-bell shape. 

1 or 1 

………… Max 2 

………... 5(c) (i)  Hund`s rule:

When electrons are placed in a set of orbital with equal energies, the

electrons must occupy them singly with parallel spin before they can

occupy the orbital in pairs. 

Pauli exclusion principle: Each orbital can only be filled with two electrons with opposite spins. 

Aubau principle:

Electrons occupy orbitals in order of the energy levels of the orbitals.

Orbitals with the lowest energy are always occupied first. 

5(c) (ii)  Number of electrons in O2-

ion = 8+2=10 Step 1: Apply Pauli exclusion principle and Aufbau principle.Fill 1s

orbital with two electrons. 

1s 2s 2p Step 2:Fill 2s orbital with two electrons 

1s 2s 2p Step 3: Apply Hund`s rule. 

Fill 2p x,and 2p y orbitals wih three electrons. The electrons must be in parallel spins. 

1s 2s 2p Step 4: Fill the remaining three electrons in 2p x,2p y  and 2pz orbitals .

 Each pair of electrons must be in parallel spins. 

1s 2s 2p 

1 Describe or

show the filling

using a

diagram Any 3 

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12

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 5(d) (i)  Atomic radius is defined as half the distance between the nuclei of two

closest and identical atoms.

5(d) (ii)  When across a period(from lithium to neon), atomic radius decreases. Because the nuclear charge increases(number of protons) But the screening effect remains almost constant as the number of 

shells remain the same. The attraction of the valence electrons by the nucleus increases. 

1 1 1 

Total 15

6 (a)  NH3 Valence electron, N : 5e 

3H : 3e 

------------- 8e 3 bond pair, 1 lone pair 

------------- 

.. N 

Or trigonal pyramidal 1070

NH4+ 

Valence electron, N : 5e 4H : 4e 

Positive charge : -1e (less 1 electron) 

------------- 8e 4 bond pair, no lone pair 

-------------- 

Or tetrahedral 

109.5) 

The bond angle of NH3 is 1070

wheares the bond angle of NH4+ 

Draw or state

the shape 1 

1

Draw or state

the shape 1 

Any 4

H H H

N

H+

H H H

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13

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 6 (b) (i)  H2O 

.. 

Overlapping

1

label

1

6 (b) (ii)C2H4

Overlapping 1 

label 1 

6 (c)  Boiling point of H2O is higher than HF because each HF molecule forms 2 intermolecular hydrogen bonds compared to 4 intermolecular hydrogen bonds formed by each H2O

molecule. 

1 1 

6 (d) (i) 

From the graph, the half- life of the reaction is 13.5 minutes. 

Label the axis

and unit 1 

Plot the graph 1+1 

Indicate on thegraph

t1/2 

H(1 s) 

H(1 s) 

sp3

orbitals 

H(1s)

H(1s)

H(1s)

H(1s)

π 

δ 

C  C 

Volume of KMnO4 

10 20 30 t/min 

40 

30 

20 

10 

O

t1/2= 13.5  t1/2= 13.5 

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14

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 6 (d) (ii)  From the graph, half life doesnot depend on the concentration of 

hydrogen peroxide. Thus, the reaction is first order. 1

Total 15

7(a) When an equilibrium is disturbed then equilibrium will shift in thedirection that will reduce the effect of the disturbance.

2 m

7(b)(i) Yield of NO decreases.

When pressure is increased, equilibrium will shift to the left because

the backward reaction will reduce the number of gaseous molecules.

1 m

1 m

7(b)(ii) Value of Kc increases.

The forward reaction is exothermic, the Kc value decreases with

increase in temperature.

1 m

1 m

7(c) Let x be the number of mol of NO produced

Total number of moles at equilibrium = 0.40 – x + 0.50 – 1.25x + x +1.5x or

1.5xx1.25x0.5x0.4

x

+++

= 30 % 

x =0.29 mol

1 m

1 m

7(d) (ii) (ii)

]

7(e)(i)

]A[

1= kt +

0]A[

k =25

0.20.12 −= 0.40 mol

–1dm

3min

–1 

1 m

1 m

7(e)(ii)

0]A[

1= 2.0

[A]0 = 0.50 mol dm–3 1m

7(f) A + A → M slow step

M → 2B fast step

1 m

1 m

Total 15

Rat

 

[A

 0

A

t0

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15

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 8 (a) Boiling point of tetrachlorides increases down the group

All the tetrachlorides are non-polar and are simple molecules.

Molecular size increases from top to bottom.

Strength of van der Waals forces increases down the group.

1

1

1

1

8 (b) SiC14 undergoes hydrolysis whereas CCI4 does not.

This is because the Si atom has empty d orbitals which can be attacked

by lone pair electrons from water molecules. The product of 

hydrolysis is silicon(IV) oxide and hydrochloric acid.

, C atom in tetrachioromethane does not contain empty

d orbitals, hence does not undergo hydrolysis.

1

1

11

8 (c) (i) Semiconductor

Semimetal, electrical conductivity increases when temperature is

raised.

Ceramic

The Si—O covalent bond in the giant covalent network is strong.

In fire extinguishers. Does not support combustion

1

1

1

8 (d) An aluminium factory has to be located near a port so that the raw

material, bauxite ore, can be easily transported to it. This will reduce

cost.

As the electrolysis of molten aluminium oxide involves the use of a

large amount of electricity, the aluminium factory must be located

close to a hydroelectric dam for a supply of cheap electricity.

These two factors can help to reduce the cost of producing aluminium.

One bad effect on the environment caused by the extraction of 

aluminium is the release of poisonous hydrogen fluoride and fluorine

gas into the air, which are formed during electrolysis of a mixture of aluminium oxide and cryolite.

1

1

1

1

Total 15

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16

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 

9(a) CH3-CH=CH-CH3 

CH3- CH- CH

2-CH

Cl

2m

9(b) A: CH3CH2Cl D : CH2Br- CH2Br

B : CH3CH2OH E : CH2OH – CH2OH

C : CH2=CH2 5m

9(c) Ethanolic KOH 1m

9(d) monomer for polyester 1m

9 (e) - heat with ethanolic silver nitrate

1-iodohexane : yellow precipitate of AgI

1- chlorohexane : white precipitate of AgCl

- heat with ethanolic silver nitrate

Chlorocyclohexane : white precipitate

Chlorobenzene : no precipitate

- add acidified KMnO4 

1- chloro-1- butane : decolourisation of KMnO4 

1- chlorobutane : KMnO4 solution is not decolourised.

1 + 1

1+ 1

1+ 1

Total 15

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17

QUESTION

NO SUGGESTED ANSWERS  SUGGESTED

MARKS 10 (a) Tetrahydroisoquinoline is the stronger base. In tetrahydroquinoline, the

unshared electron pair is delocalised into the aromatic ring, making it

less available to accept a proton, whereas tetrahydroisoquinoline

resembles an alkylamine.

1

1

10 (b) 3,4-Difluorophenylamine is a weaker base than phenylamine because

the two F atoms are electron-withdrawing and they reduce the

availability of the lone pair of electrons on the N atom to accept a

proton.

1

1

10 (c) An ethyl group, CH3CH2 is transferred from triethylaluminium to

titanium(IV) chloride to produce a complex,

Ethene molecules act as Lewis bases and are bonded to titanium which

has vacant d orbitals. These ethene monomers are inserted between the

titanium and the ethyl group to form a polymer.

The process is terminated when a hydrogen atom is added to a titanium

atom, and the poly(ethene) chain is separated from titanium.

1

1

1

1

1

1

10 (d) The poly(ethene) polymer formed by the addition polymerisation

process using the Ziegler-Natta catalyst is a linear polymer and of high

density and high melting point.

1

1+1

10 (e) Proteins and polypeptides.

Nylon has peptide linkages, CONH, which are also found in

proteins and polypeptides.

1

1

Total 15