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Gan Eng Seng School

Pure Chemistry Prelim Exam Paper 1-2011

Q1: Diffusion

�Option C is correct—Ans

Note: random collision of gases particles scatter the Br2 molecules throughout

the cylinder.

Q2: BP to decide formation of liquid phase, to get O2, cool to below the BP of O2

but above BP of argon & N2: between -183 to -186 exclusive

�Option B is correct--Ans

Note that mr of the 3 gases increase from:

N2 (14) to O2 (16) to Ar(18) so there is no easy logic to remember why Ar BP is

lower than O2…supposing larger molecule has larger intermolecular Van Der

Waals forces.

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Q3:

A) False: some element can form 2 oxides eg CuO, Cu2O, H2O, H2O2

B) False: some element can dissolve in water eg O2, Cl2 gas.

C) False: pure element or compounds had fixed melting point.

D) True: element cannot be further separate into substance by electrolysis.

� D is correct—Ans

Q4:

A) True:

Br2 + 2KI� 2KBr + I2 (aq), SO2 react with K2Cr2O7 as RA, O2 insoluble in water

thus collected as final gas

B) False:

Cl2 + 2KI� 2KCl + I2 (aq), SO2 react with K2Cr2O7 as RA, NH3 very soluble in

water thus no gas was collected

C) False:


Cl2 + 2KI� 2KCl + I2 (aq), CO2 does not react with K2Cr2O7, HCl very soluble in

water CO2 partially soluble thus little gas was collected

D) False:

CH4 & N2 insoluble in water and do not react with both soluble thus both will be

collected.

�Option A is correct—Ans

Q5:

Both H & D have same charge level of +1 thus similar chemical properties but

slightly different physical properties� Ba(OH)2 & Ba(OD)2

�Option B is incorrect—Ans

Q6: YCl2

Gaseous compound with chlorine� simple covalent compound� Y is non-metal

element� Group 6 element

�Option A is correct—Ans, B & C form solid ionic compounds, D is uncommon.


Q7:

A: False: only metal had this metallic bond structure

B: False: should not compare metal and metal oxide as the structure and bond

type are different.

C: False: CaO is not a macromolecular covalent compound

D: True: Ionic bond or called electrovalent bonds consist of lattice of + & - ions

held together by electrostatic forces.

�Option D is correct

Q8:

Counting the number of bond to achieve noble gas configuration

X=3 shared pairs + 1 lone pair�group 5 element such as N,P

Y= 4 shared pair�group 4 element such as C, Si

Z=1 shared pair + 3 lone pairs�group 7 element such as Cl, Br or Hydrogen H

�Option B is correct –Ans


Q9:

Assume 100% dissociation, mole of H+ = 3* 0.05=0.15 mol/dm3, however, the pH

value will not be according to the formula –Log10[H+] where [H+] in 10^pH.


Q10:

Mol of NaOH=3*0.01=0.03 mol

Mol of xCln=1.5*0.01=0.015 mol

Mol xCln : Mol NaOH=1: 2 � n=2



Q11:

Mol H2=10/2=5 mol

Mol O2=5/2=2.5 mol=2.5*24=60 dm3

Mol H2O= 5 mol =5 *(16+2)=90g


Q12:

Reaction 1 endothermic as there is only bond breaking

Reaction 2 exothermic as there is only bond forming

Reaction 3 is exothermic as there is only bond forming

�Option D is correct—Ans

Q13:The process is exothermic (NOT endothermic as the question stated)


Adding catalyst reduces activation energy but not affect the net enthalpy change.


Q14:

A: True as indicated by the displacement reaction

B: True as the process is exothermic

C: False: X is oxidized by loosing 2e-

D: True: temp increase as the process is exothermic

�Option C is incorrect—Ans

Q15:

At Cathode: where e- were received by +ions thus R must be Positive ion with

reactivity less than hydrogen.

�Option B is correct—Ans

Q16:

A: 4*3=12 mol cathode

B: 5*1=5 mol Anode

C: 4*2=8 mol Anode

D: 10*1=10 mol Cathode

�Option D is largest—Ans


Q17: Product of electrolysis

A: CuSO4 cathode: Cu(s) anode: O2

B: Conc ZnCl2 cathode: H2 anode: Cl2

C: HNO3 cathode: H2 anode: O2 volume of H2:O2=2:1

D: Propanol is non-conductive


Q18:

A: Mg oxidized thus it is RA

B: Not a redox but acid/base reaction

C: Cl reduced thus it is an OA

D: ZnO reduced thus it is an OA


Q19:

A: Cl from -1 to +5

B: Se from +6 to +6

C: Cr from + 3 to +2

D: Mn from +4 to +2



Q20:

2H2O2� 2H2O + O2

X: Mol of H2O2 = 0.02*1.5=0.03 mol

A: same concentration thus same SOR lower O2 produced

B: higher conc higher SOR with 0.0125 mol thus lower O2 produced

C: lower conc lower SOR with 0.015 mol thus lower O2 produced

D: higher conc thus higher SOR with 0.015 mol thus lower O2 produced

�Option B/D possible but D is better as mol ratio is ½ so O2 produced is ½ as

well.--Ans

Q21: compare g/cc concentration

A) 0.04

B) 0.03

C) 0.1

D) 0.08

�Option C is highest conc thus highest initial SOR –Ans

Q22: Amphoteric

A: Alkali property

B: React with KOH to form salt + H2O, Amphoteric


C: Acid/base reaction

D: Acid/base reaction

�Option B is Amphoteric—Ans

Q23:

A) True: Acid + carbonate� salt + water+CO2

B) False: Alkali + Ammonia salt� NH3…..

C) False: acid has pH<7

D) False: some concentrated acid react with metal to produced other gases

Reactions of Metals with Acids:

* nonoxidizing acids (HCl, dilute H2SO4) result in a single replacement reactions

Metal + acid → salt + H2(g)

e.g. Zn + H2SO4 → ZnSO4(aq) + H2(g)

net = Zn(s) + 2 H+

(aq) → Zn 2+(aq) + H2(g)

* oxidizing acids (hot concentrated H2SO4, HNO3) form additional products

besides

the salt + water.

w/ dilute HNO3 → NO

conc. HNO3 → NO2

conc. H2SO4 → H2S or SO2



Q24:

A) NH4OH + HNO3 �NH4NO3 + H2O titration method

B) Ba(NO3)2 + Na2SO4 � BaSO4 (s)+ 2 NaNO3 ppt method

C) Fe2(CO3)3 (s) + 6 HCl (aq) �2 FeCl3 (aq) + 3 CO2 (g) + 3 H2O (l)

D) 2K + 2HCl� 2KCl + H2 explosive reaction not suitable

Option D is incorrect—Ans

Q25: Soluble basic oxides

A) CO2 acidic

B) MgO + H2O� Mg(OH)2 marginally soluble less than Ca(OH)2

C) CuO insoluble

D) SO2 acidic


Note:

For each metal oxides, there is some degree of solubility given by the solubility

constant Ksp, here is an example of calculating pH for Mg(OH)2

solution….however it seems to be beyond O level scope


Q26:

�Option B 30% O2 in the bottle air sample—Ans

Q27:

Methane & CO2: Greenhouse gas

SO2 & CO2 : acid rain (effect of CO2 much less than SO2)


Q28: (NH4)2SO4 is an important fertilizer


�Option A is correct

Q29:

Fe(s) + CuCl2 (aq)� FeCl2 (aq) + Cu(s)

(Blue )

Fe2+ � light blue color Cu �pinkish brown solid

�Option A is correct (Transition metal ions typically has color except Zn )--Ans


Q30:

Mg as sacrificial metal act as anode to protect iron pipe

Mg(s) � Mg2+ + 2e at Anode, it is oxidized thus act as RA


Q31:

Valance electron of U= 5�Group 5 or (V) eg N, P…



Q32:

A) False : Metal oxide is ionic in nature

B) False: group 1 oxide is alkali

C)True: group 1 carbonate dissolve in water

Eg K2CO3

K= ions are strong base stay as free ions but CO32- ions are weak acid thus tend to

combine with H+ ions to form H2CO3 (low dissociation %), the overall pH value

increases.

Salt Hydrolysis, acidity and alkalinity of salt solutions

• Despite being taught at lower academic levels that salts e.g. sodium chloride, dissolve in water to form neutral solutions of pH 7.

• In reality, and looking at a wider variety of 'salts', the picture is much more complicated and a 'salt' solution may be acid, neutral or alkaline depending on the nature of the interaction of the salt ions with water.

• The reasons are quite clear when you consider the possible Bronsted-Lowry interactions that can take place between the ions of the salt and water.

• 6.1.1 Examples of acidic salt solutions: pH <7 o 6.1.1a: Hexa-aqa metal cations often show acidic behaviour particularly with their salts

with strong acids. � e.g. the hydrated aluminium ion from aqueous aluminium

chloride/sulphate/nitrate. � [Al(H2O)6]

3+(aq) + H2O(l) [Al(H2O)5(OH)]

2+(aq) + H3O

+(aq)

� or more simply: [Al(H2O)6]3+

(aq) [Al(H2O)5(OH)]2+

(aq) + H+

(aq) � The hydrated metal ion acts as an acid (proton donor) and water acts as the

base (proton acceptor) and so aqueous hydrogen/oxonium ions are formed.


� The greater the charge on the central metal ion, the stronger the hexa-aqua ion acid. e.g.

� [Al(H2O)6]3+

(aq) > [Mg(H2O)6]2+

(aq) > [Na(H2O)6]+

(aq) (the cations of Gps 1-3 on

Period 3) � or [Fe(H2O)6]

3+(aq) > [Fe(H2O)6]

2+(aq) (in the 3d-block transition metal example)

� From left to right, the trend is due to a decreasing charge density effect of the central metal ion on the O-H bond of a co-ordinated water molecule. The charge density decreases as the positive charge of the central metal ion decreases and its ionic radius increases.

� The sodium ion shows virtually no acidic behaviour. � Further discussion of this situation will be on the Transition Metals

Appendix page section 1 (currently under production). � However, the anion of the salt must not be neglected for a full explanation. The

anions derived from the very strong hydrochloric/sulphuric/nitric acids are all very weak bases and so have little tendency to interact with water in an acid-base reaction. Its a general, and logical rule, that the conjugate base of a very strong acid is very weak.

o 6.1.1b: Salts of weak bases and strong acids give acidic solutions. � e.g. ammonium chloride. The chloride ion is such a weak base that there is no

acid-base reaction with water, but the ammonium ion is an effective proton donor. As a general rule, the conjugate acid of a weak base is quite strong. The result here is that ammonium salt solutions have a pH of 3-4.

� NH4+

(aq) + H2O(l) NH3(aq) + H3O+

(aq) � In zinc-carbon batteries an acidic ammonium chloride paste dissolves the zinc in

the cell reaction, though an oxidising agent must be added (MnO2) to oxidise the hydrogen formed into water, or batteries might regularly explode!

� If you place a piece of magnesium ribbon or a zinc granule in ammonium chloride or ammonium sulphate solution you will see fizzing as hydrogen gas is formed.

� 2H3O+

(aq) + M(s) ==> M2+

(aq) + H2O(l) + H2(g) � M = zinc or magnesium

• 6.1.2 Examples of nearly neutral salt solutions: pH approx. 7 o 6.1.2a: Salts of strong acids and strong bases e.g. sodium chloride

� Here the hexa-aqua sodium ion shows no acidic behaviour and the chloride ion no base behaviour, so little or no interaction with water to produce either H

+(aq) or

OH-(aq) to change the pH.

o 6.1.2b: Salts of weak acids and weak bases: e.g. ammonium ethanoate � Here the ammonium ion can act as an acid to form H

+(aq) with water, but the

ethanoate ion acts as a base to give OH-(aq) with water, so they effectively

neutralise each other. • 6.1.3 Examples of alkaline salt solutions: pH>7

o 6.1.3a: Salts of a weak acid and a strong base e.g. sodium ethanoate � The hydrated sodium ion shows no acidic character but the ethanoate ion is a

strong conjugate base of a the weak ethanoic acid (pKa = 4.76, Ka = 1.74 x 10-5

mol dm

-3), so an acid-base hydrolysis reaction occurs to generate hydroxide ions

to raise the pH to about pH 9. � CH3COO

-(aq) + H2O(l) CH3COOH(aq) + OH

-(aq)

o 6.1.3b: Potassium cyanide: is the salt of the very strong base potassium hydroxide and the very weak hydrocyanic acid (pKa = 9.31, Ka = 4.9 x 10

-10 mol dm

-3). The hydrated

potassium ion shows no acidic behaviour, but the cyanide ion is a strong conjugate base of the very weak hydrocyanic acid (HCN) which interacts with water to generate hydroxide ions. Hydrocyanic acid (pKa = 9.4) is weaker than ethanoic acid (pKa = 4.76) , so the equilibrium is more on the right, more OH

-, and so the pH is more alkaline, i.e.

over 9. � CN

-(aq) + H2O(l) HCN(aq) + OH

-(aq)


o 6.1.3c: Sodium carbonate is the 'salt' of the strong base sodium hydroxide and the very weak 'carbonic acid'

� Again the hydrated sodium ion shows no acidic character but the carbonate ion is a strong conjugate base of a the weak 'carbonic' acid, so an acid-base hydrolysis reaction occurs to generate hydroxide ions to raise the pH.

� CO32-

(aq) + H2O(l) HCO3-(aq) + OH

-(aq)

D) False: strong RA as they are good donar of e-

�Option C is correct--Ans

Q33:

A: False: Bauxite is for extraction of aluminum not iron

B: True

C: False: Molten iron heavier than slag thus collection at E

D: False: Iron ore reduced by Coke & CO instead of oxidized.


Q34: charge carrier

Z metal�sea of electrons

ZO molten state � + & - ions

ZSO4 aqueous state� + & - ions


Q35: C4H9Cl


Possible isomers are:

1) CH3-CH2-CH2-CH2Cl

2) CH3-CH2-CHCl-CH3

3) CH3-CH-CH2Cl

|

CH3

4) CH3-CCl-CH3

|

CH3


Q36:

2C2H5COOH + 2Na� 2C2H5COO-Na + H2

Propanoic acid


Q37: Fermentation should be carried up in anaerobic condition, P(Oxygen) must

be kept out and Q (CO2) must be relief of excessive pressure



Q38:

Miscible with water� the molecule must be polarized, hydrocarbon (A) is not

polarizes

A: Non-polar ==Insoluble in water, decolorize Br2

B: Weak acid==Soluble in water and decolorise Br2 due to C=C

C: Weak acid==Soluble in water, cannot decolorise Br2

D: Nonpolar== Insoluble in water, decolorize Br2


Q39:

C2H5COOH + C2H5OH � C2H5COO-CH2CH3

Propanoic acid ethanol Ethyl-Propanoate


Q40:

Tracing back the monomer based on the attachment group to the 2 Carbon atoms

C=C

F and CH3 attached to same C

H & C2H5 attached to the other C



Gan Eng Seng School

Pure Chemistry Prelim Exam Paper 2-2011

A1: Note A is sulphur in S8 allotropes

a) A & B are elements

b) D is a simple molecular compound, E is giant molecular structure

c) D will have the lowest MP due to weak intermolecular bonding of the small

molecule (eg H2O). (A) is larger molecule thus Van De Waal force is larger than (D).

d) B represents metal iron

e) C is ionic compound which conduct electricity in molten state or aqueous

solution


A2:

a)

X=group 4 XH4

Y=group 6 H2Y

Z=group 1 ZH=hydride

b)

H2S2O7

H2S2O7+ H2O � 2H2SO4

A3:


a) X is non-metallic as metal is not a powder at rtp and the oxide is not gaseous at

rtp.

b) reaction equation is :

2X + mO2 � 2XOm

Mol ratio of O2 / XOm=3/2=m/2

� M=3

Oxides produces is XO3� X from group 6 as well eg SO3, SeO3 etc

A4:

Tube A

Zn (s) + H2O (g)� ZnO (s) + H2 (g) ZnO yellow when it is hot


Tube B

H2 (g) + CuO (s) Black � Cu(s) pinkish brown + H2O water vapor appear and

condensation at end of tube

A5:

a) 2I- (aq) + 2H+ (aq) + H2O2 (aq) � I2 (aq) + 2H2O

b) Iodine was oxidized from -1 to 0 thus H2O2 (hydrogen peroxide) is the OA

c) Brown solution ( with black particles ) appear due to formation of I2 diatomic

molecule in the solution.

A6:

a) H2SO4 is di-protic compared to HCl is mono-protic, thus higher concentration

of H+ in H2SO4 resulting in high SOR

b) The transition metal Cu ion ( exist as CU(I) & CU(II) )act as catalyst for the

reaction speeding up the reaction by reducing the activation energy.

c) The total volume of H2 will be the same as catalyst will not affect the amount

of products formed.


A7: Alkynes

CHCH ethyne C2H2

CH3CCH Propyne C3H4

CH3CH2CCH Butyne C4H6

C4H6 + 2H2O� C4H8(OH)2 or HO-CH2CH2CH2CH2-OH

A8:


a) |-OCHCH3COO-|-CH2CO-| ester linkage

b) Condensation polymerization = smaller molecules such as H2O or HCl were

eliminated during the process

c) The other monomer is: HO-CH2-COOH

A9:

a)

Electrode Y provide electrons thus is anode, electron flow from Y to X in the

external circuit.

The NaOH is the electrolyte in this fuel cell which allow free movement of the ions.

Overall reaction is:

2H2 (g) + O2 (g)� 2H2O(l) exactly reverse of electrolysis of weak acid.


Volume of O2=1/2 of H2=1/2*9.6=4.8 dm3

The fuel cells take in H2/O2 continuously thus is able to provide electricity

continuously for long duration.

b) Cathode : 2 H+ (aq) +2e- � H2 (g)

Anode: 4OH- (aq) � 2 H2O (l) + O2 (g)+ 4e-

Oxidation of graphite with O2 produced at anode causes reduction in mass of the

anode electrode C + O2�CO2

CO2 + Ca(OH)2� CaCO3 (s) + H2O (l)

A10:

a) CuCO3 (s) � CuO(s) + CO2 (g) decomposition of carbonate under heat

CuO ionic compound is insoluble in water and is non-conductor….only conduct

when heated to molten state

b)

C (s) + 2CuO (s) � CO2 (g) + Cu(s)

CuO was reduced by carbon which is more reactive than copper.


c)

Reaction III

CuCO3 (s)+ H2SO4 (aq)� CuSO4 (aq) + H2O (aq) + CO2 (g)

use excess carbonate�evaporate filtrate to obtain saturated

solution�Crystallize to get CuSO4� filter� dry

Reaction IV:

CuO (s)+ H2SO4 (aq)� CuSO4 (aq) + H2O (aq)

use excess CuO�evaporate filtrate to obtain saturated solution�Crystallize to

get CuSO4� filter� dry

Reaction V:

Cu(s) + (Ag)2SO4 (aq)� CUSO4 (aq) + 2Ag (s)

use excess Cu�evaporate filtrate to obtain saturated solution�Crystallize to get

CuSO4� filter� dry

Note : making CuSO4 from conc H2SO4 + Cu is not safe in the lab due to SO2

formation.

d) Cu2+ -SO4 2- The bonding between Copper ion and SO4 ion is ionic, however,

within the SO4 ion, the bonding between S & O is covalent.


A11:

a) Experiment (II) involved insoluble CaSO4 formation which will deposit over the

carbonate preventing full reaction of the carbonate with H2SO4.

b)

step 1: CaCO3 (1 mol) + HNO3 (excess) � Ca (NO3)2 (aq) + H2O (aq)+ CO2 (g)

step 2: add excess K2SO4 (aq) solution to the solution to precipitate out all the

CaSO4

step 3: filter the solid CaSO4 and rinse and dry

c) 1 mol of CaCO3 contain 1 mol of Ca2+ ion , maximum yield means 100%

recovery of the Ca2+ ion in the form of CaSO4

�mass of CaSO4=40+32+16*4=136 g

d) CaO dissolve into water to form Ca(OH)2 which react with ammonia salt to

give NH3 gas

Overall reaction:

2NH4NO3 + CaO + H2O � Ca(NO3)2 (aq)+ 2NH3 (g) + 2H2O (l)


Most NH4+ will be lost in the form of NH3 gas instead of available for plant.

OR

A11

a) NH3 gas (A), AgI (B), Fe(OH)3.

b) M contain Fe3+, NO3-, I- & Zn or Al ions

c) I- (aq) + Ag+(aq) � Ag-I yellow ppt

Fe3+ + 3OH- +xH2O �Fe(OH)3.xH2O brown ppt.

Zn(OH)2 or Al(OH)3 re-dissolve into excess NaOH


d) 2Fe(OH)3 �Fe2O3 +3H2O dehydration reaction �Rusting

e) The original solution contain a mixture of ZnI2 & Fe(NO3)3 salt.

End of paper


8 gan eng seng school 2011 - olevelmathsandscience.com · gan eng seng school ... net = zn(s) + 2...

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