8 Gan Eng Seng School 2011 - olevelmathsandscience.com · Gan Eng Seng School ... net = Zn(s) + 2...
Transcript of 8 Gan Eng Seng School 2011 - olevelmathsandscience.com · Gan Eng Seng School ... net = Zn(s) + 2...
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:
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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.
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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
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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.
�Option C is correct—Ans
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
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�Option C is correct—Ans
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
�Option C is correct—Ans
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)
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Adding catalyst reduces activation energy but not affect the net enthalpy change.
�Option B is correct--Ans
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
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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
�Option C is correct—Ans
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
�Option A is correct—Ans
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
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�Option B is correct—Ans
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
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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
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�Option A is correct—Ans
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
�Option B is correct—Ans
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
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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)
�Option D is correct—Ans
Q28: (NH4)2SO4 is an important fertilizer
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�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
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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
�Option B is correct—Ans
Q31:
Valance electron of U= 5�Group 5 or (V) eg N, P…
�Option D is correct—Ans
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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.
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� 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)
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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.
�Option B is correct--Ans
Q34: charge carrier
Z metal�sea of electrons
ZO molten state � + & - ions
ZSO4 aqueous state� + & - ions
�Option B is correct—Ans
Q35: C4H9Cl
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Possible isomers are:
1) CH3-CH2-CH2-CH2Cl
2) CH3-CH2-CHCl-CH3
3) CH3-CH-CH2Cl
|
CH3
4) CH3-CCl-CH3
|
CH3
�Option C is correct—Ans
Q36:
2C2H5COOH + 2Na� 2C2H5COO-Na + H2
Propanoic acid
�Option D is correct—Ans
Q37: Fermentation should be carried up in anaerobic condition, P(Oxygen) must
be kept out and Q (CO2) must be relief of excessive pressure
�Option B is correct—Ans
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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
�Option B is correct—Ans
Q39:
C2H5COOH + C2H5OH � C2H5COO-CH2CH3
Propanoic acid ethanol Ethyl-Propanoate
�Option C is correct—Ans
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
�Option D is correct—Ans
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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
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A2:
a)
X=group 4 XH4
Y=group 6 H2Y
Z=group 1 ZH=hydride
b)
H2S2O7
H2S2O7+ H2O � 2H2SO4
A3:
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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
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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.
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A7: Alkynes
CHCH ethyne C2H2
CH3CCH Propyne C3H4
CH3CH2CCH Butyne C4H6
C4H6 + 2H2O� C4H8(OH)2 or HO-CH2CH2CH2CH2-OH
A8:
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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.
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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.
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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.
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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)
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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
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