Chemistry Paper 3 (Mon, 5 Nov 07, 8-10am)

This is a reminder for students taking GCE A Level H2 Chemistry (9746).

Chemistry Paper 3

Date: 5 Nov 07, Mon

Time: 0800 - 1000

Venue: Refer to notice boards

Format of Paper: Free Response, answer 4 out of 5 questions (80 marks)

Items to Bring:

• Entry Proof (Not confirmation slip! Duh!)
• EZ Link card
• Graphic / Scientific calculator (optional: extra batteries)
• Stationery (including curve rule)

Data Booklet will be provided by examiners. DO NOT scribble anything on the booklet.

Some Advice:

• Do invest 2 to 5 min to read through the whole paper and decide which question you want to answer first.
• Plan your answer. Save time by not writing excessively or in a haphazard manner.
• Start each question on a fresh sheet of paper. No correction tape or fluid.
• Circle the question you have answered and enter the question number on the cover page (if applicable).
• Spend about 1.5 min for every 1 mark given.

Have confidence in yourselves. All the best!

2007 NJC Prelims Paper 3 FAQs

A Note for Q1b
2 mol of OH- reacts with 1 mol of A to give a white ppt B. The thermal decomposition of B yields a white ppt C. Thus M is likely to be a Group II element.

NOTE: If M = Zn, you will also get similar observations (except that the mass of B formed is NOT 0.0215 g).

Explanation for Q3b(i)
The pH of the solution is 2.5, so [H+] = 10^(-2.5) mol/dm3.
The initial concentration of CrO42- is 1.00 x 10^(-3) mol/dm3.
2 mol of CrO42- react with 2 mol of H+ to form 1 mol of Cr2O72-.
Thus, [CrO42-] : [Cr2O72-] = (1.00 x 10^(-3) - 2y) : y = 200 : 1
[Cr2O72-] = y = 4.98 x 10^(-6) mol/dm3
[CrO42-] = 1.00 x 10^(-3) - 2[4.98 x 10^(-6)] = 9.90 x 10^(-4) mol/dm3
[H+] = 10^(-2.5) - 2[4.98 x 10^(-6)] = 3.15 x 10^(-3) mol/dm3

2007 VJC Prelims Paper 2 FAQs

Explanation for Q2c
The temperature at which Rb melts is 39.5 deg C or (273 + 39.5) K. This is the temperature at which an eqm is established between Rb(s) and Rb(l). When the two species are at eqm, delta G of fusion is zero. delta G = delta H - T x delta S. Hence, delta H = T x delta S.

Use delta H from the table and substitute T = 273 + 39.5 to find delta S.

When delta G is +ve (T < g =" 0" t =" 39.5"> 39.5 deg C), melting of Rb takes place spontaneously.

A Note for Q3
The question told you that Ni(OH)2 and Cd(OH)2 are formed during discharge (i.e. when the cell behaves as a voltaic cell). Thus, electrode Y (Cd electrode) must be corroded to form Cd(OH)2 and since Cd is oxidised, electrode Y is the anode.

When the battery is being recharged, Cd(OH)2 must change back to Cd. Hence, Cd2+ is reduced at electrode Y. The cell has become an electrolytic cell! When all Cd(OH)2 has changed back to Cd, the electric current will bring about the electrolysis of KOH(aq). In this case, electrode Y will be where H+ ions get reduced.

2007 TJC Prelims Paper 3 FAQs

A Note for Q2e
Fe(III) is stabilised by CN- ligands, so it is less likely to reduce to Fe(II) in aqueous solution.
Another example: [Ni(NH3)6]2+ vs [Ni(H2O)6]2+

2007 PJC Prelims Paper 3 FAQs

A Note for Q3c
"Do I have to memorise the colours of all transition element complexes?"
The answer is NO!

Only for Cr, Mn, Fe and Cu. Examples: Cr3+, CrO42-, Cr2O72-, Mn2+, MnO2, MnO4-, Fe2+, Fe3+, Cu2+, [Cu(NH3)4]2+.

Explanation for Q5b
NaX + H2SO4 --> NaHSO4 + HX, X = Cl, Br or I
This is NOT a redox reaction!!!

SO42- does not oxidise Cl- to Cl2.
SO42- + 4H+ + 2e- <--> SO2 + 2H2O, E = +0.17V
Cl2 + 2e- <--> 2Cl-, E = +1.36V
Ecell = +0.17 - (+1.36) = -1.19V (< 0)

SO42- oxidises HBr to Br2,
eqn: 2HBr + H2SO4 --> Br2 + SO2 + 2H2O

SO42- oxidises HI to I2,
eqn: 2HI + H2SO4 --> I2 + SO2 + 2H2O or 8HI + H2SO4 --> 4I2 + H2S +4H2O
SO2 has an irritating (or pungent) smell while H2S smells like rotten eggs.

2007 HCI Prelims Paper 3 FAQs

Explanation for Q4b(i)
CrCl3.6H2O is the general formula for a series of complexes formed.
The complexes can be [Cr(H2O)6]Cl3, [Cr(H2O)5(Cl)]Cl2 . H2O, etc.
Ag+ ions can only precipitate AgCl with the Cl- ions that are NOT bonded to Cr(III).

Explanation for Q5a(i)
Imagine that you have a "half-cell" containing Fe2+ ions (in acidic medium) initially. This is connected to a standard hydrogen electrode via a salt bridge. Now, add Cr2O72- dropwise to Fe2+ so that Fe3+ is formed. Cr2O72- added before equivalence point will change to Cr3+; there is no excess of Cr2O72-. The mixture contains Fe2+ and Fe3+. At X, [Fe2+] = [Fe3+], so we can ESTIMATE the E value to be +0.77 V using "Fe3+ + e- <--> Fe2+, E = +0.77V". After equivalence point (at 16.5 cm3), there is no more Fe2+ (all oxidised to Fe3+). When excess Cr2O72- is added into the mixture, there will be Cr3+ and Cr2O72-. At Y, [Cr3+] = [Cr2O72-], so we can ESTIMATE the E value to be +1.33 V using "Cr2O72- + 14H+ + 6e- <--> 2Cr3+ + 7H2O, E = +1.33V".

Do note that all concentrations (inclusive of H+) must be 1 mol/dm3 to measure the suggested E values at X and Y.

2007 ACJC Prelims Paper 3 FAQs

A Note for Q1d(ii)
Hot alkaline aqueous iodine (i.e. I2(aq), Na(OH)(aq), heat) will not distinguish the compounds because there is neither methyl ketone nor RCH(OH)CH3 after alkaline hydrolysis.

Explanation for Q2a(i)
M3+ + 3e- <--> M, E= x V
A more positive x means that the reduction of M3+ to M is more feasible. It also implies that M3+ oxidises other substances more readily (while M3+ itself is reduced).

Explanation for Q3b(ii)
Relative oxidising power of halogens can be compared using:
(1) E values of X2/X-, e.g. E of F2/F- is more +ve than Cl2/Cl-, so F2 is more oxidising than Cl2
(2) delta H of "X2 + 2e- --> 2X-", e.g. delta H is more negative for X = Cl than X = Br, so the reduction of X2 to X- is more energetically feasible for X = Cl, i.e. Cl2 oxidises other substances more readily because Cl2 is reduced more readily
(3) displacement, e.g. Cl2 displaces Br-(aq) from solution
(4) oxidation state of product formed, e.g. Cl2 oxidises S2O32- to SO42- but I2 oxidises S2O32- to S4O62-

2007 NJC Prelims Paper 2 FAQs

Explanation for 2c
Total pressure = Partial pressure of O2 + Partial pressure of H2O
Partial pressure of O2 = 101 kPa - 42.3 kPa = 58.7 kPa
Let n be no. of mol of O2 collected in jar.
(58.7 x 10^3)(87 x 10^-6) = n(8.31)(30+273)
n = 2.03 x 10^-3

Explanation for 3b(ii)
In 3b(i), you found amount of heat required to raise the temperature of the calorimeter by 1 degree Celsius.
In 3b(ii), temperature rise = 2 degree Celsius. Find amount of heat evolved. That is for 0.5 g of aspirin burnt. How much heat is evolved if 1 mol of aspirin is burnt?

Explanation for 5c(ii)
0.12 mol of Cl2 take part in both reactions (1) and (2). Let x mol of Cl2 react in (1) and 0.12-x mol of Cl2 react in (2). Solve for x and hence evaluate 0.12-x.

2007 MJC Prelims Paper 2 FAQs

Explanation for Q3b(ii)
Electron to be removed from K+ is from 3p orbital whereas electron to be removed from the other elements shown is from 4s orbital, which is of higher energy. Hence, K+ has highest 2nd IE.

Explanation for 3d(iv) no. 2
A is unlikely to be HOOC-C(triple bond)C-COOH.
Too difficult to dehydrate (HOOC)CH=C(OH)(COOH). (NOT in syllabus)

Explanation for 4e
K = [butylparaben (organic)] / [butylparaben (aq)] = 1 x 10^3
Partition coefficient is referring to the equilibrium constant, K.

Explanation for 5e
Each "building block" of the protein contains 6 amino acids, which must be linked by 6 amide groups (5 in between + "1/2" at each end).
Hence, 6 molecules of H2O are lost to form one "building block".
Thus, Mr of each "building block" is 2 x 75 + 146 + 3 x 105 - 6(18) = 503.
Each amino acid has one side chain, which is the "residue".

2007 HCI Prelims Paper 2 FAQs

Explanation for Q1c
Hydromagnesite decomposes to give MgO, CO2 and H2O. MgO dissolves in excess HCl(aq) to give MgCl2 and H2O. Excess HCl(aq) added is neutralised using NaOH(aq).

Explanation for Q3a(iii)
The buffer in blood has a large reservoir of H2CO3 (weak acid) and HCO3- (conjugate base of weak acid).
When H+ is added, HCO3- reacts with all the excess H+ and converts them to H2CO3.
Thus, your task is to find the amounts of H2CO3 and HCO3- before and after the reaction. Then use pH = pKa + lg{[HCO3-]/[H2CO3]}. Note that [HCO3-] = 20[H2CO3] from 3a(ii).

Explanation for Q4b(iii)
A colourless complex is formed. [Ag(S2O3)2]3-.
Note: Ag+ also forms colourless [Ag(CN)2]- complex in excess CN-.

Explanation for Q5d(i)
When [phospholipid] is high, the rate is constant because there are not enough enzyme molecules to make the reaction go faster (enzyme = catalyst, i.e. small amt used).

Explanation for Q6a
HCOOH and (COOH)2 are easily oxidised to CO2 by strong oxidising agents.

Explanation for Q6d(i)
CH3O-Na+ is a strong base. CH3O- is like the HO- (hydroxide) ion. Thus, one of the products of the acid-base reaction is methanol, CH3O--H (HO--H if hydroxide is used).

Explanation for Q6d(ii)
RCOO- is charged and is nucleophilic. CH3I has a slightly electron deficient C atom, so CH3I will undergo (SN2) nucleophilic substitution.

2007 VJC Prelims Paper 1 FAQs

Explanation for Q3
ClO2- is bent, i.e. O=Cl--O-, where Cl has 10 e- around it (inclusive of 2 lone pairs) and O bears the negative charge. Bond angle of ClO2- is similar to that of H2O. About 105 degrees.

Explanation for Q32
Fix T. Then draw a vertical line upwards to cut both curves and extrapolate horizontally to get values for fraction of X. At a fixed T, more X is present when pressure is lower (2 x 10^7 Pa). By LCP, X must be on the side of the equation with more gas molecules.

The shapes of the curves tell us that when T increases, more X is present. By LCP, an increase in temperature favours an endothermic reaction. Thus, X must be the product of an endothermic reaction.

Explanation for Q37
Ethene must attack Br2 first to give a carbocation intermediate. After that, Cl-, H2O and Br- (from the original Br2) can attack the carbocation. Hence, the products must have Br in them.

2007 PJC Prelims Paper 1 FAQs

Explanation for Q32 (Errata)
Electronegativity: Na < Mg < Al << O
Difference in electronegativity of elements in oxides: Na2O > MgO > Al2O3 (i.e. third statement is TRUE)
Answer is A.

2007 NJC Prelims Paper 1 FAQs

Explanation for Q7
p x 1.01 x 10^5 x V x 10^-6 = (m/M)RT
p x 1.01 x 10^-1 = (m/V)(RT/M)
density = m/V = 0.101pM/(8.31T) = 273pM/22400T

Explanation for Q8
An acid dissociation reaction is shown. It is a heterogeneous eqm where H2O is a solvent (i.e. in great excess), so [H2O] is NOT in Ka expression. Ka = ([H3O+][S2-])/[H2S].

NH4+ + H2O <--> NH3 + OH- and OH- reacts with H3O+ of the eqm.
CO32- also reacts with H3O+.
PbS is precipitated as a black solid, so [S2-] decreases.

Adding more water DOES result in greater dissociation of H2S. Ka has to be remain constant at the same temperature! When you add more water, more H2S must dissociate to keep Ka constant. HOWEVER, we can assume that changes in concentrations of H3O+, S2- and H2S are so small that they are negligible. So, A is the most suitable answer.

You can also say that there is NO ANSWER.

Explanation for Q10
Suppose 1 mol of e- passes through the solutions. 1/m mol of X and 1/n mol of Y are formed.
Let M be molar mass of X and 3M be molar mass of Y (since Mr(X) : Mr(Y) = 1 : 3).
Mass of X formed : Mass of Y formed = (1/m x M) : (1/n x 3M) = 1 : 2
(1/m) / (1/n) = (1/2) x (M/3M)
n/m = 3/2
Thus, n = (3/2)m

A Note for Q13
Solubility of Group II hydroxides and oxides increases down Group II.
Solubility of Group II sulphates decreases down Group II.

A Note for Q22
The intermediate is a diazo compound (azo = N, so diazo = 2 N atoms). The loss of N2 is favoured because of the strong N-N triple bond formed. Hence, adding I- results in nucleophilic substitution of the diazo compound and the formation of N2.

A Note for Q23
H2, in THIS CASE, reduces the benzene ring and the C=C double bond only. (NB. Not enough information about catalyst or conditions used)
If H2/Pt is used, ALL the functional groups shown will be reduced.

Explanation for Q26
CH3CHBrCH2CH3, when heated in NaOH(aq), gives but-1-ene or but-2-ene.
Trans-but-2-ene is most stable because the two -CH3 groups are further away from each other while the cis-isomer is less stable because the two -CH3 groups experience steric repulsion.
But-1-ene is the least stable isomer and is formed as the minor product (apply Saytzeff's rule).

Combustion is exothermic, so delta H is negative.


Explanation for Q37 (Errata)
HNO3 is NOT an electrophile because all its atoms are NOT electron deficient and have either duplet (i.e. H atom) or octet structures (N and O).
Na+ is NOT electron deficient because it has an octet structure.
Answer is D.

2007 HCI Prelims Paper 1 FAQs

For Q4, see previous post by Mr WTY.

Explanation for Q12
O2 (i.e. gas K) is formed at anode.
4OH- --> O2 + 4H+ + 4e-
For 1 mol of O2 formed, 4 mol of e- are released.
These e- must be used to discharge the cations, Jn+.
1 mol of O2 is formed when 2 mol of Jn+ are discharged.
2Jn+ + 4e- --> 2J
Thus, n = 2.

Gas L is Cl2.
2Cl- --> Cl2 + 2e-
Note that the same quantity of electricity is used, i.e. 4 mol of e- must be involved.
Hence, the equation becomes 4Cl- --> 2Cl2 + 4e-.

Thus, ratio of J:K:L = ratio of O2:J:Cl2 = 2:1:2.

Explanation for Q14
[Br-] = [Cl-] = [I-] = [SO42-] = 1 x 10^-3 mol dm-3
Use the concentrations and the Ksp of the lead(II) compounds to find out the concentrations of lead(II) ions required for precipitation.

Explanation for Q15
[Ba(OH)2] = [HCl] = 1 mol/dm3
1 mol of Ba(OH)2 reacts with 2 mol of HCl.
20 cm3 of Ba(OH)2 is used. Thus, 40 cm3 of HCl is needed to reach equivalence point. No such titration curve is given.

Explanation for Q21
The "OH" group at the bottom can only give the cis-alkene after dehydration.

The "OH" on top lead to 2 possibilities. If the double bond forms at the carbon of the cyclohexyl ring, there are 2 isomers, i.e. cis and trans. If the double bond forms between carbon-2 and carbon-3 of the side chain (carbon-1 is bonded to Cl), then there is a chiral centre (carbon-4, counting from carbon-1 bonded to Cl). Hence, there are 2 optical isomers and each optical isomer can have cis or trans configuration at the double bond. Total of 4 isomers.

Hence, grand total = 2 + 4 = 6 isomers.

A Note for Q37
The concentrations of SO2, O2 and SO3 at equilibrium are 0.08, 0.04 and 0.02 (in mol/dm3) respectively. Kc = [SO3]2/{[SO2]2 x [O2]}.

Acidity of Organic Compounds

"Is CH3CH2OH deprotonated when pH>7, i.e. alkaline?"
"Does ethanol neutralise aqueous sodium hydroxide?"

The answers are: NO!

Acidity decreases in the order: carboxylic acid > phenol > water (it's not acidic!) > alcohol.


"What is the product formed when CH3CHBrCH2CH2COOH is heated with NaOH dissolved in ethanol?"

The product is CH3CH=CHCH2COO-Na+. To get the carboxylic acid, you need to add H2SO4(aq).

Ethanolic NaOH favours elimination while NaOH(aq) favours nucleophilic substitution. That's why ethanolic NaOH is used in the elimination of HBr to form an alkene.

Do note that ethanolic NaOH still contains some water. Elimination of HBr requires OH- ions. In pure ethanol, NaOH does not ionise. Hence, the OH- ions will neutralise the acidic -COOH group.


"Will there be an acid-base reaction between NaOH and HCl in CCl4?"

NO. For NaOH to neutralise HCl, there must be H+ and OH- ions. However, NaOH and HCl do not ionise in CCl4.

NOTE: NaOH is insoluble in CCl4. Thus, the question is actually invalid.

A Level Chemistry Paper (Nov 2006 / Jun 2007)

For students who are using the Dyna ten-year series (2007 Edition), please take note of the following errata.

Nov 2006 Paper 1 (MCQ)
Q4 Answer is A.
Q8 Answer is A.
Q32 Answer is A.

In addition, please take note of the following.

Nov 2006 Paper 1
Q33 Option 3 reads "The enthalpy change of combustion of diamond is greater than that of graphite". The word "greater" is used to describe the magnitude of the enthalpy change.

Nov 2006 Paper 3
Q1a(ii) When NaBr(aq) is electrolysed using inert electrodes, H2 is evolved at the cathode while Br2 (not O2) is evolved at the anode. (See explanation below)

------------------------------------------------------

CASE 1: Consider the electrolysis of NaCl(aq).

Cl2 + 2e- <--> 2Cl-, E=+1.36V ---(1)
O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V ---(2)
O2 + 2H2O + 4e- <--> 4OH-, E=+0.40V ---(3)

In NaCl(aq), there is Na+(aq), Cl-(aq) and H2O(l) [or write as "H+(aq) and OH-(aq)"].

At the anode, one of these reactions occur.
From (1), 2Cl- --> Cl2 + 2e-, E=-1.36V
From (2), 2H2O --> O2 + 4H+ + 4e-, E=-1.23V ---(a)
From (3), 4OH- --> O2 + 2H2O + 4e-, E=-0.40V ---(b)

From (a) and (b), the E values tell us that H2O and OH- are more easily oxidised than Cl-. Thus, O2 (and not Cl2) is produced at the anode.

NOTE: Cl2 is produced at the anode if concentrated aqueous NaCl is used.

CASE 2a: Consider the electrolysis of NaBr(aq).

Br2 + 2e- <--> 2Br-, E=+1.07V ---(4)
O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V ---(5)
O2 + 2H2O + 4e- <--> 4OH-, E=+0.40V ---(6)

In NaBr(aq), there is Na+(aq), Br-(aq) and H2O(l).

At the anode, one of these reactions occur.
From (4), 2Br- --> Br2 + 2e-, E=-1.07V
From (5), 2H2O --> O2 + 4H+ + 4e-, E=-1.23V ---(c)
From (6), 4OH- --> O2 + 2H2O + 4e-, E=-0.40V ---(d)

From (c), the E value tells us that Br- is more likely to be oxidised than H2O. Thus, Br2 (and not O2) is produced at the anode.

IMPORTANT NOTE:
(d) will give us the wrong conclusion, i.e. O2 is produced at the anode.
Thus, we should use "O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V" whenever we deal with the electrolysis of aqueous metal halides.

-------------------------------------------------------

O2 + 4H+ + 4e- <--> 2H2O, E=+1.23V OR O2 + 2H2O + 4e- <--> 4OH-, E=+0.40V ?

"O2 + 4H+ + 4e- <--> 2H2O" is used when we discuss the electrolysis of aqueous metal halides, e.g. NaBr(aq), KI(aq). Also applies when the electrolyte is H2SO4(aq).

"O2 + 2H2O + 4e- <--> 4OH-" is NOT used for the electrolysis of metal halides.

In general, we will consider using "O2 + 2H2O + 4e- <--> 4OH-" for the anodic reaction when the electrolyte is an aqueous solution which does NOT contain halide ions, e.g. CuSO4(aq), NaNO3(aq), KOH(aq).

-------------------------------------------------------

CASE 2b: Consider the electrolysis of NaI(aq).
The answer is: I2 is produced at the anode because I- is more easily oxidised than H2O.
2I- --> I2 + 2e-, E=-0.54V
2H2O --> O2 + 4H+ + 4e-, E=-1.23V

CASE 2c: Consider the electrolysis of NaF(aq).
The answer is: O2 is produced at the anode because H2O is more easily oxidised than F-.
2F- --> F2 + 2e-, E=-2.87V
2H2O --> O2 + 4H+ + 4e-, E=-1.23V

-------------------------------------------------------

Jun 2007 Paper 4
Q1b examines the electrolysis of molten ZnCl2, concentrated ZnCl2(aq) and dilute ZnCl2(aq).

At cathode:
2H+ + 2e- --> H2, E=0.00V --- (7)
Zn2+ + 2e- --> Zn, E=-0.76 V ---(8)

In the case of molten (or liquid) ZnCl2, there are only Zn2+ and Cl- ions. Thus, Zn is produced at the cathode and Cl2 is produced at the anode.

In the case of concentrated ZnCl2(aq), Cl2 is produced at the anode. From (7) and (8), we conclude that H+ is more easily reduced than Zn2+. Hence, H2 should be formed at the cathode.

In the case of dilute ZnCl2(aq), O2 (and not Cl2) is produced at the anode. H2 is produced at the cathode.

-------------------------------------------------------

There is no need to panic! The troublemaker seems to be metal chlorides. Hence, just be more careful when you see questions involving the electrolysis of metal chlorides.

-------------------------------------------------------

Jun 2007 Paper 2 Q3a (Erratum)
The electronic configuration of Sr2+ is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d0.
(The electronic configuration of Sr is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d0 5s2.)

Answers to MOCK MCQ

Q1-5 C C C D C
Q6-10 D B C D C
Q11-15 C C B C B
Q16-20 D B B A C

HCI 07 prelims P1 Q4

This is a brilliant question. Now the key is, water and Iodine will condense (to liquid and solid respectively) when the temperature is cooled to room temp (298K). So all the gas thats remaining is Argon. Therefore the no of mols of gas, n, is totally accounted by argon.
using the eqn, nR=p1V1/T1=p2V2/T2, where p1 is 101kPa, V1 is 4000 cm3(thats the volume occupied initailly by argon), T1 is 423K (thats 150 C), p2 is new pressure, V2 is 14090 cm3(the argon expands into all 3 containers) and T2 is 298K
we get p2 = 101(4000)(298)/(423)(14090), => p2 is 20.2 kPa. essentially, this p2 is the new pressure of argon after it has been allowed to expand to all 3 containers, followed by cooling to room temp. We do not consider the other cpds because they no longer exist as gas at the end of the day anyway.

YJC Paper 2 Q1 d(iii) Answer Key

The answer to YJC Paper 2 Q1 d(iii) was left out of the printed answer key. The answer is:

CaCO3 is insoluble in water, hence the presence of solid CaCO3 would indicate that all excess acid has reacted/been removed.

HCI H3 Prelim Answers

Hi! The answers for the HCI H3 Prelims are out. Please do a copy for yourself. Take care yah. =)

Errata (H2 Chem Specimen Paper Solutions)

Hi All.

Please take note of the errors in the "worked solutions" of H2 Chem Specimen Paper 1 (MCQ).

Q22 The third step of the reaction scheme shows sodium benzoate as the product after alkaline hydrolysis. That is incorrect. The product should be C6H5CH2COO-Na+.

Q27 There are five (NOT four) -OH groups in a molecule of glucose.

Thanks to Natalie for pointing out the errors.

The Last Lap...

H3 Prelim Papers from Other JCs

The H3 prelim papers from HCI, NJ and VJ are at the photocopy shop. Please do the photocopying yourself. Answers for the HCI paper are not out yet, we will update you when we receive it.

We really need you to exercise true Rafflesian spirit..by preserving the original copy as it is!!

There will be NO mock paper for H3, as the revision time table is already very tight. Implement the exam conditions on your own, OK?


All the best and take care...

H2 Chemistry ACJC, TJC Suggested Solutions

Dear students, at the time of printing, the suggested solutions of ACJC P1, TJC P1 and TJC P3 were not available. A copy has been placed in the photocopy store.

ACJC PAPER 1
BADCB CDBBD ACBBB ABADA ACDBA DACAC AACDD BBDAB

TJC PAPER 1
BAADC CDCDC DBADC AAADC CCBBD CADCB CBCDD CDBDC

H2 Chem Revision Lectures / Mock Papers

Students who scored E, S or U for H2 Chem in the Prelims must attend revision lectures and sit for Mock Papers. Please see the details below. Direct all queries to your respective tutors.

8 Oct 07 (Mon), 0755 - 0945, LT1 and LT3
H2 Specimen Paper debrief

22 Oct 07 (Mon), 1330 - 1515, LT1 and LT3
2007 June Paper debrief + Mock Paper 1 (MCQ)

24 Oct 07 (Wed)
Session 1 (A-D graders): 1045 - 1330, LT1
Session 2 (ESU graders): 1235 - 1545, LT2
Mock Paper 2

25 Oct 07 (Thu), 1330 - 1630, LT1 and LT5
Mock Paper 3

H2 Chem Revision Tutorials

Students who scored E, S or U for H2 Chem in the Prelims must attend revision tutorials. Your tutors will tell you what you should prepare for each tutorial. Please direct all queries to your respective tutors.

Dates: 11 Oct 07 (Thu), 16 Oct 07 (Tue)

Time: 0755 - 0945

Venues:
A43 - Ms Chua LC
A44 - Mrs Chung AL
A45 - Mr Eric Gwee
A46 - Ms Jolene Tan
A47 - Ms Lee LS
A48 - Mrs Poh SJ (aka Mdm Lee)
B41 - Mrs Olivia Ang
B42 - Mr Paul Cheong
B43 - Ms Soh XF
B44 - Ms Tang LL
B45 - Mr Tham ZS
B46 - Dr Wong PL
B47 - Mr Wong TY