Post-Results 2016

Post-Results 2016

Chemistry, JC Chemistry, JC General Paper, JC Mathematics, JC Physics, Mathematics, Studying Tips, University Mathematics

Let’s face it. Some of us will not get the dream results we want. Don’t give up and let fear conquer you.

For students unsure of the available courses, they can check out the following post. It contains the grade profile for local universities.

Our Team will be here if you need help/ advice. Feel free to text us.

P.S. Today, I saw an image shared by Mr Wee, which said that “You’re the architect of your own life”. So let’s not let the grades define us.

2016 A’level Suggested Solutions

2016 A’level Suggested Solutions

JC Chemistry, JC General Paper, JC Mathematics, JC Physics

Congratulations on the completion of A’levels for the 2016 batch!

As for those who still, are taking A’levels 2017, we hope you find this site helpful. Read the comments (ignore the trolls who have yet grown up) and learn from our mistakes.

It has been a pleasure for all us to interact with you guys. Do check back after your release of A’levels, as we will love to hear from you! In the meantime, you can always read through some of the posts regarding undergraduate & postgraduate courses in Mathematics (KS), Economics (KS), Physics (Casey) as we share some of our past experiences, along with some of our works today.

You’ve come a long way, now party hard. We wish you all the best in your future endeavours.

H1 General Paper (By Christine)


H2 Mathematics (By KS)

Read KS’ thoughts about the upcoming paper here.


H1 Mathematics (By KS)

H2 Physics (By Casey)

 

100 days more…

100 days more…

JC Chemistry, JC General Paper, JC Mathematics, JC Physics, Studying Tips

So the Midyear results are all out and Prelims are known to be in 4-5 weeks’ time. Many students are frantically searching for help and attempting to salvage their results. We are sorry to say that we aren’t able to take any more private students due to time constraints, and only the group classes are available. Our classes are all held in Newton Apple Jurong East.

Importance of June Holidays

Importance of June Holidays

JC Chemistry, JC General Paper, JC Mathematics, JC Physics, Studying Tips

June holidays are coming and JC2 students should really grab this last opportunity that you have to study. This is effectively your time you have a full month to study for your upcoming A’levels.

Credits: Calvin and Hobbes by William Boyd “Bill” Watterson II
Credits: Calvin and Hobbes by William Boyd “Bill” Watterson II

Through the years, we see many students who have their efforts paid off for working hard during June Holidays. So make sure you plan out your June Holiday well to study and catch up as much work as you can. Doing well in Mid-Years is a good start and also confidence boosting to your A’levels preparation. Some advice…

Self-discipline goes a long way. Not just in studies.

Start waking up early. Get your brains tuned to functioning in the mornings. Students who claim to be only functioning/ thinking only at night, should realise that your exams are mostly in the MORNINGS. And since we know that the exams are in the morning, maybe you want to start making your brain work in the morning.

Give yourself timed practice to train your time management and speed. This is very crucial in Mathematics. Putting yourself under time constraints is the best. And if its a 3 hours paper, sit there and do for 3 hours. Its amazing how students get restless after 2 hours of studying. How can one take a 3 hours paper when you can barely study for 2 hours. You can always find a friend and do the practice paper together. Peer studies are really helpful.

Try not lead a sedentary lifestyle. Time and time, we see some students burning out from studying. Do some sports. C’mon, you still have PROM to attend and its right after A’levels (first week of December, maybe?). You don’t want to be buying a new uniform too. And if you don’t like to exercise, then slot in 30 minutes of TV, YouTube (Just don’t get carried away.) every 3-4 hours of studying. Our brain needs some rest too. Be sure to sleep. Sleeping reinforces your memory and sharpens your thinking.

Seek help soon. Really, your teachers need a life too. So settle consultation hours with them early. Don’t flood them in the last week of June. They don’t owe you a living. Be considerate.

All the best for your results!

JC Chemistry, JC General Paper, JC Mathematics, JC Physics

All of us from theculture.sg would like to wish all the readers or ex-students the best of luck to the release of A’levels results 2015!

Enjoy the open houses and do read up a bit on all the available university courses. Narrow some choices down and clear up as much doubts during the open houses as possible. 🙂

To our ex-students: Do update us about your results and feel free to ask us about the university courses that you are interested in!

Really a big congratulations to all our students! For those who are disappointed, remember not to let a letter grade define you or limit your future! Chin up!

Side note: For the students that need to retake.
2016 A’levels Timetable

O’levels Results 2016!

JC Chemistry, JC General Paper, JC Mathematics, JC Physics, Secondary Chemistry, Secondary Math

All of us wish the students receiving the O’levels Results 2016 the best! And do not let grades define you. 🙂

If you’re keen to meet up with us for the JC Talk, you may contact Newtonapple Learning Hub at +65 9222 3423 for more details.

This talk will be opened freely to the public and existing students. There will be discussion about Subject Combinations, discussion and introduction with different subjects. Come down to find out more 🙂

2015 A Level H1 Chemistry (8872) Paper 1 Suggested Solutions

JC Chemistry

All solutions here are SUGGESTED. Mr. Lee will hold no liability for any errors. Comments are entirely personal opinions.

Many thanks to the student who sent us the question paper. 🙂

 

Screen Shot 2015-11-27 at 6.05.29 pm

You can post any questions regarding the answers but please do not ask me how many marks to secure an A for the examination. Nobody will know so there is no point in discussing this.

2015 A Level H2 Chemistry (9647) Paper 1 Suggested Solutions

JC Chemistry

All solutions here are SUGGESTED. Mr. Lee will hold no liability for any errors. Comments are entirely personal opinions.

Screen Shot 2015-11-27 at 12.24.13 am

 

Question 1:   B

The element must be from Group IV so the electronic configuration must be either 1s22s22p2 or 1s22s22p63s23p2. The four electrons of highest energy refer to the electrons at shells with higher principal quantum number. The best answer is B.

 

Question 2:   B

Sigma bonds can be formed via head-on overlap between two s orbitals, two p orbitals and between one s and one p orbitals. In hydrocarbons, the carbon atoms have hybrid orbitals consisting of s and p orbitals. Hence, sigma bonds can be formed by either s or p orbitals while the pi bonds can only be formed by side-way overlap between two p orbitals.

 

Question 3:   C

Oxygen atom has 8 electrons. Hydrogen atom has 1 electron. Since hydroxide anion has an extra electron gained, the total number of electrons is 10. Oxygen atom has 8 neutrons while hydrogen atom has no neutrons. Hence, the answer is C.

 

Question 4:   C

A higher charge on Ca2+ would mean that the electrostatic forces of attraction between the cation and the sea of mobile electrons are stronger as compared to Na+. Option D is wrong because the electrons in the calcium ion are not mobile and even if calcium ion has more electrons than sodium ion, it does not explain the stronger metallic bonding. The correct explanation should be calcium metal has more mobile electrons than sodium metal, instead of ion.

 

Question 5:   B

In zwitterion, –NH2 (trigonal pyramidal, 107°) group becomes –NH3+ (tetrahedral, 109.5°).

 

Question 6:   D

Lattice energy is the enthalpy changed when 1 mole of solid ionic compound is formed from its constituent gaseous ions.

 

Question 7:   B

Number of moles of NaOH = 12.5 / 1000 x 0.0500 = 6.25 x 10-4 mol (limiting reagent)

Number of moles of HCl = 25.0 / 1000 x 0.100 = 2.50 x 10-3 mol (in excess)

Number of moles of HCl remaining = 1.875 x 10-3 mol

Concentration of HCl remaining = 1.875 x 10-3 / (37.5 / 1000) = 0.0500 mol dm-3

 

Question 8:   C

CH4 + 2O2 → CO2 + 2H2O

CH4 + 3/2 O2 → CO + 2H2O

 

9CH4 + 18O2 → 9CO2 + 18H2O

CH4 + 3/2 O2 → CO + 2H2O

 

Therefore, 10 moles of CH4 react with 19.5 moles of oxygen to produce 9 moles of CO2 and 1 mole of CO (9 : 1 ratio)

Hence, 1 dm3 of CH4 reacts with 1.95 dm3 of oxygen.

 

 

Question 9:   A

Formation of hydrogen bonds indicates that the reaction is exothermic. ΔH is negative.

The product (1 mole) is more ordered as compared to the reactants (1 mole of CO2 and n moles of H2O). Hence, ΔS is negative.

 

Question 10: B

Standard conditions means that all concentrations must be 1 mol dm-3.

 

Question 11: D

Equilibrium concentration for CH3CO2H = C(1 – α)

Equilibrium concentration for CH3CO2 = Cα

Equilibrium concentration for H+ = Cα

Kc = (Cα)2 / C(1 – α) = α2C / (1 – α)

 

Question 12: C

Increasing the pressure will shift the equilibrium to the right because the forward reaction produces fewer number of moles of gaseous substances. Decreasing the temperature for the exothermic reaction will shift the equilibrium to the right because the forward reaction will release heat energy to counteract the decrease in temperature.

 

Question 13: D

Total number of moles = 2.0 mol

Number of moles of H2 = 33.3 / 100 x 2.0 = 0.666

Number of moles of CO2 = 33.3 / 100 x 2.0 = 0.666

 

Percentage of H2O = (100% – 33.3% – 33.3%) / 2 = 16.7%

Number of moles of H2O = 16.7 / 100 x 2.0 = 0.334

Number of moles of CO = 16.7 / 100 x 2.0 = 0.334

Kc = (0.666)2 / (0.334)2 = 3.98 ≈ 4.0

 

Question 14: B

The reaction is second order with respect to NO. Hence, when concentration decreases by 2, the initial rate will decrease by 4. Hence, x = 6.0 / 4 = 1.5

The reaction is first order with respect to H2. Hence when concentration increases by 2, the initial rate will increase by 2. Hence, y = 1.5 x 2 = 3.0

The reaction is second order with respect to NO. Hence, when the initial rate decreases by (3.0 / 0.75) = 4, the concentration of NO will decrease by 2. Hence, z = 0.5.

 

Question 15: D

Thermal stability of nitrates increases down Group II due to lower charge density of Group II metal cations as the size of the cations increases.

 

Question 16: D

Phosphorus exists as P4. Option A is wrong because sulfur can also form two acidic oxides, SO2 and SO3. Option B is wrong because argon should have the highest ionisation energy. Option C is wrong because chlories of silicon (SiCl4) and sulfur (S2Cl2) can react with water to form HCl solution which is acidic.

 

Question 17: A

Only Option A is correct. Option B is wrong because it is not a precipitate. Option C is wrong because it is a blue precipitate instead of a pale blue solution. Option D is wrong because Cu+ complex should not have any colour because it has completely filled 3d orbitals so no d-d transition is possible.

 

Question 18: C

Atomic radius should decrease across the period from Mg to P while for melting point, Si has the highest, followed by Al, Mg and P.

 

Question 19: B

Only dilute sulfuric acid is used to acidify aqueous potassium dichromate(VI). Ethanol can be oxidised to form ethanoic acid.

 

Question 20: A

PCl5 will react with hydroxyl groups and carboxylic acid groups. The Br atom does not react with PCl5

 

Question 21: B

The 3 structural isomers are:

1

 

 

 

2

 

 

 

3

 

 

 

 

Question 22: D

4

 

 

 

 

 

 

Question 23: B

NaBH4 will reduce aldehyde to alcohol but it does not reduce alkene to alkane.

 

Question 24: A

Molecular formula of citric acid = C6H8O7

Condensation reaction will lose one molecule of water.

Hence, C6H8O7 + C3H6O2 → C9H12O8 + H2O

Hence, option A is the best answer as the structural formula of C2H5CO2H fits the molecular formula of C3H6O2

 

Question 25: D

Ethanoyl chloride is an acid chloride. It will react with water to produce ethanoic acid and Cl  and the Clion will form a white precipitate (AgCl) with Ag+

 

Question 26: D

 

pH 11 is alkaline so the functional groups will be deprotonated.

 

 

Question 27: D

Addition of molecular mass of 14 suggests that one oxygen atom is added but two hydrogen atoms are removed. This indicates that primary alcohol (–CH2OH) is converted to carboxylic acid group (–COOH).

 

Question 28: C

Cold NaOH(aq) is able to react with the acidic phenol group but not with the alcohol group.

 

Question 29: B

The phenol group is an activating group and is 2,4,6-directing.

 

Question 30: A

The carbon-carbon double bond will attack the Br–Br first to form a carbocation containing one Br atom.

 

 

Question 31: D

Mn2+: [Ar] 3d5 (no paired 3d electrons)

Fe2+: [Ar] 3d6 (one paired 3d electrons)

Co3+: [Ar] 3d6 (one paired 3d electrons)

 

Question 32: A

Simple covalent compounds have low boiling point. AlBr3 has a simple molecular structure because it has a low boiling point even though it has a metal and non-metal. If AlBr3 is ionic, the boiling point should be 4 digit. Hence, all 3 compounds are covalent.

 

Question 33: C

The molecules of the gas have mass, so option 1 is wrong.

 

Question 34: A

Zinc will be oxidised while NH4+ will be reduced. So the overall potential will be +1.50V (energetically feasible). Option 2 and 3 are correct also.

 

Question 35: B

Rate constants are affected by activation energy and temperature based on Arrhenius equation k = Ae–Ea / RT. Introducing a catalyst will increase both kf and kb equally while increasing the temperature for an exothermic reaction will increase BOTH kb and kf but will increase kb more than kf so equilibrium shifts to the left. Rate constant does not depend on concentration of the reactants.

 

Question 36: A

Methanal is trigonal planar with respect to carbon and the carbon has an oxidation number of 0 (hydrogen has an oxidation number of +1 while oxygen has an oxidation number of –2). The equation for complete combustion of methanal is:

CH2O + O2 → CO2 + H2O

Hence, 1 mole of oxygen is required for 1 mole of methanal.

 

Question 37: C

Radicals do not have a lone pair of electrons. They have unpaired electrons and can be formed by homolytic fission of a covalent bond.

 

Question 38: C

Geometric isomerism is possible if there are two different groups attached to the sp2 carbon in the carbon-carbon double bond. Hence, Y cannot be H and can be Br so that the carbon-carbon double bond with Y can exist as geometric isomers. The carbon-carbon double bonds with Z and X cannot exist as geometric isomers since the groups attached to the sp2 carbon in the carbon-carbon double bond are the same.

 

Question 39: A

Thymol has a phenol group so the acidic phenol group can react with the alkali present in alkaline KMnO4 via neutralisation. The phenol group can also react with ethanoyl chloride to form an ester via condensation and it can also react with sodium metal to form phenoxide ion.

 

Question 40: B

Option 1 and 2 have no line of symmetry while option 3 has a line of symmetry. Hence, option 3 is not optically active.

 

Please do let me know of any mistakes or typing errors that I made while rushing this. Much appreciated and thanks!