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Visit from ex-students

Mr Khemistry

Visit from ex-students

Visit from ex-students – So happy to see my ex-students coming back. Still remember the day i saw their family, two of them together with their parents at my former B2 unit. Hope we can get a chance to take a group picture together with their parents again.

Woon Siblings

School term has restarted, are you still in holiday mood? 🙂

Time to hit the road running!

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Year End Miri Trip

Mr Khemistry

Year End Miri Trip

Felix had lots of fun at his cousin’s indoor playground. The children are 2-6 years old, literally 1 every year. Having them all in the car together is a rare event.

We visited The Nest, a new stylist cafe in a atas area of Miri. The food there was quite alright, it’s pricey in ringgit but cheap in sing dollars.

How was your year end holidays? Was it a time of reflection and planning for the new year?

Here’s wishing everyone a happy new year!

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drowning in ball poolOn the way to dinnerNice decoChicken sandwichEgg benedict

2020 Class Schedule

Mr Khemistry

2020 Class Schedule

2020 Class Schedule: Our chemistry tuition class schedule for the new year is up 🙂

JC 2 students: Do you struggle with Chemistry in JC 1 or feel that you have not fully grasped all the important foundation concepts? Did poorly for your Promos and want to start scoring As and Bs for your tests?

JC 1 students: Realised that you don’t understand Chemical Bonding, Thermochemistry and Reaction Kinetics? Can’t do gases calculations or don’t understand chemical equilibrium questions? How to tell if a reaction is a redox reaction?

The fact is, a lot more is demanded from JC students but not many students adapted quickly enough.

The jump in content required from O level Pure Chemistry to A level H2 Chemistry is a big one, quite easily 2~3 times! And if you are already struggling to master the content in JC 1, the lecture pace in JC 2 will be 50% faster.

If you realised that you need help, reach out to us and let us help you! If you don’t understand the basic concepts fundamentally, reading the notes over and over again seldom helps. Time is of essence in JC, don’t waste any more time!

Image result for he who do the same thing and expect different results quote

Start the year right by taking action to change your grade! For early sign-ups for both J1 and J2, we will waive the registration and material fees 🙂

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Merry Christmas!

Mr Khemistry

Merry Christmas!

Dear all, it’s the time of the year to be jolly! 3 more days to Christmas day and 10 more days to the new year 🙂

Better brace yourselves for some hindsight 2020 jokes and memes haha!

Mr Khemistry would like to take this opportunity to wish everyone a Merry Christmas and Happy New Year!

We will see you again in 2020!

Image result for Christmas

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Reaction Kinetics

Mr Khemistry

Reaction Kinetics

Reaction Kinetics – A team of scientists at Harvard university have built an apparatus which can achieve ultra cold temperatures. This allows them to capture reaction mechanisms previously elusive as they occur too quickly.

Now, they not only performed the coldest reaction yet, they discovered their new apparatus can do something even they did not predict. In such intense cold — 500 nanokelvin or just a few millionths of a degree above absolute zero — their molecules slowed to such glacial speeds, Ni and her team could see something no one has been able to see before: the moment when two molecules meet to form two new molecules.”

Coldest reaction


Wow! Scientists are able to actually replicate such low temperatures in their labs! Nano is 10^-9 times, that’s like 0.000000005 K.

With such low temperatures, the kinetic energy in the molecules will be so low they will literally be “crawling”. Imagine Maxwell Boltzmann distribution with just one thin line near zero Kelvin.

This will enable scientists to capture images of what actually happens in the transition state. Previously, the transition state cannot be captured. With this new breakthrough, we will be able to validate or disprove the transition state theory. Exciting times! 🙂

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Preparations for A level

Mr Khemistry

Preparations for A level

This seems like a pretty odd time to be advocating students to start preparing for their A levels next Nov. However, winter time is always the best time to prepare for a major expedition. In this time, almost everyone is flying overseas for holidays, we need to start our preparations.

What do we need to prepare for? Clarification of doubts for J1 topics as most of them will not be revisited until late in Sept when we are revising for prelims. Even then at that crucial period, there will not be enough time to correct all the misconceptions in the foundation topics. For your information, the pace of lectures will be 1.5 times faster in J2 as there will be an extra period allocated to all the H2 content subjects. This is due to the completion of PW in J1.

As GCE A levels is mainly a two year program, this mid-program break of about 1.5 months is extremely crucial. Sadly, many students were not able to plan a consistent study schedule. Having a goal in mind is essential to taking actions. We do not want to be busy just for the sake of being busy.

For preparations for A level H2 Chemistry, it’s essential that you get adequate practise for physical chemistry topics. Topics such as Mole Concept, Energetics, Chemical Equilibrium, Kinetics and Gases cannot just be scanned through. The process of calculating the answer has to be practised regularly.

Victory does not go to the smarter nor to the fast, it belongs to those who diligently prepare for it and take prompt action.

If you are looking for weekly tuition for H2 Chemistry, feel free to contact us at 98537960 or click on the whatsapp icon at the bottom of the page.

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MCQ Answers for 2019 H2 Chemistry Paper 1

Mr Khemistry

MCQ answers

  1. B
  2. C
  3. B
  4. C
  5. C
  6. A
  7. D
  8. B
  9. D
  10. D
  11. A
  12. B
  13. D
  14. D
  15. D
  16. A
  17. B
  18. A
  19. A
  20. D
  21. D
  22. C
  23. A
  24. C
  25. B
  26. B
  27. C
  28. C
  29. B
  30. B

The standard for this MCQ paper is pretty easy, there wasn’t really any trick questions or questions which are ambiguous. IMO.
Only the last question will take a bit longer to calculate, but you should have plenty of buffer from the easy organic questions earlier.

Weightage of Paper 1 is 15% so each question is worth 0.5%. Accounting for careless mistakes, most students should score  more than 24 to
have a decent chance to score distinction.

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Nomenclature of anions

Mr Khemistry

Nomenclature of anions

One of the common questions that pops up from time to time is how do i write the formula of sodium nitrite? And what’s the difference between sodium nitrite and sodium nitrate? Or sodium nitride? This has to do with the nomenclature of anions.

The convention for naming anions is as follows:

ide ending is for a monoatomic anion of an element. e.g. hydride, fluoride, oxide, sulfide, nitride, phosphide etc.

ite ending is for polyatomic anions containing oxygen, oxyanions. -ite is for the one with less oxygen atoms. eg, nitrite, sulfite

ate ending is for polyatomic anions containing more oxygen atoms. eg nitrate, sulfate

If there’s a series of 4 oxyanions, hypo prefix refers to the anion with less oxygen and per- prefix refers to the one with more oxygen. e.g. hypochlorite, perchlorate.

Hopefully now you have a clearer understanding of the nomenclature of anions. For H2 Chemistry, this is not super crucial but it’s good to know. Usually questions will give you the formula of the compound, not just its name.

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Suggest type question

Mr Khemistry

Suggest Type Questions

Firstly, suggest type questions took up 12/75 (16%) marks in 2018 paper 2 and ~24/80 (30%) marks in paper 3. So we cannot ignore its importance.

 Secondly, most “suggest” questions aren’t totally novel questions, they are just contextual. Meaning we can look at the question and derive the answer from the information provided.

 In this article, I wish to highlight some ways to think about how to approach such questions. Please refer to the 2018 paper when reading through this document, it will give you some context.

Some simple general questions

  • What is the difference between A & B?
  • Is this linked to the earlier parts of the qn? What is the direction/flow of the qn?
  • Why do we need to do ______? What if we DON’T do _____? What will happen?
  • To explain given observations, think about which topic and then narrow down.
  • For organic mechanisms, look for similarities with the 6 mechanisms that we have in our syllabus.
  • Suggest questions are usually 1-2 marks, try not to vomit out a long answer, it wastes time and doesn’t answer the qn.
  • How does the change affect the eqm? Are reactants/pdts removed or added?
  • To answer qns on suggesting alternative reagents, think abt the properties of the orginal reactant that is important, that defines its role. Then look for the same in the replacement.
  • To answer qns about the advantages or disadvantages, think abt what would be an ideal substance/reactant. Usually, it has to do with ease of storage, transportation, non-reactive, not pollutive, not corrosive/dangerous/poisonous or costs. Really depends on the context. E.g. what means a gd fuel (easy to be fully combusted), battery (storage, current), solvent (not reactive with reactants)
  • To answer qns how to improve the expt, think abt your aim and assumptions. What would be the ideal expt and what would cause errors?

Suggest Type Questions in 2018 Paper 2 and 3

2018 P2 Q1c(iii) Suggest why, at room temperature, methyl isocyanate is a gas but A is a solid.

CH3NCO           ->       C6H9N3O3

Methyl isocyanate                                   A

Thought process: Given physical properties, we will need to explain this using our knowledge of chemical bonding. A is simply a trimer of methyl isocyanate, so what’s the difference between the two molecules?

Take note of the earlier parts of the question, usually they will “lead” you to a certain direction when answering the question. In part (ii), we are told A has no dipole moment, so the higher IMFs is probably not due to pd-pd interactions.

Q2f(ii) Ammonia can be formed from its elements.

N2 (g) +  H2 (g)  <-> NH3 (g)           ∆Gf0 = -16.6 kJ mol-1

Suggest whether the ratio of [products]/[reactants] at equilibrium for the formation of ammonia at 298K will be less than, equal to or greater than 1. Give a reason for your answer.

Question to ask yourself:

What did part (i) ask about? Is it a clue to how we can answer part (ii)?

Given the negative Gibbs free value, what can you tell about the equilibrium position?

What does the equilibrium position tell you about the ratio of the amount of products vs the amount of reactants?

Standard questions commonly posed as suggest type questions

P2 Q4 (b)(ii) Methylbenzene reacts with fuming sulfuric acid, H2S2O7, to form a mixture of three isomers, G, H and J, in equilibrium.

Suggest why there is a lower concentration of G than of J in the equilibrium mixture.

Qn to ask: What’s the difference between G and H/J?

How might this difference hinder the electrophilic attack? (since it is formed in lower conc)

c(iii) K also undergoes a reaction with fuming sulfuric acid.

Suggest the order of reactivity of benzene, methylbenzene and K with fuming sulfuric acid. Explain your answer.

Qn to ask: Order of reactivity depends on whether the substituent group is activating or deactivating. Think about how the substituents increase/decrease the electron density on the benzene ring.

Is –COCH3 activating or deactivating? In all fairness, this is really a very standard qn.

The importance of reading and underlining key ideas in the question stem…

“When the external voltage is increased beyond Ecell, current will flow and a chemical reaction takes place.”

AgCl + e <-> Cl + Ag           E0 = -0.22V

O2 + 4H+ + 4e <-> 2H2O     E0 = +1.23V

Qn 5a(i) Suggest why there is a flow of electrons when the external voltage is +1.45V.

Qn to ask: E0cell is exactly 1.45V and since the question stem stated that the current will only flow when external voltage is greater than 1.45V. What does this tell us about the Ecell?

Under what kind of conditions will the E­cell drop?

(vi)  Suggest why the water sample is stirred constantly during the measurement.

Qn: What do we want to measure in this electrolysis?

What would happen if we do NOT stir the water sample constantly? Can this idea be applied to other electrolysis/galvanic cell?

(c)(iii) Suggest why the actual value for the pressure recorded in vessel Y was slightly different from the value calculated by this method. Assume oxygen gas behaves as an ideal gas under these conditions.

Qn:  What’s the difference between vessel X and Y?

The question did not specify whether the pressure is more or less in vessel Y so we can either explain it is lower or higher.

We cannot say due to IMF (which is what we do normally) as qn already assumes oxygen behaves ideally.

If explaining it is lower: think in terms how there could be less gaseous oxygen molecules

If explaining it is higher: think maybe how other dissolved gases in water might have escaped.

Can this be applied to other scenarios?

Paper 3

Q1e(ii) In some circumstances, double bonds will undergo a nucleophilic addition reaction. Suggest reasons to explain Fig. 1.2.

Suggest reasons to explain Fig 1.2. Use the concepts of electronegativity, electronic and steric effects, and delocalization in your answer.

Qn to ask: why doesn’t 1,3-butadiene undergo nucleophilic addition with HCN? (same reason why LiAlH4 cannot reduce alkenes) Think about the difference between alkene and carbonyl compounds. Hint: Electronegativity, polarity.

As for why B cannot be formed, but C can be formed: Think in terms of steric and electronic effects, the difference between the attack position.

What is the difference between the C=C in 1,3-butadiene and the C=C in 4-methyl-1-penten-3-one? (Only delocalization hasn’t been used)

Q2b(i) HCN can be oxidized to cyanogen, C2N2.

C2N2 (g) + 2H+ (aq) + 2e <-> 2 HCN (aq)                 E0 = + 0.37 V

Suggest a suitable oxidant to oxidise HCN to C2N2 in acidic solution, using the Data Booklet. Write an equation for the reaction and calculate the E0cell.

Note: Use of Data Booklet means you need to quote data from it.

Qn: If the above half equation is the oxidation half cell, what must be the E0 of the reduction half cell in order to have a spontaneous reaction?

(c) (ii) Suggest why the units of k2 are s-1, despite its rate-determining step being a bimolecular reaction.

Qn: What is the difference between reaction 2 and 3? In order to have units of rate constant as s-1, what must be the total units of the reactants? (hint: think about H2O)

(e) Cyanogen bromide, BrCN, undergoes addition reactions with alkenes. With propene, isomer D is produced rather than isomer E.

Suggest a mechanism for this reaction and use it to explain the preferential production of isomer D.

Qn: Think about which portion of BrCN will get the partial positive charge. Which intermediate carbocation formed will be more stable?

Important exam skill alert!

3c(ii) The addition of NH3(aq) to a solution containing Cu2+(aq) produces a deep blue solution. This solution changes from deep blue to purple when ethane-1,2-diamine, H2NCH2CH2NH2, is added.

Suggest an explanation for this observation and write an equation for the reaction occurring. [The formula of ethane-1,2-diamine can be shortened to en in the formulae and equations of your answer.]

Qn: Change of colour is due to change of ligands, so you need to state en is the stronger ligand. This qn requires an eqn and is only 2 marks hence there’s no need to explain using ∆E and wavelength etc.

Usually when the qn wants you to account for the colour, it is about 3 marks. (eg. Qn 5a(i))

Thus an important skill is also to know how much to write and is all the details required. Time is precious, don’t regurgitate everything you know about the topic unless required.

Is this a novel qn? Not really, but it’s a “suggest” qn as you are not required to remember the ligands’ strength.

Section B

Q4d(v) The entropy change for the exothermic oxidation of powdered charcoal (carbon) and tetranitratoethane is large and positive. Suggest two reasons why the entropy change for the reaction is large and positive. Explain your answer.

Thought process: We are again presented with observation in this suggest type questions and asked to explain it using our content knowledge.

This is very common and forms the bulk of “suggest” type of questions. The context is novel but the concept is not. Usually an analogous example can be found in our syllabus or in the question context.

So we will need to explain two parts, why entropy is large and why it’s positive. Part iv requires us to write the equation so that we can see how the entropy has changed. How many ways are there to increase entropy? This is an easy qn.

Q5b(i) When a solution of sodium vanadate (V) is gradually acidified, the colour changes from colourless to orange to red, and finally to a yellow solution which contains the ion VO2+.

The standard reduction potentials, E0, for the electrode reactions VO3/VO2+ and VO2+/VO2+ are both +1.00 V.

Suggest how the oxidising abilities of VO3 and VO2+ will change as pH increases, and hence suggest which will be the more powerful oxidant at pH 7. Explain your answer.

Qn: [H+] DECREASES as pH increases. How will that affect the two equilibriums? For oxidising power, is it better to have a lower or higher E0?

Write out the two eqms to see clearly, since the E0 for both eqms is the same, you will need to differentiate which eqm is more affected by a change in [H+].

Parting words…

Hopefully, you now have a good understanding of suggest type questions and how to approach them.

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Dilution of Buffer Solutions

Mr Khemistry

Dilution Effects on pH changes – Buffer

For avoidance of doubts, in this post, we are referring to acidic buffer.

In Acid-base Equilibria, there is a portion of the syllabus on buffers. Normal questions will ask for an outline of how the buffer works. They might ask you what is the mass of strong acid or base to a weak acid/base to add to create a buffer at a certain pH.

A more advanced question will ask you what effects will dilution have on the pH. Generally, dilution has NO effect on the pH. Why is this so? If you look at the buffer formula, pH = pKa + lg [salt]/[acid], dilution does not affect the [salt]/[acid] ratio. i.e. they are diluted to the same extent or volume would cancel out.

But the issue comes when the question further asks you will the pH change be larger if you dilute the buffer? We have to be careful here to watch out for the conditions. It is important to note whether the total volume of the buffer changed. Why is this important?

3 Scenarios

Firstly, if the volume changed drastically, i.e. large volumes of water was added to the buffer, the pH will tend to move towards 7. Which is the pH of water. This is because concentration of H+ ions tends to be closer to the amount from auto-ionization of water.

Second, if the total volume remained constant, it means that dilution AND removal of some of the solution had happened. (since we have to add water which would increase the total volume) In that case, although the [salt]/[acid] ratio will remained constant, there will be a larger change in ratio when a small amount of strong acid is added. Note that since some of the solution was removed, the number of moles of salt and acid decreased.

Before dilution: [salt]/[acid] ratio is 5/5. So when a small amount of strong acid is added, say 0.1 mol, the ratio becomes (5-0.1)/(5+0.1),

After dilution: ratio becomes 0.5/0.5. So when a small amount of acid is added, the ratio becomes 0.4/0.6

It should be apparent that the salt to acid ratio is significantly different after dilution. Thus the pH change will also be larger.

Third, if the total volume DID NOT remain constant, it means that total volume has increased. If so, then the number of moles of salt and acid remained the same after dilution. So when a small amount of strong acid is added, the ratio is 4.9/5.1, as if there was no dilution! Why?

e.g. for a buffer with 5 mol of both salt and acid in 1 dm3 of water, we dilute it with another 1 dm3 of water. The ratio will remain the same since the volume will cancel out.


Hence you see that what’s crucial is actually the number of moles of salt and acid, as these will determine how large is the pH change.

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Feel free to browse through the blog for more posts on other interesting topics 🙂 eg check out this post on Energetics!

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