Wow so late haven't sleep ah?

barely did any revision the whole day so have to do it now haha. usually i only have the drive to do revision at midnight

Let Ann's money at first be $ 7x

Since the ratio is 7 : 5, Ben's money at first
= $ 5x

After each spent $60, the ratio becomes 3 : 2.

But the difference in their savings is unchanged. (This is a Primary 5/6 concept)


Ann's money now = $(7x - 60)
Ben's money now = $(5x - 60)


Since the ratio is 3 : 2 now,.

(7x - 60) / (5x - 60) = 3/2

Cross multiply,

2(7x - 60) = 3(5x - 60)

(Or you can skip the previous step and write this immediately. We are just swapping the numbers

i.e if Ann : Ben = 3 : 2, then 2 Ann = 3 Ben)


14x - 120 = 15x - 180

Bring over ,

180 - 120 = 15x - 14x

x = 60

Ann's money now

= $(7x - 60)

= $(7(60) - 60)

= $(6(60))

= $360

This question is more of a secondary 1 standard question (problems in a single variable)

Burn midnight oil is okay, just get enough rest.

No more school lessons already right?

thx :) nope

do you think it is necessary to finish the entire tys? o levels are getting closer but i barely finished half of the entire tys...

Just look through the questions and see which ones you find difficult to do.

Those that are similar to what you've done before or easy, can skip.

But for E Math, do as many of those real-world application questions (the last one of every paper). Very helpful. Gets you to think more criitically.

The one in 2018 (Huan and her box) was a good one.

thx :) will do so

Why do I have a feeling that there is at least one "primary school" question each year for the past few years.?

It's free marks, basically.

Primary school uses models, units and parts or just units alone.

Secondary transition to algebra.

"do you think it is necessary to finish the entire tys? o levels are getting closer but i barely finished half of the entire tys...'

After doing many of these, you should be familiar with what could possibly go on in every question. However, expect some variations in the storylines and the phrasing of the questions used, especially in the application question in Q10/Q11.

The same goes for the application question in Chemistry. You never know what will come out for the application question. Unlike other school papers in which the application questions seem to be extracted from JC topics, the O Level ones seem to be "out of the blue".

hi, is calcium is added to separate samples of a salt solution, why will it react with water before reacting with the salt solution?

What kind of salt? Any specific salt or generally?

Anyway, if it's a solution, the solvent (water) is in excess. So if there are effective collisions, the probability of a calcium atom colliding with water molecules is much higher than with the salt ions (eg. Sulphate or chloride anions)

Calcium, being a reactive Group II metal ,will quickly react with the water to form calcium hydroxide and hydrogen gas.

Ca + 2H2O → Ca(OH)2 + H2

This is so as to attain the stable octet/noble gas configuration quickly for stability.

copper(ii) sulfate, magnesium sulfate, cobalt(ii) sulfate, chromium (iii) sulfate.

the qn is "the student added calcium to separate samples of each of the four salt solutions. the student observed fizzing. explain this observation"

the answer key wrote : the calcium which is a reactive metal reacted with the water in the solutions and produced hydrogen gas.

i dont understand the answer, isnt the calcium supposed to displace the 4 cations?

See the previous reply.

if calcium reacts with water, calcium hydroxide is formed, wont it react with the salts to form another type of salt? when they say "separate samples of each of the four salt solutions" , do they mean that they just want to separate the cation from the anion in the respective salts?

Separate samples means :

In sample 1, you have copper(II) sulphate(CuSO4) solution

This is a soluble salt so you will have Cu²+ and SO4²- ions in water.

In sample 2, you have magnesium sulphate (MgSO4)

This is a soluble salt so you will have Mg²+ and SO4²- ions in water


In sample 3, you have cobalt(II) sulphate (CoSO4)

This is a soluble salt so you will have Co²+ and SO4²- ions in water


In sample 4, you have chromium (III) sulphate, Cr2(SO4)3

This is a soluble salt so you will have Cr³+ and SO4²- ions in water.

The only sparingly soluble (which is 'insoluble' for O level chemistry) sulphate salts you learn at your level are BaSO4, CaSO4, PbSO4


After Ca(OH)2 is formed, it dissociates in the water to form a strongly alkaline solution.

You have Ca²+ ions and OH- ions.


The Ca²+ then can displace those cations in each respective solution, (It is above them in the reactivity series) to form the sparingly soluble/'insoluble' CaSO4, which precipitates.

BUT, the question is asking you to explain the observation of fizzing.

That comes from calcium's reaction with water.

If you were asked to explain " a white solid was formed later", then it would be the displacement.

thankss :)! by the way if chlorine atoms acts as a catalyst, does the chlorine in ClO acts as a catalyst too despite it being a chloride ion now?

for example :
Cl + O3 ~> ClO + O2
ClO + O3 ~> 2O2 + Cl

I would count “Cl” before the first reaction as being the catalyst, while “ClO” is just an intermediate form taken during the process.

I would not actually say that the “Cl” inside “ClO” is a catalyst.

Remember the definition of catalyst in O level chemistry.

It is chemically unchanged at the end of the reaction.

Here we're looking at catalytic ozone destruction.


The Cl● radical reacts with O3 (ozone) to form the ClO● radical and O2 (oxygen gas)

The ClO● radical then reacts with an O● radical to form O2 gas, and this regenerates the Cl● radical.

The ClO● can also react with another O3 molecule to form two O2 molecules and regenerate the Cl● radical


ClO● is not a chloride ion.

So to answer your question, the ClO● radical is an intermediate compound and therefore we should only consider Cl● radical as the catalyst.


You learnt Chapman Cycle before?

https://www.compoundchem.com/2018/01/19/solomon-cfcs/

For your reference. It's an excellent site for learning chem.

thx :) i have not learn about the chapman cycle before

Oh okay. Keep in view for next time then :D