<![CDATA[Chemistry teacher - Post your ''A'' level questions here]]>Hi all,


I am opening this thread to assist students who have difficulties in their ‘‘O’’ level and ‘‘A’’ level chemistry.
I will try my best to answer your questions.

Cheers,
Chemistry teacher

]]>https://forum.kiasuparents.com/topic/23389/chemistry-teacher-post-your-a-level-questions-hereRSS for NodeFri, 08 May 2026 07:07:01 GMTFri, 24 Jun 2011 19:54:25 GMT60<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Thu, 03 Oct 2013 17:42:04 GMT]]>

equink:
Hi Mr David,

Can you please help me solve the qn below? I've been trying really long to figure it out already. 😞

\"2.8g of metal M reacts a metal oxide with 0.8g of oxygen to form MO. Metal M also reacts with oxygen to form a second oxide with oxygen in which the metal and oxygen present is in a ratio of 7:3. What is the formula of the metal oxide? \"

A:MO2 , B: M2O , C:M2O3 , 😧 M3O2
Ar of O = 8
no. of moles of O = 0.8/8 = 0.1
no. of moles of M = 0.1
Ar of M = 2.8/0.1 = 28
M = Ni

Let the no. of moles of M be x and no. of moles of O be y
mass of metal = 28x
mass of oxygen = 8y
Since metal and oxygen is present in the ratio of 7:3,
28x/8y = 7/3
x/y = (8/28)(7/3) = 2/3
Formula of metal oxide = M2O3

]]>https://forum.kiasuparents.com/post/1112397https://forum.kiasuparents.com/post/1112397Thu, 03 Oct 2013 17:42:04 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Sat, 21 Sep 2013 04:26:40 GMT]]>Hi Mr David,

Can you please help me solve the qn below? I’ve been trying really long to figure it out already. 😞

"2.8g of metal M reacts a metal oxide with 0.8g of oxygen to form MO. Metal M also reacts with oxygen to form a second oxide with oxygen in which the metal and oxygen present is in a ratio of 7:3. What is the formula of the metal oxide? "

A:MO2 , B: M2O , C:M2O3 , 😧 M3O2

]]>https://forum.kiasuparents.com/post/1100138https://forum.kiasuparents.com/post/1100138Sat, 21 Sep 2013 04:26:40 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Sat, 21 Sep 2013 04:23:06 GMT]]>Can someone pleaaase help me solve this qn!! I’ve been trying really long to figure it out already. 😞


"2.8g of metal M reacts a metal oxide with 0.8g of oxygen to form MO. Metal M also reacts with oxygen to form a second oxide with oxygen in which the metal and oxygen present is in a ratio of 7:3. What is the formula of the metal oxide? "

A:MO2 , B: M2O , C:M2O3 , 😧 M3O2

]]>https://forum.kiasuparents.com/post/1100136https://forum.kiasuparents.com/post/1100136Sat, 21 Sep 2013 04:23:06 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Sat, 21 Sep 2013 04:22:25 GMT]]>Can someone pleaaase help me solve this qn!! I’ve been trying really long to figure it out already. 😞


"2.8g of metal M reacts a metal oxide with 0.8g of oxygen to form MO. Metal M also reacts with oxygen to form a second oxide with oxygen in which the metal and oxygen present is in a ratio of 7:3. What is the formula of the metal oxide? "

A:MO2 , B: M2O , C:M2O3 , 😧 M3O2

]]>https://forum.kiasuparents.com/post/1100135https://forum.kiasuparents.com/post/1100135Sat, 21 Sep 2013 04:22:25 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Mon, 06 May 2013 04:13:42 GMT]]>

Jennifer:
Hi, my elder boy ran into an issue with the following.


Hope someone can enlighten him. Many thanks in advance.
******************************************************

We need to calculate the enthalpy change of a certain reaction, usually acid-base, acid-carbonate, or dissolution reactions. So we take the two substances and mix them together and record the temperature change. Using Q=mcT, we calculate the amount of thermal energy given off or taken in by the reaction, and then using H=Q/n we find the enthalpy change of the reaction. However, the confusion is what value of n (number of moles) to use.

Examples of reactions:

HCl + NaOH -> H20 + NaCl (here n is the mol of any one of the reactants or products, no confusion.)

2HNO3 + Na2CO3 -> H2O + 2NaNO3 (here teacher says take mol of H2O, because it is the one carrying off the thermal energy. Or take it as the mol of Na2CO3, because that is the reactant that is changing, not the acid.)

H2SO4 + KOH -> 2H2O + K2SO4 (here following the same rule as above, taking the number of mol of water, is for some reason wrong.)

Is there a general rule or concept that allows you to figure out exactly which number of mol to use?
'N' is defined as the number of moles of limiting reagent. You will need to calculate which reactant is in excess and the other is simply the limiting reagent.

Any enthalpy change that occurs during the reaction is due to the number of moles of limiting reagent reacting with the other reactant(s). Excess reactants will not undergo the reaction as the limiting reagent would have been used up and hence excess reactants do not contribute to the enthalpy change.

Do take note of the reacting ratio as well.

]]>https://forum.kiasuparents.com/post/998456https://forum.kiasuparents.com/post/998456Mon, 06 May 2013 04:13:42 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Sun, 05 May 2013 17:04:31 GMT]]>can someone help with this chemistry Q

The equilibrium constant Kp for the Haber Process at 500 deg C is 1.5*10^-5 atm^-2.
Calculate the pressure required to convert half f the mixture of nitrogen and hydrogen in molar ratio 1:3 into ammonia at equilibrium at 500 deg
C.
thank you

]]>https://forum.kiasuparents.com/post/998265https://forum.kiasuparents.com/post/998265Sun, 05 May 2013 17:04:31 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Sun, 31 Mar 2013 09:57:56 GMT]]>Hi, my elder boy ran into an issue with the following.


Hope someone can enlighten him. Many thanks in advance.
******************************************************

We need to calculate the enthalpy change of a certain reaction, usually acid-base, acid-carbonate, or dissolution reactions. So we take the two substances and mix them together and record the temperature change. Using Q=mcT, we calculate the amount of thermal energy given off or taken in by the reaction, and then using H=Q/n we find the enthalpy change of the reaction. However, the confusion is what value of n (number of moles) to use.

Examples of reactions:

HCl + NaOH -> H20 + NaCl (here n is the mol of any one of the reactants or products, no confusion.)

2HNO3 + Na2CO3 -> H2O + 2NaNO3 (here teacher says take mol of H2O, because it is the one carrying off the thermal energy. Or take it as the mol of Na2CO3, because that is the reactant that is changing, not the acid.)

H2SO4 + KOH -> 2H2O + K2SO4 (here following the same rule as above, taking the number of mol of water, is for some reason wrong.)

Is there a general rule or concept that allows you to figure out exactly which number of mol to use?

]]>https://forum.kiasuparents.com/post/977862https://forum.kiasuparents.com/post/977862Sun, 31 Mar 2013 09:57:56 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Wed, 27 Feb 2013 13:05:00 GMT]]>Oh... I see. Okay,thanks so much.

HAHAHA. Yeah, qn 10 of chem tutorial? 😛

]]>https://forum.kiasuparents.com/post/962852https://forum.kiasuparents.com/post/962852Wed, 27 Feb 2013 13:05:00 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Tue, 26 Feb 2013 10:32:54 GMT]]>

equink:
Can you please help me to solve this question?


10cm^3 of a hydrocarbon C4Hy was allowed to react with an excess of oxygen at 150°c and 1atm. There was an expansion of 10cm^3 in volume. Deduce the value of y.

Is the \"expansion\" stated in comparison to the volume of the hydrocarbon AND excess oxygen or to the volume of hydrocarbon AND reacted oxygen?
hahaha I think we might be in the same school... Q10?

Anyway, I think the expansion of 10cm^3 refers to the total volume of gaseous products after complete combustion. (Ie, Vol of CO2 & H2O (g) + vol of excess unreacted O2 = 20 cm^3)

To solve,
vol of CO2 + vol of H2O (g) + vol of unreacted O2 - (10 cm^3 of C4Hy + vol of reacted & unreacted O2) = 10 cm^3

10 cm^3 = vol of CO2 + vol of H2O - 10 cm^3 - vol of reacted O2
= 40 + 10 (y/2) - 10 - 10 (4+y/4) [based on chemical equation, multiply by 10 cm^3 since 1 'mol of C4Hy gives 10 cm^3 -- since it's gas, volume can multiply throughout].

y = 8

Took me quite some time to get it too, hope this helps!

]]>https://forum.kiasuparents.com/post/962101https://forum.kiasuparents.com/post/962101Tue, 26 Feb 2013 10:32:54 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Sat, 23 Feb 2013 01:06:05 GMT]]>Can you please help me to solve this question?


10cm^3 of a hydrocarbon C4Hy was allowed to react with an excess of oxygen at 150°c and 1atm. There was an expansion of 10cm^3 in volume. Deduce the value of y.

Is the "expansion" stated in comparison to the volume of the hydrocarbon AND excess oxygen or to the volume of hydrocarbon AND reacted oxygen?

]]>https://forum.kiasuparents.com/post/960681https://forum.kiasuparents.com/post/960681Sat, 23 Feb 2013 01:06:05 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Tue, 03 Jul 2012 08:44:22 GMT]]>Dear friends


I like Chemistry teacher - Post your ''A'' level questions here very much.

Very useful for me.

If you have some time, pls visit my blog at: http://teacherinterviewquestions.info/chemistry-teacher-interview-questions

Rgs

]]>https://forum.kiasuparents.com/post/788248https://forum.kiasuparents.com/post/788248Tue, 03 Jul 2012 08:44:22 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Fri, 22 Jul 2011 10:21:36 GMT]]>

Mr David:
Hi.

As far as ''O'' level is concerned, there are 3 standard reactions for acids i.e. 1) acid + metal, 2) acid + base and 3) acid + metal carbonate. Meanwhile, there are 2 standard reactions for bases i.e. 1) base + acid (same as that for acid) and 2) base + ammonium salt. Therefore, we have the so-called 4 standard reactions since the acid-base neutralization repeats itself in both categories. Of course, there are cases when a given reaction doesn't come under any of these 4 categories. Then, we just have to adopt the displacement reaction i.e. exchanging partners, such as the example which you had given. Sodium is a more reactive metal than iron(II), therefore it will displace iron(II) from its sulfate. If you cannot categorize a given reaction under any of these 4 categories, it is ''safe'' to adopt the displacement reaction, as far as ''O'' level is concerned. Actually, there is one more reaction too i.e. metal + water to give salt and hydrogen.

Hope it makes chemistry simpler for you.
Chemistry Teacher
Sorry, but I strongly disagree with your answer.

This reaction is clearly a 'precipitation' reaction under preparation of salts.

True, a more reactive pure metal displaces a less reactive metal from its solution, but this applies only to e.g. Fe+CuSO4 ->FeSO4 + Cu, but its compound does not, e.g. NaCl + CuSO4 do not react with each other.

The reaction stated is a precipitation(when an aqueous acid/alkali/salt and another aqueous salt are mixed, if one of the compound is insoluble in water after exchanging the ions, a reaction will occur. Otherwise there will be no reaction), precipitation reaction has no link to reactivity series.

Examples of precipitation:

HCl + AgNO3 - > HNO3 + AgCl (s)
2KOH + ZnCl2 -> 2KCl + Zn(OH)2 (s)
CuSO4+BaCl2 -> CuCl2 + BaSO4 (s)

Some examples of no reaction:

HCl + CuSO4 (after ions are exchanged, no insoluble compound will form)
KOH + CaCl2 (after ions are exchanged, no insoluble compound will form)
CuSO4+FeCl2(after ions are exchanged, no insoluble compound will form)

]]>https://forum.kiasuparents.com/post/503313https://forum.kiasuparents.com/post/503313Fri, 22 Jul 2011 10:21:36 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Thu, 21 Jul 2011 22:10:45 GMT]]>Hi Mr David, thanks. Do you have good reference book for O and A lvl?

]]>https://forum.kiasuparents.com/post/501610https://forum.kiasuparents.com/post/501610Thu, 21 Jul 2011 22:10:45 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Thu, 21 Jul 2011 20:56:43 GMT]]>Hi.

As far as ‘‘O’’ level is concerned, there are 3 standard reactions for acids i.e. 1) acid + metal, 2) acid + base and 3) acid + metal carbonate. Meanwhile, there are 2 standard reactions for bases i.e. 1) base + acid (same as that for acid) and 2) base + ammonium salt. Therefore, we have the so-called 4 standard reactions since the acid-base neutralization repeats itself in both categories. Of course, there are cases when a given reaction doesn’t come under any of these 4 categories. Then, we just have to adopt the displacement reaction i.e. exchanging partners, such as the example which you had given. Sodium is a more reactive metal than iron(II), therefore it will displace iron(II) from its sulfate. If you cannot categorize a given reaction under any of these 4 categories, it is ‘‘safe’’ to adopt the displacement reaction, as far as ‘‘O’’ level is concerned. Actually, there is one more reaction too i.e. metal + water to give salt and hydrogen.

Chemistry Teacher

]]>https://forum.kiasuparents.com/post/501606https://forum.kiasuparents.com/post/501606Thu, 21 Jul 2011 20:56:43 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Tue, 19 Jul 2011 00:44:48 GMT]]>Hi Mr David, i have another question. We know there are 4 basic types of chemical reaction and we need to understand which one before makinh up the equation.


Eg. When sodium hydroxide solution is added to aqueous iron(II) sulfate, iron(II) hydroxide precipitate is formed.
FeSO4(aq) + 2NaOH(aq) –> Na2SO4(aq) + Fe(OH)2(s)

How to know which type of reaction the above belongs to, if we are only given the left sides equation and ask to form the right equation.

Thanks again.

]]>https://forum.kiasuparents.com/post/495195https://forum.kiasuparents.com/post/495195Tue, 19 Jul 2011 00:44:48 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Tue, 28 Jun 2011 14:48:01 GMT]]>Sure, no problem.


Cheers,
Chemistry Teacher

]]>https://forum.kiasuparents.com/post/460663https://forum.kiasuparents.com/post/460663Tue, 28 Jun 2011 14:48:01 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Tue, 28 Jun 2011 13:10:40 GMT]]>Thanks for your explanation. The 12 & 1 are total electrons. :thankyou:

]]>https://forum.kiasuparents.com/post/460572https://forum.kiasuparents.com/post/460572Tue, 28 Jun 2011 13:10:40 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Mon, 27 Jun 2011 14:52:24 GMT]]>Hi. Firstly, we have to be clear on one thing: Are you talking about valence electrons or the total number of electrons? For instance, if A has 3 electrons in total, then A has to be from Group 1 ( Remember your electronic configuration-2,1) If A has 3 valence electrons then, A has to be from Group 3.


In this question, there is no ambiguity because there is no way an element has 12 valence electrons (octet rule), although elements in Period 3 can expand their octet configuration. Therefore, it is safe to declare that W has 12 electrons in total. If that is the case, then W has to be from Group 2 i.e. magnesium (electronic configuration-2,8,2). Logically, Y has to be hydrogen.

The result is an ionic compound which is magnesium hydride. In this special scenario, the hydrogen takes on a -1 charge i.e. H-. H+ is more common, but in the presence of a highly electropositive element such as magnesium, H takes on a -1 charge. Therefore, you have to draw the dot and cross diagram for magnesium hydride (ionic compound) i.e. MgH2 or WY2

Chemistry Teacher

]]>https://forum.kiasuparents.com/post/459062https://forum.kiasuparents.com/post/459062Mon, 27 Jun 2011 14:52:24 GMT<![CDATA[Reply to Chemistry teacher - Post your ''A'' level questions here on Mon, 27 Jun 2011 05:56:54 GMT]]>Hi, I have a question here. Really appreciate your help. 😄


W has 12 electrons
Y has 1 electron
What are the elements make up W & Y, it can be covalent or ionic using dot and cross diagrams. (cannot remember the exact words used in the question, but something like that).

Thank you so much!

]]>https://forum.kiasuparents.com/post/457964https://forum.kiasuparents.com/post/457964Mon, 27 Jun 2011 05:56:54 GMT