Chemistry: Post your doubts here!

[Wei Ling]

ooh i see but are Te and Si in the same group?

 

[Metanoia]

ooh i see but are Te and Si in the same group?

Apologies, I should have typed S (sulfur) instead of Si (silicon)

I'll amend the post

 

[Metallic9896]

When the element used in a question is unfamiliar, we try to use known equations as a comparison.
We know that SCl4 + 2H2O --> SO2 + 4HCl, since Te and S are in the same group, we can expect TeCl4 to follow the reaction

Edit: Typed Si instead of S

Interesting. However, I have a question. Our CIE syllabus mentions in section 9.2 that the only chlorides we need to be familiar with are of sodium to phosphorus, not up to sulfur. And there's no mention of it in the Nitrogen and sulfur section (section 13) either. So on what grounds is this question a valid question? The official book even says, "Sulfur does form chlorides, such as SCl2 and S2Cl2, but you do not need to cover these for your examination." And I don't find anything like this, despite the greater detail on sulfur, in the 2015 or 2014 syllabi either. So assuming a student strictly followed the syllabus, how else would he/she have been able to solve this question? Or am I missing something?

 

[Metanoia]

Interesting. However, I have a question. Our CIE syllabus mentions in section 9.2 that the only chlorides we need to be familiar with are of sodium to phosphorus, not up to sulfur. And there's no mention of it in the Nitrogen and sulfur section (section 13) either. So on what grounds is this question a valid question? The official book even says, "Sulfur does form chlorides, such as SCl2 and S2Cl2, but you do not need to cover these for your examination." And I don't find anything like this, despite the greater detail on sulfur, in the 2015 or 2014 syllabi either. So assuming a student strictly followed the syllabus, how else would he/she have been able to solve this question? Or am I missing something?

Yes, that's a valid concern. That's when we would need to search within the collection of equations (a non-metal tetrachloride) which are familiar to us and find a close match

I've originally typed in SiCl4 (within the syllabus) instead of SCl4 (outside the syllabus). I've just checked the examiners's report and it seemed that their intention was for students to draw inspiration from SiCl4

 

[Metallic9896]

Yes, that's a valid concern. That's when we would need to search within the collection of equations (a non-metal tetrachloride) which are familiar to us and find a close match

I've originally typed in SiCl4 (within the syllabus) instead of SCl4 (outside the syllabus). I've just checked the examiners's report and it seemed that their intention was for students to draw inspiration from SiCl4

View attachment 62568

Very, very interesting. CIE can be weird at times. But yes the equation in the answer mirrors what our book quotes for SiCl4, and yeah I noticed you mentioned SiCl4 before. It's the "non-metal" and "tetrachloride" part which they want us to think about here.

 

[Wei Ling]

Thanks a lot guys , that was really helpful !

 

[Metallic9896]

Thanks a lot guys , that was really helpful !

No problem at all, feel free to ask more questions.

 

[Bishnu Dev]

Please solve this and explain

 

[Metallic9896]

Test is for ammonia gas so gas is ammonia. You can confirm this from the qualitative analysis notes too for p3, which you should have on your fingertips for all papers, not just p3. If ammonia has released then cation must be ammonium ion, NH4+. Finally, since during test we got rid of the ammonium ion as ammonia gas, and I assume heating released the water of crystalization, so Fe2+ and SO22- were left, which would mean that the residue is FeSO4. Is it correct? Hope I helped.

 

[Bishnu Dev]

Test is for ammonia gas so gas is ammonia. You can confirm this from the qualitative analysis notes too for p3, which you should have on your fingertips for all papers, not just p3. If ammonia has released then cation must be ammonium ion, NH4+. Finally, since during test we got rid of the ammonium ion as ammonia gas, and I assume heating released the water of crystalization, so Fe2+ and SO22- were left, which would mean that the residue is FeSO4. Is it correct? Hope I helped.

It is correct. Thanks!

 

[Metallic9896]

No problem! Feel free to ask more via pm or on here.

 

[syed babar]

Hi guys,
As there is less than a month left in the result, the anxiety is off the roof. Can any of you guys predict the threshold for A* in chemistry 9701/42 & 52 taken this year and 12, 22, 33 taken last year. I had an A in AS, and p5 went amazingg almost 27 or above. P4 even was okayish. Expecting a 75 in p4. Will that be enough for an A*?

 

[Bishnu Dev]

No problem! Feel free to ask more via pm or on here.

Here's another one(chemistry)... Can you please explain me the answer for question number 4 --> B --> (i) ..... Paper 9701/21/M/J/09

 

[Metallic9896]

Here's another one(chemistry)... Can you please explain me the answer for question number 4 --> B --> (i) ..... Paper 9701/21/M/J/09

First of all let's identify what C and D are. I assume you got that, but for the sake of clarity, I'll explain it again. For D, we reduced an aldehyde, an ethanal, so we get an alcohol, ethanol. For C, we first reacted C with HCN to get propanenitrile, which upon reacting with sulphuric acid became propanoic acid as hydrolysis via acid of nitriles gives us carboxylic acids. That upon reacting with concentrated sulphuric acid forms prop-2-enoic acid, which is something not part of our course as far as I know, which is why they gave it themselves. Finally, this when reacted with a cold oxidizing agent must form a diol where the double bond exists, which is C. So C must be 2,3-dihydroxy-propanoic acid (I may have gotten the name wrong but you know what I mean): HOCH2CH(OH)COOH

C is: HOCH2CH(OH)COOH
D is: CH3CH2OH

C is a diol and a carboxylic acid, D is an alochol. Obviously when C and D react, the alcohol part of D will react with the carboxylic acid part of C. Thus, the product should be an ester:

HOCH2CH(OH)COOH + HOCH2CH3 = HOCH3CH(OH)COOCH2CH3

Is this correct? Hope I helped.

 

[Bishnu Dev]

First of all let's identify what C and D are. I assume you got that, but for the sake of clarity, I'll explain it again. For D, we reduced an aldehyde, an ethanal, so we get an alcohol, ethanol. For C, we first reacted C with HCN to get propanenitrile, which upon reacting with sulphuric acid became propanoic acid as hydrolysis via acid of nitriles gives us carboxylic acids. That upon reacting with concentrated sulphuric acid forms prop-2-enoic acid, which is something not part of our course as far as I know, which is why they gave it themselves. Finally, this when reacted with a cold oxidizing agent must form a diol where the double bond exists, which is C. So C must be 2,3-dihydroxy-propanoic acid (I may have gotten the name wrong but you know what I mean): HOCH2CH(OH)COOH

C is: HOCH2CH(OH)COOH
D is: CH3CH2OH

C is a diol and a carboxylic acid, D is an alochol. Obviously when C and D react, the alcohol part of D will react with the carboxylic acid part of C. Thus, the product should be an ester:

HOCH2CH(OH)COOH + HOCH2CH3 = HOCH3CH(OH)COOCH2CH3

Is this correct? Hope I helped.

But how to solve the second one? When compound C reacts with E?

 

[Metallic9896]

Haha your original message said i) but anyway, in ii):

Compound C as we established is 2,3-dihydroxy-propanoic acid, HOCH2CH(OH)COOH. Compound E is an alkanal/ethanal after oxidization, so it's ethanoic acid, CH3COOH.

C: HOCH2CH(OH)COOH
E: CH3COOH

Now when C reacts with E, BOTH of it's alcohol groups will react with a total of 2 molecules of E to form two ester linkages where the OH groups occur. So the question will look something like:

CH3COOH + HOCH2CH(OH)COOH = CH3COOCH2CH(OH)COOH

Then

CH3COOCH3CH(OH)COOH + HOOCCH3 = CH3COOCH3CH(OOCCH3)COOH

Thus, the overall equation is:

2 CH3COOH + HOCH2CH(OH)COOH = CH3COOCH3CH(OOCCH3)COOH

I hope I helped. Basically the two diols of C form esters with one ethanoic acid molecule each.

 

[syed babar]

Hi guys,
As there is less than a month left in the result, the anxiety is off the roof. Can any of you guys predict the threshold for A* in chemistry 9701/42 & 52 taken this year and 12, 22, 33 taken last year. I had an A in AS, and p5 went amazingg almost 27 or above. P4 even was okayish. Expecting a 75 in p4. Will that be enough for an A*?

 

[Metallic9896]

Yeah more than enough bud don't worry about it.

 

[Dukula Jayasinghe]

Hi people,
I have prepared a lattice enthalpy examination paper (A2 only). If you would like to get it marked, do the paper and scan or take photos and send it back to me personally through [email protected]

 

[Wei Ling]

Can anyone explain to me why doesn't HCL react with KI to produce chlorine gas?