We are currently struggling to cover the operational costs of Xtremepapers, as a result we might have to shut this website down. Please donate if we have helped you and help make a difference in other students' lives!
Click here to Donate Now (View Announcement)
ooh i see but are Te and Si in the same group?
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?
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
Thanks a lot guys , that was really helpful !
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.
Here's another one(chemistry)... Can you please explain me the answer for question number 4 --> B --> (i) ..... Paper 9701/21/M/J/09No 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
But how to solve the second one? When compound C reacts with E?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.
For almost 10 years, the site XtremePapers has been trying very hard to serve its users.
However, we are now struggling to cover its operational costs due to unforeseen circumstances. If we helped you in any way, kindly contribute and be the part of this effort. No act of kindness, no matter how small, is ever wasted.
Click here to Donate Now