Chemistry: Post your doubts here!

[Browny]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w03_qp_1.pdf

Can anybody please explain question 16 and 28?

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf

Can anybody please explain question 16 and 28?

Q16. Check that the oxidation state of silver before and after the reaction is +1, it means that it is neither oxidized or reduced. This effectively eliminates options B, C and D. silver chloride is insoluble and the silver complex formed is soluble.

Q28. Since this is a paper from long ago, I think this concept might be out of the syllabus?

The idea is that the H on the hydroxyl (OH) group can be exchanged with the D from deuterium.

R-OH + D-O-D --> R-O-D + H-O-D

If there are 3 hydroxyl groups, then 3 H can be exchanged with D.

 

[IGCSE13]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s10_qp_11.pdf. Q10 ans D Q29 ans B Q35 ans C

 

[omeraz]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Question 28 please

 

[Browny]

Q16. Check that the oxidation state of silver before and after the reaction is +1, it means that it is neither oxidized or reduced. This effectively eliminates options B, C and D. silver chloride is insoluble and the silver complex formed is soluble.

Q28. Since this is a paper from long ago, I think this concept might be out of the syllabus?

The idea is that the H on the hydroxyl (OH) group can be exchanged with the D from deuterium.

R-OH + D-O-D --> R-O-D + H-O-D

If there are 3 hydroxyl groups, then 3 H can be exchanged with D.

Thanks so much!

 

[Ch.Walid]

Notes on organic chemistry please!

 

[Metanoia]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Question 28 please

You can check out page 462.

#9228 and #9239

 

[Gehad Mohamed]

Q4. Use any of the alkane as an example and construct a balance equation for complete combustion, for here, I'll use CH4

I prefer to use table, but not sure if the alignment shows when typed out.
CH4 (g) + 2O2 (g) --> CO2 (g) + 2H2O (l)
10 .............70 ................0 .................. - .......... initial
-10 ............-20 ............ +10 ............. - .......... change
0 ............ 50 .............. 10 ............ - ............ final

Total gases at the end = 50 + 10 = 60 cm3, this fits option D.

Q5. A pi bond is when there is a sideway overlap of orbitals, which is reflected in option B.

Q14. Group II nitrates decompose based on the equation below:

2 X(NO3)2 (s) --> 2 XO (s) + 4 NO2 (g) +O2(g)
The loss in mass is due to the formation of NO2 and O2 gas, the left over mass is XO.

Mass of XO = 1.71 g
Moles of XO = 1.71/ (Mr of X + 16)

Mass of X(NO3)2 = 5 g
Moles of X(NO3)2 = 5/(Mr of X + 62)

since moles of X(NO3)2 = moles of XO
5/(Mr of X + 124) = 1.71/ (Mr of X + 16)

We can solve for Mr of X using the normal maths approach, but it might be faster to do trial and error from the Mr of the four options.

Mg : 5/(24 + 124) does not equal to 1.71/ (24 + 16)
Ca: 5/(40 + 124) = 1.71/ (40+ 16)
Therefore, answer is Ca.

Q39. Sulfuric acid will turn the alcohols in options 1 and 2 into alkenes, which will undergo oxidative cleaving with acidified KMnO4 (decolourises)

Thanks soooòoooooooooooo much

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s10_qp_11.pdf. Q10 ans D Q29 ans B Q35 ans C

Q10. Taking X as monobasic acid and Y as monoacidic base...

If X was strong acid, [H+] = 2M , pH = -log(2) = -0.3 . So student P is wrong.

For student Q, I'll use reasoning rather that complicated maths formula.

Imagine adding both X and Y into seperate glasses of neutral water (pH=7).

Note that for the same concentration (2M) , X (acid) changes the pH by only 1 unit, while Y (base) changes the pH by 2 units. This implies more moles of Y dissociates compared to X.

So student Q is correct.


Q29.
A) butan-1-ol forms CH3CH2CH2=CH2
B) butan-2-ol forms CH3CH2CH2=CH2 , CH3CH2=CH2CH3 (which exists as cis and trans). So total 3 possible alkenes
C) 2-methylpropan-1-ol forms (CH3)2C=CH2
D) 2-methylpropan-2-ol also forms (CH3)2C=CH2


Q35. Is a bit controversial for me. I'll copy and paste the examiners report below.

" 21% of candidates chose the correct answer, C. The most popular answer was A, chosen by 32% of candidates. It is clear from this that for the majority of candidates the key point was whether or not statement 1 was true, i.e., does the oxidation of CO “occur in the atmosphere”?

While CO can be oxidised to CO2 in fires or lightning strikes, this is a combustion reaction, it is not “a reaction that occurs in the atmosphere”, unlike the oxidation of NO to NO2, and the oxidation of SO2 to SO3, both of which do occur “in the atmosphere"

For myself, I am tempted to choose statement 2 and 3, but for different a reason. I took the phrase "non-metallic element X' literally and thought C is in the form of a hydrocarbon compound and not a element.

Again, its one of those debatable questions and interpretation.

 

[Gehad Mohamed]

Hey! Can someone please solve Q 31 in http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s08_qp_1.pdf

 

[Metanoia]

Hey! Can someone please solve Q 31 in http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s08_qp_1.pdf

hmm...Try creating 3 balanced equation for their combustion. If you are still stuck, then I'll provide the answers.

 

[Gehad Mohamed]

hmm...Try creating 3 balanced equation for their combustion. If you are still stuck, then I'll provide the answers.

Got it thank you

 

[Minion96]

I'll discuss this as a theoretical thought procedure, rather than an attempt to predict what experiment is expected.

Some of the procedure are straightforward in theory, but not easy to carry out accurately with limited lab apparatus.

The amount of CaCO3 could be measured by
1) heating mixture to decompose CaCO3 and measuring loss in mass (hard to achieve the decomposition temp)
2) heating mixture to decompose CaCO3 and measuring volume of CO2 (very inaccurate, hard to set up apparatus, temperature affects volume, need to assume no CO2 lost)
3)Dissolve mixture in water, filter the mixture and collect CaCO3 as residue. Heat to dry CaCO3 and weigh it. (most straightforward of all 3 methods).

Thanks alot sir, You helped me out

 

[ZaqZainab]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w03_qp_1.pdf
Q2

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w03_qp_1.pdf
Q2

Out of every 100 g of fertilizer, 30 g is P205.

Out of 30 g of P2O5 ,

mass of P
= (2 x Mr of P/Mr of P2O5) x 30 g
= (62/142) x 30 g
=13.1 g

 

[ZaqZainab]

Out of every 100 g of fertilizer, 30 g is P205.

Out of 30 g of P2O5 ,

mass of P
= (2 x Mr of P/Mr of P2O5) x 30 g
= (62/142) x 30 g
=13.1 g

Thanks

 

[FarahMJ]

Help please, how do we solve this? Answer is C

 

[ZaqZainab]

Help please, how do we solve this? Answer is C

You have
ΔHc of C2H6 = -1560
ΔHc of H2 = -286
ΔHf of C2H6= -158
Write an equation of formation of C2H6 because the student was trying to calculate ΔHf
C + H2 ---> C2H6
but this isn't balanced
2C+ 3H2---> C2H6
there better
we want ΔHc of C lets take it as x
2x+(3*-286)+1560= -158
+ because C2H6 is in the products and its combustion
make x the subject
x=((-158)-(1560)-(3*-286))/2

 

[Metanoia]

Help please, how do we solve this? Answer is C

Using a cycle, it can look like this.

 

[FarahMJ]

Thank you ZaqZainab and Metanoia so much