Chemistry: Post your doubts here!

[omarjaved619]

HELP! Answer is C. Does it have something to do with the proportions of the acid and base?

 

[Metanoia]

because the concentration of the limiting reactant is not changing

Actually, since the limiting reactant here is zinc (solid), it so would be more correct to use "mass or moles" being the same between the two experiments rather than "concentration"

Its B now am asking why B not D

The is no change in the

1) particle size of solid reactant (zinc)
2) concentration of the aqueous reactant (acid)
3 )initial temperature of reaction

 

[Metanoia]

View attachment 44033

HELP! Answer is C. Does it have something to do with the proportions of the acid and base?

Heat of neutralization for strong acids and strong alkali is estimated to be -57kJ PER MOLE of H2O formed.

For both experiments. (Involving strong acid and bases) , it reflects twos moles of water formed .

So heat of reaction is -57 x 2 for both equations.

 

[Gehad Mohamed]

because the concentration of the limiting reactant is not changing

Ok thanks a lot

 

[Gehad Mohamed]

Actually, since the limiting reactant here is zinc (solid), it so would be more correct to use "mass or moles" being the same between the two experiments rather than "concentration"


The is no change in the

1) particle size of solid reactant (zinc)
2) concentration of the aqueous reactant (acid)
3 )initial temperature of reaction

Thank you so much

 

[omarjaved619]

Heat of neutralization for strong acids and strong alkali is estimated to be -57kJ PER MOLE of H2O formed.

For both experiments. (Involving strong acid and bases) , it reflects twos moles of water formed .

So heat of reaction is -57 x 2 for both equations.

Thanks a lot!

 

[omarjaved619]

Answer is B. Cant quite figure out the intermediates.

 

[Metanoia]

View attachment 44035

Answer is B. Cant quite figure out the intermediates.

W: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3

 

[♣♠ Magnanimous ♣♠]

W: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3

CAn you explain of X again

 

[Metanoia]

CAn you explain of X again

Notice that the H on the centre Carbon is ultimately replaced by CN?

This would require us to change the H with a halogen (either Br or Cl), one way to do it would be free radical substitution.

Then the halogen alkane intermediate will undergo a nucleophilic substitution by adding CN- (using NaCN or KCN)

 

[♣♠ Magnanimous ♣♠]

Notice that the H on the centre Carbon is ultimately replaced by CN?

This would require us to change the H with a halogen (either Br or Cl), one way to do it would be free radical substitution.

Then the halogen alkane intermediate will undergo a nucleophilic substitution by adding CN- (using NaCN or KCN)

ok

 

[omarjaved619]

W: reacts with Br2 and uv to form intermediate CH3CHBrCH3 , which then reacts with CN-
X: reacts with Br2 and uv to form intermediate (CH3)2C(OH)Br ,which then reacts with CN-
Y: reacts with HBr to form intermediate CH3CHBrCH3 , which then reacts with aq. NaOH
Z: undergoes hydrolysis to form intermediate CH3CH2CH2OH , which then reacts with PBr3

Daaaang. Thanks again mate

 

[Metanoia]

Daaaang. Thanks again mate

No problem, I'm actually a tutor based in Singapore, so I guess we are neighbors.

Would like to get an understanding of students exams schedule in other countries, so in Malaysia, both the A and O levels (?) are done in May and not November?

 

[hamzashariq]

C40H82 ---> C16H34 + 2C12H24
How would you calculate the bond energy of this reaction? (bond energy C-C 350, C=C 610, C-H 410)

 

[Gehad Mohamed]

Hello ! can someone plz explain the following : Q 4 , 5, 14,39 in
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

 

[omarjaved619]

No problem, I'm actually a tutor based in Singapore, so I guess we are neighbors.

Would like to get an understanding of students exams schedule in other countries, so in Malaysia, both the A and O levels (?) are done in May and not November?

Oh I'm glad to hear that. Well, as far as I know, students from most colleges and schools (including my school) sit for the May/June exams. Then those who wish to retake sit for the Oct/Nov session.

 

[sadiaali]

Please anyone explain me Q5
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s07_qp_1.pdf

 

[Metanoia]

C40H82 ---> C16H34 + 2C12H24
How would you calculate the bond energy of this reaction? (bond energy C-C 350, C=C 610, C-H 410)

Is this the actual question?
We'll need to make some reasonable assumptions here that the C12H24 is an alkene rather than a cycloalkane. The C12H24 would be having one C=C bond as it is "missing" 2 hydrogen

Energy taken in to break bonds of C40H82
82 (C-H) and 39 (C-C) bonds
= 82 (410) + 39 (350)
=47270 kJ

Energy given out in forming bonds of C16H34 and 2 C12H24
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 34 (410) + 15 (350) + 2 (610) + 20 (350) + 48(410)
=46680 kJ

Enthalpy change = +47270 - 46680 = +590kJ/mol

Quite a bit of numbers at this late hour, hope I didn't make a mistake. There is a slightly shorter way, but I think this way should be more understandable.

 

[hamzashariq]

Is this the actual question?
We'll need to make some reasonable assumptions here that the C12H24 is an alkene rather than a cycloalkane. The C12H24 would be having one C=C bond as it is "missing" 2 hydrogen

Energy taken in to break bonds of C40H82
82 (C-H) and 39 (C-C) bonds
= 82 (410) + 39 (350)
=47270 kJ

Energy given out in forming bonds of C16H34 and 2 C12H24
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 34 (410) + 15 (350) + 2 (610) + 20 (350) + 48(410)
=46680 kJ

Enthalpy change = +47270 - 46680 = +590kJ/mol

Quite a bit of numbers at this late hour, hope I didn't make a mistake. There is a slightly shorter way, but I think this way should be more understandable.

Thanks. The actual question is Q4b(iii) in the following link. http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_qp_4.pdf
In the mark scheme (http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w08_ms_4.pdf)
they used a method i don't understand to get a different answer. Can you check again please?

 

[Metanoia]

Hello ! can someone plz explain the following : Q 4 , 5, 14,39 in
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w06_qp_1.pdf

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)