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Chemistry: Post your doubts here!

Metanoia

Metanoia

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hamzashariq said:
Thanks. The actual question is Q4b(iii) in the following link. http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w08_qp_4.pdf
In the mark scheme (http://papers.xtremepapers.com/CIE/Cambridge International A and 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?
Click to expand...

Ok, I spotted my mistake, it was ok till one part.

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)
=47090 kJ (instead of 46680 kJ in my first attempt)

Enthalpy change = +47270 - 47090 = +180kJ/mol

The marking scheme actually skips a lot of steps , unless we are very used to solving the question, it is not likely to do it in 1 single step like they did.

Number of bonds broken
82 (C-H) and 39 (C-C) bonds broken

Number of bonds formed
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 82 (C-H) and 35(C-C) and 2 (C=C) formed

Comparing the bonds broken vs formed, we have an overall of 4 (C-C) bonds broken and 2 (C=C) bonds formed.
 
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hamzashariq

hamzashariq

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Metanoia said:
Ok, I spotted my mistake, it was ok till one part.

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)
=47090 kJ (instead of 46680 kJ in my first attempt)

Enthalpy change = +47270 - 47090 = +180kJ/mol

The marking scheme actually skips a lot of steps , unless we are very used to solving the question, it is not likely to do it in 1 single step like they did.

Number of bonds broken
82 (C-H) and 39 (C-C) bonds broken

Number of bonds formed
34 (C-H) and 15 (C-C) bonds and 2 (C=C) and 20 (C-C bonds) and 48 (C-H)
= 82 (C-H) and 35(C-C) and 2 (C=C) formed

Comparing the bonds broken vs formed, we have an overall of 4 (C-C) bonds broken and 2 (C=C) bonds formed.
Click to expand...

Thanks a lot :)
 
Metanoia

Metanoia

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MariamMalik said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_11.pdf
Please can someone explain Q.6, 9, 14, 18 and 30?
Click to expand...

s12qp11

Q6. Its a matter of using a gas constant of your choice and then converting the rest of the variables to the correct units.

For me, I usually use R = 8.31 , so pressure has to be expressed in Pa, volume in m3, temperature in K

n = 0.56/Mr of ethene = 0.56/28 = 0.02

V= nRT/p = 0.02 x 8.31 x 303 / 102 000 = 4.937 x 10-4 m^3 = 493.7 cm^3

Q9. Easiest approach I would use is to do trial and error with the 4 options, which is fast after some practice.
Since B is the answer, I'll use it as example

2P --> 2Q+ R
2.........0 .....0 ..... initial
-2x ...... +2x .... +x ..change
2-2x..... 2x ..... x..... final

Refer to the steps below if you are not sure how the info in the above table is filled up.
1st step : write down info for the initial row
2nd step write down that R will be x moles as stated in question
3rd step: fill in the info in green
4th step: fill in the info in yellow
5th step: fill in the row for final
Check that the info in the final step adds up to 2 + x . ( 2 -2x + 2x + x )

Q14.
I'm guess the hurdle is getting the balanced equation?

10Al + 3Ba(NO3)2 --> 5Al2O3 + 3BaO + 3N2

moles of barium nitrate = 0.783 / 261 = 0.003
moles of N2 = 0.003
vol of N2 at RTP = 0.003 x 24000 = 72 cm^3

Q18.
1st reaction:
conc H2SO4 + KCl --> HCl + K2SO4.
HCl will not react with KI(aq), so observation X is “colourless solution”.

2nd reaction:
Adding Ag+ and Cl- will form a ppt of AgCl if NH3(aq) was not present. Since NH3 (aq) is present, the AgCl will dissolve to form a colourless solution.
Therefore, observations for both reactions are colourless solutions.
 
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Metanoia

Metanoia

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Q30.
pvc_formation.jpg

A: (Correct) PVC gives off toxic gases when burnt, among them are HCl which is the acid gas.
B: (Correct) C=C bonds forms C-C bonds
C: (Correct) Empirical formular of both PVC and monomer are the same (addition polymer) C2H3Cl
D: (Wrong) Repeating unit is C2H3Cl
 
Metanoia

Metanoia

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aliciaa said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_11.pdf
Explanations for no. 9, 18, 36 please! :) The answers are D, C, A.
For no. 36, what is X?
Click to expand...

w11qp11

Q9. I usually try to create the eqm table

Ag+ .........+ ......... Fe2+.........--> .........Ag.........+ Fe3+
1...............................1...............................-..............0.... initial conc
-0.56........................-0.56................. -..........+0.56.... change conc
0.44...........................0.44...................- .............0.56.... final conc

Common student mistake: It is important to realise that Ag is a solid, so we do not fill in any info for it.

Refer to the steps below if you are not sure how the info in the above table is filled up.
1st step : Fill in info for the initial concentration
2nd step Fill in info that Ag+ is 0.44 moles for final concentration
3rd step: fill in the info in green
4th step: fill in the info in yellow
5th step: fill in the row for final concentration

Kc = [Fe3+]/[Ag+][Fe2+]
= (0.56)/(0.44)(0.44)
= 2.89

Q18. The reaction is Ca(OH)2 + SO2 --> CaSO3 + H2O


Q36.
X is N2 (element)
Y is NO (N2 + 0.5 O2 --> NO2)
Z is NO2

It might be tempting to think of X as C, but C does not really exist in the form of carbon element in the engine, but rather as part of a hydrocarbon fuel.

aliciaa said:
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s12_qp_12.pdf
Help with no. 28, answer is D.
Click to expand...

s12q12

I'll try to explain this more with diagram when I figure out what is the best way to attach my sketches to a post. :confused:

For now, understand that using the original dibromine to react with NH3
NH2 will replace one of the Br
then the NH2 will join to the carbon that contains the other Br. The Br is removed and a ring is formed.

So, if we work backwards from the structure of coniine
1) cut the single bond between N and C, thus opening up the ring
2) C is now missing a bond, add a Br to it.
3) Replace the N-H group with a Br
4) That is the original structure of X
 
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Metanoia

Metanoia

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immie.rose said:
Please explain, anyone?
View attachment 44079
Answer: C
Click to expand...

An autocatylsed reaction is when one of the products is actually a catalyst.

The rate of reaction :
starts off slow (no catalyst)
then increases (catalyst)
then decreases (concentration of reactants decreases)

immie.rose said:
View attachment 44080
Answer: D
Click to expand...

In general, to estimate the "Mr of a gas mixture containing A and B", we use (% of A x Mr of A) + (% of B x Mr of B)

Using D as example,
D. (0.825 x 2) + (0.152 x 4) + (0.023 x 16) = 2.626

Doing a quick calculation and comparing to the other options will give D as the highest "Mr", meaning most dense gas mixture.

immie.rose said:
View attachment 44081
Answer : D (Why not B? :unsure:)
Click to expand...

The two structures on the right are more obvious since they are acids which reacts with NaOH to form a carboxylate salt.
The two structures on the left are left out by students. They are esters which can undergo alkali hydrolysis to form a carboxylate salt and alcohol.

immie.rose said:
View attachment 44082
Answer: C ( the concept of no. 3 pls!)
Click to expand...

Carbon is more obvious as it has four valance electrons to form 4 covalent bonds.
Nitrogen can also form 4 bonds (3 single bonds and 1 dative bond), for example in NH4+
 
K

kingo44

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strangerss said:
SOSO247 OKay , in question 21 the only way to solve it is by drawing the displayed formula of the the compound. the first thing I did is I drew the cyclohexene ring you know how right? in the cyclo hexene there is 12 hydrogen atoms and 6 carbon atoms , then from the sixth carbon atom I drew a bond and counted till I reached the 11th carbon atom where I put a double bond because in the question the cis isomer is between the 11th and 12th carbon atom right? the I drew the remaing carbon atoms after the 11th one which are 9 at the last carbon atom I put the aldehyde group because it's pretty obvious it's always at the last. now the last and probably most important step is the number of double bonds between the carbon atoms , now we know that there are 12 hydrogen in the cyclohexene and two for the cis isomer and the total number of hydrogen atoms is 28(from the question) so 28-14=14 carbons atoms are left to be bonded to the carbon chain , so if you give each carbon atom a hydrogen till you used 14 , you add a double bond between the carbon atoms with only three bonds and you will get six double bonds ..I will put the pic. I drew if you didn't get it ok?...question 23 it's C because we get a polychroalkene when we polymerise the monomer given right? then by reacting with water a nucleaphilic substitution reaction takes place in which the hydroxide ions which acts as a nucleaphile displaces the weaker nucleaphile the chlorine atom , it's exactly like when reacting a halognoalkane with sodium hydroxide and we get an alcohol , in A we cannot hydrate an alkane because there are no double bonds , and in B and D it's the same case we cannot oxidise an alkane , so C is correct. question 28 I really got no clue :/ . question 29 it's B because we know that a halogenoalkane plus CN will give a nitrile and when we hydrolyse a nitrile we get a carboxylic acid , when CN displaces Br in B , the final product formed will have a carbon atom attached to an ethyl group which is shown in the diagram given , notice how all the other products formed will not have an ethy group that is the C2CH3 being attached to the carbon atom that was attached to the br , did you get it ?and question 31 I don't know :/ ..hope I helped :)
Click to expand...
where is the pic please cant understand
 
♣♠ Magnanimous ♣♠

♣♠ Magnanimous ♣♠

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Metanoia said:
An autocatylsed reaction is when one of the products is actually a catalyst.

The rate of reaction :
starts off slow (no catalyst)
then increases (catalyst)
then decreases (concentration of reactants decreases)



In general, to estimate the "Mr of a gas mixture containing A and B", we use (% of A x Mr of A) + (% of B x Mr of B)

Using D as example,
D. (0.825 x 2) + (0.152 x 4) + (0.023 x 16) = 2.626

Doing a quick calculation and comparing to the other options will give D as the highest "Mr", meaning most dense gas mixture.



The two structures on the right are more obvious since they are acids which reacts with NaOH to form a carboxylate salt.
The two structures on the left are left out by students. They are esters which can undergo alkali hydrolysis to form a carboxylate salt and alcohol.



Carbon is more obvious as it has four valance electrons to form 4 covalent bonds.
Nitrogen can also form 4 bonds (3 single bonds and 1 dative bond), for example in NH4+
Click to expand...
Can you explain the 32 ques again
 
MariamMalik

MariamMalik

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Metanoia said:
Q6. Its a matter of using a gas constant of your choice and then converting the rest of the variables to the correct units.

For me, I usually use R = 8.31 , so pressure has to be expressed in Pa, volume in m3, temperature in K

n = 0.56/Mr of ethene = 0.56/28 = 0.02

V= nRT/p = 0.02 x 8.31 x 303 / 102 000 = 4.937 x 10-4 m^3 = 493.7 cm^3

Q9. Easiest approach I would use is to do trial and error with the 4 options, which is fast after some practice.
Since B is the answer, I'll use it as example

2P --> 2Q+ R
2.........0 .....0 ..... initial
-2x ...... +2x .... +x ..change
2-2x..... 2x ..... x..... final

Refer to the steps below if you are not sure how the info in the above table is filled up.
1st step : write down info for the initial row
2nd step write down that R will be x moles as stated in question
3rd step: fill in the info in green
4th step: fill in the info in yellow
5th step: fill in the row for final
Check that the info in the final step adds up to 2 + x . ( 2 -2x + 2x + x )

Q14.
I'm guess the hurdle is getting the balanced equation?

10Al + 3Ba(NO3)2 --> 5Al2O3 + 3BaO + 3N2

moles of barium nitrate = 0.783 / 261 = 0.003
moles of N2 = 0.003
vol of N2 at RTP = 0.003 x 24000 = 72 cm^3

Q18.
1st reaction:
conc H2SO4 + KCl --> HCl + K2SO4.
HCl will not react with KI(aq), so observation X is “colourless solution”.

2nd reaction:
Adding Ag+ and Cl- will form a ppt of AgCl if NH3(aq) was not present. Since NH3 (aq) is present, the AgCl will dissolve to form a colourless solution.
Therefore, observations for both reactions are colourless solutions.
Click to expand...

Thank you so much for the help! :D
 
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