Chemistry: Post your doubts here!

[MariamMalik]

Hi, to make it easier for us to check, could you also put the marking scheme answers next to the 5 questions?

The answers for the questions are B,B,B,C and D

 

[Metanoia]

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?

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.

 

[Metanoia]

Q30.

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

 

[aliciaa]

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?


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.

 

[immie.rose]

Please explain, anyone?

Answer: C


Answer: D


Answer : D (Why not B? )


Answer: C ( the concept of no. 3 pls!)

 

[Metanoia]

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?

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.

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.

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.

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

 

[Metanoia]

Please explain, anyone?
View attachment 44079
Answer: C

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)

View attachment 44080
Answer: D

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.

View attachment 44081
Answer : D (Why not B? )

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.

View attachment 44082
Answer: C ( the concept of no. 3 pls!)

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+

 

[kingo44]

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

where is the pic please cant understand

 

[♣♠ Magnanimous ♣♠]

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+

Can you explain the 32 ques again

 

[Metanoia]

Can you explain the 32 ques again

Is your confusion on the statement 3 for nitrogen? Using NH4+ as example, we can see that a N atom can actually form 4 covalent bonds (where 1 of them is a dative bond).



So it makes statement 3 of the question true.

 

[♣♠ Magnanimous ♣♠]

Is your confusion on the nitrogen? Using NH4+ as example, we can see that a N atom can actually form 4 covalent bonds (where 1 of them is a dative bond).

View attachment 44097

So it makes statement 3 of the question true.

Can you explain statement 2 and 1

 

[Metanoia]

Can you explain statement 2 and 1

I've attached the electron in boxes configuration for C and N.
C has and empty p orbital but N does not, so statement 1 is incorrect.

The last occupied shell is 2 for both C and N, so statement 2 is correct.

 

[♣♠ Magnanimous ♣♠]

View attachment 44098

View attachment 44099 View attachment 44100

I've attached the electron in boxes configuration for C and N.
C has and empty p orbital but N does not, so statement 1 is incorrect.

The last occupied shell is 2 for both C and N, so statement 2 is correct.

thanks

 

[MariamMalik]

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.

Thank you so much for the help!

 

[aliciaa]

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.



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.

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

Alright thank you soo much!
For no. 9, I used to correct method, just that i calculated the mol for Ag (solid) as well, so i got the incorrect answer.

 

[Snowysangel]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_12.pdf 4,10 and 18

 

[Metanoia]

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.

This is one way to visualise the answer to Q28, I think a diagram is easier than the chunk to text I typed in #9228

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_12.pdf 4,10 and 18

Hi, for ease of discussion, can put up the suggested answers for your 3 selected questions?

w11qp12

Q4.
N2 + 3H2 --> 2NH3

Mass of H2 reacted = 12 000 - 96 000 = 24 000 g
Moles of H2 reacted = 24 000 / 2 = 12 000 mol

Moles of NH3 produced = (12 000/3 ) x 2 = 8 000 mol
Mass of NH3 produced = 8 000 x 17 = 136 000 g = 136 kg

Q10.
Draw a cycle based on heat of formation of CH4 as below, we can see that all the values in option A is needed to calculate the BE of (C-H)



Q18.
Old lime mortar produces a gas when reacted with HCl, so it should be CaCO3 (to produce CO2) , instead of CaO and Ca(OH)2.

This leaves us with Options A and D.
A: CaO is harder than CaCO3
D: Ca(OH)2 is softer than CaCO3

So option D is more likely

 

[redd]

Does anyone know anything about practical 34?

 

[Minion96]

Does anyone know anything about practical 34?

My teacher says that there is some titration performance and we will be provided with a mixture of calcium carbonate and potassium chloride! they have also asked for a crucible, but what will we be doing with the crucible because the mixture isn't decomposing. please ask ur teacher or search that at which temp does it decompose or how much time it must take. we couldnt work out any experiment!