Chemistry: Post your doubts here!

[RHB :3]

It is an endothermic reaction which means the bond broken will be more than bonds formed.
from the data booklet we know that the N-N triple bond is 994 kJ/mol and the O=O bond is 496 kJ/mol
Therefore:
994 + 496 - 2xNO = 180
by solving you get NO bond energy is 655kJ/mol

Thank you

 

[Browny]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
Can anybody please explain in detail questions 11, 32, 33 and 35, including all the the points in the questions from 30-40?

 

[MariamMalik]

excuse me --> the real answer of 10 is C

Oh, really? If its C then how is it so?

 

[MariamMalik]

http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_s03_qp_1.pdf
Can anybody please explain in detail questions 11, 32, 33 and 35, including all the the points in the questions from 30-40?

Q11. A) Ammonium ethanoate is not completely ionised in water and it's not acidic in aqueous solution. Water is more polar than ammonia, so the only choice left is that ammonia is a stronger base than water


Q30. D) When propanone reacts with hydrogen cyanide, nucleophilic addition takes place and a hydroxynitrile is formed.
CH3COCH3 + HCN → CH3C(OH)CH3CN
2-hydroxybutanenitrile is hydrolysed under acidic conditions to Butanoleic acid
CH3C(OH)CH3CN + H2O + H+ → (CH3)2C(OH)CO2H + NH4

Q31. C) Silicon tetrachloride doesn't have co-ordinate bonding because it follows the octet rule, sharing all of its valence electrons with the chlorine atoms.
Both silicon and chlorine are non-metals, so, it has covalent bonding.
There are instantaneous dipole-induced dipole forces between the molecules (Van der Waals forces)

Q32. A) The right-hand side of the structure is polar and since water is a dipole, it is attracted to water.
The alkyl chain is non-polar and attracted to other alkyl chains by Van der Waals forces. Since oil is of a similar character to this alkyl chain, the alkyl chain is soluble in oil droplets.
In alkanes, each carbon atom forms a tetrahedral structure (due to sp3 hybridisation), so the C-C-C bond angles are tetrahedral

Q33. A) The reaction is endothermic, which means that diamond has more energy than graphite. The enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from one mole of the element in the standard state. Since diamond has more energy than graphite, it requires a smaller enthalpy change of atomisation.
Since the enthalpy change of atomisation is smaller in diamonds, it means that the C-C bonds in diamond are weaker than in graphite because it requires less energy to change into the gaseous state.
Since diamond has more energy than graphite, there is a higher energy requirement to break the C-C bond to form new C=O bonds (in carbon dioxide) in combustion.


Q34. C) The electronegativity difference decreases between the elements (3.05, 2.13, 1.43, 0.65)
All of the compounds fulfill the octet rule and are isolelectronic.
The compounds become increasingly covalent (starting from ionic)

Q35. B) Sulphur dioxide is a reducing agent which prevents oxidation.
Since it's an anti-oxidant, it prevents alcohols from oxidizing to carboxylic acids (prevents sour-tasting acids).
It does smell and is toxic in large quantities.

Q36. C) Iodide ions are strong reducing agents and so they reduce the sulphuric acid, first to sulphur dioxide, then to sulphur and then hydrogen sulphide. Barely any hydrogen iodide is formed because it's displaced by the sulphuric acid.
Iodide ions are reducing agents, and become oxidised to iodine.
The majority of the products of the reaction are sulphur compounds (as explained above)

Q37. B) A chiral centre is an atom bonded to four different groups.
An optical isomer (geometric isomer) occurs when there's a chiral centre.
Chiral carbon atoms DO NOT need to have structural isomers.


Q38. B) Step X is a nucleophilic substitution because the reagent is hot aqueous sodium hydroxide (OH- being the nucleophile).
A chloroalkane cannot be formed by reacting sodium chloride with alcohol. It can be done with phosphorus (III) chloride or phosphorus (V) chloride.

Q39. C) Only an aldehyde forms a brick-red precipitate with Fehling's solution. Aldehydes are formed by the oxidation (in acidified dichromate) of primary alcohols. The two primary alcohols are CH3CH2CH2OH and CH3OH

Q40. B) It only has one chiral carbon (The one in the middle).
It has a carboxylic acid group, so it can be esterified by ethanol. It has an OH group, so it can be esterified by ethanoic acid.
The molecule contains tertiary and primary alcohols, not secondary.

 

[redd]

Calcium carbonate does decompose. There will be a loss in mass due to release in CO2.

(Note that this is not an clue to the experiment, I have no idea what it is.)

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!

Thanku minion!
Metanoia
Yes, it must decompose, but can u include the temperature or for how long it must be heated, because even after heating for an hour, we couldnt find any loss in mass!

 

[Minion96]

Calcium carbonate does decompose. There will be a loss in mass due to release in CO2.

(Note that this is not an clue to the experiment, I have no idea what it is.)

Please can u ask ur teacher about the temp and time it will be taking for decomposing?
We couldnt get to anything

 

[sadiaali]

Tukkay lagao musalmanoo Q9
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

 

[kitkat <3 :P]

Tukkay lagao musalmanoo Q9
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

B :/?

 

[Metanoia]

Thanku minion!
Metanoia
Yes, it must decompose, but can u include the temperature or for how long it must be heated, because even after heating for an hour, we couldnt find any loss in mass!

The decomposition temperature of CaCO3 is above 800 plus degrees Celsius if I remember correctly.

Time taken to reach that temperature? Well, it depends on how strongly its being heated?

This procedure that you are carrying out, is it based on the actual test question or is actually a self designed experiment?

Please can u ask ur teacher about the temp and time it will be taking for decomposing? We couldnt get to anything

Well, I'm actually a teacher myself..

 

[immie.rose]

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+

JazakAllah khairan!! One small doubt, tho. In the autocatalysed reaction, yeah the conc. of reactant decreases, but the conc. of product doesn't right? so since the product is the catalyst here, shouldn't the graph become a horizontal line like that of B.

 

[Metanoia]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s03_qp_1.pdf

Please can you explain questions 1, 2, 7, 10, 21, 22, 23, 37 and 39?
The answers are A,B,D,B,D,D,B,B,C.

Q1. Find out the number of moles of O2, since 1 mole of gas occupies 24000 cm^3 at RTP.

moles of O2 = 500/24 000
molecules of O2 = moles x 6.02 x 10^23 = (500/24000) x 6.02 x 10^23 = 1.25 x 10^22

Q2. Comparing the original structure to the resulting structure, we see that 5 C=C double bonds "gone".
So 5 moles of H2 are added.

If unable to visualise the above explanation, the diagram can show how H could be added in these locations.


Q7 and Q10



Q21. Is the answer really D? I got B.


Q22. The C=C undergoes oxidative cleaving, and the ends are oxidised to COOH.



Q23.
If confident, this question can be approached in a mathematical way

CnH(2n+2) + (3n+1)/2 O2 --> n CO2 + (n+1)H2O

when n increases, moles of O2 increases linearly. So its a line with a positive gradient.

 

[Browny]

Q11. A) Ammonium ethanoate is not completely ionised in water and it's not acidic in aqueous solution. Water is more polar than ammonia, so the only choice left is that ammonia is a stronger base than water


Q30. D) When propanone reacts with hydrogen cyanide, nucleophilic addition takes place and a hydroxynitrile is formed.
CH3COCH3 + HCN → CH3C(OH)CH3CN
2-hydroxybutanenitrile is hydrolysed under acidic conditions to Butanoleic acid
CH3C(OH)CH3CN + H2O + H+ → (CH3)2C(OH)CO2H + NH4

Q31. C) Silicon tetrachloride doesn't have co-ordinate bonding because it follows the octet rule, sharing all of its valence electrons with the chlorine atoms.
Both silicon and chlorine are non-metals, so, it has covalent bonding.
There are instantaneous dipole-induced dipole forces between the molecules (Van der Waals forces)

Q32. A) The right-hand side of the structure is polar and since water is a dipole, it is attracted to water.
The alkyl chain is non-polar and attracted to other alkyl chains by Van der Waals forces. Since oil is of a similar character to this alkyl chain, the alkyl chain is soluble in oil droplets.
In alkanes, each carbon atom forms a tetrahedral structure (due to sp3 hybridisation), so the C-C-C bond angles are tetrahedral

Q33. A) The reaction is endothermic, which means that diamond has more energy than graphite. The enthalpy change of atomisation is the enthalpy change when one mole of gaseous atoms is formed from one mole of the element in the standard state. Since diamond has more energy than graphite, it requires a smaller enthalpy change of atomisation.
Since the enthalpy change of atomisation is smaller in diamonds, it means that the C-C bonds in diamond are weaker than in graphite because it requires less energy to change into the gaseous state.
Since diamond has more energy than graphite, there is a higher energy requirement to break the C-C bond to form new C=O bonds (in carbon dioxide) in combustion.


Q34. C) The electronegativity difference decreases between the elements (3.05, 2.13, 1.43, 0.65)
All of the compounds fulfill the octet rule and are isolelectronic.
The compounds become increasingly covalent (starting from ionic)

Q35. B) Sulphur dioxide is a reducing agent which prevents oxidation.
Since it's an anti-oxidant, it prevents alcohols from oxidizing to carboxylic acids (prevents sour-tasting acids).
It does smell and is toxic in large quantities.

Q36. C) Iodide ions are strong reducing agents and so they reduce the sulphuric acid, first to sulphur dioxide, then to sulphur and then hydrogen sulphide. Barely any hydrogen iodide is formed because it's displaced by the sulphuric acid.
Iodide ions are reducing agents, and become oxidised to iodine.
The majority of the products of the reaction are sulphur compounds (as explained above)

Q37. B) A chiral centre is an atom bonded to four different groups.
An optical isomer (geometric isomer) occurs when there's a chiral centre.
Chiral carbon atoms DO NOT need to have structural isomers.


Q38. B) Step X is a nucleophilic substitution because the reagent is hot aqueous sodium hydroxide (OH- being the nucleophile).
A chloroalkane cannot be formed by reacting sodium chloride with alcohol. It can be done with phosphorus (III) chloride or phosphorus (V) chloride.

Q39. C) Only an aldehyde forms a brick-red precipitate with Fehling's solution. Aldehydes are formed by the oxidation (in acidified dichromate) of primary alcohols. The two primary alcohols are CH3CH2CH2OH and CH3OH

Q40. B) It only has one chiral carbon (The one in the middle).
It has a carboxylic acid group, so it can be esterified by ethanol. It has an OH group, so it can be esterified by ethanoic acid.
The molecule contains tertiary and primary alcohols, not secondary.

Thank you so so much!!!
You didn't actually get my point though I needed explanation only on 11, 32, 33 and 35.
But anyway thanks so much for taking your time and explaining everything it might be of good use in the end.

 

[Metanoia]

JazakAllah khairan!! One small doubt, tho. In the autocatalysed reaction, yeah the conc. of reactant decreases, but the conc. of product doesn't right? so since the product is the catalyst here, shouldn't the graph become a horizontal line like that of B.

Notice that the Y-axis is measuring speed of reaction, not amount of products.

When the reactants gets used up towards the end, the speed of reaction drops back towards zero.

 

[Browny]

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

Can anybody please explain question 30?
Why can't D be the answer as it also react with NaOH as hydrolysis happens?

 

[Metanoia]

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

Can anybody please explain question 30?
Why can't D be the answer as it also react with NaOH as hydrolysis happens?

Ethyl ethanoate is not very soluble in water.

 

[Metanoia]

Tukkay lagao musalmanoo Q9
http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w07_qp_1.pdf

w07qp1

Qn 9 :

moles of SO3 2- : moles of electrons : moles of metal
0.0025 : ? : 0. 005
1 :? : 2
1 : 2 : 2

2 mol of metal gained 2 mol of electrons
1 mol of metal gained 1 mol of electron (oxidation state will thus decrease by 1)
Original oxidation state of metal = +3
Final oxidation state = +3 - 1 = +2

 

[Browny]

Ethyl ethanoate is not very soluble in water.

No it is soluble I found that in another paper.

 

[immie.rose]

Notice that the Y-axis is measuring speed of reaction, not amount of products.

When the reactants gets used up towards the end, the speed of reaction drops back towards zero.

Got it! Thanks!

 

[immie.rose]

Answer: B Howww?

 

[Minion96]

The decomposition temperature of CaCO3 is above 800 plus degrees Celsius if I remember correctly.

Well, I'm actually a teacher myself..

ohh ohokay, i never knew sir, sir its this that we are provided with Caco3 and Kcl mixture and have been provided with crucible and heating apparatus and some Hcl also, my teacher is saying that it must be something related to percenatage impurity or purity, so by the loss in mass we can deduce it, but whenever we are heating it it is not decomposing.... can u deduce any experiment with that? or help us out through this?