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Chemistry: Post your doubts here!

Metanoia

Metanoia

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redd said:
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!
Click to expand...

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?

Minion96 said:
Please can u ask ur teacher about the temp and time it will be taking for decomposing? We couldnt get to anything :(
Click to expand...

Well, I'm actually a teacher myself..:)
 
immie.rose

immie.rose

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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...

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

Metanoia

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MariamMalik said:
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.
Click to expand...

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.
Picture 18.png

Q7 and Q10

Picture 5.png

Q21. Is the answer really D? I got B.
Picture 6.png

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

Picture 7.png

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.
 
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B

Browny

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MariamMalik said:
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.
Click to expand...

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

Metanoia

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immie.rose said:
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.
Click to expand...

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.
 
Metanoia

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Minion96

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

Well, I'm actually a teacher myself..:)
Click to expand...

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?
 
redd

redd

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Metanoia said:
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?
Click to expand...

Jazak Allah sir :)
Well we just guessed it this way, its not any kind of test question. Can u design it anyway?
 
A

Abdel Moniem

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immie.rose said:
View attachment 44145
Answer: B Howww? :sleep:
Click to expand...
number of moles of Tl+NO3-=10*0.30/1000
=3*10^-3
Number of moles of NH4+VO3-=20*0.10/1000
=2*10-3
Divide by the smalles figure, we would obtain the ratio of 1.5:1 *2 to get rid of a fraction so the mole ration will 3:2 so for every 3 moles of Tl+( six electrons are removed) only two moles are reduced so divide the six electrons by 2 to figure out how many electrons are gained per mole
 
Metanoia

Metanoia

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Metanoia

Metanoia

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Minion96 said:
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?
Click to expand...

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).
 
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