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Chemistry: Post your doubts here!

Metanoia

Metanoia

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Minion96 said:
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! :(
Click to expand...

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.)
 
Metanoia

Metanoia

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♣♠ Magnanimous ♣♠ said:
in 29 why did

you go for isomerism rather than simple structure o_O
How can we know that this time we will go for this structure like you drawn in 29
Click to expand...

Do you mean to ask why X is not a straight chain butene?

If it was a straight chain butene, we would get butane when it reacts with H2.

since the question stated that we get methylpropane upon hydrogenation, it means the C4H8 is branched at the start.
 
O

omeraz

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MariamMalik

MariamMalik

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Browny said:
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?:)
Click to expand...


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

redd

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Metanoia said:
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.)
Click to expand...


Minion96 said:
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! :(
Click to expand...

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!
 
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