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Can someone explain this to me? I have done all the calculation but answer is coming different.
Summer 2011, Variant 13, Q12.
My calculation
Al2O3 + 6HCl = 2AlCl3 + 3H20
As 1 mol of Al2O3 reacts with 6 mol of HCl, 0.02 mol react with 0.12 mol.
0.12/2= 0.06 dm3 of HCl 60 cm3. Whats wrong with this?
View attachment 28684
Q18: It says that when it was heated without any air so no compound was added or removed from the Ammonium Cynate. Only the addition of CNO will balance both sides Ammonium will be Nh4. The compound was NH4CNO
Q20: I have no idea about that.
Q22: Hydration and oxidation all occur at the double bond but after polymerisation the double bond will be gone so they can not occur. Hydrolysis with Chloro ethene is neucholophillic substitution so it occurs ar Chlorine not double bond and can happen in Polymer.
Q40: In Polymer atleast 2 molecules join together to form them. 1 mol of monomer producing 1 mol of polymer is impossible as monomer has to join, so it will reduce. Any other number is possible so the last 2 are correct. Answer C.
B and C can be easily eliminated. The answer is D or A? And can you please tell me the year and number of this question?
are u from bss hyderabadCan someone explain this to me? I have done all the calculation but answer is coming different.
Summer 2011, Variant 13, Q12.
My calculation
Al2O3 + 6HCl = 2AlCl3 + 3H20
As 1 mol of Al2O3 reacts with 6 mol of HCl, 0.02 mol react with 0.12 mol.
0.12/2= 0.06 dm3 of HCl 60 cm3. Whats wrong with this?
View attachment 28684
Atomic crystal are metals and those have very high boiling points. This has boiling point of just over 100 so it has to be Molecular crystal.My apologies. My "elimination" I wrote incorrectly up there.
Yeah it IS either A or D.
Answer is D - but why? That is the main question.
Yeah it's from the "early 90's".
J93 to be precise.
Thanx.
From the start:
4Al+3O2 will give 2Al2O3
so the ratio of aluminium to aluminium oxide is 2:1.so Al2O3 moles will be 0.01moles.so the ratio of aluminium oxide to HCl is 1:6,so HCL moles will be 0.06moles.divide these moles by 2.00 and u will get the answer.hope u got it.
Q21. B) Citric acid (from top to bottom): The first carbon is bonded to 2 hydrogens, so it's not chiral. The second carbon is bonded to the top and bottom chains which are the same, so it's not chiral. The third carbon is bonded to 2 hydrogens, so it's not chiral.
Isocitric acid (from top to bottom): The first C is bonded to 2 hydrogens, so it's not chiral. The second carbon is bonded to the top chain, an H, a CO2H, and the bottom chain, so it is chiral. The third carbon is bonded to the upper part of the molecule, an OH, an H, and a CO2H, so it is chiral.
Conclusion: Citric acid has 0 chiral centres and isocitric acid has 2 chiral centres
Q22. D) P is an alkene and a primary alcohol. Alkenes are oxidized by hot concentrated KMnO4 to carboxylic acids with two carboxylic acid groups. Primary alcohols are oxidized to a carboxylic acid. This means that the products of this oxidation are CH3CH2CO2H (from primary alcohol oxidation) and HO2CCH2CO2H (from alkene oxidation)
Q23. B) Write down some equations for the complete combustion of simple alkanes so as to determine how the ratio of carbon atoms to moles of oxygen changes as there are more carbon atoms in the alkane.
CH4 + 2O2 → CO2 + 2H2O
2C2H6 + 7O2 → 4CO2 + 6H2O
C3H8 + 5O2 → 3CO2 + 4H2O
The ratio of carbon atoms to moles of oxygen in each equation respectively is:
1:2, 2:7, 3:5
The number of carbon atoms is directly proportional to the number of moles of oxygen, so the answer has to be B
Q24. B) When an alkene reacts with cold liquid bromine, dibromoalkene is formed. Since the compound above is made up of two alkenes, then four bromine atoms are added on to the molecule. This reaction is an electrophilic addition reaction. Each bromine atom adds on to a carbon engaged in the double bond. The answer is 1,3,4,6 - tetrabromocyclohexane
Q25. D) Reaction D is a nucleophilic addition (CN- is the nucleophile)
Q26. A) When a halogenoalkanes reacts with KCN in ethanol, a nucleophilic substitution reaction takes place, in which the CN- ions substitute the halogen atoms (in this case the bromine atoms from 1,4-dibromobutane)
Q27. C) The weakest bond is the C-Cl bond. The C-H bond is very strong because it is an ionic bond. The C-F bond is strong because fluorine is very electronegative. The C-Cl bond however, is covalent and Cl is not as electronegative as Fluorine, which means that it is a relatively weak bond (relative to the other bonds in the molecule). This is why in the reaction, the most susceptible atom to leave the molecule is chlorine, and so the radical in option "C" is formed
Q28. D) When an alcohol reacts with sodium an alkoxide ion is formed (O-Na+). When a carboxylic acid reacts with sodium, since it's an acid, a salt is formed (COONa). The only compound that forms one mole of hydrogen is the last one.
CH3CH(OH)CO2H + 2Na → CH2CH(ONa)COONa + H2
Q29. C) An alcohol and concentrated sulphuric acid under reflux will produce an alkene which can be purified by dilute sodium hydroxide (base hydrolysis of esters).
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.
P.S :- Others please post work solutions for all P1 yearly like this..
u mean sis nope it does notb
Bro will acidic KMnO4 react with alkene?
u mean sis nope it does not
May June. 2005 q1
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