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ALL CIE CHEMISTRY DOUBTS HERE !!
moon:
plz can u help me in ppr 1 Nov.03 Q2,6,13,31,40. thanx in advance :)
Deadly_king:
--- Quote from: moon on November 10, 2010, 10:56:00 pm ---plz can u help me in ppr 1 Nov.03 Q2,6,13,31,40. thanx in advance :)
--- End quote ---
2. First you need to find the % of phosphorus in P2O5
Mr of P2O5 = 2(31) + 5(16) = 142
Hence % of phosphorus in P2O5 = 2(31)/142 x 100 = 43.6%
Since P2O5 forms part of only 30% of the fertiliser, % of phosphorus in P2O5 in the fertiliser will be (30/100 x 43.6) = 13.1%
Answer is B
6. That's kind of obvious. ;)
Six bonding electrons imply a molecule which is composed of only 3 bonds since a bond carries 2 electrons.
A C2H4 ------> 6 bonds (12 bonding electrons)
B C2F6 ------> 7 bonds (14 e)
C H2O -------> 2 bonds (4 e)
D NF3 -------> 3 bonds (6 e)
NOTE : A double bond carries 2 pairs of electrons.
Hence Answer is D
13. You should be aware of the trends of ionisation energies for period 3. Take a look here.
There is a fall at Aluminium since there is a change in sub-shells namely from 3s to 3p. That's why aluminium has a lower 1st I.E. ;)
Answer is B
31. So let's examine the answers one by one.
1. An acid- base reaction : That's true since SO3 and water combines to form sulfuric acid. Sulfuric acid then reacts with basic NH3 to form the corresponding salt. ;)
2. Ionic bond formation : That's true as well. The salt exists in the form of NH4+ and SO42- which are ions involved in ionic bonds.
3. Oxidation and reduction : You need to calculate the oxidation numbers of the different elements in order to determine this.
If we just look at the equation, we might have the impression that SO3 is being oxidised but this is not the case.
Let oxidation number of sulfur in SO3 be and take oxidation number of O as -2.
+ 3(-2) = 0 ----> = +6
Now for SO42-.
+ 4(-2) = -2 ----> = +6
Since oxidation number remains the same, there has been neither oxidation nor reduction. I'll let you calculate the oxidation numbers of the other elements yourself for confirmation. ;)
Hence answer is B since only 1 and 2 are good.
40. Heating under reflux with NaOH is an alkaline hydrolysis, i.e there is the breaking of bonds usually COO.
Therefore the ester bond will be broken and two compounds, namely an alcohol and a carboxylic acid will be formed.
Since NaOH is used the carboxylate will be formed instead of the carboxylic acid, but upon distillation the carboxylate will get back to the acid.
Product 3 cannot be formed since it involves the breaking of a C=O which is not possible.
Hence answer is D.
Hope it helps :)
moon:
Ur explanation really helped me a lot. thanx so much 4 ur co-operation :) ;)
Deadly_king:
--- Quote from: moon on November 11, 2010, 03:59:33 pm ---Ur explanation really helped me a lot. thanx so much 4 ur co-operation :) ;)
--- End quote ---
Hehe.........Anytime buddy ;)
moon:
plzzzzzzzzz help me in these questions. :-[ :-[June 07 Q 1, 9
June 01 Q6,11,37
June 2002 Q 38
I would really apreciate any help. thanx a lot. :)
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