ALL CIE CHEMISTRY DOUBTS HERE !!

[Deadly_king]

does anyone know why some books say
1) E(std)cell = E(+ve terminal) - E(-ve terminal)
2) E(std)cell = E(right hand terminal) - E(left hand terminal)

in my class, we used the first one. but when i attempt June 2008 paper4 Q1, they used the 2nd one to get the answer. im confused :S

Usually both equations should result in the same answer, since the conventional way of drawing an electrode potential cell is to draw the positive terminal to the right.

The question however, seems to be telling us otherwise. Personally I think the marking scheme is wrong.

This is the only CIE question that arose some concerns. The rest are all pretty easy and abide by the laws of electrode potential.

For current to flow in the wires, the reactions in the half cells should occur. And for the reactions to occur, the standard electrode potential value MUST be positive. In this case the value is negative but still it is asking the direction for the flow of electrons. So it appears to be contradicting the rule of electrode potential.

Hence the answer for this question should be +0.59V.

Nevertheless this is my theory. As a student, I may make mistakes.
This is why i'll advice you to consult someone more experienced in chemistry, like your teacher.

[moon]

Thanks a lot Deadly_King for ur effort ,explanation and advice.

[Deadly_king]

Thanks a lot Deadly_King for ur effort ,explanation and advice.

My pleasure buddy

[Greed444]

Usually both equations should result in the same answer, since the conventional way of drawing an electrode potential cell is to draw the positive terminal to the right.

The question however, seems to be telling us otherwise. Personally I think the marking scheme is wrong.

This is the only CIE question that arose some concerns. The rest are all pretty easy and abide by the laws of electrode potential.

For current to flow in the wires, the reactions in the half cells should occur. And for the reactions to occur, the standard electrode potential value MUST be positive. In this case the value is negative but still it is asking the direction for the flow of electrons. So it appears to be contradicting the rule of electrode potential.

Hence the answer for this question should be +0.59V.

Nevertheless this is my theory. As a student, I may make mistakes.
This is why i'll advice you to consult someone more experienced in chemistry, like your teacher.

Thankyou Dk. eventhough u say that you are a student, but still your explanations and advices are like coming from a true expert

[Deadly_king]

Thankyou Dk. eventhough u say that you are a student, but still your explanations and advices are like coming from a true expert

Hahaha.............sorry to disappoint you but I am a student.

Anyway thanks for the kind words

[moon]

plz can u help me in ppr 1 Nov.03 Q2,6,13,31,40. thanx in advance

[Deadly_king]

plz can u help me in ppr 1 Nov.03 Q2,6,13,31,40. thanx in advance

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 SO₄^2- 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 SO₄^2-.
+ 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]

Ur explanation really helped me a lot. thanx so much 4 ur co-operation

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.

[Hypernova]

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.

June 07 Q 1, 9

2.

Relative atomic mass is the average mass of an element compared to that of carbon 12

You need to find the average mass of the element using its reative abundance, so

Ar=10 x 1/5 + 11 x 4/5
=10.8

9.

Kc of a reaction is equal to the product of consentration of products over the product of concentration of reactants all raised to their stochiometric coefficients

so Kc for reaction 1 is 2

Kc=[X₂Y]²      
   [X₂]²x[Y₂]

Know you have an equation for kc of reaction 1

The queation is find Kc for reaction 2

    ?     =   [X₂]x[Y₂]^1/2
                 [X₂Y]


Take your equation for kc of reaction 1 and make it look like the Kc equation for reaction 2

2=[X₂Y]²            
  [X₂]²x[Y₂]

the reciprocal looks like          

   1     = [X₂]²x[Y₂]    
2          [X₂Y]²

now square root it

   1     =   [X₂]x[Y₂]^1/2
sqrt(2)           [X₂Y]

so Ans is A

[Hypernova]

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.

I dont have June 01 paper 1, could you please upload it

Q38

For this reaction you need to know what is happening

Br₂ will be attracted to the double bonds and it will become polar. Then 1 Br atom binds across the double bond, and the other has a negative charge.

This is the bromination mechanism used in the test for alkenes, I assume you know it, I will elaborate if you ask.

So then the transition state (attached) for alkene is formed.

There are charged particles in solution:
the Cl- from NaCl
The other Br- atom from Br₂

these atoms are negatively charged

Notice the positively charged C+ atom in the ethene transition state (attached). It will bind to one of these charged particles.

So

1 CH₂ClCH₂Cl   impossible

2 CH₂BrCH₂Cl   possible

3 CH₂BrCH₂Br   possilbe

Ans is C

[moon]

Thank u so much hypernova By the way for june 2001 i have hard copy so i can't upload it sorry .

[Deadly_king]

Thank u so much hypernova By the way for june 2001 i have hard copy so i can't upload it sorry .

Post the whole question then

We'll try to help as mush as we can.

[moon]

I have attached June 2001 as well as November 2001.It seems that these exams are difficult to find so i thought these might help you practice more questions. Best wishes