Author Topic: ALL CIE CHEMISTRY DOUBTS HERE !!  (Read 107736 times)

Offline Deadly_king

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Re: CIE Chemistry doubts.
« Reply #30 on: September 27, 2010, 07:37:55 am »
Nov 08 P4 No7(b)

Condensation reaction occurs between the NH2 group of one amino acid and the CO2H group of another amino acid to form a peptide linkage as described in the first picture.

However the questions ask about the formation about a tripeptide. Hence it will be a reaction involving 3 amino acids. The displayed formula of a tripeptide is shown in the second picture.

NOTE : The R groups of atoms may be replaced by almost any atom. However it will be much easier and simpler to replace it by H atom. Your answer should not consist of R.

Offline moon

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Re: CIE Chemistry doubts.
« Reply #31 on: September 27, 2010, 09:20:46 pm »
Thank u very much for ur reply as well as ur explanation. It is really helpful, but can u explain Q4(b)(iii), of the same year Nov. 2008? You have explained Q3 not Q4 and what is the formula of the tripeptide as shown in the markscheme?....I really do appreciate ur co-operation.

Offline Deadly_king

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Re: CIE Chemistry doubts.
« Reply #32 on: September 28, 2010, 05:25:14 am »
Thank u very much for ur reply as well as ur explanation. It is really helpful, but can u explain Q4(b)(iii), of the same year Nov. 2008? You have explained Q3 not Q4 and what is the formula of the tripeptide as shown in the markscheme?....I really do appreciate ur co-operation.

Oops ....... sorry for the confusion dude.

4(b)(iii)

That will depend on the equation you wrote for part (b)(i).

Let's take it to be : C8H18 -----> 2C2H4 + C4H10
Now you need to find the enthalpy change for this reaction.

Number of bonds to be broken in the reactant : 18C-H bonds and 7C-C bonds.
Total number of bonds to be formed in the products : 2C=C bonds + 8C-H bonds and 3C-C bonds + 10C-H bonds.

Use the values from the Data Booklet to find :
(i) Energy required for bond-breaking : 18(410) + 7(350) = 9830 KJ/mol
(ii) Energy given out for bond formation : 2(610) + 8(410)  + 3(350) + 10(410) =  9650 KJ/mol

Standard enthalpy change of reaction : 9830 - 9650 = +180 KJ/mol

NOTE : Bond breaking is endothermic(+ve) while bond forming is exothermic(-ve).

Same method is applied whatever the equation you wrote in (b)(i).

(iv)

The conditions include a temperature of about 600oC. Hence this suggests that the reaction is endothermic since heat must be supplied for the reaction to occur and this is confirmed by the entahlpy change obtained in (b)(iii).

7(b)
The formula of the tripeptide can vary. This is why the marking scheme only mentioned 'correct tripeptide'. The product in the second picture I uploaded earlier is the right answer. Except that it is asking for displayed formula, so you need to elaborate on the bonds and replace all the R groups by H for the simplest tripeptide. :)

Offline moon

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Re: CIE Chemistry doubts.
« Reply #33 on: September 28, 2010, 12:27:33 pm »
Thank you very much for ur help. I got it... :)

Offline Deadly_king

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Re: CIE Chemistry doubts.
« Reply #34 on: September 29, 2010, 04:42:36 am »
Thank you very much for ur help. I got it... :)

You are welcome pal :)

Offline HUSH1994

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Re: CIE Chemistry doubts.
« Reply #35 on: October 01, 2010, 01:50:32 pm »
Anyone can answer this question urgently:
State the factors affecting the ionisation energy and state the action of each one of them

Offline Twinkle Charms

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Re: CIE Chemistry doubts.
« Reply #36 on: October 01, 2010, 02:15:51 pm »
Anyone can answer this question urgently:
State the factors affecting the ionisation energy and state the action of each one of them
1) Size of the positive nucleur charge -- as the atomic number increases, there is an increase in the nucleur charge which causes an increase in the ionisation energy.

2)Distance of the electron from the nucleus -- when the distance of electron increases from the nucleus, the attractive force decreases therefore ionisation energy decreases.

3)Shielding effect of the inner electrons -- all the electrons are negatively charged and repel each other. Electrons in filled inner shells repel electrons in the outer shell and reduce the effect of positive nucleur charge. This is called the sheilding effect. The greater the shielding effect on an electron, lower the ionisation energy.



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Offline Arthur Bon Zavi

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Re: CIE Chemistry doubts.
« Reply #37 on: October 01, 2010, 02:57:17 pm »
1) Size of the positive nucleur charge -- as the atomic number increases, there is an increase in the nucleur charge which causes an increase in the ionisation energy.

2)Distance of the electron from the nucleus -- when the distance of electron increases from the nucleus, the attractive force decreases therefore ionisation energy decreases.

3)Shielding effect of the inner electrons -- all the electrons are negatively charged and repel each other. Electrons in filled inner shells repel electrons in the outer shell and reduce the effect of positive nucleur charge. This is called the sheilding effect. The greater the shielding effect on an electron, lower the ionisation energy.


 8) 8) This!!

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Offline Twinkle Charms

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Re: CIE Chemistry doubts.
« Reply #38 on: October 01, 2010, 04:03:31 pm »
8) 8) This!!
You have to but in everything yeah?  :P :D
You See - You like - You try - You fail meanwhile I see - I like - I want - I get!

La Fata Illa Ali, La Saif Illa Zulfikar . (:

Offline Deadly_king

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Re: CIE Chemistry doubts.
« Reply #39 on: October 01, 2010, 05:37:26 pm »
1) Size of the positive nucleur charge -- as the atomic number increases, there is an increase in the nucleur charge which causes an increase in the ionisation energy.

2)Distance of the electron from the nucleus -- when the distance of electron increases from the nucleus, the attractive force decreases therefore ionisation energy decreases.

3)Shielding effect of the inner electrons -- all the electrons are negatively charged and repel each other. Electrons in filled inner shells repel electrons in the outer shell and reduce the effect of positive nucleur charge. This is called the sheilding effect. The greater the shielding effect on an electron, lower the ionisation energy.





rightly said Twinkle :)
+rep

Offline HUSH1994

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Re: CIE Chemistry doubts.
« Reply #40 on: October 02, 2010, 04:58:28 am »
rightly said Twinkle :)
+rep
Thank You twinkle,+rep as well ;)

Offline ruby92

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Re: CIE Chemistry doubts.
« Reply #41 on: October 03, 2010, 04:45:14 pm »
given that the bond energy needed to break an O2 bond is 150 what is thae standard enthalpy of atomisation?

Offline Deadly_king

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Re: CIE Chemistry doubts.
« Reply #42 on: October 04, 2010, 05:23:30 am »
given that the bond energy needed to break an O2 bond is 150 what is thae standard enthalpy of atomisation?
According to your statement, when breaking the O2 molecules, we get 2 oxygen atoms. For this 150KJ/mol energy is required. (Bond energy)

The definition of enthalpy change of atomisation :
It is the heat energy change when  one mole of separate gaseous atom of the element is formed from the element under standard state conditions.

By breaking the O2 bond we obtain 2 moles of oxygen atom. But enthalpy change of atomisation requires only one mole of gaseous atom to be formed.

Hence the answer will be :150/2 = 75 KJ/mol


Offline Adzel

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Re: CIE Chemistry doubts.
« Reply #43 on: October 05, 2010, 03:43:05 pm »
Can someone plz answer these questions for me  :)...

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Re: CIE Chemistry doubts.
« Reply #44 on: October 05, 2010, 03:51:16 pm »
All of these? in a few hours.