Chemistry Paper 1 problems...

[nox_fjmoony]

I was solving the may/june 2008 question paper yesterday and I had problems with these questions:

6,10,13,17,20,23,25,26,27,28,32 and 40

Anyhelp would definitely be appreciated, along with an explanation of the solution, yeah. Thanks all.
Ciao

[nox_fjmoony]

Anyone? Need some help here, please 

[nid404]

oh my!

too many...I'll do them by tom 

[tmisterr]

6. 1cm³ of 1g/cm³=1g of water. you know that at 25 degrees and 1 atm it occupies 24 dm³. Use pV=nRT to find R (T in K, Volume in m³ and p in Pa 1atm=101kpa). R is 8.134.....
now use this value of R to find the volume of the water at 596k. V= 2.67dm³

10. using hess's law, you can either first form TiO₂ and C then form Ti and CO(this second part is the enthalpy change of the reaction) or you can directly form Ti and CO from the raw elements Ti, C, and O₂. the two enthalpy changes should be equal. therefore since 2 moles of CO are formed, 2*-110=-940 + enthalpy change. so enthalpy change is equal to +720

13. this is Na₂O  Mg(OH)₂ is only partially soluble so it will be a weak alkali and the other two will form acidic solutions. (revise chemical periodicity, it will help you on this type of questions)

17. when an ammonium salt is heated with an alkali, ammonia is liberated do lime-water is calcium hydroxide which is an alkali and the only alkali on the list

20. number 20 should be B. alkenes undergo electrophilic substitution, then the haloalkane will undergo neucleophilic addition.

23. this is clearly D, pentane cannot be cracked to a hydrocarbon with a longer carbon chain!!

24. Sn1 will be for tertiary halogenoalkanes since these are stabilised by three methyl groups so the reaction can take part in two stages.

I have to go though

[nox_fjmoony]

@tmisterr: Hey thankyou soo much for all your help man Really appreciate it, but could you answer one question? Its like, isn't R a constant? And so why do we have to manually find out the value of R, when we already know it? Just a bit confused really. Thanks.

And can anyone help me with the remaining questions?

Right, thanks all.

[CHEMMASTER6000]

26. ok so if you put the Oh on the 2nd or 3rd carbon after oxidation you will not get 4 oxygen . either a 3 02 OR 2 O2 SINCE YOU ONLY FORM KETONE. itbasically to math the C4H6o4 given

27. dehydration is the same as elimination and thus an alchahol forms and alkene. so you must try to deduce which one could have a tertairy alchahol . in this case D would have form the tertiary alchahol

28. i think for this case you might have put B but hte answe is C those sneaky cie peple . the H- indicates the CN and in the machanism its the Cn which attacks first.

32. its D because catalysts dont eveft KP or KC cause they worl both ways in the reaction . 3 indicates increasing activation enerygy which is wrong cause strictly speaking, catalst lowers activatoin energy.

40. 1 is a base so reacts with the acid. the rest reacts with alkene

[CHEMMASTER6000]

  hope that clears things up

[CHEMMASTER6000]

by the way R is fixxed at 8.31. thing abotu this questoin is that even by using Pv=nrt the answer cant be produce

[nox_fjmoony]

@:CHEMMASTER6000: Thanks a lot for all your help man, really appreciate it. And if you're giving any exams or whatever, best of luck to you:D Thanks again...and indeed those sneaky CIE folks XD

Yeah, thats what I was wondering, since R is a constant, and yeah using pv=nRT you can't get that answer, though you do get it with the method outlined by tmisterr, though I still don't understand why we have to find a separate value of R for that :\
Oh well.

Thanks all.
Good night.
And Good luck ^ ^

[ruby92]

can someone explain o/n 2006 q 9 and 21
o/n 2008
22 and how is 3 posibble for 32?

[ayesha.]

m/j/03 q 27,28,34

[Vin]

*Buzz* People need help though it's too late. I'm helpless I've not done the portion. :\

Good job tmisterr and CHEMMASTER6000! +REP!

[Deadly_king]

m/j/03 q 27,28,34

27. Normally the strength of halogen bonds get weaker as we move down the group. In other words C-F bond is much stronger than C-Cl bond. So in order for a radical to be formed, homolytic bond fission should undergo across a C-Cl bond.
A : CHFClC*FCl..........it shows that a C-F bond has been broken as one F atom has been lost. This cannot be the case since fluorine forms very stable molecules which are unlikely to break.
B : Same explanation as for A.
D : A C-H bond is also much stronger than a C-Cl bond. It is said that it is stronger tahn even a C-F bond. So its not likely to occur.  

Therefore the answer is C.

28. Sodium reacts with acids or alcohols to release hydrogen gas. However one mole of an acid or alcohol releases only half mole of hydrogen. Therefore X should either contain either 2 acidic groups. 2 alcohol groups or 1 acidic and 1 alcohol group.

Only compound D corresponds to that. One mole of all the other compounds will release only half mole of hydrogen gas.

34. All of these are compounds are seen to be ionic ones and are all across the same period in the periodic table. Therefore they form ions with the same number of electrons.
As we move along the period 3 from sodium to argon the covalent character of the compounds increases. Sodium(the first element in the period) is a metal while chlorine is a gas and exists as diatomic molecules with simple covalent structure.
But its electonegativity difference does not increase since fluorine has a much higher electronegativity value than any other element.

As we move from sodium to argon the electronegativity value increases.

Therefore answer is C since only 2 and 3 are correct.

Reference : http://www.chemguide.co.uk/atoms/bonding/electroneg.html

[Arthur Bon Zavi]

27. Normally the strength of halogen bonds get weaker as we move down the group. In other words C-F bond is much stronger than C-Cl bond. So in order for a radical to be formed, homolytic bond fission should undergo across a C-Cl bond.
A : CHFClC*FCl..........it shows that a C-F bond has been broken as one F atom has been lost. This cannot be the case since fluorine forms very stable molecules which are unlikely to break.
B : Same explanation as for A.
D : A C-H bond is also much stronger than a C-Cl bond. It is said that it is stronger tahn even a C-F bond. So its not likely to occur.   

Therefore the answer is C.

28. Sodium reacts with acids or alcohols to release hydrogen gas. However one mole of an acid or alcohol releases only half mole of hydrogen. Therefore X should either contain either 2 acidic groups. 2 alcohol groups or 1 acidic and 1 alcohol group.

Only compound D corresponds to that. One mole of all the other compounds will release only half mole of hydrogen gas.

34. All of these are compounds are seen to be ionic ones and are all across the same period in the periodic table. Therefore they form ions with the same number of electrons. as we move along the period 3 from sodium to argon the covalent character of the compounds increases. But its electonegativity difference does not increase since fluorine has a much higher electronegativity value than any other element.

As we move from sodium to argon the electronegativity value increases.

Therefore answer is C since only 2 and 3 are correct.

Reference : http://www.chemguide.co.uk/atoms/bonding/electroneg.html

+rep...dead brottha!!

[Deadly_king]

can someone explain o/n 2006 q 9 and 21
o/n 2008
22 and how is 3 posibble for 32?

Nov 06 No 9 and 21.

9. Standard enthalpy change of reaction = -286 + 44 - (-283) = 41 KJmol^-1
First the hydrogen gas needs to be converted to water in liquid form and then into gas. This whole process requires (44 - 286) KJmol^-1.
Then Carbon dioxide must be converted to carbon monoxide which will require 283 KJmol^-1.
addition of all these energies leads to answer C.

21.
When n=1 ---------> C₂H₅Cl (1 type only) Chloroethane

When n=2 ---------> C₂H₄Cl₂ (2 types) 1,2 dichloroethane and dichloroethane

When n=3 ---------> C₂H₃Cl₃ (2 types) 1,2 trichloroethane and trichloroethane.

When n=4 ---------> C₂H₂Cl₄ (2types) tetrachloroethane and 1tricholro 2 chloroethane.

In all there are 7 different choroethanes that can be formed. Hen answer is C.
It is easier to understand when you draw their displayed formula. Unfortunately i dont know how to draw them here.

Nov 08 No 22 and 32

22. The hydrocarbon has to undergo a monochloro-substitution which means that this reaction will not affect the carbon-carbon double bond. One hydrogen atom needs to be replaced by a chlorine atom. Chirality refers to  one carbon atom being attached to four different group of atoms.

A is rejected since it will not show cis-trans isomerism even with the substitution of chlorine as the same carbon atom carries two hydrogen atoms. Moreover a double bond is not indicative of a chiral centre.

B and C are also rejected as one carbon atom contains two methyl groups. Again no cis-trans isomerism and no chiral centres even upon substitution.

Answer is D.
Upon substitution of a hydrogen atom on the third carbon atom by a chlorine atom, the carbon atom becomes a chiral centre being attached to four different group of atoms namely methyl group, hydrogen atom, cholrine atom and propene.

Drawing the respective compounds will allow you to eliminate those not showing cis-trans isomerism.


32. A Brosted-Lowry acid is one which can donate an H⁺ molecule that is a proton donor.

1. H₃0⁺ <-------> H⁺ + H₂0

2. NH₄⁺ <-------> NH₃ + H⁺

3. H₂O <--------> H⁺ + OH_-
Water is an acid as well as an alkali according to Bronsted-Lowry. It can as well give H⁺ to form OH^- and also accepts a proton to form H₃0⁺

Since all these reactions are reversible (<-------->) all three of them can donate a proton. Hence answer is A.


If you dont understand something do not hesitate to ask. We'll try to clear your doubts. And if i said something wrong do tell me