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ALL CIE CHEMISTRY DOUBTS HERE !!

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moon:
Can someone please explain me nov.2008 q3,39 ??? ::) i need help in these questions urgently :'( I can't get them at all I would really appreciate any help.....thanx

$!$RatJumper$!$:
W07 Q 9, 10
W06 Q 4
W04 Q 9, 10, 23

Any help would be humbly appreciated :) Thank you very much :)

Dania:
The answer is D

Deadly_king:

--- Quote from: moon on November 18, 2010, 04:16:17 pm ---Can someone please explain me nov.2008 q3,39 ??? ::) i need help in these questions urgently :'( I can't get them at all I would really appreciate any help.....thanx

--- End quote ---

3. Only one electron in s-orbital -----> should be a group I metal.

So everyone apart D is rejected.

Li is a Group I metal. So it has its outermost electron in an s-orbital. Cr is a transition metal. One of its s-orbital goes into it's partially filled d-orbital for stability. ;)

39. From the information given above, you can note that for methane which contains only one carbon atom, ethane with two carbon atoms can be obtained as product. In other words the number of carbon atoms double.

So for propane which contains 3 carbon atoms, the resulting product may contain maximum 6 carbon atoms with as reactants two radicals carrying 3 carbon atoms.

The third one indeed contains 6 carbon atoms. But if you break it to form two radicals, you'll find that you'll be getting two compounds one with 4 carbon atoms and the other with only two carbon atoms.

This cannot be since propane contains 3 carbon atoms. So both radicals should contain 3 carbon atoms as well.  ;)

Answer is B.

Deadly_king:

--- Quote from: Dania on November 18, 2010, 04:34:07 pm ---The answer is D

--- End quote ---

Ammonium nitrate exists as ions : NH4+ and NO3-

So now you can calculate its oxidation number.

Let oxidation number of nitrogen be x.
NH4+ => x + 4(+1) = +1 -----> x = -3

NO3- => x + 3(-2) = -1 -----> x = +5

N2O = > 2x + (-2) = 0 ----> x = +1

So change in oxidation number is from -3 to +1 ---> +4 and from +5 to +1 ---> -4

SO answer is D.

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