Author Topic: ALL CIE CHEMISTRY DOUBTS HERE !!  (Read 119165 times)

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #255 on: November 18, 2010, 02:23:18 pm »
W04 Q3 pls
:)

It's just like Ari explained the previous number.

10CH2SH + 30O2 -----> 10CO2 + 10SO2 + 20H2O

When cooled, the water will condense. Hence it will no more be in gaseous state.

NOTE: 60cm3 of oxygen was available and only 30cm3 were used . So there are 30cm3 in excess.

In all it makes 50cm3

Therefore answer is C.
« Last Edit: November 18, 2010, 03:16:24 pm by Deadly_king »

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #256 on: November 18, 2010, 02:24:57 pm »
Equation should be X + Cl2 => XCl2
Right so , X is supposed to be 2.920g
and the Xcl2 is 5.287
subtract the two to get the mass of Chloride used which will come out to be = 2.367g
Mass/Mr = 2.367/71 will get you moles of Chloride which are = 0.333333
Mole Ratio is 1:1
if 0.3333 mole of X is 2.920
then how much will be in one mole = 2.920/0.3333 = 87.9g
Thus Ans is C.
Strontium

Offline cs

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #257 on: November 18, 2010, 02:40:07 pm »
An urgent one:

June 2010 P11, Q1 please...

Offline Hypernova

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #258 on: November 18, 2010, 02:41:52 pm »
then how much will be in one mole = 2.920/0.3333 = 87.9g
Thus Ans is C.
Strontium

Nah, thats what i did. You get the MR=87.9, but the Mr of strontium is 38
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Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #259 on: November 18, 2010, 02:45:03 pm »
buddy Mr is 88 , atomic number is 38.

Offline Hypernova

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #260 on: November 18, 2010, 02:49:03 pm »
buddy Mr is 88 , atomic number is 38.

holywillywonker.

your right, +rep

Sorry, using a diff periodic table,and not used to it. I read it wrong.
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Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #261 on: November 18, 2010, 02:50:01 pm »
An urgent one:

June 2010 P11, Q1 please...

Ans is C.
Due to the fact that it will have the config 1s2 2s2 2p6 3s2 3p3 in each atom the p orbital has 3 unpaired!

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #262 on: November 18, 2010, 02:56:15 pm »
Plz anyone J04 Q28 P1

Offline komailnaqvi

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #263 on: November 18, 2010, 03:11:11 pm »
Q10 N04 P1

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #264 on: November 18, 2010, 03:14:12 pm »
Plz anyone J04 Q28 P1

Compound X is unreactive to mild oxidising agents ----> It should be a tertiary alcohol which will produce an alkene upon dehydration.

Only D offers this possiblity. ;)
« Last Edit: November 18, 2010, 03:44:55 pm by Deadly_king »

Offline moon

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #265 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

Offline $!$RatJumper$!$

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #266 on: November 18, 2010, 04:27:38 pm »
W07 Q 9, 10
W06 Q 4
W04 Q 9, 10, 23

Any help would be humbly appreciated :) Thank you very much :)
« Last Edit: November 18, 2010, 04:30:33 pm by $!$RatJumper$!$ »

Offline Dania

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #267 on: November 18, 2010, 04:34:07 pm »
The answer is D
:)

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #268 on: November 18, 2010, 05:00:09 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

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.

Offline Deadly_king

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Re: ALL CIE CHEMISTRY DOUBTS HERE !!
« Reply #269 on: November 18, 2010, 05:11:51 pm »
The answer is D

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.