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Chemistry: Post your doubts here!

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please anybody can help me in Q#11 May /june 2011 component 12 with complete solution? any help would be highly appreciated.
 
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http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w09_qp_11.pdf

November 2009, P11. Questions 19, 20 and 21 please. :)

In 21, I don't understand why there are 5 C=C bonds instead of 6.

19)--> its A because C and D are automatically out as they both do not represent cyanate now in Ammonium cyanate charge of NH4 is 1 which means that charge on CNO must be -1.
20) D because Carbon atoms connect to CH3 (there are two such Carbon atoms) and to OH also posses chiral center as they are connected to 4 different groups...
21) its confusing 4 me too
 
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19)--> its A because C and D are automatically out as they both do not represent cyanate now in Ammonium cyanate charge of NH4 is 1 which means that charge on CNO must be -1.
20) D because Carbon atoms connect to CH3 (there are two such Carbon atoms) and to OH also posses chiral center as they are connected to 4 different groups...
21) its confusing 4 me too
u seem to b doing the same paper....do u get q 40??..
 
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2 and 3 should be correct. Because there are very large number of monomers combined in a polymer, the number of moles of a polymer will be much lesser than 1.
the statement sounds very logical...that if 1 is wrong the answer is C...but i was just wondering that how is the 2nd statement correct?...10^-6??...
 
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2 and 3 should be correct. Because there are very large number of monomers combined in a polymer, the number of moles of a polymer will be much lesser than 1.
okk..after reading ur statement again...i hav understood a bit....thnk u!...and the video ..perfectly matches the situation.........:p
 
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u seem to b doing the same paper....do u get q 40??..
option 2 and 3 are correct as compound with molecular mass 88 can be formed using given conditions
1 is omitted because saturated refers to double and triple bonds both, however molecular mass of 88 could be obtained using double bonds but using triple bonds its never possible as carbon numbers is restricted to 4. I think this is the concept
 
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option 2 and 3 are correct as compound with molecular mass 88 can be formed using given conditions
1 is omitted because saturated refers to double and triple bonds both, however molecular mass of 88 could be obtained using double bonds but using triple bonds its never possible as carbon numbers is restricted to 4. I think this is the concept
thx but i guess the variant was different ....its 12.....thx anyway....:)
 
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wat does variant in this case refer to??
the mcq paper u were solving is paper 11...i.e paper 1 ,first variant....
i did 12 ie..........paper 1 ,second variant...the questions are same but the order is different in both papers....
 
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the mcq paper u were solving is paper 11...i.e paper 1 ,first variant....
i did 12 ie..........paper 1 ,second variant...the questions are same but the order is different in both papers....

oh...no problem
whatever the order be
concept is same(y)
 
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I still don't get why you're even mentioning carbonic acid. :/
If you've studied bio, you'd know it's usually lactic acid that that forms in the teeth. I feel I'm missing something?
Option 3 says that Ca ions react with acids ( hence the equilibrium will shift to the right, decaying the tooth)
That statement is, theoretically, false. That is because any acid (such as sulphuric acid) will 'react' with the ions and the Ca salt formed will be soluble, hence there will be no overall change.
That is okay, but what if you add carbonic acid: there will be carbonate ions in solution, which will react with the Ca ion forming insoluble CaCO3. Hence the equilibrium will shift.
Yes, you are missing something: the syllabus assumes you do not study Biology.
 
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Q 14 A 5.00 g sample of anhydrous Group II metal nitrate loses 3.29 g in mass on strong heating
which metal is present ?

A magnesium
B calcium
C strontium
D barium

can some one explain

answer is B
 
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group II metal nitrate is going to produce metal oxide and nitrogen dioxide gas,the ratio between metal nitrate and metal oxide would be 1:1
5/x+28+96 = (5-3.29)/x+16
equate it u will get the answer
 
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group II metal nitrate is going to produce metal oxide and nitrogen dioxide gas,the ratio between metal nitrate and metal oxide would be 1:1
5/x+28+96 = (5-3.29)/x+16
equate it u will get the answer

Please take a look at my questions as well
 
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