Post by: ruby92 on April 06, 2010, 11:26:15 am
Q 11, 24,26
also anyone have any worksheets etc with a summary of all organic chem reactions??
Post by: ruby92 on April 06, 2010, 12:12:59 pm
U mean paper 1, it is simple. It doesn't matter if it is a new compund dat u haven't studied all what u need to do is to count da number of oxygen atoms u will find them four, carbon atoms nd they r also four nd the hydrogen atoms nd they are also 4. hence the molecular formula is C4H4O4. It's B. U don't need to simplify since the question is about the molecular formula.which answer is this?
Post by: ruby92 on April 06, 2010, 12:30:33 pm
ive attached the papers here
Post by: nid404 on April 06, 2010, 12:42:48 pm
Q11..try n understand....total no of moles=4
partial pressures= no.of moles/ total no of moles X pressure
so the no of moles of products=x each
pp of h2= x/4 X p
pp of I2= x/4 X p
pp of HI= (b-x)2/4 X p2
Kp= x2 p2/16
(b-x)2/4 X p2
= x_2__
4(b – x)2
Post by: nid404 on April 06, 2010, 12:45:09 pm
B
Because only primary n secondary alcohols can be oxidised.
so C n D r out
Not A because it's primary and should form carboxylic acid on further oxidation but this one doesn't, so B
Post by: nid404 on April 06, 2010, 12:56:09 pm
U have to c which one of them form diols whose groups are secondary....secondary diols will form diketones on oxidation
Post by: ruby92 on April 06, 2010, 06:15:09 pm
Q:3 specifically; b(ii) (iii), C(ii) and d
Post by: nid404 on April 06, 2010, 06:20:12 pm
oct/nov 2003 paper 2
Q:3 specifically; b(ii) (iii), C(ii) and d
In a while
Post by: 3ishakay on April 06, 2010, 06:32:04 pm
oct/nov 2003 paper 2:(
Q:3 specifically; b(ii) (iii), C(ii) and d
yo 3)b)i)
is
mc(chnage in temp.)
so its
200 x 4.2 x 12.2 = 10.2 kJ
Post by: nid404 on April 06, 2010, 06:32:35 pm
Post by: 3ishakay on April 06, 2010, 06:34:57 pm
u jst told me u didn't study mmuch :P
lol trust me .. i jus got bak frma 4 month vacation ..
this stuff ws in IGz .. :D
Post by: Saladin on April 06, 2010, 07:16:58 pm
lol trust me .. i jus got bak frma 4 month vacation ..
this stuff ws in IGz .. :D
ya it was actually.
Post by: ruby92 on April 06, 2010, 07:41:25 pm
oct/nov 2003 paper 2
Q:3 specifically; b(ii) (iii), C(ii) and d
also m/j 2002 Q5 a(iii) iv and v
Post by: nid404 on April 07, 2010, 10:09:58 am
Post by: ruby92 on April 07, 2010, 10:47:12 am
its m/j 2002 paper 2 By the way
Post by: nid404 on April 07, 2010, 11:07:18 am
CH2OH CH(OH) CH3
with bromide cylcohexene's double bond breaks and Br joins to one of the carbons of the bond.
(http://www.chemada.com/cat1/items/BCH.gif)
with hot KMNO4
bond breaks, forming dicarboxylic acid....
Post by: tmisterr on April 07, 2010, 11:27:45 am
b(iii) is sulphur trioxide (it forms sulphuric acid).....it cant be any of the others because the first two are basic, Aluminium oxide is insoluble and Sulphur dioxide form sulphurous acid
c(ii) its an acid base reaction to form a salt and water
SO2 + NaOH ---> NaHSO3 + H2O (Unbalanced)
d i 9O2 and 6SO2 to balance
ii 4x+(6*-2)=0 solve for x, x=+3
iii Work with mole ratios here.
10 moles of Sb2S3 will produce 10/2=5 moles of Sb4O6, carry down the 5 moles to the next equation and five mole of Sb4O6 will produce 5*3= 15 moles of Carbon dioxide. at rtp one mole of a gas occupies 24 dm3 so 15*24 will be 360dm3 of carbon dioxide
Post by: ruby92 on April 07, 2010, 11:42:19 am
b(ii) is aluminium oxide
b(iii) is sulphur trioxide (it forms sulphuric acid).....it cant be any of the others because the first two are basic, Aluminium oxide is insoluble and Sulphur dioxide form sulphurous acid
c(ii) its an acid base reaction to form a salt and water
SO2 + NaOH ---> NaHSO3 + H2O (Unbalanced)
d i 9O2 and 6SO2 to balance
ii 4x+(6*-2)=0 solve for x, x=+3
iii Work with mole ratios here.
10 moles of Sb2S3 will produce 10/2=5 moles of Sb4O6, carry down the 5 moles to the next equation and five mole of Sb4O6 will produce 5*3= 15 moles of Carbon dioxide. at rtp one mole of a gas occupies 24 dm3 so 15*24 will be 360dm3 of carbon dioxide
im sorry i attached the wrong paper its actually october/nov 2003 :D
here's the paper :P
Post by: ruby92 on April 07, 2010, 08:09:33 pm
Q23
26
28
35
39 (why only the first one why not the second one; theyre both aldehydes)
Post by: nid404 on April 08, 2010, 07:25:28 am
m/j 2004 paper 1
Q23
26
28
35
39 (why only the first one why not the second one; theyre both aldehydes)
23) straight D
Cracking of a nonane can nvr give u a hydrocarbon with 10 carbon atoms...
26)straight D again....
slower reaction=high Ea
faster reaction=low Ea
28) D agn
monosubstituted alkene, one R group is attached to the carbon-carbon double bond. Not oxidized by mild oxidising agents
35) B
The temperature of the process causes the decomposition of the calcium carbonate
into calcium oxide.
Hence not 3
39) D straight
When you see a benzene ring, u knw it isn't soluble in water. So, 2 and 3 are out
Post by: tmisterr on April 08, 2010, 10:34:52 am
b iii 10.248kj is the enthalpy change when 1 g of calcium is dissolved, but one mole of calcium is 40.1g so 40.1*10.248=410.9 = 411 kj mol-1
c ii you need to use a hess cycle
We know the enthalpy change of the reaction to form Calcium Hydroxide and Hydrogen (calculated in biii) and we have been given the enthalpy change of formation of water.
Ca + 2H2O ----> Ca(OH)2 + H2
Ca + 2H2 + O2
so applying Hess law, starting from the lower equation, you can either first form calcium and water then form calcium hydroxide or directly form Calcium Hydroxide(this is the enthalpy change we are looking for). therefore:
2*-286+-411= enthalpy change of formation of Calcium hydroxide = -983kj mol-1
d) mole ratios (from equation in b)
1 mole of calcium forms one mole of hydrogen so 1/40.1 moles of calcium will form 1/40.1 moles of hydrogen.
so volume of oxygen is 1/40.1 * 24= 0.5985 dm3 or 598.5 cm3
Post by: ruby92 on April 08, 2010, 11:14:41 am
monosubstituted alkene, one R group is attached to the carbon-carbon double bond. Not oxidized by mild oxidising agents
could you explain this a bot further please?
Post by: chanella on April 08, 2010, 11:58:49 am
This website has notes,worksheets, tests
Post by: ruby92 on April 08, 2010, 12:09:40 pm
Thank You ppl :D
Post by: ruby92 on May 08, 2010, 05:41:03 pm
4 9 13 19 39(how is 2 correct) 32 ( water acts as a base rite?) 29 (why not A), 26(why not B)
Post by: nid404 on May 08, 2010, 06:40:47 pm
H has 1
C has 6
O has 8 (times 3)
and the overall charge is -1 (which means there's one more electron)
1+6+24+1=32
Q9) A
just the reverse right....so greater activation energy (endothermic reaction) and so
Q13) Phosphoric acid is more acidic as compared to the others. D
Q19) ammonium ion acts as an acid ( proton donor) H+ ion donor. hence D
Q39) 2nd one is an isomer of the first...so it's absolutely ok.
Q32) water is amphoteric. it can act as an acid and also a base. It can donate H+ ions and can also accept them forming(hydronium ion)
Q26) count the no of Carbon atoms in the product formed...lesser than the one they are asking for
Q29) again count the no of carbon atoms. In A the C from CN is not present. The count of C atoms should increase by 1 due to addtion of CN- ion
Post by: ruby92 on May 08, 2010, 08:13:01 pm
m/j 2008
34
39(why is 3 not correct?)
38
29 why not c
21 15 9
Post by: nid404 on May 08, 2010, 08:20:39 pm
Post by: ruby92 on May 08, 2010, 10:12:37 pm
o/n 2007
4 isnt it D group 3
5
9
37
Post by: nid404 on May 09, 2010, 07:21:46 am
Post by: ruby92 on May 09, 2010, 09:29:45 am
could you answer the rest of the questions as soon as possible :S ?
Post by: nid404 on May 09, 2010, 09:54:26 am
Post by: nid404 on May 09, 2010, 10:28:17 am
o/n 2007
4 isnt it D group 3
5
9
37
Q4) The greatest difference in I.E is between the 2nd and 3rd I.Es. That means it belongs to group 2
Q5) A polar molecule. Forms dipole, and can form hydrogen bonds with HCHO
Q9) The metal has to form a salt with sulphate ion. SO4 has a ox state of -2....so that of metal has to be +2 (only then will it form a stable salt compound)
Q37) I'll try get the mechanism...