Chemistry P1 Help

[flyingcrayons]

November 2009 Variant 11 Q21
How do you find the total number of C=C double bonds?

Any help is much appreciated.

 

[beacon_of_light]

Cyclohexene contains one double bond. while the second double bond is b/w 11th n 12th hydrocarbon.so total double bonds become 2. since total hydrogen atoms are 28 , 9

go in cyclohexen and 2 in the cis double bond.arrange the rest 17 hydrogens so that you get C20H28O......it will require 3 more double bonds...

So the total double bonds become 5!

 

[flyingcrayons]

Thanks! I got it!

 

[flyingcrayons]

June 2008 Q6 & Q22 please, anyone?

 

[beacon_of_light]

here is the solution...

Mass of 1cm^3 of H2O = 1g (got from the density of water)

Moles of H2O = 1/18 mol

Now,
since 1 mol of a gas occupies 24dm^3 at 298K,
so 1/18 mol of steam will occupy (1/18) x 24 = 1.33dm^3 at the same temperature.

now use V1 /T1 = V2/T2
1.33/ 298 = V2/ 596 .....
V2= 2.66dm^3

 

[beacon_of_light]

for mcq 22,

That is so simple...you join the monomers and make a polymer ...in the polymer(PVC) all C-C-C will be 109.5 since C is attached two four atoms with no lone pairs!!!! So the answer is A

Hope u got that!!!

 

[flyingcrayons]

Thanks again. I was confused cause it said the C-C-C bond.

 

[flyingcrayons]

June 2006 Q10?

 

[beacon_of_light]

Now work out step by step!

N204 ---> 2NO2
initial moles: 1 0
moles reacted: 0.5 0
equilibrium moles
or moles left: ( 1-0.5) = 0.5 1( since 1 mole of n2o4 forms 2 moles of no2 so 0.5 moles will form 0.5 x 2)

Nw find the partial pressure of each gas by dalton's law of partial pressure ka formula...
PP would be 0.5/1.5 1/1.5

so Kp = (1/1.5)^2 / (0.5/1.5) = 4/3 n u get the answer!!!

 

[flyingcrayons]

Thank you so much! You've helped a lot!

 

[flyingcrayons]

Can anyone help me out with June 2004 Q18? Thanks!

 

[Amanina]

Can anyone help me out with June 2004 Q18? Thanks!

let me try.

NaCl + H20 --> (1/2)H2 + (1/2)Cl2 + NaOH

find num of mole for NaCl, 58.5kg/(23+35.5) = 1 mol
so, 1 mol of NaCl will produce 0.5 mol of H2, 0.5 mol of Cl2 and 1 mol of NaOH.

for Cl2, 0.5 mol X (35.5+35.5) = 35.5kg
for H2, 0.5 mol X (1+1) = 1kg
for NaOH, 1mol X (23+16+1) = 40kg

hope it helps u.

 

[flyingcrayons]

Thanks!

And Q28, Q35, Q36, Q39. Can someone kindly explain those to me? Oh I suck...

 

[beacon_of_light]

which paper do these questions refer to?

 

[flyingcrayons]

June 2004, my bad.

 

[beacon_of_light]

Thanks!

And Q28, Q35, Q36, Q39. Can someone kindly explain those to me? Oh I suck...

Don't get tensed..transfer that to the examiner...hehehe just kidding!
(my badluck! i typed the whole explanation and due to bogus server breakdown, all got deleted!)

well for q28:

Only tertiary alcohols do not undergo oxidation...So the question asks for the structure AFTER dehydration which means that before dehydration, the alcohol must be a tertiary one...try adding OH and H across the double bond in all the options ...u'll only get a tertiary alcohol in D...

q35: Option C is wrong because CO is neutral and neutral compounds donot react with basic oxides...

q36: HI is not coloured, its colourless...while 2 & 3 are facts!

q39: As a matter of fact, ketones do not undergo further oxidation while aldehydes do..buh the question also says that carboxylic acid must be soluble in H20 and readily....buh option 2 has a benzene ring which is insoluble!!!

Hope u got that!!!

 

[flyingcrayons]

For Q36, what's the solid produced? Wow, thanks so much! I'm sure I have more questions lol, but for the mean time, thanks a lot! I appreciate all the help. Much love

 

[beacon_of_light]

Ur welcome!

Iodine is produced which is a solid!

 

[British Council]

Hey... Can anyone please explain me the following mcqs??? plzzz.. do reply.

June 2007: Q.10, Q.34 and Q.40..

 

[British Council]

Q.37 as well.