[Alice453]

(Original post by Marcus4004)
Sorry about how long this has taken me.... Any explanations just ask...

Also please tell me where i went wrong thanks

ADDAB BDCCA CDDBA AACCD CCBDB ACDAB BABDB BCADD

1 a)i) 295.6 ii) 1s2 2s2 2p6 3s2 3p6 3d9
b) Photoconductive/voltaic
c)Positive holes

2a)i)5 ii) 3
b)2HNO2 + 4e- +4H+ ----->>>> H2N2O2 + 2H2O
c) dicyanocuprate(I)

3a) ionic radii
b) 6 chlorine surrounding 1 sodium vice versa
c) KF
d)i) hydration of both ions and lattice enthalpy
ii) Increase in entropy

4a) Trigonal Planar
b) Lone pair of electrons from oxygen donated to form bond, ie only oxygen electrons in bond.
c) ethers

5a) 3.95g
b) Tap water was used which contains metal ions, using deionised would have removed this problem and hence more accurate
c) Copper formed a complex as well as nickel.

6a) Separating Funnel
b)0.13
c) The temperature was different
d) Ethanol is soluble in water due to hydrogen bonding and hence will not partition

7a) 0.575 mol l^-1
b) Hard to see the colour change as the solution is dark brown and hence the equivalence point will be inaccurate
c) CH3COO- (aq)

8a) 500K-> 2000K
b) Since gradient is -entropy, a downwards line represents an increase in entropy.
c)Boiling point of Mg

9)1.45

10) a) 1
B)0
C) s^-1

11)a)i) Hydrogen ii) Hydrogen Chloride
b)i) Acid Chlorides
ii)meh cba
iii) Faster and goes to completion as it is not an equilibrium. Also means greater yield

12a) C8 H14 S2 O2
bi) The one with 3 lines coming out of it
ii) 4 different functional groups and hence asymmetric. Hence Chiral ----> optically active

13a) To make sure that none of the reactants/ products escape on heating
b) Hydrolysis
c) Anti bumping Granules
d) No oily residue remains
e) 3.36g

14a)i) Electrophilic Substitution ii) Phenylamine ( The funny thing is that this question is incorrect as NaOH is required to turn the salt into the phenlyamine but oh well... )
b)i) Check the internet for structure
ii) One double bond no overall charge

15a) Secondary alcohol

b) meh cba

c) Two peaks excluding TMS at 1 and 2 with the peak at 1 having a height of 6 and the peak at 2 having a height of 4.

dont think 15a is a secondary alcohol because its molecular formula is c4 h9 cl

[Student666]

damn. i wrote that then for some reason scored it out! WHY.

[tompy123]

(Original post by Student666)
ah for 4c) wasn't it aldehydes?

I put ketones ... May have been aldehydes though?

[Niamh-94]

(Original post by tompy123)
I put ketones ... May have been aldehydes though?

It's an ether

[Alice453]

(Original post by victoria-x)
the temperature may have been different

yay same `

[tompy123]

Thinking about it, I've not actually done as well as I first thought in a few of the questions.

[Kitttty]

Kicking myself over how many stupid mistakes I made!

Anyway, for the rate question I thought it was 0 and 1, but my teacher was saying it was 1 and 2? Hoping she's wrong! Wouldn't be the first time :P

[tompy123]

(Original post by Niamh-94)
It's an ether

So it is! How silly of me! :P

[AeroLB]

For 4c I think I put monohalogenoaldehyde 'cuz I know what I'm talking about

[jordan0141]

The empirical formula is always the lowest whole number ratio, the empirical formula for benzene is CH and that isnt a real molocule so i still think the answer was C2H5

[Marcus4004]

(Original post by Alice453)
dont think 15a is a secondary alcohol because its molecular formula is c4 h9 cl

How not? A primary would not have worked for the next part of the question because if the OH was at the end no isomers would be formed on dehydrating..

[Marcus4004]

(Original post by jordan0141)
The empirical formula is always the lowest whole number ratio, the empirical formula for benzene is CH and that isnt a real molocule so i still think the answer was C2H5

But that's not a molecular formula. (try drawing C2H5)

[jordan0141]

Just reread the question and realised it asked for molecular formula, excuse my ignorance

[Niamh-94]

(Original post by Marcus4004)
How not? A primary would not have worked for the next part of the question because if the OH was at the end no isomers would be formed on dehydrating..

The molecular formula is C4H9Cl

so it must be with either 1-chlorobutane or 2-chlorobutane

There is NO oxygen in the molecular formula, so it can't be any alcohol...

[rp1401]

I got one peak of six and one peak of two (or 3:1) in the nmr sketch. Surely there would be 2CH3s at the end of the butene, identical distances from the double bond so 6, and 1 hydrogen on each of the carbons in the double bond, so 2. Lots of people seem to have different things for that, I'm probably wrong

[Marcus4004]

(Original post by Niamh-94)
The molecular formula is C4H9Cl

so it must be with either 1-chlorobutane or 2-chlorobutane

There is NO oxygen in the molecular formula, so it can't be any alcohol...

Oops, soz i meant a secondary halogenoalkane. :/ will edit

[adam_johnson]

(Original post by Marcus4004)
Sorry about how long this has taken me.... Any explanations just ask...

Also please tell me where i went wrong thanks

ADDAB BDCCA CDDBA AACCD CCBDB ACDAB BABDB BCADD

1 a)i) 295.6 ii) 1s2 2s2 2p6 3s2 3p6 3d9
b) Photoconductive/voltaic
c)Positive holes

2a)i)5 ii) 3
b)2HNO2 + 4e- +4H+ ----->>>> H2N2O2 + 2H2O
c) dicyanocuprate(I)

3a) ionic radii
b) 6 chlorine surrounding 1 sodium vice versa
c) KF
d)i) hydration of both ions and lattice enthalpy
ii) Increase in entropy

4a) Trigonal Planar
b) Lone pair of electrons from oxygen donated to form bond, ie only oxygen electrons in bond.
c) ethers

5a) 3.95g
b) Tap water was used which contains metal ions, using deionised would have removed this problem and hence more accurate
c) Copper formed a complex as well as nickel.

6a) Separating Funnel
b)0.13
c) The temperature was different
d) Ethanol is soluble in water due to hydrogen bonding and hence will not partition

7a) 0.575 mol l^-1
b) Hard to see the colour change as the solution is dark brown and hence the equivalence point will be inaccurate
c) CH3COO- (aq)

8a) 500K-> 2000K
b) Since gradient is -entropy, a downwards line represents an increase in entropy.
c)Boiling point of Mg

9)1.45

10) a) 1
B)0
C) s^-1

11)a)i) Hydrogen ii) Hydrogen Chloride
b)i) Acid Chlorides
ii)meh cba
iii) Faster and goes to completion as it is not an equilibrium. Also means greater yield

12a) C8 H14 S2 O2
bi) The one with 3 lines coming out of it
ii) 4 different functional groups and hence asymmetric. Hence Chiral ----> optically active

13a) To make sure that none of the reactants/ products escape on heating
b) Hydrolysis
c) Anti bumping Granules
d) No oily residue remains
e) 3.36g

14a)i) Electrophilic Substitution ii) Phenylamine ( The funny thing is that this question is incorrect as NaOH is required to turn the salt into the phenlyamine but oh well... )
b)i) Check the internet for structure
ii) One double bond no overall charge

15a) Secondary alcohol

b) meh cba

c) Two peaks excluding TMS at 1 and 2 with the peak at 1 having a height of 6 and the peak at 2 having a height of 4.

for 1a)ii) i got 1s2 2s2 2p6 3s2 3p6 3d10
because in the ground state it is 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p1
and for 13c) i put HCl

[rp1401]

(Original post by Marcus4004)
How not? A primary would not have worked for the next part of the question because if the OH was at the end no isomers would be formed on dehydrating..

You've given compound B which is a secondary alcohol, it was asking for the structure of A

[c.macarthur]

(Original post by victoria-x)
This is my first post on TSR so am not sure how this whole thing works but I have the answers my teacher gave me for the multiple choice. He did them pretty quick so there may be a couple mistakes he said.

ADDAB BDCCA CDDBA AACCD CCBDB ACDAB BABDB BDADD

i like the look of that i'd take 34!!

[Marcus4004]

(Original post by rp1401)
You've given compound B which is a secondary alcohol, it was asking for the structure of A

Yeh, but i dont have the patience to draw the structure on TSR so i just described it