Chemistry Paper 4-theory- doubts =D

[AlishaK]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_41.pdf
Q2 a iv) What is that?!
PhyZac
xhizors
knowitall10

 

[leosco1995]

Can someone please tell me how to solve this?

 

[leosco1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_41.pdf
Q2 a iv) What is that?!

For a relative rate of 1.00, you need

0.20 CH3CHO
0.10 CH3OH
0.05 H+

Now they want you to predict the rate when all of them have a concentration of 0.20.

The first one is already 0.20 so leave that.
For the second one you need to 2x the concentration to get 0.20, so the rate becomes 2x.
FOr the third one you need to 4x the concentration to get 0.20, so the rate becomes 4x.

In total its 2 * 4 = 8 times the initial rate.

 

[AlishaK]

For a relative rate of 1.00, you need

0.20 CH3CHO
0.10 CH3OH
0.05 H+

Now they want you to predict the rate when all of them have a concentration of 0.20.

The first one is already 0.20 so leave that.
For the second one you need to 2x the concentration to get 0.20, so the rate becomes 2x.
FOr the third one you need to 4x the concentration to get 0.20, so the rate becomes 4x.

In total its 2 * 4 = 8 times the initial rate.

Thanks!
Cheers

 

[AlishaK]

Can someone please tell me how to solve this?

what year is that?

 

[leosco1995]

what year is that?

November 2005
http://papers.xtremepapers.com/CIE/...d AS Level/Chemistry (9701)/9701_w05_qp_4.pdf

 

[Zari]

lol dont say sorry
overload the questions on me i would love that
look due to other functional groups the 2 4 (and 6) position of ring is activated so u don't need any further reagants to Bromonize rings

ok sure JazakAllah for the reply

 

[AlishaK]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_41.pdf
Q2 a i) how is H+ order 1 ?
xhizors leosco1995

 

[AlishaK]

Sorry i w

Can someone please tell me how to solve this?

as solving another ppr so took longer. Hey it's gon be the same as for Cl2
Sn+2I2---> SnI4

 

[leosco1995]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_41.pdf
Q2 a i) how is H+ order 1 ?
xhizors leosco1995

If you know the first 2 reactants are 1, then compare equations 1 and 4:

CH3CHO stays the same, so ignore that.
CH3OH increases by 1.6, so the rate also increases by 1.6. So the new rate will be 1.6.

Now see H+,
H+ concentration doubles and the rate changed from 1.6 to 3.2 (also doubled). What does this imply? H+ has an order of 1.

 

[leosco1995]

Sorry i w

as solving another ppr so took longer. Hey it's gon be the same as for Cl2
Sn+2I2---> SnI4

Yeah this is what I don't get.. I mean this is what I did:

I2/I- system is +0.54V and Sn2+/Sn4+ is +0.15, tin gets oxidised to Sn4+ by I2, so can't you just write:

I2 + Sn2+ -> 2I- + Sn4+

Like you normally do in redox equations?

 

[AlishaK]

If you know the first 2 reactants are 1, then compare equations 1 and 4:

CH3CHO stays the same, so ignore that.
CH3OH increases by 1.6, so the rate also increases by 1.6. So the new rate will be 1.6.

Now see H+,
H+ concentration doubles and the rate changed from 1.6 to 3.2 (also doubled). What does this imply? H+ has an order of 1.

Oh so we consider the new rate!?
Thanks a load!!

 

[AlishaK]

Yeah this is what I don't get.. I mean this is what I did:

I2/I- system is +0.54V and Sn2+/Sn4+ is +0.15, tin gets oxidised to Sn4+ by I2, so can't you just write:

I2 + Sn2+ -> 2I- + Sn4+

Like you normally do in redox equations?

what u r writing isn't wrong, u r just writing an ionic eqn, but they didnt ask u to write ionic. they asked for a simple one.

 

[leosco1995]

what u r writing isn't wrong, u r just writing an ionic eqn, but they didnt ask u to write ionic. they asked for a simple one.

Oh right.. yeah I get it now. Thanks.

 

[AlishaK]

Oh right.. yeah I get it now. Thanks.

No worries. We help each other and that is what makes us XP Family.

 

[AlishaK]

Okay so is NH3 with a + sign abv 'N' an e- withdrawing grp?
PhyZac

 

[PhyZac]

Okay so is NH3 with a + sign abv 'N' an e- withdrawing grp?
PhyZac

Sorry..dint get you.

 

[AlishaK]

http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_w11_qp_41.pdf
Q4 b ii, see the third diag. it has a very low pka value so obvio it has a high ka value that means it's acidity is quite high.
so it's cause of the NH3 grp. therefore, is NH3 an electron withdrawing grp?

 

[PhyZac]

http://papers.xtremepapers.com/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_w11_qp_41.pdf
Q4 b ii, see the third diag. it has a very low pka value so obvio it has a high ka value that means it's acidity is quite high.
so it's cause of the NH3 grp. therefore, is NH3 an electron withdrawing grp?

I am not sure..but

NH3+ has lone pair...and is is delocalised with benzene electrons..and making NH bond even weaker..so act as an acid

 

[Oliveme]

Can u gimme a few more examples for a) metal ions in which H2O is imp b) for which it aint imp.

Jazak Allah Khair!

There is no specific transition ions that you need to know about that H2O form ligands with.

The most common ones you'll come across are --->
[Fe(H2O)6]^2+ and this can go on to a ligands exchange/ replacement. If, for example, you add concentrated ammonia solution, you'll get [Fe(H2O)2(NH3)4]^2+ and this happens because this complex is more stable than the first one.

Another common one is copper ions. [Cu(H20)6]^2+ ----> this is when there are Cu^2+ ions in solution. This complex gives a pale blue colour.
If sodium hydroxide is added, a blue ppt is formed of Cu(OH)2(H2O)4 is formed.
On adding conc. ammonia solution, you get [Cu(H2O)2(NH3)4]^2+ and this complex is dark blue.

Let's start with [Cu(H20)6]^2+ again. This is, again, blue. Adding conc. hydrochloric acid gives the complex [CuCl4]^2+ which is yellow solution. Notice that the no. of ligands reduces from 6 to 4 and this is because the chloride ions are bigger in size than the H2O molecules.
Hope this was what you were looking for.