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For a relative rate of 1.00, you need
Thanks!
what year is that?
November 2005what year is that?
ok surelol dont say sorry
overload the questions on me
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
JazakAllah for the reply
as solving another ppr so took longer. Hey it's gon be the same as for Cl2
If you know the first 2 reactants are 1, then compare equations 1 and 4:
Yeah this is what I don't get.. I mean this is what I did:
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.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.
No worries. We help each other and that is what makes us XP Family.Oh right.. yeah I get it now. Thanks.
Sorry..dint get you.
I am not sure..buthttp://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?
There is no specific transition ions that you need to know about that H2O form ligands with.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!
Jazak Allah Khair! I knew all those Cu reacns but now, after u explained, everything seems so Clear!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.
just a small mistake here.. the complex [CuCl4] is actually written as [CuCl4]^2- and not as [CuCl4]^2+Jazak Allah Khair! I knew all those Cu reacns but now, after u explained, everything seems so Clear!
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