[YB101]

(Original post by PlatypusPigeon)
If it's the ocr chem book then they have loads of mistakes so don't be too surprised :P

Well that's just brilliant (!)

cheers anyways happy revising!

[Bright]

(Original post by chemicalX)
ahhh ok thanks i get it know

and yeah full marks as always

Umm I still don't get it..
can you explain please? :/

[:):D]

Okay so another question im stuck with:
BrO3- +...Br-+...H+---->...Br2+...H2O
the (...) is the space left for me me to balance. How do you balance this, the markscheme says there are
5Br- and 3Br2. Why is it not 1Br- and 1Br2?

This question is taken from the june 2010 new spec paper
Thankss

[chemicalX]

(Original post by otrivine)
are you sure u got it , difficult to explain ?
ok

Explain why the H-O-H bond angle in water ligand is 107 and not 104.5 ? (3)

because the electrons are close together so they repel slightly? sorry i have no idea about this question

[seamen]

CGP OCR chemistry book is terrible stay away from it if possible, the way they work out the entropy change is different to what the exam board shows it, just use the OCR chemistry by Hienemann.

[otrivine]

(Original post by chemicalX)
because the electrons are close together so they repel slightly? sorry i have no idea about this question

close 1/3
you have to use your F321 knowledge so you say like you said electrons repel/ lone pairs repel more than bonding pairs, in water there are 2 lone pairs and 2 bonding pairs, and have to mention electron pairs repel as into maximise separation and minimise repulsion

[seamen]

(Original post by chemicalX)
because the electrons are close together so they repel slightly? sorry i have no idea about this question

A lone pair of electrons in the ligand is donated to the transition metal ion resulting in 1 lone pair and 2 bonded pair, the lone pair has more repulsion than bonded pairs, but is donated so the angle is 107.

[Aa234]

(Original post by :):D)
Okay so another question im stuck with:
BrO3- +...Br-+...H+---->...Br2+...H2O
the (...) is the space left for me me to balance. How do you balance this, the markscheme says there are
5Br- and 3Br2. Why is it not 1Br- and 1Br2?

This question is taken from the june 2010 new spec paper
Thankss

look at the oxidation number of br in bro3 and compare with the oxidation number in br2, goes from +5 to 0 so you need 5 electrons so it has to be 5 br-


This was posted from The Student Room's iPhone/iPad App

[PlatypusPigeon]

(Original post by :):D)
Okay so another question im stuck with:
BrO3- +...Br-+...H+---->...Br2+...H2O
the (...) is the space left for me me to balance. How do you balance this, the markscheme says there are
5Br- and 3Br2. Why is it not 1Br- and 1Br2?

This question is taken from the june 2010 new spec paper
Thankss

The oxidation state of Br in BrO3- is 5 (O3 is 6- so Br must be 5+ to balance it out as a - charge overall), and it's going from +5 to 0 (Br2) so 5 electrons must be transferred via the Br-. Now you have 6 Br on the left hand side so you need 6 on the right, thus 3Br2.

Any easier?

[otrivine]

(Original post by seamen)
A lone pair of electrons in the ligand is donated to the transition metal ion resulting in 1 lone pair and 2 bonded pair, the lone pair has more repulsion than bonded pairs, but is donated so the angle is 107.

but wait water has 2 lone pairs and 2 bonding pairs

[mack94]

Any tips on working out the rate determining step?

[Hi_:)]

Hey just wondering if anyone could help me with this question? Its question 4(e) from the June 2011 paper. I have no clue how you identify which acid and its conjugate base you are meant to chose. I beg anyone help me, I am so stuck!!
Here's the link to the paper cause I dunno how to attach images..

http://pdf.ocr.org.uk/download/pp_11...gce_uf325.pdf?

Thank you so so so much anyone!

[seamen]

(Original post by otrivine)
but wait water has 2 lone pairs and 2 bonding pairs

Yh talking about the ligand, a ligand donates a lone pair of electrons to make a dative covalent bond, this results in less repulsion so bigger angle.

[The Illuminati]

Here is a nontypical Kw/pH question I found:

Calculate the pH when you have
25cm3 of 0.005moldm-3 H2SO4 and 10cm3 of 0.1moldm-3 NaOH

open to anyone.

EDIT: FORGOT TO STRESS THAT YOU NEED TO FIND THE pH OF THE MIXTURE NOT THE SEPERATE COMPONENTS
ANOTHER EDIT: I made a typo for cocn of the acid. H2SO4 should be 0.05 moldm-3 to get an answer of 1.37.
If you use 0.005 then your answer should be 12.3

Sorry x

[seamen]

(Original post by Hi_:))
Hey just wondering if anyone could help me with this question? Its question 4(e) from the June 2011 paper. I have no clue how you identify which acid and its conjugate base you are meant to chose. I beg anyone help me, I am so stuck!!
Here's the link to the paper cause I dunno how to attach images..

http://pdf.ocr.org.uk/download/pp_11...gce_uf325.pdf?

Thank you so so so much anyone!

Lactic acid because it has the closest PKa value to the PH, this means equal amounts of conjugate base and acid resulting better minisation of acid addition or alkali addition. The pH is 3.55 so the conjugate base of any other acid will react with H+ ions in the buffer, to prevent that the closest PKa value is chosen.

Lactic acid , sodium lactate.

[PlatypusPigeon]

(Original post by mack94)
Any tips on working out the rate determining step?

I asked this a few pages up and sorta understand it now, question 1 of Jan 2012 is a great one to practice!

H2(g) + 2ICl(g) 2HCl(g) + I2(g) and rate = k[H2(g)] [ICl(g)]

Start by putting H2 and ICl on the left hand side of your first step - this is the slowest.
So you now have H2 + ICl ---->
You can see that you need to HCl in the final equation and you can make that here, so put that on the right hand side, then balance it with the remaining HI (this is the intermediate).

Next, add the intermediate to the lhs of the second equation. You see you need 2ICl to make up the reactants and you only have one, so add another one!
So now you have H2 + ICl ---> HCl + HI and
HI + ICl --->
Now just complete by making the remaining products (the I2 and the other HCl) and your done!

[seamen]

(Original post by The Illuminati)
Here is a nontypical Kw/pH question I found:

Calculate the pH when you have
25cm3 of 0.005moldm-3 H2SO4 and 10cm3 of 0.1moldm-3 NaOH

open to anyone.

pH 2 for H2SO4, 25/1000 x 0.005=1.25x10^-3, 2.5x10^-3 moles of H+ you got volume etc then pH and 13 for NaOH

[Bright]

(Original post by The Illuminati)
Here is a nontypical Kw/pH question I found:

Calculate the pH when you have
25cm3 of 0.005moldm-3 H2SO4 and 10cm3 of 0.1moldm-3 NaOH

open to anyone.

Is the answer 1.60 by any chance?

[otrivine]

(Original post by seamen)
Yh talking about the ligand, a ligand donates a lone pair of electrons to make a dative covalent bond, this results in less repulsion so bigger angle.

but how u know there is 2 ligands?

[PlatypusPigeon]

(Original post by seamen)
pH 2 for H2SO4, 25/1000 x 0.005=1.25x10^-3, 2.5x10^-3 moles of H+ you got volume etc then pH and 13 for NaOH

For NaOH did you do kw/[H+]??