[master_blaster66]

(Original post by Jasmine_777)
do we need to know about solubility product, Ksp ?

no

[paul.e.rose]

I was wondering what IMB was stronger Ion-dipole or hydrogen bonds. Hence which would improve the solubility in a dye to a greater extent, adding an NH2 group or an SO3-?

[master_blaster66]

(Original post by paul.e.rose)
I was wondering what IMB was stronger Ion-dipole or hydrogen bonds. Hence which would improve the solubility in a dye to a greater extent, adding an NH2 group or an SO3-?

so3 i think

[SimpleGirl]

(Original post by paul.e.rose)
I was wondering what IMB was stronger Ion-dipole or hydrogen bonds. Hence which would improve the solubility in a dye to a greater extent, adding an NH2 group or an SO3-?

Ionic groups make dyes more soluble. So adding SO₃^- will allow it to dissolve in water.

[Venomilys]

messed up jan 2012 paper... only got 84/120 raw. Damn enzymes!

[blingmonkeys]

(Original post by Ilyas)
messed up jan 2012 paper... only got 84/120 raw. Damn enzymes!

Two questions:

1)How did u calculate the pH for the last question?

2) (Different paper) can u explain how to complete the reduction half equation?

SO₄^2- +....H⁺+ .....e- ---> H₂S +......

[kishenp]

Can anyone explain in the June 2010 paper
where the mark scheme got 190 in the reactants side
Thanks
here is the link for the paper
http://pdf.ocr.org.uk/download/pp_10..._gce_f335.pdf?

[kishenp]

and has anyone got any tips about naming compounds
i always get them wrong =/
thanks

[master_blaster66]

(Original post by kishenp)
Can anyone explain in the June 2010 paper
where the mark scheme got 190 in the reactants side
Thanks
here is the link for the paper
http://pdf.ocr.org.uk/download/pp_10..._gce_f335.pdf?

what question

[Venomilys]

(Original post by blingmonkeys)
Two questions:

1)How did u calculate the pH for the last question?

2) (Different paper) can u explain how to complete the reduction half equation?

SO₄^2- +....H⁺+ .....e- ---> H₂S +......

1)skipped it

2) SO₄^2- +10H⁺+ 8e- ---> H₂S +4H₂O, I just did it by considering what else could be eliminated (we only have H's abd O's left, immediately thought about water so I worked on that basis)

[kishenp]

(Original post by master_blaster66)
what question

Hi
sorry
It is question 1e(ii)
and i just came across the question of calculating 2(f)
how did they get CH3CH2OCH2CH3
i dont understand how they obtained this
would massively appreciate the help
Thanks alot

[blingmonkeys]

(Original post by Ilyas)
1)skipped it

2) SO₄^2- +10H⁺+ 8e- ---> H₂S +4H₂O, I just did it by considering what else could be eliminated (we only have H's abd O's left, immediately thought about water so I worked on that basis)

Oh kk, but why 8 electrons and not 4?

[kishenp]

i also have real difficulty naming compounds
any tips ?
thank you

[Venomilys]

(Original post by blingmonkeys)
Oh kk, but why 8 electrons and not 4?

one SO4 with -2 charge
ten H+ ions with charge +1
8 electrons with charge -1

one(-2) + ten(+1) + eight(-1) = 0.

[Venomilys]

hmmm... start off always by counting the longest possible string of C's, then start naming functional groups.

[nosh1994]

apart from when specifically asked about the atomic emission spectrum, when else do we mention electrons drop back down a level releasing a photon? i used to just mention this in the normal 6 markers asking why chromophores are coloured, but after re-doing june 2010 (new spec) i realised in the mark scheme it would con a lot of marks (only allowed 2/6)?

thanks in advance!

[blingmonkeys]

(Original post by Ilyas)
one SO4 with -2 charge
ten H+ ions with charge +1
8 electrons with charge -1

one(-2) + ten(+1) + eight(-1) = 0.

Ahh thanks man that just helped so much !


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[SimpleGirl]

(Original post by kishenp)
Can anyone explain in the June 2010 paper
where the mark scheme got 190 in the reactants side
Thanks
here is the link for the paper
http://pdf.ocr.org.uk/download/pp_10..._gce_f335.pdf?

190 is not from the reactant side.
This is the reaction given to you:
C₆H₁₂O + 2HNO₃ --> C₆H₁₀O₄ + 2H₂O + N₂O
They've told you that the equation represents the production of nylon (C₆H₁₀O₄) and nitrous oxide (N₂O).

You need to calculate the atom economy for this reaction. The formula for atom economy is:
(Total molar mass of desired products / total molar mass of all products) x 100

Desired products = C₆H₁₀O₄ and N₂O
All Products = C₆H₁₀O₄, 2H₂O and N₂O

They've given you the M_r values you need so just add them together.
Total M_r of desired products = 146 + 44 = 190
Total M_r of all products = 146 + 2(18) + 44 = 226

Atom economy = (190/226) x 100 = 84.1%

[examtalk]

(Original post by kishenp)
Can anyone explain in the June 2010 paper
where the mark scheme got 190 in the reactants side
Thanks
here is the link for the paper
http://pdf.ocr.org.uk/download/pp_10..._gce_f335.pdf?

The 190 comes from adding 146 (Mr of hexanedioic acid as it is used in the production of nylon) and 44 (Mr of nitrous oxide as it is used as an aerosol propellant) so together 190 is the mass of useful products. Does this help?

[Hazel247]

(Original post by kishenp)
i also have real difficulty naming compounds
any tips ?
thank you

If you count the longest chain of carbons first then you know what the end part will be, and you can then add on the functional groups in alphabetical order. And remember you always number the functional groups as low as you possibly can (so think about which end you start counting from!)

Hope this helps