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OCR B F335 - Chemistry by Design - 13th June 2012

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Calculate the iodine number of DHA.
Mr: DHA, 328; Ar: I, 127
Give your answer to a suitable number of significant figures.
Why is it that some molecules absorb UV light and some absorb in the visible?
Reply 322
hey, just been flicking through my book and we don't need to learn the colours of the azo compounds do we? on pages 330-331? Because there's only three colours....
Do we need to know about solubility Ksp? for the new spec/paper?

for example question 4(d) in the legacy paper Jan 10....
Original post by Bi0logical
Calculate the iodine number of DHA.
Mr: DHA, 328; Ar: I, 127
Give your answer to a suitable number of significant figures.


I2: 254

254/328=0.77439
0.77439 x 100 = 77.439
= 77

Is that the way it goes? :smile:
Typical exam answer when asked if temperature is increased for an endothermic reaction:

(Forward) reaction is endothermic (1); 3
Equilibrium (position) moves to oppose change/ Increased temperature moves
equilibrium (position) in endothermic direction(1);
i.e. in direction consistent with stated endo/exo of forward reaction(1); (you can then go onto saying it's increased yield of product if asked)
Original post by Princess Kawaii


Use the formula for pH. You need to rearrange it to work out the value of [H+]. All the other values you need are given to you.

Check the spoiler for the full method. :smile:

Spoiler

(edited 11 years ago)
Original post by Princess Kawaii
I2: 254

254/328=0.77439
0.77439 x 100 = 77.439
= 77

Is that the way it goes? :smile:

I'm just marking the paper, once I come to this questions I'll tell you if you did it right though, thanks
Describing why ionic compounds e.g. NaCl have a high melting point and can conduct electricity:

ionic (1); strong electrostatic forces/bonds (lead to high melting point) (1);
(dissolves because) ions are hydrated/ ions form (ion-dipole) bonds/ ions attract
water moleules (1);
conducts because (free) ions can move (1)
Original post by Princess Kawaii
I2: 254

254/328=0.77439
0.77439 x 100 = 77.439
= 77

Is that the way it goes? :smile:


Mark scheme:

6 x 254 g of iodine (per 328 g DHA)*
Iodine no. = 1524 *(ecf) x 100/328 = 465 **

I dont get this, can someone take a look it's June 2006 I think you have to look at the structural formula too.
Original post by SimpleGirl
Use the formula for pH. You need to rearrange it to work out the value of [H+]. All the other values you need are given to you.

Check the spoiler for the full method. :smile:

Spoiler



Excellent explanation, do you have method like these for each calculation they can ask us? It would be VERY useful
Form B is coloured while form A absorbs in the UV. Heres why:

Form B absorbs (certain frequencies of) visible light;
(when) electrons excited (to higher energy-level);
Form B is more delocalised/larger chromophore/ larger conjugated system/ more
conjugated (NOT extra double bond) (ora for Form A)
because of alternating single and double bonds joining (benzene) rings (ora for
Form A)
Form B needs less energy to excite electrons (ora for Form A)/visible light has
lower energy than u.v.;
energy level difference measures frequency/wavelength absorbed/
(∆)E = hv;
max 2 if emission of light (rather than transmission) is implied
Guys, how does a buffer solution actually work? is this right
Addition of H+ moves the equilibrium to the left, this removes H+ and maintains the pH
The addition of salt, also makes the equilibrium shift to the left by reacting A- with H+ to form acid (maintaining pH?)
Original post by Bi0logical
Mark scheme:

6 x 254 g of iodine (per 328 g DHA)*
Iodine no. = 1524 *(ecf) x 100/328 = 465 **

I dont get this, can someone take a look it's June 2006 I think you have to look at the structural formula too.


Oh okay...I did it like that for another question before and got it right.
Yeh probably have to look at the formula to get the "6" part of it.

I don't have that paper...can u put a link or attach it please :smile:
Original post by Bi0logical
Guys, how does a buffer solution actually work? is this right
Addition of H+ moves the equilibrium to the left, this removes H+ and maintains the pH
The addition of salt, also makes the equilibrium shift to the left by reacting A- with H+ to form acid (maintaining pH?)



- When small amounts of acid is added the equilibrium shifts to the left to oppose the change/ remove the H+.
- When small amounts of a alkali is added the equilibrium shiftsto the right to oppose the change
- pH remains constant because there is a large amount of HA and A-
Original post by Bi0logical
Excellent explanation, do you have method like these for each calculation they can ask us? It would be VERY useful


Glad it helped :smile:

What other calculation questions do you need help on? I need an example to work through.
when does the rate constant increase?
I know it increases when the yield of a reaction increases.
and does it always increase with temp, increased pressure, conc ?
for something to be an arene, must it have a benzene right or aromatic ring?
do we need to know about solubility product, Ksp ?

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