[otrivine]

(Original post by oPJ_Lad)
Generally yes, but not when electrolytes are involved

not sure ? as well which page in book is it?

can shall we ask questions

[like_a_star]

does anybody have the jan 12 past paper and mark scheme? the link on page 2 isn't working anymore

[LifeIsGood]

This exam is actually starting me to me feel sick, I'm feel so anxious for tomorrow.

[otrivine]

just to confirm the exam is in the morning right?

[pete clark]

(Original post by dizzeedollee)
^
I didn't know this either! Doesn't say it in either of my textbooks and my teacher had no clue either :/

But I still don't understand why. Surely you want an acid with a pKa value that's close to the equivalence not the half equivalence point? :/

I am not 100% here, so someone correct me if I am wrong..

the pKa of an acid, is what its pH will be in PURE water.

E.g -
[H+] = 10^-7
[OH-] = 10^-7

I believe that is the case, but someone please correct me if I am wrong.

[PriyankaN]

I have a feeling they're going to ask some stupid application questions thats going to make me want to cry!

Do we only need to know the colours of the transition metal ions in the precipitation reactions? And do we need to know the variable oxidation states of a few?

[LifeIsGood]

HELP!!!!!
The Dissolved Oxygen Concentration (DOC) in rivers and lakes is important for aquatic life. If the
DOC falls below 5 mg dm–3, most species of fish cannot survive.
Environmental chemists can determine the DOC in water using the procedure below.
• A sample of river water is shaken with aqueous Mn2+ and aqueous alkali.
The dissolved oxygen oxidises the Mn2+ to Mn3+, forming a pale brown precipitate of
Mn(OH)3.
O2(aq) + 4Mn2+(aq) + 8OH–(aq) + 2H2O(l) 4Mn(OH)3(s)
• The Mn(OH)3 precipitate is then reacted with an excess of aqueous potassium iodide,
which is oxidised to iodine, I2.
2Mn(OH)3(s) + 2I–(aq) I2(aq) + 2Mn(OH)2(s) + 2OH–(aq)
• The iodine formed is then determined by titration with aqueous sodium thiosulfate,
Na2S2O3(aq).
2S2O3
2–(aq) + I2(aq) S4O6
2–(aq) + 2I–(aq)

Jan2011 Paper Q7

I don't understand how to get the moles of Oxygen:

I got 2.46*10^-5 mol of S2O32-
Then I2 = 1.23 * 10^-5 mol
2Mn(OH)3 = 2.46*10^-5 mol
Then multiplied it by 2 to get the first equation then divided it by 4 and got 1.23 * 10^-5

I don't understand!?

[LifeIsGood]

(Original post by otrivine)
just to confirm the exam is in the morning right?

Yes

[otrivine]

thanks

anyone up for some questions and answer as revision

[kimmey]

i think things that could come up are the equations to do with H2-O2 half cells

theyre quite easy to remember

just remember the overall equation is 0.5O2 + H2 ----> H2O

so at one electrode you have H2 splitting H2+2OH- ----> 2e- +2H20 (the hydrogen part)

now the products get combined with O2 from the other half cell 0.5O2 + 2e- + H2O ----> 2OH- (oxygen part)

[annony]

(Original post by moh.alt)
Sure.

An experiment was carried out to find the enthalpy change of neutralisation of ethanoic acid with sodium hydroxide.

CH3COOH + NAOH >>> CH3COONA + H20

50 cm3 of 2 mol dm-3 of Naoh was placed in a plastic cup and its temperature was recorded. The same amount of ethanoic acid was placed in a measuring cylinder and its temperature was recorded. The acid was added to Naoh solution and the temperature was recorded at 1 minute intervals for 5 minutes.

Results-

Temperature of Naoh solution- 19.4 degrees Celsius
Temperature of ethanoic acid solution- 18.8 degrees Celsius

time (mins) - 1 2 3 4 5
temp of mixture- 31.5 31.0 30.5 30.1 29.6

1g cm-3 and c=4.18.

(a) Estimate the maximum temperature which would have been reached in the absence of any heat loss.

(b) calculate the total amount of heat evolved during this reaction.

(c) calculate the amount of heat evolved per mole of water formed during neutralisation.

thank you!

[annony]

(Original post by The Illuminati)
It's a very long question so I'll just write the key things

An experiment was carried out to find the enthalpy of neutralisation of ethanoic acid with NaOH

50cm3 of 2moldm-3 NaOH
50cm3 of ethanoic acid

Temp of NaOH solution=19.4 degrees celsius
Temp of ethanoic acid solution=18.8

The 2 solutions were added together and temp was recorded at 1 min intervals for 5 mins

1 min, temp=31.5 2min temp=31 3 min, temp=30.5 4 min, temp=30.1 5 min, temp=29.6

Assume that the density of all solutions is 1 g cm-3 and their specific heats are all 4.18Jg-1 K-1

a) estimate the maximum temperature that would be reached in the absence of any heat losses
b)calculate the total amount of heat evolved (heat change) during this reaction

thank you!

[chemicalX]

(Original post by otrivine)
thanks

anyone up for some questions and answer as revision

ok ill be on in 30 mins, need to get some fresh air LOL see you in a bit

[otrivine]

(Original post by chemicalX)
ok ill be on in 30 mins, need to get some fresh air LOL see you in a bit

sure just quote me

[like_a_star]

(Original post by like_a_star)
does anybody have the jan 12 past paper and mark scheme? the link on page 2 isn't working anymore

please please please please anyone?

[xthelasthorcrux]

Does anyone have a link to the january 2012 paper? I can't get hold of it anywhere...
also, what do you guys predict will come up? So nervous.

[StrawberryKoi]

Has anyone else reached the freaking out stage of their revision?

[StrawberryKoi]

(Original post by xthelasthorcrux)
Does anyone have a link to the january 2012 paper? I can't get hold of it anywhere...
also, what do you guys predict will come up? So nervous.

Here it is with the mark scheme.

[xthelasthorcrux]

(Original post by StrawberryKoi)
Here it is with the mark scheme.

Thanks so much! Let's hope the grade boundaries of our paper aren't as high as the jan ones.

[ChrisJ]

(Original post by otrivine)
when they say allow ECF what it means are they still giving you the marks ? cause sometimes i see they even give full marks for ECF?

You'll still drop one mark for the error, but everything after the error you can pick up if you do the rest of the calculations correctly using the error answer