The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

ocr a f325 revision thread

Scroll to see replies

Original post by samzurai
It's comparing the cl- ions to f- so f- has greater attraction to na compared to cl-


I know, hence I said the Na+ ions were more attracted to F- than Cl-, would I get that mark then?

Would I also get the mark for saying F- is more attracted to Mg2+ (is that the same as saying Mg2+ has greater attraction for F-)?
Original post by otrivine
Do we have to know about the vanadium colours, or in our syllabus do we only need to know the colours for the precripitation and ligand sub.


We don't. We got told it was violet.
Reply 2482
Avatar for HSR08
HSR08
Original post by Ronak134
10cm3 of the 50cm3 of solution produced was titred.. it says in the question.




it says they are oxidised when titred with Mno4

it goes from +2 to +5 when titred with Mno4. 'V(n+) ions are oxidised back to VO3(-) ions'

V(2+) goes to VO3(-)

oxygen can't come from nowhere, so the water in the solution most also react. then the only thing that can happen to the hydrogen is that it becomes H+. this seems to be the case as in the other half equation there are H+ ions as well.


Can you do Jan 13 6e please?
Reply 2483
Avatar for Namod
Namod
10
Just to be 100% sure,

we don't need to

know the indicator

range and colour

change?
Reply 2484
Avatar for kuku2013
kuku2013
in dynamic equilibrium do concentration of reactant and product the same?
so scared, i bombed my bio...i cant fail this too
Original post by DudeBoy
Do we need to know the other ones, e.g.
Cr2+ blue
Cr3+ green

And what not?


The ones I would say DEFINITELY learn;
Crr 6+ orange
Cr 3+ green
Mn 2+ pale pink
Mn 7+ purple
Fe 2+ pale green
Fe 3+ pale yellow
Co 2+ pink
Cu 2+ blue
Original post by Namod
Just to be 100% sure,

we don't need to

know the indicator

range and colour

change?


Surely not! They usually give us the table to use as reference so I don't think we need to worry about that. We basically need to know the role of the indicator.
Reply 2488
Avatar for rival_
rival_
10
Original post by wl1
When you work out the concentration, the volume you divide by is 25cm3. Although you have scaled up to 250cm3, that 250cm3 was originally made up with water from 25cm3.


Thanks dawg, I owe you.
Original post by HSR08
Any chance you can do jan 13 6e please?


At chromium electrode Cr3++3e−→CrCr^{3+} + 3e^- \rightarrow Cr
At X electrode X→X2++2e−X \rightarrow X^{2+} + 2e^-
Overall 2Cr3++3X→2Cr+3X2+2Cr^{3+} + 3X \rightarrow 2Cr + 3X^{2+}
Amount of Cr = 1.45652=0.028mol\frac{1.456}{52}=0.028mol
Ratio Cr:X = 2:3 so amount of X = 0.042 mol
Mr(X)=1.0210.042=24.3M_r(X)=\frac{1.021}{0.042}=24.3
So X is magnesium.
Reply 2490
Avatar for Liamb147
Liamb147
Hi guys, I would be eternally grateful if someone could link me to some of the legacy papers for the exam... I feel that I'd be gaining little by doing all of the f325 papers again. Good luck to everyone for tomorrow !!!
Original post by kuku2013
in dynamic equilibrium do concentration of reactant and product the same?


Dynamic equilibrium
Rate of foward reaction = rate of backwards reaction
The concentrations of both are constantly interchanging
Original post by kuku2013
in dynamic equilibrium do concentration of reactant and product the same?


Yep. Book says so.

1) Closed system
2) Forward and backward reaction at same rate
3) Conc of reactants and products are the same
Original post by X44
Lool damn right, it took my teacher 20 mins to figure it out but on the examiners report some geniuses still managed full marks in that question


Posted from TSR Mobile


LOL same! He even saw the mark scheme beforehand but when he was explaining it to me, it took him like 5 minutes just to work out the balanced equation. I don't think I even fully understand it still, need to go back and check >_>
anyone up for some revision?



Describe why catalysts are used (3)
Original post by DavidH20
I know it's a little late, but I've got some notes here if anyone wants something concise for last-minute revision :smile:

http://www.thestudentroom.co.uk/showthread.php?t=2378761


they're great! thanks for sharing :wink:
Original post by X44
ImageUploadedByStudent Room1370970222.513793.jpg Can someone please explain how c (ii) can be answered I don't understand the markscheme :s thanks


Posted from TSR Mobile


Ok, firstly lets look at what we're given:

•

50cm3 of 0.250 moldm-3 butanoic acid

•

50cm3 of 0.0500 moldm-3 sodium hydroxide

•

Ka of butanoic acid is 1.51x10-5 moldm-3



Firstly we know that butanoic acid and sodium hydroxide have different quantities and thus one is in excess, thus we need to work out how much of each chemical reacts.

To do this you world out the moles for each chemical:

Concentration x Volume / 1000 = Moles.

So let's work out the moles of each substance to see which one is in excess:

For butanoic acid:

50 x 0.250 / 1000 = 0.0125 moles of butanoic acid

For Sodium Hydroxide:

50 x 0.0500 / 1000 = 2.5x10-3 moles of sodium hydroxide.

From these to mole values we can see that butanoic acid is in excess, thus we need to work out how many moles of it react with sodium hydroxide.

0.0125 reacts with 2.5x10-3 moles
0.0125 - 2.5x10-3 = 0.01 moles of butanoic acid that has reacted with ALL of the NaOH.

Now that we have the moles of Butanoic acid and Sodium Hydroxide we can finally work out the answer in a few more steps:

First let's look at the PH buffer formula for reference:

H+ = Ka x [HA (moldm-3)] / [A- (moldm-3)]

Keep in mind that we need the values in concentration, yet we have moles. So we convert Moles to concentration.

There is only one difference though, in the buffer solution the volume of both chemicals are added together, thus the volume is actually 100cm3

So the concentration of Butanoic acid in the buffer solution is:

0.01 x 1000 / 100 = 0.1moldm-3 of Butanoic acid in the buffer solution.

And NaOH: 2.5x10-3 x 1000 / 100 = 0.025 moldm-3 of NaOH.

So we enter these values into the equation:

H+ = 1.51x10-5 moldm-3 x [0.1 (moldm-3)] / [0.025 (moldm-3)]

H+ = 6.04x10-5 moldm-3

pH = -log(H+)

pH = 4.2189

pH = 4.22


That took time to do, hopefully you get it.
Reply 2497
Avatar for Namod
Namod
10
Original post by eggfriedrice
The ones I would say DEFINITELY learn;
Crr 6+ orange
Cr 3+ green
Mn 2+ pale pink
Mn 7+ purple
Fe 2+ pale green
Fe 3+ pale yellow
Co 2+ pink
Cu 2+ blue


WHERE IN the specification does it say you have to know the colours of them!
And can you tell me if in any of the past papers WE had to know the colour to do the question.
Original post by otrivine
anyone up for some revision?



Describe why catalysts are used (3)


Lower activation energy, increase yeild, increase rate of reaction, some catalysts are need to make specific products.

The of life of a first order reaction is independent of ...... (1)
What is the rate of a reaction? (1)
What does the rate equation tell us about the rate determining step? (1)
Reply 2499
Avatar for HSR08
HSR08
Original post by DavidH20
At chromium electrode Cr3++3e−→CrCr^{3+} + 3e^- \rightarrow Cr
At X electrode X→X2++2e−X \rightarrow X^{2+} + 2e^-
Overall 2Cr3++3X→2Cr+3X2+2Cr^{3+} + 3X \rightarrow 2Cr + 3X^{2+}
Amount of Cr = 1.45652=0.028mol\frac{1.456}{52}=0.028mol
Ratio Cr:X = 2:3 so amount of X = 0.042 mol
Mr(X)=1.0210.042=24.3M_r(X)=\frac{1.021}{0.042}=24.3
So X is magnesium.


Hw do you know the chemical reaction at the X electrode?

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you