ocr a f325 revision thread

Reply 2060

eggfriedrice

16

Original post by Namod

The exam is tomorrow right? Because today we are tuesday the 11th.

Yes

Reply 2061

Namod

10

What is the shape of a 4 coordinate bond complex ion e.g CuCl2- and what is the bond angle?

Reply 2062

eggfriedrice

16

Original post by Namod

What is the shape of a 4 coordinate bond complex ion e.g CuCl2- and what is the bond angle?

You mean CuCl4-? It's tetrahedral, bond angle 109.5.

Oh gawd, auto correct changed bond to boob, luckily I noticed it before I posted. >.>

Reply 2063

MathsNerd1

17

Original post by AyshaK

My F325 exams on Wednesday lool

Posted from TSR Mobile

That's what I meant sorry, keeping getting C3 and this one mixed up :tongue:

Posted from TSR Mobile

Reply 2064

BrightStar57

can some one please post how to do the last question on the jan 13 paper i dont evnknow how to start it :confused:

Reply 2065

2013leaver

5

Original post by BrightStar57

can some one please post how to do the last question on the jan 13 paper i dont evnknow how to start it :confused:

Work out the moles of Mno4- used. then remember to multiply this figure by the factor, as a sample was taken overall ( cant remember the ratio offf the top my head)

In the passage, it states the g of the initial vanadium, so work out the moles of V.

Divide the the moles of V by the moles of Mno4- you have just worked out. The ratio will come up as 1.6:1 of V:Mno4- which is 5:3 in whole numbers

multiply the Mno4 equation by 3, and you will get 15 electrons. (Also set up a simple V equation and multiply by 5, 5Vn+ ----> 5V5+ +5xe- where x is a value to be solved)

The electrons should be equal for them to be balanced. so (Mno4)15=5x (v)
x=3 so vanadium losed 3 electron. So it goes from +2 to + 5

sorry its a bit long winded

Reply 2066

otrivine

14

Original post by 2013leaver

Work out the moles of Mno4- used. then remember to multiply this figure by the factor, as a sample was taken overall ( cant remember the ratio offf the top my head)

In the passage, it states the g of the initial vanadium, so work out the moles of V.

Divide the the moles of V by the moles of Mno4- you have just worked out. The ratio will come up as 1.6:1 of V:Mno4- which is 5:3 in whole numbers

multiply the Mno4 equation by 3, and you will get 15 electrons. (Also set up a simple V equation and multiply by 5, 5Vn+ ----> 5V5+ +5xe- where x is a value to be solved)

The electrons should be equal for them to be balanced. so (Mno4)15=5x (v)
x=3 so vanadium losed 3 electron. So it goes from +2 to + 5

sorry its a bit long winded

Hi

How can you use oxidation numbers to balance complex equations? thanks

so if you knew which is oxidised and which is reduced, how can we balance a equation

Reply 2067

Better

8

92% on Jan 13.

Come at me Chem 5. Time to start pumping myself. Have got over 86% on every paper I did, and over 90% in 4 of them.

This exam is now just purely psychological.

Come at me bro.

If anyone could Explain

•

E=mcT

•

and Initial rates that would help though.

(edited 10 years ago)

Reply 2068

otrivine

14

Original post by Better

92% on Jan 13.

Come at me Chem 5. Time to start pumping myself. Have got over 86% on every paper I did, and over 90% in 4 of them.

This exam is now just purely psychological.

Come at me bro.

If anyone could Explain

•

E=mcT

•

and Initial rates that would help though.

Well done

Reply 2069

RH1994

Original post by Pride

sorry I didn't reply, I just did the question and left to revise something else without checking.

Had a look at the mark scheme and I've got the same answer. I'll try to explain it.

right so you calculate the no. of moles of sodium thio. Then you know that half that number of moles of I2 reacted right? Now all of that I2 was produced from the previous reaction. So double the number of moles of Mn(OH)3 must have been reacted.

Again all of that Mn(OH)3 was produced in the previous reaction, so a quarter of that number of moles of O2 must have been reacted.
Then that's m = n x Mr, and x1000 to convert to mg.

Hi can someone please help me with this question. I'm trying to understand why you don't multiply number of moles of 2Mn(OH)3 by 2 to get to 4Mn(OH)3 BEFORE you divide number of moles by 4 to get moles of O2?
Thanks

Reply 2070

lukenoel123

the molar ratios of the same molecules do not matter - you cant have a certain amount of a molecule and then suddenly have more, molar ratios only occur between different molecules

Reply 2071

ankyloser

9

Original post by Pride

Well I wouldn't worry about that, although the maths can be difficult in chem, understanding log laws isn't required at a-level.

I'm amazed that's only 1 mark, maybe there's a much simpler way of doing it that I missed, or maybe there's a formula for it in that syllabus :dontknow:

I worked out how to do it, using kw again but I got the right answer this time. I dunno how I didn't get this answer first time, must have forgotten to log it or something.
So you're finding pH, [H+]= kw/[OH-]
[OH-]=100x[H+], kw=1.00x10^-14
so [H+]=1.00x10^-14/100[H+]
[H+]^2= 1.00x10^-14/100
[H+]= square root of 1.00x10^-14/100
Then you minus log the [H+], -log10(1x10-8
which gives 8 :biggrin:

Reply 2072

username938518

1

Original post by RH1994

Hi can someone please help me with this question. I'm trying to understand why you don't multiply number of moles of 2Mn(OH)3 by 2 to get to 4Mn(OH)3 BEFORE you divide number of moles by 4 to get moles of O2?
Thanks

Right I did this question yesterday :smile:

So you calculated the number of mole of S2O3 2- used.
Then you divide that by 2 to get the number of moles of I2 formed.
You then multiply this by 2 again to get the number of moles of Mn(OH)3 formed
You divide this number by 4 to get the number of 02 moles

The reason you divide by 4 straight away is that the number of moles of Mn(OH)3 that you calculated is the number of moles formed in the first reaction. It is 4 times more moles than there is O2, but only 2 times more moles than there was I2. The number of moles of Mn(OH)3 is the same in reaction 1 and reaction 2. The 4 and 2 are the ratios of it? Does that help you?

Reply 2073

BrightStar57

Original post by 2013leaver

Work out the moles of Mno4- used. then remember to multiply this figure by the factor, as a sample was taken overall ( cant remember the ratio offf the top my head)

In the passage, it states the g of the initial vanadium, so work out the moles of V.

Divide the the moles of V by the moles of Mno4- you have just worked out. The ratio will come up as 1.6:1 of V:Mno4- which is 5:3 in whole numbers

multiply the Mno4 equation by 3, and you will get 15 electrons. (Also set up a simple V equation and multiply by 5, 5Vn+ ----> 5V5+ +5xe- where x is a value to be solved)

The electrons should be equal for them to be balanced. so (Mno4)15=5x (v)
x=3 so vanadium losed 3 electron. So it goes from +2 to + 5

sorry its a bit long winded

from the markscheme we have to divide the moles of V by five how do you know that you have to divide by 5

Reply 2074

username878045

17

Original post by Better

92% on Jan 13.

Come at me Chem 5. Time to start pumping myself. Have got over 86% on every paper I did, and over 90% in 4 of them.

This exam is now just purely psychological.

Come at me bro.

If anyone could Explain

•

E=mcT

•

and Initial rates that would help though.

What do you want to know about them?

Basically, E=mcT is based around how much energy you need to heat a substance. The energy needed to raise by 1K is the same at all temperatures, and is equal to the constant c. Given that most reactions at A-Level occur in heavily diluted solutions, we tend to only focus on the energy needed to heat the water.

Initial rates can be calculated in a number of ways. Either a concentration/time graph can be drawn, and a tangent drawn at t=0. The gradient of the line is the rate, because gradient = conc/time = rate. In 'clock reactions', the initial concentration is approximated by using 1/t, where t is the time for the reaction to complete. This isn't exact, it's just an approximation.

Reply 2075

username938518

1

Can anyone tell me where you've got the Jan 2013 paper from? I did it back in march for my mock but I think I could do much better on it now. If it's the hardest paper I want to see what I might be up against tomorrow haha.

Reply 2076

ofudge

For a half cell, why when an electrode loses electrons to the half cell does that electrode become negative? surely it would be positive if its lost electrons?

Reply 2077

Brad0440

Original post by ofudge

For a half cell, why when an electrode loses electrons to the half cell does that electrode become negative? surely it would be positive if its lost electrons?

If the electrode is producing electrons, then it is counted as the negative electrode as it is supplying electrons to the more positive electrode. Not sure if I explained this very well. :s-smilie:

Reply 2078

chignesh10

15

Original post by Better

92% on Jan 13.

Come at me Chem 5. Time to start pumping myself. Have got over 86% on every paper I did, and over 90% in 4 of them.

This exam is now just purely psychological.

Come at me bro.

If anyone could Explain

•

E=mcT

•

and Initial rates that would help though.

was that your first attempt at JAN 2013?