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A-level Chemistry Revision Squad!

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This is a June 2012 paper. The answer is B not C. :smile:

Oh yeah it is sorry. It's no wonder I was so confused lol. I was stuck on that answer for ages but I was checking the wrong answer all along hahahaa. Thank you so much.

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Reply 621

Farmerjj

Original post by TeaAndTextbooks

Oh yeah it is sorry. It's no wonder I was so confused lol. I was stuck on that answer for ages but I was checking the wrong answer all along hahahaa. Thank you so much.

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Would someone be able to name this for me: CH3(CH2)2CH2CH2OH? It's just that the brackets confuse me.
Thanks

Reply 622

dpoojaraa

Original post by Farmerjj

Would someone be able to name this for me: CH3(CH2)2CH2CH2OH? It's just that the brackets confuse me.
Thanks

Try drawing it out so you can see what it looks like

Reply 623

Sandy_Vega30

Original post by Farmerjj

Would someone be able to name this for me: CH3(CH2)2CH2CH2OH? It's just that the brackets confuse me.
Thanks

Anything in the brackets belongs to the carbon in the left. But in your question, the carbon in the left already has 3 hydrogens. I think you made a mistake somewhere, could you recheck? I can help then. :smile:

Reply 624

Farmerjj

Original post by Sandy_Vega30

I'm not sure, that's what it said in the book.

Reply 625

CERC

Original post by Farmerjj

Hi,
On mass spectrometry how do do you know which one the Molecular ion is? For example on this one the book says the Mr would be 86, but I would of said 85 because that's the large peak? Or is it always the peak that is the most to the right hand side?
Cheers

The tallest peak is called the base peak and its the most STABLE fragment.
Yes, the molecular ion peak is always the one furtherest to the right. This represents the UN-FRAGMENTED molecule minus one electron.

However, be careful as there could be isotopes present and this could give you an M+1 peak. However, you will know if this is the Molecular ion peak or not as the the molecular ion peak should be greater than the M+1 peak (isotope peak)

Hope this helped, ask me to clarify more if it doesn't!

:smile:

Reply 626

CERC

Original post by Marco1000

For ligand exchange, how do I know how many ligands get exchanged. For example, if I had Fe(H^2O)6 reacting with excess OH-, is there a way to find out if only 1 ligand is exchanged, maybe 2 or 3 etc?

It really depends if the substance you are adding is in excess or not.

If only a few drops of OH ions are added then some of the H20 ligands will be replaced, if however, the OH ions are in excess, then all the ligands will be replaced.

Eg. In [Fe(H20)6] the charge is +2. Therefore you add 2OH, and you end up with [Fe(H20)4(OH)2] which is a precipitate and 2H20 is also produced. (You add the same amount of OH ions as the charge of the ion and you make this number of water molecules also)

As the Iron precipitate does not dissolve in excess OH ions, then this precipitate will remain even in excess.

However, the likes of Cobalt produces a blue precipitate in OH ions and dissolves in excess giving a yellow solution. The equations look like this:

[Co(H20)6] +2 + 2OH ------------- [Co(h20)4(oh)2] (solid) + 2H20

in excess the equation is:

[Co(H20)4(oh)2] + 4OH -------the-----[Co(OH)6] -4 charge + 4H20

The overall charge is -4 as Cobalt has a charge of +2 and we added 6OH- ions and so +2-6 = -4!

Hope i helped!

Reply 627

Sandy_Vega30

Original post by Farmerjj

I'm not sure, that's what it said in the book.

Which paper is that?

Reply 628

PlayerBB

Original post by CERC

Hey, can you please explain a little further on the M+1peak, like for Cl2, there is Cl(35) and Cl(37) so which one would be for the M+1 peak ?

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Reply 629

shiney101

Can someone please help me with this question?

The concentration of iron(II) ions in aqueous solution can be determined by titrating the solution, after acidification, with a standard solution of potassium manganate(VII).

(i) Explain, by reference to the data given in the table above, why hydrochloric acid should not be used to acidify the solution containing iron(II) ions.

(ii) Explain, by reference to the data given in the table above, why nitric acid should not be used to acidify the solution containing iron(II) ions.

So the answers are:
i) MnO4 -/Mn2+ has a more positive Eο value than Cl2/Cl– and will oxidise Cl– or change Cl– to Cl2

Allow converse answers

(ii) NO3-/HNO3 has a more positive Eο value than Fe3+/Fe2+ and will oxidise Fe2+ or changeFe2+ to Fe3+

BUT what I'm confused on is can't the answers be the other way round too since the electrode potential for Cl2/cl- is greater than that of Fe3+/Fe2+ so could have the same answer as part ii, and similarly Mno4-/mn2+ has a great a potential than No3-/HNO3 so could have the same answer to part i?

If anyone can help me out i'd greatly appreciate it, so confused about this right now

Reply 630

Marco1000

Original post by CERC

Thanks a lot that was very helpful. So do I just need to learn the individual complex ions that will cause the ppt to dissolve with excess OH-? If the overall charge or the complex ion is neutral is that when I know theres a ppt? Do I need to know about the NH3 ligand and when it causes things to dissolve?
thanks

Reply 631

n2697

Hey guys i was wondering if someone could help me on this quick question:
Q11
https://110d0b1a7b584219c71ba4cfe896ad1711fa3ef7.googledrive.com/host/0B1ZiqBksUHNYR00tYjdZN052azQ/June%202014%20QP%20-%20Unit%201%20Edexcel%20Chemistry.pdf

And this one (they're pretty much similar i just don't get how to to do it - i know there must be a simple way of doing it)
Q11
https://110d0b1a7b584219c71ba4cfe896ad1711fa3ef7.googledrive.com/host/0B1ZiqBksUHNYR00tYjdZN052azQ/January%202014%20(IAL)%20QP%20-%20Unit%201%20Edexcel%20Chemistry.pdf

Reply 632

youreanutter

https://110d0b1a7b584219c71ba4cfe896ad1711fa3ef7.googledrive.com/host/0B1ZiqBksUHNYR00tYjdZN052azQ/June%202013%20QP%20-%20Unit%201%20Edexcel%20Chemistry.pdf

can someone help with enthalpy change 21ii and percentage uncertainty 21 c

Reply 633

ayvaak

I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.

Reply 634

username986184

Original post by ayvaak

I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.

Look at this data and look for the big jumps. C is fairly 'flat.'

https://en.wikipedia.org/wiki/Molar_ionization_energies_of_the_elements

Elements are Na Ca Mn Ne

Tricky question though.

Reply 635

man i

Original post by youreanutter

for 21ii) you get the value you worked out above which should be 1149.5 now remember this is the heat of the SURROUNDINGS so if energy has been released in the atmosphere so its + then from the SOLLUTION energy has been lost so its negative

now remember standard enthalpy states that the enthaply change is the saem whatever route is taken so the for the solution the value will be -1149.5/0.010 so its -114950J but it needs to be in Kj so dived by 1000 and you get -114.95Kj and so 3 significant will be -115kJmol-1

Reply 636

ayvaak

I am currently stuck on question 5 of the photo. I was hoping someone could help. The answers C but i cant work out why. Thanks in advance.

Reply 637

Junaidc122

Original post by n2697

10% reacted, so 30cm^3 had been converted in to ozone, leaving 270cm^3 of oxygen. Since there is a 3:2 molar ratio, the volume of ozone produced from 30cm^3 oxygen = 30 x (2/3) = 20
Therefore, the total volume of gas left in the mixture is now 270 + 20 = 290cm^3.

With the second question, 100cm^3 decomposed, so 50cm^3 of nitrogen and 150cm^3 must have been formed, respectively. (Due to the molar ratios). This leaves you with 400 + 150 + 50 =600cm^3.
Hope this made sense, and just remember, the ratio of the volumes in which gases react is the same as how they react in terms of moles!!!

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(edited 7 years ago)

Reply 638

CERC

Original post by Marco1000

Yes, well depending on your exam board.

I do CCEA, so you need to know what elements dissolve in excess and what don't.
You also need to know ligands with ammonia as well and colours of precipitates and their formula of the ion.

And yes, if the overall charge is neutral then it is likely it will be a precipitate! :smile:

Hope i helped and if you have any more questions, im happy to help, if i can! :smile:

Reply 639

CERC

Original post by PlayerBB

Hey, can you please explain a little further on the M+1peak, like for Cl2, there is Cl(35) and Cl(37) so which one would be for the M+1 peak ?

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Ok, so i'm just going to type out what my revision guide says about this:

1) The M+1 peak is a small peak one m/e value above the M+ peak. It is caused by the presence of ONE Carbon-13 atom in the molecular ion.

2) A MAJOR peak at M+2 is caused by the presence of either Cl or Br atom in the molecule. Both Chlorine and Bromine have isotopes that differ by two mass units. Chlorine has 35 and 37 whilst Bromine has 79 and 81.

IMPORTANT: If the ration of the M+ peak to the M+2 peak is 3:1 i.e. the M+ peak is three times as abundant as the M+2 peak, then a chlorine atom is present.

However, the Bromine 79 and Bromine 81 have the same abundance, in a ration of 1:1. So, if the M+ peak has the same abundance as the M+2 peak then a Bromine atom is present in the molecule!

Really hope this helps you! :smile: