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A2 chemistry
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shengoc
I am doing a degree. I quoted my previous statement from an inorganic text.
Ohh, i see. Good luck with that mate, i am pretty sure you're gonna ace it and if you get time please take a look at my next question
boromir9111
http://www.thepaperbank.co.uk/papers/AS-A%20Levels/Chemistry/Trends%20and%20Patterns/2815-01_2005_jan_w.pdf
question 2 e (i) - would the answer just be that it would reduce the iron(III) ions to iron(II) ions and the titration wouldn't work????
question 2 e (i) - would the answer just be that it would reduce the iron(III) ions to iron(II) ions and the titration wouldn't work????
otherwise, you'd have competing reactions between
1) Mn7+ oxidising your Fe2+ to Fe3+
2) The Cu2+ reducing your Fe3+ formed back to Fe2+
This would mean that your average titre of Mn7+ won't be accurate.
It is just as simple as that.
shengoc
otherwise, you'd have competing reactions between
1) Mn7+ oxidising your Fe2+ to Fe3+
2) The Cu2+ reducing your Fe3+ formed back to Fe2+
This would mean that your average titre of Mn7+ won't be accurate.
It is just as simple as that.
1) Mn7+ oxidising your Fe2+ to Fe3+
2) The Cu2+ reducing your Fe3+ formed back to Fe2+
This would mean that your average titre of Mn7+ won't be accurate.
It is just as simple as that.
Ahh of course, it's the same as i as was saying but you summed it up right. I can't find the mark scheme for this but what i said would it be right?
http://www.thepaperbank.co.uk/papers/AS-A%20Levels/Chemistry/Trends%20and%20Patterns/2815-01_2004_jun_w.pdf
question 2 d (ii) ....... tbh i got no idea there???? i calculated the empirical formula to be, KFe4C6N6 but not sure what to do next.
question 2 d (ii) ....... tbh i got no idea there???? i calculated the empirical formula to be, KFe4C6N6 but not sure what to do next.
anyone?
boromir9111
anyone?
Your EF doesn't look right to me...
K 547/39 = 14.02 = 4
Fe 195/56 = 3.48 = 1
C 252/12 = 21 = 6
N 294/14 = 21 = 6
So your formula is K4FeC6N6
You are told that it's a potassium salt so the potassium is just balancing cations. This leaves the iron, carbon and nitrogen
This is potassium hexacyanoferrate (II)
The octahedral complex ion is [Fe(CN)6]4-
charco
Your EF doesn't look right to me...
K 547/39 = 14.02 = 4
Fe 195/56 = 3.48 = 1
C 252/12 = 21 = 6
N 294/14 = 21 = 6
So your formula is K4FeC6N6
You are told that it's a potassium salt so the potassium is just balancing cations. This leaves the iron, carbon and nitrogen
This is potassium hexacyanoferrate (II)
The octahedral complex ion is [Fe(CN)6]4-
K 547/39 = 14.02 = 4
Fe 195/56 = 3.48 = 1
C 252/12 = 21 = 6
N 294/14 = 21 = 6
So your formula is K4FeC6N6
You are told that it's a potassium salt so the potassium is just balancing cations. This leaves the iron, carbon and nitrogen
This is potassium hexacyanoferrate (II)
The octahedral complex ion is [Fe(CN)6]4-
Do you mind explaining the part where you mentioned about where we can leave out potassium again please.
Also from there would that mean since K is +1 and now that it's removed because it has a total charge of +4 and therefore the overall charge becomes -4 for the complex ion?
Thanks in advance.
shengoc
I would say you have 4 K+, 1 Fe(II) and 6 x CN-, that balances the charge nicely. The potassium is left out as it is just a counter ion to balance the charge in the compound. Complexes are formed with the transition metal as they coordinate to ligand better.
Okay, thanks for your reply mate
..... yeah so it's (CN-)*6 and 2+ for Fe, therefore K is 4+...... but we leave out it because of course if we had that the shape would be more than the octrahedral needed. So since we have left out K, the overall would have to be 4-?boromir9111
Okay, thanks for your reply mate ..... yeah so it's (CN-)*6 and 2+ for Fe, therefore K is 4+...... but we leave out it because of course if we had that the shape would be more than the octrahedral needed. So since we have left out K, the overall would have to be 4-?
..... yeah so it's (CN-)*6 and 2+ for Fe, therefore K is 4+...... but we leave out it because of course if we had that the shape would be more than the octrahedral needed. So since we have left out K, the overall would have to be 4-?The ions are separate species and should never be included together, unless you wish to show a part of the whole ionic lattice.
For example sodium chloride
Although it has the formula NaCl this just represents the simplest formula unit of a giant lattice. In reality the Na+ ions and the Cl- ions are separate species with separate identities especially in solution.
The compound in this question is a salt formed by some positive ions (potassium) and a negative ion, which is complex.
The crystal lattice then has a ratio of four potassium ions to each hexacyanoferrate (II) ion.
charco
The ions are separate species and should never be included together, unless you wish to show a part of the whole ionic lattice.
For example sodium chloride
Although it has the formula NaCl this just represents the simplest formula unit of a giant lattice. In reality the Na+ ions and the Cl- ions are separate species with separate identities especially in solution.
The compound in this question is a salt formed by some positive ions (potassium) and a negative ion, which is complex.
The crystal lattice then has a ratio of four potassium ions to each hexacyanoferrate (II) ion.
For example sodium chloride
Although it has the formula NaCl this just represents the simplest formula unit of a giant lattice. In reality the Na+ ions and the Cl- ions are separate species with separate identities especially in solution.
The compound in this question is a salt formed by some positive ions (potassium) and a negative ion, which is complex.
The crystal lattice then has a ratio of four potassium ions to each hexacyanoferrate (II) ion.
Maybe i am going about this the wrong way.... i think my question is why is the oxidation state in 4-? i think that would be a better place to start. Sorry about that mate.
lol, i just figured out now lol, K is 4+ so the other Iron complex has to be 4-. Thanks for your help mate!!!
"state two functions of the phospholipid bilayers that form cell membranes" For this would the answer just be that it controls the chemicals going in and out????
Seperates cytoplasm from external environment and controls entry/exit of substances...chemistry?!
nope - part of my chem module (biochemistry)... this is for two marks. So would the answer just be it controls what comes in and goes out?
http://www.thestudentroom.co.uk/showthread.php?t=1088479&page=2
How do we know that a brown solution is being formed?
How do we know that a brown solution is being formed?
boromir9111
http://www.thestudentroom.co.uk/showthread.php?t=1088479&page=2
How do we know that a brown solution is being formed?
How do we know that a brown solution is being formed?
Read through the thread...
It says quite specifically:
KI is a reducing agent because the iodide ion can get oxidised to iodine. It can lose electrons and give them to something else.
Sodium is never going to displace anything except in direct reactions between solid sodium and solid metal ores (very unlikely)
Sodium ferrate(VI) is an oxidising agent as the iron VI can turn to iron (III) or iron (II)
So let's just suppose you are discussing the reaction between KI and sodium ferrate(VI) Iin acidic solution ??
equation 1: 2I- --> I2 + 2e
equation 2: FeO42- + 8H+ + 3e--> Fe3+ + 4H2O
MTB 3 in equation 1 and MTB 2 in equation 2 and add together >>
6I- + 2FeO42- + 16H+ --> 3I2 + 2Fe3+ + 8H2O
you will observe an orange/brown colour appearing as iodine is formed...
Sodium is never going to displace anything except in direct reactions between solid sodium and solid metal ores (very unlikely)
Sodium ferrate(VI) is an oxidising agent as the iron VI can turn to iron (III) or iron (II)
So let's just suppose you are discussing the reaction between KI and sodium ferrate(VI) Iin acidic solution ??
equation 1: 2I- --> I2 + 2e
equation 2: FeO42- + 8H+ + 3e--> Fe3+ + 4H2O
MTB 3 in equation 1 and MTB 2 in equation 2 and add together >>
6I- + 2FeO42- + 16H+ --> 3I2 + 2Fe3+ + 8H2O
you will observe an orange/brown colour appearing as iodine is formed...
charco
Read through the thread...
It says quite specifically:
It says quite specifically:
But no where does it talk about the colour? that is what's getting me, where is this colour coming from? you understand where i'm coming from mate?
I know iodine is brown now but i don't see that anywhere in my book about the colour i don't know why they expect you to know that then.
boromir9111
I know iodine is brown now but i don't see that anywhere in my book about the colour i don't know why they expect you to know that then.
A question like this can come up,
1) Why is no indicator used for the reaction?
or
2) How would you determine the end point of titration/reaction?
Then obviously, knowing that there is a colour change is going to answer those questions!
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