Edexcel Chemistry A2 Unit 5 ~ Wednesday 19th June 2013 (Now Closed) Watch

Poll: How pumped up are you for this exam?-(warning)-(bad jokes arene this poll!)
"Titanium-I'm not going to corrode (even at high temperatures)" (A*) (22)
16.67%
"Benzene's my middle name, give me the paper in a week and I'll ace it!" (A) (27)
20.45%
"Yeah, I'm fairly electrophillic (positively charged) about the exam" (B) (27)
20.45%
"I'm in the middle of the salt bridge, but I will pass-eventually" (C) (21)
15.91%
"I'm feeling rather electroNegative about this exam" (D) (18)
13.64%
"Benzene, what's that?" (E) (6)
4.55%
"Chemistry, what's that?" (F) (11)
8.33%
This discussion is closed.
F1's Finest
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#1941
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#1941
(Original post by AT95)
Can some one please tell me how to do jan 12 q21bii? On what basis do we select the metal?
Step 1 shows Cr3+ going to Cr2+

So that should be one half equation.

Use your data booklet to find a half equation involving Cr3+ and Cr2+.

The second half equation should involve a metal, by which when you combine the two half equations, you get a positive E cell, hence making it a feasible reaction.
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AT95
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#1942
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#1942
So u mean we can take any value that is more negative than the value for Cr3+??

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jojo1995
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#1943
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#1943
(Original post by SuziieB)
Am I the only one who genuinely prefers Unit 4 over Unit 5? I can't stand this unit.

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im the sAme
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F1's Finest
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#1944
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#1944
(Original post by AT95)
So u mean we can take any value that is more negative than the value for Cr3+??

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Yeah but it has to be a metal. Read the question carefully.
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AT95
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#1945
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#1945
(Original post by James A)
Step 1 shows Cr3+ going to Cr2+

So that should be one half equation.

Use your data booklet to find a half equation involving Cr3+ and Cr2+.

The second half equation should involve a metal, by which when you combine the two half equations, you get a positive E cell, hence making it a feasible reaction.
Oh, but the MS says we can take Zn as well but the Ecell for that would be -0.02 then. God this isnt making sense :facepalm:

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F1's Finest
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#1946
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#1946
(Original post by AT95)
Oh, but the MS says we can take Zn as well but the Ecell for that would be -0.02 then. God this isnt making sense :facepalm:

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No, you've done the wrong working out.

The ms clearly says next to Zn '0.35', so that's the value you should have obtained when combining the two half equations.

Remember the metal (in it's solid state) and Cr3+ has to be on the left hand side of the overall equation when you combine then.
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posthumus
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#1947
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#1947
Ah I over slept

This exam is going to brutal.... still needa learn the separating techniques & complex reactions/colours ! -______-
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AT95
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#1948
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#1948
(Original post by James A)
No, you've done the wrong working out.

The ms clearly says next to Zn '0.35', so that's the value you should have obtained when combining the two half equations.

Remember the metal (in it's solid state) and Cr3+ has to be on the left hand side of the overall equation when you combine then.
Can you PLEASEEEE show me the working?:dontknow:

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AtomicMan
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#1949
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#1949
If a burette reads to the nearest 0.05cm3, you get a result of say 10.00cm3. So the titire could be either 10.025cm3 or 9.975cm3. Is the uncertainty +-0.025cm3 and error +-0.05cm3, or are the error and uncertainties the same? Or the other way around? Really confused :s
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jojo1995
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#1950
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#1950
(Original post by AtomicMan)
If a burette reads to the nearest 0.05cm3, you get a result of say 10.00cm3. So the titire could be either 10.025cm3 or 9.975cm3. Is the uncertainty +-0.025cm3 and error +-0.05cm3, or are the error and uncertainties the same? Or the other way around? Really confused :s
I thought they were both the same. I agree with the 10.025 and 9.975 so error/ unvertainty is -or+ o.25 so is o.25x2/10.00 x 100
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posthumus
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#1951
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#1951
I've only ever seen reagents for nitrobenzene to phenylamine being HCl & HNO3 conc.

But now I just saw you have to add NaOH to seperate the salt ? C6H5N3Cl ???!!!
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study beats
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#1952
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#1952
Lol those jokes were funny
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Gnome :)
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#1953
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#1953
(Original post by posthumus)
I've only ever seen reagents for nitrobenzene to phenylamine being HCl & HNO3 conc.

But now I just saw you have to add NaOH to seperate the salt ? C6H5N3Cl ???!!!
Yep- it's the same for the preparation of aliphatic amines too. The ammonium salt always forms first, so you use NaOH to liberate the amine. So it's C6H5NO2 + conc HCl (+ Sn catalyst) --> C6H5NH3+Cl- , then adding NaOH forms C6H5NH2 + H2O + Cl-
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«WD»ddd
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#1954
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#1954
(Original post by SuziieB)
Am I the only one who genuinely prefers Unit 4 over Unit 5? I can't stand this unit.

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For a very good reason i feel much more comfortable with unit 4 than with unit 5
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randyaloul
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#1955
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#1955
Good luck to all! I'd stop revising if I were you, don't want to over-cram your brain before the exam directly.
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ChooChooCherry
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#1956
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#1956
good morning all! wishing everyone the best of luck, including myself! *cries*
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F1's Finest
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#1957
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#1957
Best of luck brahs
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Gnome :)
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#1958
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#1958
Good luck all- I'm off to college now, gonna give me some time to try and cram all the complex ion colours D: A few hours and it'll all be over, and unless you did something silly like apply for chem at uni (WHY?!?!?) then you'll never have to do any of this ever again
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AS01
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#1959
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#1959
Good luck everyone! Hopefully everything goes well!
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#1960
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#1960
Hi everyone, the exam went really well.
Firstly, for the multiple choice question which asked you to select which reaction produced a polyamide, did you put answer b or answer d? It was either one of them.

The second multiple choice which said about a hydrocarbon was burnt and they gave you values. Did you put down C3H8? I think that was answer b. I took a guess on it.

Okay and um the electrode potentials at the start of section B, because one of the questions told you to select two half cells and combine them, which ones out of the four did you choose? Bearing in mind the question said it has to be in alkaline. I used the two half cells which had OH- somewhere in them, so I used 2 and 4 and combined them.

For section C on the titration, did anyone get 1.8moldm-3 for the Titanium (IV) Chloride concentration?

Also for the next question which was the last one, I said some of the Titanium Chloride is readily oxidized by oxygen in the air, hence less TiCl4 is available in the titration.
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