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# 8ch01 edexcel chemistry 2016

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Why bother with a post grad course - waste of time? 17-10-2016
1. How did everyone find it! Feel free to discuss answers, etc!
2. What did you guys get for the calculation of the concentration! I completely lost track there! I got 4 g dm-3 (it was a decimal like 4.02 then I rounded it to 4)
3. I found it to be really difficult.. The six markers killed me
4. (Original post by Florent venhari)
What did you guys get for the calculation of the concentration! I completely lost track there! I got 4 g dm-3 (it was a decimal like 4.02 then I rounded it to 4)
I got 100.53 or something like that
5. (Original post by Shk)
I got 100.53 or something like that
has to be correct, I can wave that A good bye :P
6. (Original post by Shk)
I got 100.53 or something like that
I got that!!! ))))
7. (Original post by Cryingbcmaths)
I got that!!! ))))
How did you do it, can u explain your steps if u remember please?
8. how did u guys work out the relative atomic masses
9. (Original post by yeah1106)
how did u guys work out the relative atomic masses
I did 5 - 0.409 (the mass was something like that), which gave u the mass for the isotope of Li-7.

You then did: (0.409 x 6) + ( the mass u found after subtracting from 5 , x 7) all divided by 5,

and I got like 6.9 something, cant really remember.
10. Someone get me some bleach! So annoyed with that calculation question
11. (Original post by Florent venhari)
I did 5 - 0.409 (the mass was something like that), which gave u the mass for the isotope of Li-7.

You then did: (0.409 x 6) + ( the mass u found after subtracting from 5 , x 7) all divided by 5,

and I got like 6.9 something, cant really remember.
same thx god i was stuck at that question
12. (Original post by Florent venhari)
How did you do it, can u explain your steps if u remember please?
Ok so i think the alkai was NaOH, i worked out th moles of NaOH as they gave the conc and the volume in the table. No of moles = conc x volume
Since the nitric acid and the alkali had a one to one ratio there was equal moles of acid. This was 25cm^3
The original concentrated acid was 250cm3 so i multiplied the moles by 10 to find the moles in 250cm^3.
I then divided by the volume of acid added intitally to get the inital acid concentration in mol dm^-3.
I multiplied this value by the mr of nitric acid to get the conc in g mol^-1
13. Seemed like everyone found it difficult. I'm in Yr 13, but I know someone who took up AS chemistry this year to try to boost her application for Oxford.

The new spec seems so much more difficult than the old one it seems.
14. (Original post by Florent venhari)
I did 5 - 0.409 (the mass was something like that), which gave u the mass for the isotope of Li-7.

You then did: (0.409 x 6) + ( the mass u found after subtracting from 5 , x 7) all divided by 5,

and I got like 6.9 something, cant really remember.
Oh i thought you had to find thepercentage of the isotope of the total mass
As in (0.409/5) x 100 to get the percentage, do the same for the other isotope and then times each isotopic mass by its percentage ove 100
15. (Original post by Shk)
Ok so i think the alkai was NaOH, i worked out th moles of NaOH as they gave the conc and the volume in the table. No of moles = conc x volume
Since the nitric acid and the alkali had a one to one ratio there was equal moles of acid. This was 25cm^3
The original concentrated acid was 250cm3 so i multiplied the moles by 10 to find the moles in 250cm^3.
I then divided by the volume of acid added intitally to get the inital acid concentration in mol dm^-3.
I multiplied this value by the mr of nitric acid to get the conc in g mol^-1
Okay so this is what I did:

moles of NaOH = 19.95 x 0.08 / 1000 = 0.001596
moles of HNO3 in 25 cm3 = same as NaOH = 0.001596
moles of HNO3 in 250 cm3 = 0.001596x10 = 0.01596
concentration of HNO3 = (0.01596 x 1000 / 250 ) = 0.06384 mol dm-3
concentration of HNO3 in g dm-3 = 0.06384 x 63 = 4.02192 g dm-3 ????
16. (Original post by Florent venhari)
Okay so this is what I did:

moles of NaOH = 19.95 x 0.08 / 1000 = 0.001596
moles of HNO3 in 25 cm3 = same as NaOH = 0.001596
moles of HNO3 in 250 cm3 = 0.001596x10 = 0.01596
concentration of HNO3 = (0.01596 x 1000 / 250 ) = 0.06384 mol dm-3
concentration of HNO3 in g dm-3 = 0.06384 x 63 = 4.02192 g dm-3 ????
Ok so i dont think you were supposed to divideby 250 as this the volume after the water has been added, i think it said at the starof the question100cm3 of acid before topping up with water
Tbh though everything youdid was right apart from that so i reckon you would havlost 1 mark, two max
17. (Original post by Shk)
Ok so i dont think you were supposed to divideby 250 as this the volume after the water has been added, i think it said at the starof the question100cm3 of acid before topping up with water
Tbh though everything youdid was right apart from that so i reckon you would havlost 1 mark, two max
Ohhh, okay thanks
18. If anyone wants to make a mark scheme for the short questions, feel free to! I did one for question one based on my answers! Correct me if I'm wrong.

Mark scheme:1.
(a) Chlorine is a gas and Iodine is a solid
(b) Astatine would be a solid because it is less volatile than Iodine because the London forces strengthen due to there being more electrons down the group
(c) Cl2 + 2NaOH -> NaCl + NaClO + H2O It's disproportionation
(d) white smoke
(e) Electronegativity is the ability of an atom to withdraw electron density within a covalent bond. ClF3 is a non symmetrical shape, so the bond polarities do not cancel out, therefore it is a polar molecule
(f) 1.81x10^23
19. but wasn't (e) referring to the bond polarity? or is that the same as the molecule being polar?
20. (Original post by nmaster)
but wasn't (e) referring to the bond polarity? or is that the same as the molecule being polar?
The electrons in the ClF3 would be more attracted to the Fluorine atoms, so there's polarity. If you draw out ClF3, you have a seesaw type of shape with 2 lone pairs, therefore it isn't symmetrical and because fluorine is more electronegative than Chlorine, the electrons would be more attracted to the F atoms (withdrawn from Chlorine in the covalent bond, hence the definition of ELECTRONEGATIVITY)

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