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Official AQA AS Chemistry Unit 1 - 23Rd May 2013

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Reply 320
Can someone pleasesend the jan 13 mark scheme please
Reply 321
Original post by addzyx
Thanks a lot so isw it a decrease of 2.5 for each lone pair


Yep, the general rule of thumb is 2 degrees or 2.5 degrees decrease for each lone pair! (Notice you won't get penalised for using either 2 degrees or 2.5 degrees in the exam, in the markschemes they allow you a range of answers)
Original post by addzyx
how does Boron deviate from the trend of first ionisation energies of period 2 from lithium to nitrogen i dont get it


It decreases, just like aluminium for period 3.. This is because of its 2p subs shell, an electron is being removed from it, which is of higher energy than 2s (or further from the nucleus) so requires less energy to remove
Original post by addzyx
how does Boron deviate from the trend of first ionisation energies of period 2 from lithium to nitrogen i dont get it


Decrease,

Electron being removed from P orbital
Which is further to the nucleus
So less electrostatic forces of attraction
So easier to remove


Hope this helps
Good luck everyone, hope the paper is decent!
Original post by beth196
'The weighted mean mass of an atom of an element compared with one-twelfth of the mass of an atom of carbon-12'

Posted from TSR Mobile



Theres a few different answers you can use to define RAM

Average mass of 1 atom of an element x 12 / mass of an atom of 12C

is what I've learnt
Reply 326
image.jpgCan someone do part 2 to this Q please
Reply 327
Anyone got any notes that explain clearly the stuff about atomic radius and ionisation energies across and down groups. + Why. Thanks!! Dont worry iv already revised this stuff so i know it but my mate needs notes on it and i cant send him my notes because there not digital.
Reply 328
Reply 329
Original post by Bixel
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In this question we're asked for the number of isomers of an alkane, Hexane.

Firstly, as it's an alkane, there is only CHAIN ISOMERS.
I began by drawing out the isomers, to which I found 4.
Then, I considered Hexane as an isomer of itself (this is confusing, but also true!)
Hence, hexane has 5 structural isomers!
BK5eyCrCIAAmQRz.jpg
Again sorry for the quality of the picture, having to upload these directly to twitter and relink it here from my phone!


Ohhh that's great! I understand it now! Thanks mate
Original post by Inamxa
Can someone pleasesend the jan 13 mark scheme please

Follow the link
https://www.dropbox.com/sh/4m2j1fye6mpcvbz/PlhQvakB_z (posted earlier in the thread)
Its the 2nd file

Original post by Manni
image.jpgCan someone do part 2 to this Q please


You worked out the number of moles needed to react. They gave you volume in cm3 that reacted. Convert volume to dm3, then divide number of moles by volume to give you concentration, remember, concentration in mol dm-3 = mol/dm3
Reply 331
is carbon and carbon particles(particulates)
are they the same thing?
Reply 332
Original post by mynameisntbobk
Follow the link
https://www.dropbox.com/sh/4m2j1fye6mpcvbz/PlhQvakB_z (posted earlier in the thread)
Its the 2nd file



You worked out the number of moles needed to react. They gave you volume in cm3 that reacted. Convert volume to dm3, then divide number of moles by volume to give you concentration, remember, concentration in mol dm-3 = mol/dm3

Still not managing to get the right answer
Original post by Manni
Still not managing to get the right answer


what did you get, and what's the right answer?
Reply 334
I get 0.0151 but the answer is 0.938
Reply 335
In the questions where it says 'the following table gives the successive ionisation energies of an element in period 3 identify this element' how do we do these? Are we supposed to memorise them or what?
Original post by PettitN
Anyone got any notes that explain clearly the stuff about atomic radius and ionisation energies across and down groups. + Why. Thanks!! Dont worry iv already revised this stuff so i know it but my mate needs notes on it and i cant send him my notes because there not digital.


basically

atomic radius across a period: decreases. increased nuclear charge meaning stronger nuclear attraction. Shielding remains the same.
(so due to the increased nuclear attraction, electrons are pulled in closer towards the nucleus, so the size decreases)

atomic radius down a group: increases. increased shielding so electrons are further away from the nucleus meaning a weaker nuclear attraction


ionisation energy down a group: decreases. Increased shielding so electrons are further away from the nucleas meaning a weaker nuclear attraction
(Which makes the outer electron more easier to remove)

ionisation energy across a period: increases. Increased nuclear charge meaning stronger nuclear attraction. Same Shielding.

note: ionisation energy across a period can decrease for certain elements.
for example, if an element enters a higher energy level, there the outer electron is further away from the nucleas as has more shielding (so is easier to remove)
so, Be with electronic structure of 1s2 2s2 and B with electron structure 1st 2s2 2p1.
ionisation energy decreases for B, as it is easier to remove. (see above explanation)

finally, ionisation energy can also decrease across a period due to electrons pairing. For example,
electronic structure of N: 1s2 2s2 2p3 and of O: 1s2 2s2 2p4
electron in the 2p orbital has paired, meaning there is repulsion


hope this helped...
Original post by Aqib7
How did you find it compared to the other papers?


I mean, it wasn't hard (altogether) but the yield question threw me off :P and I made stupid mistakes... very stupid mistakes.
Original post by hhopkin26
In the questions where it says 'the following table gives the successive ionisation energies of an element in period 3 identify this element' how do we do these? Are we supposed to memorise them or what?


look at the jumps and count the number of electrons removed before the first jump. this will tell you the group number

e.g. period 2 element with successive ionisation energies of 1203 2304 3046 6078
the jump occurs after the third electron is removed, so it is in group 3 period 2, therefore it is Boron.
Good luck everyone
:biggrin:

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