Aqa gcse papers 2015?

Original post by KryptoModz

Hi guys!

By any chance does anyone have the official AQA chemistry C2 paper + mark scheme! If you do please can you give a link.

Thanks in advanced!

im not exactly sure but is this the one
FIRST 2 PAGES 2Cl- ---> Cl2 + 2e-Oxidation state of Mn = +7Mn04- + 5e- +8H+ ---> Mn2+ + 4H20Cl2 + 2Br- ----> 2Cl- + Br2Solution turns orange/brownCl2 + 2H20 ----> 4HCl + O2oxidation state of chlorine containing compound = -1Chlorine radical = CF3 with radical . on the Carbon NOT the Fluorine(s) Why Cl lower b.p. = Cl is a smaller molecule than Br (1) or has smaller atomic radius and therefore has a smaller surface area for vdws meaning weaker van der waals BETWEEN MOLECULES (1) 3RD PAGEReasons for atomic radius = As you go down the group size of the atoms increases as extra shells of electrons are being added, meaning more shielding from outer s electrons to the nuclear charge, meaning electrons are situated less closely to the nucleus Trend in reactivity = Increases down the groupSr + 2H20 ---> Sr(OH)2 + H2Most soluble = Ba(OH)24th PAGE = REAGENTS (alcohols)Acidified K2Cr207 --> color change orange to green --> no visible change(Alkane and alkene) Bromine water/ Br2 ---> No visible change ----> color change orange/brown to colorless (ketone and aldehyde) --> Tollens reagent of Fehlings --> Tollens = silver mirror Feelings = red ppt ---> no visible change (AgNO3 and NaN03) Any halide ---> x precipitate (x = color) ---> no visible change (BaCl2 and something else) Concentrated H2S04 ---> White precipitate ---> no visible change (Ill edit this if i remember the other reagent you had to give)CALCULATIONS 1st = -1651 kjmol-12nd = -1110 kjmol-1 (or 1114.6) but i think 3 sig figs was desired 3rd = -1615 kjmol-1 Balancing equation 2 3 3 3 (i put a multiple of x2 by accident so I'm sure multiples will be excepted)MECHANISMS Not going to draw them out but P -Q was Electrophilic addition --> reagent = Na or K OH --> conditions to favor = conc. base, ethanol as the solvent or high temperatures Q-P was Elimination Q - Z was N.S. (REMEMBER N+H3 WAS FORMED THEN THE BREAKAGE OF THE H BOND TOWARDS THE N CATION)POLYMERS ---> Propanoic acid -- Prop-2-en-1-ol = 2HC=CH-CH2OH--> ADDITION POLYMERISATION Mr 48.1 idea was 4(12.00000) + 10(1.00797) (probs wrong number but you needed to write out full number) ANS = 3S.F = 48.1Carbon 12.00000 because it is the standard reference value/ or by definition DRAWING AND NAMING OF COMPOUNDS Butan-2-ol = secondary alcohol -- Propanoic acid -- Prop-2-en-1-ol = 2HC=CH-CH2OH --> ADDITION POLYMERISATION Names = 3,3-dimethylbut-1-ene (ONLY) and 2-chloro-3,3dimethylbutane (ONLY)Propanone OXIDATION EQUATION CH3CHOHCH3 [O] ---> CH3COCH3 + H2OEXTRACTING METALS Equation = -1615 kjmol-1Pure Ca is expensive (may except other credible things) Vanadium Q = VCL4 + 2H2 ---> V + 4HCLRisk = H2 is explosive and V is radioactive Reason for purity = H2 is reducing agent
i am really sorry because of the way it is poorely structured and do u think this is the one

Reply 2

Original post by KryptoModz

Thanks

However I did manage to find the official paper + mark scheme so I don't need it! Luckily enough I did do well in my paper! Just waiting on the results which will be revealed on Wednesday!

ok no problem i since u stated that u have the unofficial mark scheme and paper would u plz show me the link or anything how you have got it since i need it too thanx

Reply 3

Original post by KryptoModz

Please private message me :wink:

im really sorry how am i meant to do that

Reply 4

Original post by honey55497

im not exactly sure but is this the one
FIRST 2 PAGES 2Cl- ---> Cl2 + 2e-Oxidation state of Mn = +7Mn04- + 5e- +8H+ ---> Mn2+ + 4H20Cl2 + 2Br- ----> 2Cl- + Br2Solution turns orange/brownCl2 + 2H20 ----> 4HCl + O2oxidation state of chlorine containing compound = -1Chlorine radical = CF3 with radical . on the Carbon NOT the Fluorine(s) Why Cl lower b.p. = Cl is a smaller molecule than Br (1) or has smaller atomic radius and therefore has a smaller surface area for vdws meaning weaker van der waals BETWEEN MOLECULES (1) 3RD PAGEReasons for atomic radius = As you go down the group size of the atoms increases as extra shells of electrons are being added, meaning more shielding from outer s electrons to the nuclear charge, meaning electrons are situated less closely to the nucleus Trend in reactivity = Increases down the groupSr + 2H20 ---> Sr(OH)2 + H2Most soluble = Ba(OH)24th PAGE = REAGENTS (alcohols)Acidified K2Cr207 --> color change orange to green --> no visible change(Alkane and alkene) Bromine water/ Br2 ---> No visible change ----> color change orange/brown to colorless (ketone and aldehyde) --> Tollens reagent of Fehlings --> Tollens = silver mirror Feelings = red ppt ---> no visible change (AgNO3 and NaN03) Any halide ---> x precipitate (x = color) ---> no visible change (BaCl2 and something else) Concentrated H2S04 ---> White precipitate ---> no visible change (Ill edit this if i remember the other reagent you had to give)CALCULATIONS 1st = -1651 kjmol-12nd = -1110 kjmol-1 (or 1114.6) but i think 3 sig figs was desired 3rd = -1615 kjmol-1 Balancing equation 2 3 3 3 (i put a multiple of x2 by accident so I'm sure multiples will be excepted)MECHANISMS Not going to draw them out but P -Q was Electrophilic addition --> reagent = Na or K OH --> conditions to favor = conc. base, ethanol as the solvent or high temperatures Q-P was Elimination Q - Z was N.S. (REMEMBER N+H3 WAS FORMED THEN THE BREAKAGE OF THE H BOND TOWARDS THE N CATION)POLYMERS ---> Propanoic acid -- Prop-2-en-1-ol = 2HC=CH-CH2OH--> ADDITION POLYMERISATION Mr 48.1 idea was 4(12.00000) + 10(1.00797) (probs wrong number but you needed to write out full number) ANS = 3S.F = 48.1Carbon 12.00000 because it is the standard reference value/ or by definition DRAWING AND NAMING OF COMPOUNDS Butan-2-ol = secondary alcohol -- Propanoic acid -- Prop-2-en-1-ol = 2HC=CH-CH2OH --> ADDITION POLYMERISATION Names = 3,3-dimethylbut-1-ene (ONLY) and 2-chloro-3,3dimethylbutane (ONLY)Propanone OXIDATION EQUATION CH3CHOHCH3 [O] ---> CH3COCH3 + H2OEXTRACTING METALS Equation = -1615 kjmol-1Pure Ca is expensive (may except other credible things) Vanadium Q = VCL4 + 2H2 ---> V + 4HCLRisk = H2 is explosive and V is radioactive Reason for purity = H2 is reducing agent
i am really sorry because of the way it is poorely structured and do u think this is the one

Oh Lord.
Not sure what sane person would be willing to go through this. Glad OP found the original paper.

Reply 5

Original post by KryptoModz

I'll private message you then lol

lmao just done that found out

Reply 6

Original post by KryptoModz

Thanks

However I did manage to find the official paper + mark scheme so I don't need it! Luckily enough I did do well in my paper! Just waiting on the results which will be revealed on Wednesday!

So you cheated?

Reply 7

Original post by KryptoModz

I did my mocks over 3 weeks ago, someone in my class group chat posted the link a few days ago, So I used it to basically see how I did on the paper. The wait was to long! I just had to lol.

Ah right, so you got access to the paper once you had already done the mock - in that case you didn't cheat...well done :P

Reply 8

Original post by KryptoModz

Yeah, I didn't even think AQA would let anyone release the papers.

They wouldn't, AQA's 2015 papers won't be released til 2016! I'm sure someone had access to it via a teacher or some other form of "illegal" method, it would definitely not be on the AQA website as far as my knowledge goes.

Reply 9

Sindy2000

Can u please send me it

Reply 10

Sindy2000

hey could u send me the paper please?

Reply 11

Sindy2000

Original post by KryptoModz

I did my mocks over 3 weeks ago, someone in my class group chat posted the link a few days ago, So I used it to basically see how I did on the paper. The wait was to long! I just had to lol.

Coupd i please see this paper

Reply 12