AQA A Level Chemistry Paper 1 4th June 2019 Unofficial Markscheme Paper 1

I think they’re gonna be standard cos from what I’m seeing on Twitter and student room some people did good here others there 😂 kinda thing. It’s quite a range of where people felt they did good or bad

Original post by iamjosephvella

thourghts on boundaries lower or higher?

Reply 81

username3241486

4

The new equation was half of that for the original and the Kp value they gave was for the original, so you half the one they gave you to get the Kp for the new equation

Original post by Khushi.S

Feel free to add yours answers and then I can order them

1.1) Equations- Cd(s)+ 1/2I2(s) and then Cd+(g)+e-+I
1.2) 107 (bit of a debate about this one)
1.3) Perfect ionic model
1.4) 333kJmol-1 and since G>0, its not feasible

Acids and Bases Question:
4.57 * 10^-14
4.30
The third box on the table
For Ka- 1.26*10^-5
HA= A-

Question with the Fe complexes
Fe(OH)3(H2O)3
Brown Colour
2[Fe(H2O)G]3+ + 3CO32- => 2[Fe(H2O)3(OH)3] + 3CO2 +3H20
FeCl3? and then equation with 3 Cl-
Fe(H2O)4(OH)2 and green precipitate

Question about Period 4
Most electronegative- Bromine
Potassium had the smallest atomic radius due to least number of protons- least attraction between outer shell of electrons and the nucleus
Amphoteric

Titration
Percentage by mass of carbon is 46.4%
To increase the accuracy- use a larger mass of solid or less concentrated thiosulphate
Iodine acts as the oxidising agent?
Its white because it has a full d orbital and so, cannot absorb any light and emits all of it so looks white

Shapes of molecule
The drawing between the molecules with two lone pairs on the oxygen to show a dashed line with the hydrogen
The ethanol has hydrogen bonding but the methoxymethane has dipole-dipole/ van der Waals so as the hydrogen bonding is stronger, more energy is required to overcome the attraction and so, the boiling point is higher
The Pocl3 was tetrahedral and the other one was squareplanar (don't know WHY i thought it was trigonal bipyramid) with a bond angle of 90 degrees

Electrospray Ionisation
Vapourise the sample, pass through hypodermic and hollow syringe, a high voltage terminal and proton gained and I think the equation was P(g) + H+(g) => PH+ (g)
Mass number was 155 because the H+ added an extra one?
Mass of molecule was 169

Kp question
Overall pressure- ranges have been from around 190,000-195,000
Kp was higher with the lower temperature and since a lower temperature favours the endothermic reaction, the forward reaction must be endothermic
Kp doesn't change and units remain the same? People square rooted Kp but I thought that since there wasn't a change in temperature, the Kp couldn't change?

Reply 82

Trapmoneybenny

15

Original post by Nina_85262737

I thought the bond angle was 86 degrees cos I thought you’re meant to take away from the original bond angle if there are lone pairs ????

nah. the two lone pairs are on different planes because of their repulsion so it was 90

Reply 83

Headieoneishere

5

Original post by Khushi.S

I thought atomisation was already set for 1/2?

Same I thought it was 1/2

Reply 84

Nina_85262737

8

That’s what I was thinking... but they asked for NaBr reaction and they already stated the observation of misty fumes so it can’t have been HBr so I assumed that it had gone to the redox reaction, as in the acid/base already happend ... if you get what I mean?

Reply 85

Ssss129

8

What were the mol fractions I got total moles 2.12 so divided all them by 2.12?

Reply 86

Trapmoneybenny

15

Original post by vergilsbee

I got:

2NaBr + H2SO4 ----> Na2SO4 + Br2 + H2O

to form SO2?

Reply 87

Dreametuber

6

No you need to square root it

Original post by vergilsbee

The new equation was half of that for the original and the Kp value they gave was for the original, so you half the one they gave you to get the Kp for the new equation

Reply 88

Trapmoneybenny

15

Original post by Ssss129

What were the mol fractions I got total moles 2.12 so divided all them by 2.12?

that is how you calculate mole fractions

Reply 89

YellowYellow2

OP

16

Original post by Nina_85262737

That’s what I was thinking... but they asked for NaBr reaction and they already stated the observation of misty fumes so it can’t have been HBr so I assumed that it had gone to the redox reaction, as in the acid/base already happend ... if you get what I mean?