@Bulletzone This is my submission for your first post. I have confidence in all my answers here; I have left out things I'm unsure on.

Here are the answers I can remember which I also have confidence are correct. (due to checking with friends)
[NB: dA means delta A]

- lattice dissociation: -3888
- entropy change: +181
- dG: -1080 (to 3 SF); increasing the temperature caused dG to decrease because dS is positive so "-TdS" is negative
- BaCl₂ in excess; 10cm³ of Na₂SO₄ required
- isomerism was cis-trans (or e/z); the other isomer had the two Cl ligands opposite each other
- bond angle was 104.5 degrees. The central N had 6 of its own electrons (5, +1 due to the overall negative charge). Two electrons used in bonding to H atoms so two lone pairs. Shape is tetrahedral but lone pairs repel more than bonding pairs, reducing the bond angle from 109.5 by about 5 degrees (2.5 per lone pair).
- x = 1

ALUMINIUM QUESTION:
a) AlCl₃ + 6 H₂O --> [Al(H₂O)₆]³⁺ + 3 Cl^-
b) with sodium carbonate a white precipitate forms and there is effervescence:
2[Al(H₂O)₆]³⁺ + 3 CO₃^2- --> 2 Al(H₂O)₃(OH)₃ + 3 CO₂ + 3 H₂O
c) initially a white precipitate would form:
[Al(H₂O)₆]³⁺ + 3 OH^- --> Al(H₂O)₃(OH)₃ + 3 H₂O
the white precipitate would re-dissolve when the OH- is in excess:
EITHER Al(H₂O)₃(OH)₃ + OH^- --> [Al(OH)₄]^- + H₂O OR Al(H₂O)₃(OH)₃ + 3 OH^- --> [Al(OH)₆]3^- + 3 H₂O (both accepatble)

6 MARK QUESTION:
- NaBr and NaI are both giant ionic lattices with strong electrostatic attractions between positive and negative ions which require a lot of energy to be overcome. However, the I- ion has a larger radius than Br- and therefore a smaller charge density. Therefore the ionic bonding in NaI is weaker than in NaBr, requiring less energy to be overcome.
- Na is a metal with metallic bonding between Na+ ions and a sea of delocalised electrons. However this is not as strong as the ionic bonding in NaBr/NaI because metallic Na does not exist in a giant lattice.

ELECTROCHEMICAL CELLS QUESTION:
a) Ions in the salt bridge are free to move, carrying a charge and completing the electrical circuit.
b) V = 0.34 - EMF
c) Because the concentration of Cu²⁺ ions in the LHS cell is lower than 1 mol/dm³ which is what's required for the standard electrode potential.
d) The concentration of Cu²⁺ ions would increase over time
e) EITHER the concentrations of Cu²⁺ become equal OR all the metallic copper on the LHS runs out (both acceptable)

K_P QUESTION:
- partial pressures were 7.5, 22.5, 120.
- Expression for K_p = ( p.N^₂ * p³.H₂ ) / ( p².NH₃ )
- units for K_p = kPa^-2 (however I can't remember the numerical value)
- since the reaction is exothermic, increasing the temperature moves the position of the equilibrium to the left, reducing the partial pressure of NH₃ and thereby decreasing K_p.

(edited 5 years ago)

Reply 86

harooney123

Original post by 67823

If you remember any questions &amp;amp; answers, please comment below

Lattice formation: -3888 kJ mol (3marks)
Kp: 2.99x10-3
Entropy: 181
Gibbs free: -1063
TOF distance: 1.54m (5 marks)
PH: 13.1
Ph: 4.28
Electrode potential: +0.18 (1 mark)
Ideal gas volume: 133cm
Concentration: 0.0218 (1 mark)
Type of isomerism: optical
Bond angle: 104.5 (3 marks)
Transition metals: complex ion Al3+
Al3+ with Co32-: white precipitate & effecversence (2 marks)
Back titration: 1 (7 marks)

Wasn't the bond angle 107 because there was only one lone pair of electrons

Reply 87

Lovearmy

Original post by sas00

i got 267 too but lots of ppl say 133 do you knwo why?

Would not G be -1060 cos of theee significant figures. I got -1063 but then changes it to -1060 cos i verything was upto 3 sF in question

Reply 88

negzi

grade boundary predictions?

Reply 89

brendonurine

Was the rfm of sr 87.3 or 87.7? Because I think I wrote it down wrong and used 87.3, but I did the correct method but with 87.3. Do you think I would have got any marks?

Got that too

Original post by Lovearmy

Would not G be -1060 cos of theee significant figures. I got -1063 but then changes it to -1060 cos i verything was upto 3 sF in question

I did that as well

Reply 92

Jadejay

Original post by rexs2000

Wasn’t there a minimum temp for reaction to occur question

I don’t remember it... did I miss a question out? There was one that asked how delta G is affected when temp is increased I think

Reply 93

brownkagaroo

Original post by AlkynesOfTruble

How many marks would I get if I forgot to do the subtraction??? :frown:

I got 10 too and now I'm literally kicking myself

Reply 94

Jadejay

Original post by yy9

transition metal complexes with four of the same ligands and two of different ligands show cis-trans isomerism, it’s on the specification bit of transition metals

I called it geometric would I get the mark?

Reply 95

Buffoon18

For kp I got 120, 19.95 and 10.05

Reply 96

JJarvisCocker

Original post by AlkynesOfTruble

How many marks would I get if I forgot to do the subtraction??? :frown:

Probably two I would guess, one for the method of subtracting and one for the right answer

Reply 97

Alex kore

Original post by Tommy59375

Here are the answers I can remember which I also have confidence are correct. (due to checking with friends)
[NB: dA means delta A]

- lattice dissociation: -3888
- entropy change: +181
- dG: -1080 (to 3 SF); increasing the temperature caused dG to decrease because dS is positive so "-TdS" is negative
- BaCl₂ in excess; 10cm³ of Na₂SO₄ required
- isomerism was cis-trans (or e/z); the other isomer had the two Cl ligands opposite each other
- bond angle was 104.5 degrees. The central N had 6 of its own electrons (5, +1 due to the overall negative charge). Two electrons used in bonding to H atoms so two lone pairs. Shape is tetrahedral but lone pairs repel more than bonding pairs, reducing the bond angle from 109.5 by about 5 degrees (2.5 per lone pair).
- x = 1

ALUMINIUM QUESTION:
a) AlCl₃ + 6 H₂O --> [Al(H₂O)₆]³⁺ + 3 Cl^-
b) with sodium carbonate a white precipitate forms and there is effervescence:
2[Al(H₂O)₆]³⁺ + 3 CO₃^2- --> 2 Al(H₂O)₃(OH)₃ + 3 CO₂ + 3 H₂O
c) initially a white precipitate would form:
[Al(H₂O)₆]³⁺ + 3 OH^- --> Al(H₂O)₃(OH)₃ + 3 H₂O
the white precipitate would re-dissolve when the OH- is in excess:
Al(H₂O)₃(OH)₃ +OH^- --> [Al(OH)₄]^- + H₂O

6 MARK QUESTION:
- NaBr and NaI are both giant ionic lattices with strong electrostatic attractions between positive and negative ions which require a lot of energy to be overcome. However, the I- ion has a larger radius than Br- and therefore a smaller charge density. Therefore the ionic bonding in NaI is weaker than in NaBr, requiring less energy to be overcome.
- Na is a metal with metallic bonding between Na+ ions and a sea of delocalised electrons. However this is not as strong as the ionic bonding in NaBr/NaI because metallic Na does not exist in a giant lattice.

ELECTROCHEMICAL CELLS QUESTION:
a) Ions in the salt bridge are free to move, carrying a charge and completing the electrical circuit.
b) V = 0.34 - EMF
c) Because the concentration of Cu²⁺ ions in the LHS cell is lower than 1 mol/dm³ which is what's required for the standard electrode potential.
d) The concentration of Cu²⁺ ions would increase over time
e) EITHER the concentrations of Cu²⁺ become equal OR all the metallic copper on the LHS runs out (BOTH ACCEPTABLE)