Edexcel chemistry a level 20th of June 2018 paper 3 unofficial markscheme

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Unofficial markscheme
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Inorganic
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Q1 London forces form as instantaneous dipole is induced in halogen molecule (one end delta positive and one end delta negative), causing temporary dipoles to be induced in neighbouring molecules. Process continues throughout entire substance, resulting in intermolecular London forces.

Br2 higher boiling temp as more electrons so stronger London forces require more energy to break down.

Reactivity: reactivity decreases as group descends. First KI + Br2, bromine is more reactive so displaces iodide : 2I- + Br2 - - 2Br- + I2 , iodine is formed, which dissolves in cylcohexane to form purple bottom layer (more dense) adding bromine to KCl has no effect (Cl2 is a stronger oxidising agent than Br2) so when cyclohexane is added, bottom bromine layer is orange-brown. Adding iodine to KCl has no effect. Bottom purple I2 layer.

Q2 Synthesis is ethanal + HCN and KCN at pH=8 to form 2-hydroxypropanitrile. This is nucleophilic addition.
Then hydrolyse with sulfuric acid to produce lactic acid.
2-hydroxypropanitrile + 2H2O + H+ - - lactic acid + HCN + H+ + H2

Condensation polymerisation

Vanadium, cba to write all equations but reduced from VO3- to VO 2+ (colourless to blue) then reduced to V 3+ (green) then reduced to V 2+ (lilac), but is not reduced to V (s) as Ecell is negative.

Acts as heterogenously catalyst in contact process : SO2 and O2 adsorbed then bonds break, new bonds form to form SO3

Copper 6 marker

Deprotonation with NaOH to form blue ppt
Ligand exchange with excess NH3 to form blue PPT then feel blue solution
Ligand exchange with Cl- to form yellow solution due to CuCl4 2-

Titration can't remember but lowest % uncertainty was burette

Titre decreased for both mistakes

NMR CH3 split into 3 due to adjacent CH2
CH2 split into 4 due to adjacent CH3 (n + 1 rule) etc. Singlet was due to the CH3 at the right and, as the adjacent carbon had no hydrogen attached (only oxygen)

Thermal decomposition

Remember the mass for something was 1.4

MCO3 - - MO + CO2

Errors resulted in less CO2 volume so higher calculated Mr

Mr at first was 87.04 , metal was magnesium
Second Mr (using decomposition) was 82.

pH was 13.32 (5 marker) found by getting pOH then 14 - pOH.

pKA = pH when half of the acid is neutralised (12.5 cm3 of alkali on graph) so get pKA then do 10 to the power of -(pKa) to get Ka

Can't remember the last ester question, ****ed it up though.

Paper was challenging, but found it manageable up until the last question. Glad chem is over 8)
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