Unofficial Chemistry AQA Paper 1 Mark scheme:
- lattice dissociation: -3888
- entropy change: +181
- dG: -1063 (to 3 SF); increasing the temperature caused dG to decrease because dS is positive so &quot;-TdS&quot; is negative
in excess; 10cm3
- The least soluble Hydroxide is Mg(OH)2
+ 6 H2
O --&gt; [Al(H2
+ 3 Cl-
b) with sodium carbonate a white precipitate forms and there is effervescence:
+ 3 CO32-
--&gt; 2 Al(H2
+ 3 CO2
+ 3 H2
c) initially a white precipitate would form:
+ 3 OH-
+ 3 H2
the white precipitate would re-dissolve when the OH- is in excess:
+ 3 OH-
+ 3 H2
O (both accepatble)
-Reactants adzorb, reaction takes place, desorb.
-Lowers Ea held by intermolecular forces so higher probability of successful collisions with correct orientation (don't think that bits relevant I was just rambling) The mechanism of how it lowers Ea is a bit fringey on the spec, weakens bonds in reactant so less energy required for reaction.
-V2O5 + SO2 =&gt; SO3 + V2O4
-V2O4 + 1/2O2 =&gt; V2O5
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.
Comparison of second IE of K and Ca
-K has highest second IE
-Electron removed from 3p rather than 4s
-So less shielding, greater attraction to nucleus, requires more energy to overcome.
Highest first IE in s-block:
-Beryllium as it has joint least shielding with Lithium
-But an extra proton so stronger attraction to the nucleus, requires more energy to overcome
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
the concentrations of Cu2+
become equal OR
all the metallic copper on the LHS runs out (both acceptable)
- partial pressures were 7.5, 22.5, 120.
- Expression for Kp
= ( p.N2
) / ( p2
- units for Kp
- 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 Kp
Isomerism and VSEPR Shapes:
- isomerism is
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
Don't even ask how I did the last question, it just worked.
Contributors (Make sure to rep them when you see them
):TruthfulHoax 67823 Psychology109 Alof209 Tommy59375
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