Chemistry Paper 2 June 2017

1. Ais bonded to four other carbon atoms
2. Buses only three of its four outer electrons in bonding
3. Ccarries a small positive charge
4. Dis held by weak van der Waals forces

1. Athe layers are held together weakly and slide over one another
2. Bit is a good conductor of heat
3. Cthe covalent bonds within a layer are weak
4. Dit has a low melting point

1. Aoccupy a large fraction of the volume of the container
2. Blose energy at every collision
3. Chave negligible volume and exert no forces on one another
4. Dattract one another strongly

1. Aused to form a fourth strong covalent bond to a neighbouring layer, holding the layers rigidly together.
2. Btransferred completely to an adjacent carbon atom, forming an ionic bond within the layer.
3. Cdelocalised, free to move between the layers, giving graphite its electrical conductivity.
4. Dpaired with a lone electron on the layer above, forming a hydrogen bond between the layers.

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1. Agoverned mainly by strong intermolecular attractive forces.
2. Brare, because gas particles hardly ever collide with one another.
3. Cperfectly elastic, so no kinetic energy is lost overall.
4. Dinelastic, so the particles gradually lose kinetic energy as heat.

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1. Aone mole of water is formed from an acid and a base under standard conditions
2. Bone mole of hydrogen ions is produced by an acid
3. Cequal volumes of an acid and an alkali are mixed
4. Done mole of an acid reacts completely with an alkali

1. A$-109.2$
2. B$-54.6$
3. C$-27.3$
4. D$-13.0$

1. A$-786$
2. B$-564$
3. C$-282$
4. D$-222$

1. Aa reaction carried out in the presence of a catalyst always has a different overall enthalpy change from the same reaction without one.
2. Bthe total enthalpy change for a reaction is the same regardless of the route taken between the same reactants and products.
3. Centhalpy changes can only ever be measured directly by careful experiment in a calorimeter, never calculated indirectly from other data.
4. Dthe enthalpy change of a reaction always equals the sum of the activation energies of every step in its mechanism.

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1. Ais a poor conductor of heat, so it minimises heat loss from the reacting solution.
2. Bhas a very high heat capacity of its own, absorbing any excess heat released safely.
3. Creacts with the hydrochloric acid used in the experiment, buffering the solution's pH throughout.
4. Dis transparent, letting any colour change in the reacting solution be observed directly.

1. A$3Cl_2 + 6KOH \rightarrow 5KCl + KClO_3 + 3H_2O$
2. B$Cl_2 + 2KOH \rightarrow KCl + KClO + H_2O$
3. C$Cl_2 + 2KOH \rightarrow 2KCl + H_2O + \frac{1}{2}O_2$
4. D$Cl_2 + KOH \rightarrow KCl + HOCl$

1. A$-1$ and $+3$
2. B$-1$ and $+5$
3. C$+1$ and $+5$
4. D$-1$ and $+1$

1. Aa purple colour in the organic layer
2. Ban orange colour in the aqueous layer
3. Ca white precipitate in the aqueous layer
4. Deffervescence at the surface

1. Adecreases, because the increasing nuclear charge down the group holds each molecule's own electrons more strongly.
2. Bincreases, because the covalent bond within each diatomic molecule becomes weaker and breaks more easily.
3. Cincreases, because the molecules become smaller in size and so evaporate more readily from the liquid.
4. Ddecreases, because the increasing number of electrons increases the van der Waals forces between molecules.

1. A-1 and +5
2. B+1 and +5
3. C-1 and +3
4. D-1 and +1

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1. Aester, alcohol and alkyne
2. Baldehyde, ketone and alkene
3. Camide, phenol and alkene
4. Dcarboxylic acid, phenol and alkene

1. Aan alkene
2. Ban ester
3. Can acyl chloride
4. Dan aldehyde

1. Anucleophilic substitution
2. Belectrophilic addition
3. Celectrophilic substitution
4. Dfree radical substitution

1. Ano isomerism at all, since the whole molecule has a plane of symmetry running through the double bond.
2. Bcis-trans (geometric) isomerism, since each doubly-bonded carbon carries two different groups.
3. Coptical isomerism, since the molecule contains a carbon atom bonded to four completely different groups.
4. Dstructural (chain) isomerism, since the carbon skeleton itself can be rearranged in different ways.

1. Acarboxylic acid group(s), cleaving the chain at the double bond.
2. Ba diol formed at the double bond, without breaking the carbon chain at all.
3. Can epoxide ring bridging the two carbons that were joined by the double bond.
4. Dno product at all, since hot acidified KMnO4 does not oxidise carbon-carbon double bonds.

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1. Aconcentrated $HNO_3$ and concentrated $H_2SO_4$; X is nitrobenzene
2. B$NH_3$ in ethanol; X is phenylamine
3. C$Sn$ and concentrated HCl; X is nitrobenzene
4. D$Br_2$ and $FeBr_3$; X is bromobenzene

1. Athe nitrogen atom has no lone pair
2. Bthe molecule is much larger
3. Cthe $-NH_2$ group is attached to a saturated carbon
4. Dthe lone pair on nitrogen is delocalised into the benzene ring

1. Aan orange precipitate forms
2. Ba violet colour develops
3. Cthere is no visible change
4. Dthe bromine is decolourised and a white precipitate forms

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