M of (COOH)₂.2H₂O = (2 x 12) + (6 x 16) + (6 x 1) = 126 g mol⁻¹, so the amount of (COOH)₂.2H₂O = 0.1 mol
Each (COOH)₂.2H₂O contains six oxygen atoms, so the number of O atoms =  6 x 0.1 x 6.02 x 10²³ = 3.6 x 10²³

CaCO₃(s) + 2HCl(aq) → CO₂(g) + H₂O(l) + CaCl₂(aq)
Amount of CaCO₃ = 4/(40 + 12 + 48) = 0.04 mol; Amount of HCl = 20/1000 x 2.0 = 0.04 mol so HCl is the limiting reagent as 2 mol of HCl are required to react with 1 mol of CaCO₃ to produce 1 mol of CO₂.
Amount of CO₂ produced = 0.02 mol, so volume of CO₂ evolved = 0.02 x 22.7 = 0.454 dm³ = 454 cm³

Increasing the temperature increases the volume whereas increasing the pressure decreases the volume. The temperature must be measured in Kelvin so 36 ^oC and 63 ^oC is 309 K and 336 K respectively.

P₁V₁/T₁ = P₂V₂/T₂ so V₂ = V₁ x T₂/T₁ x P₁/P₂ = 100 x (336 ÷ 309) x (9.63 x 10⁴ ÷ 9.89 x 10⁴) = 100 x (336 ÷ 309) x (9.63 ÷ 9.89)

Tin has an atomic number of 50, so contains 50 protons. This isotope has a mass number of 118 which is equal to the number of protons and neutrons, so it contains 68 neutrons. A neutral tin atom contains the same number of electrons as protons but the Sn²⁺ ion has lost two electrons, so it contains 48 electrons.

E = hν = hc/λ for one electron = hcL/λ in Joules for one mole of electrons = hcL/1000λ kJ mol⁻¹

If it is in the fourth period the outer level of electrons must be occupying the 4s, 3d or 4p sub-levels and as it is in group 6 there must be six electrons in the outer level. This might be expected to be [Ar]3d⁴4s² but it is more energetically favourable for the configuration to be [Ar]3d⁵4s¹ as all the outer electrons are unpaired which lowers the repulsion between them. The element is chromium.

Water is a neutral ligand and each hydroxide ion has a charge of minus 1. Since the overall charge on the complex is zero the metal must have an oxidation state of +2.

The spectrochemical series lists ligands according to the energy difference they produce between the two sets of d-orbitals in an octahedral complex, so I⁻ causes the d orbitals to split the least and ammonia will split the d orbitals more than water. They are all monodentate ligands as they only form one coordinate bond to the metal. Bidentate ligands, such as H₂N-CH₂-CH₂-NH₂ can form two coordinate bonds to the metal. The colour of a transition metal complex also depends on the identity of the metal and its oxidation state as well as the nature of the ligand(s).

In carbon monoxide the triple bond between the C and O atom is formed by the carbon atom sharing two electrons with two electrons from the O and the O donating a further pair of electrons. In the other three species all the bonds are normal covalent single or double bonds, even in the carbonate ion which forms a resonance hybrid.

There are five electron domains around the central bromine atom consisting of three bonding pairs and two non-bonding pairs. This results in a T-shaped molecule.

Both propan-2-ol and propanoic acid have a hydrogen atom bonded directly to an electronegative oxygen atom so form hydrogen bonds between molecules whereas in propanal the hydrogen atom is bonded to a carbon atom and so only dipole-dipole attractions occur.

The two carbon atoms in the −CH=CH₂ group are sp² hybridized as are the six carbon atoms in the phenyl group. C₆H₅−  so all are sp² hybridized.

The nitrile carbon atom is sp hybridized so forms one sigma and two pi bonds with the nitrogen atom and the four bonds formed by the sp³ hybridized carbon atom on the methyl group are all sigma bonds.

Temperature is not a part of the definition of standard state (although 298 K is commonly given as the temperature of interest).
ΔH^⦵ reaction = Σ(ΔH^⦵_f products) − Σ(ΔH^⦵_f reactants) not Σ(ΔH^⦵_f reactants) − Σ(ΔH^⦵_f products).
Although ΔH^⦵_f and ΔH^⦵_c  are normally given in kJ mol⁻¹ as they refer to one particular substance, ΔH^⦵reaction is normally given in kJ, not kJ mol⁻¹ as, for example in the reaction N₂(g) + 3₂(g) ⇌ 2NH₃(g), it is not clear if the mol⁻¹ is referring to N₂(g), H₂(g) or NH₃(g).

x = [(6 x −394) + (8 x − 286) − (2  x − 318)] kJ

But x is the enthalpy change for the combustion of 2 mol of propan-2-ol

so ΔH = ½ [(6 x −394) + (8 x − 286) − (2  x − 318)] kJ mol⁻¹

Removing electrons is always endothermic. Adding one electron to an atom is always exothermic but adding another electron to a negative ion is endothermic.

There is no change in the amount (in mol) of gas in the reaction between hydrogen and iodine whereas in the other three reactions the number of moles of gas either increases or decreases.
The actual values in J K⁻¹ are:
H₂(g) + I₂(g) → 2HI(g) ΔS = +21.5
N₂O₄(g) →2NO₂(g)  ΔS = +177
2SO₂(g) + O₂(g) → 2SO₃(g) ΔS = −188
3H₂(g) + N₂(g) → 2NH₃(g) ΔS = −199

Increasing the temperature increases the average energy of the reacting particles so T₂ > T₁ but does not change the number of reacting particles so the area under the graph remains constant. Adding a catalyst does not change the energy of the reacting particles but does provide a different pathway for the reaction with a lower activation energy.

100 cm³ of 2.0 mol dm⁻³ HCl(aq) = 0.20 mol, 1.0 g of Mg = 1.0/24.3 = < 0.05 mol so the acid is in excess and Mg is the limiting reagent. If only 0.5 g of magnesium is used half the volume of hydrogen will be evolved so C and D cannot be correct. Using powdered Mg instead of a single piece of Mg will increase the rate whereas using 1.0 mol dm⁻³ HCl(aq) instead of 2.0 mol dm⁻³ HCl(aq) will decrease the rate but the same volume of hydrogen will eventually be evolved.

From experiments 1 and 2 keeping [A] constant and doubling [B] doubles the rate so the reaction is first order with respect to [B]. From experiments 2 and 3 keeping [B] constant and doubling [A] quadruples the rate so the reaction is second order with respect to [A].

The units of the rate constant are given as s⁻¹. This means that the reaction is first order as the units of rate are mol dm⁻³ s⁻¹ and the units of concentration to the power of 1 are mol dm⁻³. Since the only reactant is dinitrogen pentoxide the reaction must be first order with respect to N₂O₅. The slow step must involve the breakdown of a molecule of N₂O₅ rather than a collision between two N₂O₅ molecules but the total number of separate steps in the mechanism cannot be deduced from the information provided. The molecularity of an elementary step is the number of reactant particles taking part in that step.

All of the changes will alter the position of equilibrium but the equilibrium constant is a constant at a particular temperature so only a change in temperature will change the value of K_c. As ΔH^⦵ has a negative value the reaction is exothermic so lowering the temperature will increase the value of K_c.

If K_c >> 1 then the equilibrium mixture will contain mainly products and if K_c <<1 then the equilibrium mixture will consist mainly of reactants. ΔG = − RTlnK_c so if K_c is << 1 lnK_c will be negative and ΔG will be positive whereas if K_c is >>1 lnK_c will be positive and ΔG will be negative

In the original solution with a pH of 12 [H⁺(aq)] = 1 x 10⁻¹² mol dm⁻³ and [OH⁻(aq)] = 1 x 10⁻² mol dm⁻³. Adding 90 cm³ of water dilutes the solution ten times so [OH⁻(aq)] = 1 x 10⁻³ mol dm⁻³ and [H⁺(aq)] = 1 x 10⁻¹¹ mol dm⁻³ so pH = 11.

Water is amphiprotic so depending upon the reaction can act as an acid or a base by losing or gaining a proton. In this reaction it is not acting as an acid as it is not losing a proton to form OH⁻ but is acting as a Brønsted-Lowry base by gaining a proton to form H₃O⁺. Since the gas is dissolving and the ions formed are hydrated it is also acting as a solvent.

Lewis acids can accept a pair of electrons. Brønsted-Lowry acids can donate a proton so H₃O⁺ and NH₄⁺ are Brønsted-Lowry acids and Lewis acids. F⁻ cannot accept a pair of electrons but can donate a pair so is a Lewis base. AlCl₃ can accept a pair of electrons but has no proton to donate so is a Lewis acid but not a Brønsted-Lowry acid.

An alkaline buffer solution contains a weak base and the salt that the base forms with a strong acid. This can be achieved by adding excess ammonia to hydrochloric acid so that all the acid has been turned into the salt and the excess weak base remains.

The only one of the reactions in which where there is a change in oxidation state is the reaction between chlorine and methane. Elemental chlorine is reduced from 0 to  − 1 in hydrogen chloride and methane is oxidized to chloromethane. The reaction between silver ions and chloride ions to form silver chloride is precipitation. The reaction between boron trifluoride and fluoride ions is a Lewis acid - base reaction and the reaction with hydrated copper ions is a ligand exchange reaction.

Ni²⁺(aq) is a better oxidizing agent than Mn(s) as it can remove electrons from Mn(s) to form Ni(s), but Pb²⁺(aq) is an even stronger oxidizing agent as it can remove electrons from Ni(s). Mn(s) is the best reducing agent.

The overall reaction in the cell is Fe(s) + Cu²⁺(aq) → Cu(s) + Fe²⁺(aq) E^⦵_total = 0.79 V so the EMF from each of the two cells is 0.79 V. When they are connected in series the EMF will be 2 x 0.79 V = 1.58 V.

In the copper(II) sulfate cell copper is deposited at the cathode so the cathode gains in mass, Cu²⁺(aq) + 2e⁻ → Cu(s). An equal amount of copper is lost from the anode, Cu(s) → Cu²⁺(aq) + 2e⁻ so [Cu²⁺(aq)] remains constant and the solution retains its blue colour. 3.2 g of copper was lost, i.e. 3.2/64 = 0.05 mol of copper. This means that the charge passed through both cells was 2 x 0.05 F.  2H⁺(aq) + 2e⁻ → H₂(g) so the amount of hydrogen formed at the cathode in the sulfuric acid cell will be 0.05 mol and the amount of oxygen formed at the anode will be 0.025 mol since 2H₂O(l) → 4H⁺(q) + O₂(g) + 4e⁻. 

Tertiary alcohols contains three R− groups bonded to the carbon atom to which the −OH is bonded. Compounds such as propan−1,2,3−triol (also known as glycerol), which contain three −OH groups, are known as trihydric alcohols.

A homologous series is a series of compounds of the same family, with the same general formula, which differ from each other by a common structural unit. Ethene, CH₂CH₂ and propene, CH₂CHCH₃ are both members of the alkene homologous series with the general formula C_nH_2n. Ethyne, CHCH is an alkyne whereas ethene, CH₂CH₂ is an alkene. Ethoxyethane, CH₃CH₂OCH₂CH₃ and butan-1-ol, CH₃CH₂CH₂CH₂OH are an ether and an alcohol respectively.  They are isomers as they both have the same molecular formula, C₄H₁₀O but are members of different classes of compounds.  Butanal, CH₃CH₂CH₂CHO and butan-2-one, CH₃CH₂COCH₃ are also isomers, one is an aldehyde and the other is a ketone.

The initiation step is the homolytic fission of the Cl−Cl bond by ultraviolet light to give chlorine free radicals. These radicals then react to produce more radicals in propagation steps such as CH₄ + Cl· → CH₃· + HCl and CH₃· + Cl₂ → CH₃Cl + Cl·. At no time are hydrogen, H· radicals formed. Termination steps such as CH₃· + Cl· → CH₃Cl result in a product that is not a free radical.

In the S_N2 mechanism the OH⁻ nucleophile bonds to the carbon atom before the bromine atom leaves as the bromide ion so the product has the opposite optical activity in terms of rotating the plane of plane polarized light to the reactant (this is known as Walden inversion). Iodine is a better leaving group than bromine so iodoalkanes react faster than bromoalkanes.

Carboxylic acids are reduced by lithium aluminium hydride firstly to the aldehyde then to the primary alcohol. Ketones can also be reduced, but to form secondary alcohols, not primary alcohols.

Iodine monochloride is polar, ^δ+I−Cl^δ− as chlorine is more electronegative than iodine. When iodine monochloride adds by electrophilic addition the electrophile adds to the carbon atom already bonded to the most hydrogen atoms to form the more stable tertiary carbocation intermediate which then bonds to Cl⁻ to give 2-methyl-2-chloro-3-iodobutane.

If some of the product was lost, not all the magnesium burned or if some of the magnesium burned was already magnesium oxide then the mass of the weighed product would be less than expected, so the amount of oxygen combining with the magnesium would be less and the empirical formula would be richer in Mg compared to O. If the crucible reacted with some of the magnesium then the product might have a mass more than if it had just combined with oxygen so the formula would be richer in O compared to Mg.

When determining IHD O atoms count as zero and for a N atom add one to the number of C atoms and add one to the number of H atoms. This gives the equivalent of C₁₀H₁₂ which needs 10 more hydrogen atoms (5 units of H₂) to become saturated as C₁₀H₂₂, hence the IHD is 5.  This can also be seen from the phenyl group which has an IHD of 4 due to the 'three double bonds' and the ring plus the IHD of 1 for the C=O in the carboxyl functional group to give a total IHD of 5.

C₂H₅− is actually an ethyl group, CH₃−CH₂−, so pentan-2-one, C₂H₅−CH₂−CO−CH₃ will give 4 signals. The other three compounds will each give three signals as their hydrogen atoms are in three different chemical environments. The ratio of the integration traces will be 6:1:1 for propan-2-ol, CH₃CH(OH)CH₃. Both propanoic acid CH₃CH₂COOH and but-2-one, CH₃COCH₂CH₃ will give a singlet, triplet and quartet but their integration traces will be different. For propanoic acid it will be 1:3:2 and for but-2-one it will be 3:3:2.