Water, H₂O and sucrose, C₁₂H₂₂O₁₂ are both pure single substances. Steel is an alloy in which iron is bonded chemically to other elements such as carbon. Sand is a mixture as its different constituents are not bonded chemically to each other and retain their individual properties.

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)

In an emission spectrum light is emitted when excited electrons fall to lower levels. When they fall to the n = 1 level the light emitted is in the ultraviolet region of the spectrum.  When they fall from higher levels to the n = 2 level the light emitted is in the visible region of the spectrum.

Chlorine has an atomic number of 17, so contains 17 protons. This isotope has a mass number of 37 which is equal to the number of protons and neutrons, so it contains 20 neutrons. A neutral chlorine atom contains the same number of electrons as protons but the Cl⁺ ion has lost one electron, so it contains 16 electrons.

If it is in the third period the outer level of electrons must be occupying the 3s or 3s and 3p sub-levels and as it is in group 13 there must be three electrons in the outer level, so the correct electron configuration is 1s²2s²2p⁶3s²3p¹ (which is the electron configuration of aluminium).

Atomic radius increases as the outer electrons become further from the nucleus. ionization energy decreases as it requires less energy to remove the outermost electron and melting point increases as the intermolecular forces get stronger due to an increase in mass and number of electrons.

The oxidation state of Eu in Eu₂(SO₄)₃ is +3 so Eu forms the Eu³⁺ ion. The formula for the nitrate ion is NO₃⁻ so the formula of Eu(III) nitrate is Eu(NO₃)₃.

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.

Water contains two non-bonding pairs of electrons which repel the electrons in the two bonding pairs making the bond angle less than 109^o. BeCl₂ and CO₂ both have two electron domains making the molecules linear with bond angles of 180^o and BCl₃ has three electron domains so has a trigonal planar shape with bond angles of 120^o.

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.

Enthalpy of formation refers to the enthalpy change when one mole of the compound is formed from its elements with all the substances being in their standard state at 298 K (25 ^oC).

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

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 H₂ will eventually be evolved.

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.

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.

Oxidation states should be represented with the sign given before the number, e.g. +5 not 5+. Roman numerals, e.g. (V) refer to oxidation numbers. When the plus sign is after the number it refers to the charge on a positive ion e.g. Fe²⁺.

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

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.

Bromine adds across the C=C double bond of alkenes such as but-2-ene and the organic product is colourless. Benzene does not readily undergo addition reactions. Butane is saturated so cannot undergo addition reactions and butanal is an aldehyde and has no C=C double bond.

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.

Straight line graphs follow the general equation y = mx + c, where m is the gradient and c is the intercept.

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 give three signals as the hydrogen atoms are in three different chemical environments. The ratio of the integration traces will be 3:2:1 for propanoic acid, CH₃CH₂COOH, 6:1:1 for propan-2-ol, CH₃CH(OH)CH₃ and 3:3:2 for butan-2-one, CH₃COCH₂CH₃.