The balanced equation is 4NH₃(g)  +  ­5O₂(g)     →      4NO(g)  +  6H₂O(l)

Amount of HCl in 150 cm³ of 0.20 mol dm⁻³ HCl(aq) = 150/1000 x 0.20 mol
Amount of HCl in 250 cm³ of 0.30 mol dm⁻³ HCl(aq) = 250/1000 x 0.30 mol
Amount of HCl in 400 cm³ of mixture = (150/1000 x 0.20) + (250/1000 x 0.30) mol
Amount of HCl in 1000 cm³ (1 dm³) of mixture = [(150/1000 x 0.20) + (250/1000 x 0.30)] x 1000/400 = 0.26 mol

CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) so 1.60 g (0.1 mol) of methane requires 0.2 mol of oxygen (6.40 g) to combust completely. This means there will be 6.40 g (0.2 mol) of excess oxygen remaining and 0.1 mol of carbon dioxide product present (water is a liquid at STP) which will occupy 3 x 2.27 = 6.81 dm³.

The atomic number of Fe is 26 so it will contain 26 protons. Its mass number is 56 so there will be 30 neutrons in the isotope. Fe contains 26 electrons.  Three electrons have been removed to form Fe³⁺ so there will be 23 electrons remaining.

Emission spectra involve excited electrons falling from higher to lower energy levels. An electron transition from ∞ to the n = 1 level will emit the same amount of energy as needs to be absorbed when the atom is ionized and the electron moves from its ground state of n =1 to n=∞.

Once the second electron has been removed the species together with their electron configuration will be Ne²⁺ 1s²2s²2p⁴, O²⁺ 1s²2s²2p², Li²⁺ 1s¹ and Be²⁺ 1s². In the same way that He has a higher 1st ionization energy than H, Be²⁺ will have a higher ionization energy than Li⁺ as the electron being removed from Be²⁺ is spin-paired and there is an extra proton in the nucleus of Be²⁺ compared to Li⁺.

The oxides of period 3 (Na → Cl) change from basic to acidic moving from left to right across the period with aluminium oxide having both basic and acidic properties.

[Zn(H₂O)₆]²⁺ is the only one of the four ions that does not contain an incomplete d sub-level (the 3d sub-level is full in [Zn(H₂O)₆]²⁺) so no transitions can occur within the split d level.

All four compounds are ionic but the nitrate, sulfate and ammonium ions, (NO₃⁻, SO₄²⁻ and NH₄⁺) are complex ions which also contain covalent bonding. Potassium nitride, K₃N only contains K⁺ and N³⁻ ions.

CF₄ and BF₃ have four electron domains around the central atom. CF₄ has a regular tetrahedral shape with bond angles of 109.5^o, PF₃ has one non-bonding pair of electrons so has a trigonal pyramid shape with bond angles less than 109.5^o  as the non-bonding pair of electrons exerts a greater repulsion than the bonding pairs. SF₆ has six electron domains giving a regular octahedral shape with bond angles of 90^o and 180^o. PF₅ has five bonding electron domains so its shape is trigonal bipyramid with bond angles of 90^o, 120^o and 180^o.

The three factors which increase the strength of metallic bonding are a decrease in the radius of the metal ion, an increase in the charge on the metal ion (these two factors increase the charge density) and the number of delocalized electrons. An increase in the number of electrons is not the same as an increase in the number of delocalized electrons. Potassium, for example, has 39 electrons whereas lithium only has 3 electrons but both have only one electron in the outer s orbital that can delocalize and the much smaller lithium ion means lithium has a much higher boiling point than potassium. 

A π bond is formed when two atomic orbitals combine 'sideways' on. When one of the atomic orbitals is an s orbital a sigma (σ) bond will always be formed and when two p_x atomic orbitals combine they meet 'head' on so also form a σ bond.

The six carbon atoms in the phenyl group are sp hybridized and the carbonyl carbon atom is sp² hybridized. Only the carbon atom in the methyl group is sp³ hybridized.

Removing an electron from a gaseous atom or ion is always endothermic. Adding an electron to a gaseous atom is always exothermic but adding an electron to a gaseous negative ion is always endothermic.

Average bond enthalpies refer only to the gaseous state and additional enthalpy terms involving change of state are required for all the reactions except for the reaction between methane and chlorine where all the reactants and products are in the gaseous state.

ΔG = ΔH − TΔS = +101 − T(136/1000) kJ mol⁻¹. At 298 K ΔG has a positive value so the reaction is non-spontaneous but as T increases eventually T(136/1000) will become greater than 101 and ΔG will have a negative value at which point the reaction becomes spontaneous.

E_a for the forward catalysed reaction is the difference between the starting energy and the peak of the dotted red line (catalysed pathway), ΔH is the difference in energy between the initial and final states and E_a for the reverse catalysed reaction is the difference between the new starting energy and the peak of the catalysed pathway.

The larger the particle size the smaller the surface area of the magnesium which will lower the number of collisions. The lower the temperature the lower the number of successful collisions as less particles will collide with the necessary activation energy. The lower the concentration of the acid the lower the number of collisions between the acid and the magnesium.

For a second order reaction rate = k x concentration², so k = (mol dm⁻³ s⁻¹) ÷ (mol dm⁻³)² = mol⁻¹ dm³ s⁻¹

A is the frequency factor although it is also sometimes known as the Arrhenius constant or pre-exponential factor. When 1/T is plotted against lnk the gradient will be equal to −E_a/R. k is the value of the rate constant at temperature T (in Kelvin) and the value of the activation energy, E_a is independent of the temperature.

The equilibrium constant for the reverse reaction will be (K_c)⁻¹, i.e. 1 ÷ (1.60 x 10²).

K_c = [HI(g)]² ÷ ([H₂(g)] x [I₂(g)] = (1.2/V)² ÷ [(0.9 − 0.6)/V x (0.7 − 0.6)/V] = 1.44 ÷ 0.03 = 48

Brønsted-Lowry acids donate a proton to form their conjugate base, so NH₂⁻ is the conjugate base of NH₃ whereas ammonia itself is the conjugate base of NH₄⁺.

Sulfuric acid is a strong diprotic acid so [H⁺(aq)] will be equal to 2 x 10⁻² mol dm⁻³. pH = − log₁₀[H⁺(aq)] so the pH will be less than 2 but more than 1 (as [H⁺(aq) would need to be 10 x 10⁻² mol dm⁻³ for it to be 1). This has assumed that H₂SO₄ is a strong dibasic acid. In fact the first dissociation to HSO₄⁻ is strong but the second dissociation is not completely strong. Even so, the actual concentration of H⁺(aq) will still be more than 1 x 10⁻² (but less than 2 x 10⁻² mol dm⁻³) so the pH will between 1 and 2.

NH₄Cl is the salt of a weak base and a strong acid so like CH₃COOH (which is a weak acid) will have a pH less than 7. KCl is the salt of a strong acid with a strong base so will have a pH = 7. Na₂CO₃ is the salt of a strong base with a weak acid so will have a pH greater than 7 as the CO₃²⁻ ions combine with H⁺ ions from the water to form H₂CO₃ leaving an excess of OH⁻ ions.

A Lewis acid can accept a pair of electron. AlCl₃ is electron deficient as there are only six electrons around the Al atom so it can accept a pair of electrons to form, for example AlCl₄⁻. In all the other species there is an octet (or a share in an octet) of electrons around each atom.

The oxidation state of sulfur in sulfuric acid, H₂SO₄ is +6. In sodium thiosulfate, Na₂S₂O₃ and in sulfur dioxide, SO₂ it is +4 and in hydrogen sulfide, H₂S it is −2.

Oxidation always occurs at the anode. In a voltaic cell the anode is the negative electrode and in a electrolytic cell the anode is the positive electrode. For a voltaic cell to produce electrical energy the spontaneous chemical process must be exothermic, not endothermic.

The spontaneous reaction between the two half-reactions is Ni(s) + Sn²⁺(aq) → Sn(s) + Ni²⁺(aq) E^⦵_total = 0.12 V,  so Ni is the strongest reducing agent.

With inert electrodes (such as platinum) oxygen is evolved at the anode and copper is deposited at the cathode.  During this process the solution loses its blue colour as the concentration of copper(II) ions decreases. With copper electrode the copper anode dissolves forming Cu²⁺(aq) ions and the blue colour remains constant as copper is simultaneously deposited at the cathode.

Carboxyl (−COOH), benzyl (C₆H₅CH₂−) and amide (−CO−NH₂) functional groups are absent.

The compound is an isomer of hexanol, C₆H₁₃OH, but the longest chain contains four carbon atoms and there is a hydroxyl group on the second carbon atom. It is also bonded to two methyl groups, one on the second carbon atom and one on the third, so it is 2,3-dimethylbutan-2-ol.

Tertiary halogenoalkanes proceed by an S_N1 mechanism that is faster than the S_N2 mechanism followed by primary halogenoalkanes and iodine is a better leaving group than chlorine.

The equation for this esterification reaction is HCOOH + CH₃OH ⇌ HCOOCH₃ + H₂O

Sulfuric acid protonates the nitric acid to form H₂NO₃⁺ which then dissociates to form water and the nitronium ion, NO₂⁺. It is the NO₂⁺ ion that then acts as the electrophile by being attracted to the delocalised π electrons on the benzene ring.

A chiral carbon atom is asymmetric, i.e. contains four different groups bonded to it.
The chiral carbon atom is shown below with a red asterisk, *.

The ideal gas equation is pV = nRT. Since M_r of O₂ is 32, n = 1/32 mol. P = 1.0 x 10⁵ Pa = 1.0 x 10⁵ Nm⁻² and I J = 1 Nm

32 (mol⁻¹) x 1.0 x 10⁵ (Nm⁻²) x V (m³) = R (J K⁻¹ mol⁻¹) x T (K)

a plot of (3.2 x 10⁶)V against T will give a gradient of R in J K−1 mol−1)

When calculating IHD O atoms count as zero and Cl atoms can be considered to behave as H atoms so the formula can be reduced to C₁₂H₈. If it was fully saturated it would be C₁₂H₂₆ so it is 18 H atoms (or 9 H₂) short which means that the IHD Is 9. This can also be seen from the structural formula as the two benzene rings each have an IHD of 4 (three for the 'double bonds' and one for the ring) and the additional ring makes a total of 9.

The triplet and quartet will be shown by an ethyl group −CH₂CH₃ with no other hydrogen atoms on the neighbouring carbon atom it is bonded to as the −CH₃ protons are split into a triplet by the −CH₂ protons and the −CH₂ protons are split into a quartet by the −CH₃ protons. A singlet will be shown by all the other hydrogen atoms in the molecule being in the same chemical environment with no hydrogen atoms on neighbouring carbon atoms.  C(CH₃)₃COCH₂CH₃ will give this splitting pattern with an integration trace of 9 for the singlet, 2 for the quartet and 3 for the triplet.