SASS Chem P1 Revision Ans

Page 1 ST ANDREW’S SECONDARY SCHOOL NAME CLASS CHEMISTRY 6092/01 Paper 1 REVISION PACKAGE + MULTIPLE CHOICE QUESTIONS Solutions 1 B Since forward rxn represents exothermic, the reverse / backward rxn hence represent endothermic. Ea is labelled for the energy difference between the reactants and the apex / peak of the graph. 2 D NbClx + x Na ® Nb + x NaCl No. of moles of Nb produced = mass ÷ molar mass = 18.58 g ÷ 92.9 g mol–1 = 0.2 mol Mass of Cl = 54.08 g – 18.58 g = 35.50 g No. of moles of Cl lost = No. of moles of NaCl formed = 35.50 g ÷ 35.5 g mol–1 = 1.0 mol Nb : NaCl 1 : x 0.2 : 1.0 x = 5 3 C NaOH + HCl ® NaCl + H2O No. of moles of NaOH added = vol × conc = 0.0125 dm3 × 0.05 mol dm–3 = 0.000625 mol No. of moles of HCl added = vol × conc = 0.025 dm3 × 0.1 mol dm–3 = 0.0025 mol ∴HCl is the ER. No. of moles of HCl remain unreacted = 0.0025 mol – 0.000625 mol = 0.001875 mol Combined volume of solution = 0.0125 dm3 + 0.025 dm3 = 0.0375 dm3 Conc of remaining HCl = mol ÷ vol = 0.001875 mol ÷ 0.0375 dm3 = 0.05 mol dm–3 4 C Na2O reacts with water to form NaOH (strong alkali) NaCl dissolves in water to form its solution (neutral) SO2 reacts with water to form H2SO3 (sulfurous acid, strong acid) SiCl4 reacts with water to form HCl (strong acid): SiCl4 + 2 H2O ® SiO2 + 4 HCl [Hydrolysis = Breaking apart with water] Even if you are unaware of the hydrolysis rxn, the other 3 pairs still does not give a pH of 7 each. 5 C Total no. of moles of gases in gaseous mixture = 0.12 dm3 ÷ 24 dm3 mol–1 = 0.005 mol No. of moles of O2 = 0.096 g ÷ 32 g mol–1 = 0.003 mol No. of moles of N2 = 0.056 g ÷ 28 g mol–1 = 0.002 mol 6 B No. of moles of 13C = 0.13 g ÷ 13 g mol–1 = 0.01 mol Each mole of 13C contains L atoms, of which each 13C atom contains 7 neutrons. 7 C Oxide of P = P2O5 (OS of P = +5) Oxide of S = SO2 (OS of S = +4) P4S3 + 8 O2 ® 2 P2O5 + 3 SO2 8 C No. of moles of H2O formed = 3.6 g ÷ 18 g mol–1 = 0.2 mol No. of moles of HCl required = No. of moles of NH3 produced = vol × conc = 0.1 dm3 × 1.0 mol dm–3 = 0.1 mol ⇒ x = 1, y = 2

Page 2 9 D S (Group VI, forms a molecule S8) : simple molecular structure Si (Group IV, forms a lattice similar to C) : giant covalent structure Al (Group III, metal) : giant metallic structure 10 A Atomisation: Forms discreet atoms from molecules (e.g. X2: X–X ® 2 X) through bond breaking = Endothermic rxn Combustion = Burning with O2 = Exothermic rxn Hydration = Addition of water, e.g. water of crystallisation = Exothermic rxn Neutralisation = Acid + Alkali = Exothermic rxn 11 D No. of moles of KMnO4 titrated = vol × conc = 0.0274 dm3 × 0.02 mol dm–3 = 0.000548 mol MnO4– : Fe2+ 1 : 5 0.000548 mol : 0.00274 mol Conc of FeSO4 reacted = mol ÷ vol = 0.00274 mol ÷ 0.025 dm3 = 0.1096 mol dm–3 12 B Q contains NH4+ that accounts for the NH3 evolved. At least 4 H in Q. Molecular formula of Q = 2(NH2O) = N2H4O2 or NH4NO2 (ammonium nitrite) Anion in Q = NO2– (additional 1 e–) Each N has 7 e–; O has 8 e–. Total e– present in NO2– = 7 + (2 × 8) + 1 = 24 13 B Na2CO3 is thermally stable, it does not decompose upon heating. However, when heated strongly, solid Na2CO3 melts to a liquid. 14 C Catalyst provides the reaction with an alternative pathway with a lower Ea, thereby resulting in a larger proportion/frequency of successful collisions for the increased rate of reaction. 15 A Endothermic energy profile diagram: Enthalpy change = Energyproducts – Energyreactants = +ve 16 D OS of Cr in Cr2O72– = +6 OS of Cr in Cr2O3 = +3 OS of Cr decreases from +6 to +3, reduction. 17 A Cl– is the LR. After precipitation, the residue (containing Cl–) is removed from this mixture by filtration. 18 C Al3+ and Pb2+ both form white ppt that dissolve in excess NaOH (aq). 19 C Oxidation: Fe (s) ® Fe2+ (aq) + 2 e– Reduction: Fe3+ (aq) + e– ® Fe2+ (aq) Overall: Fe (s) + 2 Fe3+ (aq) ® 3 Fe2+ (aq) As the final mixture contains equimolar amounts of Fe2+ (aq) and Fe3+ (aq), there is an excess of 3 mol Fe3+ (aq). Hence, the starting mixture should contain 2 + 3 = 5 mol Fe3+ (aq). 20 A 8 bonding e– = 4 covalent bonds formed CO2 : 4 covalent bonds C2H4 : 6 covalent bonds C3H6 : 9 covalent bonds NH3 : 3 covalent bonds 21 A No. of moles of Ag coated = 0.216 g ÷ 108 g mol–1 = 0.002 mol No. of moles of Ag per cm3 = 0.002 mol ÷ 150 cm2 = 1.333 × 10–5 mol/cm2 No. of Ag atoms per cm3 = (1.333 × 10–5) mol/cm2 × (6 × 1023) atoms/mol = 8.0 × 1018 atoms/cm2 22 C Cl in ClO– = +1 x + (–2) = –1 ClO3– = +5 x + 3(–2) = –1 Cl– = –1 14 Chemical bonding and structure506 Potential energy (enthalpy) resonance energy – – O O O C – –O O O C 2–O O O C When resonance occurs, the resonance hybrid is more stable than any of the resonance structures. The difference between the energy of the most stable form and the hybrid is known as the resonance energy. This concept is illustrated in Figure 14.45 for the carbonate ion. For significant resonance stabilization to occur within a molecule or ion, the suggested resonance structures must meet all of the following requirements, as exemplified by the carbonate ion: ■ All resonance forms must have the same distribution of atoms or nuclei in space, that is, the same molecular shape. All three resonance structures of the carbonate ion have the three oxygen atoms distributed around a central carbon atom in a trigonal planar arrangement. ■ No resonance form may have a very high energy. In particular, no resonance form containing carbon, oxygen or nitrogen (or other atom from the second row of the periodic table) may have more than eight valence electrons (or for hydrogen, two valence electrons). In other words, the octet (and duplet) rules must be obeyed. The three oxygen atoms and one carbon atom of the carbonate ion each obey the octet rule – in all three resonance forms they each have eight valence electrons. ■ All resonance forms must contain the same number of electron pairs. All the resonance structures of the carbonate ion have three σ pairs, one π pair and eight lone (non-bonded) pairs. ■ All resonance forms must carry the same total or net charge. All the resonance structures of the carbonate ion have a total charge of −2. To understand the importance of resonance in the stabilization of a structure, and the contribution of a particular resonance form, the following points must be considered: ■ In the case of ions, the most stable resonance structures will have negative charges on the electronegative atoms, usually oxygen and sulfur, and positive charges on the less electronegative atoms, usually carbon and nitrogen. ■ The most important resonance structures are those where as many atoms as possible obey the octet rule, have the maximum number of bonded electrons and have the fewest charges. ■ Resonance effects are always stabilizing and generally the greater the number of resonance structures that can be drawn, the greater the stability of the resonance hybrid. The concept of formal charge can be used to identify the correct Lewis structure for a molecule or the most important resonance structure for a resonance-stabilized molecule or ion. The concept of formal charge is discussed later in this chapter. Examples of resonance-stabilized ions and simple molecules Two other inorganic ions that are stabilized by resonance and described by a series of symmetrical resonance structures are the nitrate(iii) (‘nitrite’) ion, NO2 − , and the nitrate(v) (‘nitrate’) ion, NO3 − (Figure 14.46). Curly arrows may be used to show the movement of electron pairs to generate resonance structures. ■Figure 14.46 Resonance structures and hybrids for the nitrite and nitrate ions + +– O O N + N – –O O O N –O – – O O ON – OO N –OO nitrate ionnitrite ion ■Figure 14.45 A diagram illustrating the concept of resonance e