EJC 2023 8873 'A' Level H1 Chemistry Suggested Solutions P1

Eunoia Junior College 9729 H1 Chemistry 2023 Paper 1 Suggested Solution 1 A ✓: The relative charge of a proton, +1, is –1 times the relative charge of an electron, –1. B : A proton has a relative charge of +1, while a neutron has no charge. C : The relative mass of an electron is 31 27 9.11 10 1 1 1.67 10 1833 2000 − −  = times the relative mass of a neutron . D : The relative masses of protons and neutrons are the same, which are about 200 times the relative mass of an electron.  A 2 3s 3p Al 1s2 2s2 2p6   1 Si 1s2 2s2 2p6    2 P 1s2 2s2 2p6     3  A 3 Rate of reaction is affected by the rate constant which in turn, is affected by the activation energy of the reaction. The lower the activation energy, the higher the rate constant, the faster the reaction. The C–Cl bond is more polar than the C–I bond as Cl is more electronegative than I. We would expect chloroethane to react faster with the negatively charged OH – in view of the polarity, which is not the case. Thus, the weaker C–I bond (240 kJ mol –1) compared to C –Cl bond (340 kJ mol –1) must have lowered the activation energy sufficiently to render the reaction of iodoethane faster  D 4 P4 + aHNO3 → 4H3PO4 + …H2O + …NO2 P4 + aHNO3 → 4H3PO4 + …H2O + aNO2 P4 + aHNO3 → 4H3PO4 + (a–16) H2O + aNO2 Balancing the number of H: a = 12 + 2(a–16) a = 20 HNO3 + bHBr → 2 b Br2 + …H2O + …NO HNO3 + bHBr → 2 b Br2 + …H2O + 1NO HNO3 + bHBr → 2 b Br2 + 2H2O + 1NO Balancing the number of H: b + 1 = 4 b = 3  C 5 To calculate the relative molecular mass, which is the average mass of one molecule of the compound, relative to 1 12 the mass of 12C, the molecular formula which shows the actual number of each atom is needed and not the empirical formula which shows the simplest ratio. Since the average mass of one molecule of the compound is required, the relative atomic masses of the elements in the compound, which is averaged over the isotopic distribution, is needed and not the exact relative isotopic masses.  C 6 X is a 3s orbital and Y is a 3p orbital. For the same principal quantum shell, energy of the orbitals: ns < np < nd < nf There are 1 ns orbital and 3 np orbitals for each principal quantum number n  2.  A 7 1 ✓: r 60.1 69 39.9 71 69.8100A  + = 2 ✓: The atomic number of Ga is 31, hence both isotopes have 31 protons. 3 ✓: 69Ga has 69 nucleons (neutrons + protons), while 71Ga has 71.  A 8  A 9 pH = –lg [H+] = 1.4 1.4 3H 10 0.0398 mol dm+ − − == H 0.0398 4 0.159 moln + =  = H2X → 2H+ + X2– 2HX H 1 0.0796 mol2nn +==  B 10 Halogen exists as non-polar diatomic X 2 molecules held together by instantaneous dipole-induced dipole attractions. Volatility, i.e. boiling, involves overcoming the intermolecular forces of attraction and not breaking of the X–X bond. Instantaneous dipole -induced dipole attractions get stronger from F2 to I2 due to the increasingly number of electrons in the X2 molecules and hence higher polarisability of the electron cloud.  C 11 A ✓: Both Al2O3 and SiO2 are insoluble in water, hence the pH are both 7. B : Na2O reacts exothermically with water to give NaOH. C : Both Al2O3 and SiO2 are insoluble in water, hence no reaction. D : Only Na2O is fully soluble in water to give a clear solution of NaOH. MgO is only sparingly soluble, while Al2O3 and SiO 2 are insoluble in water, resulting in a clear solution with undissolved solid.  A 12 114Fl is in Group 14, with 4 valence electrons. Hence, F l is able to loss all 4 valence electrons to give the F l4+ cation, with the highest oxidation state of +4. Thus, the formula of the oxide is Fl4+(O2–)2.  D 13 1 ✓: Group 13, with 3 valence electrons 2 : Element has electronic configuration of 1s22s22p63s23p3  Group 15 3 : 1s22s1  Group 1 4 ✓: Element has electronic configuration of 1s22s22p63s23p1  Group 13  B 14 A : KOH should be soluble like NaOH. B ✓: Al reacts with Cl2 to give solid AlCl3. Ga should react similarly. C : GeO2 should have a giant molecular structure, similar to SiO2. D : AsCl3 should have a simple molecular structure like PCl3.  B 15 A : Oxidation state of H is +1, S is –2. H2 is oxidised to H2S. B ✓: Contains the H– ion where the oxidation state of H is –1. H2 is reduced to H–. C : Oxidation state of H is +1, N is –3. H2 is oxidised to NH3. D : Oxidation state of H is +1, P is –3. H2 is oxidised to PH3.  B 16 CH4 + 2O2 → CO2 + 2H2O CH4 + 3 2 O2 → CO + 2H2O Adding up, 2CH4 + 7 2 O2 → CO + CO2 + 4H2O 4CH 1.0 0.0625 mol12.0 1.0 4n == + 2O 7 0.0625 0.1094 mol22n =  = 2 3 O 0.1094 24 2.63 dmV =  =  B 17 fH ( )methanol : C + 2H2 + 1 2 O2 → CH3OH ----------- (1) fH ( )CO : C + 1 2 O2 → CO ------------------------- (2) cH ( )ethanol : C2H5OH + 3O2 → 2CO2 + 3H2O ---- (3) cH ( )CO : CO + 1 2 O2 → CO2 --------------------- (4) (2) + (4): C + O2 → CO2 cH ( ) fC H= ( ) cCO H+ ( )CO  A 18 L.E. qq rr +− +−  + |L.E.| is greater for M 2+X2– (Group 2 and Group 16) than M+X– (Group 1 and Group 17) |L.E.| is greater when r– is smaller (top of Group 16)  D

19 A ✓: Lattice energy is the energy released when one mole of a solid ionic lattice is formed from its constituent gaseous ions. B : The ions should be in the gaseous state. C : One mole of FeC l3(s) instead of two and not formed from the elements. D : Opposite of lattice energy of Na2O.  A 20 Since X remains unchanged and it speeds up the rate of reaction, X is a catalyst. Catalyst increases the rate of reaction by providing an alternate pathway with lower activation energy. Activation energy is the difference between the energy of reactants and the highest point in the energy profile diagram.  C 21 1 : The average energy (kinetic energy) of the molecules depends only on the temperature, which is constant in this case. 2 : The Boltzmann distribution depends on the mass of the particles and the temperature, which are both constant in this case. 3 ✓: A catalyst, e.g. enzyme, speeds up a reaction by providing an alternative pathway with lower activation energy.  D 22   31 3 25 1 rate 0.00276 mol dm s N O 0.040 mol dm 0.0690 s k −− − − == = Since the [N2O5] is halved after every 10 s, the half-life of the reaction is 10 s.  C 23 From expt 1 and 2, when [Z] is doubled, the rate is doubled, hence first order w.r.t. Z. From expt 1 and 3, when [X] is doubled, the rate is doubled, hence first order w.r.t. X. From expt 3 and 4, when [Y] is doubled, the rate remains constant, hence zero order w.r.t. Y. Thus, rate = k[X]1[Y]0[Z]1 = k[X][Z]  D 24       2 2 c 2 2 2 13 NOBr 0.0049 0.12 0.13NO Br 0.0128 mol dm K − ==  =  C 25 Since poly(ethene) is non-polar, the most likely type of bonding between a polymer of methyl cyanoacrylate and poly(ethene) is instantaneous dipole-induced dipole attractions.  C 26 Thermoplastics has low melting points and becomes mouldable at a certain elevated temperature and solidifies upon cooling. Hence, they can be recycled. Thernosets generally do not melt when heated, but typically decompose at high temperature and do not reform upon cooling. Hence, they cannot be recycled. Thermosets are generally stronger than thermoplastics due to the 3-dimensional network of bonds (crosslinking) and keep their shape as strong covalent bonds between polymer chains cannot be broken easily.  A 27  D 28 A ✓: C5H8O3 B ✓: C5H8O3 C ✓: C5H8O3 D : C5H10O3  D 29 1 ✓: 2 : 3 ✓:  B 30 Each –CH2OH → –CO2H, –2H and +1O Four –CH2OH  –8H and +4O Hence, C5H12O4 → C5H4O8  A