2025 DHS H1 Chem Prelim Paper 1 Worked Solutions

2025 DHS Preliminary Examination H1 Chemistry 8873 Paper 1 Suggested Solutions © DHS Chemistry Unit Page 1 of 4 Answer Key 1 2 3 4 5 6 7 8 9 10 D D B B D A C D C B 11 12 13 14 15 16 17 18 19 20 D B A A A B C A C C 21 22 23 24 25 26 27 28 29 30 B D A B A D C D B B 1 D Since the gas is diatomic, the number of molecules of this gas in the 5 g sample = (3.76 x 1022)  2 = 1.88 x 1022 Therefore, the number of moles of this gas = (1.88 x 1022)  (6.02 x 1023) = 0.03123 Thus, Mr of the gas = 5  (0.03123) = 160 2 D  A Since both particles have a positive angle of deflection, they are deflected in the same direction towards the same plate. Similar to 4He2+, X must be positively charged.  B The beam of positively charged particles X travels in a curved path towards the negatively charged plate.  C Charge to mass ratio of 12C+ = + 1 12 Charge to mass ratio of 4He2+ = + 2 4 = + 1 2 Since angle of deflection is proportional to charge to mass ratio, the angle of deflection of 12C+ is expected to be 1 6(1) = +0.17  D None of the above options are correct. 3 B  A Element with el ectronic c onfiguration of 1s22s22p63s23p1 is Al. Al can lose three electrons to form Al3+.  B Element with el ectronic c onfiguration of 1s22s22p63s23p3 is P. P can gain three electrons to form P3–.  C Element with el ectronic c onfiguration of 1s22s22p63s23p63d14s2 is Sc. Sc can lose three electrons to form Sc3+.  D Element with el ectronic c onfiguration of 1s22s22p63s23p63d74s2 is Co. Co tends to lose electrons to form simple ions such as Co2+ or Co3+. 4 B 1. Determine the number of neutrons in Sr (find proton number of Sr using Data Booklet.) ⇒ 84 – 38 = 46 2. Find number of electrons present in Sr 2+ (using proton number of Sr given in Data Booklet.) ⇒ 38 – 2 = 36 ⇒ W has 36 electrons 3. Find number of protons present in W W: 36 protons 4. Calculate the nucleon number of W W: 36 + 46 = 82 5 D  A Electronic c onfiguration of Cr atom is 1s22s22p63s23p63d54s1. There are six unpaired electrons.  B Electronic c onfiguration of C u+ ion is 1s22s22p63s23p63d10. There are no unpaired electrons.  C There are no unpaired electrons.  D There is one unpaired electron on the N atom in the NO2 molecule. 6 A (1 and 2 only)  1 C=O and C=S are polar covalent bonds due to the difference in electronegativity between the atoms.  2 The dipole moments of the polar bonds are equal and opposite in O=C=O and S=C=S molecules so they cancel out and the molecules are non -polar. There exists only instantaneous dipole –induced dipole (id –id) interactions between these non –polar molecules. As O is more electronegative than S, C=O bond is more polar than C=S bond and there is a net dipole moment in O=C=S. Hence, COS molecule is polar and it has both id –id and permanent dipole –permanent dipole interactions between molecules.  3 O=C=S has a smaller, less polarisable electron cloud than S=C=S. Hence, O=C=S has weaker id-id interactions between molecules than S=C=S. 7 C  A Ice has a simple molecular structure with hydrogen bonds between water molecules. There are covalent bonds between atoms in the water molecules.  B Iodine, I2, has a simple molecular structure and is a non -polar molecule. There exists instantaneous dipole –induced dipole (id –id) interactions between molecules and covalent bond between iodine atoms in a molecule.  C Sodium nitrate, NaNO 3, has a giant ionic structure with ionic bonds between oppositely charged Na + and NO 3− ions. Within the NO 3− ion, there are covalent bonds between N and O atoms.  D Manganese has a giant metallic structure with metallic bonds between the positively charged metal cations and sea of delocalised electrons. S H H N S O O

Dunman High School 2025 DHS Preliminary Examination – H1 Chemistry 8873/01 Solutions © DHS Chemistry Unit Page 2 of 4 8 D  A Both CH3CH2CH2NH2 and (CH3)3N are polar molecules.  B Both CH3CH2CH2NH2 and (CH3)3N are chain isomers. Hence they have the same number of electrons.  C CH3CH2CH2NH2 is a straight chain molecule while (CH3)3N is a branched chain molecule. Hence CH3CH2CH2NH2 has a greater surface area of interaction. However, this explanation is only used to explain the difference in boiling points when the predominant intermolecular forces of interaction for both molecules is id- id interactions .  D The predominant forces of attraction for CH3CH2CH2NH2 is intermolecular hydrogen bonding while that for (CH3)3N is permanent dipole-permanent dipole interactions. Hence more energy is required to overcome the hydrogen bonding between CH3CH2CH2NH2. 9 C Volume of gas before burning = 60 cm3 Volume of gas used = 10 + 30 = 40 cm3 Volume of gas produced after burning = 30 cm3 Volume of gas remaining = 60 – 40 + 30 = 50 cm3 CO2 and SO2 are both acidic gases hence both will react with aq NaOH. Hence vol of gas after adding aq NaOH = 50 – 30 = 20 cm3 10 B Oxidation half-equation: Fe2+ → Fe3+ + e– Fe2+ ≡ e– amount of electrons transferred = 30.0 1000  0.200 = 0.00600 mol amount of electrons amount of nitrate ions = 0.00600 0.00200 = 3 1 The oxidation state of nitrogen decreases by 3 from +5 in NO3– to +2.  A The oxidation state of N in N2 is 0.  B The oxidation state of N in NO is +2.  C The oxidation state of N in NO2– is +3.  D The oxidation state of N in NO2 is +4. 11 D Reduction half-equation (from Data Booklet): Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O Combine both half-equations by balancing the number of electrons transferred: ( Cr2O72– + 14H+ + 6e– → 2Cr3+ + 7H2O ) x 2 ( CH3CH2OH + H2O → CH3COOH + 4H+ + 4e– ) x 3 Overall balanced redox equation: 2Cr2O72– + 16H+ + 3CH3CH2OH → 4Cr3+ + 11H2O + 3CH3COOH 2Cr2O72– ≡ 3CH3CH2OH 1 mol of Cr2O72– reacts with 1.5 mol of CH3CH2OH 12 B (2 and 3 only) C(s) + O2(g) CO2(g) CO(g) + O2(g) 1 2 DH DH1 DH2 DH1 = the standard enthalpy change of formation of carbon dioxide DH2 = the standard enthalpy change of combustion of carbon monoxide By Hess’ Law, DH = DH1 − DH2 so values of DH1 and DH2 are needed to calculate DH. 13 A CH H H H + O H H C O H H+ 3 DH = [4(410) + 2(460)] – [1077 + 3(436)] = +175 kJ mol−1 14 A  A 1 mol of PCl5(s) is formed from its constituent elements, P(s) and Cl2(g).  B The correct equation is H+(aq) + OH−(aq) → H2O(l)  C The correct equation is 2Na+(g) + O2−(g) → Na2O(s)  D The correct equation is Cl2(g) → 2Cl(g) 15 A Removal of aspirin x (100%) t1/2 → x 2 (50%) t1/2 → x 4 (25%) t1/2 → x 8 (12.5%) 3 t1/2 = 6 h t1/2 = 2 h = 120 min 16 B rate = k[BrO3–][Br–][H+]2 units of rate = units of k x (units of concentration)4 units of k = units of rate  (units of concentration)4 = 𝑚𝑜𝑙 𝑑𝑚−3 𝑠−1 (𝑚𝑜𝑙 𝑑𝑚−3)4 = 𝑠−1 (𝑚𝑜𝑙 𝑑𝑚−3)3 = 𝑚𝑜𝑙−3 𝑑𝑚9 𝑠−1

Dunman High School 2025 DHS Preliminary Examination – H1 Chemistry 8873/01 Solutions © DHS Chemistry Unit Page 3 of 4 17 C  A Addition of catalyst does not shift the position of equilibrium as it increases the rate of both forward and backward reaction to the same extent. Kc remains unchanged as it is temperature dependent.  B Addition of ammonia gas increases its concentration. The position of equilibrium shifts left to remove the added ammonia gas. Kc remains unchanged as it is temperature dependent.  C Decrease in temperature will shift the position of equilibrium to the right to favour exothermic reaction to release heat. Hence, there is an increase in [products] and decrease in [reactants]. Kc will increase.  D Increasing the pressure of the vessel causes the position of equilibrium to shift right to decrease the n