NJC 2022 H1 Chem Promo P1 Ans

NJC SH1 Promotional Examination 8873/01/2022 [Turn over 1 NATIONAL JUNIOR COLLEGE SH1 Promotional Examination Higher 1 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper 1 Multiple Choice Additional Materials: Optical Answer Sheet Data Booklet 8873/01 29 September 2022 40 mins READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. Write your name, subject class and registration number on the Answer Sheet in the spaces provided unless this has been done for you. There are twenty questions on this paper. Answer all questions. For each question there are four possible answers A, B, C and D. Choose the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Read the instructions on the Answer Sheet very carefully. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. The use of an approved scientific calculator is expected, where appropriate. Instructions on how to fill in the Optical Mark Sheet Shade the index number in a 5 digit format on the optical mark sheet: 2nd digit and the last 4 digits of the Registration Number. Example: Student Examples of Registration No. Shade: 2205648 25648 This document consists of 8 printed pages.

NJC SH1 Promotional Examination 8873/01/2022 2 Answer key for 2022 SH1 H1 Chemistry Promo Paper 1 1 C 11 D 2 B 12 B 3 D 13 A 4 B 14 D 5 A 15 A 6 B 16 C 7 D 17 A 8 C 18 C 9 D 19 C 10 C 20 A 1 How many hydrogen atoms are present in 4.0 g of methane? (L = Avogadro constant) A L 16 B L 4 C L D 4L Ans: C Amount of methane, CH4 = 4.0 16.0 = 0.25 mol Amount of H atoms in CH4 = 4 × 0.25 = 1 mol No. of H atoms = 1 × L = L 2 The table shows the relative abundance of a sample of naturally occurring isotopes of zinc. isotope relative abundance Zn30 64 10 Zn30 65 8 Zn30 67 2 Zn30 68 1 What is the relative atomic mass of this sample of zinc? A 64.0 B 64.9 C 68.1 D 72.2 Ans: B Ar = 64×10 + 65×8 +67×2+68×1 21 = 64.9

NJC SH1 Promotional Examination 8873/01/2022 [Turn over 3 3 Analysis of a mixture of two sulfur-containing gases show that H2S and CS2 are present in a 3 : 1 mole ratio. This mixture is burned in excess oxygen to give CO2 and SO2 gas. What is the mole ratio of the gases CO2 : SO2 obtained after complete combustion? A 1 : 2 B 1 : 3 C 1 : 4 D 1 : 5 Ans: D Let amount of H2S be 3x mol and CS2 be x mol, H2S + 3 2O2 → SO2 + H2O 3x 3x Amount of SO2 produced by H2S = 3x mol CS2 + 3O2 → 2SO2 + CO2 x 2x x Amount of SO2 produced by CS2 = 2x mol, Amount of CO2 produced = x mol Hence, mole ratio of CO 2 : SO2 obtained after complete combustion is x : (3x + 2x) 1 : 5 4 2 moles of an oxidising agent, XO4−, in the presence of excess acid, oxidised 96.0 dm3 of nitrogen dioxide gas to NO3− at room temperature and pressure. What is the number of moles of electrons accepted by one mole of XO4−? A 1 B 2 C 3 D 4 Ans: B Amount of NO2 at r.t.p = 96.0 24.0 = 4 mol [O]: NO2 + H2O → NO3− + 2H+ + e− (from Data Booklet) 4 mol of NO2 donates 4 mol of e− to 2 mol of XO4− Hence, 1 mol of XO4− gains 2 mol of e−

NJC SH1 Promotional Examination 8873/01/2022 4 5 The first six ionisation energies of an element, Y, in kJ mol‒1 are shown. 1st 2nd 3rd 4th 5th 6th Ionisation Energy / kJ mol−1 738 1451 7733 10543 13630 18020 Y forms an oxide by heating Y with oxygen gas. What is the molecular formula of the oxide of Y formed? A YO B YO2 C Y2O D Y2O3 Ans: A The sharp increase from the second to the third ionisation energy indicates that the third most loosely held electron is removed from an inner principal quantum shell, hence there are 2 valence electrons, Y is in Group 2. Therefore, the formula of its oxide is YO. 6 The radioactive isotope Ra223 88 decays to give Q and emits a high energy α-particle, He2 4 . No other particle is produced. Ra223 88 ⎯→ He2 4 + Q How many neutrons are present in Q? A 86 B 133 C 135 D 219 Ans: B Ra223 88 Q219 86 + He4 2 Number of neutrons in Q = 219 – 86 = 133 7 Why is the second ionisation energy of fluorine lower than that of oxygen? A There are more paired electrons in the 2p orbitals of fluorine than in oxygen. B The ionic radius of O+ is greater than F+. C Fluorine has a lower nuclear charge compared to oxygen. D All 2p orbitals of O+ are singly filled but one of the 2p orbitals of F+ is doubly filled. Ans: D The second I.E. involves the removal of the most loosely held electron in F+ and O+. F+(g): 1s22s22p4 O+(g): 1s22s22p3 With electronic configuration of 2p 4, t he paired electrons in the 2p subshell of F + experiences inter -electron repulsion and hence it is easier to be removed despite the increase in nuclear charge. Option A refers to the electronic configuration of F and O atoms, it does not explain the 2nd I.E.

NJC SH1 Promotional Examination 8873/01/2022 [Turn over 5 8 Chlorine atoms in the PCl5 molecule can be successively replaced by fluorine atoms, with the axial chlorine atoms replaced before the equatorial ones. Which of the possible molecules formed in the above reaction does not have a net dipole moment? A PClF4 B PCl2F3 C PCl3F2 D PCl4F Ans: C F P F F F Cl dipole moments do not cancel out F P F F Cl Cl dipole moments do not cancel out Cl P F F Cl Cl dipole moments cancel out Cl P Cl F Cl Cl dipole moments do not cancel out 9 Q has the following physical properties. • It is non-volatile. • It does not conduct electricity in its standard state. • It dissolves in water. What is the identity of Q? A Magnesium B Carbon dioxide C Silicon dioxide D Sodium chloride Ans: D Q is non-volatile (does not vapourise easily) eliminates carbon dioxide which has a simple covalent structure with weak instantaneous dipole-induced dipole interactions and is a gas at rtp. Q does not conduct electricity in its standard state eliminates magnesium as metals can conduct electricity. Q dissolves in wa ter eliminates silicon dioxide as it is insoluble in water due to its giant covalent structure with extensive covalent bonds.

NJC SH1 Promotional Examination 8873/01/2022 6 Sodium chloride is the only option that • is non-volatile (due to strong ionic bonds holding the giant ionic lattice), • does not conduct electricity in its standard state (no mobile ions as charge carriers in the solid state), and • dissolves in water (by forming ion-dipole interactions with H2O molecules). 10 Which statement best explains why the boiling point of butanone (80 C) is higher than that of pentane (36 C)? A The covalent bonds in the butanone molecule are stronger than those in the pentane molecule. B The relative molecular mass of butanone is higher than that of pentane. C There are permanent dipole-permanent dipole forces between butanone molecules, but not between pentane molecules. D There are hydrogen bonds between butanone molecules, but not between pentane molecules. Ans: C Butanone, CH 3COCH2CH3 (Mr = 72.0) , is polar molecule with permanent dipole – permanent dipole