2026 H2 Timed Practice Paper 1 MCQ Combined + answer (finalised)

1 NJC SH2 Timed Practice 9476/1/26 [Turn over NATIONAL JUNIOR COLLEGE SH2 TIMED PRACTICE Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper A Multiple Choice Additional Materials: Optical Answer Sheet Data Booklet 9476/01 16 July 2026 1 hour READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, 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 thirty 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. A Data Booklet is provided. 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: 2505648 55648 This document consists of 14 printed pages.

2 NJC SH2 Timed Practice 9476/1/26 1 Use of the Data Booklet is relevant to this question. Which sample of gas contains twice the number of atoms as 4 g of helium gas, He? A 4 g of hydrogen, H2 B 8 g of methane, CH4 C 12 g of steam, H2O D 22 g of carbon dioxide, CO2 Answer: C Molar mass / g mol⁻¹ Moles No. of atoms per molecule total moles of atoms 4 g of He 4 4 4 = 1 1 1 Twice the no.of He atoms = 2 moles of atoms. A. 4 g H₂ 2 4 2 = 2 2 2 × 2 = 4 ❌ B. 8 g CH₄ 15 8 16 = 0.5 5 0.5 × 5 = 2.5 ❌ C. 12 g H₂O 18 12 18 = 2 3 3 2 3 × 3 = 2 ✅ D. 22 g CO₂ 12 + 32 = 44 22 44 = 0.5 3 0.5 × 3 = 1.5 ❌

3 NJC SH2 Timed Practice 9476/1/26 [Turn over 2 The isotope of an element has a nucleon number of 112. It forms an ion with a charge of +2 with the electron configuration [Kr]4𝑑10. How many protons, neutrons, and electrons are present in this ion? A 46 protons, 66 neutrons, 44 electrons B 46 protons, 66 neutrons, 46 electrons C 48 protons, 64 neutrons, 46 electrons D 48 protons, 64 neutrons, 44 electrons Answer: C [Kr] = 36 electrons, plus 4𝑑10 = 10 electrons → ion has 46 electrons. • Charge +2 → neutral atom has 48 electrons → atomic number (proton no.) 48 (cadmium). • Nucleon number 112 → neutrons = 112 − 48 = 64. • Ion electrons = 46 (as above). 3 An element R consists of four isotopes. The table below shows the percentage abundance of the isotopes. Relative isotopic mass Percentage abundance /% 90.0 50.0 91.0 10.0 92.0 20.0 94.0 20.0 What is the relative atomic mass of R? A 91.00 B 91.30 C 91.75 D 92.00 Answer: B Relative atomic mass of R = 90 ( 50 100) + 91 ( 10 100) + 92 ( 20 100) + 94 ( 20 100) = 91.30

4 NJC SH2 Timed Practice 9476/1/26 4 An element X has the following successive ionisation energies. 1st 2nd 3rd 4th 5th Ionisation energy /kJ mol⁻1 786 1580 3230 4360 16000 Given that t he oxide of X is acidic and is insoluble in water, w hich statement about X is correct? A X is a gas at room temperature. B X forms a chloride with a formula XCl5. C The oxide of X has a linear shape with no dipole moment. D The oxide of X has a giant covalent lattice with a high melting point. Answer D is correct. • From IE data: A large jump after 4th IE (4360 → 16000) indicates that X has 4 valence electrons → Group 14. Hence option B incorrect where X is in Group 15. oxide data strongly points to Silicon (Si), whose oxide SiO₂ is: - Acidic (reacts with bases to form silicates) - Insoluble in water - Giant covalent lattice with high melting point • Option A is incorrect as silicon is a solid at room temperature (metalloid, melting point 1414°C). • Option C is incorrect as the oxide of silicon is SiO₂. - SiO₂ does not exist as discrete molecules. SiO₂ has a giant covalent lattice. - Therefore, it does not have a "linear shape" or "dipole moment". - (CO₂ is linear and non-polar, but CO₂ is soluble, so it's excluded by the oxide data.)

5 NJC SH2 Timed Practice 9476/1/26 [Turn over 5 Which option correctly describes the shape and polarity of the species? species shape polarity A AlCl3 trigonal planar polar B SiF4 square planar non-polar C BrF3 trigonal pyramidal polar D BeCl2 linear non-polar Answer: D species shape polarity A AlCl3 trigonal planar (3 b.p.) non-polar B SiF4 tetrahedral (4 b.p.) non-polar C BrF3 T shaped (3 b.p. + 2 l.p.) polar D BeCl2 Linear (2 b.p.) non-polar 6 When 10 cm3 of a gaseous hydrocarbon was sparked with excess oxygen gas and cooled to room temperature, the gaseous mixture contracted by 30 cm3. When the residual gas was passed through aqueous potassium hydroxide, there was a further contraction of 40 cm3. What is the hydrocarbon? A C3H8 B C4H8 C C4H6 D C4H10 Answer: B CxHᵧ + (𝑥 + 𝑦 4 ) O2 → x CO2 + 𝑦 2 H2O Second contraction (40 cm³) = volume of CO₂ absorbed by KOH CO₂ volume = 40 cm³ = (10x) cm³ of hydrocarbon x = 4 First contraction (30 cm³) = volume decrease after burning and cooling (H2O, water condenses) [V(CxHᵧ ) + V(O2)] – [V(CO2) + 𝑉(H2O)] = 30

6 NJC SH2 Timed Practice 9476/1/26 [10 + V(O2)] – [40 + 0)] = 30 V(O2) = 60 hence 𝑥 + 𝑦 4 = 6 4 + 𝑦 4 = 6 ⇒ 𝒚 = 𝟖 Formula: C₄H8 7 Which diagram correctly describes the behaviour of a fixed mass of an ideal gas? (T is measured in K.) A B C D Answer: D pV= nRT for a fixed mass of an ideal gas, number of moles of gas is constant, n is constant A B C D p = 𝑛𝑅𝑇 𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉 since constant n, constant R, constant T → nRT = constant pV= nRT = constant since constant n, constant R, constant T → nRT = constant pV = nRT 𝑉 = 𝑛𝑅𝑇 𝑃 = 𝑛𝑅 𝑃 𝑇 since constant n, constant R, constant P 𝑉 ∝ 𝑇 Correct graph Correct graph Correct graph Answer

7 NJC SH2 Timed Practice 9476/1/26 [Turn over 8 Liquefaction can be defined as a process that turns a gas into a liquid by increasing the pressure at a constant temperature. In the liquefaction of CH4 and NH3, the pressure needed for NH3 is less than CH4. Which reason best explains this observation? A NH3 has stronger intermolecular forces of attraction than CH4. B NH3 has lower bond energy than CH4. C NH3 molecules are bigger than CH4 molecules. D NH3 molecules possess less kinetic energy than CH4 molecules. Answer: A The process of liquefaction is making the gas molecules come closer together via pressurisation. If the intermolecular forces between the molecules are stronger, the pressure required would be less. Since NH3 has hydrogen bonds between molecules and is stronger than instantaneous dipole-induced dipoles interactions between CH4 molecules, less pressure needed for NH3 to undergo liquefaction. 9 Use of the Data Booklet is relevant to this question. Hexamine has an enthalpy change of combustion of –4288 kJ mol–1. 12.4 g of hexamine tablets were burnt to heat up 850 g of water. Given tha