NJC 2025 H2 Timed Practice2 Paper A and B Answer booklet

1 NJC SH2 Timed Practice 9729/A1/25 [Turn over NATIONAL JUNIOR COLLEGE SH2 TIMED PRACTICE Higher 2 CANDIDATE NAME SUBJECT CLASS REGISTRATION NUMBER CHEMISTRY Paper A1 Multiple Choice Additional Materials: Optical Answer Sheet Data Booklet 9729/A1 30 May 2025 30 min 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: 2405648 45648 This document consists of 8 printed pages.

2 NJC SH2 Timed Practice 9729/A1/25 2025 SH2 H2 Chem Timed Practice Paper A1 Answer Key 1 D 6 D 11 C 2 C 7 D 12 C 3 C 8 D 13 D 4 B 9 C 14 D 5 B 10 D 15 C 1 The equilibrium percentage of Z varies according to varying pressures and temperatures as shown in the graph. Which row in the table shows the correct information about the equilibrium? equilibrium reaction sign of H for the forward reaction A Y(g) + X(g) 3Z(g) Positive B 4X(g) + Y(g) 2Z(g) Positive C Y(g) + X(g) 3Z(g) Negative D 4X(g) + Y(g) 2Z(g) Negative Ans: D Graph shows decreasing % of product Z with increasing temperature. As increasing temperature favours endothermic reaction, since less Z is formed, the backward reaction is endothermic, and the forward reaction is exothermic (C or D) Graph also shows increasing % of Z when pressure increases. As increasing pressure favours the side of the equation with fewer moles of gaseous particles, the RHS of the equation must have fewer gaseous moles than the LHS, so answer is D.

3 NJC SH2 Timed Practice 9729/A1/25 [Turn over 2 X decomposes on heating according to the following equation 2X(g) 2Y(g) + B(g) When 4 mol of X were put into a 2 dm3 container and heated, the equilibrium mixture contained 0.8 mol of B. What is the numerical value of the equilibrium constant, Kc? A 0.8 × 0.8 3.2 B 1.6 × 1.6 × 0.8 2.4 × 2.4 C 1.6 × 1.6 × 0.8 2.4 × 2.4 × 2 D 0.4 × 0.4 × 0.8 0.4 × 0.4 Ans: C 2X 2Y + B Initial / mol 4 0 0 Change / mol −2×0.8 +2×0.8 +0.8 Equilibrium / mol 2.4 1.6 0.8 Kc = [Y]2[B] [X]2 = 1.6 2 × 1.6 2 × 0.8 2 2.4 2 × 2.4 2 = 1.6 × 1.6 × 0.8 2.4 × 2.4×2

4 NJC SH2 Timed Practice 9729/A1/25 3 The concentration and pH of each solution, P, R and S are shown below. Solutions P, R and S could be either a monobasic acid or a monoacidic base. Which statements are correct? 1 1. R is a weak base. 2 1. P is a weak acid while S is a strong acid. 3 1. Mixing 100 cm3 of R and 100 cm3 of S produces a buffer solution. 2. 4 3. Mixing 500 cm3 of P and 500 cm3 of R produces a buffer solution. A 1 and 2 B 1 and 4 C 2 and 4 D 2 and 3 Ans: C pH = 2.4 means [H+] = 10−2.4 = 0.00398 mol dm−3 < 1.0 mol dm−3 This means that P is a monobasic weak acid. pH = 12.0 means pOH = 2 and [OH−] = 10−2 = 0.01 mol dm−3 = [R] so R is a monoacidic strong base. Statement 1 is wrong. pH = 3.0 means [H+] = 10−3 = 0.001 mol dm−3 = [S] This means that S is a monobasic strong acid. Statement 2 is correct. Mixing 100 cm 3 of 0.01 mol dm−3 R (0.001 mol of strong base) and 100 cm 3 of 0.001 mol dm −3 S (0.0001 mol of strong acid) produces 0.009 mol of excess strong base and 0.001 mol of neutral salt which is not a buffer solution). Statement 3 is wrong. Mixing 500 cm3 of 1.0 mol dm−3 P (0.5 mol of weak acid) and 500 cm3 of 0.001 mol dm−3 R (0.0005 mol of strong base) produces 0.4995 mol of excess weak acid and 0.0005 mol of conjugate base of weak acid which is a buffer solution. Statement 4 is correct. P 1.0 mol dm–3 pH = 2.4 R 0.010 mol dm–3 pH = 12.0 S 0.001 mol dm–3 pH = 3.0

5 NJC SH2 Timed Practice 9729/A1/25 [Turn over 4 Equal volumes of 1.0  10–2 mol dm –3 aqueous BaSiF6 and 2.5  10–1 mol dm –3 aqueous CsBrO3 are mixed in a test-tube. The numerical values of the solubility products are: Ksp of Ba(BrO3)2 = 2.43  10–4 Ksp of Cs2SiF6 = 1.3  10–5 Which statement best describes what will be observed in the test tube? A No precipitation is observed. B Precipitation of Cs2SiF6 only. C Precipitation of Ba(BrO3)2 only. D Precipitation of both Cs2SiF6 and Ba(BrO3)2. Ans: B IP of Ba(BrO3)2 = [Ba2+][BrO3−]2 = (½  1.0  10–2)(½  2.5  10–1)2 = 7.81 x 10–5 < Ksp (no ppt formed) IP of Cs2SiF6 = [Cs+]2[SiF62−] = (½  2.5  10–1)2(½  1.0  10–2) = 7.81 x 10–5 > Ksp (ppt formed) 5 In the free radical substitution of 3 -methylpentane with chlorine, a mixture of products was obtained. How many constitutional isomers of mono-chlorinated products are theoretically possible? A 3 B 4 C 5 D 6 Ans: B CH3 CH2 C H CH3 CH2 CH3 CH3 CH2 C H CH3 CH2 CH2Cl CH3 CH2 C H CH3 CHCl CH3 CH3 CH2 C Cl CH3 CH2 CH3 CH3 CH2 C H CH2Cl CH2 CH3

6 NJC SH2 Timed Practice 9729/A1/25 6 Nitrobenzene is a precursor to many reagents. It can be prepared by reacting benzene with concentrated HNO3 and concentrated H2SO4. What is the role of concentrated H2SO4? A To generate NO2− for the reaction B To act as a dehydrating agent C To direct the electrophile to a specific position on the benzene ring D To act as a catalyst Ans: D Concentrated H2SO4 reacts with HNO3 to generate +NO2 (A is wrong) H2SO4 is not a substituent on the benzene ring, so it does not have any directing effect (C is wrong) As H2SO4 is regenerated and +NO2 is a strong electrophile to react with benzene with a lower E a, Concentrated H2SO4 is a catalyst. (D is correct and B is wrong) 7 Equal amounts of compounds W, X, Y and Z are added separately to four test tubes containing equal concentrations of ethanolic silver nitrate solution in a heated water bath. No precipitate forms in two of the tubes. In the two other tubes, precipitates form at different rates. Which row is correct? Compounds which do not form a precipitate Colour of the precipitate which forms the fastest A W cream B W and X yellow C X and Z white D W and Y yellow Ans: D Compound W and Y do not form a precipitate when heated with ethanolic silver nitrate. C -Cl and C-Br bonds have partial double bond character due to the lone pair electrons on C l and Br atom being delocalised into the π electron cloud of benzene. C-X bond is strengthened hence resistant to substitution. Comparing compounds X and Z, compound Z will give a yellow ppt of Ag I more readily due to the weaker C-I bond as compared to C-Cl bond in X.

7 NJC SH2 Timed Practice 9729/A1/25 [Turn over 8 Which compound is less acidic than phenol? A B C D Ans: D Acid strength increases with stability