NYJC 2021 Prelim 9729 P1, P2, P3 Answers

Nanyang Junior College 2021 J2 H2 Chemistry Prelim Answers. 1 J2 H2 Chemistry Prelim Answers Paper 1 Answer Key Paper 1 Worked Solutions 1 (B) 100 g of mixture contains 21.25 g of Mg. Mr MgCl2 95.3 Mg(NO3)2 148.3 Let mass of MgCl2 present in 100 g of mixture be x. x 100 - x × 24.3 + × 24.3 = 21.2595.3 148.3 x = 53 g 2 (C) n(M2On) = 0.500 u 10.00 1000 = 0.00500 mol n(KMnO4) = 0.300 u 20.00 1000 = 0.00600 mol Given the reduction half equation: MnO4 + 8H+ + 5e o Mn2+ + 4H2O n(e-) involved in the redox = 0.00600 u 5 = 0.0300 mol 0.00500 mol of M2On contain 0.0100 mol of Mn+. Hence, 0.0100 mol of Mn+ loses 0.0300 mol of e to form M6+. 1 mol of Mn+ loses 3 mol of e to form M6+. n = 3 1 B 11 C 21 A 2 C 12 A 22 A 3 C 13 C 23 D 4 D 14 D 24 C 5 D 15 B 25 C 6 B 16 B 26 A 7 A 17 C 27 B 8 D 18 A 28 C 9 D 19 B 29 D 10 B 20 B 30 C

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 2 3 (C) 1st 2nd 3rd 4th 5th 6th 660 1267 2218 3313 7863 9500 607 951 1095 4550 1637 Largest increase from 4th I.E.to 5th I.E. indicates that the 5th electron is removed from an inner electron shell. Element Q is a Group 14 element with 4 valence electrons and reacts with chlorine to form QCl4. Other statements cannot be inferred from the data given (while option D is incorrect). Since Q is ns2np2, element preceding it is ns2np1, hence it should follow the general increasing trend). 4 (D) Statement 1 is incorrect. van der Waals radius of chlorine is smaller than that of argon. This is because Cl2 molecules (Mr = 71.0) have stronger instantaneous dipole-induced dipole (id-id) interactions than id-id forces between Argon atoms (Ar = 39.9). Statement 2 is incorrect. CH3CH2CHO do not have hydrogen bonds between its molecules. Statement 3 is correct. C-C bond length in diamond is longer and weaker than graphite due to the smaller s character of the sp3−sp3 overlap in diamond as compared to the sp2−sp2 overlap in graphite. 5 (D) Step 1. Determine R under standard conditions: R = p nT V= 5- 3 1 × 273 10 × 22.7 × 10 Step 2. Substitute R into the equation P =RT rM U = 5- 3 8 × (300 + 273) 10 × 22.7 × 10 × 2.0 1 × 273 = 573 × 4 × 22.7 × 100 273 6 (B) NH3(g) + HCl(g) o NH4Cl(s) ½N2(g) + 2H2(g) + ½Cl2(g) ∆Hʅf (HCl) + (–92/2) + (–176) = –629/2 ∆Hʅf (HCl) = –92.5 kJ mol1 –176 –629/2 ∆Hʅf (HCl) + (–92/2) Alternative: ∆Hʅf (NH4Cl) = –629/2 ∆Hʅf (NH3) = –92/2 –176 NH3(g) + HCl(g) o NH4Cl(s) –176 = –629/2 – (–92/2) – ∆Hʅf (HCl) ∆Hʅf (HCl) = –92.5 kJ mol1

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 3 7 (A) Statement 1 is correct: ∆HrƟ = ∑n∆HfƟ (products) − ∑m∆HfƟ (reactants) = −1273 – [6×(−394) + 6×(−286)] = +2807 kJ mol−1 Statement 2 is correct. The photosynthesis process has no change in number of particles of gases, but there is a change of state (liquid water in the reactant to solid C6H12O6 in the product). There is a decrease in overall entropy and thus, ∆S has a negative sign (e.g. less disordered). Statement 3 is correct. Since ∆G = ∆H (+ve) – T∆S (–ve), the ∆G for the reaction will always be positive at all temperatures as ∆H and – T∆S is always positive. 8 (D) t1/2 = ln2/k = ln2/7.70 x 104 = 900 s = 15 min no. of half-lives = 80 / 15 = 5.33 n t 0 A1= A2 §· ¨¸©¹ §· ¨¸©¹ 5.33 0 0.0315 1= A2 Conc = 1.27 mol dm3 9 (D) Statement 1 is correct. Comparing Expt 1 and 3, when [HCl] is halved (0.20/0.10), the rate is halved (0.192/0.096). Hence it is a first order with respect to HCl. Comparing Expt 2 and 3, when [methyl propanoate] is increased 1.5 times (0.15/0.10), the rate increased 1.5 times (0.144/0.096). Hence it is a first order with respect to methyl propanoate. Since rate = k[methyl propanoate][HCl], when [methyl propanoate] doubles and [HCl] increases 1.5 times from expt 2 to expt 4, the rate = 0.144 u 2 u 1.5 = 0.432. Statement 2 is incorrect as units are incorrect. Using Expt 1: 0.192 = k(0.10)(0.20) k = 9.6 mol1 dm3 min1 Statement 3 is correct. rate = k[methyl propanoate][HCl] rate = k’[methyl propanoate] where k’ = k[HCl]. Since HCl is a catalyst which gets regenerated in the reaction, the [HCl] remains constant. t1/2 = ln2/k’ = ln2/k[HCl] Since [HCl] is decreased 0.15 3 = 0.20 4 , the time taken will be 4 × 63 = 8 min.

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 4 10 (B) N2O4 ⇌ 2NO2 Initial amt/ mol x 0 Change/ mol 0.5x +1.0x Eqm amt/ mol 0.5x 1.0x Total amount = 1.5x mol 24NOP = (0.5x 1.5x) u 1 = 1 3 atm 2NOP = (x 1.5x) u 1 = 2 3 atm Kp for the reverse reaction =

24 2 2 NO NO P P = 2 1()3 23 = 3 4atm1 11 (C) Given [OH] = 107 12 (A) A buffer solution’s pH will not change significantly when some alkali is added. Statement 1 is correct. When 0.003 mol of CH3COOH (excess) was mixed with 0.002 mol of KOH (limiting), 0.002 mol of CH3COONa+ (salt) and 0.001 mol of CH3COOH (remaining) will be formed at equilibrium. This is a buffer solution. Statement 2 is correct. 0.0025 mol of HPO42 (conjugate base) is mixed with 0.00075 mol of H2PO4(acid). This is a buffer solution. Statement 3 is correct. 0.002 mol phenylalanine exist as a zwitterion O NH3+ O . It can act as a buffer as it has both an acidic (–NH3+) and a basic (–COO) group. 13 (C) Highest Lowest Melting point Si Al Mg P First IE P Si Mg Al 5.681 u 103 -3 -3 -3 5.681 × 10 5.65 × 105.681 × 10 +1 -35.681 × 10 1

Nanyang Junior College 2021 J2 H2 Chemistry Prelims Answers. 5 14 (D) Option A is incorrect. Down the group, the number of electron shell increases, shielding effect increases significantly, hence the outermost electron is further away and is less attracted by the nucleus. Hence first ionisation energies points of elements decreases. Option B is incorrect. With increasing proton number, cationic radius increases while charge remains the same. Charge density decreases with increasing proton number of Group 2 ions. The acidity of solutions of the chlorides decreases due to the decrease in charge density. Option C is incorrect. From MgSO4 to BaSO4, the |∆Hʅhydration| of Group 2 sulphates decreases faster than the |lattice energy| of Group 2 sulphates, hence the solubility decreases down the group. We can also infer that MgSO4 (soluble salt) while BaSO4 (insoluble salt). Option A is correct. The decomposition temperature of carbonates increases since the polarising power of Group 2 cations decreases with decreasing charge density. The C–O bond in carbonates is less weakened. 15 (B) Statement 1 is incorrect. The higher the boiling point, the lower its volatility. HF has the highest boiling point as most energy is needed to overcome the strong hydrogen bonds between HF molecules. Boiling point of HI > HBr > HCl. Down the group, the size of electron cloud of hydrogen halides increases and become more polarisable. More energy needed to overcome the stronger instantaneous dipole – induced dipole interactions between molecules. Statement 2 is incorrect. The thermal stability of the hydrogen halides decreases down the group. This is due to the decreasing strength of the H–X bond. The weaker the bond, the lesser the energy required for decomposition. Statement 3 is correct. Reducing strength of halide ions increases down the group. E{(X2/X) becomes less positive, hence X is more likely to be oxidised. 16 (B) Cl Br OH ethanolic NaOH OH * There is only 1 chiral centre (marked with *), and 1 C=C that can exhibit cis-trans isomerism. Note that the C=C in the ring cannot exhibit cis-trans isomerism. Hence, total number of stere