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 proce