SUGGESTED ANSWER 2022/2023 1. a) 24Mg = 23.985 u = 78.99% 25Mg = 24.986 u 26Mg = 25.983 u Assume : 25Mg= x, thus 26Mg: 21.01%-x 24.3 = (23.985 78.99) + (24.986 ) + (25.983)(21.01 − ) 100 X=10.52 25Mg= 10.52% 26Mg= 10.49% b) 2H2O + SO2 SO4 2- + 4H + +2e 2e + Br 2 2Br - 2H2O + SO2 + Br 2 SO4 2- + 2Br - + 4H + c) i. nNaOH = 0.2 M X (250/1000)L = 0.05 mol mass NaOH = 0.05 mol X 40 gmol-1 = 2.0 g ii. M1V1=M2V2 (0.2M) (250mL) = 0.1M (V2) V2 = 500 mL ( V solution) Vsolution = Vsolute, NaOH + Vwater V water needed = (500-250)mL = 250 mL
d) O3 + NO O2 + NO2 3 = 0.740 48 / = 0.670 30 / =0.0154 mol = 0.0223 mol 1 mol O3 ≡ 1 mol NO 0.0154 mol O3 ≡ 0.0154 mol NO (needed) nNO needed < nNO available Thus O3 is a limiting reactant 1 mol O3 ≡ 1 mol NO2 0.0154 mol O3 ≡ 0.0154 mol NO2 Mass NO2 = 0.0154 mol X 46 gmol-1 = 0.7084 g 2. a) n=1 n=2 n=3 n=4 n=5 n=6 Energy ↑∆E = hv↑ ↑∆E : transition from n=6 to n=1 thus, 1 ʎ = ( 1 1 2 − 1 2 2 ) n1<n2 1 ʎ = ( 1 1 2 − 1 6 2 ) = 93.76 nm
b) 13X = 1s2 2s2 2p6 3s2 3p1 l= 0; 1s2 = 2e 2s2 = 2e 3s2 = 2e Total = 6 electrons Draw the orbital when l=1 orbital is 3p x y z 3pz 3px or 3py also accepted. only draw 1 orbital because orbital 3p contain 1 electron only. 3. a) i. Lewis structure O S O octet or O S O expended octet ii. SeO2 has 2 bonding pair and 1 lone pair. The basic shape is Trigonal planar. According to VSEPR theory, the repulsion between: lone-pair bonding pair > bonding pair-bonding pair Due to the repulsion, the molecular shape become Bent/ V-shape iii. O S O Se has 2 sigma bond and 1 lone pair. The hybridisation is sp2 iv. The bond between Se---O is polar bond because oxygen is more electronegative than Se. The molecule is unsymmetrical, thus the bond dipole cannot cancel out. µ≠0. Molecules is a polar molecule. Se O O
b) Calcium atom donate 2 electrons to formed Ca2+ ion. The electrostatic attraction between positive ion (Ca2+) with delocalised electrons will form a metallic bonding and can conduct electricity. c) AlCl3 is a covalent molecule. No free moving electrons to conduct electricity. Al has 3 valence electrons. Free moving electrons/ delocalized electrons sea can conduct electricity. 4. a) Zn + HCl ZnCl2 + H2 H2 : V= 7.80 L, P + 0.980 atm T= 298.15 K = 2 + 2 PH2 = 0.980 atm – 0.0313 atm = 0.9487 atm PV = nRT n H2= 0.3025 mol 1 mol H2 ≡ 1 mol Zn nZn = 0.3025 mol mass Zn = 0.3025 mol x 65.5 g mol-1 = 19.78 g Ca 2+ Ca 2+ Ca 2+ Ca 2+ e e e e e e e e delocalised electrons sea
b) i. ii. water is denser than ice because ice has an open structure. The volume of ice is higher than volume of water. Thus, the density of ice is lower than water. 5. a) Given : Kp =8.3 x 10-3 Kp = Kc(RT)∆n ∆n = 1 = (0.08203 673.15) = 1.503 x 10-4 b) How mass HCl affected by increasing temperature? - This is an endothermic reaction. Increasing the temperature will shift the equilibrium position to the right. Thus, the mass of HCl increase. c) Predict the direction of reaction; = [] 4 [2] [2][2] 2 = (0.25) 4 (0.05) (0.7)2(0.35) 2 = 3.25 x 10-3 Qc > Kc, the nett reaction/equilibrium position will shift from right to left/backward.
6. a) i. C6H5COOH (aq) + H2O (l) H3O + (aq) + C6H5COO - (aq) [ ] i 0.1 M 0 0 [ ] change -x +x +x [ ] f 0.1-x x x = [3 + ][65−] [65−] 3.5 10−8 = 2 (0.1 − ) x = 5.916 x 10-5 [H3O + ] = 5.916 x 10-5 M, [C6H5COO- ] = 5.916 x 10-5 M, C6H5COOH] = 0.099M ii. pH = -log [H3O + ] = - log 5.916 x 10-5 = 4.228 b) i. Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) Ksp x 2x 5.61 x 10-12 = 4x3 X = 1.1193 x 10-4 M Molar solubility : 1.1193 x 10-4 M ii. Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq) when Mg(NO3) 2 added Mg(NO3) 2 (aq) Mg 2+ (aq) + 2OH - (aq) non reversible arrow because Mg(NO3) 2 is a soluble salt Mg 2+ is a common ion. According to Le Chatelier's principle, addition Mg(NO3) 2 to Mg(OH) 2 will increase the concentration of Mg 2+ ion. The equilibrium position will shift from right to left. Thus the solubility of Mg(OH) 2 decrease. Prepared by : Marziah binti Mohamad