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B 33 B 33 A 33 D33 D 33 C33 Positive marking from question 2.1.3 +5+x=─3 3 OR x = ─ 8 nC 2.5 Copyright reserved = 5 x 1010 3 electrons ܳ ݊ൌ ݁ ͺିͲͳݔଽ ൌ 3 ͳǡିͲͳݔଵଽ Q3 3 2.4 = ( + 5 + (─ 11) 2 Q = ─ 3 nC3 ܳଵ ܳଶ ݂ܳܽ ݎ݁ݐൌ ʹ ─11 ─ x = ─ 33 x = ─ 8 nC [10] (3) (1) (2) (2) Please turn over The net/total charge of an isolated system remains constant during any physical process33 2.3 They attract each other3because they have opposite charges. 3 2. 2 (2) [12] June 2022 Common Test 2.1 QUESTION 2 1.1 1.2 1.3 1.4 1.5 1.6 QUESTION 1 2 NSC- Grade 11 ଼ି 3 NSC- Grade 11 ݇ܳଵ ܳଶ ݎଶ 3 9 163 = 9 x10 x Q Q = 4,22 x10-5 C3 ܨൌ 2 3 3 Marking Rubric : Triangle of forces Criteria Mark allocation Forces correctly drawn and 3x1=3 labelled angle correctly shown 1 NB : Ignore the sizes of the force vectors. F / Fattraction / Felectrostatic FTension / T 60 0 The net force acting on sphere B is zero. Copyright reserved 4.1.2 4.1.1 3 ǡହି Positive marking from question 3.3 QUESTION 4 3.4 = = 16 3 (Accept range 15-17) Please turn over (4) FTension /T 300 F / Fattraction / Felectrostatic (1) (4) [9] (2) (1) (2) June 2022 Common Test QUESTION 3 3.1 The magnitude of the electrostatic force exerted by one point charge (Q1) on another point charge (Q2 ) is directly proportional to the product of the magnitudes of the charges 3and inversely proportional to the square of the distance (r) between them 3. 3.2 They are directly proportional 3 ௱ி 3.3 Gradient = భ ௱ మ Physical Sciences Fgravity / Fg / W Physical Sciences Fgravity / Fg / W 4 NSC- Grade 11 ி௩௧௬ ݇ܳ ݎଶ 3 3 ENet = 45000 ─ 33 7503 = 11 250 NC-1 3 Accept : 33 750 ─ 45 000 3 = ─ 11 250 NC-13(Ignore sign of answer) = 9x109 x 2 x 10-9 (0,020)2 = 45000 NC-1 ܧሺܳJܴሻ ൌ = 9 x109 x 6 x 10-9 (0,040)2 = 33750 NC-1 ݇ܳ ܧሺܲJܴሻ ൌ ଶ ݎ Please turn over (5) (3) Marking Rubric : Sketch of Electric field Criteria Mark allocation Correct direction of field lines 1 Shape of electric field 1 No field lines crossing each other/ 1 or in the sphere NB: if shape incorrect 0/3 (2) (5) June 2022 Common Test The electric field at a point is the electrostatic force experienced per unit positive charge placed at that point 33 ( 2 or 0) Ͳǡʹͷ Ͳ 3 ൌ 3 ݃ܨ ݃ܨൌ Ͳǡͳͷͻܰ3 ߠ ൌ ி ݇ܳଵ ܳଶ ݎଶ ሺͻ ൈ ͳͲଽ ሻሺ͵ǡʹ ൈ ͳͲି ሻሺ ൈ ͳͲି଼ ሻ3 ൌ ሺͲǡͲʹͷሻଶ 3 ൌ 0,2765 N ܨൌ Copyright reserved 4.2.3 4.2.2 4.2 4.2.1 4.1.3 Physical Sciences (Positive marking from question 4.2.3) F = E·Q3 = (11 250)(1,6x10-19) 3 = 1,8 x 10-15 N3 4,76 4,76 0,68 7 14 No of moles Divide by smallest Ratio 9 18 6,16 0,68 6,16 6,16 1 H 6,16g 1 2 0,68 0,68 0,68 9,52 14 M = 14(12) + 18(1) + 2(14) + 5(16) = 294 g.mol-1 Empirical Formula: C14 H18 N2 O5 9 57,14 12 C 57,14g n = m/M Mass Simplest ratio of elements in a compound. 99 Copyright reserved 6.3 6.2 6.1 N 9,52g 9 [8] (1) (5) 9 X2 9 9 9 (2) (2) [8] (4) Please turn over 2,5 5 1,70 0,68 1,70 27,18 16 O 27,18g Water. 3 Higher boiling point. 3 5.3 QUESTION 6 Ammonia has ONE site for hydrogen bonding 3while water has TWO sites for hydrogen bonding.3 Therefore, force of attraction between water molecules are stronger than that of ammonia molecules. 3 Thus more energy is needed to overcome intermolecular forces in water, 3 therefore higher boiling point. 5.2 (2) [23] (3) June 2022 Common Test The temperature at which the vapour pressure of a substance equals atmospheric pressure. 33 5 NSC- Grade 11 5.1 QUESTION 5 4.2.4 Physical Sciences 6 NSC- Grade 11 : NO(g) : 1 0,01 = = x 75% 3 0,48 0,64 100 1 actual yield theoretical yield Mass of O2 = nM = (0,02)(32) 3 = 0,64g 3 Positive marking from 7.3 Percentage yield = O2 (g) 1 0,02 = 0,75 30 = 0,03mol3 n = m/M NO 3 x 100 1 = nNO + nO2 + nNO2 = 0,01 3 + 0,02 + 0,02 3 = 0,05 mol 3 O3(g) 1 0 Total moles of gas at end Ratio Moles at end n O3 : nO2 1 : 1 3 Therefore nO2 = n O3 = 0,02mol3 Therefore O3 is limiting3 = 0,8 48 = 0,02mol 3 n = m/M 3 O3 : NO2(g) 1 0,02 Please turn over [15] (4) (3) (2) (4) (2) June 2022 Common Test The reactant that is completely used up in a chemical reaction.33 Copyright reserved 7.5 7.4 7.3 7.2 7.1 QUESTION 7 Physical Sciences 7 NSC- Grade 11 = C x V = 1,0 x V 3 = 0,016 dm-3 3 % purity = Pure sample x 100 Impure sample 1 = 1,10 x 100 3 1,5 1 = 73,33% 3 m = nM = (0,011)(100) 3 = 1,10g n CaCO3 : nCO2 1 : 1 3 Therefore: n CaCO3 = 0,011mol n CO2 = m/M = 0,5 3 44 = 0,011 mol n NaHCO3 : nCO2 1 : 1 3 n = m/M 0,016 = m 3 44 m = 0,70g 3 Copyright reserved 8.2 8.1.3 n 0,016 V 8.1.2 n NaHCO3 = m/M = 1,35 3 84 = 0,016 mol n NaHCO3 : nHCI 1 : 1 3 Therefore, nHCI = 0,016 mol 3 (3) (5) (2) Please turn over TOTAL MARKS: 100 M = 40 + 12 + 3(16) = 100 g.mol-1 M = 12 + 2(16) = 44 g.mol-1 M = 23 + 1 + 12 + 3(16) = 84 g.mol-1 (5) [15] June 2022 Common Test 8.1.1 The number of moles per unit volume of a substance. 33 QUESTION 8 Physical Sciences