Physical Sciences Grade 11 P2 KwaZulu Natal June 2019 Memo

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Physical Sciences P2 Downloaded from hlayiso.com, 2019 common Test INSTRUCTIONS AND INFORMATION TO CANDIDATES ae 2. © While your name on the ANSWER BOOK Answer ALL the questions in the answer book. You may use a non-programmabie calculator. You may use appropriate mathemalical instruments. Number the answers correctly according to the numbering system used in this question paper YOU ARE ADVISED TO USE THE ATTACHED DATA SHEETS. Give brief motivations, discussions, et cetera where required. Show the formulae and substitutions in ALL calculations. Round off answers to a minimum of TWO decimal places Copynght Reserved Prease Tum Over

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Downloaded from hlayiso.com Physical Sciences P2 NSC June 2019 Common Test 1.5 2,50 mol of SO2 and 1 mol of O; are sealed in a 1 dm’ flask and allowed to react completely at STP according to the following balanced equation. 2 SO2 (g)+ O2(g) -> 2 SOs (g) The TOTAL number of moles of gas in the flask at the END of the reaction is: A 8 c D 2 3,50 2,50 0,50 (2) 1.6 The gas law that expresses the relationship between pressure and temperature of a gas is known as: A 8 c D Charle's Law Gay- Lussac Law Boyle's Law Avogadro's Law (2) 1.7 The type of intermolecular force involved when CO; is added to water is : A B Cc D Dipole- induced dipole forces London forces Covalent bonds lon — induced dipole forces (2) TOTAL : SECTION A [14] Copynght Reserved Please Turn Over

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Downloaded from hlayiso.com Physical Sciences P2 NSC June 2019 Common Test QUESTION 3 Grade 11 learners investigated the effect of intermolecular force on capillarity. They pour 100ml each of water; glycerine and nail polish remover in separate beakers. A capillary tube is inserted into each liquid and after a while, the level of liquid in the capillary tube is measured. am Capillary tube They recorded their results in a table as follows: je Liquid és Height (mm) ] | Water site (eat | aT | Glycerine _ oes 5 | [Nail polish remover 26 | 3.1 State the dependent variable in the above investigation. (1) 3.2 Which liquid displayed the greatest degree of capillarity? (1) 3.3. Explain the answer to question 3.2 above. (3) 3.4 __ Use the results in the table and arrange the liquids in order of INCREASING STRENGTH of intermolecular force that is, from the weakest intermolecular force to the strongest intermolecular force. (2) 3.5. Identify the liquid with the highest boiling point. (1) [8] Copynght Reserved Please Turn Over

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Downloaded from hlayiso.com Physical Sciences P2 NSC June 2019 Common Test QUESTION 5 A compound contains the elements carbon, hydrogen and oxygen only. It consists of 54.56% carbon and 36.36% hydrogen. The molar mass of the compound is 132 g. mol". 5.1 State the definition of empirical formula. (2) 5.2 Calculate the empirical formula of the compound. (6) 5.3 Determine the molecular formula of the compound. (2) [10} QUESTION 6 tron is recovered from iron ore (Fe2O3) in a blast furnace. The following reaction takes place. Fe203(s) + 3CO(g) — 2 Fe(s) + 3C02g) In one such reaction, 160g of impure iron ore was reacted and 63 dm’ of CO2 was produced at STP. 6.1 Write down the definition of the mole. (2) 6.2 Calculate the number of CO2 molecules that formed at STP (4) 6.3. Calculate the maximum no of moles of iron that will be formed in the above reaction. (2) 6.4 Calculate the percentage purity of the iron ore sample used. ia QUESTION 7 Industrially, vanadium metal, (V) which is used in steel alloys can be obtained by reacting vanadium pentoxide( V2Os) with calcium at high temperatures. The balanced equation for the reaction is: 5Ca+V20s5 +~5CaO +2V During an industrial process 31850 g of V2Os reacts with 2 x 10* g of Ca. 7.1 State the definition of a limiting reagent. (2) 7.2 Calculate the theoretical yield of vanadium. (6) 7.3 Calculate the percentage yield if 8,67 x 10°g of vanadium is obtained. (2) [10} Please Turn Over

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Downloaded from hlayiso.com J0AO WN] OSE%qg TABLE 3: THE PERIODIC TABLE OF ELEMENTS 5 6 7 Atomic number Atoomgetal KEY/SLEUTEL 4 9 10 © 120:«13—C—“‘“ HK 61 18 ay (yy) (vil) (vill) Electronegativity Elektronegatiwiteit_y| 29 2 Cu = 63,5 Symbol |_—Simbool Approximate Benaderde relatiewe atoommassa atomic mass 241 25 2 Crj2Mn 52| 55 26 = Fe 56 27] 28 2Co/2 Ni 59 | 59 42| 43 2 Mo|? Te 96 44 = Ru 101 45| 46 a Rhit Pd 103| 106 74 75 W)| Re 184| 186 76 Os 190 7 78 Ir| Pt 192| 195 Pr | Nd 141 | 144 61 Pm Sm 150 91 92 Pa | U 238 93 Np 94 Pu QQUENS feoeAUg S01 VOWWOD 6102 OUNr

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Downloaded from hlayiso.com Physical Sciences P2 NSC-Marking Guideline Common Test June 2019 1 PHYSICAL SCIENCES P2 (CHEMISTRY) COMMON TEST JUNE 2019 MARKING GUIDELINE NATIONAL SENIOR CERTIFICATE GRADE 11 TIME: 2 hours MARKS: 100 This marking guideline consists of 7 pages. Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Physical Sciences P2 NSC-Marking Guideline Common Test June 2019 2 QUESTION 1 1.1. C  1.2. B  1.3. A  1.4. D  1.5. C  1.6. B  1.7. A  7x2 = (14) QUESTION 2 2.1.1 A and I (2) 2.1.2. (B and F)   OR (D and F) (2 or 0) (2) 2.1.3. G  (1) 2.1.4. G (1) 2.1.5. C  (1) 2.1.6. E  (1) 2.2 H2O has hydrogen bonding and H2S has dipole- dipole forces.  The intermolecular forces in water are stronger Therefore more energy is required to break the IMF in water.  (4) 2.3. H x ● H x● C ●x H ● x H  (2) 2.4.1 NON- POLAR - CO2 is a symmetrical molecule with even distribution of electrons There is no net dipole moment/dipoles cancel out/. There is no distinct opposite positive and negative ends. (3) 2.4.2 London forces  (1) [18] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 NSC-Marking Guideline Common Test June 2019 3 QUESTION 3 3.1. Capillarity / height  (1) 3.2. Nail polish remover (1) 3.3. It has the weakest intermolecular forces /cohesive forces  molecules separate easily.  The adhesive forces are stronger than the cohesive forces/IMF  (3) 3.4. Nail polish remover; water ; glycerine  (2) 3.5 Glycerine (1) [8] QUESTION 4 4.1.1 Charle’s Law -- The volume of an enclosed gas is directly proportional to its Kelvin temperature provided the pressure is kept constant.  (3) 4.1.2. Amount/mass of gas (1) 4.1.3. volume is directly proportional to Kelvin temperature/ Vα T / volume is linearly proportional to Celsius temperature ACCEPT As temperature increases, volume increases. (1) 4.1.4. P3  (1) - 4.1.5. From pV = nRT , gradient which is V/T is equal to nR/P. Since R and n are constant,  Gradient is inversely proportional to the pressure/ higher pressure represents smaller gradient.  (3) OR From Boyles law --- at a constant temperature,  the gas with the highest volume will have the lowest pressure pV = k Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Physical Sciences P2 NSC-Marking Guideline Common Test June 2019 4 4.2. OPTION 2 OPTION 1 Change in pressure = 100-55 = 45kPa pV = nRT 100 x 103 x 10 x 10-3 = n x 8.31 x 298 pV = nRT n= 0.404mol ( original no. of 45 x 103 x 10 x 10-3 = n x 8.31 x 298 moles of oxygen in vessel) n = 0.182mol pV = nRT n = m/M [20] 55 x 103 x 10 x 10-3 = n x 8.31 x 298 0.182 = m/32 n= 0.222mol ( no of mol in vessel m = 5.824g  of O2 is lost after leak is repaired) no of moles of gas leaked NB: ACCEPT p values in kPa with V values in dm3 =0.404 – 0.222 = 0.182mol n = m/M 0.182 = m/32 m = 5.824g  of O2 is lost (8) [17] QUESTION 5 5.1 Simplest whole number ratio in which elements in a compound combine  (2) 5.2. Element Mass per 100 g n=m/M(mol) Simplest ratio C 54.56 54.56/12 = 4.547 4.547/2.2725 = 2 H 9.08 9.08/1 = 9.08 9.08/2.2725 = 4  O 36.36 36.36/16 = 2.2725 2.2725/2.2725 = 1 Empirical formula is C2H4O (6) 5.3 n = True Mr/ Empirical Mr = 132/44 = 3 Molecular formula is C6 H12 O3 (award both marks if answer correct without calculation) (2) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 NSC-Marking Guideline Common Test June 2019 5 QUESTION 6 6.1. The amount of substance having the same number of particles as there are atoms in 12g of C-12. / Amount of substance having 6,02 x 1023 elementary particles. (2) 6.2. n = V/Vm  N = 2.8125 x 6.02 x 1023 = 63/22.4  = 1.693 x 10 24 molecules of CO2 = 2.8125 mol (4) 6.3 POSITIVE MARKING FROM 6.2 Fe : CO2 2:3  nFe = 2.8125 x2/3 (4) = 1.875mol (2) 6.4. Positive marking form 6.2/6.3 CO2 : Fe2O3 OR Fe : Fe2O3 3 :1 2 :1 nFe2O3 = 2.8125/3 nFe2O3 = 1.875/2 = 0.9375 mol = 0.9375 mol mFe2O3 = nM = 0.9375 x 160 (4) = 150 g [12] % purity = 150/160 x 100 = 93,75%  Copyright reserved Please turn over

Downloaded from hlayiso.com Downloaded from Stanmorephysics.com Physical Sciences P2 NSC-Marking Guideline Common Test June 2019 6 QUESTION 7 7.1 Substance that is used up completely in a reaction.   (2) 7.2 Ratio 5 1 5 2 Moles Ca V2O5 CaO V Initial 500 175  0 0 Change 500 100 500 200 End 0 75 500 200 Ca V2O5 n = m/M n = m/M = 20 000/40 = 31 850/182 = 500 mol = 175mol nV formed = 200 mol m = nxM = 200 x 51 = 10 200 g  (6) 7.3 % yield = 8670/10 200 x 100 = 85 % (2) [10] Copyright reserved Please turn over

Downloaded from hlayiso.com Physical Sciences P2 NSC-Marking Guideline Common Test June 2019 7 QUESTION 8 8.1 Amount of solute per litre of solution.   (2) 8.2. n = m/M = 4.14/69  = 0.06mol  (2) 8.3 Positive marking from 8.2. c = n/V 0.05 =0.06/V V = 1,20 dm3  (3) 8.4 Positive marking from 8.3 Positive marking from 8.2. and 8.3 c1V1 = c2V2  c = n/V 0.05 x 1.20 = c2 x 1.45 =0.06/1.45 C2 = 0.04 mol.dm-3 c= 0.04 mol.dm-3 (4) [11] TOTAL MARKS : 100 Copyright reserved Please turn over