Physical Sciences Grade 11 P2 2018 QP & Memo

MARKS: TIME Copyright Reserved Please Turn Over | education Department: Education PROVINCE OF KWAZULU-NATAL NATIONAL SENIOR CERTIFICATE GRADE 11 PHYSICAL SCIENCES P2 (CHEMISTRY) “ COMMON TEST i JUNE 2018 a Pa 100 : 2 Hours This question paper consisis of 9 pages AND 2 data sheets.

Physical Sciences P2 Downloaded from hlayiso.com NSC Common Test June 2018 INSTRUCTIONS AND INFORMATION TO CANDIDATES 1. Write your name on the ANSWER BOOK. Answer ALL the questions in the answer book. You may use a non-programmable calculator. You may use appropriate mathematical 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. Copyright Reserved Please Turn Over

Downloaded from hlayiso.com, NSC Physical Sciences P2 mmon Test June 2018 SECTION A QUESTION 1: MULTIPLE- CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Write only the letter (A - D) next to the question number (1.1 — 1.6) in the ANSWER BOOK. e.g. 1.7 A 1.1. Which one of the following is the best explanation for the high degree of solubility of NaCl in water? A Both molecules are polar. B Both molecules are ionic. Cc The electrostatic forces between the ions in NaCl are of the same order as that of the intermolecular forces in water. D Water is ionic and NaCl is polar. 1.2. The factor that does NOT affect bond length between two atoms is : A Size of the atoms. B Difference in electronegativity between the atoms. Cc Number of bonds between the atoms. D Valencies of the atoms. 1.3. The phenomenon that allows insects such as mosquitoes to walk on water is: A Capillarity B Adhesion Cc Surface tension D Evaporation 1.4. Which ONE of the following statements regarding the Kinetic Molecular Theory of gases is INCORRECT? All molecules have the same kinetic energy. Gas molecules collide elastically. Gas molecules are in random motion. UO OW PY Attractive and repulsive forces between the gas molecules can be neglected. (2) Copyright Reserved Please Turn Over

Downloaded from HlayiSO.COMonmon Tost June 2018 NSC Physical Sciences P2 1.5. The number of nitrate ions in 2 moles of Ca(NOs)z is A 4 x 6.02 x 1078 B 2x 6.02» 10 Cc 4 D 8 x 6.02 x 109 1.6. | Consider the combustion of methane gas in a closed rigid container containing 30cm° of oxygen gas. CHa) + 202) — CO2g + 2H20@) What will be the total volume of gas in the container if 20cm? of oxygen gas reacts with all the methane at a given temperature and pressure? A 40 cm? B 30 cm® Cc 20 cms D 10 cms (2) [12] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences P2 Common Test June 2018 NSC SECTION B INSTRUCTIONS AND INFORMATION 1. Answer all questions. 2. Show the formulae and substitutions in ALL calculations. 3. Round off your numerical answers to a minimum of TWO decimal places. QUESTION 2 2.1. Use only the substances from the list below when answering questions 2.1.1 to 2.1.7 HClig) HF) = PHaq) = Cus) )~— He =~ KFis)_—LiCKaqy «= CClaqy Select one substance from the list that has: (Write only the question number and the Formula of the substance next to it. You may use a substance more than once). 2.1.1 Pure covalent bonds. (1) 2.1.2 A high melting point due to strong electrostatic attraction between the cations and anions in the crystal lattice. (1) 2.1.3 Hydrogen bonding. (1) 2.1.4 Has dipole-dipole intermolecular forces. Gh) 2.1.5 Has ion-dipole forces. (1) 2.1.6 Has a trigonal pyramidal shape. (1) 2.1.7 Has polar covalent bonds, but the molecule is non-polar. (1) 2.2 Consider the table below, which shows the boiling points of the halogens. | Halogen Fe Cle [ Bre [ [Boilingpoint(°C) 188 -35 59d 2.2.1 Name the type of Van de Waal force that exists between molecules of Cle. (1) 2.2.2 State the trend in the boiling points of the Halogens. (1) 2.2.3 Explain the trend in question 2.2.2 by making reference to the strength of the intermolecular forces and the energies involved. (3) 2.2.4 Define vapour pressure. (2) 2.2.5 Which halogen has the lowest vapour pressure? Give a reason for your answer. (2) [16] Copyright Reserved Please Turn Over

Physical Sciences P2 Downloaded from hlayiso.com Common Test June 2018 NSC QUESTION 3 Molecules such as HzO and HCN are formed by covalent bonding 3.1 3.2 3.3 3.4 35 Define a covalent bond. (2) Use the VSEPR theory to compare the shape of the water molecule to that of the HCN molecule. (4) Draw the Lewis structure for the HCN molecule. (2) Water is unique in that its solid state (ice) floats in its liquid state. Explain fully why ice floats in water. (3) Sodium chloride dissolves in water. The ions are surrounded by water molecules as shown below. é 9 o®., : 3.5.1 Is the ion shown in the diagram a sodium ion (Na*) or a chloride ion (Cl)? Explain your choice. (2) 3.5.2. A relatively large amount of water is needed to dissolve a small amount of sodium chloride. Explain why this is so by referring to the structure of sodium chloride and the strength of the forces involved. (3) [16] Copyright Reserved Please Turn Over

Downloaded from hlayiso.com Physical Sciences P2 Common Test June 2018 NSC QUESTION 4 The graph below indicates the relationship between the pressure (p) and the reciprocal of volume (1/V) of an enclosed mass of helium (He) gas at 25°C. Assume that helium behaves as an ideal gas. 450 p (kPa) 105 > Xo K (dm?) 4.1. Give the name and state in words the Law that is illustrated by the graph. (3) 4.2 Calculate the volume of the He gas at a pressure of 450 kPa. (4) 4.3. Use the graph to calculate the mass of helium that was used. (5) 4.4 Copy the graph in your answer book. Label this graph A. On the same set of axes draw the graph that will be obtained if: 4.4.1. The experiment is conducted at 35°C (Label this graph B). (2) 4.4.2 Areal gas such as ammoria is used instead of helium (Label this graph C). (2) 4.5 Explain fully why the shape of the graph is different in question 4.4.2. Copyright Reserved Please Turn Over

Downloaded fram hlayiso.com NSC Physical Sciences P2 Common Test June 2018 QUESTION 5 5.1. Agrade 11 learner dissolves 13,995 g of Cu(NOe)2 in water to make up a standard solution with a concentration of 0,15 mol-dm®. 5.1.1. Define a standard solution. (2) 5.1.2 Calculate the volume of water (in cm) required to prepare this solution. (4) The learner then transfers some of the above solution to a volumetric flask, and adds 150cmé® of water to dilute it to a concentration of 0.05mol-dm°*. 5.1.3 Calculate the volume of the original solution that must be transferred to the volumetric flask to prepare the diluted solution. (4) 5.2 50g of a substance contains 24,68 g of K, 10,13 g of S and 15,19 g of O. 5.2.1 Define empirical formula. (2) 5.2.2 Determine the empirical formula of the compound. (4) [16] Copyright Reserved Please Turn Over

Physical Sciences P2 Downloaded from hlayiso.com NSC ommon Test June 2018 QUESTION 6 6.4 6.2 6.3 An impure sample containing tin dioxide (SnOz) is tested to determine how much tin dioxide it contains. The sample weighs 4,50 g Excess sulfuric acid is added to the sample and tin (IV) sulfate forms. The equation for the reaction is: SnOas) + 2H2SO4 (aq) Sn (SOa)2 (s) + 2 H2O The mass of tin sulfate produced is 4,90 g. Calculate the percentage purity of this sample. (6) Magnesium burns in oxygen to form magnesium oxide according to the following balanced reaction: 2Mgis) + O2g) > 2MgQOc) The percentage yield of the reaction is 75%. Calculate the mass of magnesium that must be burnt in excess oxygen to produce 20 g of magnesium oxide. (6) 30 g of Al and 1,50 moles of oxygen gas at STP are placed in a reaction chamber and allowed to react. The balanced equation for the reaction is: 4Al + 302 ~~ 2Al20s 6.3.1 Explain what is meant by a limiting reactant. (2) 6.3.2 Calculate the volume of excess oxygen in the reaction chamber. (7) [21] TOTAL: 100 Copyright Reserved Please Turn Over

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Physical science P2 Downloaded from hlayiso.com une 2018 Common Test TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES NAME/NAAKi SYMBOL/SIA SL | VALUEIWAARDE Standard pressure i 4 5, Standaarddruk P ge a EAP Vm 22.4 dm*-mol’ Ni gas te 284 JeKmok! Molar gas constant x acSihontr iat Standard temperature x 479 | Standaarctemperatuts 7 ana Avogadro s Constant N offor Ne 6.022102? mor Charge on Electron . 5 sae Lading op elektron © “LGC TABLE 2: FORMULAE/TABEL 2: FORMULES orlof _f V _ Mm MV P4V1_ P2V2 Ty pV=nRT To Copyright Reserved

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Downloaded from hlayiso.com Physical Sciences P2 NSC-Marking Guideline-GR. 11 Comm. est June 2018 3 QUESTION 3 3.1 Sharing of electrons between 2 atoms to form a molecule. ¥ ¥ (2) 3.2 Water has an angular shape” whereas hydrogen cyanide has a linear shape.” Water has two lone pairs on the central atom whilst hydrogen cyanide has no lone pairs on the central atom. ” The repulsion between the lone pairs and bond pairs is stronger than that of the bond pairs and bond pairs, hence giving water an angular shape.” (4) 33 HXCMNE wv @) He 3.4 The same mass of liquid water has a smaller volume than that of icev because of (the predominance of) Hydrogen bonding in ice’, hence ice has a smaller density than liquid water.” (3) OR Ice has a lower density than water. “ The molecules are loosely arranged in a tetrahedral structure with lots of space” within the structure. There are fewer molecules per volume of ice than per volume of water. “ 3.5. Sodium ion (Na*).” The water molecules are orientated in such a way that the negative ends of the dipoles are attracted by the positive ion. (2) 3.5.2. The electrostatic force holding the ions in the crystal lattice are much stronger than the ion-dipole forces” between ions and the water molecule. Large numbers of water molecules “ need to surround the ions in order to overcome the strong electrostatic forces Y and pull the ions apart. (3) [16] Copyright reserved Please tum over Physical Sciences P2 NSC-Marking Guideline-GR. 11 Common Test June 2018 4 QUESTION 4 4.4 Boyle’s Law. The pressure exerted by an enclosed gas is inversely proportional to the volume ” it occupies at constant temperature. ¥ (3) 42 pw = p2V2 ¥ 105x50Y = 450¥ x V2 V2 = 11.67 dm ¥ (4) POSITIVE MARKING FROM 4.2. 43 Gradient pV 450-105 iol 11,67 50 = 5251. 957 J y pV = nRTV (450 x 10 x 11.67x107 ) W=nx 8.31% 298¥ n=2.121 mol 5251.957 = n x8.31 x 298” n= 2.121 /2.12 mol n =miM Can use 105 x 10x 50 x 10? 2.121=mi4v m = 8.484/ 8.483 /8.48 g v (5) 1/V (dv) Please turn over Copyright reserved

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Downloaded from hlayiso.com Physical Sciences P2 NSC-Marking Guideline-GR. 11 Comm. est June 2018 7 6.2 OPTION2 20 g Mg represents a 75% yield. ¥ Therefore, 100% yield would be 26,667 g. ~ Ngo = m/M 26,667/40 v 0,667 mol Aig = Ngo = 0,667 mol Nig = m/M 0,667 = m/i24v m= 16g (16,03 g) (6) 6.3.1 The reactant that gets used up completely in a reaction. vv (2) 6.3.2 Criteria for marking. « Converting 30 g of Al to moles (1,11 mol). “ « Establishing that Al is the limiting reagent. v « Applying ratio (4:3) to determine moles of O2 used. “ + Subtracting moles used from initial moles to determine excess moles of O2. e Using the formulan = V/Vm ~ ¢ Substituting 22,4 dm® in the above formula. ~ e Final answer. Physical Sciences P2 NSC-Marking Guideline-GR. 11 Common Test June 2018 8 OPTION 2 nal =m/M = 30/27 v (any one) = 1.11 mol aaa Noz required to react = es 3 = 0.8325 mal (< 1.50 mol). Limiting reactant is Al. v OR n(O2) = 1.50 mol n(Al) required = =°x4 = 2 mol > 1.21 mo!) Limiting reactant is Al. ¥ OPTION 1 4Al + 302 2ALOs 4 3 411v 1,5 Change in moles | 1,11” 0,8325” End moles 0 00,6675” n=VNm ¥ 0,6675 = V/22.4v (7) V = 14,952 dm? v Copyright reserved Please turn over Therefore, no of moles of O2 in excess = 1,5—0,8325 = 0,6675 mol” n=VNn¥ 0,6675 = V/22.4v V= 14,952 dms v Copyright reserved (7) (21] TOTAL : 10