Physical Sciences Grade 11 P2 Mpumalanga June 2025 QP & Memo

Testpapers.co.za Downloaded from Stanmorephysics.com education MPUMALANGA PROVINCE £ REPUBLIC OF SOUTH AFRICA MARKS: 50 TIME: 1 hour This question paper consists of 7 pages and 1 data sheet Copyright reserved Please turn over

Testpapers.co.za Downloaded from Stanmoreph sics. com Physical Sciences P2 ade 1 MDE/JUNE 2025 INSTRUCTIONS AND INFORMATION 1; 2. 10. 11. 12. Write your name in the appropriate space on the ANSWER BOOK. This question paper consists of FIVE questions. Answer ALL the questions in the ANSWER BOOK. Start EACH Questionon a NEWypage in the ANSWER BOOK. Number the ans\ question paper. ding to the numbering system used in this Leave one line be’ and QUESTION 2. You may use a non Jestions, for example between QUESTION 2.1 Iculator. You may use appropriate mathematical instruments. You are advised to use the attached DATA SHEET. Show ALL formulae and substitutions in ALL calculations. Round off your final numerical answers to a minimum of TWO decimal places. Give brief motivations, discussions, etc, where required. Write neatly and legibly. Copyright reserved Please turn over

Testpapers.co.za Downloaded from Stanmoreph sics. com Physical Sciences P2 ade 1 MDE/JUNE 2025 QUESTION 1: MULTIPLE CHIOICE QUESTIONS Various options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write only the letter (A—D) next to the question number (1.1 to 1.3) in the ANSWER BOOK, eg. 1.4E 1.1. Which ONE of the foligaigg compoupgs has the highest ionic character? A NaCt B RbC? C MgC D BaCte (2) 1.2 Three chemical bonds are given as; H—N, Cl-Cl and H—F. Consider the following statements in regard to these bonds: (i) All of three bonds are polar covalent (ii) H—N is the most polar bond. (iii) H—F is the most polar bond. Which of the above statements are/ is TRUE? A i and ii only B i and iii only Cc ii only D iii only (2) 1.3 In which ONE of the following will hydrogen bonding NOT occur between molecules? A HF(s) B NHa(s) Cc HBr(s) D —-Ha20(s) (2) [6] Copyright reserved Please turn over

Testpapers.co.za Downloaded from Stanmoreph ics.com Physical Sciences P2 FET- Grade 11 MDE/JUNE 2025 QUESTION 2 Two fluorine atoms, F(g), combine by a chemical bond to form a fluorine molecule, F2(g), as shown in the equation below. F(g) + F(g) — Fa(g) 2.1 Define the term chemical (2) 2.2 How does the stability ( i ioms, F2(g) compare to that of individual atoms, F(g)? Choo ER or EQUAL TO. Write down a reason (2) for the answer. 2.3 Consider F2(g) molecul 2.3.1 Draw a Lewis do’ FF (2) of lone pairs of electrons in the molecule. (1) ed by the reaction of ammonia, NHa(g), with to the following equation: ams to show how NH4’ is formed. (4) he molecular shape of NHs. (1) [12] Copyright reserved Please turn over

Testpapers.co.za Downloaded from Stanmoreph ysics.com Physical Sciences P2 MDE/JUNE 2025 QUESTION 3 The bond lengths and bond energies of the chemical bonds are given in the table below. CHEMICAL BOND BOND LENGTH BONG ENERGY (nm) (Kj.mol") 432 413 358 347 3.1 Define the term bond energy of the compound. (2) 3.2 Write down the: 3.2.1 Relationship between bond length and bond energy of the bonds. (1) 3.2.2 Explanation for the answer in QUESTION 3.2.1 (2) 3.3 Carbon dioxide molecule, COz, has polar covalent bonds, C=O, but the molecule is non polar. Write down a reason why the: 3.3.1 Bond energy of C=O bond is higher than that of C—O bond. (2) 3.3.2 C=O bonds are polar but COz molecule is non polar. (3) [10] Copyright reserved Please turn over

Testpapers.co.za Downloaded from Stanmoreph ics.com FET- Grade 11 Physical Sciences P2 MDE/JUNE 2025 QUESTION 4 4.1. In the diagram below, two methane molecules,CHs, are linked together by intermolecular force. Intermolecular force H [*71_Interatomic force 4.1.1 Type of intermolecular force een the two methane molecules. (1) 4.1.2 Type of interatomic force between carbon and hydrogen atoms. (1) 4.1.3. Which forces require more energy to separate, intermolecular forces or interatomic forces? (1) 4.2 Letters A to F represent six compounds used by grade 11 learners to investigate the relationship between molecular mass and boiling point. The results obtained are shown in the table below. | (2) Consider experiment I, Write down the: 4.2.2 |Independent variable. (1) 4.2.3. Conclusion that is drawn from the results obtained. (1) Copyright reserved Please turn over

Testpapers.co.za Downloaded from Stanmoreph Sics.com Physical Sciences P2 FET- Grade 11 MDE/JUNE 2025 4.2.4 Explain the answer to QUESTION 4.2.3. (2) A learner makes a conclusion based on the results obtained in experiment II: The vapour pressure of compound E is lower than that of F. 4.2.5- |s the learner CORRECT or WRONG? Explain the answer. (4) [13] QUESTION 5 . w! The effect of intermolecu y of solutes is investigated using iodine, lz, and sodium chloride crystal qu jounts of each solute are placed in equal volumes of water at the sam ure Of 22°C, as shown in the diagram below. The mixture in each test tu with a stirring rod. lodine crystals j Sodium chloride —\ crystals <i 2) Test tube P | Test tube Q <= ‘ oS Wate = ZT Aimnorenp © 5.CO 5.1 Define the term solubility. (2) 5.2 Write down the observation made in each test tube. Choose from CRYSTALS ARE SOLUBLE or CRYSTALS ARE NOT SOLUBLE. 5.2.1 Test tube P. (1) 5.2.2 Test tube Q. Explain the answer. (4) 5.3 Write down the: 5.3.1 Type of intermolecular forces between iodine crystals. (1) 5.3.2 Conclusion made from observation in test tube P. (1) [9] TOTAL: 50 Copyright reserved Please turn over

— MDE/JUNE 2025 TABLE 3: THE PERIODIC TABLE OF ELEMENTS SiICS.com ade 11 ph FE re Testpapers.co.za Downloaded from Stanmo Physical Sciences P2 0 KEY 1 2 «H| U se number iM Iv Vv VI Vill He 1 4 3 4 29 5 6 7 8 9 | 10 = Li 2 Be, — Electronegativity ——>2 Cy*— symbol = Blé C\&3 N14 O|F F| Ne 1 9 11| 12| 14] 16] 19| 20 i! 2 ‘ 3) 14{ 15] 16] 17{| 18 2 Na|3Mg F : : 2 ACS Silk P12 S |% Ce} Ar J ” a a a 3 24 Relative atomic mass (approximately) 7 28 31 32| 355| 40 19 | 20| 227] 22] 23] 267. 25). 26] 27] 28] 29] 30] 31| 32) 331 341 351 36 = Ca|2 Se |2 Ti|2 V |S Cr 2Mn/2 Fe 2 Co|2 Ni|2 Cu Z Zn|2 Ga|% Ge '& As | Se 2 Br| Kr 40) 45° 55| 56| 59| 59| 635) 65| 70| 73] 75!/ 79| $01] 84 3) 3 4 43/ 44] 45/ 46[ 47] 48[ 49[ So] S1| 52] 531 54 =Sr/S Y = Mo)? Te | Ru|& Rh/ Pd 2 Ag'= Cd|5 In 2 Sn|2 Sb Te G I | Xe 88) 89 101} 103; 106| 108) 112] 115} 119] 122) 128! 127| 131 56 | 57 j= 5{ 76| 77| 78| 79) 80| 81| 82] 83] 84| 9851 86 2 Ba) La’ fa; “W) Re) Os| Ir| Pt) Au Hg\= Tl|2 Pb|2 Bi|% Po/G At) Rn 137| 139 1 186| 190| 192| 195| 197) 201| 204] 207/ 209 88 | 89 a s = Ac 5s8/ 59| 60] otf 62] 63] 64] 65] 66] 67] 68] 69] 20] 71 | Ce) Pr| Nd| Pm) Sm} Eu) Gd| Tb) Dy) Ho! Er! Tm) Yb) Lu 140) 141) 144 150| 152| 157} 159] 163) 165} 167| 169| 173| 175 9 | | 92] 93] 94] 951 96{ 97] 98] 991 100] 101] 102) 103 Th; Pa} U)} Np) Pu; Am) Cm Bk! Cf! Es, Fm! Md! No! Lr 232 238 Copyright reserved

Testpapers.co.za Downloaded from Stanmorephysics.com education MPUMALANGA PROVINCE _ REPUBLIC OF SOUTH AFRICA MARKS/PUNTE 50 These marking guidelines consist of 10 pages. Hierdie nasienriglyne betstaan uit 10 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P FET — Gr 11 — Marking guidelines/Nasienriglyne QUESTION / VRAAG 1 11 BYY 12 DY 1.3 C¥v QUESTION / VRAAG 2 2.1 A mutual attraction be ato: esulting from the simultaneous ; outer electrons.v¥¥ ‘n Wedersydse aantrekking tussen twee atome as gevolg van die gelyktydige aantrekking tussen hul kerne en die buite / valenselektrone. [If any of the underlined key words/phrases in the correct context are omitted; deduct 1 mark. / Indien enige van die onderstreepte woorde / frases in die korrekte konteks uitgelaat is, trek 1 punt af.] 2.2 HIGHER / HOéR ¥ The (potential) energy of F2/combined atoms is lower than that of individual (F) atoms.¥ Die (potensiéle) energie van F2/gekombineerde atome is ler as die individuele fe re io electrons shown as e bole van beide F atome met twee elektrone as kolletjies / kruisies tussen hulle. Six lone pairs placed around F atoms. Ses alleenpare rondom die F atome. MDE/MOD?/ June /Junie 2025 (2) (2) (2) [6] (2) (2) (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 2 MDE/MOD/ June /Junie 2025 FET — Gr 11 — Marking guidelines/Nasienriglyne 2.3.2 6¥ (1) 2.4.1 d F f Hlvv Correct formula of nule van H* ¥ Correct shape of / NHa* vv (4) 2.4.2 Trigonal pyramidal / Trigonaal piramidaal ¥ (1) [12] QUESTION / VRAAG 3 3.1 The energy needed to break one mole of the compound’s molecules into separate atoms. / Die energie benodig om 1 mol van ‘n verbinding se molekules op te breek in atome. ¥¥ [2 or 0 marks / 2 of 0 punte] (2) 3.2.1. As the bond length increases, the bond energy decreases. Vv OR: As the bond length decreases, the bond energy increases. Soos die bindingslengte toeneem, sal die bindingsenergie afneem. OF: Indien die bindingslengte afneem, sal die bindingsenergie toeneem. (1) 3.2.2 Bonds with longer bond lengths have lesser forces of attraction,“ and require lesser energy to break.v OR: Bonds with shorter bond lengths have greater forces of attraction, “and require more energy to break.” Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 3.3.1 3.3.2 Testpapers.co.za MDE/MOD/ June /Junie 2025 FET — Gr 11 — Marking guidelines/Nasienriglyne Bindings met langer bindingslengtes het swakker aantrekkingskragte, en benodig minder energie om te breek. OF: Bindings met korter bindingslengtes het sterker aantrekkingskragte, en (2) benodig meer energie om te breek. $ are closer / bond length is shorter. ~ iv In C=O there is a\do Therefore, the bond s In C=O is daar ‘n dubé korter. Daarom is die 1e is nader aan mekaar / bindingslengte is (2) The electronegativity difference in C=O ( 3,5 —2,5) is 1,0 (which makes the bonds polar).~” The charge (electron density) distribution in a molecule is even/ the charge distribution is symmetrical.” There is no net dipole moment/ net dipole moment is zero.~ Die verskil in elektronegatiwiteit in C=O (3,5 —2,5) is 1,0 (daarom is die bindings polér.) Die ladingsverspreiding (elektrondigtheid) is simmetries. Daar is geen netto dipoolmoment nie / die netto dipoolmoment is zero. (3) [10] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 2 MDE/MOD/ June /Junie 2025 FET — Gr 11 — Marking guidelines/Nasienriglyne QUESTION / VRAAG 4 4.1.1. Induced dipole forces / London forces ¥ Geinduseerde dipoolkragte / London kragte (1) h 4.1.2 Covalent bonding 4.1.3 Interatomic forces /_ 4.2.1 The temperature at w he vapour pressure of a substance equals atmospheric pressure.¥ ¥ Die temperatuur waarby die dampdruk van ‘n stof gelyk is aan die omringende atmosferiese druk. [If any of the underlined key words / phrases in the correct context are omitted; deduct 1 mark / Indien enige van die onderstreepte woorde / frases in die korrekte konteks uitgelaat is, trek 1 punt af.] (2) 4.2.2 Molecular mass / Molekulére massa “ (1) 4.2.3 OPTION / OPSIE 1 Boiling point increases with an increase in molecular mass. ¥ Kookpunt neem toe met ‘n toename in molekulére massa. OPTION / OPSIE 2 Boiling point decreases with a decrease in molecular mass. ¥ Kookpunt neem af met ‘n afname in molekulére massa. (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 2 MDE/MOD/ June /Junie 2025 FET — Gr 11 — Marking guidelines/Nasienriglyne 4.2.4 OPTION/ OPSIE 1 e As the molecular mass increases, the strength of intermolecular forces / van der Waals forces / induced dipole forces / London forces increases. ¥ e The energy needed to separate th molecules / to overcome the i pe intermolecular forces 21 Waals forces / induced dipole forces / London forces e As die molekulére sal die sterkte van die intermolekulére kragte / van der W\ duseerde dipoolkragte / London kragte toeneem. e Die energie wat die molekules benodig om die intermolekulére kragte / van der Waals kragte / geinduseerde dipoolkragte / London kragte te oorkom neem toe. OPTION / OPSIE 2 e As the molecular mass decreases, the strength of intermolecular forces / van der Waals forces / induced dipole forces / London forces decreases. ¥ sal die sterkte van die intermolekulére induseerde dipoolkragte / London kragte e The energy 2 molecules / to overcome the Is forces / induced dipole forces / e Die energie wat die molekules benodig om die intermolekulére kragte / van der Waals kragte / geinduseerde dipoolkragte / London kragte te oorkom neem af. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 2 MDE/MOD/ June /Junie 2025 FET - Gr 11 — Marking guidelines/Nasienriglyne 4.2.5 OPTION/ OPSIE 1 CORRECT / REG ¥ e Between molecules of E / H2O there are hydrogen bonding, and between molecules of F / H2S there are dipole-dipole forces. “ e Dipole-dipole forces are weaker than hydrogen bonding. ¥ « Tussen die mole van E/ H20 is daar waterstofbinding, en tussen die molekules van F/ H ol-dipool kragte. ¢ Dipool-dipool kragte waterstofbinding. OR / OF: Intermolecula H2S are weaker than between E / H20. Intermolekulére kragte tussen F / H2S is swakker as tussen E / H20. e Less energy is needed to overcome dipole-dipole forces (than hydrogen bonding) “ eMinder energie word benodig om dipool-dipoolkrgate te oorkom (as waterstofbinding) OR / OF: Less energy is needed to overcome Intermolecular forces in F / H2S lig om die intermolekulére in F / H2S te oorkom as ded to separate the molecules of F / H2S (than g om molekules van F / H2S te skei, as die molekules van E / H20. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 2 MDE/MOD/ June /Junie 2025 FET — Gr 11 — Marking guidelines/Nasienriglyne OPTION / OPSIE 2 CORRECT/REG ¥ e Between molecules of E/ H2O there are hydrogen bonding, and between molecules of F/ H2S there are dipole-dipole forcesY Hydrogen bonding are stronger than dipole-dipole forces” eTussen die molekul é : jaar waterstofbinding, en tussen die molekules van e Waterstofbinding is ORIOF: Intermolecul ces in E/.0 are stronger than in F/ H2S. Intermolekulére kragte tussen E / H20 is sterker as tussen F / H2S. e More energy is needed to overcome hydrogen bonding (than dipole-dipole forces)“ e Meer energie word benodig om waterstofbinding te oorkom (as dipool- dipoolkragte) O /OF: More energy is needed to overcome Intermolecular forces in E/H20 (than Intermolecular forces in F/ H2S) Meer energie is nodig om Intermolekulére kragte in E / H2O te oorkom (as Intermolekulére kragte in F / H2S) OR/ OF: More energy is needed to separate the molecules of E/ H2O (than (4) the molecules of F / H2S.) Meer energie word benodig om die molekules van E / H20 te skei (as die molekules van F / H2S.) [13] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 2 MDE/MOD/ June /Junie 2025 FET — Gr 11 — Marking guidelines/Nasienriglyne QUESTION /VRAAG 5 5.1 The property of a solid, liquid or gaseous chemical substance (solute) to dissolve in a solid, liquid or gaseous solvent to form a homogenous solution.“ ¥ \ 2 - Die eienskap va gas (opgeloste stof) om op te los in ddel) om ‘n homogene oplossing te vorm. ases in the correct context are omitted; ‘n vaste stof, vloeist: [If any of the under! ned (2) deduct 1 mark / Indien enige var die onderstreepte woorde / frases in die korrekte konteks uitg: 5.2.1 CRYSTALS ARE NOT SOLUBLE / KRISTALLE LOS NIE OP NIE ¥ (1) 5.2.2 CRYSTALS ARE SOLUBLE / KRISTALLE LOS OP ¥ «Water molecules are held ogether by strong hydrogen bonding. ¥ a NOK rsterk waterstofbinding bymekaar gehou. are held together by strong ion-dipole forces. “ ord deur ioon-dipoolkragte bymekaar gehou. C water because the forces are h/ are of comparable strength. ¥ rin water want die kragte is van dieselfde grootte (4) 5.3.1 Induced dipole forces / London forces. ~ Geinduseerde dipool kragte / London kragte. (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Testpapers.co.za Downloaded from Stanmorephysics.com Physical sciences/Fisiese Wetenskappe P 2 MDE/MOD/ June /Junie 2025 5.3.2 FET — Gr 11 — Marking guidelines/Nasienriglyne Intermolecular forces / van der Waals forces between iodine and water molecules are not of comparable strenghths.” Intermolekulére kragte / van der Waals kragte tussen jodium en watermolekules is NIE van vergelykbare sterkte NIE. ORI OF: Induced. d ion forces between iodine and hydrogen bonding in water are not of comparable strenghth. Geinduseerde dipool kragt on kragte tussen jodium en waterstof binding in water is NIE . ikbare sterkte NIE. OR/ OF: Substances dissolve in each other only if their Intermolecular forces / van der Waals forces are of comparable strenghths. Stowwe sal oplos indien hul intermolekulére kragte / van der Waals kragte van vergelykbare grootte is. (1) [9] GRAND TOTAL 50 GROOTTOTAAL 50 Copyright reserved/Kopiereg voorbehou