Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 THE STRUCTURE OF THE ATOMS PAPER 2 : STRUCTURE Question Explanation Marks (a)(i) Diffusion Molecules - Made of tiny / discrete particles/ molecules (iii) - Move faster / rapidly ++ - Between air / another particles (b)(i) 8 o C Remain, because heat absorbed is used to overcome the forces between particles. + (iii) Move faster Particles gain kinetic energy. 2(a)(i) White fume Ammonium chloride (iii) Diffusion (b) Ammonia, because ammonia is lighter than hydrogen chloride. + (c) NH + HCl NH 4 Cl (a)(i) Number of proton in the nucleus of an atom. (b) 2 X and Y. Because they have same proton number but different neutron or nucleon number. (c) 2.8.7 (d)(i) + Y + (iii) (e) 4(a)(i) (iii) (iv) 7 Y - 5 X 7 - Both axis are labeled with unit and consistent scales - All points transfer correctly - Size of graph at least ½ of the page of the graph paper - Curve of the graph is correct and smooth 80 o C Heat is absorbed is used to overcome the forces of attraction between particles. 2 Chemistry Perfect Score Module Form 4 200 Marking Scheme Set
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 (b) To ensure uniform heating (c) Solid X is highly flammable 5(a) Iodine and ethanol 2 (b) copper (c) copper (d) ion PAPER : ESSAY No. 6 Rubric Marks (a) Particle Relative charge Relative mass Proton + + Neutron 0 (neutron) + Electron - /840 + (b)(i) (c) Any 2 pairs. Nucleus contains 7 protons and 7 neutrons. 2. Electrons moves around the nucleus.. Two shells filled with electrons. 4. 5 valance electrons / electrons arrangement 2.5 Nucleon number Proton number 4 X 7 Stage/Time State of matter Particle arrangement Change in energy t o t Solid Closely packed Kinetic energy increases 4 4 2 ++ t t 2 Solid liquid Close together but not in orderly arrangement Higher kinetic energy ++ t 2 t Liquid Further apart Highest kinetic energy ++ 0 Total 20 Chemistry Perfect Score Module Form 4 200 Marking Scheme Set 2
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 No. 7 Rubric Marks 7 (a) Melting point is the temperature at which a solid changes into a liquid at a particular pressure. Because the energy absorbed by the naphthalene is used to overcome the forces of attraction between the molecules of naphthalene. 4 7(b) Before condensation The kinetic energy is high The particles are very far apart from each other The attraction forces between particles are very weak. During condensation The kinetic energy decreases The particles begin to move closer toward one another / the distance between the particles decrease The attraction forces between particles become stronger. After condensation The kinetic energy is low The particles are packed closely together NOT in an orderly manner The attraction forces between particles are strong. 7(c)(i) Proton number = Nucleon number = 4 + = 27 7(c) 0.8 Number of moles of Y = = 0.4 27 Number of moles of Y 2 O = 0.4 2 = 0.2 Relative formulas mass of Y 2 O = 2(27) + (6) = 02 Mass of Y 2 O = 0.2 02 = 20.4 g + 0 2 4 Total 20 PAPER : STRUCTURE 8 Rubric Marks (a) To study the rate of diffusion in the three states of matter. (b) The rate of diffusion in matter decreases in the order of gas > liquid > solid (c)(i) (iii) Manipulated : medium of diffusion gas, liquid and solid Responding : rate of diffusion Fixed : temperature of each medium (d) Solid, liquid, gas (e) Solid the particles are very closely pack Liquid the particles closely but there are more space between them Gas the particles far apart from each other (f) Smell perfume / smell gas from leaked pipe or gas cylinder Chemistry Perfect Score Module Form 4 200 Marking Scheme Set
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 CHEMICAL FORMULA AND EQUATION. Ion Chloride Nitrate Hydroxide Sulphate Carbonate oxide Sodium NaCl NaNO NaOH Na 2 SO 4 Na 2 CO Na 2 O Magnesium MgCl 2 Mg(NO ) 2 Ma(OH) 2 MgSO 4 MgCO MgO Lead(II) PbCl 2 Pb(NO ) 2 Pb(OH) 2 PbSO 4 PbCO PbO Copper (II) CuCl 2 Cu(NO ) 2 Cu(OH) 2 CuSO 4 CuCO CuO Iron (II) FeCl 2 Fe(NO ) 2 Fe(OH) 2 FeSO 4 FeCO FeO Iron (III) FeCl Fe(NO ) Fe(OH) Fe 2 (SO 4 ) Fe 2 ( CO ) Fe 2 O Aluminium AlCl Al(NO ) Al(OH) Al 2 (SO 4 ) Al 2 ( CO ) Al 2 O 2. (a) CuCO CuO + CO 2 (b) HNO + NaOH NaNO + H 2 O (c) 2HCl + Zn ZnCl 2 + H 2 (d) Cu(NO ) 2 + Mg Mg(NO ) 2 + Cu (e) Cl 2 + 2LiOH LiCl + LiOCl + H 2 O. (a) 0. x 6.02 x 0 2 = 6.02 x 0 22 atoms (b).5 x 6.02 x 0 2 = 9.05 x 0 2 atoms (c) 2.0 x 6.02 x 0 2 =.2 x 0 24 molecules (d).5 x 6.02 x 0 2 = 9.05 x 0 2 atoms (e) 2.0 x 6.02 x 0 2 =.2 x 0 24 molecules 4. (a) 6.02 x 0 2 6.02 x 0 2 =.0 mol (b).8 x 0 2 6.02 x 0 2 = 0.0 mol (c).2 x 0 2 6.02 x 0 2 = 0.2 mol (d) 2.4 x 0 20 6.02 x 0 2 = 4.0 x 0-4 mol (e).0 x 0 2 6.02 x 0 2 = 0.5 mol 5. (a).5 x 7 = 06.5 g (b) 2.5 x 2 = 80 g (c) 2 x 98 = 96 g (d) 0.5 x 7 = 8.5 g (e) 2.5 x 267 = 667.5 g (f) 0.5 x 88 = 94 g 6. (a) 0.5 mol x 22.4 dm mol - =.2 dm (b) 0.2 mo x 24 dm mol - = 4.8 dm (c).5 mol x 24 dm mol - = 6 dm (d) 0.5 mol x 24 dm mol - = 2.0 dm (e) 2.5 mol x 22.4 dm mol - = 56 dm 7. (a) 250 cm 24000 cm mol - = 0.0 mol (b) 500 cm 22400 cm mol - = 0.02 mol (c) 200 cm 24000 cm mol - = 8. x 0 - mol (d) 750 cm 24000 cm mol - = 0.0 mol (e) 00 cm 22400 cm mol - = 0.0 mol Chemistry Perfect Score Module Form 4 200 Marking Scheme Set 4
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 8. (a) Mg + 2HCl MgCl 2 + H 2 (b) (i). Mol of Mg = 2.4 /24 = 0. mol 2. ration : mol of Mg produced mol MgCl 2 0. mol Mg produced 0. mol MgCl 2. mass of MgCl 2 = 0. x 95 = 9.5 g. Ratio : mol of Mg produced mol of H 2 0. mol of Mg produced 0. mol of H 2 2. volume of H 2 = 0. mol x 24 dm mol - = 2.4 dm 9. (a) C + 2PbO CO 2 + 2Pb (b) 2 mol of PbO produced 2 mol of Pb 0.5 mol of PbO produced 0.5 mol of Pb (c) 2 mol of PbO reacts with mol of C 0.5 mol of PbO reacts with 0.25 mol of C (d) (i) mol of PbO = 44.6 / 22 = 0.2 mol mass of Pb produced = 0.2 x 207 = 4.4 g 0. (a) CuCl 2 + Na 2 CO 2NaCl + CuCO (b) (i) mol of CuCl 2 = 0.5 x 50 / 000 = 0.025 mol mol of CuCl 2 produced mol of CuCO 0.025 mol of CuCl 2 produced 0.025 mol of CuCO (iii) mass of salt = 0.025 x 24 =. g. (a) CaCO + 2HCl CaCl 2 + CO 2 + H 2 O (b) (i) mol CaCO = 5 /00 = 0.05 mol mol of CaCO produced mol of CO 2 0.05 mol of CaCO produced 0.05 mol of CO 2 (iii) Volume of CO 2 = 0.05 mol x 24 dm mol - =.2 dm 2. (a) 2NaHCO Na 2 CO + CO 2 + H 2 O (b) (i) mol of NaHCO = 8.4 /84 = 0. mol 2 mol of NaHCO produced mol of CO 0. mol of NaHCO produced 0.05 mol of CO Volume of CO 2 = 0.05 x 24 dm mol - =.2 dm Mass of Na 2 CO = 0.05 x 06 = 5. g. (i) Zn + 2HCl ZnCl 2 + H 2 mol of HCl = 2 x 00 / 000 = 0.2 mol (iii) volume of H 2 = 0. x 24 dm mol - = 2.4 dm 4. (i) CaCO + 2HNO Ca(NO ) 2 + CO 2 + H 2 O mol of CaCO = 5/00 = 0.05 mol (iii) mol of CaCO reacts with 2 mol of HNO 0.05 mol of CaCO reacts with 0. mol of HNO (iv) mol of CaCO produced mol of CO 2 0.05 mol of CaCO produced 0.05 mol of CO 2 Volume of CO 2 = 0.05 mol x 24 dm mol - =.2 dm 5. (i) Na 2 CO + 2 HCl 2 NaCl + CO 2 + H 2 O mol of HCl = x 00 /000 = 0. mol (iii) 2 mol of HCl reacts with mol Na 2 CO 0. mol of HCl reacts with 0.05 mol Na 2 CO mass of Na 2 CO = 0.05 x 06 = 5. g Chemistry Perfect Score Module Form 4 200 Marking Scheme Set 5
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 6. (i) CuO + 2 HCl CuCl 2 + H 2 O mol of CuO = 0 / 80 = 0.25 mol mol CuO reacts with 2 mol HCl 0.25 mol CuO reacts with 0.25 mol HCl Molarity of HCl = 0.25 x000 / 00 = 2.5 mol dm - 7. (a) mol of NaOH = 8/40 = 0.2 mol Molarity of NaOH = 0.2 mol dm - (b) M V = M 2 V 2 0.2 x 50 = M 2 x 50 M 2 = 0.67 mol dm - 8. (a) 2NaOH + H 2 SO 4 Na 2 SO 4 + 2H 2 O (b) mol NaOH = 2 x 25 /000 = 0.05 mol (c) 2 mol of NaOH reacts with mol of H 2 SO 4 0.05 Mol NaOH racts with 0.025 mol of H 2 SO 4 (d) molarity of H 2 SO 4 = 0.025 x 000 / 8.5 =.5 mol dm - 9. (a) 2HCl + Na 2 CO 2NaCl + CO 2 + H 2 O (b) mol of Na 2 CO = x 25 /000 = 0.025 mol (c) (i) mol of HCl = 0.025 x 2 = 0.05 mol volume of HCl = 0.05 x 000 /.25 = 40 cm PAPER 2 :STRUCTURE Question Explanation Mark (a) (b)(i) (iii) (iv) (c) Chemical formula that shows the simplest whole ratio of atom of each elements in the compound. Mass of Cu = 20.5 8.75 =.6 g Mass of O = 20.75 20.5 = 0.4 g Mol of Cu =.6 / 64 = 0.025 mol Mol of O = 0.4 / 6 = 0.025 mol Mol ratio : Cu : O = 0.025 : 0.025 = : Empirical formula CuO CuO + H 2 Cu + H 2 O.flow dry hydrogen 2.collect the gas. place lighted splinter at the mouth of the test tube. Question Explanation Mark 2 (a) (i) 46 89 (b) (i) 2.408 0 24 molecules 4.86 0 24 atoms (c) 0.005 mol (d) 44 g Chemistry Perfect Score Module Form 4 200 Marking Scheme Set 6
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 Question Explanation Mark (e) 8.28 = 0.0 mol Ag 2 CO 276 2 mol of Ag 2 CO produce 2 mol of CO 2 0.0 mol of Ag 2 CO produce 0.0 mol of CO 2 0.0 24 dm = 0.72 dm Total 9 Question Explanation Mark (a). green turns black 2. lime water turns cloudy 2 (b) CuCO CuO + CO 2 (c) Mol of CuCO =. / 24 = 0.025 mol (d) mol of CuCO produced mol of CO 2 0.025 mol of CuCO produced 0.025 mol of CO 2 Volume of CO 2 = 0.025 mol x 22.4 dm mol - = 0.56 dm (e)(i) Black turns brown CuO + H 2 Cu + H 2 O Mass of Cu = 0.025 x 64 =.6 g 0 ESSAY 4. (a) (i) C H mol 85.70 4.0 2 7.4 4.0 7.4 7.4 ratio 2 The empirical formula CH 2.. ( CH 2 ) n = 56 [ 2 + 2() ]n = 56 56 // 4 4 The molecular formula C 4 H 8.. Chemistry Perfect Score Module Form 4 200 Marking Scheme Set 7
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 (iii) Empirical formula Molecular formula The formula shows that carbon and hydrogen are present The formula shows that the ratio of carbon to hydrogen is :2 The formula shows that carbon and hydrogen are present The formula shows that one molecule of X consists of 4 carbon atoms and 8 hydrogen atoms 2 (b) An empty crucible and its lid are weighed and the mass are recorded 2 Magnesium ribbon is cleaned with sandpaper, placed in crucible and weighed again. The mass are recorded The crucible and its contents are heated over a strong flame 4 The crucible lid opened once in a while during the experiment 5 When the magnesium does not burn anymore, the crucible and its contents are cooled in room temperature, 6 and then weighed. The mass is recorded 7 The heating, cooling and weighing is repeated until the final mass becomes constant 8 Result Mass of crucible + lid = a g Mass of crucible + lid + magnesium ribbon = b g Mass of crucible + lid + magnesium oxide = c g 9 Mass of magnesium = (b a) g Mass of oxygen = (b c) g 0 Mol of magnesium atom = b-a 24 Mol of oxygen atom = b-c 6 Simplest ratio mol of magnesium atom to mol of oxygen atom = x:y/ : 2 Empirical formula : Mg x O y / MgO PAPER : STRUCTURE Chemistry Perfect Score Module Form 4 200 Marking Scheme Set 8
Set Marking Scheme: The Structure of the Atom & Chemical Equation 200 5(a) Rubric [Able to state three inferences according to the observations correctly] Example: Observations Inferences i) white fumes is released. magnesium oxide is produced // magnesium has been oxidised. ii) Bright burning. magnesium is a reactive metal. iii) The mass increases. magnesium has combined with oxygen. Score 5(b) Rubric [Able to record the data correctly] Example: Description The crucible and lid. The crucible, lid and magnesium powder. The crucible, lid and magnesium oxide. Mass / g 2.6 28.6.6 Score 5(c) Rubric [Able to calculate the mass of magnesium, mass of oxygen and show the steps to determine the empirical formula of magnesium oxide accurately] Example: i) Mass of magnesium = 28.6 2.6 = 4.80 g ii) Mass of oxygen =.6 28.6 =.2 g iii)the empirical formula of magnesium oxide: Element Mg O No. of moles 4.8 24 = 0.2.2 6 = 0.2 Simplest ratio Empirical formula : MgO Score 5(d) Rubric Score Cannot. Lead is less reactive metal Chemistry Perfect Score Module Form 4 200 Marking Scheme Set 9