sch3uc_practice_test_answers.pdf - Chemistry SCH3U-C Practice Test Time 2 hours Total Marks 119 Final Test Score � 119 × 100 = Instructions • There

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Unformatted text preview: Chemistry SCH3U-C Practice Test Time: 2 hours Total Marks: 119 Final Test Score: ____ ÷ 119 × 100 = ____% Instructions • There is a label attached to this page. Compare the course code on the label with the course code printed on the Final Test to make sure that they are the same. Inform the Final Test supervisor immediately if they are not the same. • The Final Test pages are numbered 1 through 13. Check to see that all 13 pages are attached. Inform the Final Test supervisor immediately if there are any pages missing. • You may use a non-marked dictionary during the Final Test if one is available at the test site. You may not bring any books or notes into the test. • You will need a pen and a scientific calculator. • There is a periodic table of the elements included at the beginning to assist you with the chemistry questions. • You must write your answers in the space provided. • There are five (5) parts to this Final Test. A breakdown of the marks and the approximate time required is given below. Look over the test carefully before you begin (5 minutes). Manage your time carefully and leave some time at the end to review your work (5 minutes). Part Category Marks Time (minutes) Preview Part A Matter, Chemical Trends, and Chemical Bonding 30 25 Part B Chemical Reactions 22 20 Part C Quantities in Chemical Reactions 17 20 Part D Solutions and Solubility 21 20 Part E Gases and Atmospheric Chemistry 29 25 Review Total 5 5 119 120 • At the end of the Final Test, return this test paper and any other paper you write on to the Final Test supervisor. Please note that, for security reasons, marked tests are not returned to students. • Mathematical solutions require proper problem-solving format. Final answers must include the correct units and the correct number of significant figures. Formula Sheet and Periodic Table of Elements Common anions Cl ,Br ,I – High solubility > or = 0.1 mol/L (at SATP) Low solubility < or = 0.1 mol/L (at SATP) – – most Ag+, Pb2+, Tl+, Hg2+, Hg+, Cu+ S OH– Groups I & II, NH4+1 Group I, NH4+, Sr2+, Ba2+, Tl+ 2– most most SO42– most Ag+, Pb2+, Ca2+, Ba2+, Sr2+, Ra2+ CO32–,PO44– C2H3O2– NO3– Group I & NH4+ most all most Ag+ none Avogadro’s number = 6.02 × 1023 universal gas constant = 8.31 kPa·L/mol·K Moles pH formula: pH = –log [H+(aq)] n=m/M Charles’s law: V1/T1 = V2/T2 Mass Boyle’s law: P1V1 = P2V2 n = N / NA Number of Particles Combined gas law: P1V1/T1 = P2V2/T2 Polyatomic ions hydroxide nitrate hydrogen carbonate carbonate sulphate phosphate ammonium OH– NO3– HCO3– CO32– SO42– PO43– NH4+ -ide table C: carbide –4 N: nitride –3 O: oxide –2 F: fluoride –1 P: phosphide –3 S: sulphide –2 Cl: chloride –1 Br: bromide –1 1: mono (I) 2: di (II) 3: tri (III) 4: tetra (IV) 5: penta (V) 6: hexa (VI) 3 3 2 2 1 1 * The systematic names and symbols for elements greater than 111 will be used until the approval of trivial names by IUPAC. 3 3 2 2 1 1 2 1 Potassium 2 1 1 –1 3 6 5 4 5 4 4 7 –1 6 7 2 3 4 3 4 6 5 4 6 5 4 3 2 6 3 4 5 4 Tantalum 5 3 5 4 3 4 4 3 Average Atomic Mass Name Symbol Atomic Number 101.07 4 –4 2 3 2 4 3 5 5 4 3 4 2 3 3 2 4 2 3 3 4 2 3 3 4 6 2 3 Ds 110 4 2 111 Rg 3 1 1 2 1 112 Uub 3 4 5 3 2 3 3 Darmstadtium Roentgenium Ununbium (269) (272) (275) 3 2 4 2 3 Common Oxidation States 3 4 3 4 2 1 2 2 114 Uuq 2 4 4 2 115 Uup 3 5 3 –3 5 3 –3 5 5 –3 3 4 4 3 –3 5 4 –4 2 116 Uuh 4 2 4 –2 6 4 –2 6 6 –2 2 –2 117 Uus Fluorine –1 1 3 –1 1 5 –1 1 5 –1 1 3 –1 118 Uuo 3 3 3 3 3 2 3 2 Ununtrium Ununquadium Ununpentium Ununhexium Ununseptium Ununoctium (278) (282) 113 Uut 1 3 3 3 3 3 3 Practice Test Chemistry SCH3U-C Part A: Matter, Chemical Trends, and Chemical Bonding (30 marks) (approximate time: 25 minutes) 1. Use the periodic table provided and the information in the following table to fill in the missing spaces. (6 marks) Name Z A Number of protons Number of electrons Number of neutrons Mercury 80 202 80 80 122 Copper 29 65 29 29 36 Antimony 51 123 51 51 72 Nuclide symbol 202 80 Hg 65 29 Cu  Sb 2. Place the atoms C, Ne, Na, and Si in decreasing order of atomic radius and give an explanation for the order. (3 marks) Na > Si > C > Ne All atoms in the third period are larger than those in the second. This is because the third period has an extra electron energy level. Na has a lower number of protons so its nuclear charge is smaller, making its attraction for the electrons weaker. With weaker attraction, the electrons can get farther away from the nucleus. Therefore, Na > Si. The same holds true for C and Ne, making C > Ne. 3. Draw a neat Lewis structure for each of the following compounds. Make sure your diagram clearly distinguishes the type of bonding and accounts for all the valence electrons. (4 marks: 2 marks each) a) LiBr 1 Practice Test Chemistry SCH3U-C b) NCl3 4. a) In general, how does the melting and boiling data give you information about the bonding in a substance? (3 marks) In general, the higher the melting point, the more difficult it is to separate the molecules of solid and allow them to change to the liquid state. That means that when attractive forces are larger, the melting point is higher. The same holds true for the boiling point. Since attractive forces are larger in ionic bonding than in covalent bonding, melting and boiling points for ionic substances will be very high but very low for covalent molecules. Since dipole forces are larger than London force attraction, polar molecules will have higher melting and boiling points than non-polar molecules. b) In general, how does the data about solubility in water give you information about the bonding in a substance? (3 marks) Substances that dissolve in water must be similar to water and this will be either polar or ionic. Water is polar, so it will accept a polar molecule into it, and since it is polar it will be attracted to the ions in ionic bonds, allowing them to dissolve too. Non-polar molecules without dipoles will not dissolve in water. 2 Practice Test Chemistry SCH3U-C c) Identify specifically the bonding in each of the three substances as either ionic, polar covalent, or non-polar covalent. (3 marks: 1 mark each) Type of bond Substance State at room temperature Melting point (°C) Boiling point (°C) Solubility in water 1 gas –115 –85 soluble polar covalent 2 solid 801 1580 soluble ionic 3 liquid –63 61 insoluble 5. 6. non-polar covalent Ethanol, C2H5OH, is very soluble in water. Ethane, C2H6, and hexanol, C6H13OH are both insoluble in water. Explain why? (5 marks) Ethanol has a very polar site where the OH group is found on the molecule. Polar water molecules are strongly attracted to that OH site, allowing water to attach to the ethanol molecule. Ethanol is a small molecule since it only has two carbons in its carbon chain, so it is small enough that the attached water molecules can tow it around in the solution, making ethanol very soluble. Ethane, C2H6, has no polarity on the molecule, so there is no place for water molecules to attach to the ethane and tow it. Ethane cannot dissolve in water as a result. Hexanol is similar to ethanol in that they both have the same OH group (site) on the molecule that is quite polar. This means that polar water molecules can readily attach to hexanol molecules in the same way that water can attach to ethanol molecules. The difference with hexanol is that this molecule has a chain of six carbons in it, making it too large for the attached water molecules to tow the hexanol. The hexanol and the water will remain as two layers and not form a solution. Hexanol is insoluble in water. Describe the process of dissolving an ionic compound, such as NaCl, in water. What is this process called? (3 marks) Polar water molecules are attracted to the positive (Na+) and negative (Cl-) ions on the surface of the ionic solid. The positive side of the water molecule (the H side) is attracted to the negative, Cl-, ion and the negative side of the water molecule (the O side) is attracted to the positive, Na+, ion on the solid. If the water has sufficient kinetic energy (is warm enough), the water molecule will tear off the ions from the surface, one at a time. The ion, after being ripped off the surface, will be totally surrounded by additional water molecules. An aqueous (aq) ion has been created. This process of breaking up the ionic crystal as it dissolves in water is called dissociation. Part B: Chemical Reactions (22 marks) (approximate time: 20 minutes) 7. Give the name of each of the following. (2 marks: 1 mark each) iron(II) sulphate a) FeSO4 _____________________________________________________ mercury(II) chloride b) HgCl2 _____________________________________________________ Write the formula of each of the following. (2 marks: 1 mark each) AgNO3 a) silver nitrate ________________________________________________ 8. P2O3 b) diphosphorus trioxide ________________________________________ 9. Balance the following chemical equation: (2 marks) 3 Ag2SO4 + CoCl 6 2 3 → Co2(SO4)3 + AgCl 3 Practice Test Chemistry SCH3U-C 10. Identify the correct products for each chemical equation. Identify the reaction type for each equation. (8 marks) Equation a) AgNO3 + BaCl2 → Ba(NO3)2 + AgCl b) Li3N → Li + N2 c) Zn + PbCl2 → ZnCl2 + Pb d) Rb + O2 → Rb2O Reaction type double displacement ________________________ decomposition ________________________ single displacement ________________________ synthesis ________________________ 11. Identify the correct products for the following neutralization reaction: HCl + Mg(OH)2 → MgCl2 + H2O (2 marks) 12. a) Write a balanced chemical equation for the complete combustion of C5H12 in oxygen, O2. (3 marks) C5H12 + 8O2 --> 5CO2 + 6H2O b) Use the same reactants and write a balanced chemical equation for the incomplete combustion. (3 marks) C5H12 + 4O2 --> 2CO + 3C + 6H2O Part C: Quantities in Chemical Reactions (17 marks) (approximate time: 20 minutes) 13. Calculate the following using complete solutions and indicating proper units and appropriate significant digits: a) The mass of 0.250 mol of phosphorous trichloride. (2 marks) phosphorous trichloride --> PCl3; M(PCl3) = 30.97 g/mol + 3(35.45 g/mol) = 137.32 g/mol mass = n x M = 0.250 mol x 137.32 g/mol = 34.3 g 4 Practice Test Chemistry SCH3U-C b) The mass of 3.14 × 1031 molecules of carbon dioxide. (3 marks) n = N/NA = 3.14 x 10^31 molecules / 6.02x10^23 molecules/mol = 5.22 x 10^7 mol mass = n x MCO2 = 5.22x10^7 mol x (12.01 + 2x16.00) g/mol = 2.30 x 10^9 g 14. In the reaction between aluminum ribbon (a silver grey solid) and copper(II) chloride solution (clear blue solution), one of the possible products is a clear, colourless solution and the other is a reddish brown solid. Describe what physical observations can be used to identify the limiting reagent in this reaction. (2 marks) - If the blue colour disappears, then CuCl2 is the limiting reagent. - If the aluminum disappears, the Al is the limiting reagent. 15. A 100.0 g sample of a compound is composed of 16.3 g of carbon, 32.1 g of chlorine, and the remainder is fluorine. Determine the empirical formula of the compound. (5 marks) 5 Practice Test Chemistry SCH3U-C 16. Determine the mass of lead(II) chloride produced when 2.57 g of sodium chloride reacts with excess aqueous lead(II) sulphate. What would the percent yield be if only 5.71 g of lead(II) chloride were actually produced? (5 marks) 2NaCl(aq) + PbSO4(aq) --> Na2SO4(aq) + PbCl2(s) 2.57 g moles NaCl = 2.57 g / (22.99 + 35.45) g/mol = 0.0440 mol 2 mol NaCl --> 1 mol PbCl2 0.0440 mol NaCl --> 0.0220 PbCl2 mass PbCl2 = 0.0220 mol x (207.2 + 2 x 35.45) g/mol = 6.12 g PbCl2 (theoretical yield) % yield = 5.71 g / 6.12 g x 100% = 93.3 % Part D: Solutions and Solubility (21 marks) (approximate time: 20 minutes) 17. Consider a clear, colourless solution that could contain Al3+ ions or Pb2+ ions. Show, using sequential analysis, how you could identify which ions are present. (6 marks) 6 Practice Test Chemistry SCH3U-C 18. a) What is the molar concentration of a solution that contains 165 g of sodium sulphate dissolved in 3.00 L of solution? (3 marks) n = moles of sodium sulphate = 165 g/(142.01 g/mol) = 1.16 mol c=n/v = 1.16 mol / 3.00 L = 0.387 mol/L b) How much water would be required to decrease the concentration to 0.111 mol/L? (2 marks) c1v1 = c2v2 v2 = c1v1 / c2 = (0.387 mol/L)(3.00 L) / (0.111 mol/L) = 10.5 L (final volume) Therefore, 7.5 L of water would need to be added to the 3.00 L of the concentrated stock solution. 19. Fill in the following table: (5 marks: 1 mark each) Formula Name Concentration (mol/L) pH HCl (aq) hydrochloric acid 2.3 × 10–2 1.16(?) NaOH sodium hydroxide 3.32 × 10–4 10.5 HNO3 nitric acid 2.5 x 10^-6 7 5.6 Practice Test Chemistry SCH3U-C 20. Given the following data, determine the concentration and pH of the acid if 25.0 mL of the hydrochloric acid was titrated using 0.120 mol/L sodium hydroxide. (5 marks) Trial 1 Trial 2 Trial 3 final burette reading (mL) 17.9 36.0 70.3 initial burette reading (mL) 0.1 18.0 52.3 17.8 18.0 amount of NaOH used average volume NaOH = (17.8 + 18.0 + 18.0) ml / 3 = 17.9 mL HCl (aq) + NaOH (aq) --> H2O (aq) + NaCl (aq) moles NaOH = 0.120 mol/L x 0.0179 L = 2.148 x 10^-3 mol 1 mol HCl --> 1 mol NaOH 2.148 x 10^-3 mol HCl --> 2.148 x 10^-3 mol NaOH [HCl(aq)] = 2.148 x 10^-3 mol / 0.0250 L = 0.0859 mol/L pH = -log(0.0859) = 1.07 8 18.0 Practice Test Chemistry SCH3U-C Part E: Gases and Atmospheric Chemistry (29 marks) (approximate time: 25 minutes) 21. 2.75 g of magnesium metal mixes with 125 mL of 1.00 mol/L HCl. The balanced equation for the chemical reaction is Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) a) Determine which is the limiting reagent. (3 marks) Moles of Mg present = 2.75 g/(24.32 g/mol) = 0.113 mol Moles of HCl present = 0.125 L x 1.00 mol/L = 0.125 mol 1 mol Mg --> 2 mol HCl 0.113 mol Mg --> 0.226 mol HCl Since we were given only 0.125 mol of HCl, the HCl is insufficient and will be the limiting reagent. b) Determine how many moles of the gas are formed in the reaction. (2 marks) 2 mol HCl --> 1 mol H2 0.125 mol HCl --> 0.0625 mol H2 Therefore, 0.0625 mol of H2 gas will be formed. 9 Practice Test Chemistry SCH3U-C c) Determine the volume of the hydrogen gas if the conditions are SATP. (3 marks) PV = nRT V = nRT / V n = 0.0625 mol R = 8.31 (kPa)(L)/(mol)(K) T = 25 degrees Celcius + 273 = 298 K P = 100. kPa V = (0.0625 mol)(8.31 kPa x L)(298 K) / (100. kPa)(mol x K) = 1.55 L At SATP, 1 mol of any gas has a volume of 24.8 L (molar volume at SATP): V = 0.0625 mol x 24.8 L/mol = 1.55 L 22. An automobile tire has an internal volume of 27 L at 225 kPa and 18°C. a) What would the volume of the air be if it escaped from the tire into a balloon at standard atmospheric pressure but at constant temperature? (2 marks) V1P1 / T1 = V2P2 / T2 but here T1 = T2 so these will cancel. V1P1 = V2P2 V2 = (27 L)(225 kPa) / 100 kPa = 61 L b) Describe what happens to the air in a tire as a car is driven on a hot day. (2 marks) - On a hot day, the particles will increase their kinetic motion as they heat up. - This increases the force of the collisions with the walls of the tire. - This causes an increase in pressure. 10 Practice Test Chemistry SCH3U-C 23. A balloon is brought to the top of Mt. Logan, where it occupies a volume of 775 mL at a temperature of –28°C and a pressure of 92.5 kPa. What is the pressure at the bottom of the mountain if the same balloon has a volume of 825 mL at a temperature of 15°C? (3 marks) V1P1 / T1 = V2P2 / T2 P2 = V1P1T2 / T1V2 V1 = 775 mL P1 = 92.5 kPa T1 = - 28 degrees C +273 = 245 K V2 = 825 mL P2 = (92.5 kPaa)(775 mL)(288 K) / (245 K)(825 mL) = 102 kPa 24. What volume of hydrogen gas can be produced by the downward displacement of water (Pvap = 3.17 kPa) when 3.55 g of zinc metal is reacted with 555 mL of 1.80 × 10–1 M hydrochloric acid at SATP? (6 marks) Zn(s) + 2HCl(aq) --> H2(g) + ZnCl2(aq) 3.55 g 555 mL 1.80x10^-1 M moles Zn = 3.55 g / 65.41 g/mol = 0.0543 mol moles HCl = 0.555 L x 1.80 x 10^-1 M = 0.0999 mol/L 1 mol Zn --> 2 mol HCL 0.0543 mol Zn --> 0.109 mol HCl Since we only have 0.0999 mol of HCl, the HCl will run out so it is limiting. 2 mol HCl --> 1 mol H2 0.0999 mol HCl --> 0.04995 mol H2 PV = nRT --> V = nRT / P (taking into account vapour pressure of water) volume H2 = (0.04995 mol)(8.31 kPa x L/mol x K)(298 K) / (100 - 3.17 kPa) = 1.28 L 11 Practice Test Chemistry SCH3U-C 25. Describe how chlorofluorcarbons used as refrigerants or propellants can contribute to global warming. (4 marks) - CFCs can be released into the atmosphere and then make their way into the upper atmosphere. - Free chlorine atoms from the CFC molecules are able to collide with and splice the ozone molecules into smaller molecules. - This thinning of the ozone layer allows more UV light and energy to enter and heat up the atmosphere. - This causes an overall increase in the temperature of the earth. 26. Describe the evaporation of a liquid in terms of kinetic molecular theory. (2 marks) In a liquid, the molecules can move around from place to place somewhat freely. At the surface, a molecule can be bumped out of the liquid into the air space above, and thus it evaporates. 12 Practice Test Chemistry SCH3U-C 27. Explain any difference between a real gas and an ideal gas if both are cooled. (2 marks) When a real gas is cooled, its molecules move more slowly. As the molecules slow down, their attractive forces become more significant. Eventually, if they move slowly enough they can begin to stick together in a liquid state and thus condense. An ideal gas will also slow down, but since there are no attractive forces between molecules it will never condense to a liquid. 13 ...
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