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AP Chemistry Assessment 2

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AP Chemistry Assessment 2 AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 – January 12 TOPICS COVERED: Foundational Topics, Reactions, Gases, Thermochemistry, Atomic Structure, Periodicity, and Bonding. MULTIPLE CHOICE KEY AND CROSSWALK Question Correct Essential Learning Science Number Answer Knowledge Objective Practice Q1 C 2.D.4 2.32 6.2 Q2 D 1.C.1 1.11 5.1 Q3 A 1.C.1 1.9 6.4 Q4 A 1.B.2 1.7 5.1 Q5 A 3A-3C 3.1 7.1 Q6 D 1.A.2 1.2 2.2 Q7 B 1.D.3 1.16 5.1 Q8 D 1.D.2 1.14 1.4 Q9 C 5.B.2 5.5 1.4 Q10 B 3.A.2 3.4 2.2 Q11 B 2.D.1 2.24 6.2 Q12 C 5.B.3 5.6 2.2 Q13 A 3.A.2 3.4 2.2 Q14 C 3.A.1 3.2 1.5 Q15 B 1.A Prior knowledge Q16 C 3.A.2 3.4 2.2 Q17 D 2.C.4 2.21 1.4 Q18 A 2.A.2 2.6 2.3 Q19 C 2.A.2 2.4 6.4 Q20 B 2.C.4 2.21 1.4 Copyright © 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org 1 Q21 B 2.D.1 2.24 7.1 Q22 D 1.C.1 1.9 6.4 Q23 B 2.C.4 2.21 1.4 Q24 C 2.A.3 2.9 1.4 Q25 C 2.D.2 2.25 1.4 3.A.2 3.4 2.2 Rubric Q26 2.C.4 2.21 1.4 2.A.2 2.6 2.2 Rubric Q27 5.C.2 5.8 2.3 Q28 Rubric 1.A.2 1.2 2.2 FREE RESPONSE SCORING GUIDELINES AND CROSSWALK CH4(g) + 2 Cl2(g) → CH2Cl2(g) + 2 HCl(g) 26. Methane gas reacts with chlorine gas to form dichloromethane and hydrogen chloride, as represented by the equation above. (a) A 25.0 g sample of methane is placed in a reaction vessel containing 2.58 mol of Cl2(g). (i) Identify the limiting reactant when the methane and chlorine gases are combined. Justify your answer with a calculation. Cl2 is the limiting reactant 1 point for the correct claim 1 mol CH 1 mol CH Cl 25.0 g CH × 4 × 2 2 =1.56 mol CH Cl 4 16.0 g CH 1 mol CH 2 2 4 4 1 point for proper 1 mol CH Cl 2.58 mol Cl × 2 2 =1.29 mol CH Cl justification 2 2 mol Cl 2 2 2 (ii) Calculate the total number of moles of CH2Cl2(g) in the container after the limiting reactant has been totally consumed. 1.29 mol CH2Cl2 is produced when 2.58 1 point consistent with part a(i); work mol of Cl2 is consumed. may be shown in part a(i) or proper explanation in part (ii) Copyright © 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org 2 Initiating most reactions involving chlorine gas involves breaking the Cl–Cl bond, which has a bond energy of 242 kJ mol–1. (b) Calculate the amount of energy, in joules, needed to break a single Cl–Cl bond. 1 point for correct 242 kJ 1000 J 1mol bonds −19 × × 23 = 4.02×10 J/bond answer mol of bonds 1 kJ 6.02×10 bonds (c) Would a red laser, wavelength of 656 nanometers, supply the needed energy to break the Cl–Cl bond? Justify your answer. Red light does not have enough energy to 1 point for claim that red light would break the Cl−Cl bond since the red photon only not be enough contains 3.03x10-19 J. hC 1 point for supporting evidence E = λ (6.626×10−34 Js)(3.0×108 m/s)(109 nm/m) E = (656 nm) −19 E = 3.03×10 J (d) (i) Draw the Lewis structure for dichloromethane, CH2Cl2(g), in the box provided. H 1 point for a correct Lewis structure H C Cl Cl (ii) Identify the molecular structure for this molecule. tetrahedral 1 point for a correct molecular structure Copyright © 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org 3 (e) Consider the following Boltzmann distribution curve representing the products of this reaction. Identify the peak that represents HCl. Explain your reasoning. HCl Number of Molecules 0 2200 4000 6000 8000 10000 12000 14000 Speed (m/s) HCl should be assigned to the peak with 1 point for identification of the correct peak greater number of particles with greater speed (the lowest peak on this graph) Both of the products are at the same 1 point for appropriate reasoning temperature and possess the same average kinetic energy; however, CH2Cl2 has a mass of 84.9 g/mol while HCl has a mass of 36.5 g/mol. The greater the mass, the slower the particles because KE = ½ mv2 27. Propane, C3H8, is a hydrocarbon that is commonly used as fuel for cooking. (a) Write a balanced equation for the complete combustion of propane gas, which yields CO2(g) and H2O(l). C3H8 + 5 O2 → 3 CO2 + 4 H2O 1 point for correct reactants and products 1 point for correct balancing Copyright © 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org 4 (b) (i) Calculate the volume of air at 30°C and 1.00 atmosphere that is needed to burn completely 10.0 grams of propane. Assume that air is 21.0 percent O2 by volume at sea level. 1 mol C H 5 mol O 1 point for moles of oxygen 10.0 g C H × 3 8 × 2 = 1.14 mol O 3 8 44.0 g C H 1 mol C H 2 3 8 3 8 1 point for correct nRT (1.14 mol)(0.0821 L atm/mol K)(303 K) substitution and calculation VO = = 2 P 1.00 atm using ideal gas equation VO = 28.3 L 2 1 point for volume of oxygen 28.3 L in air Vair = =135 L of air 0.21 (b) (ii) If the reaction were performed at a higher altitude, would the volume of air needed to completely react the propane be greater than, less than or eQual to the calculated volume at sea level? Explain. The volume of air needed will be greater than the 1 point for a correct claim with calculated value at sea level. The amount of oxygen appropriate reasoning present is less than 21%; dividing by a smaller value will result in a larger volume. (c) The heat of combustion of propane is –2,220.1 kJ/mol. Calculate the heat of formation, ΔH°f, of propane given that ΔH°f of H2O(l) = –285.3 kJ/mol and ΔH°f of CO2(g) = –393.5 kJ/mol. o 1 point for correct substitution H H o H o Δ comb = [Δ f (CO ) + Δ f (H O ) ] 2 2 using coefficients o o − Δ H + Δ H [ f (C3 H 8 ) f (O 2 ) ] 1 point for correct answer with -2220.1 = [3(-393.5) + 4(-285.3)] - [X+ 0] sign and significant digits X = ΔH°comb = -101.7 kJ/mol (d) Assuming that all of the heat evolved in burning 30.0 grams of propane is transferred to 8.00 kilograms of water (specific heat = 4.18 J/g °C), calculate the increase in temperature of water. Copyright © 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org 5 1 mol 2220.1 kJ 1 point for kJ released by q = 30.0 g C H × × = 1514 kJ propane released 3 8 44.0 g 1mol heat released by propane = heat absorbed by water 1 point for correct temperature change qabsorbed = (m)(Cp )(ΔT) 1,514,000 J = (8000 g)(4.18 J/g.°C)(ΔT) ΔT = 45.3°C Answer the following questions relating to gravimetric analysis. 28. A student is assigned the task of determining the number of moles of water in one mole of MgCl2•n H2O. The student collects the data shown in the following table. Mass of empty container 22.347 g Initial mass of sample and container 25.825 g Mass of sample and container after first heating 23.982 g Mass of sample and container after second heating 23.976 g Mass of sample and container after third heating 23.977 g (a) Explain why the student can correctly conclude that the hydrate was heated a sufficient number of times in the experiment. A negligible change in mass between the 1 point for correct explanation second and third heating indicates that all of the water has been removed. (b) Use the data above to (i) calculate the total number of moles of water lost when the sample was heated, and 25.825 g hydrate + container 1 point for correct calculation of –23.977 g anhydrate + container moles of water 1.848 g water = 0.1026 mol water Copyright © 2017 National Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us online at www.nms.org 6 (ii) determine the formula of the hydrated compound 23.977 g anhydrate + container 1 point for correct formula with –22.347 g container appropriate calculation 1.630 g MgCl2 = 0.01712 mol MgCl2 ratio is 1:6; therefore, the formula is MgCl2•6H2O (c) A different student heats the hydrate in an uncovered crucible, and some of the solid spatters out of the crucible.
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