AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 January 12 TOPICS COVERED: Fundatinal Tpics, Reactins, Gases, Thermchemistry, Atmic Structure, Peridicity, and Bnding. MULTIPLE CHOICE KEY AND CROSSWALK Questin Number Crrect Answer Essential Knwledge Learning Objective Science 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 Prir knwledge 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 Cpyright 2017 Natinal Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us nline at www.nms.rg 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 CH 4 (g) + 2 Cl 2 (g) CH 2 Cl 2 (g) + 2 HCl(g) 26. Methane gas reacts with chlrine gas t frm dichlrmethane and hydrgen chlride, as represented by the equatin abve. (a) A 25.0 g sample f methane is placed in a reactin vessel cntaining 2.58 ml f Cl2(g). (i) Identify the limiting reactant when the methane and chlrine gases are cmbined. Justify yur answer with a calculatin. Cl 2 is the limiting reactant 25.0 g CH 4 1 ml CH 4 16.0 g CH 4 1 ml CH 2 Cl 2 1 ml CH 4 =1.56 ml CH 2 Cl 2 2.58 ml Cl 2 1 ml CH 2 Cl 2 2 ml Cl 2 =1.29 ml CH 2 Cl 2 1 pint fr the crrect claim 1 pint fr prper justificatin (ii) Calculate the ttal number f mles f CH2Cl2(g) in the cntainer after the limiting reactant has been ttally cnsumed. 1.29 ml CH 2 Cl 2 is prduced when 2.58 ml f Cl 2 is cnsumed. 1 pint cnsistent with part a(i); wrk may be shwn in part a(i) r prper explanatin in part (ii) Cpyright 2017 Natinal Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us nline at www.nms.rg 2
Initiating mst reactins invlving chlrine gas invlves breaking the Cl Cl bnd, which has a bnd energy f 242 kj ml 1. (b) Calculate the amunt f energy, in jules, needed t break a single Cl Cl bnd. 242 kj ml f bnds 1000 J 1ml bnds 1 kj 6.02 10 23 bnds = 4.02 10 19 J/bnd 1 pint fr crrect answer (c) Wuld a red laser, wavelength f 656 nanmeters, supply the needed energy t break the Cl Cl bnd? Justify yur answer. Red light des nt have enugh energy t break the Cl Cl bnd since the red phtn nly cntains 3.03x10-19 J. E = hc λ 1 pint fr claim that red light wuld nt be enugh 1 pint fr supprting evidence E = (6.626 10 34 Js)(3.0 10 8 m/s)(10 9 nm/m) (656 nm) E = 3.03 10 19 J (d) (i) Draw the Lewis structure fr dichlrmethane, CH2Cl2(g), in the bx prvided. H 1 pint fr a crrect Lewis structure H C Cl Cl (ii) Identify the mlecular structure fr this mlecule. tetrahedral 1 pint fr a crrect mlecular structure Cpyright 2017 Natinal Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us nline at www.nms.rg 3
(e) Cnsider the fllwing Bltzmann distributin curve representing the prducts f this reactin. Identify the peak that represents HCl. Explain yur reasning. Number f Mlecules HCl 0 2200 4000 6000 8000 10000 12000 14000 HCl shuld be assigned t the peak with greater number f particles with greater speed (the lwest peak n this graph) Speed (m/s) 1 pint fr identificatin f the crrect peak Bth f the prducts are at the same temperature and pssess the same average kinetic energy; hwever, CH 2 Cl 2 has a mass f 84.9 g/ml while HCl has a mass f 36.5 g/ml. The greater the mass, the slwer the particles because KE = ½ mv 2 1 pint fr apprpriate reasning 27. Prpane, C 3, is a hydrcarbn that is cmmnly used as fuel fr cking. (a) Write a balanced equatin fr the cmplete cmbustin f prpane gas, which yields CO 2 (g) and H 2 O(l). C 3 + 5 O 2 3 CO 2 + 4 H 2 O 1 pint fr crrect reactants and prducts 1 pint fr crrect balancing Cpyright 2017 Natinal Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us nline at www.nms.rg 4
(b) (i) Calculate the vlume f air at 30 C and 1.00 atmsphere that is needed t burn cmpletely 10.0 grams f prpane. Assume that air is 21.0 percent O2 by vlume at sea level. 10.0 g C 3 1 ml C 3 44.0 g C 3 5 ml O 2 1 ml C 3 = 1.14 ml O 2 V O2 = nrt (1.14 ml)(0.0821 L atm/ml K)(303 K) = P 1.00 atm V O2 = 28.3 L V air = 28.3 L 0.21 =135 L f air 1 pint fr mles f xygen 1 pint fr crrect substitutin and calculatin using ideal gas equatin 1 pint fr vlume f xygen in air (b) (ii) If the reactin were perfrmed at a higher altitude, wuld the vlume f air needed t cmpletely react the prpane be greater than, less than r equal t the calculated vlume at sea level? Explain. The vlume f air needed will be greater than the calculated value at sea level. The amunt f xygen present is less than 21%; dividing by a smaller value will result in a larger vlume. 1 pint fr a crrect claim with apprpriate reasning (c) The heat f cmbustin f prpane is 2,220.1 kj/ml. Calculate the heat f frmatin, ΔH f, f prpane given that ΔH f f H2O(l) = 285.3 kj/ml and ΔH f f CO2(g) = 393.5 kj/ml. Δ H cmb = [ Δ H f (CO + Δ H 2 ) f (H 2 O )] [ Δ H f (C + Δ H 3 ) f (O 2 )] -2220.1 = [3(-393.5) + 4(-285.3)] - [X+ 0] X = ΔH cmb = -101.7 kj/ml 1 pint fr crrect substitutin using cefficients 1 pint fr crrect answer with sign and significant digits (d) Assuming that all f the heat evlved in burning 30.0 grams f prpane is transferred t 8.00 kilgrams f water (specific heat = 4.18 J/g C), calculate the increase in temperature f water. Cpyright 2017 Natinal Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us nline at www.nms.rg 5
q released = 30.0 g C 3 1 ml kj 2220.1 44.0 g 1ml = 1514 kj heat released by prpane = heat absrbed by water q absrbed = ( m)(c p )(ΔT) 1,514,000 J = ( 8000 g)(4.18 J/g. C)(ΔT) ΔT = 45.3 C 1 pint fr kj released by prpane 1 pint fr crrect temperature change Answer the fllwing questins relating t gravimetric analysis. 28. A student is assigned the task f determining the number f mles f water in ne mle f MgCl 2 n H 2 O. The student cllects the data shwn in the fllwing table. Mass f empty cntainer Initial mass f sample and cntainer Mass f sample and cntainer after first heating Mass f sample and cntainer after secnd heating Mass f sample and cntainer after third heating 22.347 g 25.825 g 23.982 g 23.976 g 23.977 g (a) Explain why the student can crrectly cnclude that the hydrate was heated a sufficient number f times in the experiment. A negligible change in mass between the secnd and third heating indicates that all f the water has been remved. 1 pint fr crrect explanatin (b) Use the data abve t (i) calculate the ttal number f mles f water lst when the sample was heated, and 25.825 g hydrate + cntainer 23.977 g anhydrate + cntainer 1.848 g water = 0.1026 ml water 1 pint fr crrect calculatin f mles f water Cpyright 2017 Natinal Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us nline at www.nms.rg 6
(ii) determine the frmula f the hydrated cmpund 23.977 g anhydrate + cntainer 22.347 g cntainer 1.630 g MgCl 2 = 0.01712 ml MgCl 2 1 pint fr crrect frmula with apprpriate calculatin rati is 1:6; therefre, the frmula is MgCl 2 6H 2 O (c) A different student heats the hydrate in an uncvered crucible, and sme f the slid spatters ut f the crucible. What effect will the spattering have n the calculated mass f the water lst by the hydrate? Justify yur answer. Splattering will result in a greater lss f mass that is calculated as water, this will prduce a higher water/salt rati 1 pint fr crrect claim with prper justificatin Cpyright 2017 Natinal Math + Science Initiative, Dallas, Texas. All rights reserved. Visit us nline at www.nms.rg 7