_ap06_chemistry_sg - AP® Chemistry 2006 Scoring Guidelines...

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Unformatted text preview: AP® Chemistry 2006 Scoring Guidelines The College Board: Connecting Students to College Success The College Board is a not-for-profit membership association whose mission is to connect students to college success and opportunity. Founded in 1900, the association is composed of more than 5,000 schools, colleges, universities, and other educational organizations. Each year, the College Board serves seven million students and their parents, 23,000 high schools, and 3,500 colleges through major programs and services in college admissions, guidance, assessment, financial aid, enrollment, and teaching and learning. Among its best-known programs are the SAT®, the PSAT/NMSQT®, and the Advanced Placement Program® (AP®). The College Board is committed to the principles of excellence and equity, and that commitment is embodied in all of its programs, services, activities, and concerns. © 2006 The College Board. All rights reserved. College Board, AP Central, APCD, Advanced Placement Program, AP, AP Vertical Teams, Pre-AP, SAT, and the acorn logo are registered trademarks of the College Board. Admitted Class Evaluation Service, CollegeEd, connect to college success, MyRoad, SAT Professional Development, SAT Readiness Program, and Setting the Cornerstones are trademarks owned by the College Board. PSAT/NMSQT is a registered trademark of the College Board and National Merit Scholarship Corporation. All other products and services may be trademarks of their respective owners. Permission to use copyrighted College Board materials may be requested online at: www.collegeboard.com/inquiry/cbpermit.html. Visit the College Board on the Web: www.collegeboard.com. AP Central is the official online home for the AP Program: apcentral.collegeboard.com. AP® CHEMISTRY 2006 SCORING GUIDELINES Question 1 1. Answer the following questions that relate to solubility of salts of lead and barium. (a) A saturated solution is prepared by adding excess PbI2(s) to distilled water to form 1.0 L of solution at 25°C. The concentration of Pb2+ (aq) in the saturated solution is found to be 1.3 × 10 − 3 M . The chemical equation for the dissolution of PbI2(s) in water is shown below. → PbI2(s) ← Pb2+(aq) + 2 I −(aq) (i) Write the equilibrium-constant expression for the equation. K sp = [Pb 2+ ][I− ]2 One point is earned for the correct expression. (ii) Calculate the molar concentration of I −(aq) in the solution. By stoichiometry, [I− ] = 2 × [Pb2+] , thus [I− ] = 2 × (1.3 × 10 − 3 ) = 2.6 × 10 − 3 M One point is earned for the correct concentration. (iii) Calculate the value of the equilibrium constant, Ksp . Ksp = [Pb2+ ][I−]2 = (1.3 × 10−3)(2.6 × 10−3) 2 One point is earned for a value of Ksp that is consistent with the answers in parts (a)(i) and (a)(ii). = 8.8 × 10−9 (b) A saturated solution is prepared by adding PbI2(s) to distilled water to form 2.0 L of solution at 25°C. What are the molar concentrations of Pb2+ (aq) and I −(aq) in the solution? Justify your answer. The molar concentrations of Pb2+ (aq) and I −(aq) would be the same as in the 1.0 L solution in part (a) (i.e., 1.3 × 10 − 3 M and 2.6 × 10 − 3 M , respectively). The concentrations of solute particles in a saturated solution are a function of the constant, Ksp , which is independent of volume. One point is earned for the concentrations (or stating they are the same as in the solution described in part (a)) and justification. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 2 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 1 (continued) (c) Solid NaI is added to a saturated solution of PbI2 at 25°C. Assuming that the volume of the solution does not change, does the molar concentration of Pb2+ (aq) in the solution increase, decrease, or remain the same? Justify your answer. [Pb2+] will decrease. The NaI(s) will dissolve, increasing [I− ] ; more I−(aq) then combines with Pb2+(aq) to precipitate PbI2(s) so that the ion product [Pb2+][I− ] 2 will once again attain the value of 8.8 × 10− 9 (Ksp at 25°C). One point is earned for stating that [Pb2+] will decrease. One point is earned for justification (can involve a Le Chatelier argument). (d) The value of Ksp for the salt BaCrO4 is 1.2 × 10−10. When a 500. mL sample of 8.2 × 10− 6 M Ba(NO3)2 is added to 500. mL of 8.2 × 10− 6 M Na2CrO4 , no precipitate is observed. (i) Assuming that volumes are additive, calculate the molar concentrations of Ba2+ (aq) and CrO42−(aq) in the 1.00 L of solution. New volume = 500. mL + 500. mL = 1.000 L , therefore [Ba2+] in 1.000 L is one-half its initial value: [Ba2+] = 500. mL × (8.2 × 10− 6 M ) = 4.1 × 10− 6 M 1,000. mL [CrO42−] = One point is earned for the correct concentration. 500. mL × (8.2 × 10− 6 M ) = 4.1 × 10− 6 M 1,000. mL (ii) Use the molar concentrations of Ba2+ (aq) ions and CrO42−(aq) ions as determined above to show why a precipitate does not form. You must include a calculation as part of your answer. The product Q = [Ba2+][CrO42−] One point is earned for calculating a value of Q that is consistent with the concentration values in part (d)(i). = (4.1 × 10− 6 M)(4.1 × 10− 6 M) = 1.7 × 10− 11 Because Q = 1.7 × 10− 11 < 1.2 × 10−10 = Ksp , no precipitate forms. One point is earned for using Q to explain why no precipitate forms. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 3 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 2 CO(g) + 1 O (g) → CO2(g) 22 2. The combustion of carbon monoxide is represented by the equation above. (a) Determine the value of the standard enthalpy change, ∆H rxn , for the combustion of CO(g) at 298 K using the following information. C(s) + 1 O (g) → CO(g) 22 ∆H 298 = − 110.5 kJ mol−1 ∆H 298 = − 393.5 kJ mol−1 C(s) + O2(g) → CO2(g) Reverse the first equation and add it to the second equation to obtain the third equation. CO(g) → 1 O (g) + C(s) 22 + C(s) + O2(g) → CO2(g) ∆H 298 = + 110.5 kJ mol−1 1 O (g) → CO2(g) 22 One point is earned for the correct answer (with sign). ∆H 298 = − 393.5 kJ mol−1 _______________________________ CO(g) + One point is earned for reversing the first equation. OR ∆H rxn = 110.5 + (− 393.5) = −283.0 kJ mol−1 Two points are earned for determining ∆H rxn from the enthalpies of formation. OR ∆H rxn = ∆H f of CO2(g) − ∆H f of CO(g) = −393.5 kJ mol−1 − (−110.5 kJ mol−1) = − 283.0 kJ mol−1 (If sign is incorrect, only one point is earned.) (b) Determine the value of the standard entropy change, ∆S rxn , for the combustion of CO(g) at 298 K using the information in the following table. S2 98 Substance (J mol−1 K−1) CO(g) 197.7 CO2(g) 213.7 O2(g) 205.1 © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 4 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 2 (continued) One point is earned for ∆S rxn = 213.7 J mol−1 K−1 − (197.7 J mol−1 K−1 + = − 86.5 J 1 (205.1 J mol−1 K−1)) 2 mol−1 K−1 taking one-half of S2 98 for O2(g). One point is earned for the answer (with sign). (c) Determine the standard free energy change, ∆Grxn , for the reaction at 298 K. Include units with your answer. One point is earned for substituting the values from parts (a) and (b) into the equation. ∆Grxn = ∆H rxn − T ∆S rxn = −283.0 kJ mol−1 − (298 K)(− 0.0865 kJ mol−1 K−1) ∆Grxn = − 257.2 kJ mol−1 One point is earned for the answer (with sign and units). (d) Is the reaction spontaneous under standard conditions at 298 K ? Justify your answer. Yes, the reaction is spontaneous because the value of ∆Grxn for the reaction is negative (− 257.2 kJ mol−1). One point is earned for an answer with justification (consistent with the answer in part (c)). (e) Calculate the value of the equilibrium constant, Keq , for the reaction at 298 K. ∆Grxn = −R T ln Keq ⇒ -257,200 J mol -1 -(8.31 J mol -1 K -1 )(298 K) ∆Grxn = ln Keq − RT = ln Keq ⇒ One point is earned for correct substitution into the equation. Keq = 1.28 × 1045 One point is earned for the answer. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 5 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 3 3. Answer the following questions that relate to the analysis of chemical compounds. (a) A compound containing the elements C , H , N , and O is analyzed. When a 1.2359 g sample is burned in excess oxygen, 2.241 g of CO2(g) is formed. The combustion analysis also showed that the sample contained 0.0648 g of H. (i) Determine the mass, in grams, of C in the 1.2359 g sample of the compound. 2.241 g CO2(g) × 1 molCO2 12.011 g C 1 mol C × × 1 mol C 1 mol CO2 44.01 g CO2 One point is earned for the correct answer. = 0.6116 g C (ii) When the compound is analyzed for N content only, the mass percent of N is found to be 28.84 percent. Determine the mass, in grams, of N in the original 1.2359 g sample of the compound. 1.2359 g sample × 0.2884 = 0.3564 g N One point is earned for the correct answer. (iii) Determine the mass, in grams, of O in the original 1.2359 g sample of the compound. Because the compound contains only C , H , N , and O , mass of O = g sample − ( g H + g C + g N ) = 1.2359 − (0.0648 + 0.6116 + 0.3564) = 0.2031 g One point is earned for the answer consistent with the answers in parts (a)(i) and (a)(ii). (iv) Determine the empirical formula of the compound. Converting all masses to moles, 1 mol C 12.011 g C 1 mol H 0.0648 g H × 1.0079 g H 1 mol N 0.3564 g N × 14.007 g N 1 mol O 0.2031 g O × 16.00 g O 0.6116 g C × = 0.05092 mol C One point is earned for all masses converted to moles. = 0.06429 mol H Note: Moles of C may be shown in part (a)(i). = 0.02544 mol N = 0.01269 mol O © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 6 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 3 (continued) Divide all mole quantities by the smallest number of moles: 0.05092 mol 0.06429 mol 0.02544 mol 0.01269 mol ÷ ÷ ÷ ÷ 0.01269 0.01269 0.01269 0.01269 mol mol mol mol = = = = One point is earned for dividing by the smallest number of moles. 4.013 5.066 2.005 1.000 One point is earned for the empirical formula consistent with the ratio of moles calculated. ⇒ Empirical formula is C4H5N2O (b) A different compound, which has the empirical formula CH2Br , has a vapor density of 6.00 g L− 1 at 375 K and 0.983 atm. Using these data, determine the following. (i) The molar mass of the compound PV = nRT ⇒ PV =n RT (0.983 atm)(1.00 L) (0.0821 L atm mol -1K -1 )(375 K) molar mass of gas (M) = One point is earned for applying the gas law to calculate n. = 0.0319 mol One point is earned for calculating the molar mass. 6.00 g = 188 g mol−1 0.0319 mol OR OR M= DRT 6.00 g L−1 × 0.0821 L atm mol−1 K −1 × 375 K = 0.983 atm P = 188 g mol−1 Two points are earned for calculating the molar mass using M = DRT P (ii) The molecular formula of the compound Each CH2Br unit has mass of 12.011 + 2(1.0079) + 79.90 = 93.9 g, 188 g and = 2.00, so there must be two CH2Br units per molecule. 93.9 g Therefore, the molecular formula of the compound is C2H4Br2 . One point is earned for the molecular formula that is consistent with the molar mass calculated in part (b)(i). © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 7 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 4 4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations described below. Answers to more than five choices will not be graded. In all cases, a reaction occurs. Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged by the reaction. You need not balance the equations. General Scoring: Three points are earned for each reaction: 1 point for correct reactant(s) and 2 points for correct product(s). Designation of physical states is not required. (a) Solid potassium chlorate is strongly heated. KClO3 → KCl + O2 (b) Solid silver chloride is added to a solution of concentrated hydrochloric acid. AgCl + Cl− → [AgCl2 ]− (c) A solution of ethanoic (acetic) acid is added to a solution of barium hydroxide. HC2H3O2 + OH − → H2O + C2H3O2− (d) Ammonia gas is bubbled into a solution of hydrofluoric acid. NH3 + HF → NH4+ + F − (e) Zinc metal is placed in a solution of copper(II) sulfate. Zn + Cu2+ → Zn2+ + Cu (f) Hydrogen phosphide (phosphine) gas is added to boron trichloride gas. PH3 + BCl 3 → H 3PBCl 3 Note: PH3BCl3 also acceptable as a product. (g) A solution of nickel(II) bromide is added to a solution of potassium hydroxide. Ni2+ + OH− → Ni(OH)2 (h) Hexane is combusted in air. C6H14 + O2 → CO2 + H2O © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 8 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 5 5. Three pure, solid compounds labeled X , Y , and Z are placed on a lab bench with the objective of identifying each one. It is known that the compounds (listed in random order) are KCl , Na2CO3 , and MgSO4 . A student performs several tests on the compounds; the results are summarized in the table below. Compound pH of an Aqueous Solution of the Compound Result of Adding 1.0 M NaOH to a Solution of the Compound Result of Adding 1.0 M HCl Dropwise to the Solid Compound X >7 No observed reaction Evolution of a gas Y 7 No observed reaction No observed reaction Z 7 Formation of a white precipitate No observed reaction (a) Identify each compound based on the observations recorded in the table. Compound X ______ Na2CO3 _______ Compound Y ______ KCl _________ Compound Z _____ MgSO4 _________ One point is earned for one correct identification, and a second point is earned for a second correct identification. (No points are earned if all three identifications are the same compound; no second point is earned if two identifications are the same compound.) (b) Write the chemical formula for the precipitate produced when 1.0 M NaOH is added to a solution of compound Z . One point is earned for the correct formula. Mg(OH)2 (c) Explain why an aqueous solution of compound X has a pH value greater than 7. Write an equation as part of your explanation. One point is earned for identifying CO32− as a base. CO32− reacts with water to form OH −. CO32−(aq) + H2O(l) → OH −(aq) + HCO3−(aq) One point is earned for a correct equation. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 9 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 5 (continued) (d) One of the testing solutions used was 1.0 M NaOH . Describe the steps for preparing 100. mL of 1.0 M NaOH from a stock solution of 3.0 M NaOH using a 50 mL buret, a 100 mL volumetric flask, distilled water, and a small dropper. 1,000 mL of 1.0 M NaOH contains 1.0 mol NaOH; therefore, 100. mL of 1.0 M NaOH contains 0.10 mol NaOH (i.e., 0.10 mol NaOH is needed for the solution) volume of 3.0 M NaOH needed = 0.10 mol NaOH × 1,000 mL 3.0 mol NaOH = 33 mL Step 1: Use the buret to deliver 33 mL of the 3.0 M NaOH stock solution into the clean 100 mL volumetric flask. Step 2: Add distilled water to the flask until the liquid level is just below the calibration line in the neck of the flask; swirl gently to mix. Step 3: Use the small dropper to add the last amount of distilled water, drop by drop, until the bottom of the meniscus in the flask neck is level with the calibration line. Insert the stopper, and invert the flask to mix. One point is earned for using the buret to dispense 33 mL of NaOH(aq). One point is earned for adding distilled water to the calibration mark. (e) Describe a simple laboratory test that you could use to distinguish between Na2CO3(s) and CaCO3(s). In your description, specify how the results of the test would enable you to determine which compound was Na2CO3(s) and which compound was CaCO3(s) . A water solubility test would work. Put a small amount of one substance in a beaker of distilled water. If the substance dissolves readily when stirred, then it is Na2CO3 ; if it does not dissolve, it is CaCO3 . OR A flame test would work. Dip a moistened wire into a sample of one of the substances and place the wire in the flame of a bunsen burner. If a bright orange-yellow color is observed, then the sample is Na2CO3 ; if a brick red color is observed, it is CaCO3 . Note: The student does NOT have to perform a confirmatory test on the other substance if one has already been identified with a test. One point is earned for any reasonable test. One point is earned for interpreting the results that will identify one compound. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 10 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 6 6. Answer each of the following in terms of principles of molecular behavior and chemical concepts. (a) The structures for glucose, C6H12O6 , and cyclohexane, C6H12 , are shown below. Identify the type(s) of intermolecular attractive forces in (i) pure glucose Hydrogen bonding OR dipole-dipole interactions OR van der Waals interactions (London dispersion forces may also be mentioned.) One point is earned for a correct answer. (ii) pure cyclohexane One point is earned for London dispersion forces. London dispersion forces (b) Glucose is soluble in water but cyclohexane is not soluble in water. Explain. The hydroxyl groups in glucose molecules can form strong hydrogen bonds with the solvent (water) molecules, so glucose is soluble in water. In contrast, cyclohexane is not capable of forming strong intermolecular attractions with water (no hydrogen bonding), so the water-cyclohexane interactions are not as energetically favorable as the interactions that already exist among polar water molecules. OR • Glucose is polar and cyclohexane is nonpolar. • Polar solutes (such as glucose) are generally soluble in polar solvents such as water. • Nonpolar solutes (such as cyclohexane) are not soluble in the polar solvent. One point is earned for explaining the solubility of glucose in terms of hydrogen bonding or dipole-dipole interactions with water. One point is earned for explaining the difference in the polarity of cyclohexane and water. OR One point is earned for any one of the three concepts; two points are earned for any two of the three concepts. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 11 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 6 (continued) (c) Consider the two processes represented below. ∆ H ° = + 44.0 kJ mol−1 Process 1: H2O(l) → H2O(g) Process 2: H2O(l) → H2(g) + 1 O (g) 22 ∆ H ° = + 286 kJ mol−1 (i) For each of the two processes, identify the type(s) of intermolecular or intramolecular attractive forces that must be overcome for the process to occur. In process 1, hydrogen bonds (or dipole-dipole interactions) in liquid water are overcome to produce distinct water molecules in the vapor phase. One point is earned for identifying the type of intermolecular force involved in process 1. In process 2, covalent bonds (or sigma bonds, or electron-pair bonds) within water molecules must be broken to allow the atoms to recombine into molecular hydrogen and oxygen. One point is earned for identifying the type of intramolecular bonding involved in process 2. (ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation. When water boils, H2O molecules break apart to form hydrogen molecules and oxygen molecules. I disagree with the statement. Boiling is simply Process 1, in which only intermolecular forces are broken and the water molecules stay intact. No intramolecular or covalent bonds break in this process. One point is earned for disagreeing with the statement and providing a correct explanation. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 12 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 6 (continued) (d) Consider the four reaction-energy profile diagrams shown below. (i) Identify the two diagrams that could represent a catalyzed and an uncatalyzed reaction pathway for the same reaction. Indicate which of the two diagrams represents the catalyzed reaction pathway for the reaction. Diagram 1 represents a catalyzed pathway and diagram 2 represents an uncatalyzed pathway for the same reaction. One point is earned for identifying the correct graphs and indicating which represents which pathway. (ii) Indicate whether you agree or disagree with the statement in the box below. Support your answer with a short explanation. Adding a catalyst to a reaction mixture adds energy that causes the reaction to proceed more quickly. I disagree with the statement. A catalyst does not add energy, but provides an alternate reaction pathway with a lower activation energy. One point is earned for disagreeing with the statement and providing an explanation. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 13 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 7 7. Answer the following questions about the structures of ions that contain only sulfur and fluorine. (a) The compounds SF4 and BF3 react to form an ionic compound according to the following equation. SF4 + BF3 → SF3BF4 (i) Draw a complete Lewis structure for the SF3+ cation in SF3BF4 . One point is earned for the correct Lewis structure (the structure must include lone pairs of electrons, which may be represented as dashes). (ii) Identify the type of hybridization exhibited by sulfur in the SF3+ cation. sp3 One point is earned for the correct hybridization. (iii) Identify the geometry of the SF3+ cation that is consistent with the Lewis structure drawn in part (a)(i). One point is earned for the correct shape. Trigonal pyramidal (iv) Predict whether the F–S–F bond angle in the SF3+ cation is larger than, equal to, or smaller than 109.5°. Justify your answer. The F–S –F bond angle in the SF3+ cation is expected to be slightly smaller than 109.5° because the repulsion between the nonbonding pair of electrons and the S –F bonding pairs of electrons “squeezes” the F–S –F bond angles together slightly. One point is earned for stating that the angle is smaller, with justification. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 14 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 7 (continued) (b) The compounds SF4 and CsF react to form an ionic compound according to the following equation. SF4 + CsF → CsSF5 (i) Draw a complete Lewis structure for the SF5− anion in CsSF5 . One point is earned for the correct Lewis structure (the structure must include lone pairs of electrons, which may be represented as dashes). (ii) Identify the type of hybridization exhibited by sulfur in the SF5− anion. sp3d2 One point is earned for the correct hybridization. (iii) Identify the geometry of the SF5− anion that is consistent with the Lewis structure drawn in part (b)(i). One point is earned for the correct shape. Square pyramidal (iv) Identify the oxidation number of sulfur in the compound CsSF5 . +4 One point is earned for the correct oxidation number. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 15 AP® CHEMISTRY 2006 SCORING GUIDELINES Question 8 8. Suppose that a stable element with atomic number 119, symbol Q , has been discovered. (a) Write the ground-state electron configuration for Q , showing only the valence-shell electrons. One point is earned for the electron configuration. 8s 1 (b) Would Q be a metal or a nonmetal? Explain in terms of electron configuration. One point is earned for the correct answer and explanation, which must include reference to the valence electron. It would be a metal (OR an alkali metal). The valence electron would be held only loosely. (c) On the basis of periodic trends, would Q have the largest atomic radius in its group or would it have the smallest? Explain in terms of electronic structure. One point is earned for the correct answer and explanation; the size must refer to number of electron shells. It would have the largest atomic radius in its group because its valence electron is in a higher principal shell. (d) What would be the most likely charge of the Q ion in stable ionic compounds? One point is earned for the correct charge. (Must be consistent with configuration in part (a).) +1 (e) Write a balanced equation that would represent the reaction of Q with water. 2 Q(s) + 2 H2O(l) → 2 Q+ (aq) + 2 OH −(aq) One point is earned for H2 as a product. + H2(g) One point is earned for balancing the equation. (f) Assume that Q reacts to form a carbonate compound. (i) Write the formula for the compound formed between Q and the carbonate ion, CO32− . Q2CO3 One point is earned for the formula consistent with the charge given in part (d). (ii) Predict whether or not the compound would be soluble in water. Explain your reasoning. One point is earned for the answer consistent with the identification of Q. It would be soluble in water because all alkali metal carbonates are soluble. © 2006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 16 ...
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