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AP Chemistry Exam Spring 2015 Released FRQ’s

AP Chemistry Exam Spring 2015 Released FRQ’s. West Valley High School AP Chemistry Mr. Mata. 1a. (10 points) ( i ) E cell = 0.34 V – (-1.31 V) = 1.65 V (1 point) (ii) The arrow should point to the left ( 1 point ) b. ( i ) The mass increases. ( 1 point )

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AP Chemistry Exam Spring 2015 Released FRQ’s

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  1. AP Chemistry Exam Spring 2015 Released FRQ’s West Valley High School AP Chemistry Mr. Mata

  2. 1a. (10 points) (i) Ecell = 0.34 V – (-1.31 V) = 1.65 V (1 point) (ii) The arrow should point to the left (1 point) b. (i) The mass increases. (1 point) (ii) O2 gas from air reacts with Zn(s) in the cell, producing ZnO(s) which has more mass than the original Zn(s). (1 point) c. (i) The cell potential will be lower. (1 point) (ii) O2(g), a reactant, will be at a lower partial pressure at higher elevation; therefore the reaction has a greater Q value (closer to K). Partial pressure deviations that take the cell closer to equilibrium will decrease the size of the cell potential. (1 point) Problem 1 Solution

  3. d.Na: 1.0 g Na x 1.0 mol Na x 1.0 mol e- = 0.043 mole e- 22.99 g Na 1.0 molNa Ca: 1.0 g Ca x 1.0 mol Ca x 2.0 mol e- = 0.050 mol e- 40.08 g Ca 1.0 mol Ca (1 point) The cell with the Ca anode would transfer more electrons. (1 point) e. (i) 1s22s22p63s23p64s23d10 or [Ar] 4s23d10(1 point) (ii) 4s sublevel (1 point) Problem 1 (cont’d)

  4. a. (i) 35.7 torr x 1 atm= 0.0470 atm 760 torr Pethene = Ptotal – Pwater = 0.822 atm – 0.0470 atm = 0.775 atm (1 point) n = PV = (0.775 atm)(0.0854 L) = 0.00264 mol (1 point) RT (0.8206 L atmmol K)(305 K) (ii) 0.200 g C2H5OH x 1 mol C2H5OH x 1 mol C2H4 = 0.00434 mol C2H4 46.1 g C2H5OH 1 mol C2H5OH (1 point) b. % yield = actual yield x 100 = 0.00264 molx 100 = 60.8% (1 point) max yield 0.00434 mol c. Yes, the data supports the student’s claim. ΔG = Δ H = T Δ S = 45.5 kJ/mol – (298K)(0.126 kJ/K*mol) = 8.0 kJ/mol (1 point) Because Δ G > 0, the value of kp <1.00. (1 point) d. Diagram should include all bonding pairs plus two nonbonding pairs on the O atom (a line may be used to represent an electron pair.) (1 point) e. Approximately 109 degrees (1 point) f. Ethene is slightly soluble in water due to weak dipole/induced dipole IMF between nonpolar ethene and polar water molecules which are weaker than hydrogen bonds between water molecules. Ethanol molecules are soluble in water because they are polar and form hydrogen bonds with water molecules as they dissolve. (1 point for solubility & polarity) (1 point for IMF force comparison) Problem 2 (10 points)

  5. H+ + C6H7O2- <-> HC6H7O2(1 point) b. 1.25 molHCl= x molHClx = 0.0374 molHCl 1000 mL 29.95 mL (1 point) 0.0374 mol C6H7O2-= x mol C6H7O2- -> 0.832 M(1 point) 45.0 mL 1000 mL c. Thymol blue; it has a pKa close to the pH at the equivalence point, so ti will change color near the equivalence point. (1 point for correct ind. )(1 point for correct just.) d. pH = pKa = - log (1.7 x 10 -5) = 4.77 (1 point) e. The ph curve should have the correct shape (reverse sigmoid) (1 point for correct half-equivalence point consistent with d) (1 point for curve that levels off horizont. Through half equiv. point) (1 point for relatively steep neg. slope through equiv. point) f. [HC6H7O2] > [C6H7O2 -] (1 point) Problem 3 10 points

  6. (4 point value) • Ca(OH)2 <-> Ca 2+ + 2 OH -(1 point) b. Ksp = [Ca 2+] [OH-]2 1.3 x 10 -6 = (0.10 + x) (2x)2 ~ (0.10)4x2 1.3 x 10 -5 = 4x2(1 point correct setup) X = 0.0-018 M Molar solubility of Ca(OH)2 = 0.0018 M(1 point for final answer) c. Diagram should show the oxygen side of the water molecules oriented closer to the Ca 2+ ion (1 point) Problem 4

  7. First order (1 point) b. Increasing the concentration of the food coloring should be circled. (1 point) If the initial concentration of blue food coloring is increased, then more time is required for the bleach to oxidize the additional blue food coloring. (1 point) c. The spectrophotometer should be set to a different wavelength. (1 point) Problem 5 (4 points)

  8. LiI and KI. LiI has a small cation and a large anion and KI has a large cation and the same large anion. The melting point of LiI (with its smaller cation) is lower than that of KI. or LiI and LiF or LiI and NaF (1 point for correct pairing) (1 point for explaination) b. Either LiF or NaF is acceptable (1 point correct compound) F - + H2O <-> HF + OH -(1 point correct bal. eq.) Problem 6 4 points

  9. q = 24 J x 1.00 mol x 635 K = 15000 J = 15 kJ (1 point) It takes 10.7 kJ to melt the Al at 933 K. 15 kJ + 10.7 kJ = 26 kH(1 point) b. For extracting Al from ore: 1675 kJ/mol x 1 mol reaction = 837.5 kJ per mol Al 2 mol Al (1 point) Producing 1.00 mol of Al from Al2O3 requires 837.5 kJ. Because 26 kJ <837.5 kJ, recycling requires less energy. (1 point) Problem 7 4 points

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