FREE RESPONSE QUESTIONS Directions : SHOW YOUR WORK/WRITE IN THE SPACES BELOW AND BOX YOUR ANSWERS. FRQ 1 (10 Points) Mr.Dries, the AP Biology teacher, decided to troll all of his AP Biology classes with an experiment that is not included in the AP Biology curriculum. The biology experiment requires the preparation of a water bath at 37.0C (body temperature). The temperature of the cold tap water is 22.0C, and the
temperature of the hot tap water is 55.0C. If a student starts with 90.0 g of cold water, what mass of hot water must be added to reach 37.0C? (Specific Heat of Water: 4.18 J/g C ) Work Explanation : -Q system = Q surroundings -[4.18 J/g ℃ x m hotwater x (37.0 ℃ -55.0 ℃ )] = 4.18 J/g ℃ x 90.0g coldwater x (37.0 ℃ -22.0 ℃ ) 75.0g Hot water needed Scoring Guidelines For FRQ 2 FRQ 2 (10 points) Subsections (a) - 2 pts, (b) - 5 pts, (c) - 3 pts Answer the following questions relating to the elements Aluminum and Phosphorus a) Write the ground-state electron configuration for an atom of each of the following. i) Al : [Ne] 3s2 3p1 (1 point) ii) P : [Ne] 3s2 3p3 (1 point) b) Consider the information in the table below. First Ionization Energy (kj mol -1 ) Second Ionization Energy (kj mol -1 ) Aluminum 580 1980 Phosphorous 1060 1890 i) Explain, in terms of atomic structure, why P has a higher first ionization energy than Al. Answer: P has a higher first ionization energy because it is affected by a
stronger nuclear charge since P has more protons than Al. ( 2 points) ii) Explain, in terms of atomic structure, why Al has a higher second ionization energy than P. Answer: Electrons are being removed from the p and s sublevels. However, in phosphorous, electrons are being removed from only the p sublevel. Since the s sublevel is closer to the nucleus, it takes more energy to overcome the attractive force to remove an electron. ( 3 points) c) Consider the Al+ ion. i) Identify an ion of P that is isoelectronic with Al+ Answer: P 3 + ¿ ❑ ¿ (1 point) ii) Which species has a larger radius : Al+ or the ion you identified in part (c) (i)? Al+ would be larger. Consider the protons and electrons each ion has: Al+ has 13 protons and 12 electrons, and P 3+ ion has 15 protons and 12 electrons. As one can see, both of these ions have the same number of electrons, yet a different number of protons. The one with a larger number of protons will have exert more of an attractive force on its electrons pulling towards the highly positive nucleus. In this case the Phosphorus ion has more protons and thus will have its electrons closer to the nucleus. Therefore, its atomic radius will be smaller than that of Al+. (2 point) EXTRA CREDIT Phosphorus reacts with fluorine to form SbF 5 (i) (3 pts) Draw in the space the complete Lewis electron-dot diagram for the SbF 5 molecule.
ii) (2 pts) Are all F-Sb-F bond angles in the SbF 5 molecule the same? Explain in the space below. No, this molecule’s geometry is trigonal bipyramidal. It consists of axial and equatorial symmetry. The angles along axial and equatorial are 90 degrees while angles along equatorial only are 120 degrees. Thus all F-Sb-F bond angles are not all same.
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