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KEY E3_FA03 - Answer Key with Detailed Solutions for C111...

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Answer Key with Detailed Solutions for C111 Exam 3 FA 03 1. According to Avogadro's and Guy-Lussac's laws, equal volumes of gases at the same temperature and pressure contain equal moles. Therefore the volumes of reacting gases are in the same ratio as their coefficients in the balanced chemical equation. 2 N 2 N 2 H 2 H n V n V = solve for V H2 = 2 2 2 H mol 3 x N mol 1 N mL 0 . 15 = 45.0 mL ans. (e) 2. P, V and T are all changing, however n is constant. Convert o C to kelvins. 2 2 2 1 1 1 T V P T V P = Solve this equation for V 2 and substitute in the known values. There is no need to convert the pressure units. Atmospheres work fine. V 2 = ) atm 75 . 2 )( K 307 ( ) K 203 )( L 28 . 7 )( atm 15 . 1 ( = 2.01 L ans (b) 3. If you memorized the formula, M = P DRT , substitute the values and solve for M = 91.9 g/mol Otherwise, one easy way to approach this is to assume 1 liter, having a mass of 28 g. This assumption is a direct consequence of the density, being 28 g/L. Then, PV = nRT and you can find n for this volume. n = RT PV = ) K 100 )( K mol atm L 082057 . 0 ( ) L 00 . 1 )( atm 5 . 2 ( = 0.30466 moles, which have a mass of the 28 g. Therefore molar mass = mol 30466 . 0 g 28 = 91.9 g/mol ans (a) 4. The mole fraction is X halothane = gas moles total halothane moles . The number of moles of each gas is proportional to its own partial pressure, therefore X halothane = gases all halothane P P = ) 570 170 ( 170 + = 0.23 ans (a) 5. Find frequency as ν = c/ λ = m 10 x 514 s / m 10 x 998 . 2 9 8 - = 5.83 x 10 14 s -1 (note units, this is frequency) Find E = h ν = (6.626x10 - 34 J s)( 5.83 x 10 14 s -1 ) = 3.86 x 10 - 19 J ans (a) 6. Consider only the CO 2 gas. P 1 V 1 = P 2 V 2 , where P 1 and V 1 are given, and V 2 is V total , for the entire apparatus. P 2 = ) mL 250 mL 50 ( ) torr 720 )( mL 250 ( V V P 2 1 1 + = = 600 torr. Notice it is not necessary to take time to convert the pressure units to atm. They cancel out, and torr are legitimate pressure units, in any case. 7. There are 3 separate steps, with a H (or q p ) for each. One approach is to compute each step separately and add to find q process . q for each temperature change within a phase has the form q p = C sp x mass x T note: the entire temperature range is a change of 14 degrees, but only a portion of this change pertains to the solid (+ 4 o C), a portion pertains to the liquid (+ 10 o C). These T’s must be used separately
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with their phase-appropriate Csp’s. Each step has its own T and Csp. They apply only to ONE particular phase and must not be mixed up or used interchangeably. note: It is not wrong, but it is a waste of time to convert these temps to kelvins because you are calculating a difference in two temps , for each phase, and this difference has the same value regardless of which kind of degrees we choose, because o C are the same size as kelvins. A transition from one phase to another phase has a q p = ( H phase change )( mass ) Heating (or cooling) within a phase has a q p = (C sp , phase )( mass )( T) There are three processes in this problem: (a) heat the ice from –4 o C to melting point 0.0 o C T = 0 – (-4) = +4 o C q (a) = (2.09 C g J o ) (5 g) (+4 o C) = 41.8 J (b) melt the ice (phase change) q (b) = (334 J/g)(5 g) = 1670 J (c) heat the water from melting point to +10 o C T = 10 – 0 = +10 o C q (c) = (4.18 J/g-
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