practice exam 2 solutions

practice exam 2 solutions - 1. Start this kind of problem...

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1. Start this kind of problem with a statement of the law of conservation of energy. For this particular problem, we write heat lost by hot water = heat gained by cold water + heat gained by the surroundings Then write the statement using symbols: –q hot = q cold + q surroundings –50.0 g(4.184 J / g K)(25.0 – 60.5)°C = 50.0 g(4.184 J / g K)(25.0 – 18.5)°C + q surroundings Observe the units and solve to obtain q surroundings = 6070 J. 2. Refer to the three equations that are given with values of ∆H as equations (1), (2), and (3). The overall equation is ClF + F 2 → ClF 3 . Look for substances that appear only once in equations (1), (2), and (3): ClF appears only in equation (1); F 2 appears only in equation (3); and ClF 3 appears only in equation (2). Divide equation (1) and its ∆H by 2; divide equation (3) and its ∆H by 2; reverse equation (2) and divide it by (2), and change the sign of ∆H of equation (2) and divide by 2. Then, for the overall equation, ∆H = 83.7 kJ – 170.7 kJ – 21.7 kJ = –108.7 kJ. 3. q = ms∆T = 100g(4.184 J / g K)(10.0°C) = 4184 J E = hν = hc / λ = hc / 0.122 m = 1.63 x 10 –24 J 4184 J / 1.63 x 10 –24 J = 2.57 x 10 27 4. hν = φ + ½ mv 2 The minimum energy is φ. Calculate φ = hν – ½ mv 2 = (hc / λ) ½ mv 2 = (hc / 221 x 10 –9 m) – (100 kJ/mol)(1 mol / 6.022 x 10 23 )(1000 J / kJ) = 8.89 x 10 –19 J – 1.66 x 10 –19 J = 7.33 x 10 –19 J. 5. E n = –R h c Z 2 / n 2 ∆E = –R h c ( (1 / n f 2 ) – (1 / n i 2 ) ) = hν = hc / λ Cancel hc on both sides of the equation to obtain –R( 0 (1 / 9) ) = 1 / λ , where n f = ∞ and n i = 3 were used. Then, λ = 9 / R = 8.20 x 10 –7 m = 820 nm. 6. λ = h / mv gives m = h / λv = h / (3.33 x 10 –15 m)(3.00 x 10 7 m /sec) = 6.63 x 10 –27 kg or 6.63 x 10 –24 g. Do not confuse the symbol v with ν. 7.
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practice exam 2 solutions - 1. Start this kind of problem...

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