First Practice Solutions 160

First Practice Solutions 160 - 1. A heating or cooling...

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1. A heating or cooling curve (page 351 of the textbook) is a graph of temperature vs. heat added or removed. The slope of the graph is ΔT / q. Statement X in the question: q = mcΔT and slope = 1 / mc = (1 / 25.0 g) (1 / (2.09 J/g°C) ) = 0.0191°C / J Statement Y in the question: slope is zero. 2. ln P = (– H / RT) + C ln (P 2 / P 1 ) = – ( H / R)( (1 / T 2 ) – (1 / T 1 ) ) ln (P 2 / 1) = – (26.2 kJ / R) ( (1/ 210.15 K) – (1/ 194.65 K) ) R = 8.3145 x 10 –3 kJ/mol K ln P 2 = –3151(–0.00037892) = 1.194024 P 2 = 3.3 atm 3. I 2 is non-polar. Benzene is non-polar. I 2 is soluble in benzene. KCl is ionic. CCl 4 is non-polar. KCl is not soluble in CCl 4 . CH 3 OH is polar. H 2 O is polar. CH 3 OH is soluble in H 2 O. 4. d = (x cell M) / (V cell N A ) V cell = (x cell M) / (d N A ) V cell = (2)(262 g/mol) / (4.62 g/cm 3 ) N A = 1.883 x 10 –22 cm 3 V cell = a 3 a = V cell 1/3 = 5.73 x 10 –8 cm r = (3 1/2 ) a / 4 = 248 pm. 5. packing fraction = f = N V * / V = (x cell V * ) / V cell d = (f M) / (V * N A ) d 1 = (f 1 M 1 ) / ( (V * 1 ) N A ) d 2 = (f 2 M 2 ) / ( (V * 2 ) N A ) But M 2 = M 1 = 107.87 g/mol and V 2 * = V 1 * = volume of a silver atom. Therefore, d 2 / d 1 = f 2 / f 1 and d 2 = (0.68)(10.52 g/cm 3 ) / 0.74 = 9.66 g/cm 3 . Alternatively, this problem can be solved by starting with d = (x cell M) / (V cell N A ). Then d 2 / d 1 = ( (x cell ) 2 / (x cell ) 1 ) ( (V cell ) 1 / (V cell ) 2 ). But (V cell ) 1 / (V cell ) 2 = (a 1 / a 2 ) 3 and using the expressions for a in terms of r gives (a 1 / a 2 ) 3 = ( (3/2) 1/2 ) 3 . Thus, d 2 / d 1 = (2/4)(1.8371) = 0.9186 and d 2 = 0.9186(10.52 g/cm 3 ) = 9.66 g/cm 3 . 6.
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First Practice Solutions 160 - 1. A heating or cooling...

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