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The key concepts and assigned problems for chapter 1 of the ziegler chemistry textbook, 9th edition. Topics covered include identifying elements, compounds, and mixtures; unit conversions; significant figures; and density. Students are expected to read specific page ranges and complete assigned problems.
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CH 221 - Ziegler KEY CONCEPTS Fall 04 Chapter 1 9 th^ edition
Class period 1 – Reading pp. 1-12.
Class period 2 – Reading pp. 13(b)-17, 20-24.
Class period 3 – Reading pp. 22(b) - 28.
CH 221 - Ziegler KEY CONCEPTS Fall 04 Chapter 1 9 th^ edition
Answers to assigned problems:
1.7 See text A-1;
1.4 , 1.5 See table 1.2;
1.2 a) homogeneous mixture, b) heterogeneous mixture, c) pure substance,
d) heterogeneous mixture.
1.14 a) chemical b)physical c) physical d) physical.
1.22 a) volume b) area c) volume d) density e) time f) length e) temperature
1.30 a) 31°C b) 1390 °F c) -140 °F, 178 K d) 297 K, 77 °F e) mp= 24.6 K, bp =27.
K
1.34 140 °C
1.36 a)4 b) 3 c) 4 d) 5 e) 6
1.38 a)1.44 x 10^5 b) 9.75 x 10-2^ c)8.90 x 10^5 d) 6.76 x 10^4 e) 3.40 x 10^4 f) -6.
1.40 -2.3 x 10^3 b) 8.260 x 10^7 c) 3.4 x 10^4 d) 7.62 x 10^5
1.44 b) 1.35 x 10^21 L
1.46 a) 44.39 m b) 539 mL c) 1.35x 10-5^ km/hr d) 3.041 m^3 e) $6.59/kg
f) 0.156 g/mL.
1.54 There are lots of ways to do this. Here’s one way: Select a common unit for comparison,
say, the cm. Then recall that 1 in ~ 2.5 cm, 1 m = 100 cm. Also, it is helpful to notice that 57 cm
is close to 50 cm which is about half as big as the 1.1 m length. 14 in. is about 1.5 times 10 inches
which is easier to convert to cm, (25 cm) so 14 in is around (25 + 12) cm. Sometimes you have to
do the conversions (like when the estimation method isn’t good enough), and sometimes you
don’t (as was the case here). It is useful to make these ball-park estimates so that you can judge
whether your answer is right when you do go through with the more formal conversions.
1.51 See answer in your text.
1.63 peat: 0.13 g/cm^3 ; top soil: 2.5 g/cm^3. No. The densities tell us that a certain volume of
peat moss is “lighter” (weighs less) than the same volume of top soil. Volume must be specified in
order to compare mass.
1.67 a) 4.67 g/cm^3 b) 2.54 L of mercury.