application/ and andogie/
edlted by RON DELORENZO
Mlddle Georgle College Cochran Georgia 31014
6.67 X lo3 krn X 1.76 X 10"ingle single file
The Liquid Silver Parade Robert Perkins
files
Mernorlal University Sw Wilfred Grenfeli College Corner Brook. Newfoundland
Canada AEH 6P9
Everyone knows that atoms are small, but just how small are they? This is a question which I ask my students durmg the first week of classes. The following example helps to illustrate just how small atoms are. The symbol of mercury (Hg) is derived from hydrargyrum which means "liquid silver." I ask the students to imagine dumping 1.00 cm3 of mercury into the source of the Nile River. If they then race to the outlet of the river in the Mediterranean Sea, they would discover mercury atoms (if mercury could float). If we imagine that, mercury atoms line up in a single file just touching one another, we may calculate:
which is a ~ ~ r o x i m a t etwice l v the distance from the Sun to the planet ~ l ; d o . ~ When we finallv come to the end of the comuutations evemphasizes the fact that 1.00 cm" of ~g contains an extremely large number of atoms and may help students appreciate the size of Avogadro's Number.
' Mahan, Bruce, "University Chemistry,'' 3rd Ed., Addison Wesley,
1975, p. 586.
"The World Almanac," Newspaper Enterprise Association, 1980, p. 446,764.
(a) the number of single files of mercury atoms extending over the length of the river. (b) the length that the Nile would have to be to ensure that the parade of mercury atoms is restricted t o one single file, which is one atom deep. Additional inf~rrnation',~ (a) radius of a mercury atom 0.144 nm (b) density of mercury 13.6 g cm-" (c) length of the Nile 6.67 X 10"km
Robert Perkins
Initially, the length of the Nile River must be converted to nanometers
Corner Brook. Newfoundland
.
This number then represents the length of a single file of Hg atoms (6.67 X lOI5 nm4ayer-'). Now the diameter of one mercury atom is 0.288 nm (twice the radius). Thus, the number of atoms required to stretch out over the length of the river will be: 6.67 X IOl5 nm X
1Hg atom
0.288 nm
The Spilled Can of Paint Memorial University Sir
Wilfred GrenfeliCollege
Canada A2H 6P9
In order for my beginning chemistry students to understand better the concepts of unit conversions, I have them consider the following problem. The students are asked to imagine that someone has tipped over a gallon can of paint in the center of a room of floor area 125 ft2. If the paint spreads evenly and covers the entire floor, what will the thickness of the paint layer be? Before heeinnine. ". I ask the students to trv to visualize exactly what they are looking for. Someone soon realizes that the volume of the ~ a i non t the floor must be the same as the volume which came from the paint can. Thus, we have the relationship:
-
1.00 gal = area X thickness = 125 ft" thickness
The number of atoms contained in 1.00 cm3 of mercury will he: 13.6 g Hg 1 mol Hg 6.02 X 1OZ7 atoms Hg X 1.00 cm3 Hg x --X 1mol Hg e m U 0 0 . 6 g Hg or 4.08 X 10'" Hg atoms
Therefore, the number of single files of mercury atoms will be: 4.08 X 10?' Hg atoms 2.32 X 1 0 I W g atorns/single file
or 1.76 X 10"ingle
files
In order to have one single file of mercury atoms, the Nile River would have to be:
which leads to 1.00 gal thickness = -125 ft2
I then suggest that the students try to convert two units to a common third unit-millimeters, for instance. For Canadian students a gallon is equal to 4.55 dm3 ((3.79 dm" for US students). Thus, thickness = 4.55 dm3/125 St2 The students can now see that they will be left with a unit of length for the thickness. Applying the various conversion factors and utilizing the unit-factor method they may evaluate the thickness in millimeters. Volume GO
Number 4
April 1983
343
'l'hv qutwiun mnt only rexc knoa.11.dyc~ uf t he metric prerixes 1x11;I..$> 1111. ~xmvvrsitmfrom the lmu~rial tht. . \ f v t r ~ c . \ ~Ins t ~ involving both cubic and square units. The instructor can also vary the numbers to test other prefixes. For example, students can calculate that in order to cover Lake Superior (surface area 2.17 X lo4 mi2) with a layer of oil 2.50 urn (2.50 X m) thick, only 3.09 X lo4 gallons (Imperial) of oil would be required. This is a relatively small amount considering the size of some of the more recent oil spills.
344
Journal of Chemical Education
