GSCI 1010
ASSIGNMENT 2 SOLUTIONS
30 NOVEMBER 2009
Text Problems
Chapter 3, page 156: Questions # 2; Questions for Thought # 4, 8
Questions
2. During the Moon’s waxing crescent or waning crescent phases, if you were standing on the dark
side of the Moon looking up, you would see the Earth above you, with most of it illuminated by
the Sun. Therefore, the fact that the Moon’s dark hemisphere is not completely dark but faintly
illuminated is due to reflected light from the Earth.
Questions for Thought
4. The space-time graph in Figure 3.2 shows one
dimension of “space” on the vertical axis and
“time” on the horizontal axis. Thus, moving
horizontally across the graph would correspond to
remaining stationary in space while time elapses.
Moving only vertically on the graph would be
impossible, as this would mean that time was
constant, i.e. that time had stopped!
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Space
stationary path
constant time
path
Time
8. The Earth’s seasons are actually a result of the fact that the Earth’s axis of rotation is tilted at
about 23.5 relative to the plane of rotation around the Sun. However, it is also true that the
Earth’s orbit is slightly elliptical, and therefore that we are just a little closer to the Sun in
January and just a little further from the Sun in July. Clearly, this means that summers in the
southern hemisphere (e.g. Australia, which has summer in January) will be, on average, a little
warmer than summers in the northern hemisphere (e.g. Canada, which has summer in July).
Chapter 4, page 179: Questions #4; Problems #2
Questions
4. Most of the Universe is almost completely absent of matter (i.e. a near-vacuum), so few atoms
exist to form molecules and few molecules exist to encounter one another and react chemically.
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In contrast to this, most of the matter in the Universe is found in stars. However, the vast
majority of this matter is hydrogen and helium, and helium is inert and doesn’t react with
anything. This leaves hydrogen as the primary reactive element in stars. But in the cores of
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stars, the density and temperature is so great that matter is in the form of a plasma, i.e. atoms
have been stripped of their electrons and so can’t form molecules or participate in any chemical
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reactions whatsoever. Even in the outer regions of stars, where hydrogen gas (i.e. molecular
hydrogen) and trace amounts of other elements like carbon, oxygen, etc. can be found, the
temperatures are still extreme enough that few chemical reactions occur. The Earth and other
planets or planet-like objects contain only a miniscule percentage of the Universe’s matter, but
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they are unique in that they include significant amounts of a great number of chemical elements,
and many offer a range of temperatures and pressures which enable a huge number of chemical
reactions to occur with ease.
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Problems
2. The balanced chemical equation corresponding to the chemical reaction described in
problem # 1 is:
N 2  3H 2  2 NH 3
Note that nitrogen and hydrogen gas molecules each contain two atoms of their respective
elements. Therefore, in order for the equation to “balance”, three volumes of hydrogen and one
volume of nitrogen would have to produce two volumes of ammonia.
Chapter 9, page 436: Questions #2; Problems 2(a)(b)
Questions
2. In an ionic chemical bond, one atom will actually give up one or more electrons to become a
“positive ion”, while another atom will gain those extra electrons to become a “negative
ion”. The oppositely-charged ions then come together by electrostatic attraction to form a
molecule.
For example, sodium atoms can readily lose one electron while chlorine atoms
can readily gain one electron. The resulting ionic bond is thus represented by the equation
Na+ + Cl–  NaCl .
On the other hand, a covalent chemical bond is the result of one or more pairs of electrons
being shared between two atoms (rather than being given up by one and gained by another).
For example, two atoms of hydrogen each have room for one more electron in their first (and
only) shell. At the same time, an atom of oxygen has room for 2 extra electrons in its second
shell. Without losing or gaining electrons outright, a pair of electrons can be “shared” by the
oxygen and each of the hydrogens (i.e. 2 shared pairs, or 4 shared electrons in all). This leaves
the oxygen atom “feeling” as if both its shells are filled at least part-time, and the two hydrogen
atoms will also “feel” as if their first shells are filled at least part-time. The resulting covalent
bond produces a molecule of water, or H2O, with the two shared electron pairs of the covalent
bond being represented by lines, i.e. H – O – H.
Problems
2. (a)
K+ + Br–  KBr .
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(b)
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By virtue of the atomic structures of potassium (K) and bromine (Br) suggested in the
periodic table, potassium has only 1 electron in its 4th shell and will want to give this up,
while bromine has lots of room for extra electrons in its 4th shell and will readily accept
an electron from potassium. As a result, an ionic bond between the two can form
potassium bromide, or
Referring again to the periodic table, magnesium (Mg) has only 2 electrons in its 3rd
shell, while oxygen (O) has room for exactly 2 electrons in its second shell. Therefore,
they can “exchange” two electrons according to
Mg++ + O––  MgO , which may also be written Mg2+ + O2–  MgO .
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Chapter 15, page 597: Questions # 3; Questions for Thought # 1
Questions
3. In the theory of atomic structure, the electrons belonging to any atom may be found in several
different “shells” (which have different “energy levels”). More specifically, however, the
electrons within each shell occupy what are called “orbitals”, with each orbital having room for
at most 2 electrons. The first (or smallest) shell has 1 orbital, the second shell 4 orbitals, the 3rd
shell 9 orbitals, the 4th shell 16 orbitals, and so on.
The natural tendency of each atom is to have full orbitals whenever possible. That is, if a
particular orbital has only 1 electron, the atom will look to: (i) lose that electron altogether to
empty the orbital: (ii) gain an electron outright to fill the orbital; or (iii) share its orbital electron
with another atom which also has a single-electron orbital, thereby making both orbitals seem
as if they are full part of the time.
While not all orbitals are circles or ellipses, we can conceptualize the third option above, i.e.
electron pair “sharing”, something like a “figure 8”. That is, while the two atoms (considered
separately) would each have an orbital with only 1 electron, a covalent bond between these
atoms would bring them close enough to one another that these two orbitals could essentially
“merge” into a common, “figure 8” orbital. The two electrons should then be expected to spend
part of the time orbiting the first atom and part of the time orbiting the second, with the end
result being that both atoms “feel” as if their orbitals are filled! The two diagrams below
illustrate the idea.




2 separate atoms
2 distinct 1-electron orbitals


 
covalent bond
1 merged 2-electron orbital
Questions for Thought
1. The first atomic “shell” contains only a single (spherical) orbital, capable of holding one or two
electrons. Since the first element, hydrogen, has one electron, it can be considered either to
have “just one electron in its outer shell”, making it like the other elements in Column I of the
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periodic table (lithium, sodium, etc.) or to have “room for only one more electron in its outer
shell”, making it like the other elements in Column VII (fluorine, chlorine, etc.). Hence,
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hydrogen could arguably be placed above either lithium or fluorine.
Ted’s Questions
1. Liquids and gases are both “fluids”, i.e. both can flow (e.g. through a tube or pipe). In addition
both liquids and gases conform to the shape of their containers. Solids do neither of these
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things.
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Gases (at least on a local scale) always distribute themselves evenly within their containers (e.g.
air within a balloon or propane within a tank). Moreover, gases are capable of being
compressed into a smaller volume. Solids and liquids normally don’t do either of these things.
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2. Methane, CH4, has the following molecular structure (“C”, “H” show locations of the nuclei):
H

4
H
:
 C 
:H

H
2
The “dash” notation, showing the covalent bonds is therefore:
H

HCH

H
3. Methane can burn in the presence of oxygen to produce water vapour and carbon dioxide, as
follows:
methane + oxygen  water + carbon dioxide
The unbalanced chemical equation for this is
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CH 4  O2  H 2 O  CO2 ,
while the balanced version is
CH 4  2O2  2H 2 O  CO2 .
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4. (a)
The possible distinct quark combinations for a proton (which requires two “up” quarks
and one “down” quark) are:
Down
Up
Up
Red
Blue
Green
Blue
Green
Red
Green
Red
Blue
(Note that the order of the two “up” quarks does not matter, i.e. Blue/Green is the same
as Green/Blue.)
(b)
Since a neutron requires two “down” quarks and one “up” quark, the number of distinct
combinations will again be three. This is easily explained by noting that there are three
possible choices of colour for the single “up” quark. Once that colour has been chosen,
there is no choice for the remaining two “down” quark colours. Therefore the overall
number of combinations is completely determined by the number of colour choices for
the “up” quark.
2
2
2
1
2