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Solutions to 1996 Questions
· November 1996: Avogadro's Number
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Solution to December
1996 Puzzle of the Month
The question was: Why does ice
float on water ? Aren't molecules supposed to get
closer together as they cool and lose energy?.
Solution: As water molecules cool they do get closer together,
generally making cold water more dense than warm
water. But at 4o C, water attains its maximum density. Its density
keeps decreasing,meaning
that the space between molecules start to increase until the freezing
point is reached. Why ? The combination of H-bonding
and lossof kinetic energy allows a hexagonal pattern
(though not flat) of H2O molecules to emerge. A given volume of ice,
with its repeating lawn-chairshaped hexagons, has
less water molecules than an equal volume of liquid water. The latter can
therefore support the weight of the former. The 0 to <4 degree range
represents a kind of transition state where the molecules are slowed down
enough for part of the pattern to emerge, so that such water is not as dense as
it is at 4oC.
Solution to
November 1996 Puzzle of the Month
The
question was: How many water molecules fit into the eye of a needle
?
Solution: Assuming that the eye of a needle is 0.1 by 0.1 by
0.2 cm, we have a volume of 0.002 cm3. Water's density is about 1
g/cm3. Dividing by its molar mass of 18 g/mole and multiplying by
the number of molecules per mole ( 6.02 X 1023), we obtain, paying attentiion to significant figures, 7 X 1019molecules.If
we had that number of pennies we could give each of the 6 billion people on
earth over 100 million dollars.
Solution to October Puzzle of the Month
The
question was: You're on a hot air balloon that is ascending at a constant
velocity. As you rise past the nest of an eagle you suddenly drop your keys
overboard. Will the eagle in the nest see your keys go up momentarily
? If not, why not? If so, for how long?
Solution:The keys will be seen going up, for how long depends on
the initial upward velocity,v, of the balloon. Since
displacement = 0 (the keys go up and then momentarily
back to the eagle's eye-level) and if gravity = 10 m/s2, then 0 = vt - 5t2. Solving for t, we realize the keys
will be above the eagle for t = v/5 seconds.
Solution to September Puzzle of the Month
The
question was: A man walks 12 m east and then 5.0 m south in 13 seconds. What is
his average velocity ?
Solution: Velocity is displacement over time. Although the
distance covered in 13 seconds was 15 m , the displacement from the original
point was 13 m (Pythagorean Theorem). Thus, the avg. velocity = 1.0 m/s.
Solution To August Puzzle of
the Month
The
question was: A cyclotron can be used to bombard an element with a hydrogen-2
nucleus. This isotope consists of a neutron and a proton, but the pair is split
as it approaches the nucleus of the target element. The undeviated
neutron can then be absorbed by whatever element is to be attacked, and the
neutron will be converted into a proton and a beta particle.
In other words, an element attacked by a hydrogen-2
can go up in atomic number.
An American physicist, Ernest Lawrence,
made molybdenum radioactive by bombarding it with hydrogen-2. When another
group of scientists analyzed the sample, they found several known elements
included among the mixture of products, but none of the known elements were
radioactive. If, however, manganese or rhenium (two elements chemically similar
to each other; see their position in the periodic table) were added to the
mixture, the lot came back out radioactive when isolated. What was going on?
Solution: Molybdenum is atomic number 42. Through neutron
capture and the subsequent increase in atomic number, some of the nuclei
acquire an atomic number of 43, a radioactive element (technetium) unknown to
Solution To July Puzzle of
the Month
The
question was: When salt is added to water, why does it eventually disappear ?
Solution: Water is a polar molecule. This does not imply that
it has relatives in
Solution To June Puzzle of
the Month
The
question was: Carry out a thought-experiment involving two masses and a string
to demonstrate that a heavier object in free fall cannot accelerate faster than
a lighter object.
Solution: If we assume that a heavier object accelerates
faster in free fall, we can think of attaching the two masses with a string. We
would then assume that the heavier mass will then be slowed down by the lagging
smaller mass. But if the masses are attached, then the entire system represents
a larger mass which should accelerate faster. This leads to a contradiction,which implies that
our original assumption is false. A heavier object in free fall cannot
accelerate faster than a lighter object.
Page Maintained by E.Uva
Comments: euva@retired.ca
euva@prof.emsb.qc.ca
Copyright 2007
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Born on:
April 06, 1996
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