of the metalloid silicon.
- C The freer molecules are to move around, the greater the entropy of the
state. In equation I we go from a gas (very high entropy) to a liquid (more
ordered, less entropy). That’s a decrease in entropy. Equation II also
involves an entropy decrease. Here, ions go from being able to move
throughout a solution to being restricted in the solid state. Equation III
shows an entropy increase because we are increasing the moles of gas (2
on left and 3 on right). I and II illustrate an entropy decrease, and the
correct answer is (C).
- E Choices (A) and (D) will not change water’s boiling point. Remember
that when a solute is dissolved in solution, the solution’s boiling point
will be raised (and its freezing point lowered). Only (E) involves
dissolving a solute into water. Sugar water boils at a higher temperature
than pure water under identical conditions. Why doesn’t the addition of
gasoline into the water have the same effect? Gasoline molecules are
nonpolar and will not dissolve in water. As for (C), increasing the altitude
of the water will decrease its boiling point.
- D The question concerns periodic table trends and, in particular, atomic
radius. As we move from left to right across a period, atomic radius
decreases. So, within a period, the higher the atomic number, the smaller
the atomic radius.
This question is a perfect example of the camouflage trap.
You might be thinking higher atomic number, and the correct
answer is phrased as “greater positive charge in its nucleus.”
But keep the blinders off your brain. You know the answer—
just remember that there’s more than one way of expressing
it.
- C Aluminum is a metal, and oxygen is a nonmetal. They react to form an
ionic compound. Aluminum (in the 3A group) forms a +3 ion. Oxygen (in
the 6A group) forms a −2 ion. They will produce aluminum oxide, Al 2 O 3.
When the equation is balanced, we’ll get 4Al(s) + 3O 2 (g) → 2Al 2 O 3 (s).
- E Many colored compounds contain a transition metal (an element from
the d region of the periodic table). Choices (A), (B), and (C) are ionic
