Cracking the SAT Chemistry Subject Test

(Marvins-Underground-K-12) #1
mass    of  the sample  by  the molar   mass    of  the compound.   190
g/95.3 g = 2.0 mol. Finally, we divide the moles of the
sample by the volume of the solution (in liters) to determine
the concentration of the solution. 2.0 mol/0.500 L = 4.0 M.
The correct answer is (B).


  1. C Think of the ideal gas equation: PV = nRT. What does this equation tell
    us? It means that volume is directly related to Kelvin temperature and
    inversely related to pressure. Doubling a gas’s Kelvin temperature will
    double its volume if other variables are held constant. Doubling a gas’s
    pressure will halve its volume if other variables are held constant. So the
    effect on volume of doubling pressure cancels out the effect of doubling
    Kelvin temperature, and the net result is that the gas’s volume will stay
    the same.

  2. C After 1 half-life, a 100 g sample will have a mass of 50 g. After 2 half-
    lives, it will have a mass of 25 g, so we’re talking about the expiration of
    2 half-lives. If 2 half-lives = 12.4 hours, then 1 half-life = 6.2 hours.

  3. A This is alpha decay, in which a radioactive atom loses 2 protons and 2
    neutrons. The loss of 2 protons means the atomic number decreases by 2,
    which means it was 87 before it decayed. The element with an atomic
    number of 87 is Fr (francium). The loss of 2 neutrons together with the
    loss of 2 protons means that the mass number has decreased by (2) + (2)
    = 4. The mass number is now 216, which tells us that it was 220. Notice
    that choice (C) has the right numbers, but not the right element.
    Remember that the atomic number uniquely identifies an element, so an
    atomic number of 87 must be francium, no matter what the mass number
    is. This also means that an element can have several different mass
    numbers, collectively called isotopes.

  4. C The empirical formula tells us that the ratio of carbon to hydrogen is 1:2,
    so we’re looking for an answer that reflects the same ratio. Only choices
    (B), (C), and (E) do that, so we can eliminate (A) and (D). Now, among
    (B), (C), and (E), we’re looking for the one whose molecular weight is 56
    amu. Look at the periodic table. Every carbon atom has an atomic weight
    of 12 amu, and every hydrogen atom has an atomic weight of 1 amu.
    Rather than pursue algebra, let’s just try the three choices.

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