SAT Subject Test Chemistry,10 edition

(Marvins-Underground-K-12) #1

36 . D


The most    polar   bond    is  the one that    has the greatest    difference  in  the electronegativities of
the two elements. Of the choices, H–F is the most polar since hydrogen and fluorine are the
farthest apart in the periodic table.

37 . E
This question asks you to determine the rate law of the reaction for the formation of
ammonia, NH 3 . A rate law is an equation that gives the relationship between the rate of the
reaction and the concentration of the reactants, each raised to an appropriate power that
depends on the exact reaction. For example, the rate of this reaction is equal to k [N 2 ]x[H 2 ]y,
where k is the rate constant and x and y are real numbers. k, x, and y can only be determined
experimentally by systematically varying the initial concentration of one reactant. Since there
is no experimental data, there is no way for us to know what these values are.


38 . E
10 N H 2 SO 4 is equal to 5 M H 2 SO 4 as each molecule has 2H+ ions associated with it. Use the
equation M 1 V 1 = M 2 V 2 . (100)(5) = M 2 (800). M 2 = 5/8.


39 . C
This question asks you which molecule contains both ionic and covalent bonds. Choice C is
ammonium sulfate. The bonds between nitrogen and hydrogen in the ammonium ion and
between sulfur and oxygen in the sulfate ion are all covalent. However, the bond between the
ammonium ion and the sulfate ion is an ionic bond.


40 . D
Charles’s law states that V 1 /T 1 = V 2 /T 2 , and you must convert Celsius to Kelvin. Rearrange your
equation to V 1 /V 2 = T 1 /T 2 . T 1 = 303 and T 2 = 293, and the ratio is 303/293.


41 . B
This question tests your knowledge of Le Châtelier’s principle. What will happen to this
reaction if more chlorine is added? Adding more chlorine puts a stress on the system, and the
system alleviates that stress by using up that added chlorine: The equilibrium shifts to the left.


42 . A
This is a redox reaction. We can tell this because the oxidation states of the species change
during the reaction. The oxidation state of bromine in the product, molecular bromine, is 0.

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