Cliffs AP Chemistry, 3rd Edition

(singke) #1

Scoring Guidelines


Students choose five of the eight reactions. Only the answers in the boxes are graded (unless
clearly marked otherwise). Each correct answer earns 3 points, 1 point for reactants and 2
points for products. All products must be correct to earn both product points. Equations do not
need to be balanced and phases need not be indicated. Any spectator ions on the reactant side
nullify the 1 possible reactant point, but if they appear again on the product side, there is no
product-point penalty. A fully molecular equation (when it should be ionic) earns a maximum
of 1 point. Ion charges must be correct.


Part B: Question 4


The roman numeral in each answer refers to the section in the chapter entitled “Writing and
Predicting Chemical Reactions.” For example, I is found on page 216.



  1. Restatement: Give a formula for each reaction, showing the reactants and products.
    (a) I. Sn + Cl 2 →SnCl 4
    (Usually pick the higher oxidation state of the metal ion.)
    (b) II. C 2 H 6 + O 2 →CO 2 + H 2 O
    All hydrocarbons burn in oxygen gas to produce CO 2 and H 2 O. (“Air” almost always
    means oxygen gas.) Note the use of the word “completely”. Unless this word was in the
    problem, a mixture of CO and CO 2 gases would result.
    (c) IX. Cu + H++ NO3–→Cu2++ H 2 O + NO
    This reaction is well known and is covered quite extensively in textbooks. Note how it de-
    parts from the rubric. Copper metal does not react directly with H+ions, since it has a neg-
    ative standard oxidation voltage. However, it will react with 6M HNO 3 because the ion is
    a much stronger oxidizing agent than H+The fact that copper metal is difficult to oxidize
    indicates that is readily reduced. This fact allows one to qualitatively test for the presence
    of by reacting it with dithionite (hydrosulfite) ion, as the reducing agent to produce cop-
    per:
    Cu2+(aq)+ S 2 O 4 2−+ 2H 2 O →Cu + 2SO 3 2−+ 4H+


(d) XIX. SO 4 2–+ Hg2+→HgSO 4
(e) XXI. SO 3 + CaO →CaSO 4

(f) XXIII. CuSO 4 ⋅5 H 2 O →CuSO 4 + 5 H 2 O


(g) XXIV. Zn(OH) 2 + NH 3 →Zn(NH 3 ) 4 2++ OH–
(h) XXV. C 2 H 6 + Br 2 →C 2 H 5 Br + HBr

Note: Unless the word “monobrominated” had been used in the problem, a whole host of
products would have been possible; i.e. polybrominated ethanes, polymers, and so on.

Answers and Explanations for the Practice Test
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