Download AP CHEMISTRY – Source: 1999 AP Exam CHAPTER 8 PRACTICE

AP CHEMISTRY – Source: 1999 AP Exam
CHAPTER 8 PRACTICE TEST-KEY
NO CALCULATORS MAY BE USED IN THIS SECTION OF THE EXAMINATION.
SECTION I
Part A
Directions: Each set of lettered choice below refers to the numbered statement immediately following
it. Select the one lettered choice that best fits each statement and then fill in the corresponding oval
on the answer sheet. A choice may be used once, more than once, or not at all in each set.
Questions 1-4 refer to the following types of energy.
(A) Activation energy
(D) Kinetic energy
(B) Free energy
(E) Lattice energy
(C) Ionization energy
1. The energy required to convert a ground-state atom in the gas phase to a gaseous positive ion: C
2. The energy change that occurs in the conversion of an ionic solid to widely separated gaseous
ions: E
3. The energy in a chemical or physical change that is available to do useful work: B
4. The energy required to form the transition state in a chemical reaction: A
Questions 5-8 refer to the following descriptions of bonding in different types of solids.
(A) Lattice of positive and negative ions held together by electrostatic forces:
(B) Closely packed lattice with delocalized electrons throughout
(C) Strong single covalent bonds with weak intermolecular forces
(D) Strong multiple covalent bonds (including ∏ - bonds) with weak intermolecular forces
(E) Macromolecules held together with strong polar bonds
5. Cesium chloride, CsCl(s): A
7. Carbon dioxide, CO2(s): D
6. Gold, Au(s): B
8. Methane, CH4(s): C
Directions: Each of the questions or incomplete statements below is followed by five suggested
answers or completions. Select the one that is best in each case and then fill in the corresponding
oval on the answer sheet.
9. What mass of Au is produced when 0.0500 mol of Au2S3 is reduced completely with excess H2?
SOLUTION: 0.0500 mol Au2S3 x 2 mol Au x 197g Au = 19.7 g Au
1 mol Au2S3
1 mol Au
TIP TO SOLVE W/O CALCULATOR: 0.05 x 2 = 0.10 multiply 0.10 x 197 means move decimal one
place to left.
(A) 9.85 g
(B) 19.7 g
(C) 24.5 g
(D) 39.4 g
(E) 48.9 g
10. When a solution of sodium chloride is vaporized in a flame, the color of the flame is
(A) blue
(B) yellow
(C) green
(D) violet
(E) white
C10H12O4S(s) + 12O2(g) 10CO2 + SO2(g) + 6H2O(g)
11 . When the equation above is balanced and all coefficients are reduced to their lowest whole-number
terms, the coefficient for O2(g) is
(A) 6
(B) 7
(C) 12
(D) 14
(E) 28
12. The melting point of MgO is higher than that of NaF. Explanations for this observation include
which of the following?
I. Mg 2+ is more positively charged than Na+.
II. O 2- is more negatively charged than F-.
III.The O 2- ion is smaller than the F- ion
(A) II only
(D)II and III only
(B) I and II only
(E) I, II, and III
(C)I and III only
+1 -2
+1 -1
+6 -1
+1 -1
0
13. H2Se(g) + 4 O2F2(g)  SeF6(g) + 2 HF(g) + 4 O2(g)
Which of the following is true regarding the reaction represented above?
(A) The oxidation number of O does not change.
(B) The oxidation number of H changes from -1 to +1
(C) The oxidation number of F changes from +1 to -1.
(D) The oxidation number of Se changes from -2 to +6
(D) It is a disproportionation reaction for F.
(Explanation for oxid# of O in O2F2. http://www.cs.utexas.edu/users/kbarker/halo/finalanswers.html)
a. Computing the oxidation states for O2F2.
i. Finding rules for computing the oxidation numbers for the atoms in
O2F2.
1. The oxidation number of Fluorine is always -1.
2. Therefore, the oxidation number of F in O2F2 is -1.
ii. Found rules to determine that: the oxidation number of F is -1. The
oxidation state O(x) of O in O2F2 can be computed from the formula
O(x) = C - (Z(x) / S(x)), where C is the compound charge, S(x) is the
subscript of atom x and Z(x) is the sum of the products of the other
atoms in the compound and their subscripts; the oxidation state of O in
O2F2 is thus 0 - (-2 / 2) = 1.
b. Therefore, the oxidation states for O2F2 are F:-1 and O:1.
14. A 1.0 L sample of an aqueous solution contains 0.10 mol of NaCl and 0.10 mol of CaCl2. (oops!
Was “CaCls”) What is the minimum number of moles of AgNO3 that must be added to the
solution in order to precipitate all of the Cl- as AgCl(s)? (Assume that AgCl is insoluble.)
ANALYSIS: 0.10 mol of NaCl contains 0.10 mol Cl- ions. 0.10 mol of CaCl2 contains 0.20 mol Clions. Ag & Cl combine in a 1 mol:1 mol ratio and 1 mol AgNO3 produces 1 mol Ag+ ions. SO, if
you have 0.30 mol Cl- ions, you need 0.30 mol Ag+ ions, and therefore, 0.30 mol AgNO3.
(A) 0.10 mol
(C) 0.30 mol
(E) 0.60 mol
(B) 0.20 mol
(D) 0.40 mol
Ionization Energies for element X (kJ mol-1)
First
Second
Third
Fourth
Fifth
580
1,815
2,740
11,600
14,800
15. The ionization energies for element X are listed in the table above. On the basis of the data,
element X is most likely to be
(A) Na
(C) Al
(E) P
(B) Mg
(D) Si
ANALYSIS: Ionization energy jumps 4x from the first to the second electron removed. This tells us that
there is only easily removed (valence) electron. Only sodium fits this description.
18. A molecule or an ion is classified as a Lewis acid if it
(A) accepts a proton from water
(B) Accepts a pair of electrons to form a bond
(C) Donates a pair of electrons to form a bond
(D) Donates a proton to water
(E) Has resonance Lewis electron-dot structures
19. Of the following molecules, which has the largest dipole moment?
(A) CO
(C) O2
(D) HF
(B) CO2
(E) F2
ANALYSIS: Immediately rule out B, C and E. There is no dipole moment for any of these molecules! C &
E b/c they are diatomic. There is NO difference in electronegativity, therefore, no dipole moment can
exist! For B, CO2 has a bond dipole for each of the two C=O bonds. However, this is a linear molecule
and the dipoles face in opposite directions, creating no + and – end of the molecule. Of the remaining
choices, A) CO and D) HF, electronegativity differences between the 2 atoms is greater in D. This means
that Q is greater for HF. In addition, a CO triple bond is VERY short, whereas an HF single bond is not as
short. This means r is also greater for HF. Since Dipole moment, μ = Qr, μ is greater for HF than for CO.
20. When hafnium metal is heated in an atmosphere of chlorine gas, the product of the reaction is
found to contain 62.2 percent Hf by mass and 37.4 percent Cl by mass. What is the empirical formula for
this compound?
(A) HfCl
(C) HfCl3
(E) Hf2Cl3
(B) HfCl2
(D) HfCl4
ANALYSIS: Assume 62.2 g Hf & 37.4 g Cl (this gives approx 100 g of product). Use each of these assumed
masses to determine the moles of each element present in the compound.
62.2 g Hf x 1 mol Hf ≈ 1/3 mol Hf
( you can do this in your head, or on paper)
178g Hf
37.4 g Cl x 1 mol Cl
≈ 1 mol Cl
35.5 g Cl
NOW, multiply each of these by the same number to get a whole number ration (instead of 1/3 : 1). If
you multiply by 3 you get a ratio of 1:3, so the formula will be HfCl3).
21. IN the periodic table, as the atomic number increases from 11 to 17, what happens to the atomic
radius?
(A) it remains constant
(D) It decreases only.
(B) It increases only.
(E) It decreases, then increases.
(C) It increases, then decreases.
22. Which of the following is a correct interpretation of the results of Rutherford’s experiments in which
gold atoms were bombarded with alpha particles?
(A) Atoms have equal numbers of positive and negative charges.
(B) Electrons in atoms are arranged in shells.
(C) Neutrons are at the center of an atom.
(D) Neutrons and protons in atoms have nearly equal mass.
(E) The positive charge of an atom in concentrated in a small region.
23. In which of the following processes are covalent bonds broken?
(A) I2(s)  I2(g)
(B) CO2(s) CO2(g)
(C) NaCl(s)  NaCl(l)
(D) C(diamond)  C(g)
(E) Fe(s)Fe(l)
24. In a qualitative analysis for the presence of Pb2+ , Fe2+, and Cu2+ ions in aqueous solution, which of
the following will allow the separation of Pb2+ from the other ions at room temperature?
(A) Adding dilute Na2S(aq) solution
(D) Adding dilute NH3 (aq) solution
(B) Adding dilute HCl(s) solution
(E) Adding dilute HNO3 (aq) solution
(C) Adding dilute NaOH(aq) solution
ANALYSIS: PbCl2 is insoluble, whereas FeCl2 & CuCl2 are soluble. Cl is the only ion that will allow only
the lead compound to precipitate, while leaving the other 2 (FeCl2 & CuCl2) in solution.
25. MAYBE After completing an experiment to determine gravimetrically the percentage of water in a
hydrate, a student reported a value of 38 percent. The correct value for the percentage of water in the
hydrate is 51 percent. Which of the following is the most likely explanation for this difference?
(A) Strong initial heating caused some of the hydrate sample to spatter out of the crucible.
(B) The dehydrated sample absorbed moisture after heating.
(C)The amount of the hydrate sample used was too small.
(D) The crucible was not heated to constant mass before use.
(E) Excess heating caused the dehydrated sample to decompose.
ANALYSIS: (A) if you weighed the hydrate sample before heating (say 2.00g), and then heating caused
part of the sample to be lost, your post-heating mass (say 1.50g) would be lower than the actual (say
1.70g.) When you calculated the % comp that is water, you would divide the difference between the
pre-heating mass (2.00g) and the post heating mass (1.5)by the pre-heating mass (2.00-1.50g/2.00g) x
100= 25%. In fact, the % comp water (if none of the sample was lost)should be (2.00-1.70g/2.00g) x 100
= 15%. Therefore, THE ERROR IS ON THE HIGH END, THE OPPOSITE OF THE SITUATION INDICATED
ABOVE.
(B) Use the 2.00g original mass from (A). Visualize: Heated to release the water in the hydrate. Sample
re-absorbed some moisture. Obtained mass of post-heating sample & it was higher than it should have
been, say 1.80g. Analyze % comp: (2.00-1.80g/2.00g) x 100 = 10%, which is lower than the actual (I used
15% as my “actual” in analysis of choice A, above.) IN THIS CASE, THE ERROR IS ON THE LOW END, AS
THE QUESTION INDICATES.
26. The volume of distilled water that should be added to 10.0 mL of 6.00 M HCl(aq) in order to prepare
a 0.500 M HCl (aq) solution is approximately
ANALYSIS: M1=0.600 mol/L V1=0.010L M2=0.500 mol/L
USE: M1V1=M2V2 to solve for V2.
PRACTICE SOLVING W/O CALCULATOR
(A) 50.0 mL
(B) 60.0 mL
(D) 110. mL
(C) 100. mL
(E) 120. mL
CHEMISTRY-SECTION II
(Total Time – 45 minutes)
PART A
Time – 20 minutes
YOU MAY USE YOUR CALCULATOR FOR PART A
CLEARLY SHOW THE METHOD USED AND STEPS INVOLVED IN ARRIVING AT YOUR ANSWERS. It is to
your advantage to do this, because you may earn partial credit if you do and you will receive little or
no credit if you do not. Attention should be paid to significant figures. Be sure to write all your
answers to the questions on the lined pages following each question in the booklet with the pink
cover. Do NOT write your answers on the green insert.
Part B Time - 25 minutes
NO CALCULATORS MAY BE USED WITH PART B.
Answer Question 4 below. The Section II score weighting for this question is 15 percent.
4. Write the formulas to show the reactants and the products for any FIVE of the laboratory situations
described below. Answers to more than five choices will not be graded. In all cases a reaction occurs.
Assume that solutions are aqueous unless otherwise indicated. Represent substances in solution as ions
if the substances are extensively ionized. Omit formulas for any ions or molecules that are unchanged
by the reaction. You need not balance the equations.
Example: A strip of magnesium is added to a solution of silver nitrate.
(a) Calcium oxide powder is added to distilled water.
CaO(s) + H2O(l)  Ca(OH)2 OR CaO(s) + H2O(l)  Ca2+(aq) + OH-(aq)
(b)Liquid bromine is shaken with a 0.5 M sodium iodide solution.
Br2(l) + 2Na+(aq) + 2 I- (aq)  2Na+(aq) + 2Br- (aq) + I2(s)
Br2(l) + 2 I- (aq)  2Br- (aq) + I2(s)
NOTE: equations need not be balanced.
(c)Solid lead (II) carbonate is added to a 0.5 M sulfuric acid solution.
PbCO3(s) + H2SO4(aq)  PbSO4 (s) + H2CO3(aq)
(Carbonic acid breaks down into H2O & CO2)
PbCO3(s) + 2H+(aq) + SO4 2- (aq)  PbSO4 (s) + H2O (l) + CO2(g)
OR
PbCO3(s) + H+(aq) + HSO4 - (aq)  PbSO4 (s) + H2O (l) + CO2(g)
(d) A mixture of powdered iron(III) oxide and powdered aluminum metal is heated strongly.
heat
Fe2O3 (s) + Al(s)  Al2O3(s) + Fe(s)
(e) Carbon dioxide gas is passed over hot, solid sodium oxide.
CO2 (g) + Na2O (s)  Na2CO3
Your responses to the rest of the questions in this part of the examination will be graded on the basis of
the accuracy and relevance of the information cited. Explanations should be clear and well organized.
Examples and equations may be included in your responses where appropriate. Specific answers are
preferable to broad, diffuse responses.
7. Use the following information about bond energies to answer the following questions.
Average Bond Energies (kJ/mol)
O-H
467
C=O
745
C-H
413
C=C
614
C-C
347
CΞO
1072
C-O
358
CΞC
839
a. Ethanol combustion occurs according to following reaction. C2H5OH(l) + 3O2  2CO2 + 3H2O
i. Use bond energies to predict ∆H for the combustion of ethanol
ii. Is this an endothermic or exothermic reaction? Provide support for your answer.
8. Answer the following questions using principles of chemical bonding and molecular structure.
ANSWER: http://apcentral.collegeboard.com/apc/public/repository/chemistry-released-exam-1999.pdf
The “correct” answers are shown. In addition, actual student answers are shown, with analyses of their
answers.
(a) Consider the carbon dioxide molecule, CO2, and the carbonate ion, CO3 2-.
(i) Draw the complete Lewis electron-dot structure for each species.
(ii) Account for the fact that the carbon-oxygen bond length in CO3 2-is greater than the
carbon-oxygen bond length in CO2.
(b) Consider the molecules CF4 and SF4.
(i) Draw the complete Lewis electron-dot structure for each molecule.
(ii) In terms of molecular geometry, account for the fact that the CF4 molecule is
nonpolar, whereas the SF4 molecule is polar.