Take a Diagnostic Exam 29Chapter 14
- E—Add the equations and cancel anything that
appears on both sides of the reaction arrows. - B—Definition.
- C—The table shows second order in chlorine
dioxide and first order in hydroxide ion.
Chapter 15
- D—The carbonate ion has no H+to donate to be
an acid. - D—Definition.
- E—Start with the one with a pKaas near 8.5 as
possible (H 2 PO 4 −). To go to a higher pH, the
HPO 42 −ion is needed. - B—This is an approximation.
Ka=(10−^5 )^2 /0.5- A—HBr is a strong acid.
- D—The weak acid and weak base give a nearly
neutral solution. - C—Only B and C are buffers. B is acidic and
C is basic. - B—Only B and C are buffers. B is acidic and
C is basic. - C—[OH−] =(0.0010 × 9 × 10 −^9 )1/2/ =(9 × 10 −^12 )l/2
- C—K=Ksp/Ka1Ka2
- B—I yields no change.
- D—Ksp=[La^3 +] [F−]^3 =[x][3x]^3 = 27 x^4. Solve for x.
Chapter 16
- C—10 I−(aq) + 16 H+(aq) + 2 MnO 4 −(aq)
→ 2 Mn2+(aq) + 8 H 2 O(l) + 5 I 2 (s) - D—(0.60 F)(1mol Au/3 F) =0.20 moles
- D—The cell must be nonstandard. This could be
due to variations in temperature (not 25°C) or
concentrations (1M) that are not standard. - B—A reduction is shown. Reductions take place
at the cathode. - A—Use dimensional analysis:
(0.60 coul/s)(0.75 h)(3600 s/h) (112 g S 2 O 32 −/
mol S 2 O 32 −)/(96500 coul/F) (8 F/mol S 2 O 32 −)Chapter 17
- D—The mass should be 226 −(4 + 4 + 0 +4) =
214. The atomic number should be 88 −(2 + 2 −
1 +2) =83. - C—Alpha particles are the least penetrating, and
gamma rays are the most penetrating. - C—Mass difference = 236 −4(1) − 136 =96.
Atomic number difference = 92 −4(0) − 53 =39. - C—After one half-life, 50% would remain. After
another half-life this would be reduced by^1 ⁄ 2 to
25%. The total amount decayed is 75%. Thus,
24.6 years must be two half-lives of 12.3 years
each.
Chapter 18
- B—(1.40 g) (1 mol/14 g)
- B.