Take a Diagnostic Exam ‹ 31
- B—Only B and C are buffers. B is acidic (pH < 7)
and C is basic (pH > 7). - C—[OH-] = (0.0010 × 9 × 10 -^9 )1/2 = (9 × 10 -^12 )l/2
Estimate—the square root of 10-^12 will be 10-^6.
- C—K = Ksp/Ka1Ka2
In this case, the key is setting up the calculation
but not doing the calculation.
- B—Adding Ar yields no change, as it is not part
of the equilibrium. Increasing the temperature
of an endothermic equilibrium will increase the
amount of product. - D—Ksp = [La^3 +][F-]^3 = [x][3x]^3 = 27 x^4. Solve for
x. It is only necessary to set up the problem. This
requires a knowledge of what the equilibrium is
(LaF 3 (s) La^3 +(aq) + 3 F-(aq)) and how to
write the equilibrium expression (Ksp = [La^3 +][F-]^3 ). - A—The equilibrium constant expression for the
dissolving of manganese(II) hydroxide is:
Ksp = [Mn^2 +][OH-]^2 = 1.6 × 10 -^13
If s is used to indicate the molar solubility, the
equilibrium expression becomes:
Ksp = (s)(2s)^2 = 4s^3 = 1.6 × 10 -^13
This rearranges to: s = 3 K/4
- B—K = Ksp/Ka1Ka2 = 5.0 × 10 -^18 /(9.5 × 10 -^8 )
(1 × 10 -^19 ) - B—Ksp = [Cr^3 +][OH-]^3 = [x][3x]^3 = 27 x^4 = 1.6 ×
10 -^30. Solve for x.
Chapter 16
- C—The balanced equation is:
10 I-(aq) + 16 H+(aq) + 2 MnO 4 - (aq)
→ 2 Mn^2 +(aq) + 8 H 2 O(l) + 5 I 2 (s)
- D—It is only necessary to know the mole ratio
for the reaction (Au^3 +(aq) + 3 e- → Au(s)), which
gives (0.60 F)(1 mol 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 (1 M) that are not standard. - B—A reduction is shown. Reductions take place
at the cathode.
82. A—Use dimensional analysis:
(0.60 coul/s)(0.75 h)(3,600 s/h)(112 g S 2 O 32 - /
mol S 2 O 32 - )/(96,500 coul/F)(8 F/mol S 2 O 32 - )
83. B—Recall that 5.0 amp is 5.0 C/s. The calculation
would be:
5.0 C
s
3,600 s
h (0.50h)1F
96,500C
1mol Br
10 F159.8gBr
1mol Br22
2()
( )( )
×
- D—Dimensional analysis:
(7.50 coul/s)(0.45 h)(3,600 s/h)(253.8 g I 2 /mol I 2 )/
(96,500 coul/F)(10 F)
Chapter 17
- D—The mass of an alpha particle is 4 and the mass
of a beta particle is negligible. The mass number
(superscript) should be 226 - (4 + 4 + 0 + 4) = - The charge on an alpha particle is +2 and
the charge on the beta particle is -1; therefore, the
atomic number (subscript) 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 =
- 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—The general formula simplifies to CH 2 , which
has a molar mass of 14 g/mol. This leads to (1.40 g)
(1 mol/14 g). - B.
Chapter 19
Questions on this chapter are incorporated into
the chapters concerning the specific experiments.