ANSWERS TO Top 10 questions missed - June 2012
12. What is the charge of the nucleus of an oxygen atom?
(1) 0
(2) - 2
(3) + 8
(4) + 16
41. At 50.°C and standard pressure, intermolecular forces of attraction are strongest in a
sample of
(1) ethanoic acid
(2) ethanol
(3) propanone
(4) water
42. At 101.3 kPa and 298 K, what is the total amount of heat released when one mole of
aluminum oxide, Al2O3(s), is formed from its elements?
(1) 393.5 kJ
(2) 837.8 kJ
(3) 1676 kJ
(4) 3351 kJ
47. Which ion is most easily reduced?
(1) Zn2+
(2) Mg2+
(3) Co2+
(4) Ca2+
Base your answer to question 55 on the information below.
The reaction between aluminum and an aqueous solution of copper (II) sulfate is represented
by the unbalanced equation below.
Al(s) + CuSO4(aq) → Al2(SO4)3(aq) + Cu(s)
55. Determine the total mass of Cu produced when 1.08 grams of Al reacts completely with
9.58 grams of CuSO4 to produce 6.85 grams of Al2(SO4)3.
CONSERVATION OF MASS!
1.08 g of Al + 9.58 g of CuSO4  6.85 g of Al2(SO4)3 + ? Cu
1.08 + 9.58 = 6.85 + x
x = 3.81 g
Base your answer to question 65 on the information below.
In a titration, 20.0 milliliters of 0.15 M HCl(aq) is exactly neutralized by 18.0 milliliters of
KOH(aq).
65. Determine the concentration of the KOH(aq).
MA VA = MB VB
(0.15 M) (20.0 mL) = MB (18.0 mL)
MB = 0.16 M
(HCl = acid, KOH = base)
Base your answer to question 68 on the information below.
John Dalton, an early scientist, sketched the structure of compounds using his own symbols for
the elements known at the time. Dalton’s symbols for four elements and his drawing of
potassium aluminum sulfate are represented by the diagram below.
Today, it is known that the chemical formula for potassium aluminum sulfate is
KAl(SO4)2•12H2O. It is a hydrated compound because water molecules are included within its
crystal structure. There are 12 moles of H2O for every 1 mole of KAl(SO4)2.
The compound contains two different positive ions. The gram-formula mass of KAl(SO4)2•12H2O
is 474 grams per mole.
68. Show a numerical setup for calculating the percent composition by mass of water in
KAl(SO4)2•12H2O.
% = part x 100
Whole
% = mass of water x 100
mass of compound
% = 12 (18) x 100
474
Base your answers to questions 73 through 74 on the information below.
A student constructs an
electrochemical cell
during a laboratory
investigation. When
the switch is closed,
electrons flow through
the external circuit. The
diagram and equation
to the right represent
this cell and the
reaction that occurs.
% = 45.6 %
73. Write a balanced half-reaction equation for the oxidation that occurs when the switch is
closed.
Oxidation is losing electrons & gaining charge, so e- should be on the right of the arrow.
Since this is voltaic, oxidized element is HIGHER on Table J. Remember to balance
charges!
Al  Al3++ 3e-
74. Determine the number of moles of Al(s) needed to completely react with 9.0 moles of
Ni2+(aq) ions.
2Al + 3 Ni2+  2Al2+ + 3Ni
From this we see the ratio 2 Al : 3 Ni2+ We need to keep this ratio, but with 9 Ni2+
2 : 3 as x : 9
or
2 = x
x= 6 moles Al
3 9
Base your answer to question 85 on the information below.
Nuclear fission has been used to produce electricity. However, nuclear fusion for electricity
production is still under development. The notations of some nuclides used in nuclear reactions
are shown in the table below.
85. State one potential benefit of using nuclear fusion instead of the current use of nuclear
fission to produce electricity.
Fusion produces less radioactive waste
Reactants or starting material for fusion is more readily available
Fusion produces more energy per gram of reactant (it’s more efficient)
12. (3) + 8 41. (1) ethanoic acid
55
65
42. (3) 1676 kJ
47. (3) Co2+
CONSERVATION OF MASS!
1.08 g of Al + 9.58 g of CuSO4  6.85 g of Al2(SO4)3 + ? Cu
1.08 + 9.58 = 6.85 + x
x = 3.81 g
MA VA = MB VB
(0.15 M) (20.0 mL) = MB (18.0 mL)
MB = 0.16 M
(HCl = acid, KOH = base)
% = part x 100 % = mass of water x 100
whole
mass of compound
68
% = 12 (18) x 100 % = 45.6 %
474
Oxidation is losing electrons & gaining charge, so e- should be on the right
of the arrow. Since this is voltaic, oxidized element is HIGHER on Table J.
73
Remember to balance charges! 2Al  2Al3++ 6e- or Al  Al3++ 3e2Al + 3 Ni2+  2Al2+ + 3Ni
74 From this we see the ratio 2 Al : 3 Ni2+ We need to keep this ratio, but
with 9 Ni2+
2 : 3 as x : 9
or
2 = x
x= 6 moles Al
3 9
Fusion produces less radioactive waste
85 Reactants or starting material for fusion is more readily available
Fusion produces more energy per gram of reactant (it’s more efficient)