Chem 6A 2011 (Sailor)
Name:
Student ID Number:
Section Number:
QUIZ #3
VERSION A KEY
Some useful constants and relationships:
Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45
101.325 J = 1 L.atm
1 atm = 760 Torr
1J = 1kg.m2/s2
1 eV = 1.6022 x 10-19J
.
.
-1 . -1
.
-1 . -1
R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K
Avogadro constant: 6.022 x 1023 mole-1
Planck's constant = h = 6.6261 x 10-34 J.s
8
c = speed of light: 3.00 x 10 m/s
RH = 1.097 x 10-2 nm-1
-2
.
C2 = second radiation constant = 1.44 x 10 K m
1
Emitted power (W)
Tλmax = C2
= (constan t)T 4
e = mc 2
c = λν
5
Surface area (m2 )
⎛ 1
⎛ Z 2 ⎞
1
1 ⎞
hc
= R H ⎜ 2 − 2 ⎟
E = hν
E=
E(in Joules) = −2.18 ×10−18 ⎜ 2 ⎟
λ
λ
⎝ n ⎠
⎝ n1 n 2 ⎠
€
Chem 6A 2011 (Sailor)
QUIZ #3
1. Write balanced net ionic equations for the following reactions and indicate the states
(i.e., (s) if the reaction forms a precipitate, etc.) The first one is done for you as an
example. All reactions are carried out in aqueous solution at room temperature: (5 pts
each)
Reactants
Net ionic equation:
Cu2+(aq) + S2-(aq) →CuS(s)
CuCl2 (aq) + Li2S(aq)
Sr2+(aq) + 2F-(aq) →SrF2(s) 3 pts correct balanced reaction with
no extra spectator ions, 2 pts correct states
Pb2+(aq) + CrO42-(aq) → PbCrO4(s) 3 pts correct balanced
reaction with no extra spectator ions, 2 pts correct states
Ba2+(aq) + SO42-(aq) →BaSO4(s) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states
H+(aq) + OH-(aq) →H2O(aq) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states (can have
H2O(l) or 3 in front of everything)
I-(aq) + Cu+(aq) → CuI(s) 3 pts correct balanced reaction, 2 pts
correct states
Sr(NO3)2 (aq) + KF(aq)
Pb(NO3)2(aq) + K2CrO4(aq)
BaCl2 (aq) + Na2SO4(aq)
H3PO4(aq) + CsOH(aq)
NaI(aq) + CuCl(aq)
2. Balance the following redox reactions and identify the oxidizing agent and the
reducing agent in each. The first one is done for you: (5 pts each) These equations
(except for the example) are all part of the Ostwald process to make nitric acid,
problem 4.84
Reaction
Oxidizing agent
Reducing Agent
6 Li +
N2 →
2
N2
Li3N
4NH3 + 5O2 → 4NO + 6H2O
3 pts correct balanced reaction
2NO + 1O2 → 2NO2
3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions
3NO2 + 1H2O → 2HNO3+ 1NO
3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions
Li
O2
1 pt
NH3
1 pt
O2
1 pt
NO
1 pt
NO2
1 pt
NO2
-
1 pt
3. The mass percent of Cl in a seawater sample is determined by titrating 25.00 mL of
seawater with AgNO3 solution, causing a precipitation reaction. An indicator is used
to detect the end point, which occurs when free Ag+ ion is present in solution after all
the Cl has reacted. If 53.63 mL of 0.2970 M AgNO3 is required to reach the end
point, what is the mass percent of Cl in the seawater? The density of seawater is
1.024 g/mL. Set up but do not solve. Circle your answer. (10 pts) This was problem
4.32
Balanced equation: Ag+(aq) + Cl-(aq) → AgCl(s)
Chem 6A 2011 (Sailor)
53.63 mL Ag +(aq)
×
€
QUIZ #3
0.2970 mol Ag +(aq) 1 mol Cl-(aq)
1L
35.45 g Cl×
×
×
×
L
1 mol Ag +(aq) 1000 mL 1 mol Cl-(aq)
mL seawater
×
× 100%
1.024 g seawater 25.00 mL seawater
10 points for correct expression
-2 points if all correct but they forget to multiply by 100%
-2 points if all correct but they forget to convert mL to L (1000 on bottom
or equivalent) So this problem will get 10, 8, 6, or 0 points.
Chem 6A 2011 (Sailor)
Name:
Student ID Number:
Section Number:
QUIZ #3
VERSION B KEY
Some useful constants and relationships:
Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45
101.325 J = 1 L.atm
1 atm = 760 Torr
1J = 1kg.m2/s2
1 eV = 1.6022 x 10-19J
.
.
-1 . -1
.
-1 . -1
R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K
Avogadro constant: 6.022 x 1023 mole-1
Planck's constant = h = 6.6261 x 10-34 J.s
8
c = speed of light: 3.00 x 10 m/s
RH = 1.097 x 10-2 nm-1
-2
.
C2 = second radiation constant = 1.44 x 10 K m
1
Emitted power (W)
Tλmax = C2
= (constan t)T 4
e = mc 2
c = λν
5
Surface area (m2 )
⎛ 1
⎛ Z 2 ⎞
1
1 ⎞
hc
= R H ⎜ 2 − 2 ⎟
E = hν
E=
E(in Joules) = −2.18 ×10−18 ⎜ 2 ⎟
λ
λ
⎝ n ⎠
⎝ n1 n 2 ⎠
€
Chem 6A 2011 (Sailor)
QUIZ #3
1. Write balanced net ionic equations for the following reactions and indicate the states
(i.e., (s) if the reaction forms a precipitate, etc.) The first one is done for you as an
example. All reactions are carried out in aqueous solution at room temperature: (5 pts
each)
Reactants
Net ionic equation:
Cu2+(aq) + S2-(aq) →CuS(s)
CuCl2 (aq) + Li2S(aq)
Ba2+(aq) + 2F-(aq) → BaF2(s) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states
2Ag+(aq) + CrO42-(aq) → Ag2CrO4(s) 3 pts correct balanced
reaction with no extra spectator ions, 2 pts correct states
Sr2+(aq) + SO42-(aq) →SrSO4(s) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states
H+(aq) + OH-(aq) →H2O(aq) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states (can have
H2O(l) or 3 in front of everything)
2I-(aq) + Hg22+(aq) → Hg2I2(s) 3 pts correct balanced reaction,
2 pts correct states
Ba(NO3)2 (aq) + KF(aq)
AgNO3(aq) + K2CrO4(aq)
SrCl2 (aq) + Na2SO4(aq)
H3PO4(aq) + NH4OH(aq)
NaI(aq) + Hg2F2(aq)
2. Balance the following redox reactions and identify the oxidizing agent and the
reducing agent in each. The first one is done for you: (5 pts each) These equations
(except for the example) are all part of the Ostwald process to make nitric acid,
problem 4.84
Reaction
Oxidizing agent
Reducing Agent
6 Li +
N2 →
2
N2
Li3N
3NO2 + 1H2O → 2HNO3+ 1NO
3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions
2NO + 1O2 → 2NO2
3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions
4NH3 + 5O2 → 4NO + 6H2O
3 pts correct balanced reaction
NO2
1 pt
Li
NO2
1 pt
O2
1 pt
NO
1 pt
O2
1 pt
NH3
1 pt
-
3. The mass percent of Cl in a seawater sample is determined by titrating 37.22 mL of
seawater with AgNO3 solution, causing a precipitation reaction. An indicator is used
to detect the end point, which occurs when free Ag+ ion is present in solution after all
the Cl has reacted. If 17.96 mL of 1.3838 M AgNO3 is required to reach the end
point, what is the mass percent of Cl in the seawater? The density of seawater is
1.024 g/mL. Set up but do not solve. Circle your answer. (10 pts) This was problem
4.32
Balanced equation: Ag+(aq) + Cl-(aq) → AgCl(s)
Chem 6A 2011 (Sailor)
17.96 mL Ag +(aq)
×
€
QUIZ #3
1.3838 mol Ag +(aq) 1 mol Cl-(aq)
1L
35.45 g Cl×
×
×
×
L
1 mol Ag +(aq) 1000 mL 1 mol Cl-(aq)
mL seawater
×
× 100%
1.024 g seawater 37.22 mL seawater
10 points for correct expression
-2 points if all correct but they forget to multiply by 100%
-2 points if all correct but they forget to convert mL to L (1000 on bottom
or equivalent) So this problem will get 10, 8, 6, or 0 points.
Chem 6A 2011 (Sailor)
Name:
Student ID Number:
Section Number:
QUIZ #3
VERSION C KEY
Some useful constants and relationships:
Specific heat capacities (in J/g.K): H2O(l) = 4.184; Al(s) = 0.900; Cu(s) = 0.387; Steel(s) = 0.45
101.325 J = 1 L.atm
1 atm = 760 Torr
1J = 1kg.m2/s2
1 eV = 1.6022 x 10-19J
.
.
-1 . -1
.
-1 . -1
R = Ideal gas constant: 0.08206 L atm mol K = 8.31451 J mol K
Avogadro constant: 6.022 x 1023 mole-1
Planck's constant = h = 6.6261 x 10-34 J.s
8
c = speed of light: 3.00 x 10 m/s
RH = 1.097 x 10-2 nm-1
-2
.
C2 = second radiation constant = 1.44 x 10 K m
1
Emitted power (W)
Tλmax = C2
= (constan t)T 4
e = mc 2
c = λν
5
Surface area (m2 )
⎛ 1
⎛ Z 2 ⎞
1
1 ⎞
hc
= R H ⎜ 2 − 2 ⎟
E = hν
E=
E(in Joules) = −2.18 ×10−18 ⎜ 2 ⎟
λ
λ
⎝ n ⎠
⎝ n1 n 2 ⎠
€
Chem 6A 2011 (Sailor)
QUIZ #3
1. Write balanced net ionic equations for the following reactions and indicate the states
(i.e., (s) if the reaction forms a precipitate, etc.) The first one is done for you as an
example. All reactions are carried out in aqueous solution at room temperature: (5 pts
each)
Reactants
Net ionic equation:
Cu2+(aq) + S2-(aq) →CuS(s)
CuCl2 (aq) + Li2S(aq)
3Sr2+(aq) + 2PO43-(aq) →Sr3(PO4)2(s) 3 pts correct balanced
reaction with no extra spectator ions, 2 pts correct states
Ca2+(aq) + CrO42-(aq) → CaCrO4(s) 3 pts correct balanced
reaction with no extra spectator ions, 2 pts correct states
Pb2+(aq) + SO42-(aq) →PbSO4(s) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states
H+(aq) + OH-(aq) →H2O(aq) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states (can have
H2O(l) or 3 in front of everything)
I-(aq) + Ag+(aq) → AgI(s) 3 pts correct balanced reaction, 2
pts correct states
Sr(NO3)2 (aq) + K3PO4(aq)
Ca(NO3)2(aq) + K2CrO4(aq)
Pb(NO3)2 (aq) + Na2SO4(aq)
H3PO4(aq) + KOH(aq)
NaI(aq) + AgNO3(aq)
2. Balance the following redox reactions and identify the oxidizing agent and the
reducing agent in each. The first one is done for you: (5 pts each) These equations
(except for the example) are all part of the Ostwald process to make nitric acid,
problem 4.84
Reaction
Reducing Agent
Oxidizing agent
6 Li +
N2 →
2
Li3N
2NO + 1O2 → 2NO2
3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions
4NH3 + 5O2 → 4NO + 6H2O
3 pts correct balanced reaction
3NO2 + 1H2O → 2HNO3+ 1NO
3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions
Li
N2
NO
1 pt
O2
1 pt
NH3
1 pt
O2
1 pt
NO2
1 pt
NO2
-
1 pt
3. The mass percent of Cl in a seawater sample is determined by titrating 50.00 mL of
seawater with AgNO3 solution, causing a precipitation reaction. An indicator is used
to detect the end point, which occurs when free Ag+ ion is present in solution after all
the Cl has reacted. If 76.36 mL of 0.4552 M AgNO3 is required to reach the end
point, what is the mass percent of Cl in the seawater? The density of seawater is
1.024 g/mL. Set up but do not solve. Circle your answer. (10 pts) This was problem
4.32
Balanced equation: Ag+(aq) + Cl-(aq) → AgCl(s)
Chem 6A 2011 (Sailor)
76.36 mL Ag +(aq)
×
€
QUIZ #3
0.4552 mol Ag +(aq) 1 mol Cl-(aq)
1L
35.45 g Cl×
×
×
×
L
1 mol Ag +(aq) 1000 mL 1 mol Cl-(aq)
mL seawater
×
× 100%
1.024 g seawater 50.00 mL seawater
10 points for correct expression
-2 points if all correct but they forget to multiply by 100%
-2 points if all correct but they forget to convert mL to L (1000 on bottom
or equivalent) So this problem will get 10, 8, 6, or 0 points.