Chem 6A 2011 (Sailor) QUIZ #3
Name: VERSION A KEY Student ID Number: Section Number:
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 R = Ideal gas constant: 0.08206 L.atm.mol-1 .K-1 = 8.31451 J.mol-1 .K-1
Avogadro constant: 6.022 x 1023 mole-1 Planck's constant = h = 6.6261 x 10-34 J.s c = speed of light: 3.00 x 108 m/s RH = 1.097 x 10-2 nm-1 C2 = second radiation constant = 1.44 x 10-2 K.m
€
Tλmax =15C2
Emitted power (W)Surface area (m2)
= (constan t)T 4 e = mc 2 c = λν
1λ
= RH1n1
2 −1n2
2
⎛
⎝ ⎜
⎞
⎠ ⎟ E = hν E =
hcλ
E(in Joules) = −2.18 ×10−18 Z 2
n2
⎛
⎝ ⎜
⎞
⎠ ⎟
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:
CuCl2 (aq) + Li2S(aq) Cu2+(aq) + S2-
(aq) →CuS(s)
Sr(NO3)2 (aq) + KF(aq) Sr2+
(aq) + 2F-(aq) →SrF2(s) 3 pts correct balanced reaction with
no extra spectator ions, 2 pts correct states
Pb(NO3)2(aq) + K2CrO4(aq) Pb2+
(aq) + CrO42-
(aq) → PbCrO4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states
BaCl2 (aq) + Na2SO4(aq) Ba2+
(aq) + SO42-
(aq) →BaSO4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states
H3PO4(aq) + CsOH(aq) 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)
NaI(aq) + CuCl(aq) I-
(aq) + Cu+(aq) → CuI(s) 3 pts correct balanced reaction, 2 pts
correct states 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 Li3N N2 Li
4NH3 + 5O2 → 4NO + 6H2O 3 pts correct balanced reaction O2 1 pt NH3 1 pt
2NO + 1O2 → 2NO2 3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions O2 1 pt NO 1 pt
3NO2 + 1H2O → 2HNO3+ 1NO 3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions 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) QUIZ #3
€
53.63 mL Ag+(aq)
× 0.2970 mol Ag+
(aq)
L×
1 mol Cl-(aq)
1 mol Ag+(aq)
× 1 L
1000 mL×
35.45 g Cl-
1 mol Cl-(aq)
×mL seawater
1.024 g seawater×
25.00 mL seawater
×100%
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) QUIZ #3
Name: VERSION B KEY Student ID Number: Section Number:
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 R = Ideal gas constant: 0.08206 L.atm.mol-1 .K-1 = 8.31451 J.mol-1 .K-1
Avogadro constant: 6.022 x 1023 mole-1 Planck's constant = h = 6.6261 x 10-34 J.s c = speed of light: 3.00 x 108 m/s RH = 1.097 x 10-2 nm-1 C2 = second radiation constant = 1.44 x 10-2 K.m
€
Tλmax =15C2
Emitted power (W)Surface area (m2)
= (constan t)T 4 e = mc 2 c = λν
1λ
= RH1n1
2 −1n2
2
⎛
⎝ ⎜
⎞
⎠ ⎟ E = hν E =
hcλ
E(in Joules) = −2.18 ×10−18 Z 2
n2
⎛
⎝ ⎜
⎞
⎠ ⎟
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:
CuCl2 (aq) + Li2S(aq) Cu2+(aq) + S2-
(aq) →CuS(s)
Ba(NO3)2 (aq) + KF(aq) Ba2+
(aq) + 2F-(aq) → BaF2(s) 3 pts correct balanced reaction
with no extra spectator ions, 2 pts correct states
AgNO3(aq) + K2CrO4(aq) 2Ag+
(aq) + CrO42-
(aq) → Ag2CrO4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states
SrCl2 (aq) + Na2SO4(aq) Sr2+
(aq) + SO42-
(aq) →SrSO4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states
H3PO4(aq) + NH4OH(aq) 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)
NaI(aq) + Hg2F2(aq) 2I-
(aq) + Hg22+
(aq) → Hg2I2(s) 3 pts correct balanced reaction, 2 pts correct states
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 Li3N N2 Li 3NO2 + 1H2O → 2HNO3+ 1NO 3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions NO2 1 pt NO2 1 pt
2NO + 1O2 → 2NO2 3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions O2 1 pt NO 1 pt
4NH3 + 5O2 → 4NO + 6H2O 3 pts correct balanced reaction 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) QUIZ #3
€
17.96 mL Ag+(aq)
× 1.3838 mol Ag+
(aq)
L×
1 mol Cl-(aq)
1 mol Ag+(aq)
× 1 L
1000 mL×
35.45 g Cl-
1 mol Cl-(aq)
×mL seawater
1.024 g seawater×
37.22 mL seawater
×100%
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) QUIZ #3
Name: VERSION C KEY Student ID Number: Section Number:
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 R = Ideal gas constant: 0.08206 L.atm.mol-1 .K-1 = 8.31451 J.mol-1 .K-1
Avogadro constant: 6.022 x 1023 mole-1 Planck's constant = h = 6.6261 x 10-34 J.s c = speed of light: 3.00 x 108 m/s RH = 1.097 x 10-2 nm-1 C2 = second radiation constant = 1.44 x 10-2 K.m
€
Tλmax =15C2
Emitted power (W)Surface area (m2)
= (constan t)T 4 e = mc 2 c = λν
1λ
= RH1n1
2 −1n2
2
⎛
⎝ ⎜
⎞
⎠ ⎟ E = hν E =
hcλ
E(in Joules) = −2.18 ×10−18 Z 2
n2
⎛
⎝ ⎜
⎞
⎠ ⎟
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:
CuCl2 (aq) + Li2S(aq) Cu2+(aq) + S2-
(aq) →CuS(s)
Sr(NO3)2 (aq) + K3PO4(aq) 3Sr2+
(aq) + 2PO43-
(aq) →Sr3(PO4)2(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states
Ca(NO3)2(aq) + K2CrO4(aq) Ca2+
(aq) + CrO42-
(aq) → CaCrO4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states
Pb(NO3)2 (aq) + Na2SO4(aq) Pb2+
(aq) + SO42-
(aq) →PbSO4(s) 3 pts correct balanced reaction with no extra spectator ions, 2 pts correct states
H3PO4(aq) + KOH(aq) 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)
NaI(aq) + AgNO3(aq) I-
(aq) + Ag+(aq) → AgI(s) 3 pts correct balanced reaction, 2
pts correct states 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 Li N2 2NO + 1O2 → 2NO2
3 pts correct balanced reaction. OK to leave off the 1 or to use fractions
NO 1 pt O2 1 pt
4NH3 + 5O2 → 4NO + 6H2O 3 pts correct balanced reaction NH3 1 pt O2 1 pt
3NO2 + 1H2O → 2HNO3+ 1NO 3 pts correct balanced reaction. OK to
leave off the 1 or to use fractions 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) QUIZ #3
€
76.36 mL Ag+(aq)
× 0.4552 mol Ag+
(aq)
L×
1 mol Cl-(aq)
1 mol Ag+(aq)
× 1 L
1000 mL×
35.45 g Cl-
1 mol Cl-(aq)
×mL seawater
1.024 g seawater×
50.00 mL seawater
×100%
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.