CHEM 1411
Zumdahl & Zumdahl
PRACTICE EXAM II (Chapters 4, 5, 6)
Multiple Choices: Select one best answer.
1. Which of the following is a weak electrolyte?
(a) barium hydroxide solution
(b) ammonia solution
(d) liquid aluminum chloride
(e) water
(c) argon gas
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Hint: For 9 ed.: Section 4.2.
Strong electrolytes:
(1) All ionic compound and (2) Molecular compounds which are strong acids: HNO3, H2SO4,
HCl, HBr, HI, HClO3, HBrO3, HlO3, HClO4, HBrO4, HlO4. Note: Strong bases are strong
electrolytes because they are IA and IIA metal oxides and metal hydroxides.
Weak electrolytes: Molecular compounds that are weak acids and weak bases.
Common weak acids: H2CO3, HNO2, H2S, CH3COOH, HF, H3PO4, etc.
Common weak base: NH3 (some people write it as NH4OH.
Nonelectrolytes: Molecular compounds exclude strong and weak electrolytes.
Common nonelectrolytes: sugar ( -ose), acetone ( -one), alcohol ( -ol), ether ( -O- ) etc.
2. How many grams of KOH are present in 35.0 mL of a 5.50 M solution?
(a) 5.5
(b) 10.8
(c) 15.7
(d) 17.8
(e) 21.3
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Hint: For 9 ed., see Section 4.3: p.p. 145‐150.
From M = mole solute/Liter solution and 1 L = 1000 mL
Then 5.50 = mole KOH/0.035 
mole KOH = 5.50x0.035 = 0.1925

From mole = mass in gram/molar mass 
Mass in gram = mole x molar mass =
0.1925x(39.10+16+1) = 0.1925x56.1 = 10.799 g
3. Which of the following solutions DOES contain 1.0 M nitrate ion?
(a) 1.0 M Co(NO3)2
(b) 0.5 M Al(NO3)3
(c) 0.25 M Mn(NO3)4
(d) 0.4 M Mg(NO3)2 (e) 0.5 M AgNO3
Hint: For 9th ed. See Section 4.3: p. 147: Interactive Example 4.3.
4. What is the final concentration (in M) of a solution when water is added to 25.0 mL of a
0.866 M KNO3 solution until the volume of the solution is exactly 500.0 mL?
(a) 0.0252
(b) 0.0368
(c) 0.0117
(d) 0.0534
(e) 0.0433
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Hint: 9 ed., see Section 4.3: p.p. 150‐153: Interactive Example 4.7.
Another type of dilution: see in CHEM 1412: acid-base titration.
What is the concentration (in M) of the final solution when a 46.2 mL, 0.568 M Ca(NO3)2
solution is mixed with 80.5 mL of 1.396 M Ca(NO3)2 solution?
(a) 0.568
(b) 1.09
(c) 0.953
(d) 1.545
(e) 1.874
Hint: Mixing solutions is diluting solution.
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New concentration = total mole of solute/ total volume of solutions.
Thus, M = (0.568x46.2+1.396 x 80.5) mmol/(46.2+80.5) mL = 1.09 M.
5. Which of the following pair when mixed will NOT produce a precipitate or solid?
(a) MgSO4(aq) and Pb(NO3)2(aq)
(b) NaCl(aq) and AgNO3(aq)
(c) Ca(OH)2(aq) and K2CO3(aq)
(d) Hg(NO3)2(aq) and HBr(aq)
(e) NH4OH(aq), ammonia and CH3COOH(aq) vinegar
Hint: For 9th ed. See Section 4.5: p.p. 153-158, Interactive Example 4.8. Be sure to use
Table 4.1: Simple Rules for the Solubility of Salts in Water to answer this question.
6. Which of the following is wrong concerning a net‐ionic equation?
2‐
2+
(a) S (aq) + Zn (aq) 
ZnS (s)
(b) 2Na(s) + Mg2+(aq) 
Mg(s) + 2Na+ (aq)
(c) Pb(NO3)2(aq) + 2Na(s) 
2NaNO3(aq) + Pb(s)
2‐
+
(d) CO3 (aq) + 2H (aq) H2O(l) + CO2(g)
(e) Fe(s) + Ni2+(aq) Fe2+ + Ni(s)
Hint: For 9th ed. See Section 4.6: p.p. 158‐160.
Check with the solubility table for the products, which must be solids (s), liquids (l) or
gases (g). For (e): see Fig 4.16.
7. When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed, lead (II)
sulfate precipitates. How many grams of lead (II) sulfate formed when 1.25 L of 0.0500 M
lead (II) nitrate and 2.00 L of 0.0250 M sodium sulfate are mixed?
(a) 15.2 g
(b) 19.0 g
(c) 34.1 g
(d) 34.2 g
(e) 68.3 g
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Hint: For 9 ed. See Section 4.7: p.p. 160-162: Interactive Example 4.11.
This is a comprehensive question involving chapter 2 (naming and chemical formula) and
balancing equation (chapter 3) and limiting reagent (chapter 3) and obtaining mole from
molarity and volume (chapter 4).
This is a solution-stoichiometry-with-limiting-reagent question (chapter 3):
Na2SO4 (aq) + Pb(NO3)2 (aq) PbSO4(s) + 2NaNO3(g)
Assume lead (II) nitrate Pb(NO3)2 is the limiting reagent:
(1.25 L Pb(NO3)2 x {0.0500 mole Pb(NO3)2 /1 L Pb(NO3)2 } x (1 mol PbSO4/1 mol Pb(NO3)2 ) x
(303.2 g PbSO4/ 1 mol PbSO4) = 18.95 g PbSO4
Assume sodium sulfate Na2SO4 is the limiting reagent:
(2.00 L Na2SO4) x {0.0250 mole Na2SO4/1 L Na2SO4} x (1 mol PbSO4/1 mol Na2SO4) x (303.2 g
PbSO4/ 1 mol Na2SO4) = 15.16 g PbSO4
So sodium sulfate Na2SO4 is the limiting reagent and lead (II) nitrate Pb(NO3)2 is the
excess reagent and there are 15.16 g lead (II) sulfate PbSO4 produced.
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8. What volume (in mL) of a 0.500 M HCl solution is needed to neutralize 10.0 mL of a
0.2000 M Ba(OH)2 solution?
(a) 8.00
(b) 4.00
(c) 2.00
(d) 1.00
(e) 0.50
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Hint: For 9 ed., see Section 4.8: p.p. 167‐169; also see Interactive Example 4.14: Solution
Stoichioimetry. Short‐cut without balancing equation: Formula for acid‐base
neutralization: ia x Ma x Va = ib x Mb x Vb where a refers to acid and b refers to base. ia
refers to number of H in the chemical formula of acid and ib refers to the number of OH in
the chemical formula of base.
9. Which of the following underlined atoms contains the oxidation number as ‐1?
(a) Cs2O
(b) CaC2
(c) SO42‐
(d) PtCl42‐
(e) NaO2
Hint: For 9th ed. See Section 4.9: p.p. 170‐174: Interactive Example 4.16. Be sure to
memorize Table 4.2 Rules for Assigning Oxidation States.
10. What volume (in L) does a sample of air occupy at 6.6 atm when 1.2 atm, 3.8 L of
air is compressed?
(a) 0.34
(b) 0.57
(c) 0.69
(d) 0.77
(e) 0.86
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Hint: For 9 ed., see Section 5.2: Boyle’s law: Interactive Example 5.2.
P1V1 = P2V2 1.2x3.8 = 6.6xV 
V = 0.69 L.
11.What is the final temperature (in K), under constant‐pressure condition, when a
sample of hydrogen gas initially at 88oC and 9.6 L is cooled until its finial volume is
3.4 L? (a) 31 (b) 68
(c) 94
(d) 128
(e) 261
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Hint: For 9 ed. See Section 5.2: Charles’s Law: Interactive Example 5.4. T is in Kelvin.
V1/T1 = V2/T2 
9.6/(88+273.15) = 3.4/T2 9.6/361.15 = 3.4/T2

T2 = 3.4/0.02658 = 127.907 Kelvin
12. What is the volume (in L) of 88.4 g of carbon dioxide gas at STP?
(a) 45.1
(b) 53.7
(c) 62.1
(d) 74.6
(e) 83.2
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Hint: For 9 ed., see Section 5.3: p.p. 198-203: Ideal gas law.
From PV = nRT where P is pressure in atm, V is volume in liter, n is mole, R is ideal gas
constant 0.082 atm.L/mol.K and T is temperature in Kelvin;
so V = nRT/P = {(88.4/44)x0.082x(0+273.15)}/1 = 45.1 Liters.
Application: A gas evolved during the fermentation of glucose (wine making) has a volume
of 0.78 L at 20.1oC and 1.00 atm. What was the volume (L) of this gas at the fermentation
temperature of 36.5oC and 2.00 atm pressure?
(a) 0.41
(b) 0.82
(c) 1.43
(d) 2.67
(e) 3.54
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Hint: For 9 ed., see Section 5.3: Combined Gas Law: Interactive Example 5.9: p. 202.
P1V1/T1 = P2V2/T2 1x0.78/(20.1+273.15) = 2xV/(36.5+273.15)
0.00266 = 0.00646xV V = 0.00266/0.00646 = 0.412 L
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13. What is the molar mass (g/mol) of 7.10 grams of gas whose volume is 5.40 L at 741 torr
and 40 oC?
(a) 35.0
(b) 70.3
(c) 86.2
(d) 94.6
(e) 102.3
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Hint: For 9 ed., see Section 5.4: p.p. 207-208: Formula (5.1); Interactive Example 5.14.
From PM = dRT where P is in unit of atm and T is in unit of Kelvin

Here, density is given indirectly as density = mass/volume

M= dRT/P = [(7.10/5.40)x0.082x(40+273.15)]/(741/760) = 35.0 g/mol
Application:
What is most likely the unknown gas according to Q 13?
(a) C2H4
(b) HCl
(c) C3H4
(d) CO2
Hint: From the molar mass comparison, the closest one will be most likely the unknown.
Application:
What is the density of HBr gas in grams per liter at 733 mmHg and 46 oC?
(a) 0.54
(b) 1.36
(c) 2.24
(d) 2.97
(e) 3.57
Hint: 9th ed., see Section 5.4: Formula (5.1) 
From PM= dRT

d = PM/RT = (733/760)x(1.008+79.90)/[0.082x(46+273.15)] = 2.97 g/L
14. The combustion process for methane, major component of natural gas, is
CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)
If 15.0 moles of methane are reacted, what is the volume of carbon dioxide (in L) produced
at 23.0 oC and 0.985 atm?
(a) 370
(b) 430
(c) 510
(d) 630
(e) 720
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Hint: Gas stoichiometry: For 9 ed., see Section 5.4:203-208; Interactive Examples 5.12
and 5.13.
From the given information, 15.0 mole of methane, we can calculate the theoretical yield
in mole of carbon dioxide. Then apply PV = nRT to calculate volume in liter for carbon
dioxide. Here, mole of carbonxide = mole of methane = 15 moles and thus
V = nRT/P = 15x0.082x(23+273.15)/0.985 = 369.81 L.
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15. Application: Comprehensive Question:
In alcohol fermentation, yeast converts glucose to ethanol and carbon dioxide:
C6H12O6(s) 
C2H5OH(l)) + 2 CO2(g)
If 5.97 g of glucose are reacted and 1.44 L of carbon dioxide gas are collected at 293 K and
0.984 atm, what is the percent yield of the reaction?
(a) 88.9 %
(b) 76.3%
(c) 65.9%
(d) 56.2%
(e) 47.6%
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Hint: Hint: Gas stoichiometry: For 9 ed., see Section 5.4:203-208; Interactive Examples
5.12 and 5.13.
From mass of glucose, we can calculate the theoretical yield of carbon dioxide in mole;
from 293 K and 0.984 atm, we can calculate the actual yield of carbon dioxide in mole.
Thus, the theoretical yield of carbon dioxide = 2 x mole of glucose
= 2 x (5.97/12x6+1x12+16x6) = 2 X (5.97/180) = 0.0663 mole.
From PV = nRT 
n = PV/RT we can calculate the actual yield
= 0.984x1.44/(0.082x293) = 0.05898 = 0.0590 mole.
So the percent yield = (actual yield / theoretical yield) x 100%
= (0.0590/0.0663)x100% = 88.95 %.
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16. What is the total pressure (in atm) of the mixture when a 2.5‐L flask at 15 oC contains a
mixture if nitrogen, helium, and neon gases at partial pressure of 0.32 atm for nitrogen,
0.15 atm for helium, and 0.42 atm for neon?
(a) 0.49
(b) 0.51
(c) 0.64
(d) 0.73
(e) 0.89
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Hint: Dalton’s Law of Partial Pressures: For 9 ed., see Section 5.5.
The total pressure is the sum of all partial pressures of component gases:
Total pressure = 0.32+0.15+0.42= 0.89 atm.
To use this formula, all the gases in the same container cannot react to each other (i.e.
non‐reactive gases).
Application:
See Interactive Example 5.18: When collecting gas over the water, the gas must be
insoluble or slightly soluble in water: for acidic gases, HCl(g), SO2(g), etc., and basic gas,
NH3(s), they cannot be collected over the water because they dissolve in water easily.
From Dalton’s partial pressure, Ptotal = Pwater vapor + Pgas ,
then Pgas = Ptotal - Pwater vapor
17. Nickel forms a gaseous compound of the formula Ni(CO)x. What is the value of x given
the fact that under the same conditions of temperature and pressure, methane (CH4)
effuses 3.3 times faster than the compound?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
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Hint: Graham’s law of effusion: For 9 ed., see Section 5.7: p.p. 222-224: Application of
Figure 5.24. See End-of-Chapter Exercises 111, 112, and 113.
{Rate of effusion for CH4 / Rate of effusion for Ni(CO)x} = Square root of {molar mass of
CH4 / molar mass of Ni(CO)x} = {molar mass of CH4 / molar mass of Ni(CO)x}1/2 = 3.3/ 1

Take square at the both side of equation, then
Rate of effusion for CH4 / Rate of effusion for Ni(CO)x = 10.89
Rate of effusion for Ni(CO)x = 16 x 10.89 = 174.24 = 58.69 + 28x
x = {(174.24 – 58.69)/28} = 4.1267 requiring rounding to whole number 
thus x = 4
18. What is the change in energy (J) of the gas when a gas expends and does P‐V work on
the surroundings equal to 325 J and at the same time absorbs 127 J of heat from the
surroundings?
(a) ‐198
(b) + 198
(c) + 157
(d) ‐157
(e) + 452
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Hint: For 9 ed., see Section 6.3: Example 6.1.
19. What is the heat absorbed (in kJ) when 250 grams of water is heated from 22oC to
98oC? The specific heat of water is 4.18 J/g.K (or 4.18 J/g. oC)
(a) 79
(b) 88
(c) 97
(d) 102
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(e) 137
Hint: For 9th ed., see Section 6.2: p. 255.
20. A sheet of gold weighing 10.0 g and at a temperature of 18.0oC is placed flat on a sheet
of iron weighing 20.0 g and at a temperature of 55.6oC. What is the final temperature (oC)
of the combined metals? Assume that no heat is lost to the surroundings. The specific heat
of iron and gold are 0.444 and 0.129 J/g.oC, respectively. (Hint: the heat gained by the gold
must be equal to the heat lost by the iron.)
(a) 50.7
(b) 63.1
(c) 72.4
(d) 47.2
(e) 36.8
Hint: Heat absorbed by the colder object gold + heat released by the hotter object iron = 0
or Hear absorbed by the colder object gold = − heat released by the hotter object iron.
For 9th ed., see Section 6.2: End-of-Chapter Exercises 53 & 55. From the Law of
Conservation of Energy, qhot + qcold = 0 
10.0 x 0.129 x (T‐18) + 20.0 x 0.444 x (T‐55.6) = 0
o

T = 50.7 C.
Prior to introduce Hess’s Law, learn the characteristics of Thermodynamics:
1. When reverse the reaction, the ∆Hnew = - ∆Hold
2. When the reaction coefficient is multiplied by a constant, the ∆Hnew = constant x
∆Hold
3. When the third chemical equation is the sum of the other two equations, then
∆H3 = ∆H1 + ∆H2
21. Consider the reaction below:
2CH3OH(l) + 3O2(g) 
4H2O(l) + 2CO2(g)
∆H = ‐1452.8 kJ
What is the value of ∆H for the reaction of 8H2O(l) + 4CO2(g) 
4 CH3OH (l) + 6O2
(g)?
(a) +1458.2
(b) ‐1458.2
(c) 3.46
(d) ‐2905.6
Hint: This question is a combination of Rules 1 and 2.
Thus, ∆H = 2x(+1452.8 kJ) = +2905.6 kJ
(e) +2905.6
22. Calculate the standard enthalpy change (kJ) for the reaction
2Al(s) + Fe2O3(s) 2Fe(s) + Al2O3(s)
Given that
2Al(s) + 3/2O2(g) 
Al2O3(s)
2Fe(s) + 3/2O2(g) 
Fe2O3(s)
∆H = ‐1669.8 kJ
∆H = ‐822.2 kJ
(a) ‐637.1
(b) ‐847.6
(c) ‐984.6
(d) ‐1120.3
(e) +847.6
Hint: Hess’s Law: the indirect method for calculating the ∆Hrxn. For 9th ed., see Section 6.3:
7
p.p. 260-264: Interactive Examples 6.2 & 6.8; End-of-Chapter Exercises 110, 111, and 112.
Terms used and procedures:
*Identity refers to either react or product;
*Quantity refers to coefficient;
*Given reactions: Two or more reactions with ∆H given;
*Target reaction/equation: The one that is asked; Must use it as reference. Any given
reaction that is different from it must be reversed or multiplied by a number (whole or
fraction);
*Resultant reaction/equation: The one that was obtained by adding the given reactions
that have been manipulated.
Conclusion: If the resultant reaction is identical to the target reaction, you are solving it
correctly. So add all the given reactions that have been manipulated together to get the
∆H.
23. From the following data,
C(graphite) + O2(g) 
CO2(g)
H2(g) + 1/2O2(g) 
H2O(l)
2C2H6(g) + 7O2(g) 
4CO2(g) + 6H2O(l)
∆H = ‐393.5 kJ
∆H = ‐285.8 kJ
∆H = ‐3119.6 kJ
What is the enthalpy change (kJ) for the reaction 2C(graphite) + 3H2(g) 
C2H6(g)?
(a) ‐84.6
(b) ‐42.3
(c) +84.6
(d) +42.3
(e) ‐67.2
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Hint: The Indirect Method (Hess’ Law). For 9 ed., see Section 6.3.
24. Calculate the standard enthalpy change (kJ) for the reaction
2HCl(g) + F2(g) 
2HF(l) + Cl2(g)
Given that
4HCl(g) + O2(g) 
2H2O(l) + 2Cl2(g)
1/2H2(g) + 1/2F2(g) 
HF(l)
H2(g) + ½O2(g) 
H2O(l)
(a) -637.1 (b) -847.6 (c) -984.6
∆H = -202.4 kJ
∆H = -600.0 kJ
∆H = -285.8 kJ
(d) -1015.4
8
(e) +847.6
25. Calculate the heat of combustion (kJ) for the following reaction:
2H2S(g) + 3O2(g) 
2H2O(l) + 2SO2(g)
The standard enthalpies of formation (∆Hf) for
∆Hf
H2S(g),
‐20.15,
H2O(l),
‐285.8,
SO2(g) and
‐296.4 and
O2(g) are
0.0, respectively.
kJ/mol
(a) +1124
(b) ‐1124
(c) +562 (d) ‐562
(e) +281
Hint: The Direct Method: Using the standard entropy of formation (∆Hf) to calculate the
entropy of reaction (∆Hrxn). For 9th ed., see Section 6.4: 264-271: Use formula (6.1) in p.
267.
∆Hrxn = {2∆Hf, H2O(l) + 2∆Hf, SO2(g)} – {2∆Hf, H2S(g) + 3∆Hf, O2(g)}
= {2x(-285.8) +2x(-296.4)} – {2x(-20.15) + 3x0}
= –1124 kJ
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