University of Waterloo Final Examination FALL TERM
University of Waterloo Final Examination FALL TERM Course Number Course Title Sections Instructors
CHEM 120 Physical and Chemical Properties of Matter 001, 002, 003, 004, 005 C. Bissonnette, S. Forsey, D. Gilbert, T. McMahon
Date of Exam Time Period Duration of Exam Number of Exam Pages (including the Cover & Data sheets) Exam Type Additional Materials Allowed
tba tba 3 hours (Note: Your exam is only 2 ½ hours!) 17
Student Name Student ID Number
Closed book Calculator
Given name(s)
Surname
______________________________________________________
Signature
INSTRUCTIONS 1. Calculators are permitted. Laptop or notebook computers are not permitted. 2. This examination booklet and the Data Sheet must be returned. 3. Attempt all 30 questions. Each question is worth 3 marks. Part marks may be awarded. Do NOT write in this box. 1-2:
/6
9-10:
/6
17-18:
/6
25-26:
/6
3-4:
/6
11-12:
/6
19-20:
/6
27-28:
/6
5-6:
/6
13-14:
/6
21-22:
/6
29-30:
/6
7-8:
/6
15-16:
/6
23-24:
/6
CHEM 120: Sample exam questions
1.
Page 2 of 17
Methane (CH4) reacts with oxygen (O2) to give carbon dioxide (CO2) and water (H2O). If 16 grams of methane and 32 grams of oxygen react (until one of the reactants is completely consumed) and 11 grams of carbon dioxide are produced, then what is the percent yield of carbon dioxide in this reaction?
% yield of CO2 = 2. A compound contains only the elements carbon and hydrogen. When a 1.000-g sample of this compound is burned in excess oxygen, 3.296 g CO2 and 0.899 g H2O are obtained. What is the empirical formula of this compound?
Empirical formula:
CHEM 120: Sample exam questions
3.
Page 3 of 17
A sample of MgCl2 is known to be contaminated with NaCl. A 3.47-g portion of the sample is dissolved in water and treated with an excess of AgNO3 solution. The precipitate is filtered off, dried and weighed. The dried precipitate weighs 10.3 g. Calculate the amount of MgCl2 in the original sample (in weight percent).
% MgCl2 = 4.
A solution of HCl is prepared by diluting 0.500 L of a HCl solution to 6.0 L. If 10.0 mL of the diluted HCl solution is required to titrate 20.0 mL of 0.25 mol L−1 NaOH solution, then what is the concentration, C, of the undiluted HCl solution?
C =
mol L−1
CHEM 120: Sample exam questions
5.
Page 4 of 17
In the following reaction, which element is being oxidized? Which element is being reduced? What is the net ionic equation for the reaction? 2 KCl + MnO2 + 2 H2SO4 → K2SO4 + MnSO4 + Cl2 + 2 H2O
is oxidized.
is reduced.
Net ionic equation:
6.
What is the complete and balanced chemical equation for the reaction below, assuming the reaction is carried out in basic, aqueous solution? Ag(s)
+
CrO42−(aq)
Complete and balanced equation:
→
Ag+(aq)
+
Cr(OH)3(s)
CHEM 120: Sample exam questions
7.
Page 5 of 17
The density of a gas at 273 K and 1.0 atm is 1.25 g L−1. What is molar mass of this gas?
g mol−1
Molar mass: 8.
Air bags in automobiles use the following reaction to inflate, 6 NaN3(s)
+
Fe2O3(s)
→
3 Na2O(s)
+
2 Fe(s)
+
9 N2(g).
How many grams of NaN3 are required to provide 50.0 L N2 at 298 K and 0.984 atm?
Mass of NaN3 =
g
CHEM 120: Sample exam questions
9.
Page 6 of 17
What volume of CO2(g), measured at 0.0oC and 101.3 kPa, is produced if 0.500 kg of CaCO3 is required to neutralize a sulfuric acid (H2SO4) spill? The other products of the neutralization reaction are calcium sulfate (CaSO4) and water.
Volume of CO2 =
10.
L
Exactly 1.0 L of hydrogen gas (H2) is collected over water at 292 K under a total pressure of 152 torr. If the vapour pressure of water is 16.5 torr at this temperature, what is the mole fraction of hydrogen in the gas sample?
Mole fraction of H2 =
CHEM 120: Sample exam questions
11.
Page 7 of 17
A mixture containing equal moles of N2 and Ne has a total mass of 10.0 g. What is the volume of this mixture at 101.3 kPa and 301 K?
L
Volume =
12.
How long would it take for a 0.100 kilowatt heater to warm 0.250 L of water from 20.0oC to 95.0oC in a perfectly insulated calorimeter? Heat capacity of water = 75.3 J K−1 mol−1 Density of water = 1.0 g mL−1
1 kilowatt = 1 kJ s−1
time =
s
CHEM 120: Sample exam questions
13.
Page 8 of 17
The enthalpies of combustion are −1401, −1550, and −286 kJ mol−1 respectively for C2H4(g), C2H6(g), and H2(g). What is the enthalpy change for the reaction below? (Show how the combustion reactions combine to give the reaction below!) C2H4(g) +
H2(g)
→
C2H6(g) ?
ΔH o = 14.
kJ (per mol H2)
The bond dissociation energy of H2 is 432 kJ mol−1. What is the maximum wavelength of light that can be used to dissociate a single H2 molecule into atoms, assuming that the dissociation is caused by the absorption of a single photon? Give your answer in nanometres. (1 nm = 1×10−9 m)
λ =
nm
CHEM 120: Sample exam questions
15.
Page 9 of 17
What is the wavelength of light emitted when an electron in a hydrogen atom undergoes a transition from the n = 5 to the n = 2 level?
λ =
16.
nm
Classify the following statements as true (T) or false (F). (0.5 marks each)
2
The probability of finding an electron at a point is proportional to Ψ . Each line in the line spectrum of the H atom corresponds to the energy of a specific electron orbital. For the pz orbital, there is zero probability of finding the electron in the xy plane. For an H atom, the energy of a 2p orbital is greater than that of a 2s orbital. The most probable radial distance, r, between the electron and the nucleus corresponds to the position of the maximum in a plot of R 2 versus r, where R is the radial part of Ψ . It is not possible to measure the momentum of the electron in the H atom.
CHEM 120: Sample exam questions 17.
Page 10 of 17
Answer the following questions. (0.5 marks each)
True (T) or False (F)? An acceptable set of quantum numbers for an electron in an atom is n = 3, l = 3, ml = −1, and ms = ½. How many electrons does it take to fill the 5f subshell? How many electrons does it take to fill the n = 3 shell? True (T) or False (F)? According the Aufbau procedure, the 3d orbital fills immediately following 3p. True (T) or False (F)? For an isolated atom that has an unfilled subshell, the electronic configuration of lowest energy is the one having the smallest number of unpaired electrons. The radial probability distribution for an orbital in the n = 3 shell is shown below. Which orbital is it? 4πr2 R2
r (in pm)
18.
What is the ground state electron configurations for each of the following atoms or ions? S− Ti P Fe2+
CHEM 120: Sample exam questions 19.
20.
Page 11 of 17
Arrange the elements O, As, Rb, Se, and Ca in order of increasing electronegativity values.
Draw two Lewis structures for SO2, each having a different formal charge on the sulfur atom. Which of your structures is expected to be the most important? Is this molecule linear or bent?
CHEM 120: Sample exam questions 21.
Page 12 of 17
Use bond dissociation energies to calculate ΔHo for the reaction below. Note: In CH3CHO, both H and O are bonded to the underlined carbon atom. 2 CH3CHO(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(g)
Bond dissociation energies (in kJ mol−1) H−H 432 H−C 411 H−O 459 C−C 346 C−O 358 C=C 602 C=O 799 O=O 494 C≡C 835 C≡O 1072
ΔHo =
22.
kJ
Based on their relative positions in the periodic table, which one of the following bonds is the most ionic in character? Which bond is least ionic in character? Be-O
Most ionic:
Li-F
Be-F
B-O
Least ionic:
O-F
CHEM 120: Sample exam questions
23.
Page 13 of 17
Draw the Lewis structure for IF3. According to the VSEPR model, what is the molecular geometry of IF3 ?
Geometry:
24.
Draw the Lewis structures for each of the following molecules and then state whether the molecule is nonpolar (i.e. has no permanent dipole moment) or polar (i.e., has a permanent dipole moment). Molecule
CO2
SO3
SF4
Lewis structure
Polar or nonpolar?
CHEM 120: Sample exam questions 25.
Page 14 of 17
Consider the following statements about PF3. Answer the following questions as directed.
What is the hybridization of P in PF3? (1 mark)
According to valence bond theory, what orbitals are involved in the formation of a P-F bond? (1 mark) According to VSEPR theory, what is the molecular geometry of PF3? (1 mark)
26.
Consider the following Lewis structures and then answer the questions as directed. O
O Cl
P
Cl
Cl
Cl
(i)
P
O Cl
Cl
Cl
(ii)
P Cl
(iii)
What is the formal charge of P in structure (i)? (0.5 marks) What is the formal charge of P in structure (ii)? (0.5 marks) What is the formal charge of P in structure (iii)? (0.5 marks) What is the formal charge of O in structure (ii)? (0.5 marks) What is the formal charge of Cl in structure (iii)? (0.5 marks) Which of these structures is the most important one? (0.5 marks)
Cl
CHEM 120: Sample exam questions
Page 15 of 17
27.
Draw diagram that shows how the orbitals of the atoms overlap to form the σ bonds and π bonds in the acetylene (C2H2) molecule. You may find it helpful to draw two diagrams, one for the σ bonds and one for the π bonds.
28.
Consider the Lewis structure below for CH3C2CH2CHCO. (Note: Lone pairs are not shown. The letters on the carbon atoms are used to identify them.) Fill in the blanks as appropriate.
H
H | Ca | H
Cb
Cc
H | Cd | H
H | Ce
Cf
Ca−Cb−Cc bond angle:
Hybridization of Ca:
H−Ca−Cb bond angle:
Hybridization of Cf:
Cd−Ce−Cf bond angle:
Number of π bonds:
O
CHEM 120: Sample exam questions
Page 16 of 17
29.
Sketch (and label) the molecular orbital energy level diagram for O2 showing only the atomic and molecular orbitals associated with the n = 2 shell of O. Show how the electrons are distributed among the molecular orbitals (for the ground state configuration).
30.
Complete following table.
Species C22− O22−
Ground state electron configuration
Bond order
# of unpaired electrons
CHEM 120: Sample exam questions
Page 17 of 17
DETACH THIS DATA SHEET FOR EASY REFERENCE 1 1A 1 H 1.008 3 Li 6.941 11 Na 22.99 19 K 39.10 37 Rb 85.47 55 Cs 132.9 87 Fr (223)
2 2A 4 Be 9.012 12 Mg 3 4 24.31 3B 4B 20 21 22 Ca Sc Ti 40.08 44.96 47.88 38 39 40 Sr Y Zr 87.62 88.91 91.22 56 (57-71) 72 Ba La-Lu Hf 137.3 178.5 (89-103) 104 88 Ac-Lr Rf Ra 226
5 5B 23 V 50.94 41 Nb 92.91 73 Ta 180.9 105 Db
6 6B 24 Cr 52.00 42 Mo 95.94 74 W 183.9 106 Sg
7 7B 25 Mn 54.94 43 Tc (98) 75 Re 186.2 107 Bh
8
←
26 Fe 55.85 44 Ru 101.1 76 Os 190.2 108 Hs
9 8B 27 Co 58.93 45 Rh 102.9 77 Ir 192.2 109 Mt
10
→
28 Ni 58.69 46 Pd 106.4 78 Pt 195.1 110 Uun
11 1B 29 Cu 63.55 47 Ag 107.9 79 Au 197.0 111 Uuu
12 2B 30 Zn 65.38 48 Cd 112.4 80 Hg 200.6 112 Uub
13 3A 5 B 10.81 13 Al 26.98 31 Ga 69.72 49 In 114.8 81 Tl 204.4 113 Uut
14 4A 6 C 12.01 14 Si 28.09 32 Ge 72.59 50 Sn 118.7 82 Pb 207.2
15 5A 7 N 14.01 15 P 30.97 33 As 74.92 51 Sb 121.8 83 Bi 209.0
16 6A 8 O 16.00 16 S 32.07 34 Se 78.96 52 Te 127.6 84 Po (209)
17 7A 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9 85 At (210)
18 8A 2 He 4.003 10 Ne 20.18 18 Ar 39.95 36 Kr 83.80 54 Xe 131.3 86 Rn (222)
Constants: NA = 6.022×1023 mol−1
RH = 2.179×10−18 J (Rydberg constant)
R = 0.082058 atm L K−1 mol−1
c = 2.998×108 m s−1
e = 1.602 × 10−19 C
= 8.3145 kPa L K−1 mol−1
h = 6.626×10−34 J s
π = 3.141592654
= 8.3145 J K−1 mol−1
me = 9.109×10−31 kg
Conversion factors: 1 kPa L = 1 J = 1 Pa m3
1 atm = 101.325 kPa
1 cal = 4.184 J
1 nm =
10−9 m 1 atm = 760 torr
1 atm L = 101.325 J
1 m3 = 1000 L
1 pm = 10−12 m
1 atm = 760 mmHg
0oC = 273.15 K
1 amu = 1.6605×10−27 kg
Key Equations: PV = nRT
( P + an 2 / V 2 )(V − nb) = nRT
w = − Pext ΔV
q = C ΔT
ΔU = q + w
H = U + PV
ΔH = ΔU + Δn RT
c = λν
λ = h/ p
p = mv
Δx Δp ≥ h /(4π)
E = hν
μ = δr
B.O. = 12 (nb − na )
En = − RH / n 2
CHEM 120 Physical and Chemical Properties of Matter 001, 002, 003, 004, 005 C. Bissonnette, S. Forsey, D. Gilbert, T. McMahon
Date of Exam Time Period Duration of Exam Number of Exam Pages (including the Cover & Data sheets) Exam Type Additional Materials Allowed
tba tba 3 hours (Note: Your exam is only 2 ½ hours!) 17
Student Name Student ID Number
Closed book Calculator
Given name(s)
Surname
______________________________________________________
Signature
INSTRUCTIONS 1. Calculators are permitted. Laptop or notebook computers are not permitted. 2. This examination booklet and the Data Sheet must be returned. 3. Attempt all 30 questions. Each question is worth 3 marks. Part marks may be awarded. Do NOT write in this box. 1-2:
/6
9-10:
/6
17-18:
/6
25-26:
/6
3-4:
/6
11-12:
/6
19-20:
/6
27-28:
/6
5-6:
/6
13-14:
/6
21-22:
/6
29-30:
/6
7-8:
/6
15-16:
/6
23-24:
/6
CHEM 120: Sample exam questions
1.
Page 2 of 17
Methane (CH4) reacts with oxygen (O2) to give carbon dioxide (CO2) and water (H2O). If 16 grams of methane and 32 grams of oxygen react (until one of the reactants is completely consumed) and 11 grams of carbon dioxide are produced, then what is the percent yield of carbon dioxide in this reaction?
% yield of CO2 = 2. A compound contains only the elements carbon and hydrogen. When a 1.000-g sample of this compound is burned in excess oxygen, 3.296 g CO2 and 0.899 g H2O are obtained. What is the empirical formula of this compound?
Empirical formula:
CHEM 120: Sample exam questions
3.
Page 3 of 17
A sample of MgCl2 is known to be contaminated with NaCl. A 3.47-g portion of the sample is dissolved in water and treated with an excess of AgNO3 solution. The precipitate is filtered off, dried and weighed. The dried precipitate weighs 10.3 g. Calculate the amount of MgCl2 in the original sample (in weight percent).
% MgCl2 = 4.
A solution of HCl is prepared by diluting 0.500 L of a HCl solution to 6.0 L. If 10.0 mL of the diluted HCl solution is required to titrate 20.0 mL of 0.25 mol L−1 NaOH solution, then what is the concentration, C, of the undiluted HCl solution?
C =
mol L−1
CHEM 120: Sample exam questions
5.
Page 4 of 17
In the following reaction, which element is being oxidized? Which element is being reduced? What is the net ionic equation for the reaction? 2 KCl + MnO2 + 2 H2SO4 → K2SO4 + MnSO4 + Cl2 + 2 H2O
is oxidized.
is reduced.
Net ionic equation:
6.
What is the complete and balanced chemical equation for the reaction below, assuming the reaction is carried out in basic, aqueous solution? Ag(s)
+
CrO42−(aq)
Complete and balanced equation:
→
Ag+(aq)
+
Cr(OH)3(s)
CHEM 120: Sample exam questions
7.
Page 5 of 17
The density of a gas at 273 K and 1.0 atm is 1.25 g L−1. What is molar mass of this gas?
g mol−1
Molar mass: 8.
Air bags in automobiles use the following reaction to inflate, 6 NaN3(s)
+
Fe2O3(s)
→
3 Na2O(s)
+
2 Fe(s)
+
9 N2(g).
How many grams of NaN3 are required to provide 50.0 L N2 at 298 K and 0.984 atm?
Mass of NaN3 =
g
CHEM 120: Sample exam questions
9.
Page 6 of 17
What volume of CO2(g), measured at 0.0oC and 101.3 kPa, is produced if 0.500 kg of CaCO3 is required to neutralize a sulfuric acid (H2SO4) spill? The other products of the neutralization reaction are calcium sulfate (CaSO4) and water.
Volume of CO2 =
10.
L
Exactly 1.0 L of hydrogen gas (H2) is collected over water at 292 K under a total pressure of 152 torr. If the vapour pressure of water is 16.5 torr at this temperature, what is the mole fraction of hydrogen in the gas sample?
Mole fraction of H2 =
CHEM 120: Sample exam questions
11.
Page 7 of 17
A mixture containing equal moles of N2 and Ne has a total mass of 10.0 g. What is the volume of this mixture at 101.3 kPa and 301 K?
L
Volume =
12.
How long would it take for a 0.100 kilowatt heater to warm 0.250 L of water from 20.0oC to 95.0oC in a perfectly insulated calorimeter? Heat capacity of water = 75.3 J K−1 mol−1 Density of water = 1.0 g mL−1
1 kilowatt = 1 kJ s−1
time =
s
CHEM 120: Sample exam questions
13.
Page 8 of 17
The enthalpies of combustion are −1401, −1550, and −286 kJ mol−1 respectively for C2H4(g), C2H6(g), and H2(g). What is the enthalpy change for the reaction below? (Show how the combustion reactions combine to give the reaction below!) C2H4(g) +
H2(g)
→
C2H6(g) ?
ΔH o = 14.
kJ (per mol H2)
The bond dissociation energy of H2 is 432 kJ mol−1. What is the maximum wavelength of light that can be used to dissociate a single H2 molecule into atoms, assuming that the dissociation is caused by the absorption of a single photon? Give your answer in nanometres. (1 nm = 1×10−9 m)
λ =
nm
CHEM 120: Sample exam questions
15.
Page 9 of 17
What is the wavelength of light emitted when an electron in a hydrogen atom undergoes a transition from the n = 5 to the n = 2 level?
λ =
16.
nm
Classify the following statements as true (T) or false (F). (0.5 marks each)
2
The probability of finding an electron at a point is proportional to Ψ . Each line in the line spectrum of the H atom corresponds to the energy of a specific electron orbital. For the pz orbital, there is zero probability of finding the electron in the xy plane. For an H atom, the energy of a 2p orbital is greater than that of a 2s orbital. The most probable radial distance, r, between the electron and the nucleus corresponds to the position of the maximum in a plot of R 2 versus r, where R is the radial part of Ψ . It is not possible to measure the momentum of the electron in the H atom.
CHEM 120: Sample exam questions 17.
Page 10 of 17
Answer the following questions. (0.5 marks each)
True (T) or False (F)? An acceptable set of quantum numbers for an electron in an atom is n = 3, l = 3, ml = −1, and ms = ½. How many electrons does it take to fill the 5f subshell? How many electrons does it take to fill the n = 3 shell? True (T) or False (F)? According the Aufbau procedure, the 3d orbital fills immediately following 3p. True (T) or False (F)? For an isolated atom that has an unfilled subshell, the electronic configuration of lowest energy is the one having the smallest number of unpaired electrons. The radial probability distribution for an orbital in the n = 3 shell is shown below. Which orbital is it? 4πr2 R2
r (in pm)
18.
What is the ground state electron configurations for each of the following atoms or ions? S− Ti P Fe2+
CHEM 120: Sample exam questions 19.
20.
Page 11 of 17
Arrange the elements O, As, Rb, Se, and Ca in order of increasing electronegativity values.
Draw two Lewis structures for SO2, each having a different formal charge on the sulfur atom. Which of your structures is expected to be the most important? Is this molecule linear or bent?
CHEM 120: Sample exam questions 21.
Page 12 of 17
Use bond dissociation energies to calculate ΔHo for the reaction below. Note: In CH3CHO, both H and O are bonded to the underlined carbon atom. 2 CH3CHO(g) + 5 O2(g) → 4 CO2(g) + 4 H2O(g)
Bond dissociation energies (in kJ mol−1) H−H 432 H−C 411 H−O 459 C−C 346 C−O 358 C=C 602 C=O 799 O=O 494 C≡C 835 C≡O 1072
ΔHo =
22.
kJ
Based on their relative positions in the periodic table, which one of the following bonds is the most ionic in character? Which bond is least ionic in character? Be-O
Most ionic:
Li-F
Be-F
B-O
Least ionic:
O-F
CHEM 120: Sample exam questions
23.
Page 13 of 17
Draw the Lewis structure for IF3. According to the VSEPR model, what is the molecular geometry of IF3 ?
Geometry:
24.
Draw the Lewis structures for each of the following molecules and then state whether the molecule is nonpolar (i.e. has no permanent dipole moment) or polar (i.e., has a permanent dipole moment). Molecule
CO2
SO3
SF4
Lewis structure
Polar or nonpolar?
CHEM 120: Sample exam questions 25.
Page 14 of 17
Consider the following statements about PF3. Answer the following questions as directed.
What is the hybridization of P in PF3? (1 mark)
According to valence bond theory, what orbitals are involved in the formation of a P-F bond? (1 mark) According to VSEPR theory, what is the molecular geometry of PF3? (1 mark)
26.
Consider the following Lewis structures and then answer the questions as directed. O
O Cl
P
Cl
Cl
Cl
(i)
P
O Cl
Cl
Cl
(ii)
P Cl
(iii)
What is the formal charge of P in structure (i)? (0.5 marks) What is the formal charge of P in structure (ii)? (0.5 marks) What is the formal charge of P in structure (iii)? (0.5 marks) What is the formal charge of O in structure (ii)? (0.5 marks) What is the formal charge of Cl in structure (iii)? (0.5 marks) Which of these structures is the most important one? (0.5 marks)
Cl
CHEM 120: Sample exam questions
Page 15 of 17
27.
Draw diagram that shows how the orbitals of the atoms overlap to form the σ bonds and π bonds in the acetylene (C2H2) molecule. You may find it helpful to draw two diagrams, one for the σ bonds and one for the π bonds.
28.
Consider the Lewis structure below for CH3C2CH2CHCO. (Note: Lone pairs are not shown. The letters on the carbon atoms are used to identify them.) Fill in the blanks as appropriate.
H
H | Ca | H
Cb
Cc
H | Cd | H
H | Ce
Cf
Ca−Cb−Cc bond angle:
Hybridization of Ca:
H−Ca−Cb bond angle:
Hybridization of Cf:
Cd−Ce−Cf bond angle:
Number of π bonds:
O
CHEM 120: Sample exam questions
Page 16 of 17
29.
Sketch (and label) the molecular orbital energy level diagram for O2 showing only the atomic and molecular orbitals associated with the n = 2 shell of O. Show how the electrons are distributed among the molecular orbitals (for the ground state configuration).
30.
Complete following table.
Species C22− O22−
Ground state electron configuration
Bond order
# of unpaired electrons
CHEM 120: Sample exam questions
Page 17 of 17
DETACH THIS DATA SHEET FOR EASY REFERENCE 1 1A 1 H 1.008 3 Li 6.941 11 Na 22.99 19 K 39.10 37 Rb 85.47 55 Cs 132.9 87 Fr (223)
2 2A 4 Be 9.012 12 Mg 3 4 24.31 3B 4B 20 21 22 Ca Sc Ti 40.08 44.96 47.88 38 39 40 Sr Y Zr 87.62 88.91 91.22 56 (57-71) 72 Ba La-Lu Hf 137.3 178.5 (89-103) 104 88 Ac-Lr Rf Ra 226
5 5B 23 V 50.94 41 Nb 92.91 73 Ta 180.9 105 Db
6 6B 24 Cr 52.00 42 Mo 95.94 74 W 183.9 106 Sg
7 7B 25 Mn 54.94 43 Tc (98) 75 Re 186.2 107 Bh
8
←
26 Fe 55.85 44 Ru 101.1 76 Os 190.2 108 Hs
9 8B 27 Co 58.93 45 Rh 102.9 77 Ir 192.2 109 Mt
10
→
28 Ni 58.69 46 Pd 106.4 78 Pt 195.1 110 Uun
11 1B 29 Cu 63.55 47 Ag 107.9 79 Au 197.0 111 Uuu
12 2B 30 Zn 65.38 48 Cd 112.4 80 Hg 200.6 112 Uub
13 3A 5 B 10.81 13 Al 26.98 31 Ga 69.72 49 In 114.8 81 Tl 204.4 113 Uut
14 4A 6 C 12.01 14 Si 28.09 32 Ge 72.59 50 Sn 118.7 82 Pb 207.2
15 5A 7 N 14.01 15 P 30.97 33 As 74.92 51 Sb 121.8 83 Bi 209.0
16 6A 8 O 16.00 16 S 32.07 34 Se 78.96 52 Te 127.6 84 Po (209)
17 7A 9 F 19.00 17 Cl 35.45 35 Br 79.90 53 I 126.9 85 At (210)
18 8A 2 He 4.003 10 Ne 20.18 18 Ar 39.95 36 Kr 83.80 54 Xe 131.3 86 Rn (222)
Constants: NA = 6.022×1023 mol−1
RH = 2.179×10−18 J (Rydberg constant)
R = 0.082058 atm L K−1 mol−1
c = 2.998×108 m s−1
e = 1.602 × 10−19 C
= 8.3145 kPa L K−1 mol−1
h = 6.626×10−34 J s
π = 3.141592654
= 8.3145 J K−1 mol−1
me = 9.109×10−31 kg
Conversion factors: 1 kPa L = 1 J = 1 Pa m3
1 atm = 101.325 kPa
1 cal = 4.184 J
1 nm =
10−9 m 1 atm = 760 torr
1 atm L = 101.325 J
1 m3 = 1000 L
1 pm = 10−12 m
1 atm = 760 mmHg
0oC = 273.15 K
1 amu = 1.6605×10−27 kg
Key Equations: PV = nRT
( P + an 2 / V 2 )(V − nb) = nRT
w = − Pext ΔV
q = C ΔT
ΔU = q + w
H = U + PV
ΔH = ΔU + Δn RT
c = λν
λ = h/ p
p = mv
Δx Δp ≥ h /(4π)
E = hν
μ = δr
B.O. = 12 (nb − na )
En = − RH / n 2
