2201(a)
THE UNIVERSITY OF SYDNEY
FUNDAMENTALS OF CHEMISTRY 1A - CHEM1001
FIRST SEMESTER EXAMINATION
CONFIDENTIAL
JUNE 2013
TIME ALLOWED: THREE HOURS
GIVE THE FOLLOWING INFORMATION IN BLOCK LETTERS
FAMILY
NAME
OTHER
NAMES
INSTRUCTIONS TO CANDIDATES
 All questions are to be attempted. There
are 19 pages of examinable material.
 Complete the written section of the
examination paper in INK.
 Read each question carefully. Report the
appropriate answer and show all relevant
working in the space provided.
 The total score for this paper is 100. The
possible score per page is shown in the
adjacent tables.
SID
NUMBER
TABLE
NUMBER
OFFICIAL USE ONLY
Multiple choice section
Marks
Pages
Max
2-9
28
Short answer section
Marks
Page Max
Gained
11
10
12
7
13
7
14
6
16
7
17
5
18
9
 Numerical values required for any question,
standard electrode reduction potentials, a
Periodic Table and some useful formulas
may be found on the separate data sheets.
19
4
21
8
22
3
 Pages 10, 15, 20 and 24 are for rough
working only.
23
6
Total
72
 Each new short answer question begins
with a .
 Only non-programmable, Universityapproved calculators may be used.
 Students are warned that credit may not be
given, even for a correct answer, where
there is insufficient evidence of the working
required to obtain the solution.
Gained
Check Total
Marker
CHEM1001
Page 2 of 24 pages
June 2013
The answers to the following 28 multiple choice questions should be indicated by
clearly circling the letter next to the choice you make and by filling in the
corresponding box on the computer-marked sheet provided. The marks for each
correct answer are given beside each question.
2201(a)
Marks
Instructions for use of the computer sheet. Draw a thick line through the centre
and crossing both edges of each box selected, as in this example.
Use a dark lead pencil so that you can use an eraser if you make an error. Errors
made in ink cannot be corrected – you will need to ask the examination supervisor for
another sheet. Boxes with faint or incomplete lines or not completed in the prescribed
manner may not be read. Be sure to complete the SID and name sections of the sheet.
Your answer as recorded on the sheet will be used in the event of any ambiguity.
There is only one correct choice for each question.
Negative marks will not be awarded for any question.
1.
2.
3.
How many protons (p), neutrons (n) and electrons (e) are present in the
molybdenum isotope 99
42 Mo ?
A
42 p
99 n
42 e
B
42 p
57 n
42 e
C
99 p
42 n
42 e
D
57 p
42 n
57 e
E
42 p
57 n
99 e
What is the ground state electronic configuration of phosphide anion, P3–?
A
1s2 2s2 2p6 3s2
B
1s2 2s2 2p6 3p2
C
1s2 2s2 2p6 3p5
D
1s2 2s2 2p6 3s2 3p3
E
1s2 2s2 2p6 3s2 3p6
A compound forms between a Group 2 element and a Group 17 element. Which
set of characteristics is it most likely to possess?
A
It is a shiny material that conducts electricity as a solid.
B
It is a hard, brittle material that dissolves in water.
C
It is a material with a low melting point and is insoluble in water.
D
It is a material that dissolves in water to give a non-conducting solution.
E
It is a soft material with a low melting point.
1
1
1
CHEM1001
4.
Page 3 of 24 pages
June 2013
One resonance structure of the sulfate ion is shown below.
2201(a)
Marks
1
Which statement concerning the sulfate ion is incorrect?
5.
6.
7.
A
The ion has a tetrahedral geometry.
B
The negative charges are evenly distributed over the four oxygen atoms.
C
The sulfur atom employs the d subshell in its bonding.
D
Each of the sulfur – oxygen bonds is polar.
E
Two of the sulfur – oxygen bonds are shorter than the other two.
Which one of the following pairs of atoms would form a non-polar covalent
bond?
A
Cl and O
B
N and O
C
I and I
D
Na and Cl
E
Kr and Kr
Which one of the following statements is correct?
A
Non-metals form cations when bonding with metals.
B
Ionic bonding involves bonds that are directional.
C
The elements Mg, Ca, Se and O all have a valency of 2.
D
Metallic character decreases down a group of the Periodic Table.
E
Group 15 elements contain a partially filled d subshell.
Which one of the following species contains an unpaired electron?
A
N2
B
N2O
C
NO2
D
NO3–
E
NH4+
1
1
1
CHEM1001
8.
9.
10.
11.
12.
Page 4 of 24 pages
June 2013
How many valence electrons are there in the S2– ion?
A
4
B
6
C
8
D
16
E
18
How does the volume of 1 mol of an ideal gas change if the temperature and the
pressure are both decreased by a factor of four?
A
decreases by four times
B
decreases by sixteen times
C
increases by four times
D
increases by sixteen times
E
remains unchanged
What is the name of the phase change from solid to gas?
A
Freezing
B
Melting
C
Evaporation
D
Sublimation
E
Condensation
Which one of the following pairs are isotopes?
A
15
B
C(diamond) and C(graphite)
C
S and S2–
D
CH and C2H2
E
CO2 and SiO2
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Marks
1
1
1
1
N and 14N
Which one of the following is the correctly balanced equation that represents the
combustion of methane in excess oxygen?
A
CH4(g) + 3O(g)  CO2(g) + H2O(l)
B
CH4(g) + O2(g)  CO2(g) + H2O(l)
C
CH4(g) + O2(g)  CO2(g) + 2H2O(l)
D
CH4(g) + 2O2(g)  CO2(g) + H2O(l)
E
CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)
1
CHEM1001
13.
14.
15.
Page 5 of 24 pages
June 2013
What is the formula of the compound formed between aluminium and oxygen?
A
AlO
B
AlO3
C
Al3O
D
Al3O2
E
Al2O3
The salts NaBr, Na2SO4 and BaBr2 are soluble, whilst BaSO4 is insoluble. After
mixing solutions of Na2SO4 and BaBr2, how would you proceed in order to isolate
a sample of solid NaBr?
A
Filter off NaBr(s) first, then filter off BaSO4(s).
B
Filter off BaSO4(s) first, then filter off NaBr(s).
C
Filter off NaBr(s), then evaporate the remaining solution to dryness to
recover BaSO4(s).
D
Filter off BaSO4(s), then evaporate the remaining solution to dryness to
recover NaBr(s).
E
Evaporate the resulting solution to dryness to recover both NaBr(s) and
BaSO4(s).
Which statement is correct?
A
The elements in the Periodic Table are arranged in increasing mass.
B
A mole of anything contains the same number of items as there are atoms in
exactly 12 g of 12C.
C
All atoms are the same size even though their masses may differ.
D
Equal masses of different elements contain equal numbers of atoms.
E
All isotopes of sulfur have 16 neutrons.
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY
2201(a)
Marks
1
1
1
CHEM1001
Page 6 of 24 pages
June 2013
Questions 16 - 19 refer to the diagram below and the list of reduction potentials on the
data page.
2201(a)
Marks
Voltmeter
Cd
Salt Bridge
1.0 M Cd2+
16.
17.
18.
Ni
1.0 M Ni2+
Which of the following reactions is the oxidation process occurring in this cell?
A
There will be no spontaneous oxidation process.
B
Cd2+(aq) + 2e–  Cd(s)
C
Cd(s)  Cd2+(aq) + 2e–
D
Ni2+(aq) + 2e–  Ni(s)
E
Ni(s)  Ni2+(aq) + 2e–
What would be the value of Ecell at equilibrium?
A
0.64 V
B
0.40 V
C
0.24 V
D
0.16 V
E
0.00 V
Which electrode is the anode, which is the cathode and to which electrode do the
electrons flow?
A
The Cd electrode is the cathode, the Ni electrode is the anode;
electrons flow to the cathode.
B
The Ni electrode is the cathode; the Cd electrode is the anode;
electrons flow to the cathode.
C
The Ni electrode is the cathode; the Cd electrode is the anode;
electrons flow to the anode.
D
The Cd electrode is the cathode; the Ni electrode is the anode;
electrons flow to the anode.
1
1
1
CHEM1001
19.
20.
Page 7 of 24 pages
June 2013
If the Cd2+/Cd half cell were replaced by a Pt2+(1.0 M)/Pt(s) half cell, what would
be the overall cell reaction?
A
Pt(aq) + O2(g) + 4H+(aq)  2Pt2+(s) + 2H2O(aq)
B
2Pt2+(s) + 2H2O(aq)  Pt(aq) + O2(g) + 4H+(aq)
C
2H+(aq) + Pt(s)  H2(g) + Pt2+(aq)
D
Ni2+(aq) + Pt(s)  Ni(s) + Pt2+(aq)
E
Ni(s) + Pt2+(aq)  Ni2+(aq) + Pt(s)
The equation below describes the combustion of methane to form carbon
monoxide.
CH4 + O2  CO + 2H2O
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Marks
1
1
Which one of the following statements is true?
21.
A
The oxidation number of carbon is +IV in CH4 and –IV in CO.
B
The oxidation number of carbon is –IV in CH4 and +IV in CO.
C
The oxidation number of carbon is +IV in CH4 and +IV in CO.
D
The oxidation number of carbon is +IV in CH4 and –II in CO.
E
The oxidation number of carbon is –IV in CH4 and +II in CO.
Which element undergoes reduction in the reaction between SO2 and Br2?
Br2 + SO2 + 2H2O  4H+ + SO42– + 2Br–
A
Br
B
S
C
O
D
H
E
none of them
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY
1
CHEM1001
22.
23.
Page 8 of 24 pages
June 2013
An enzyme-catalysed reaction has a thermodynamic equilibrium constant, Kc = 1
at 25 °C. If the reaction is endothermic, what will be the value of Kc at body
temperature, 37 °C?
A
Kc = 0
B
Kc > 1
C
Kc = 1
D
Kc < 1
E
Not enough information to determine
Consider the following redox equation.
2201(a)
Marks
1
1
Cl2(g) + 2Br–(aq)  Br2(aq) + 2Cl– (aq)
Which one of the following statements is correct?
24.
A
Br– is the reductant.
B
Br2 is the oxidant.
C
Cl– is the oxidant.
D
Cl2 is the reductant.
For the following reaction at equilibrium, what effect will adding a Pt catalyst have
on the amounts of reactants and/or products?
N2(g) + 2O2(g)
25.
2NO2(g)
ΔH = +66 kJ mol–1
A
Increase the amount of NO2(g) formed.
B
Increase the amount of N2(g) and O2(g) formed.
C
Increase the amount of 2NO2(g) formed.
D
Decrease the amount of 2NO2(g) formed.
E
There will be no effect on the amount of NO2(g) formed.
Which one of the following equations represents a redox reaction?
A
Pb2+(aq) + 2Cl–(aq)  PbCl2(s)
B
CaO(s) + CO2(g)  CaCO3(s)
C
Mg(s) + 2H+(aq)  Mg2+(aq) + H2(g)
D
H+(aq) + OH–(aq)  H2O(l)
E
CoCl2(s)  Co2+(aq) + 2Cl–(aq)
1
1
CHEM1001
26.
27.
Page 9 of 24 pages
June 2013
Which one of the following statements describes why chemical equilibria are
considered to be dynamic processes?
A
The forward reaction is extremely fast.
B
The reverse reaction is extremely fast.
C
The forward and reverse reactions continue to occur after equilibrium is
reached.
D
The forward and reverse reactions do not proceed after equilibrium is
reached.
E
The forward and reverse reactions always occur at different rates.
Consider the following equilibrium system.
C(s) + 2H2(g)
CH4(g)
2201(a)
Marks
1
1
∆H < 0
In which direction will the equilibrium shift in response to a decrease in temperature?
28.
A
To the right - the products will be favoured.
B
To the left - the reactants will be favoured.
C
Changing the temperature will not alter the equilibrium.
D
It is not possible to determine this unless given the value of the equilibrium
constant.
Consider the following reaction.
4NH3(g) + 7O2(g)  4NO2(g) + 6H2O(g)
1
ΔH = –1136 kJ mol–1
Which one of the following statements is correct?
A
The reaction is endothermic.
B
The reaction container would get colder during the reaction.
C
1136 kJ of heat is given off as 7.00 g of O2(g) reacts.
D
1136 kJ of heat is given off as 4.00 mol of NO2(g) is produced.
E
1136 kJ of heat is given off per mol of NH3(g) consumed.
THE SHORT ANSWER SECTION OF THE PAPER BEGINS ON PAGE 11.
CHEM1001
Page 10 of 24 pages
June 2013
THIS PAGE IS FOR ROUGH WORKING ONLY
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2201(a)
CHEM1001
Page 11 of 24 pages
June 2013
 Complete the following table, including resonance structures where appropriate. The
central atom is underlined.
Species
Lewis structure
Molecular geometry
Is the species
polar?
NF3
SO2
ClF5
BH3
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY
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Marks
10
CHEM1001
Page 12 of 24 pages
June 2013
2201(a)
Marks
 Explain the term ‘resonance structures’ and give an example.
2
 Explain why stable compounds of oxygen have 8 electrons in the valence shell, but
compounds of sulfur may have 8, 10 or 12 electrons in their valence shell.
 In the spaces provided, briefly explain the meaning of the following terms.
Valence electrons
Polar bond
Intensive properties
Page Total:
2
3
CHEM1001
Page 13 of 24 pages
June 2013
 In an experiment, 5.0 g of magnesium was dissolved in excess hydrochloric acid to
give magnesium ions and hydrogen gas. Write a balanced equation for the reaction
that occurred.
2201(a)
Marks
4
What amount of hydrogen gas (in mol) is produced in the reaction?
Answer:
What volume would the hydrogen occupy at 25 C and 100.0 kPa pressure?
Answer:
 Silicon and carbon are both in Group 14 and form dioxides. Carbon dioxide is a gas
at room temperature while silicon dioxide (sand) is a solid with a high melting point.
Describe the bonding in these two materials and explain the differences in properties
they show.
Page Total:
3
CHEM1001
Page 14 of 24 pages
June 2013
 Complete the following table by filling in the compound name or formula as required.
Name
2201(a)
Marks
2
Formula
lead(II) chloride
dinitrogen trioxide
Na2SO4
SF6
 In the Periodic Table given, hydrogen is placed at the top of Group 1. List reasons for
and against placing hydrogen in this position.
For:
Against:
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.
Page Total:
4
CHEM1001
Page 15 of 24 pages
June 2013
THIS PAGE IS FOR ROUGH WORKING ONLY
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2201(a)
CHEM1001
Page 16 of 24 pages
June 2013
 A 0.060 M solution of aluminium nitrate and a 0.080 M solution of potassium
phosphate are prepared by dissolving Al(NO3)3 and K3PO4 in water. Write the ionic
equations for these two dissolutions reactions.
Dissolution
of Al(NO3)3
Dissolution
of K3PO4
If these solutions are combined, aluminium phosphate precipitates. Write the ionic
equation for the precipitation reaction.
100.0 mL of the aluminium nitrate solution is added to 50.0 mL of the potassium
phosphate solution. What amount (in mol) of aluminium phosphate precipitates?
Answer:
What is the final concentration of aluminium ions remaining in solution after the
precipitation?
Answer:
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2201(a)
Marks
7
CHEM1001
Page 17 of 24 pages
June 2013
 By adding double bonds and lone pairs, complete the structural formulae of the
nitrogen bases adenine and thymine below.
adenine
thymine
In DNA, these two molecules interact through two hydrogen bonds. Redraw the
structures below showing the alignment of the two molecules that allows this to occur
and clearly show the hydrogen bonds.
Page Total:
2201(a)
Marks
5
CHEM1001
Page 18 of 24 pages
June 2013
2201(a)
 Rechargeable nickel-cadmium batteries normally operate (discharge) with the
following oxidation and reduction half-cell reactions.
Marks
9
Cd(s) + 2OH–(aq)  Cd(OH)2(s) + 2e–
E = 0.82 V
NiO(OH)(s) + H2O(l) + e–  Ni(OH)2(s) + OH–(aq)
E = 0.60 V
Write out a balanced overall cell reaction.
Calculate the overall cell potential.
Answer:
Using your balanced cell reaction, briefly explain why the cell potential does not
change as the battery discharges itself.
Write out the balanced overall reaction that occurs when this battery is being
recharged.
A current of 2.75 A is measured during recharging with an external potential of 2.0 V.
After 5.00 minutes charging, how many moles of Cd(s) will be redeposited?
Answer:
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CHEM1001
Page 19 of 24 pages
June 2013
 A certain mixture of gases containing 0.24 mol of He, 0.53 mol of N2 and 0.05 mol of
Ne is placed in a container with a piston that maintains it at a total pressure of 1.0 atm.
This gas mixture is now heated from its initial temperature of 290 K to 370 K by
passing 2.08 kJ of energy into it.
Calculate the volume occupied by the gas at 370 K.
Answer:
Calculate the heat capacity of the gas mixture (in J K–1 mol–1).
Answer:
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.
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2201(a)
Marks
4
CHEM1001
Page 20 of 24 pages
June 2013
THIS PAGE IS FOR ROUGH WORKING ONLY
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CHEM1001
Page 21 of 24 pages
June 2013
 Nitrogen and acetylene gases react to form hydrogen cyanide according to the reaction
N2(g) + C2H2(g)
2HCN(g)
Kc = 2.3 10–4 at 300 °C
Write out the equilibrium constant expression for Kc for this reaction as shown
above.
The value of Kp for this reaction at 300 °C is also 2.3  10–4. Why are the values of
Kp and Kc the same for this reaction?
Write a balanced equation and calculate the value of the equilibrium constant Kc' for
the formation of 1.0 mol of hydrogen cyanide gas from nitrogen and acetylene gases.
Answer:
What is the equilibrium concentration of HCN(g) if nitrogen and acetylene are mixed
so that both are at starting concentrations of 1.0 mol L–1?
Answer:
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Marks
8
CHEM1001
Page 22 of 24 pages
June 2013
 The boiling point of NH3 is –33 C and that of HF is +20 C. Explain this difference
in terms of the strengths of the intermolecular forces between these molecules.
Explain why the boiling point of water (100 C) is higher than both HF and NH3.
THE REMAINDER OF THIS PAGE IS FOR ROUGH WORKING ONLY.
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Marks
3
CHEM1001
Page 23 of 24 pages
June 2013
2201(a)
 Write the equation whose enthalpy change represents the standard enthalpy of
formation of NO(g).
Marks
3
Given the following data, calculate the standard enthalpy of formation of NO(g).
N2(g) + 2O2(g)
2NO2(g)
H° = 66.6 kJ mol–1
2NO(g) + O2(g)
2NO2(g)
H° = –114.1 kJ mol–1
Answer:
 Hydrazine, N2H4, burns completely in oxygen to form N2(g) and H2O(g). Use the
bond enthalpies given below to estimate the enthalpy change for this process.
Bond
Bond enthalpy (kJ mol–1) Bond
Bond enthalpy (kJ mol–1)
N–H
391
O=O
498
N–N
158
O–O
144
N=N
470
O–H
463
NN
945
N–O
214
Answer:
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3
CHEM1001
Page 24 of 24 pages
June 2013
THIS PAGE IS FOR ROUGH WORKING ONLY
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2201(a)
2201(b)
June 2013
CHEM1001 – FUNDAMENTALS OF CHEMISTRY 1A
DATA SHEET
Physical constants
Avogadro constant, NA = 6.022  1023 mol–1
Faraday constant, F = 96485 C mol–1
Planck constant, h = 6.626  10–34 J s
Speed of light in vacuum, c = 2.998  108 m s–1
Rydberg constant, ER = 2.18  10–18 J
Boltzmann constant, kB = 1.381  10–23 J K–1
Permittivity of a vacuum, 0 = 8.854  10–12 C2 J–1 m–1
Gas constant, R = 8.314 J K–1 mol–1
= 0.08206 L atm K–1 mol–1
Charge of electron, e = 1.602  10–19 C
Mass of electron, me = 9.1094  10–31 kg
Mass of proton, mp = 1.6726  10–27 kg
Mass of neutron, mn = 1.6749  10–27 kg
Properties of matter
Volume of 1 mole of ideal gas at 1 atm and 25 C = 24.5 L
Volume of 1 mole of ideal gas at 1 atm and 0 C = 22.4 L
Density of water at 298 K = 0.997 g cm–3
Conversion factors
1 atm = 760 mmHg = 101.3 kPa = 1.013 bar
0 C = 273 K
1 L = 10–3 m3
1 Å = 10–10 m
1 eV = 1.602  10–19 J
Decimal fractions
Fraction Prefix Symbol
10–3
milli
m
10–6
micro

10–9
10–12
nano
pico
n
p
1 Ci = 3.70  1010 Bq
1 Hz = 1 s–1
1 tonne = 103 kg
1 W = 1 J s–1
Decimal multiples
Multiple Prefix Symbol
103
kilo
k
106
mega
M
109
1012
giga
tera
G
T
2201(b)
June 2013
CHEM1001 – FUNDAMENTALS OF CHEMISTRY 1A
Standard Reduction Potentials, E
Reaction
Co3+(aq) + e–  Co2+(aq)
Ce4+(aq) + e–  Ce3+(aq)
E / V
+1.82
+1.72
MnO4 (aq) + 8H (aq) + 5e  Mn (aq) + 4H2O
+1.51
Au3+(aq) + 3e–  Au(s)
+1.50
Cl2 + 2e–  2Cl–(aq)
+1.36
O2 + 4H (aq) + 4e  2H2O
+1.23
Pt2+(aq) + 2e–  Pt(s)
+1.18
–
–
+
2+
–
+
MnO2(s) + 4H+(aq) + e–  Mn3+ + 2H2O
+0.96
NO3 (aq) + 4H (aq) + 3e  NO(g) + 2H2O
+0.96
Pd2+(aq) + 2e–  Pd(s)
+0.92
NO3–(aq) + 10H+(aq) + 8e–  NH4+(aq) + 3H2O
+0.88
–
Ag (aq) + e  Ag(s)
+0.80
Fe3+(aq) + e–  Fe2+(aq)
+0.77
Cu+(aq) + e–  Cu(s)
+0.53
Cu (aq) + 2e  Cu(s)
+0.34
BiO+(aq) + 2H+(aq) + 3e–  Bi(s) + H2O
+0.32
Sn4+(aq) + 2e–  Sn2+(aq)
+0.15
2H+(aq) + 2e–  H2(g)
0 (by definition)
Fe3+(aq) + 3e–  Fe(s)
–0.04
Pb2+(aq) + 2e–  Pb(s)
–0.126
Sn2+(aq) + 2e–  Sn(s)
–0.136
Ni2+(aq) + 2e–  Ni(s)
–0.24
Co2+(aq) + 2e–  Co(s)
–0.28
Cd2+(aq) + 2e–  Cd(s)
–0.40
Fe (aq) + 2e  Fe(s)
–0.44
Cr3+(aq) + 3e–  Cr(s)
–0.74
–
+
+
–
–
2+
–
2+
Zn2+(aq) + 2e–  Zn(s)
–0.76
2H2O + 2e  H2(g) + 2OH (aq)
–0.83
Cr2+(aq) + 2e–  Cr(s)
–0.89
Al3+(aq) + 3e–  Al(s)
–1.68
Sc (aq) + 3e  Sc(s)
–2.09
Mg2+(aq) + 2e–  Mg(s)
–2.36
Na+(aq) + e–  Na(s)
–2.71
Ca (aq) + 2e  Ca(s)
–2.87
Li+(aq) + e–  Li(s)
–3.04
–
3+
2+
–
–
–
2201(b)
June 2013
CHEM1001 – FUNDAMENTALS OF CHEMISTRY 1A
Useful formulas
Quantum Chemistry
Electrochemistry
E = h = hc/
G = –nFE
 = h/mv
Moles of e– = It/F
E = –Z2ER(1/n2)
E = E – (RT/nF)  lnQ
x(mv)  h/4
E = (RT/nF)  lnK
q = 4r2  5.67  10–8  T4
E = E –
T  = 2.898  106 K nm
0.0592
logQ (at 25 C)
n
Acids and Bases
Gas Laws
pH = –log[H+]
PV = nRT
pKw = pH + pOH = 14.00
(P + n2a/V2)(V – nb) = nRT
pKw = pKa + pKb = 14.00
Ek = ½mv2
pH = pKa + log{[A–] / [HA]}
Radioactivity
Kinetics
t½ = ln2/
t½ = ln2/k
A = N
k = Ae–Ea/RT
ln(N0/Nt) = t
ln[A] = ln[A]0 – kt
14
C age = 8033 ln(A0/At) years
ln
k2
Ea 1
=
( - 1)
k1
R T1 T2
Colligative Properties & Solutions
Thermodynamics & Equilibrium
 = cRT
G = H – TS
Psolution = Xsolvent  Psolvent
G = G + RT lnQ
c = kp
G = –RT lnK
Tf = Kfm
univS = R lnK
Tb = Kbm
RT n
Kp = Kc (
)
100
Miscellaneous
Mathematics
A = –log
I
I0
If ax2 + bx + c = 0, then x =
A = cl
E = –A
b 
ln x = 2.303 log x
e2
4 0 r
NA
Area of circle = r2
Surface area of sphere = 4r2
b2  4ac
2a
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
1
2
HYDROGEN
HELIUM
He
4.003
3
4
5
6
7
8
9
LITHIUM
BERYLLIUM
BORON
CARBON
NITROGEN
OXYGEN
FLUORINE
10
NEON
Li
Be
B
C
N
O
F
Ne
6.941
9.012
10.81
12.01
14.01
16.00
19.00
20.18
11
12
13
14
15
16
17
18
SODIUM
MAGNESIUM
ALUMINIUM
SILICON
PHOSPHORUS
SULFUR
CHLORINE
ARGON
Na
Mg
Al
Si
P
S
Cl
Ar
22.99
24.31
26.98
28.09
30.97
32.07
35.45
39.95
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
POTASSIUM
CALCIUM
SCANDIUM
TITANIUM
VANADIUM
CHROMIUM
MANGANESE
IRON
COBALT
NICKEL
COPPER
ZINC
GALLIUM
GERMANIUM
ARSENIC
SELENIUM
BROMINE
KRYPTON
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
39.10
40.08
44.96
47.88
50.94
52.00
54.94
55.85
58.93
58.69
63.55
65.39
69.72
72.59
74.92
78.96
79.90
83.80
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
RUBIDIUM
STRONTIUM
YTTRIUM
ZIRCONIUM
NIOBIUM
MOLYBDENUM
TECHNETIUM
RUTHENIUM
RHODIUM
PALLADIUM
SILVER
CADMIUM
INDIUM
TIN
ANTIMONY
TELLURIUM
IODINE
XENON
Rb
Sr
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
85.47
87.62
88.91
91.22
92.91
95.94
[98.91]
101.07
102.91
106.4
107.87
112.40
114.82
118.69
121.75
127.60
126.90
131.30
55
56
57-71
CAESIUM
BARIUM
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
HAFNIUM
TANTALUM
TUNGSTEN
RHENIUM
OSMIUM
IRIDIUM
PLATINUM
GOLD
MERCURY
THALLIUM
LEAD
BISMUTH
POLONIUM
ASTATINE
RADON
Cs
Ba
Hf
Ta
W
Re
Os
Ir
Pt
Au
Hg
Tl
Pb
Bi
Po
At
Rn
132.91
137.34
178.49
180.95
183.85
186.2
190.2
192.22
195.09
196.97
200.59
204.37
207.2
208.98
[210.0]
[210.0]
[222.0]
87
88
FRANCIUM
RADIUM
89-103 104
RUTHERFORDIUM
105
106
107
108
109
110
111
112
114
116
DUBNIUM
SEABORGIUM
BOHRIUM
HASSIUM
MEITNERIUM
DARMSTADTIUM
ROENTGENIUM
COPERNICIUM
FLEROVIUM
LIVERMORIUM
Fr
Ra
Rf
Db
Sg
Bh
Hs
Mt
Ds
Rg
Cn
Fl
Lv
[223.0]
[226.0]
[263]
[268]
[271]
[274]
[270]
[278]
[281]
[281]
[285]
[289]
[293]
LANTHANOID
S
58
59
60
61
62
63
64
65
66
67
68
69
70
71
CERIUM
PRASEODYMIUM
NEODYMIUM
PROMETHIUM
SAMARIUM
EUROPIUM
GADOLINIUM
TERBIUM
DYSPROSIUM
HOLMIUM
ERBIUM
THULIUM
YTTERBIUM
LUTETIUM
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
Lu
138.91
140.12
140.91
144.24
[144.9]
150.4
151.96
157.25
158.93
162.50
164.93
167.26
168.93
173.04
174.97
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
ACTINIUM
THORIUM
PROTACTINIUM
URANIUM
NEPTUNIUM
PLUTONIUM
AMERICIUM
CURIUM
BERKELLIUM
CALIFORNIUM
EINSTEINIUM
FERMIUM
MENDELEVIUM
NOBELIUM
LAWRENCIUM
Ac
Th
Pa
U
Np
Pu
Am
Cm
Bk
Cf
Es
Fm
Md
No
Lr
[227.0]
232.04
[231.0]
238.03
[237.0]
[239.1]
[243.1]
[247.1]
[247.1]
[252.1]
[252.1]
[257.1]
[256.1]
[259.1]
[260.1]
June 2013
ACTINOIDS
57
LANTHANUM
CHEM1001 – FUNDAMENTALS OF CHEMISTRY 1A
H
1.008
2201(b)
PERIODIC TABLE OF THE ELEMENTS