NAME:___________________________
Spring 2015
INSTRUCTIONS:
Section:_____
Student Number:__________________
Chemistry 2000 Midterm #2C
____/ 50 marks
1) Please read over the test carefully before beginning. You should have 6 pages
of questions, a blank “overflow” page and two pages of data sheets with
periodic table.
2) If your work is not legible, it will be given a mark of zero.
3) Marks will be deducted for incorrect information added to an otherwise
correct answer.
4) Marks will be deducted for improper use of significant figures and for missing
or incorrect units.
5) Show your work for all calculations. Numerical answers without supporting
calculations will not be given full credit.
6) You may use a calculator but only for the purposes of calculation. No text
capable calculators are allowed.
7) You have 90 minutes to complete this test.
Confidentiality Agreement:
I agree not to discuss (or in any other way divulge) the contents of this exam until after 8pm Mountain
Time on Tuesday, March 24th, 2015. I understand that breaking this agreement would constitute
academic misconduct, a serious offense with serious consequences. The minimum punishment would be
a mark of 0/50 on this exam and removal of the “overwrite midterm mark with final exam mark” option
for my grade in this course; the maximum punishment would include expulsion from this university.
Signature: ___________________________
Course: CHEM 2000 (General Chemistry II)
Semester: Spring 2015
The University of Lethbridge
Date: _____________________________
Question Breakdown
Q1
Q2
Q3
Q4
Q5
Q6
/6
/2
/6
/5
/ 11
/ 20
Total
/ 50
NAME:___________________________
1.
(a)
Section:_____
Student Number:__________________
The table below shows how boiling point increases as each hydrogen atom in methane (CH4) is
replaced by a chlorine atom.
[6 marks]
Molecular Formula
Boiling Point (°C)
CH4
-161.5
CH3Cl
-23.8
CH2Cl2
39.6
CHCl3
61.2
CCl4
76.7
Explain why the increase in boiling point from CH4 to CH3Cl is larger than any of the other
increases in boiling point.
[2 marks]
This is the only case in which replacing a hydrogen atom with a chlorine atom generates a polar
molecule (CH3Cl) from a nonpolar molecule (CH4).
There are three types of intermolecular forces active in polar liquids like CH3Cl (dipole-dipole,
dipole-induced dipole and induced dipole-induced dipole) while there is only one type of
intermolecular force active in nonpolar liquids like CH4 (induced dipole-induced dipole).
Since boiling point increases with the strength of the active intermolecular forces, it is reasonable
that the introduction of two new types of intermolecular force would result in a significant
increase in boiling point.
(b)
Explain why the increase in boiling point from CHCl3 to CCl4 is smaller than any of the other
increases in boiling point.
[2 marks]
This is the only case in which replacing a hydrogen atom with a chlorine atom generates a
nonpolar molecule (CCl4) from a polar molecule (CHCl3).
Thus, replacing the last hydrogen atom in CHCl3 with a chlorine atom gives a compound which
experiences fewer types of intermolecular forces (see answer to part (a) for list of active IMF).
Since boiling point increases with the strength of the active intermolecular forces, it is reasonable
that the elimination of two types of intermolecular forces would result in a smaller increase in
boiling point.
(c)
What can you conclude from the fact that the boiling point increases every time a hydrogen
atom is replaced by a chlorine atom?
[2 marks]
Based on the answer to part (b), one might expect the boiling point of CCl4 to be lower than that
of CHCl3. The fact that it is not tells us that the increase in strength of the induced dipoleinduced dipole forces is greater than the loss of the dipole-dipole and dipole-induced dipole
forces in this set of molecules. The increase in strength of induced dipole-induced dipole forces
will be due to the fact that chlorine atoms are much larger than hydrogen atoms.
This tells us that the strength of the induced dipole-induced dipole forces is significant (and that
they are likely the major intermolecular force active in most of the molecules in this series).
IMF act between molecules; physical changes like boiling DO NOT BREAK CHEMICAL
BONDS. A molecule is only polar if its overall shape is polar; POLAR BONDS DO NOT MAKE
A MOLECULE POLAR so that CH4 and CBr4 are both non-polar because tetrahedral.
Please provide short, to the point answers based on facts; waffling does not help your cause!
NAME:___________________________
2.
Section:_____
Student Number:__________________
Why does water “ball up” into spheres on wax paper but spread out (and soak in) on regular
paper?
[2 marks]
The surface of wax paper is nonpolar. The surface of regular paper is polar.
Water molecules are polar. As such, they experience stronger intermolecular forces toward other
water molecules than to a nonpolar surface such as that of wax paper. Since the water molecules
are more strongly attracted to each other than to the surface, they “ball up” into spheres.
On regular paper, the attractions of the water molecules to the surface of the paper are also
strong, so this effect is not observed.
3.
(a)
Consider the following reaction:
2 Na (s) + 2 H 2 O (l) → 2 NaOH (aq) + H 2 (g)
Is this reaction entropy-favoured? Why or why not?
[6 marks]
[2 marks]
There are more moles of gas in the products (1 moles) than in the reactants (0 moles).
Therefore, the reaction IS entropy-favoured.
(b)
If this reaction was exothermic, would it be thermodynamically favoured at all temperatures?
Why or why not?
[2 marks]
If the reaction is exothermic then it is favoured by enthalpy (∆H < 0). Since it is also favoured
by entropy (∆S < 0), it will be favoured at all temperatures.
To be favoured, a reaction must have ∆G < 0.
∆G = ∆H - T∆S
If ∆H < 0 and ∆S > 0, then ∆G < 0 at all temperatures since T is always a positive number
(Kelvin scale).
(c)
If this reaction was endothermic, would there be any conditions under which it would be
thermodynamically favoured? If yes, what would they be? If not, why not?
[2 marks]
If the reaction is endothermic then it is not favoured by enthalpy (∆H > 0).
To be favoured, a reaction must have ∆G < 0.
∆G = ∆H - T∆S
If ∆H > 0 and ∆S > 0, then ∆G < 0 at temperatures which are high enough that T∆S is greater
than ∆H.
NAME:___________________________
4.
Section:_____
Student Number:__________________
Consider a container divided into two equal compartments. One compartment contains gas
particles while the other is completely empty:
[5 marks]
Picture A
The divider is removed:
Picture B
(a)
The pictures above each have 10 dots representing gas particles.
Draw what the system looks like once it reaches equilibrium:
[1 mark]
Picture C
(b)
Draw what the system looks like 10 minutes after it reaches equilibrium.
[1 mark]
Picture D
(c)
It must be clear whether Picture D is intended to look the same as Picture C or different from it.
They must look different. The gas particles keep moving. Equilibrium is a dynamic state.
Use the kinetic molecular theory of gases to explain why the system changes from Picture B to
Picture C.
[3 marks]
According to the kinetic molecular theory, gas particles move in straight lines in random
directions until they collide with something (usually the container wall; sometimes another gas
particle).
Since the gas particles all start in the left half of the container, the ones moving generally
toward the left will encounter the container walls sooner and be redirected. The ones moving
generally toward the right will be able to continue moving in a straight line for longer before
colliding with a container wall and being redirected.
Thus, the tendency will be for more gas particles to be moving from the left half of the
container toward the right half (than from the right half to the left half), and this will continue
until both halves contain equal numbers of gas particles.
Please review the postulates of the KMT, found on slides 17 & 18 of “Lecture 15-17” notes.
Explanations based on entropy as a driving force were accepted but the postulates of the KMT
had to be used to obtain full credit for this question.
NAME:___________________________
5.
Section:_____
Student Number:__________________
Phosphorus has two common allotropes, white phosphorus and red phosphorus.
[11 marks]
(a)
Write a balanced chemical equation for the reversible conversion of white phosphorus to red
phosphorus. Note that the data is given per mole of P atoms, not P4 molecules.
[1 mark]
P (white) ⇔ P (red) or
P4 (white) ⇔ 4 P (red)
(b)
Determine the standard free energy change for the conversion of white phosphorus to red
phosphorus.
[5 marks]
The data sheet provides ∆fH° and S° values for these two species (but not ∆fG° values).
∆rG° can be calculated from ∆rH° and ∆rS° – which can, in turn be calculated from ∆fH° and S°.
Answers built on the S8 version of the question will have numerical values x 8 from what is given here.
Step 1: Calculate ∆rH° using ∆rH° = Σ ∆fH°(products) - Σ ∆fH°(reactants)
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
Δ𝑟𝑟 𝐻𝐻° = �−17.6 𝑚𝑚𝑚𝑚𝑚𝑚� − �0 𝑚𝑚𝑚𝑚𝑚𝑚� = −17.6 𝑚𝑚𝑚𝑚𝑚𝑚
1 decimal place = 3 sig. fig.
Step 2: Calculate ∆rS° using ∆rS° = Σ S°(products) - Σ S°(reactants)
𝐽𝐽
𝐽𝐽
𝐽𝐽
Δ𝑟𝑟 𝑆𝑆° = �22.80 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾� − �41.10 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾� = −18.30 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾
2 decimal places = 4 sig. fig.
Step 3: Calculate ∆rG° using ∆rG° = ∆rH° - T∆rS°
𝑘𝑘𝑘𝑘
𝐽𝐽
1𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
Δ𝑟𝑟 𝐺𝐺° = �−17.6 𝑚𝑚𝑚𝑚𝑚𝑚� − (298.15𝐾𝐾) �−18.30 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾� �1000𝐽𝐽� = −12.1 𝑚𝑚𝑚𝑚𝑚𝑚
(c)
1 decimal place = 3 sig. fig.
Which is the more stable allotrope at room temperature? Explain.
[2 marks]
Assume that “room temperature” is relatively close to standard conditions. (~25 °C)
∆rG° < 0 therefore the forward reaction is thermodynamically favoured under standard
conditions.
Therefore, P (red) is the more stable allotrope under standard conditions.
Merely stating that the allotrope with ∆fH° is the most stable was not acceptable, and is not always true!
(d)
Calculate the temperature (in K) at which these two allotropes reach equilibrium at 1 bar.
[3 marks]
∆rG = ∆rH – T∆rS
At equilibrium, ∆rG = 0. (reaction favoured in neither forward nor reverse direction)
We must assume that ∆rH and ∆rS do not vary dramatically with temperature (so that we can use
the values for ∆rH° and ∆rS° from part (b)).
1 bar is the standard pressure.
Δ𝑟𝑟 𝐺𝐺° = Δ𝑟𝑟 𝐻𝐻° − 𝑇𝑇Δ𝑟𝑟 𝑆𝑆° therefore
𝑇𝑇 =
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
�−�0
�
𝑚𝑚𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚
𝐽𝐽
�
�−18.30
𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾
�−17.6
×
1000𝐽𝐽
1𝑘𝑘𝑘𝑘
𝑇𝑇 =
= 962 𝐾𝐾
Δ𝑟𝑟 𝐻𝐻°−Δ𝑟𝑟 𝐺𝐺°
Δ𝑟𝑟 𝑆𝑆°
3 sig. fig.
NAME:___________________________
6.
(a)
(b)
Section:_____
Student Number:__________________
Consider the following reaction in the gas phase:
2 NOBr (g) ⇔ 2 NO (g) + Br2 (g)
Write the thermodynamic equilibrium constant expression for this reaction.
𝐾𝐾 =
(𝑎𝑎𝑁𝑁𝑁𝑁 )2 �𝑎𝑎𝐵𝐵𝐵𝐵2 �
(𝑎𝑎𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 )2
[20 marks]
[2 marks]
Q is rxn quotient! Activities are required for full marks here.
Calculate the equilibrium constant for this reaction under standard conditions.
[9 marks]
The data sheet provides ∆fH° and S° values for the three species in this reaction.
∆rG° can be calculated from ∆rH° and ∆rS° (which can, in turn be calculated from ∆fH° and S°).
K can be calculated from ∆rG°.
Step 1: Calculate ∆rH° using ∆rH° = Σ ∆fH°(products) - Σ ∆fH°(reactants)
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
Δ𝑟𝑟 𝐻𝐻° = �2 �90.29 𝑚𝑚𝑚𝑚𝑚𝑚� + �30.91 𝑚𝑚𝑚𝑚𝑚𝑚�� − 2 �82.13 𝑚𝑚𝑚𝑚𝑚𝑚� = 47.23 𝑚𝑚𝑚𝑚𝑚𝑚
2 decimal places = 4 sig. fig.
Step 2: Calculate ∆rS° using ∆rS° = Σ S°(products) - Σ S°(reactants)
𝐽𝐽
𝐽𝐽
𝐽𝐽
𝐽𝐽
Δ𝑟𝑟 𝑆𝑆° = �2 �210.65 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾� + �245.38 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾�� − 2 �273.53 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾� = 119.62 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾
2 decimal places = 5 sig. fig.
Step 3: Calculate ∆rG° using ∆rG° = ∆rH° - T∆rS°
𝑘𝑘𝑘𝑘
𝐽𝐽
1𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
Δ𝑟𝑟 𝐺𝐺° = �47.23 𝑚𝑚𝑚𝑚𝑚𝑚� − (298.15𝐾𝐾) �119.62 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾� �1000𝐽𝐽� = 11.57 𝑚𝑚𝑚𝑚𝑚𝑚
2 decimal places = 4 sig. fig.
Step 4: Calculate K using ∆rG° = –RTlnK
Δ𝑟𝑟 𝐺𝐺° = −𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅
𝐾𝐾 = 𝑒𝑒
−
therefore
𝑘𝑘𝑘𝑘
�11.57
�
1000𝐽𝐽
𝑚𝑚𝑚𝑚𝑚𝑚
×
𝐽𝐽
�(298.15𝐾𝐾) 1𝑘𝑘𝑘𝑘
�8.314462
𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾
𝐾𝐾 = 𝑒𝑒 −
Δ𝑟𝑟 𝐺𝐺°
𝑅𝑅𝑅𝑅
= 𝑒𝑒 −4.665 = 9.42 × 10−3
Many students chose to calculate all the ∆fG° values, and then did so again at a different temperature
for part (c). This is a highly inefficient approach and often resulted in serious arithmetic answers. Note
also that (1) the stoichiometry of the reaction must be considered in the ∆rH° and ∆rS° calculations, and
(2) kJ and J must be rationalized when calculating ∆rG°.
(c) Alternate answer:
𝑘𝑘𝑘𝑘
𝐽𝐽
1𝑘𝑘𝑘𝑘
𝑘𝑘𝑘𝑘
Δ𝑟𝑟 𝐺𝐺 280 = �47.23 𝑚𝑚𝑚𝑚𝑚𝑚� − (280𝐾𝐾) �119.62 𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾� �1000𝐽𝐽� = +13.74 𝑚𝑚𝑚𝑚𝑚𝑚
−
𝑘𝑘𝑘𝑘
�−13.74
�
1000𝐽𝐽
𝑚𝑚𝑚𝑚𝑚𝑚
×
𝐽𝐽
�(280𝐾𝐾) 1𝑘𝑘𝑘𝑘
�8.314462
𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾
𝐾𝐾 = 𝑒𝑒
= 2.74 × 10−3
But it is not permissible to use ∆rG° with T = 280 K! Standard conditions means standard!
NAME:___________________________
6.
(c)
Section:_____
Student Number:__________________
continued…
Calculate the equilibrium constant for this reaction at 280 K.
[5 marks]
From part (b), we know that K1 = 9.42 × 10-3 at T1 = 298.15 K
We need to solve for K2 at T2 = 280 K.
We also know that ∆H = 47.23 kJ/mol.
𝑘𝑘𝑘𝑘
�
𝑚𝑚𝑚𝑚𝑚𝑚
𝐽𝐽
�
�8.314462
𝑚𝑚𝑚𝑚𝑚𝑚∙𝐾𝐾
�47.23
𝐾𝐾2
𝑙𝑙𝑙𝑙 �𝐾𝐾 � =
1
𝐾𝐾
�
1000𝐽𝐽
1𝑘𝑘𝑘𝑘
1
1
� �(298.15𝐾𝐾) − (280𝐾𝐾)�
1
𝑙𝑙𝑙𝑙 �𝐾𝐾2 � = (5680𝐾𝐾) �0.00022 𝐾𝐾� = −1.2
𝐾𝐾2
𝐾𝐾1
1
= 𝑒𝑒 −1.2
𝐾𝐾2 = (𝐾𝐾1 )(𝑒𝑒 −1.2 ) = (9.42 × 10−3 )(𝑒𝑒 −1.2 ) = (9.42 × 10−3 )(0.3) = 3 × 10−3
(d)
A gas reactor is charged with 0.39 bar NO, 0.37 bar Br2 and 0.011 bar NOBr at 280 K.
Determine whether the reaction will proceed in the forward or in the reverse reaction. [4 marks]
Calculate Q for the system. (This requires calculation of activities for all reactants and
products.)
Compare Q to K.
If Q < K, the reaction will proceed forward.
If Q > K, the reaction will proceed in reverse.
Step 1: Calculate activities of all reactants and products
𝑝𝑝
0.39 𝑏𝑏𝑏𝑏𝑏𝑏
𝑝𝑝𝐵𝐵𝐵𝐵
0.37 𝑏𝑏𝑏𝑏𝑏𝑏
𝑁𝑁𝑁𝑁
𝑎𝑎𝑁𝑁𝑁𝑁 = 1 𝑏𝑏𝑏𝑏𝑟𝑟
=
𝑎𝑎𝐵𝐵𝐵𝐵2 = 1 𝑏𝑏𝑏𝑏𝑏𝑏2 =
𝑎𝑎𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁 =
𝑝𝑝𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
1 𝑏𝑏𝑏𝑏𝑏𝑏
1 𝑏𝑏𝑏𝑏𝑏𝑏
1 𝑏𝑏𝑏𝑏𝑏𝑏
=
= 0.39
2 sig. fig.
= 0.37
0.011 𝑏𝑏𝑏𝑏𝑏𝑏
1 𝑏𝑏𝑏𝑏𝑏𝑏
2 sig. fig.
= 0.011
2 sig. fig.
Step 2: Calculate Q
𝑄𝑄 =
(𝑎𝑎𝑁𝑁𝑁𝑁 )2 �𝑎𝑎𝐵𝐵𝐵𝐵2 �
(𝑎𝑎𝑁𝑁𝑁𝑁𝑁𝑁𝑁𝑁
)2
Step 3: Compare Q to K
=
(0.39)2 (0.37)
(0.011)2
= 440 = 4.4 × 102
from part (c), K = 0.003 therefore Q > K
Therefore, the reaction will proceed in reverse.
2 sig. fig.
Some Useful Constants and Formulae
Fundamental Constants and Conversion Factors
Atomic mass unit (u)
1.660 539 × 10-27 kg
Avogadro's number (NA) 6.022 141 × 1023 mol–1
Boltzmann constant (kB) 1.380 649 × 10-23 J·K-1
Ideal gas constant (R)
8.314 462 J·mol-1·K-1
8.314 462 m3·Pa·mol-1·K-1
Kelvin temperature scale
Planck's constant (h)
Speed of light in vacuum (c)
Pressure conversions
Volume conversion
0 K = -273.15 ˚C
6.626 070 × 10-34 J·Hz-1
2.997 925 x 108 m·s-1
1 bar = 100 kPa
1 atm = 1.01325 bar
1000 L = 1 m3
Formulae
K =
3
1
mv 2 = RT
2
2
v rms = v 2 =
∆S =
S = k B ln Ω
q rev
T
o
Δr G = Δr G + RT ln Q
pK a = − log K a
Activities
Solid
Pure liquid
Ideal Solvent
Ideal Solute
Ideal Gas
3RT
M
o
∆ r G = −RT ln K
pH = − log aH +
PA = X A PAo
a =1
a =1
a=X
c
a=
c°
P
a=
P°
P (white)
P (red)
NOBr (g)
NO (g)
Br2 (g)
 kJ 
∆ f H °

 mol 
0
-17.6
82.13
90.29
30.91

n2 
 P + a 2 (V − nb ) = nRT
V 

∆ r G = ∆ r H − T∆ r S
Some Useful Thermodynamic Properties
Substance
PV = nRT
 J 
S °

 mol ⋅ K 
41.10
22.80
273.53
210.65
245.38
o
 K2  ∆ r H
 =
ln
R
 K1 
[ A] = k H PA
1
1 
 −

 T1 T2 
X =
n
∑n
1
CHEM 2000 Standard Periodic Table
18
1.0079
4.0026
He
H
2
13
14
15
16
17
6.941
9.0122
10.811
12.011
14.0067
15.9994
18.9984
Li
Be
B
C
N
O
F
Ne
3
22.9898
4
24.3050
5
26.9815
6
28.0855
7
30.9738
8
32.066
9
35.4527
10
39.948
1
2
20.1797
Na
Mg
11
39.0983
12
40.078
3
4
5
6
7
8
9
10
11
12
44.9559
47.88
50.9415
51.9961
54.9380
55.847
58.9332
58.693
63.546
65.39
K
Ca
Sc
Ti
V
Cr
Mn
Fe
Co
Ni
Cu
Zn
Ga
Ge
As
Se
Br
Kr
19
85.4678
20
87.62
21
88.9059
22
91.224
23
92.9064
26
101.07
27
102.906
28
106.42
29
107.868
30
112.411
31
114.82
32
118.710
33
121.757
34
127.60
35
126.905
36
131.29
Rb
Sr
37
132.905
38
137.327
Cs
Ba
55
(223)
56
226.025
Fr
87
Ra
Y
39
La-Lu
Ac-Lr
88
24
95.94
25
(98)
Zr
Nb
Mo
Tc
40
178.49
41
180.948
42
183.85
43
186.207
Cl
Ar
17
79.904
18
83.80
Rh
Pd
Ag
Cd
In
Sn
Sb
Te
I
Xe
46
195.08
47
196.967
48
200.59
49
204.383
50
207.19
51
208.980
52
(210)
53
(210)
54
(222)
Ta
W
Re
Os
Ir
Pt
74
(271)
75
(270)
76
(277)
77
(276)
78
(281)
Bh
S
16
78.96
45
192.22
73
(268)
Sg
P
15
74.9216
Ru
Hf
Db
Si
14
72.61
44
190.2
72
(265)
Rf
Al
13
69.723
Hs
Mt
Ds
Au
79
(280)
Rg
Hg
80
(285)
Cn
Tl
Pb
Bi
Po
At
81
(284)
82
(289)
83
(288)
84
(293)
85
(294)
Uut
Fl
Uup
Lv
Uus
104
105
106
107
108
109
110
111
112
113
114
115
116
117
138.906
140.115
140.908
144.24
(145)
150.36
151.965
157.25
158.925
162.50
164.930
167.26
168.934
173.04
174.967
La
Ce
Pr
Nd
Pm
Sm
Eu
Gd
Tb
Dy
Ho
Er
Tm
Yb
57
227.028
58
232.038
59
231.036
60
238.029
61
237.048
65
(247)
66
(251)
Ac
89
Th
90
Pa
91
U
92
Np
93
62
(240)
Pu
94
63
(243)
Am
95
64
(247)
Cm
96
Bk
97
Cf
98
67
(252)
Es
99
68
(257)
Fm
100
69
(258)
Md
101
70
(259)
No
102
Rn
86
(294)
Uuo
118
Lu
71
(262)
Lr
103
Developed by Prof. R. T. Boeré (updated 2014)