www.biosan.lv
FINAL
2011
ROUND, PROBLEMS
Solve all problems provides. Fill answer sheets and till 13:00, March 20, 2011 send them to e‐mail:
kimijas_olimpiades@inbox.lv You will receive answer that your solutions are received and will be graded in few
days.
This round consists from two parts. Part A – 60 multiple choice questions with one correct answer each.
Maximum points for this round is 30 points. Part B – 6 short answer problems. Each problem is graded with
maximum 5 points. Points for final round are added to points gained in first three rounds.
Part A – Multiple choice test
Choose one correct answer for each question and write corresponding letter in answer sheet provided.
Question 1
Question 2
Question 3
Question 4
Question 5
Question 6
Question 7
Question 8
Question 9
Time, s
Question 10
Question 11
Question 12
Question 13
Question 14
Question 15
Question 16
Question 17
Question 18
Question 19
Question 20
Question 21
Question 22
Question 23
Question 24
Question 25
Question 26
Question 27
Question 28
Question 29
Question 30
Question 31
Question 32
Question 33
Question 34
Question 35
Question 36
Question 37
Question 38
Question 39
Question 40
Question 41
Question 42
Question 43
Question 44
Question 45
Question 46
Question 47
Question 48
Question 49
Question 50
Question 51
Question 52
Question 53
Question 54
Question 55
Question 56
Question 57
Question 58
Question 59
Question 60
www.biosan.lv
FINAL
2011
ROUND, PROBLEMS
Part B – Short answer questions
Problem 1
Chemical crossword – inorganic chemistry (8 points)
Letters A, B, D ect. stands for unknown inorganic compounds, which react as shown in table. Formulas for all
organic compounds are known.
•
•
•
•
•
•
•
Compound A contains transition metal and transition metal content is approximately 72% (by mass)
Solutions of compounds U, N, Ģ are strongly basic. There are no information about solutions of other
compounds.
Solutions of compounds O and N colors gas flame yellow, while compounds U un Ō colors flame purple
red.
Water solutions of compounds V, P and L are blue.
Substances E, G, H, M, R, S, Ā contain only one chemical element.
Three transition metals are mentioned in this crossword.
Equations in crossword are not balanced.
Write chemical formulas for all compounds abbreviated with letters.
A
+
E
↓
F
+
G
E
+
U
↑
F
+
T
+
+
+
←
B
↑
I
+
H
→
L
+
N
↓
O
+
P
→
S
←
+
+
G
+
J
↓
E
+
K
R
+
Q
↓
Å
+
E
+
+
+
+
→
F
←
+
C2H5Cl
→
→
Č
+
Ģ
↓
Ķ
+
F
C2H5NO2 +
D
+
H
↓
B
M
+
Q
↓
F
+
V
+
W
→
I
+
+
D
+
R
↑
B
+
Ž
F
↓
I
+
E
+
+
+
→
U
+
F
+
Ā
↓
Ē
+
E
→
Š
+
F
→
Z
+
F
+
F
→
C2H6 +
Y
+
Ō
+ C2H5OH
↑
S
C2H6
+
+
CH3Cl
I
↓
Ņ
+
D
+
+
I
F
↓
G
+
E
+
Y
C2H4
↓
C2H6
Problem 2
Use of math to describe chemical processes – physical chemistry (10 points)
Problem 3
Some organic chemistry (6 points)
Problem 4
Complex formation and red‐ox properties (6 points)
You have to investigate 1st type electrode using software MS Excel (or analogue). The first type electrode
consists from metal (M(s)) and it’s ions in solution (Mx+(aq)). This time solution also contain protolytic pair from
which conjugated base acts as ligands and form complex ions with Mx+.
Draw graph to show how reduction standard potential is dependent on pH if free ligand concentration in
solution is a) 0.1 M, b) 1.0 M and 5.0 M.
Given electrode:
Cu2+(aq) + 2e → Cu(s)
Eo = + 0.345 V
+
pKa(NH4+) = 9.25
Ligands:
NH3 / NH4
Summary complex formation constants:
lg β1 = 3.99
lg β1,2 = 7.33
lg β1,2,3 = 10.06
lg β1,2,3,4 = 12.03
lg β1,2,3,4,5 = 11.43
lg β1,2,3,4,5,6 = 8.9
Hints:
1. Assume that concentration sum for conjugate acid and base and acid are constant (if base are used for
complex formation it is added to solution to maintain constant concentration or concentration of base
is huge enough and stays almost constant).
2. Write mathematical equations for calculation real reduction standard potentials depending from molar
fraction of metal ions in solution.
Example:
3. Using MS Excel program calculate real standard potential values, if pH changes from 0 to 14 (step 0.5).
Made calculation in three cases: a) conjugate acid and base concentration sum is CL = 0.1 M, b) CL = 1
M and c) CL = 5 M. All calculation results collect in table.
www.biosan.lv
FINAL
Part A – Multiple choice test
Q. number
1.
2.
3.
4.
5.
6.
7.
8.
9.
10 .
11 .
12 .
13 .
14 .
15 .
16 .
17 .
18 .
19 .
20 .
21 .
22 .
23 .
24 .
25 .
26 .
27 .
28 .
29 .
30 .
31 .
33 .
Answer
B
B
D
A
A
B
A
D
C
D
A
B
A
B
D
A
B
B
B
C
C
D
A
B
A
C
B
B
C
A
B
D
No correct answer, D , all statements are true
32 .
B
33 .
D
34 .
A
35 .
A
36 .
B
37 .
B
38 .
C
39 .
D
40 .
A
41 .
D
42 .
B
43 .
C
44 .
B
45 .
A
46 .
D
47 .
A
48 .
A
49 .
B
50 .
B
51 .
B
52 .
D
53 .
D
54 .
A
55 .
A
56 .
C
57 .
C
58 .
A
59 .
C
60.
B
2011
ROUND, SOLUTIONS
Part B – Short answer questions
Problem 1
Chemical crossword – corrections (made with red color)
A
+
E
↓
F
+
G
E
+
U
↑
F
+
T
+
+
+
←
B
↑
I
+
H
→
L
+
N
↓
O
+
P
→
S
←
+
+
G
+
J
↓
E
+
K
R
+
Q
↓
Å
+
E
+
+
+
+
→
D
+
H
↓
B
F
←
+
C2H5Cl
→
→
Č
+
Ģ
↓
Ķ
+
F
C2H5NO2 +
M
+
Q
↓
F
+
V
+
W
→
I
+
+
D
+
R
↑
B
+
Ž
F
↓
I
+
E
+
+
+
→
U
+
F
+
Ā
↓
Ē
+
E
→
C2H6
+
Ō
↑
S
+
CH3Cl
+
+
Š
+
F
→
Ņ
+
F (not I)
↓
Ģ (not G)
Z
+
F
+
F
→
Y
Y
+ C2H5OH
+
C2H6
+
I
↓
D
+
F
E
+
C2H4
↓
C2H6
Answers:
A – Fe3O4 or other oxide which contains 72% metal, Mn3O4, Rh2O5, CaO – is not accepted wrong chemistry
B – CO
D – CO2
E – H2
F – H2O
G – Fe
H–C
I – O2
J – HCl
K – FeCl2
L – CuCl2
M – Cu
N‐ NaOH
O – NaCl
P – Cu(OH)2
Q – H2SO4
R – Zn or other metal
S – Li or other alkali metal
T – LiH or other alkali metal
U – LiOH or other alkali metal
V – CuSO4
W – SO2
Z – NO2
Y – HNO3
Š – CaH2, Ca accepted
Ž – ZnO
Ā – Al
Č – H2CO3
Ē – Li[Al(OH)4]
Ģ – Ca(OH)2
Ķ – CaCO3
Ņ – CaO
Ō – LiCl or other metal
Å – ZnSO4
Problem 2
Problem 3
Problem 4
0
ε red
= ‐ 0,764 V
Cu2+ + 2 e ‐ → Cu
Cu2+ + NH3
[Cu(NH3)]2+
lg β1 = 3,99
Cu2+ + 2NH3
[Cu(NH3)2]2+
lg β 2 = 7,33
Cu2+ + 3NH3
[Cu(NH3)3]2+
lg β 3 = 10,06
Cu2+ + 4NH3
[Cu(NH3)4]2+
lg β 4 = 12,03
Cu2+ + 5NH3
[Cu(NH3)5]2+
lg β 5 = 11,43
Cu2+ + 6NH3
[Cu(NH3)6]2+
lg β 6 = 8,9
NH4+ + H2O
H3O+ + NH3
pKA = 9,25
0
−
ε = ε red
RT
RT
1
0
= ε red
+
ln
ln [Cu 2+ ]
2+
2 ⋅ F [Cu ]
2⋅ F
⎧
[ H 3O + ] ⋅ [ NH 3 ]
⎪K A =
+
[ NH 4 ]
⎪
⎪C = [ NH + ] + [ NH ] + [{Cu ( NH ) }2+ ] + 2 ⋅ [{Cu ( NH ) }2+ ] + 3 ⋅ [{Cu ( NH ) }2+ ] +
4
3
3 1
3 2
3 3
⎪ L
⎪+ 4 ⋅ [{Cu ( NH 3 ) 4 }2+ ] + 5 ⋅ [{Cu ( NH 3 ) 5 }2+ ] + 6 ⋅ [{Cu ( NH 3 ) 6 }2+ ]
⎪
2+
⎪ K = [{Cu ( NH 3 )} ]
⎪ 1 [Cu 2+ ] ⋅ [ NH ]
3
⎪
⎪
[{Cu ( NH 3 ) 2 }2+ ]
⎪ K 2 = [Cu 2+ ] ⋅ [ NH ]2
3
⎪
⎪⎪
[{Cu ( NH 3 ) 3 }2+ ]
⎨K 3 =
[Cu 2+ ] ⋅ [ NH 3 ]3
⎪
⎪
[{Cu ( NH 3 ) 4 }2+ ]
⎪K 4 =
[Cu 2+ ] ⋅ [ NH 3 ]4
⎪
⎪
2+
⎪ K 5 = [{Cu ( NH 3 ) 5 } ]
⎪
[Cu 2+ ] ⋅ [ NH 3 ]5
⎪
2+
⎪ K = [{Cu ( NH 3 ) 6 } ]
⎪ 6 [Cu 2+ ] ⋅ [ NH 3 ]6
⎪
2+
2+
2+
2+
2+
⎪CCu = [Cu ] + [{Cu ( NH 3 )1} ] + [{Cu ( NH 3 ) 2 } ] + [{Cu ( NH 3 ) 3 } ] + [{Cu ( NH 3 ) 4 } ] +
⎪
2+
2+
⎪⎩+ [{Cu ( NH 3 ) 5 } ] + [{Cu ( NH 3 ) 6 } ]
+
C L = [ NH 4 ] + [ NH 3 ] + [{Cu ( NH 3 )1}2+ ] + 2 ⋅ [{Cu ( NH 3 ) 2 }2+ ] + 3 ⋅ [{Cu ( NH 3 ) 3 }2+ ] +
+ 4 ⋅ [{Cu ( NH 3 ) 4 }2+ ] + 5 ⋅ [{Cu ( NH 3 ) 5 }2+ ] + 6 ⋅ [{Cu ( NH 3 ) 6 }2+ ]
+
CL = [ NH 4 ] + [ NH 3 ]
⎧
[ H 3O + ] ⋅ [ NH 3 ]
K
=
⎪ A
[ NH 4+ ]
⎪
⎪
10 − pH ⋅ [ NH 3 ]
+
+
⇒
⎪CL = [ NH 4 ] + [ NH 3 ] ⇒ [ NH 4 ] = CL − [ NH 3 ] ⇒ K A =
CL − [ NH 3 ]
⎪
⎪⇒ 10 − pH ⋅ [ NH ] = K ⋅ C − [ NH ] ⋅ K ⇒ (10− pH + K ) ⋅ [ NH ] = K ⋅ C ⇒
3
3
3
A
L
A
A
A
L
⎪
K A ⋅ CL
⎪
⎪⇒ [ NH 3 ] = 10 − pH + K
A
⎪
⎪
[{Cu ( NH 3 ) }2 + ]
[{Cu ( NH 3 ) }2 + ]
⎛ K ⋅C ⎞
β
⇒ [{Cu ( NH 3 ) }2 + ] = β1 ⋅ [Cu 2 + ] ⋅ ⎜⎜ − pHA L ⎟⎟
⇒
=
⎪β1 =
1
2+
[Cu ] ⋅ [ NH 3 ]
+ KA ⎠
⎛ K ⋅C ⎞
⎝ 10
⎪
[Cu 2 + ] ⋅ ⎜⎜ − pHA L ⎟⎟
⎪
+ KA ⎠
⎝ 10
⎪
2
⎪
⎛ K A ⋅ CL ⎞
[{Cu ( NH 3 ) 2 }2 + ]
[{Cu ( NH 3 ) 2 }2 + ]
2+
2+
⎟
⎪β 2 =
⇒ [{Cu ( NH 3 ) 2 } ] = β 2 ⋅ [Cu ] ⋅ ⎜⎜ − pH
⇒ β2 =
2
[Cu 2 + ] ⋅ [ NH 3 ]2
+ K A ⎟⎠
⎪
⎛ K A ⋅ CL ⎞
⎝ 10
2+
⎟
[Cu ] ⋅ ⎜⎜ − pH
⎨
+ K A ⎟⎠
⎪
⎝ 10
⎪
3
2+
⎪β = [{Cu ( NH 3 )3} ] ⇒ K ⇒ [{Cu ( NH ) }2 + ] = β ⋅ [Cu 2 + ] ⋅ ⎛⎜ K A ⋅ CL ⎞⎟
3 3
3
⎜ 10 − pH + K ⎟
⎪ 3 [Cu 2 + ] ⋅ [ NH ]3
3
A ⎠
⎝
⎪
4
2+
⎪
⎛ K A ⋅ CL ⎞
[{Cu ( NH 3 ) 4 } ]
2+
2+
⎪β 4 =
⎟
⇒ K ⇒ [{Cu ( NH 3 ) 4 } ] = β 4 ⋅ [Cu ] ⋅ ⎜⎜ − pH
[Cu 2 + ] ⋅ [ NH 3 ]4
+ K A ⎟⎠
⎪
⎝ 10
⎪
5
⎛ K A ⋅ CL ⎞
[{Cu ( NH 3 )5 }2 + ]
⎪
2+
2+
⎪β 5 = [Cu 2 + ] ⋅ [ NH ]5 ⇒ K ⇒ [{Cu ( NH 3 )5 } ] = β 5 ⋅ [Cu ] ⋅ ⎜⎜ 10− pH + K ⎟⎟
3
A ⎠
⎝
⎪
6
⎪
2+
⎞
⎛
A ⋅ CL
⎪β 6 = [{Cu2(+NH 3 )6 } 6] ⇒ K ⇒ [{Cu ( NH 3 )6 }2 + ] = β 6 ⋅ [Cu 2 + ] ⋅ ⎜ K
⎜ 10 − pH + K ⎟⎟
[Cu ] ⋅ [ NH 3 ]
⎪
A ⎠
⎝
⎪
2+
2+
2+
2+
2+
⎪CCu = [Cu ] + [{Cu ( NH 3 )} ] + [{Cu ( NH 3 ) 2 } ] + [{Cu ( NH 3 )3} ] + [{Cu ( NH 3 ) 4 } ] +
⎪
2+
2+
⎩+ [{Cu ( NH 3 )5 } ] + [{Cu ( NH 3 )6 } ]
CCu = [Cu 2 + ] + [{Cu ( NH 3 )}2 + ] + [{Cu ( NH 3 ) 2 }2 + ] + [{Cu ( NH 3 )3}2 + ] + [{Cu ( NH 3 ) 4 }2 + ] +
+ [{Cu ( NH 3 )5 }2 + ] + [{Cu ( NH 3 ) 6 }2 + ] =
2
3
4
⎛
⎞
⎜1 + β ⋅ ⎛⎜ K a ⋅ CL ⎞⎟ + β ⋅ ⎛⎜ K a ⋅ CL ⎞⎟ + β ⋅ ⎛⎜ K a ⋅ CL ⎞⎟ + β ⋅ ⎛⎜ K a ⋅ C L ⎞⎟ + ⎟
1 ⎜
2 ⎜
3 ⎜
4 ⎜
− pH
− pH
− pH
− pH
⎟
⎟
⎟
⎟
⎜
+ Ka ⎠ ⎟
+ Ka ⎠
+ Ka ⎠
+ Ka ⎠
⎝ 10
⎝ 10
⎝ 10
⎝ 10
⎟
= [Cu 2 + ] ⋅ ⎜
5
6
⎜
⎟
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎟⎟
⎟⎟ + β 6 ⋅ ⎜⎜ − pH
⎜⎜ + β 5 ⋅ ⎜⎜ − pH
⎟⎟
+ Ka ⎠
+ Ka ⎠
⎝ 10
⎝ 10
⎝
⎠
[Cu 2 + ] =
1
2
3
4
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎟⎟ + β 2 ⋅ ⎜⎜ − pH
⎟⎟ + β 3 ⋅ ⎜⎜ − pH
⎟⎟ + β 4 ⋅ ⎜⎜ − pH
⎟ +
1 + β1 ⋅ ⎜⎜ − pH
+ Ka ⎠
+ Ka ⎠
+ Ka ⎠
+ K a ⎟⎠
⎝ 10
⎝ 10
⎝ 10
⎝ 10
5
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎟
⎟⎟ + β 6 ⋅ ⎜⎜ − pH
+ β 5 ⋅ ⎜⎜ − pH
+ K a ⎟⎠
+ Ka ⎠
⎝ 10
⎝ 10
6
,
0
= ε red
ε red
⎞
⎛
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
RT ⎜
1
⎟
+
ln
2
3
4
2⋅ F ⎜
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞ ⎟
⎟
⎜ 1 + β1 ⋅ ⎜
⎜ 10− pH + K ⎟⎟ + β 2 ⋅ ⎜⎜ 10− pH + K ⎟⎟ + β 3 ⋅ ⎜⎜ 10− pH + K ⎟⎟ + β 4 ⋅ ⎜⎜ 10− pH + K ⎟⎟ + ⎟
⎜
a ⎠
a ⎠
a ⎠
a ⎠
⎝
⎝
⎝
⎝
⎟
⎜
5
6
⎟
⎜
⎛ K a ⋅ CL ⎞
⎛ K a ⋅ CL ⎞
⎟⎟ + β 6 ⋅ ⎜⎜ − pH
⎟⎟
⎟
⎜ + β 5 ⋅ ⎜⎜ − pH
+ Ka ⎠
+ Ka ⎠
⎝ 10
⎝ 10
⎠
⎝
K a = 10 − pK a
K n = 10 lg K n
E(Cu2+), V
0,40
0,35
0,30
0,25
0,20
C= 0,1 mol/L
0,15
C = 0,5 mol/L
0,10
C = 5 mol/L
0,05
0,00
‐0,05
‐0,10
‐0,15
pH
0,0
5,0
10,0
www.biosan.lv
Final results of BCC 2011
Nr.p.k.
max
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
33
34
35
36
37
38
39
40
Name
Sigitas Mikutis
Emilis Bruzas
Jurgis Kuliesius
Kristina Smokrovich
Taivo Pungas
Rene Lomp
Jana Volaric
Julia Tseglakova
Mirna Mustapić
Aurimas Narkevicius
Kaur Aare Saar
Mindaugas Siauciulis
Eva-Lotta Kasper
Katrina Sepp
Lona-Liisa Sutt
Jānis Briška
Valdis Dakuļs
Jaan Toots
Kristers Ozols
Kert Pütsepp
Mantas Vasiliauskas
George Trenin
Ralf Ahi
Māris Seržāns
Vytaute Maciulsyte
Elza Liniņa
Kaur Karus
Greta Macileviciute
Ernestas Blazevic
Erik Tamre
Gediminas Aucina
Katrīna Krieviņa
Augustinas Juskauskas
Madis Ollikainen
Tatjana Gerasimova
Pijus Savickas
Simonas Balkaitis
j Svambaryte
y
Silvija
Dovile Meliauskaite
International jury:
State
LT
LT
LT
CR
EE
EE
CR
EE
CR
LT
EE
LT
EE
EE
EE
LV
LV
EE
LV
EE
LT
LV
EE
LV
LT
LV
EE
LT
LT
EE
LT
LV
LT
EE
LV
LT
LT
LT
LT
School
Klaipeda "Azuolynas" gymnasium
Vilniaus Žirmūnu gimnazija
V. gimnazija
Tallinn Secondary Science School
Liivalaia Secondary school
V. gimnazija
Tallinn High School of Humanities
V. gimnazija
KTUG
Tallinn English College
Mykolas Birziska gymnasium
Hugo Treffner Gymnasium
Miina Harma Gumnaasium
Miina Harma Gumnaasium
Salas secondaryy school
Viļānu secondary school
Tallinna Reaalkool
Riga State Gymnasium No.3
Tallinn Secondary Science School
Kuršėnu Lauryno Ivinskio gymnasium
Riga 95th Secondary school
Riga State Gymnasium No.1
Form TOTAL
12
12
12
11
4th
11
4th
11
10
11
11
12
10
11
11
10
11
12
10
10
Riga State Gymnasium No.3
Riga State Gymnasium No.1
11
Riga Secondary school No. 72
12
150,00
107,71
105,83
102,56
97,99
95,53
94,19
93,91
91,05
86,75
82,51
80,82
67,49
64,54
59,94
55,46
,
53,45
50,30
48,88
43,02
35,08
33,86
27,61
23,10
19,25
18,75
17,20
,
13,25
11,01
10,81
8,50
7,40
6,08
6,05
5,52
5,29
5,24
4,26
,
4,14
3,00
Prize
1st place
2nd place
3rd place
Honorary mention
Honorary mention
Honorary mention
Honorary mention
Honorary mention
Egle Maksimaviciute LT
Vladislav Ivaništšev EE
Kaspars Veldre LV
Filip Topić CRO
Students will be awarded during Chemistry Olympiad of Baltic states, 2011, April 15-17, Vilnius, Lithuania
1.
8
3,28
6,24
4,64
4,04
5,56
4
3,12
1,28
2,84
2,44
0
0
3,72
3,68
3,16
,
0,8
0
1,08
0
0,12
0,2
2.
10
9,25
9,25
7,75
6,25
9
7
6
8
4,5
6
0
4
8,5
7,75
0,75
,
6,25
0
0
8
9,25
5,25
0
0
0
0
,
1,6
0,24
3,64
0
0
0,84
0,12
0
0,6
0
0,12
0
0,52
4,5
0
6,75
0
0
0,75
0
0
0
0
0
0
0
0
0
0
0
Round 1
3.
4.
4
8
4
7,724
4
8
1
7,724
4
6,345
0
8
4
8
4
8
7,72
0
4
6,621
4
6,345
0
7,034
0
7,448
0
7,172
0
8
0
6,621
,
7,72
0
0
5,793
0
8
4
7,172
0
7,034
0
4,966
0
0
4
0
0
0
0
0
0
0
0
0
0
0
0
0
0
7,172
0
8
0
0
5,517
7,172
0
0
5,241
5,931
5,517
4,69
5,241
4,138
,
4,138
2,483
Sum
30
24,25
27,49
21,11
20,63
22,56
23
21,12
17
17,96
18,78
7,034
11,45
19,39
19,43
10,53
,
14,77
5,793
9,08
19,17
16,4
10,42
0
11,67
0
18,75
0
,
1,6
6,507
10,81
0
0
6,081
6,051
5,517
5,29
5,241
4,258
,
4,138
3,003
1.
5
4,1
3,8
4,5
3,65
5
6
4,25
2,8
3,75
4,3
4
4,15
4,38
4,13
4,45
0
3,5
0
0
4,5
2,6
2.
5
2,32
3
4,5
4,16
3,83
3,205
3,85
4
3,33
3,285
3,58
3,75
2,5
0,83
3,375
4
3,65
0
4,15
,
2,5
3,25
0
0
4,23
0
0
0
0
0
0
0
0
Round 2
3.
4.
10
5
7,5
5
7
4,5
8,9
4,5
9,3
4,5
9,2
5
8,9
5
9,4
5
4
5
8,8
5
7,5
5
8,5
5
6,5
3,5
3
0
4,5
3
5,4
5
5.
5
4,6
3,9
4,8
2,5
2
1,5
2,8
3,8
3,7
1,2
1
1,1
1,2
0,7
2
3,17
2
3,5
0,4
1,875
1
4,3
0
2,5
1
1
0,9
2,125
3
0,4
8,9
4
3
0,9
0,7
2,045
,
0,25
1,25
6
,
3,9
0
3,5
4
0
1,5
1
0
2,17
7,5
0
0
1
1
0
Sum
30
23,52
22,2
27,2
24,11
25,03
24,605
25,3
19,6
24,58
21,285
22,08
19
11,08
13,16
20,225
0
12,57
0
0
14,175
5,5
0
11,425
19,25
0
17,195
,
11,65
4,5
0
8,5
7,4
0
0
0
0
0
0
0
0
1.
5
3,5
3,5
4
3,5
4
Round 3
3.
4.
8
9
8
5,499
7,002
3,2
6,201
8
6,804
0
1,602
8
8
4,302
3,2
2,196
8
4,725
0
1,305
0
0,801
0
0,504
0
0
3,5
4,5
3,5
3,5
3
3,5
2,5
2.
8
4,83
7,17
3,54
7,17
2,33
0,375
4,08
6,5
5,21
3,17
1,5
1,5
0
1
5
0
0,833
0
0
0,594
0,594
2
1,5
0
0,999
Sum
30,00
21,83
17,67
16,94
25,47
7,93
8,38
19,88
16,40
21,44
7,98
5,30
5,50
2,50
0,00
0,00
,
0,00
1,59
6,43
0,00
4,50
0,00
0,00
0,00
0,00
0,00
0,00
,
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
,
0,00
0,00
Part A
30,00
23,50
24,50
22,50
23,00
23,50
25,00
24,00
24,00
22,50
19,50
24,50
22,00
21,50
21,50
20,00
,
21,00
21,50
23,00
19,00
0,00
17,00
21,50
0,00
0,00
0,00
0,00
,
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
,
0,00
0,00
Part B1
8,00
2,35
4,47
2,59
1,41
3,53
3,76
0,00
3,29
0,00
5,88
7,29
0,71
6,82
2,59
4,71
,
8,00
7,29
3,29
3,06
0,94
0,00
FINAL ROUND
Part B2 Part B3
10,00
6,00
6,25
6,00
7,50
6,00
7,78
4,44
0,00
3,36
7,22
5,76
4,17
5,28
3,61
0,00
5,00
5,76
0,28
0,00
5,00
4,08
8,33
5,28
4,03
4,80
0,00
3,24
2,78
0,48
0,00
0,00
,
,
5,83
3,84
0,83
0,72
7,08
0,00
0,83
0,96
0,00
0,00
0,00
0,00
6,11
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
,
,
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
,
,
0,00
0,00
0,00
0,00
Part B4
6,00
0,00
-4,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
1,00
0,00
0,00
0,00
0,00
,
0,00
0,00
0,00
0,00
0,00
0,00
Sum
60,00
38,10
38,47
37,31
27,77
40,01
38,21
27,61
38,05
22,78
34,46
46,41
31,53
31,56
27,35
24,71
,
38,67
30,35
33,38
23,85
0,00
17,94
27,61
0,00
0,00
0,00
0,00
,
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
0,00
,
0,00
0,00