Part A – 60 multiple choice questions with one correct answer each. Maximum
points ... Write chemical formulas for all compounds abbreviated with letters. A. +.
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FINAL
2011 ROUND, PROBLEMS
Solve all problems provides. Fill answer sheets and till 13:00, March 20, 2011 send them to e‐mail:
[email protected] 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
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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
+
+
C2H5NO2 +
D + H ↓ B
+
M + Q ↓ + F → + → V + + W I
+
D + R ↑ B + Ž F ↓ I + E
+
+
+ →
U + F + Ā ↓ Ē + E
F
+
C2H5Cl
→
Č + Ģ ↓ Ķ + F
Š
+
F
Z + F
+
F
←
C2H6 + Y + Ō + C H → 2 5OH ↑ 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.
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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 No correct answer, D , all statements are true A 32 . B A 33 . D B 34 . A A 35 . A D 36 . B C 37 . B D 38 . C A 39 . D B 40 . A A 41 . D B 42 . B D 43 . C A 44 . B B 45 . A B 46 . D B 47 . A C 48 . A C 49 . B D 50 . B A 51 . B B 52 . D A 53 . D C 54 . A B 55 . A B 56 . C C 57 . C A 58 . A B 59 . C D 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
+
M + Q ↓ + F → + V + → + W I
F
←
+
C2H5Cl
→
→
Č + Ģ ↓ Ķ + F
C2H5NO2 +
+
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 Cu2+ + 2 e ‐ → Cu
0 ε red = ‐ 0,764 V
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 ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎟ ⎜ 1 + β1 ⋅ ⎜ K a ⋅ CL ⎟ + β 2 ⋅ ⎜ K a ⋅ CL ⎟ + β 3 ⋅ ⎜ K a ⋅ CL ⎟ + β 4 ⋅ ⎜ K a ⋅ CL ⎟ + ⎟ ⎜ 10− pH + K ⎟ ⎜ 10− pH + K ⎟ ⎜ 10− pH + K ⎟ ⎜ 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
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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
