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7/29/2019 Bw32 Theorie l e
1/12
32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
1
28 bp 10 rp f = Task 1, 10 points
1.1. Element: silicon 1 bp
1.2. Name: silanes Formel: SinH2n+2 1 bp
1.3. formula and name: (CH3)2Si(OH)2 dimethylsilandiol1 bp
1.4. reaction equation:
n HO-Si(CH3)2-OH H-O(-Si(CH 3)2-O)n-H + (n-1) H2O
name of the product: silicones 2 bp
1.5. general formula for n: n = 4x-2y1 b
1.8. accompanying rock: lime2 bpargumentation: CaCO3 + 2 HCl CaCl 2 + H2O + CO2
CO2 + Ca(OH)2 CaCO 3 + H2O
1.9. reaction equation:
2 S3- + 2 H+ 5 S + H 2S 2 bp
1.10. LEWIS-formula: 1 bp
1.6. formula: Ag10[Si4O13]1 bp
1.7. formula of lapis lazuli: Na4[Al3Si3O12]S3 2 bp
S
S
S-
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
2
1.11. Element: selenium2 bpjustification:
6219.096.126
96.78)Se(w?SeO:onverificati
seleniummol/g96,78)X(M2)X(n
)O(n
elementreasonablenomol/g48.39)X(M1)X(n
)O(n
)X(M.16.7116
84.28)X(m).O(M)X(M).O(m
)X(n)O(n
3 ==
==
==
==
1.12. formula of HaXOb: H2SeO3 formula of HcXOd: H2SeO4 1 bp
1.13. oxidant: MnO4-
redox equation:2 MnO4- + 5 SeO32- + 6 H+ 5 SeO 42- + 2 Mn2+ + 3 H2O 2 bp
1.14. argumentation for aromaticity: Frost-Musulin-scheme:24 valence electrons
8 -bond electrons 2 charge- 8 lone-pair electrons
6 -electrons
order of -bonding: 2/8= 2 bp1 bp
1.15. Element: xenon 1 bp
1.16. oxidation number: +2, +4, +6 1 bp
7/29/2019 Bw32 Theorie l e
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
3
1.17. geometry of the three compounds:
XeF2 linearXeF4 plane squareXeF6 pentagonal pyramidal 3 bp
1.18. reaction equation:XeF6 + 3 H2O XeO 3 + 6 HF 1 bp
1.19. speciality:
exceedance of the octet rule 1 bp
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
4
Task 2, 15 points
2.7. molar initial concentration: 3 bp
TR
pc
V
nTRnVp
===
L/mol00488,0cm/mol876,4967314,8
39200)694(c 0
3 ==
=
L/mol00561,0cm/mol605,51030314,8
48000
)757(c 03
===
2.8. rate constants:
orderondsecc
1
0
0
A0AA c
1kk
t
1
c
1
c
1
==
s.mol/L842.0005605.0212
1)695(k
s.mol/L135.0004876.01520
1)695(k
=
=
=
= 3 bp
2.1. reaction equation: 2 C (s) + O2 (g) 2 CO (g) 1
2.2. BP:)O(p
)CO(pBP
2
2
= 1
2.3. Free standard enthalpie of the reaction: 2bp
kJ443J442950)101.5ln(1273314.8KlnTRG 18P1273 ====
2.4. reaction quotien + direction of the reaction: 0169.0113.0Q
2==
G1273 = -442950 + RTln(0.0169) = -486 kJG1273 < 0 left to right 2 bp
2.5. entropy of the reaction:
K/J2051273
486136225000
T
GHSSTHG =
+=
== 2 bp
2.6. reaction equation of the decomposition: 2 N2O 2 N 2 + O2 1 bp
33 bp 15 rp f =
7/29/2019 Bw32 Theorie l e
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
5
Task 3, 10 points 25 bp 10 rp f =
2.9. activation energy:
kJ241T
1
T
1
)T(k
)T(klnRE
T
1
T
1
R
E
)T(k
)T(kln
1
211
2A
21
A
1
2 =
=
=
2
2.10. resonance formulae: 2 bp
2.11. VSEPR: linear 1 bp 2.12.point group: C,v 1 bp
2.13. structure and name:CH3-CH2-OH, ethanol 1 bp
Aufgabe 2, 15 Name:
2.14. Newman-projection:
1bp
2.15. mean value of boiling point: HV,m = 86246 = 39652 J/mol
calculation (60C): K033.352T002840643.0T
12
2
==
calculation (70C): K562.351T002844445.0T
12
2
==
mean value: BP = 351.8 K = 78.7C 4bp
2.16. entropy of vaporisation: 2 bP
K/J1138.351
39652
T
HS
m,V
m,V ==
=
2.17. ebullioscopic constant:
mol/Kkg19.1396521000
46314.88.351
H1000
MRTK
2
V
2S
EB =
=
= 2
2.18. vapour pressure of the solution:n (ethanol) = 100/46 = 2.1740 mol,n (vanillic aldehyde) = 5/152 = 0.0329 mol,ng = 2.2069 mol x (ethanol) = 0.9851p (ethanol) = p0x = 46.0 kPa 2
bp
N N O N N O+
-
+
-
OH
H H
HH
H
)T(p
)T(p
lnH
R
T
1
T
1
T
1
T
1
R
H
)T(p
)T(p
ln1
2
V1221
V
1
2
=
=
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
6
3.1. solubility of BaCrO4:
L(BaCrO4) = M102.9BaCrO(K 64L
= 1 bp
solubility of SrCrO4:
L(SrCrO4) = M100.6SrCrO(K3
4L= 1 bp
3.2. concentrationen of Cr2O72- and of CrO42-:
Cr2O72- + H2O 2 H + + 2 CrO42- K= 151050.1
1
= 6.6710-16
Cr2O72- H2O H
+ CrO4-
at the beginning 0.1 55.56 - -
equilibrium 0.1-x 55.56 10-3 2x
1626
2272
224
2
1067.656.55)x1.0(
x410
]OH[]OCr[
]CrO[]H[K
+
=
=
=
x1 = 3.0410-5 (x2 = -3.0410-5)[Cr2O72-] = 1.010-1 M[CrO42-] = 6.110-5 M 6 bp
Konzentrationen von Ba2+ und Sr2+:
[Ba2+] [CrO 42-] = 8.510-11 [Ba2+] =5
11
101.6
105.8
= 1.410-6 M
[Sr2+] [CrO 42-] = 3.610-5 [Sr2+] =5
5
101.6
106.3
= 0.59 M
2 bp
3.3. mass of sodium acetate:
B
SS
n
nlgpKpH =
3.00 = -lg(1.7810-5) -x1.0lg
x = 1.7810-3 molm = n M = 1 .7810-3 mol 82.03 g/mol = 0.146 g
2 bp
7/29/2019 Bw32 Theorie l e
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
7
3.4. standard potentials x and y:
3(-0.744) =-0.408 + 2y
Y = - 0.912 V
0.55+1.34+x-30.744 = 60.293x = 2,1 V
2 bp
3.5. Is there any tendency of disproportionation of Cr(IV) to Cr(III) andCr(VI)?:
Yes!
Explaination: E=2,1-0,5(1,34+0,55) = 1,155 > 0 G < 0 1 bp
7/29/2019 Bw32 Theorie l e
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
8
3.6. half equation for Cr2O72-/Cr3+:
Cr2O72-
+ 6 e
-
+ 14 H
+
2 Cr
3+
+ 7 H2O 1 bp
potential change per pH-step:
V138.0F6
143026.2RT
])OHlog[(
F6
143026.2RT
]Cr[
]OCr[ln
F6
RTEE
)]OHln[(F6
RT
]Cr[
]OCr[ln
F6
RTEE
]Cr[
]OH][OCr[ln
F6
RTEE
323
2
72
14323
2
72
23
143
2
72
=
+=
+=
+=
++
++
+
+
3 bp
3.7. coordination number: 6 1 bp
3.8. geometry: octahedral 1 bp
3.9. name of the ion: dichlorodioxalatochromat(III) 1 bp
3.10. stereoisomers:one trans-isomertwo enantiomeric cis-isomers 3 bp
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4.1.1. numbering 1 bp
N 1 2
3
4
5
6
7
8
4.1.2 3-tropanole 1 bp
4.2.4. configurationalformula of (S)-tropic acid
2bp
4.2.5. constitutional formula ofhyoscyamine
2 bp
4.3.1. reaction forming the Mannich base
4 bp
4.4.1. configurational formulae of A and B
4 bp
4.2.1. empiricalformula of
hydrotropic acid
C9H10O2 2bp
4.2.2. constitutionalformula of
hydrotropic acid
3 bp
4.2.3. constitutionalformula of tropic acid
3bp
32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
9
Task 4, 15 points 40 bp 15 rp f =
N
OH
O
OH
HOHN
O
O
OH
C6H
5
OH
O
OH
O
OH
CHO
CHO
CH=NHCH3+
CHO
CH=NCH3
CHO
OH
CHO
N
O+ CH
3NH
2
H+
or
+
- H+
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4.4.2. isomeric relationenantiomeric 1
bp
4.4.3 type of reaction:SN2 1
bp
4.5.1. constitutional formula of tropinic acid
3 bp
4.6.1. D
2 bp
4.6.1. E
2bp
4.6.1. F
2bp
N
O
O
OOH
OMe
4.6.2. reagent ?
2 bp
4.7.1. chiral centres
2 bp
O Cl4.7.3. configuration(s)C-1: R; C-2: RC-3: S C-5: S 2 bp
4.7.2. stereoisomers 16 1
32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
10
N
C6H
5
O
N
O
C6H
5
Br- Br-
+ +
N
O
O
OMeN
O
OMe
OH
N
O
O
C6H
5
COOMe
*
*
**
NCOOH
COOH
N
COOHHOOC
or
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
11
Task 5, 5 Punkte 12 bp 5 rp f =
5.1. Leu-Val-Ser:
3 bp
5.2. flight time:
3 bp
6.3. Lebensdauer der Sonne:
g10461,17346,010988,17346,0mm 303SonneSonne,H
===+
a10697,7s10429,2s/g10014,6
g10461,1t 1018
14
33
===
1 bp
6.4. Massendefekt der Sonne:
( )kg10279,4
s/m109979,2
J10846,3
c
Em 9
28
26
2=
==
1 bp
s1025.376963
50.2vst 5===
2
mvE
2
= s/m76963J1032
m
E2v
123 mol10022.6
mol/kg610.0
15
=
==
5.3.masses of amino acids possible: 71,73,131,147,188in the case of an AS within the chain the molar mass is reduced by 18 u.
71 73 131 147 188
within the chain (+18) 89 (Ala) 91 149 (Met) 165 (Phe) 206
Therefore: Ala, Met and Phe. 3 bp
5.4.masses of amino acids possible: 73 and 188At the C-terminal the mass of the AS is reduced by 2 u, at the N-terminal by16 u.
73 188
C-terminal (+2) 75 (Gly) 190
N-terminal (+16) 89 (Ala) 204 (Trp)
AS at teh N-terminal: TrpAS at the C-terminal: Gly
3 bp
H3N
NH
NH
O
O
O
O
OH
+
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32nd Austrian Chemistry OlympiadNational Competition
Theoretical Part SolutionsJune 19th, 2006
12
Task 6, 5 Punkte 11 bp 5 rp f =
6.1. mass of the sun:
m1096.6km6960002
dr 8Sonne ===
3273
Sonne m10412.13
r4V ==
Sonne = 1.408 g/cm3=1408 kg/m3
mSonne = VSonne Sonne = 1.9881030 kg = 1.9881033 g2 bp
6.2. mass of hydrogen per second: 3 bp
E = 26.72 MeV/4 H+ = 6.68 MeV/H+ = 1.07010-12 J/H+ == 6.4451011 J/mol H+
s/mol10967.5mol/J10445.6
s/J10846.3n 14
11
26
H=
=+
s/g10014.6mol/g0078.1s/mol10967.5Mnm 1414HHH
=== +++
6.3. Lebensdauer der Sonne:
g10461,17346,010988,17346,0mm 303SonneSonne,H
===+
a10697,7s10429,2s/g10014,6
g10461,1t 1018
14
33
===
1 bp
6.4. Massendefekt der Sonne:
( )kg10279,4
s/m109979,2
J10846,3
c
Em 9
28
26
2=
==
1 bp
6.3. lifetime:
g1046117346010988.17346.0mm 3333SonneSonne,H ===+
a106977s104292s/g100146
g10461.1t 1018
14
33
==
=
6.4. mass defect:
( )kg10279.4
s/m109979.2
J10846.3
c
Em 9
28
26
2
=
==
1 bp
6.5. solar constant:
Oorbit = 4 (1.4961011 m)2 = 2.8121023 m2
2
223
26
ms/J1368m10812.2
s/J10846.3E =
= 3 bp
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