41th Latvian National Chemistry Olympiad (2000)
Theoretical answers

Na NaOH NaCl HCl ZnCl₂ ZnSO₄ Zn(OH)₂ Zn(NO₃)₂ ZnO

Only the most complex equatations:

2 Zn(NO₃)₂ 2 ZnO + 4NO₂ + O₂

m(Cr₂(SO₄)₃.15H₂O)=295g

Selenite : CaSO_4.H₄O₂ or CaSO_4. 2H₂O.
As there is no selenium in selenite , it contains 0,00 % of selenium.

a) w(CaCO₃) = 0,9465.
w(SiO₂) = 1 – w(CaCO₃) = 0,0535.

b) m(CaO) = 0,07471 g, w(CaO) = 67,2%;
m(CaCO₃) = 0,02332 g, w(CaCO₃) = 21,0%;
m(SiO₂) = 0,0131 g; w(SiO₂) = 11,8%.

5.

9

6 p.

C + O₂ CO₂H₂O + C CO + H₂
2 N₂ + 3 H₂ 2 NH₃
NH₃ + CO₂ + H₂O NH₄HCO₃

A - NaH, B - CaH_{2 .}
NaH + H₂O = NaOH + H₂
CaH₂ + 2H₂O = Ca(OH)₂ + 2H₂

Magnesium carbonate pentahydrate was heated - MgCO_3.5H₂O.

9.

9-10

6 p.

As the solution contains NaHCO₃ pH > 7 .

CO₂ Na₂CO₃ NaHCO₃ Na₂SO₄ NaBr NaNO₃ HNO₃ Fe(NO₃)₃ Fe(OH)₃

Only the most complex equtations:
Na₂SO₄ + BaBr₂ BaSO₄ + NaBr
NaBr + AgNO₃ AgBr + NaNO₃
6 HNO₃ + Fe Fe(NO₃)₃ + 3 NO₂ + 3H₂O

Fisherman should have placed the plaices in a salty water.

H₂O (18), N₂ (28), O₂ (32), Ar (40), CO₂ (44).

2Fe²⁺ + O₂ + 2H₂O 2Fe³⁺ + 4OH^-
                                       Fe²⁺ Fe³⁺ + e^-
                              O₂ + 2H₂O + 2e^- 4OH^-

Fe³⁺ + 3OH^- Fe(OH)₃

w(NH₄HCO₃) = 0,176.

m_j = (n_Ca + n_Mg)/c_M(-SO₃H) = 1000*3,0*10^-3/(10*10^-3) = 300 g.

The unknown compound is NaCl*2H₂O. NaCl can be obtained by heating in the sun.

18.

11

8 p.

CaCO₃ CaO CaCl₂ CaSO₄ CaS CaCl₂ Ca₃PO₄ Ca₃P₂ PH₃

Only the most complex equatations:
CaSO₄ + C CaS + 4 CO
or CaSO₄ + BaS CaS + BaSO₄
Ca₃PO₄ + 8 C Ca₃P₂ + 8 CO
Ca₃P₂ + 6 HCl 3 CaCl₂ + 2 PH₃

Given solutions contain 0,001mol BaCl₂ and 0,001mol Na₂SO₄, the obtained solution contains 0,002mol NaCl osmotic pressure of the obtained solution - 9,74 kPa

CO₂ (g) + H₂O CO₂ (s)
CO₂ (s) + 2H₂O HCO₃^- + H₃O⁺
HCO₃^- + 2H₂O CO₃^2- + H₃O⁺

After the reaction the solution contains 16,2% of ZnCl₂ and 12,0% of FeCl₂

The obtained mixture of acids should be diluted to 347 litres.

A – O₂
B – Mg
C - C (diamond)
D – MgO
E – CO₂
F – C (graphite)

The waterpipe was made of lead.

SO₂ + 2H₂O HSO₃^- + H₃O⁺
CO₂ + 2H₂O HCO₃^- + H₃O⁺
Cu(OH)₂ + 2H₃O⁺ Cu²⁺ + 4H₂O
CuCO₃ + 2H₃O⁺ Cu²⁺ + 3H₂O + CO₂
CO₂ + 3H₂O CO₃^2- + 2H₃O⁺
2Cu²⁺ + CO₃^2- + 4H₂O Cu(OH)_2. CuCO₃ + 2H₃O⁺
H₃O⁺ + CO₃^2- HCO₃^- + H₂O

C₂H₅OH CH₃COOH CH₃COOCH₃ CH₃OH CH₃OCH₃ CO₂ K₂CO₃ MnCO₃ Mn(NO₃)₂

Reagents:
1. CrO₃ (or dichromates), KMnO₄;
2. CH₃OH, H⁺
3. OH^-

119,0*n_cis + 114,6*n_trans = 5,18
n_cis + n_trans = 1/22,4
n_cis = 0,0145 mol, n_trans = 0,0301 mol;

φ _cis = 32,5%, φ _trans = 67,5%.

Lake Vertage [Al(OH)₃] = 0,002 %
Lake Chortock [Al(OH)₃] = 32 %

If halogen is fluorine: C_nH_2n+2 + a F₂ C_nH_2n+2-aF_a + a HF

n = 1, a = 4.

If halogen is chlorine

n = 1, a = 2,86; n = 2, a = 5,37; n = 3, a = 7,87
As chloropropanes are liquids the only matching reaction is CH₄ + 4 F₂ CF₄ + 4 HF.

CH₃CH₂Cl + KOH CH₃CH₂OH + KCl
CH₃CH₂Cl + KOH C₂H₄ + KCl + H₂O
CH₃CH₂Cl + CH₃CH₂OK CH₃CH₂OCH₂CH₃ + KCl

[1] C₆H₄ (OH)₂ + 2 NaOH C₆H₄ (ONa)₂ + 2 H₂O
[2] C₆H₄ (ONa)₂ + CH₃COCl C₆H₄ (OCOCH₃)₂ + 2NaCl
[3] C₆H₄ (ONa)_{2 +} CH₃I C₆H₄ (OCH₃)₂ + 2NaI
[4]

X – 1,4-dihydroxylbenzene (hydrohinone);
Y – X sodium salt;
Z – 1,4-dimetoxybenzene;
Q – 1,4-diacetylbenzene;
A – 1,4-benzohinone

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