Đề thi Olympic Hoá học quốc tế lần thứ 11 đến 15

... phenolphthalein), pipettes (volumes 10, and 15 or 20 cm3), 2 volumetric flasks (250 cm3), 2 titration flasks (100 – 150 cm3). 11 1111 11 thththth ... SOL UT I ON 2.1 ∆H01000 = 35040 J ∆S01000 = 32 .11 J mol-1 K-1 ∆G01000 = ∆H01000 – T ∆S01000 = 35040 – 1000 × 32 .11 = 2930 J 2.2 ∆G0 = – RT ln Kp 02930ln 0.3524188314GKpRT∆= ... 0 0.297 x2 – 0.89605 x + 0.09 = 0 x2 – 3.01703 x + 0.303030 = 0 1,21.508 515 2.275618 0.303030 1.508 515 1.972588x = ± − = ± THE 12TH INTERNATIONAL CHEMISTRY OLYMPIAD, 1979 THE...

... cm-3): m(H2O) = 150 g 1.2 After electrolysis: m(H2O) = m(solution) – m(K2SO4) = 20 g0 .15 – 20 g = 113 .3 g The mass of water decomposed on electrolysis: m(H2O) = 150 – 113 .3 = 36.7 ... potassium sulphate was dissolved in 150 cm3 of water. The solution was then electrolysed. After electrolysis, the content of potassium sulphate in the solution was 15 % by mass. Problem: What ... 23 The compound contains also oxygen, since m(C) + m(H) = 12 g + 3 g = 15 g < 23 g m(O) = 23 g - 15 g = 8 g n(O) = 0,5 mol n(C) : n(H) : n(O) = 1 : 3 : 0,5 = 2 : 6 : 1 The...

... Si: m(Si) = 4.26 mmol × 28.09 g mol-1 = 119 .7 mg 97 .111 00mg1000mg7 .119 Si%=×= Cu: m(Si + Cu) = 170 mg m(Cu) = 170 mg − 119 .7 mg = 50.3 mg (in 1000 mg of the alloy) % ... g = 0 .111 g Heating: (9) MII(OH)2 → MIIO + H2O Decrease of the mass: 0.027 g (H2O) (10) Mass of MIIO: 0.084 g In relation to (8), (9), and (10): ( )0.084( ) 18 0 .111 IIrIIrM ... corresponds: alloytheofmg150inCrmmol26.0378.0= m(Cr) = 2.6 mmol × 52 g mol-1 = 135.2 mg in 150 0 mg of the alloy % Cr = 9.013 Content of Cu and Ag: Q = 40.575 mF / 150 0 mg (1087.4 mAh)...

... via dimerization from the for-runnings of the benzene-pyrolysis fraction. This is achieved either by heating to 140 – 150 oC under pressure or by heating over several hours at 100o C. Then ... N-CH2-CH2-NHOOC-CH2HOOC-CH2CH2-COOHCH2-COOH The anion of this acid, C10H12N2O84-, forms a stable complex CaC10H12N2O82- with calcium ions. The stability constant of this complex ion is given by: ... mechanism also requires an anti-periplanar (= trans) arrangement of H and Br. The products given in this problem are only formed from a Z-configurated adduct. The bromination of A and subsequent...

... rexp = 100 × [0.0201/(0.0125 + 0.00558)] = 111 % Since 50 cm3 water dissolve 0 . 115 g of gypsum, the real quantity of Ca(H2PO4)2 is 0.856 – 0 . 115 = 0.741 mol, so that the real yield gives: ... Anhydride (acetic anhydride). Relative atomic masses: C: 12. 011; O: 15. 999; H : 1.008 Reagents 2-hydroxybenzoic acid (melting point 158 °C) Ethanoic anhydride (boiling point 140 °C) Phosphoric ... the flask to 70 – 80 °C in a beaker of near boiling water and maintain the mixture at this temperature for 15 minutes. Remove the flask from the water bath and, with gentle swirling, add dropwise...