Răspuns :
acid + alcool ⇄ ester + H2O
initial : 2 moli 1 mol 0,5 8
au reaqctionat : x x
la echilibru : (2-x) (1-x) (0,5+ x) (8+x)
ms = 2·60 + 46 + 44 + 144 = 354g md = (0,5 +x)·88 c = 17,41
(0,5+x)·88·100/354 = 17,41 x = 0,9278
Kc = 1,4278·8,9278/(1,0722·0,0722) = 164,664
2. A : 14·100/M = 11,96 M A = 117 g/mol A = C5H11NO2 nA = 0,1
B : 32·100/M = 42,66 M B = 75g/mol B = H2N-CH2-COOH nB =0,2
peptida = valil-glicil -glicina
3. maltoza (aldoza) + 2Cu(OH)2 ⇒ acid +Cu2O ↓ +...
nCu2O = 14,4/144 = 0,1 moli = n maltoza
nBr2 = 0,5·0,8 = 0,4 moli = 2·n maltoza + 2·n zaharoza
⇒ n zaharoza = 0,1 moli ⇒ n maltoza / n zaharoza = 1/1
initial : 2 moli 1 mol 0,5 8
au reaqctionat : x x
la echilibru : (2-x) (1-x) (0,5+ x) (8+x)
ms = 2·60 + 46 + 44 + 144 = 354g md = (0,5 +x)·88 c = 17,41
(0,5+x)·88·100/354 = 17,41 x = 0,9278
Kc = 1,4278·8,9278/(1,0722·0,0722) = 164,664
2. A : 14·100/M = 11,96 M A = 117 g/mol A = C5H11NO2 nA = 0,1
B : 32·100/M = 42,66 M B = 75g/mol B = H2N-CH2-COOH nB =0,2
peptida = valil-glicil -glicina
3. maltoza (aldoza) + 2Cu(OH)2 ⇒ acid +Cu2O ↓ +...
nCu2O = 14,4/144 = 0,1 moli = n maltoza
nBr2 = 0,5·0,8 = 0,4 moli = 2·n maltoza + 2·n zaharoza
⇒ n zaharoza = 0,1 moli ⇒ n maltoza / n zaharoza = 1/1