Va rooooooooooooog
Am nevoie urgenttttttțtttttt
91 si 92
Răspunsuri la întrebare
91/.
Cn = md/Eg.acid x Vs
=> md = CnxEgxVs = 0,1x40x0,02 = 0,08 g
niu.NaOH = 0,08/40 = 0,002 moli NaOH
notam acidul monocarboxilic saturat = CnH2n+1COOH
0,12 g 0,002 moli
CnH2n+1COOH + NaOH --> CnH2nCOONa + H2O
miu 1
=> miu = 60 g/mol
=> 14n+46 = 60 => n = 1
=> CH3COOH, acid etanoic (acetic)
=> b)
92/.
Cn = md/Eg.acid x Vs
=> md = CnxEgxVs = 0,1x40x0,04 = 0,16 g
niu.NaOH = 0,08/40 = 0,004 moli NaOH
notam cu a = nr moli care revin fiecarui acid din amestec
miu.HCOOH = 46 g/mol
miu.CHCOOH = 60 g/mol
deci avem in amestecul de acizi
=> md = 46a + 3a60 = 226a g
ms = 25 g
46a g n1 moli
HCOOH + NaOH --> HCOONa + H2O
46 1
180a g n2 moli
CH3COOH + NaOH --> CH3COONa + H2O
60 1
=> n1 = a , n2 = 3a
=> 4a = 0,004 => a = 0,001
=> 46a = 0,046 g ac. formic
=> 180a = 0,18 g ac. acetic
=> c%.HCCOH = 0,046x100/25 = 0,184%
c%.CH3COOH = 0,18x100/25 = 0,72%
=> c)
in liceu.. si la admitere medicina nu se mai foloseste Cn..