Bonjour,
Ex 2)
1) a)C = n/V = 0,012/0,250 = 0,0480 mol.L⁻¹
b) Cm = m/V = (n x M)/V = M x n/V = M x C = 342 x 0,0480 = 16,4 g.L⁻¹
2) a) n = C x V = 1,5.10⁻² x 20.10⁻³ = 3,0.10⁻⁴
b) M = 7xM(C) + 6xM(H) + 2xM(O) = 122 g.mol⁻¹
c) Cm = M x C = 122 x 1,5.10⁻² = 1,83 g.L⁻¹
d) m = Cm x V = 1,83 x 20.10⁻³ = 3,66.10⁻² g
Ex 9)
1) alevinage = repeuplement en alevins
2) Fe²⁺ et Fe³⁺
3) Cm(Fe²⁺) = m/V = 10.10⁻⁶/50.10⁻³ = 2,0.10⁻⁴ g.L⁻¹
et Cm(Fe³⁺) ≈ 0
C(Fe²⁺) = Cm(Fe²⁺)/M(Fe) = 2,0.10⁻⁴/56 ≈ 3,6.10⁻⁷ mol.L⁻¹
4) Cm(Fe²⁺) = 2,0.10⁻⁴ g.L⁻¹ = 2,0.10⁻¹ mg.L⁻¹ soit 0,20 mg.L⁻¹
donc < à 0,30 mg.L⁻¹
