Problème 1 : Autour de l`élément cuivre

M(S) = 32,1g.mol−1M(O) = 16 g.mol−1
Z(Cu) = 29
Z
63Cu 65Cu
x y
x+y= 1
63x+ 65y= 63,45
x= 0,775 y= 0,225
63Cu 65Cu
Cu2+
Cu2+
m= 15,3g CuSO4(s)
CuSO4(s)−→ Cu2+
(aq)+SO2− 4(aq)
M(CuSO4(s)) = 63,45 + 4 ×16 + 32,1 = 159,55 g/mol
c=n
V=m MV =15,3 255,81.0,2= 0,48 mol/L
[Cu2+] = c= 0,48 mol/L
V2= 100 mL c2= [Cu2+] =
0,12 mol.L−1c
c.V =c2.V2
V=c2V2
c=0,12.100 0,48 = 25 mL
c
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g= 9,8m.s−2
f= 23 Hz λ =
v=λf = 23 =
v1λ < 0,5h
v1=rgλ
2π=r9,8.60 2π= 9,7m/s
T=λ
v1
= 6,2s
λ > 10 h
v2=pgh =p9,8×4 = 6,3m/s
λ=v.T = 39 m
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λ=v/f
T= 6 carreaux = 6×200×10−6= 1,2ms
f= 1/T = 1/(1,2.10−3) = 833 Hz
δt1= 100 µs
∆ϕ= 2πδt/T = 2π0,1/1,2 = π/6 = 0,52 rad
x3= 21 cm
λ= 2 ∗21 = 42 cm
v=λ×f
v= 42.10−2.833 = 350 m.s−1
δt2= 200 µs x2=v×δt2= 7 cm
x4=λ= 42 cm
T0= 5 ms f0= 200 Hz
f0= 192 Hz
f1= 394 Hz
fn= (n+ 1) ×f0n n
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θ=λ
a=650.10−9 0,28.10−3= 2,3.10−3rad = 0,13 ◦
θ
θ
tan θ=d
2D⇔d∼2Dθ = 2 ×3×2,3.10−3= 1,4cm
a0=2Dλ
d= 975 µm
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