TSI1 PARTIE CHIMIE Problème 1 : Le cuivre métallique

8×1
8+ 6 ×1 2= 4
aCu = 361,5pm
M(Cu) = 63,5g.mol−1
ρCu =4M(Cu)
NA
a3
Cu ∼9000 kg/m3
CuxO
O2−
8×1
8+ 1 = 2
4
Cu2O
Cu+
• •

S0
F e2+
CH3CH2OH +H2O*
)CH3COOH + 4H++ 4e−
Cr2O2−
7+ 14H++ 6e−*
)2Cr3+ + 7H2O
3CH3CH2OH + 2Cr2O2−
7+ 16H+= 3CH3COOH + 4Cr3+ + 11H2O
F e2+
Cr2O2−
7+ 14H++ 6e−*
)2Cr3+ + 7H2O
F e2+ *
)F e3+ +e−
Cr2O2−
7+ 14H++ 6F e2+ = 2Cr3+ + 6F e3+ + 7H2O
ndS1
Cr2O2−
76F e2+ 14H+2Cr3+ 6F e3+ 7H2O
ndc2V2
nd−x c2V2−6x2x6x
nd−xeq c2Veq −6xeq 2x6x
c2V2−6nd= 0
nd=c2Veq
6
• •

na
S1
3CH3CH2OH 2Cr2O2−
716H+3CH3COOH 4Cr3+ 11H2O
nac1V1
na−3x c1V1−2x3x4x
na−3xf= 0 c1V1−2xf=nd2x6x
CH3CH2OH
na−3xf= 0 c1V1−2xf=nd
na=3
2(c1V1−c2Veq 6)
cT=na
VT
= 0,177 mol/L
c0= 10cT= 1,77 mol/L
c0
20◦C
c0.M(CH3CH2OH) = 81,42 g/L
10,3g/100g10,3◦
• •

H+
(aq)/H2(g)O2(g)/H2O(l)
H2(g)*
)2H++ 2e−
O2(g)+ 4H++ 4e−*
)2H2O(l)
H2(g)O2(g)
2H2(g)+O2(g)= 2 H2O(l)
O2
∆t= 192 h
I= 300 A
Q=I∆t=nF
n=300 ×192
96500 = 0,6mol
0,3mol H20,15 mol O2
• •

h
R
WA−→B(
→
P) =
→
AB .
→
P=AB →
ex.mg sin(α)→ ex=ABmg sin(α) = mgh
α
∆Ec=1
2mv2
B−1
2mv2
A=WA−→B(
→
P) + WA−→B(
→
N)
⇐⇒ vB=p2gh
vB=vc
∆EC=WB−→C(
→
f) = ZC
B
fdx =fd
⇐⇒ 1
2mv2
C−1
2mv2
B=fd
⇐⇒ vC=r2fd
m+v2
B
vC=√2gh g
EM=cste
EM=Ec+EP=1
2mv2+mgz =cste v z
• •