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Michonne15
@Michonne15
January 2021
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Bonsoir, j'ai un exercice de Physique à rendre rapidement!!! J'ai besoin de votre aide, s'il vous plaît! Je vous remercie infiniment... Je n'y arrive pas.
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scoladan
Verified answer
Bonjour,
Quantités de matière initiales
n(Cu2+) = C(Cu2+) x V = 2,0 x 50,0.10^-3 = 1,0.10^-1 mol
m(Al) = 5,0 g
M(Al) = 27,0 g.mol^-1
==> n(Al) = m(Al)/M(Al) = 5,0/27,0 = 1,85.10^1 mol
Tableau d'avancement
3Cu2+(aq) + 2Al(s) --> 3Cu(s) + 2Al3+(aq)
Etat Avanct.
initial x=0 0,1 0,185 0 0
en cours x 0,1-3x 0,185-2x 3x 2x
final xmax 0,1-3xmax 0,185-2xmax 3xmax 2xmax
0,1 - 3xmax = 0 ==> xmax = 0,1/3 = 0,03
0,185 -2xmax = 0 ==> xmax = 0,185/2 = 0,09
==> xmax = 0,03 mol
==> A l'état final, n(Cu2+) = 0
==> La couleur bleue initiale a disparu
2) n(Cu2+) = 0
n(Al3+) = 2xmax = 0,06 mol
==> C(Cu2+) = 0 et C(Al3+) = n(Al3+)/V = 0.06/50.10^-3 = 1,33 mol.L^1
1 votes
Thanks 1
Michonne15
Pourquoi la couleur bleue disparait-elle?
scoladan
Parce que tous les ions Cu2+ ont disparu et que ce sont les seuls à donner une coloration à la solution.
Michonne15
Ah d'accord :) Merci! Je t'en suis reconnaissante de m'aider dans les devoirs de physique que je poste^^
scoladan
pas de souci. Bonne soirée
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Verified answer
Bonjour,Quantités de matière initiales
n(Cu2+) = C(Cu2+) x V = 2,0 x 50,0.10^-3 = 1,0.10^-1 mol
m(Al) = 5,0 g
M(Al) = 27,0 g.mol^-1
==> n(Al) = m(Al)/M(Al) = 5,0/27,0 = 1,85.10^1 mol
Tableau d'avancement
3Cu2+(aq) + 2Al(s) --> 3Cu(s) + 2Al3+(aq)
Etat Avanct.
initial x=0 0,1 0,185 0 0
en cours x 0,1-3x 0,185-2x 3x 2x
final xmax 0,1-3xmax 0,185-2xmax 3xmax 2xmax
0,1 - 3xmax = 0 ==> xmax = 0,1/3 = 0,03
0,185 -2xmax = 0 ==> xmax = 0,185/2 = 0,09
==> xmax = 0,03 mol
==> A l'état final, n(Cu2+) = 0
==> La couleur bleue initiale a disparu
2) n(Cu2+) = 0
n(Al3+) = 2xmax = 0,06 mol
==> C(Cu2+) = 0 et C(Al3+) = n(Al3+)/V = 0.06/50.10^-3 = 1,33 mol.L^1