Articles
Register
Sign In
Search
Michonne15
@Michonne15
January 2021
1
102
Report
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.
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
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
More Questions From This User
See All
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Michonne15
January 2021 | 0 Respostas
Responda
Recomendar perguntas
maceocoulon
November 2022 | 0 Respostas
lblbllbl
November 2022 | 0 Respostas
naelysvannier311
November 2022 | 0 Respostas
PikaS0ra
October 2022 | 0 Respostas
andronight67
October 2022 | 0 Respostas
kim036623
October 2022 | 0 Respostas
Tocam
October 2022 | 0 Respostas
leoniecch
October 2022 | 0 Respostas
Bobjour, pourriez vous m'aider pour le tableau
lillytrsn
October 2022 | 0 Respostas
charles5022
October 2022 | 0 Respostas
×
Report "Bonsoir, j'ai un exercice de Physique à rendre rapidement!!! J'ai besoin de votre aide, s'il vous pl.... Pergunta de ideia de Michonne15"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
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