Answer:
HCl, ya que la sustancia es una base que se debe titular con un ácido.
Explanation:
¡Hola!
En este caso, teniendo en cuenta la descripción inicial de la sustancia, la cual se torna violeta cuando se le agrega la fenolftaleína, es posible inderir que esta sustancia es una base con pH básico. Ahora bien, en torno a la especificación de un proceso de titulación, es claro que dicha base debe ser titulada con un ácido, y este caso, con HCl, ácido chlorhídrico, con el fin the alcanzar el punto de equivalencia.
¡Saludos!
Every redox reaction consists of two parts, the oxidation and the reduction. Each one separately is called a half - reaction. During the redox reaction there is a transfer of electrons from the substance being oxidized to the substance being reduced.
Answer:
Wouldn't rust because zinc will lose electrons more readily than iron and will therefore oxidize first.
Explanation:
This process whereby rusting of steel is prevented by coating the steel with a layer of zinc is known as galvanization.
Now, in this process, the steel object will be coated in a thin layer of zinc. This coating will prevent oxygen and water from reaching the underneath metal since the zinc will also act as a sacrificial metal.
Now, Zinc is used because it has a lower reduction potential than iron and thus it will get easily more oxidized than iron. Which means the zinc will lose electrons more readily than iron.
Also, since zinc has a lower reduction potential, it is therefore the more active metal. Thus, even if the zinc coating is scratched and the steel is exposed to moist air, the zinc will still get to oxidize before the iron.
Answer:
0.3793 M
Explanation:
The unknown metal is zinc. So the equation of the reaction is;
Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)
From Nernst equation;
E = E° - 0.0592/n log Q
[Cu2+] = 0.050179 M
n = 2
[Zn^2+] = ?
E = 1.074 V
E° = 0.34 - (-0.76) = 1.1 V
Substituting values;
1.074 = 1.1 - 0.0592/2 log [Zn^2+]/0.050179
1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179
-0.026 = -0.0296 log [Zn^2+]/0.050179
-0.026/-0.0296 = log [Zn^2+]/0.050179
0.8784 =log [Zn^2+]/0.050179
Antilog(0.8784) = [Zn^2+]/0.050179
7.558 = [Zn^2+]/0.050179
[Zn^2+] = 7.558 * 0.050179
[Zn^2+] = 0.3793 M
