Answer:
Zn =⇒ Zn+2(0.10) + 2e- (anode)
Zn+2(?M) + 2e- === Zn(s) (cathode)
Zn + Zn+2(?M) ===⇒ Zn+2(0.10) + Zn
E = E^o -0.0592 log Q; in this case E^o is zero.
E = - 0.0592 /n logQ where n is the number of electrons transferred, in this
case n = 2
23 mV x 1 volt/1000mv = 0.023 Volts
0.023 = -0.0592 / 2 log(0.10) / [Zn+2]
0.023 = -0.0296 { log 0.10 – log [Zn+2] }
0.023 = -0.0296{ -1 - log[Zn+2] }
0.023 = +0.0296 + 0.0296log[Zn+2]
-0.0066 = 0.0296log[Zn+2]
-0.22= log[Zn+2]
[Zn+2] = 10^-0.22 = 0.603 Molar
Answer:
El termopar B presenta un mayor grado de dispersión y también es más preciso. ... (c) La estimación para T = 175 ° C es probablemente la más cercana al valor real, porque el ... (cm3). Flujo de masa. Velocidad. (kg / min). Diferencia. Duplicar. (Di). Yo y yo. 2. 1 ... atm de gas. 2. 2. 2 f. 3. 2 f f. 30 14,7 lb 20 pulg. 4 14,7 lb 24 pulg 392 lb 7,00 10 lb pulg.
The statement is true in this situation is C. The size of Ffric is the same as the size of Fapp:
From the diagram, since the body is in equilibrium, the sum of vertical forces equals zero. Also, the sum of horizontal forces equal zero.
So, ∑Fx = 0 and ∑Fy = 0
Since Fapp acts in the negative x - direction and Ffric acts in the positive x - direction,
∑Fx = -Fapp + Ffric = 0
-Fapp + Ffric = 0
Fapp = Ffric
Also, since Fgrav acts in the negative y - direction and Fnorm acts in the positive y - direction,
∑Fy = Fnorm + (-Fgrav) = 0
Fnorm - Fgrav = 0
Fnorm = Fgrav
So, we see that the size of Fapp <u>equals</u> size of Ffric and the size of Fnorm <u>equals</u> the size of Fgrav.
So, the correct option is C
The statement which is true in this situation is C. The size of Ffric is the same as the size of Fapp.
Learn more about equilibrium of forces here:
brainly.com/question/12980489
A as the lower temperature in Celsius corresponds to a lower Kelvin temperature thus reducing movement
