Answer:
B). Precision.
Explanation:
During the course of measurements, two important factors are been seen to play key part which are accuracy and also precision. Precision here can reflect to the closeness of an answer towards the main value even though accuracy can show that too. Also it can show series of values been reproduced in the measurement flow; though in some cases can be a bit far from the actual value. Also its values are swen to be different because of repeatition and alsio in some cases, observational errors too.
This means that the reaction is in dynamic equilibrium.
Hope this helps because I don't really know the question you asked :)))
Answer:
Here's what I get
Explanation:
Assume you have 0. 1 mol·kg⁻¹ solutions of A) KCl, B) CH₃OH, C) Ba(OH)₂, and D) CH₃COOH.
The formula for freezing point depression is

Where
b = the molal concentration
K_f = the freezing point depression constant
i = the van't Hoff i factor
The i-factor is the number of solute particles produced by one formula unit of the substance.
The only difference in the solutions is the i-factor of the solutes.
A) KCl
KCl(aq) ⟶ K⁺(aq) +Cl⁻(aq); i = 2
B) CH₃OH
Methanol is a nonelectrolyte.
CH₃OH(aq) ⟶ CH₃OH(aq); i = 1
C) Ba(OH)₂
Ba(OH)₂(aq) ⟶ Ba²⁺(aq) + 2OH⁻(aq); i = 3
D) CH₃COOH
CH₃COOH(aq) ⇌ CH₃COO⁻(aq)+ H⁺(aq); i ⪆ 1
The Ba(OH)₂ solution has the greatest i-value. It therefore has the greatest freezing point depression and the lowest freezing point.
Charge = 1200 * 1.6 * 10^-19 = 1920 * 10^-19
Time = 3.5 µs = 3.5 * 10^-6 s
But Current = charge/time = (1920 * 10^-19) / (3.5 * 10^-6)
So Current becomes 548.5714 * 10^(-19-(-6))
So we have 548.5714 * 10^(-19 + 6) = 548.5714 * 10^(-13)
So we have 5.485714 * 10^(-15)