Answer:
0.0295M
Explanation:
As you can see, in the mixture you have KSCN and other compounds. The KSCN in solution is dissolved in K⁺ ions and SCN⁻ ions. That means initial concentration of SCN⁻ ions is the same of KSCN, 0.0800M.
You are adding 35.0mL of this solution and the total volume of the mixture is 20.0mL + 35.0mL + 40.0mL = 95.0mL.
That means you are diluting your solution 95.0mL / 35.0mL = 2.714 times.
And the concentration of SCN⁻ is:
0.0800M / 2.714 =
<h3>0.0295M </h3>
<u>Answer:</u> The correct answer is saturated solution.
<u>Explanation:</u>
For the given options:
Dilute solutions are defined as the solutions in which solute particles are present in less very amount than the solvent particles.
Unsaturated solutions are defined as the solutions where more and more of solute particles can be dissolved in the given amount of solvent.
Saturated solutions are defined as the solutions where no more solute particles can be dissolved in the solvent. The concentration of the solute particles that can be dissolved in a solution is maximum.
Supersaturated solutions are defined as the solutions where more amount of solute particles are present than the solvent particles.
From the above information, we conclude that the given solution is saturated solution.
<span><span>4.2×1022</span>NA</span><span> N_{A} is the avagadro number</span>
Answer:
C.0.28 V
Explanation:
Using the standard cell potential we can find the standard cell potential for a voltaic cell as follows:
The most positive potential is the potential that will be more easily reduced. The other reaction will be the oxidized one. That means for the reactions:
Cu²⁺ + 2e⁻ → Cu E° = 0.52V
Ag⁺ + 1e⁻ → Ag E° = 0.80V
As the Cu will be oxidized:
Cu → Cu²⁺ + 2e⁻
The cell potential is:
E°Cell = E°cathode(reduced) - E°cathode(oxidized)
E°cell = 0.80V - (0.52V)
E°cell = 1.32V
Right answer is:
<h3>C.0.28 V
</h3>
<h3 />
pH: 1.14266750357
pOH: 12.8573324964
[H+]: 0.072
[OH-]: 1.38888888889E-13 acid
