When the first reaction equation is:
AgI(S) ↔ Ag+(Aq) + I-(Aq)
So, the Ksp expression = [Ag+][I-]
∴Ksp = [Ag+][I-] = 8.3 x 10^-17
Then the second reaction equation is:
Ag+(aq) + 2NH3(aq) ↔ Ag(NH3)2+
So, Kf expression = [Ag(NH3)2+] / [Ag+] [NH3]^2
∴Kf = [Ag(NH3)2+] /[Ag+] [NH3]^2 = 1.7 x 10^7
by combining the two equations and solve for Ag+:
and by using ICE table:
AgI(aq) + 2NH3 ↔ Ag(NH3)2+ + I-
initial 2.5 0 0
change -2X +X +X
Equ (2.5-2X) X X
so K = [Ag(NH3)2+] [I-] / [NH3]^2
Kf * Ksp = X^2 / (2.5-2X)
8.3 x 10^-17 * 1.7 x10^7 = X^2 / (2.5-2X) by solving for X
∴ X = 5.9 x 10^-5
∴ the solubility of AgI = X = 5.9 x 10^-5 M
Answer:
the conversion of a vapor or gas to a liquid
Explanation:
The cell notation of the redox reaction between Cu and ClO₃⁻ is:
<h3>What is a voltaic cell?</h3>
A voltaic cell or electrochemical cell is a device which produces electricity from chemical reactions occuring within the cell.
The reactions occuring in a voltaic cell are redox reactions.
Oxidation occurs at the anode while reduction occurs at the cathode.
In the given reaction between Cu and ClO₃⁻ as shown below:
- ClO₃⁻ (aq) + 3 Cu (s) + 6H⁺ (aq) → Cl⁻ (aq) + 3 Cu²⁺ (aq) + 3 H₂O (l)
Cu is oxidized as follows: 3 Cu (s) → 3 Cu²⁺
ClO₃⁻ is reduced as follows: ClO₃⁻ (aq) → Cl⁻ (aq)
The voltaic cell notation is as follows where Pt is used as an inert electrode in contact with the ClO₃⁻ and Cl⁻:
Therefore, the cell notation of the redox reaction between Cu and ClO₃⁻ shows that Cu is oxidized while ClO₃⁻ is reduced.
Leran more about voltaic cells at: brainly.com/question/3930479
Answer: As a result, each excited electron in an atom emits a photon of a specific wavelength. To put it another way, each excited noble gas emits a distinct hue of light. This is a reddish-orange neon light.
Answer:A) mL / s
Explanation:This is the amount of milliliters per second