Answer:
E = 1.602v
Explanation:
Use the Nernst Equation => E(non-std) = E⁰(std) – (0.0592/n)logQc …
Zn⁰(s) => Zn⁺²(aq) + 2 eˉ
2Ag⁺(aq) + 2eˉ=> 2Ag⁰(s)
_____________________________
Zn⁰(s) + 2Ag⁺(aq) => Zn⁺²(aq) + 2Ag(s)
Given E⁰ = 1.562v
Qc = [Zn⁺²(aq)]/[Ag⁺]² = (1 x 10ˉ³)/(0.150)² = 0.044
E = E⁰ -(0.0592/n)logQc = 1.562v – (0.0592/2)log(0.044) = 1.602v
Mitosis hope that helps i tried:)
Answer:
- <em>The net charge of the ionic compound calcium fluoride is </em><u><em>zero (0).</em></u>
<em>Explanation:</em>
<em>Ionic compounds,</em> such as covalent ones, have zero net charge; this is, they are neutral.
Substances with net positive charge are cations and substances with net negative charge are anions.
The charges in the <em>ionic compound calcium flouride</em> are distributed in this way:
- Calcium charge: Ca²⁺: this is, each calcium ion has a 2 positive charge
- Fluoride charge: F⁻: each fluoride ion has a 1 negative charge.
- Then, the <em>net charge</em> is: 1 × (2+) + 2 × (1-) = +2 - 2 = 0.
So, a two positve charge, from one calcium ion, is equal to two negative charges, from two fluoride tions, yielding a <u>zero net charge</u>.
if there is no carbon dioxide your test tube will be blue
if there is a medium amount of carbon dioxide your test tube is green
if there are high amounts of CO2 it will be
yellow
