Answer:
[OH-] = 6.17 *10^-10
Explanation:
Step 1: Data given
pOH = 9.21
Step 2: Calculate [OH-]
pOH = -log [OH-] = 9.21
[OH-] = 10^-9.21
[OH-] = 6.17 *10^-10
Step 3: Check if it's correct
pOH + pH = 14
[H+]*[OH-] = 10^-14
pH = 14 - 9.21 = 4.79
[H+] = 10^-4.79
[H+] = 1.62 *10^-5
6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14
Answer:
+1.03 V
Explanation:
The standard emf of the voltaic cell is the value of the standard potential of it, which is calculated by the standard reduction potential (E°).
The standard reduction potential is the potential needed for the reduction reaction happen, and it's determined by the reaction with the hydrogen cell (which has E° = 0.0V). The half-reactions of reduction of Ni⁺² and Ag⁺, are:
Ni⁺²(aq) + 2e⁻ → Ni(s) E° = -0.23 V
Ag⁺(aq) + e⁻ → Ag(s) E° = +0.80 V
The value is calculated by a spontaneous reaction, in which the cell with the greater E° is reduced (gain electrons), and the other is oxidized (loses electrons). So, Ag⁺ reduces.
emf = E°reduces - E°oxides
emf = 0.80 - (-0.23)
emf = +1.03 V
I think B
STEP-by STEP EXPLANATION
<em>This statement is false.</em> When you say full octet, it means that an element opts to have eight electrons in its valence shell to become stable. However, since a hydrogen atom only has 1 electron when neutral, it is impossible to reach a full octet. That is why Hydrogen is one of the exceptions to this octet rule.
Answer:
The minimum concentration of Cl⁻ that produces precipitation is 12.6M
Explanation:
The Ksp of PbCl₂ is expressed as:
PbCl₂(s) → Pb²⁺(aq) + 2Cl⁻(aq)
The Ksp is:
Ksp = 1.6 = [Pb²⁺] [Cl⁻]²
When Ksp = [Pb²⁺] [Cl⁻]² the solution begind precipiration.
A 0.010M Pb(NO₃)₂ is 0.010M Pb²⁺, thus:
1.6 = [0.010M] [Cl⁻]²
160 = [Cl⁻]²
12.6M = [Cl⁻]
<h3>The minimum concentration of Cl⁻ that produces precipitation is 12.6M</h3>
