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
The 2nd one I think but I need some points
I think it’s the third option but I’m not entirely sure
Mn₂O
Explanation:
The oxide that will most likely form colored solutions is Mn₂O.
This is because most transition metals form colored compounds. Manganese is a transition metal belonging to the d-block on the periodic table.
- Other examples of transition metals are scandium, titanium, iron, copper, cobalt, nickel, zinc
- They belong to the d-block on the periodic table.
- They have variable oxidation states.
- Most of their solutions are always colored.
Learn more:
Periodic table brainly.com/question/8543126
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