- Standard reduction potential of Ag/Ag⁺ is 0.80 v and that of Cu⁺²(aq)/Cu⁰ is +0.34 V.
- The couple with a greater value of standard reduction potential will oxidize the reduced form of the other couple.
Ag⁺ will be reduced to Ag(s) and Cu⁰ will be oxidized to Cu²⁺
Anode reaction: Cu⁰(s) → Cu²⁺ + 2 e⁻ E⁰ = +0.34 V
Cathode reaction: Ag⁺(aq) + e → Ag(s) E⁰ = +0.80 V
Cell reaction: Cu⁰(s) + 2 Ag⁺(aq) → Cu⁺²(aq) + 2 Ag⁰(s)
E⁰ cell = E⁰ cathode + E⁰ anode
= 0.80 + (-0.34) = + 0.46 V
Answer:
Henry Moseley
Explanation: Passed with 98
Below is an attachment of the Lewis structure with the lowest formal charges.
The formal charge is the fictitious charge that an atom in a molecule would have if the electrons in the bonds were evenly distributed among the atoms. The nonbonding electrons on a neutral atom are subtracted from its valence electron count, which is then followed by the number of bonds that bind it to other atoms in the Lewis structure, to get the formal charge. This is another way to put it. When hyponitrous acid is oxidized in the atmosphere, nitric and nitrous acids are produced. By reducing a nitrate or nitrite by sodium amalgam in the presence of water, hyponitrite salts have been created.
Learn more about formal charge here-
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The correct answer for the question that is being presented above is this one: "They are both balanced chemical equations." The <span>statement that is correct about the chemical reaction is that both chemical reaction are balanced chemical equations.</span>
Answer:
Mass = 15.20 g of KCl
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
2 KClO₃ = 2 KCl + 3 O₂
Step 1: Calculate moles of KClO₃ as;
Moles = Mass / M/Mass
Moles = 25.0 g / 122.55 g/mol
Moles = 0.204 moles
Step 2: Find moles of KCl as;
According to equation,
2 moles of KClO₃ produces = 2 moles of KCl
So,
0.204 moles of KClO₃ will produce = X moles of KCl
Solving for X,
X = 2 mol × 0.204 mol / 2 mol
X = 0.204 mol of KCl
Step 3: Calculate mass of KCl as,
Mass = Moles × M.Mass
Mass = 0.204 mol × 74.55 g/mol
Mass = 15.20 g of KCl
