Reduction Reaction:
Ag⁺ + 1e⁻ → Ag ------ (1)
Oxidation Reaction:
NO₂⁻ → NO₃⁻ + 2e⁻
Balance Oxygen atoms, by adding H₂O to side having less number of O atoms, and H⁺ to side having greater number of O atoms.
NO₂⁻ + H₂O → NO₃⁻ + H⁺ + 2e⁻
Balance the H⁺'s,
NO₂⁻ + H₂O → NO₃⁻ + 2 H⁺ + 2e⁻ ------ (2)
Multiply eq. 1 by 2 to balance the number of electrons in overall reaction.
2 Ag⁺ + 2e⁻ → 2 Ag ----- (3)
Compare eq. 2 and 3,
NO₂⁻ + H₂O → NO₃⁻ + 2 H⁺ + 2e⁻
2 Ag⁺ + 2e⁻ → 2 Ag
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NO₂⁻ + H₂O + 2 Ag⁺ → NO₃⁻ + 2 Ag + 2 H⁺
The weight of a bowling ball is 6.61 pounds.
O: palisade layer and please answer the question I am gonna post right now pleaseeee
A typical triple bond consists of one sigma and two pi bonds.
Answer:
T final = 80°C
Explanation:
∴ Q = 18000 cal
∴ m H2O = 300 g
∴ Cp H2O (15°C) = 0.99795 cal/g.K ≅ 1 cal/g.K
∴ T1 = 20°C = 293 K
∴ T2 = ?
⇒ 18000 cal = (300 g)(1 cal/g.K)(T2 - 293 K)
⇒ (18000 cal)/(300 cal/K) = T2 - 293 K
⇒ T2 = 293 K + 60 K
⇒ T2 = 353 K (80°C)