Answer:
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The reaction is:
<span>4Li(s) + O2 (g) = 2Li+ + 2O-2(s).
The oxidizing agent is the one that is being reduced which is oxygen where the charge changed from neutral to -2 while the reducing agent is the on being oxidized which is lithium where the charge change from neutral to +1.</span>
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The answer is [OH⁻] = 1 x 10⁻⁸.
To find OH⁻, divide the ionic product of water by [H₃O⁺] as :
<u>OH⁻ + H₃O⁺ = H₂O</u>
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- [OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻⁶
- [OH⁻] = 1 x 10⁻⁸
