Gold (III) nitrate in an aqueous solution is hydrolyzed with formation of gold (III) metahydroxide.
Au(NO₃)₃ → Au³⁺(aq) + 3NO₃⁻(aq)
Au³⁺ + H₂O ⇄ AuOH²⁺ + H⁺
AuOH²⁺ + H₂O ⇄ Au(OH)₂⁺ + H⁺
Au(OH)₂⁺ + H₂O → AuOOH·H₂O(s) + H⁺
Au(NO₃)₃(aq) + 2H₂O(l) = AuOOH(s) + 3HNO₃(aq)
Answer:
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Explanation:
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The correct answer is <span>Fusion Curve</span>
Answer: pH of HCl =5, HNO3 = 1,
NaOH = 9, KOH = 12
Explanation:
pH = -log [H+ ]
1. 1.0 x 10^-5 M HCl
pH = - log (1.0 x 10^-5)
= 5 - log 1 = 5
2. 0.1 M HNO3
pH = - log (1.0 x 10 ^ -1)
pH = 1 - log 1 = 1
3. 1.0 x 10^-5 NaOH
pOH = - log (1.0 x 10^-5)
pOH = 5 - log 1 = 5
pH + pOH = 14
Therefore , pH = 14 - 5 = 9
4. 0.01 M KOH
pOH = - log ( 1.0 x 10^ -2)
= 2 - log 1 = 2
pH + pOH = 14
Therefore, pH = 14 - 2 = 12
Answer:
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Explanation: