Answer:
Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + NO₃⁻(aq) → PbBr₂(s) + NO₃⁻(aq) + Li⁺(aq)
A precipitate forms
Explanation:
To solve this question we must know the general solubility rules:
All group 1A (Li, Na, K...) and NH₄⁺ ions are always soluble.
The nitrates (NO₃⁻) and acetates are always solubles.
Cl⁻, Br⁻ and I⁻ are soluble except in the presence of Ag⁺, Pb²⁺, Cu⁺ and Hg₂²⁺
That means in the mixture of ions that we have, the Br⁻ will react with Pb²⁺ to produce PbBr₂(s):
<h3>Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + NO₃⁻(aq) → PbBr₂(s) + NO₃⁻(aq) + Li⁺(aq)
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That is a tetrahedral arrangement, with an angle of 109.5°. Nothing changes in terms of the shape when the hydrogen atoms combine with the carbon, and so the methane molecule is also tetrahedral with 109.5° bond angles.
To determine the pH of a weak base, we use the equation:
pH = 14 + 0.5 log Kb
Therefore,
pH = 14 + 0.5 log 3.25
pH = 14.26
A weak base is a base which does not fully dissociates into ions when in solution. The solution would contain cations, anions and the compound itself. Hope this helps.
Gold cyanidation ( which is also known as the cyanide process or the MacArthur-Forrest process) is a hydro-metallurgical technique for extracting gold from low-grade ore by converting the gold to a water-soluble coordination complex. It is the most commonly used leaching process for gold extraction
