The reason for adding a limited amount and then an excess amount is that initially a metal hydroxide may form which becomes soluble when more base is added and the metal complex forms.
In qualitative analysis is a common to add the base in drops and then in excess. When added in drops, the metal hydroxide is formed. This metal hydroxide is often insoluble.
After this metal hydroxide is formed, the base could be added in excess such that the metal hydroxide dissolves in the excess base by forming a complex.
For instance;
CuCl2(aq) + 2NaOH(aq) -------> Cu(OH)2(s) + 2NaCl(aq)
Cu(OH)2(s) + 2OH^-(aq) -------> [Cu(OH)4]^2+(aq)
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Answer:
I think it would be C but don't quote me on it
Explanation:
It adds another obstacle for the current to go through
Letter D.
The scarier the better, and many people like loops
706.1652 g is the molar mass of 4.68 mol of sodium bromate.
Given:
Sodium bromate is used in a mixture which dissolves gold from its ores.
Moles of sodium bromate= 4.68 mol
Molar mass of sodium bromate = 150.89 g/mol number of moles × mol
= 150.89 × 4.68
= 706.1652 g
Therefore, 706.1652 g is the molar mass of 4.68 mol of sodium bromate.
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