Answer: 4.1 g of barium precipitated.
Explanation:
According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 
of particles.
To calculate the moles, we use the equation:
Given : moles of barium = 0.030
Molar mass of barium = 137 g/mol
x= 4.1 g
Thus there are 4.1 g of barium that precipitated.
We need to first find the molarity of Ba(OH₂) solution.
A mass of 3.24 mg is dissolved in 1 L solution.
Ba(OH)₂ moles dissolved - 3.24 x 10⁻³ g/171.3 g/mol = 1.90 x 10⁻⁵ mol
dissociaton of Ba(OH)₂ is as follows;
Ba(OH)₂ --> Ba²⁺ + 2OH⁻
1 mol of Ba(OH)₂ dissociates to form 2OH⁻ ions.
Therefore [OH⁻] = (1.90 x 10⁻⁵)x2 = 3.8 x 10⁻⁵ M
pOH = -log[OH⁻]
pOH = -log (3.8 x 10⁻⁵)
pOH = 4.42
pH + pOH = 14
therefore pH = 14 - 4.42
pH = 9.58
In ionic bond there is the formation of ions due to transfer of electrons from one atom<span> to </span>the other. Normally, at this link, there is an element that tends to yield electrons (metal-cation), and one that tends to receive electrons (not metal-anion).
Note: the Ionic bond is the only in the transfer of electrons<span>.</span>
Explanation:
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