Amine is the class of organic compounds that contains nitrogen
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_Mg + _HCL = _MgCl2 + H2
Separate the terms on each side:
_Mg + _HCl = _MgCl2 + H2
Mg- 1 Mg-1
H-1 H-2
Cl-1 Cl-2
Mg is balanced on both sides so move on to the next (put a 1 in the space).
1Mg
There are two H's and two Cl's on the results side, so to balance the equation put a 2 as a coefficient for HCl and it'll all balance out.
2HCl
Balamced equation will be:
1Mg + 2HCL = 1MgCl2 + H2
If you start with 0.30 m Mn₂ , at 12.5 pH, free Mn₂ concentration be equal to 4.6 x 10⁻¹¹ m
Initial molarity of Mn₂ = 0.30 M
Final molarity of Mn₂ = 4.6 x 10⁻¹¹
pH = ?
Ksp [Mn(OH)₂] = 4.6 x 10⁻¹⁴ (standard value)
Write the ionic equation
Mn(OH)₂ → Mn⁺² + 2OH⁻
[Mn⁺²] = 4.6 x 10⁻¹¹
We will calculate the concentration of OH⁻ by using Ksp expression
Ksp = [Mn⁺²][OH-]²
[Mn⁺²][OH⁻]² = 4.6 x 10⁻¹⁴
[OH⁻]² = 4.6 x 10⁻¹⁴ / 4.6 x 10⁻¹¹
[OH⁻]² = 10⁻³
[OH⁻] = (10⁻³)¹⁽²
[OH⁻] = 0.0316 M
Calculate the pOH
pOH = -log [OH⁻]
pOH = -log [0.0316]
pOH = 1.5
Now calculate pH
pH = 14 - pOH
pH = 14 - 1.5
pH = 12.5
You can also learn about molarity from the following question:
brainly.com/question/14782315
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