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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<span>when the number of moles Ca = mass of Ca / molar mass of Ca.
and we can get the molar mass of Ca, it is = 40 g/mol
and we have already the mass of Ca (given) = 9.8 g
so, by substitution: the moles Ca = 9.8 g / 40 g/mol
= 0.245 moles</span>
<span> When an </span>acid and a base<span> are placed together, they </span>react<span> to neutralize the </span>acid<span> and </span>base<span> properties, producing a salt. The H(+) cation of the </span>acid<span>combines with the OH(-) anion of the </span>base<span> to form water.</span>
Answer:8
in the respiratory system
a.
Explanation:
Answer:
Correct option is
B
5 liters of CH
4
(g)NO
2
at STP
No. of molecules=
22.4
5
mol=
22.4
5
×N
A
molecules
A) 5ℊ of H
2
(g)
No. of moles=
2
5
mol=
2
5
×N
A
molecules
B) 5l of CH
4
(g)
No. of moles of CH
4
=
22.4
5
mol=
22.4
5
N
A
molecules
C) 5 mol of O
2
=5N
A
O
2
molecules
D) 5×10
23
molecules of CO
2
(g)
Molecules of 5l NO
2
(g) at STP=5l of CH
4
(g) molecules at STP
Therefore, option B is correct.