Answer:
398 mL
Explanation:
Using the equation for molarity,
C₁V₁ = C₂V₂ where C₁ = concentration before adding water = 8.61 mol/L and V₁ = volume before adding water, C₂ = concentration after adding water = 1.75 mol/L and V₂ = volume after adding water = 500 mL = 0.5 L
V₂ = V₁ + V' where V' = volume of water added.
So, From C₁V₁ = C₂V₂
V₁ = C₂V₂/C₁
= 1.75 mol/L × 0.5 L ÷ 8.61 mol/L
= 0.875 mol/8.61 mol/L
= 0.102 L
So, V₂ = V₁ + V'
0.5 L = 0.102 L + V'
V' = 0.5 L - 0.102 L
= 0.398 L
= 398 mL
So, we need to add 398 mL of water to the nitric solution.
Answer:
Ksp = 1.07x10⁻²¹
Explanation:
Molar solubility is defined as moles of solute can be dissolved in 1L.
Ksp for NiS is defined as:
NiS(s) ⇄ Ni²⁺(aq) + S²⁻(aq)
Ksp = [Ni²⁺] [S²⁻]
As molar solubility is 3.27x10⁻¹¹M, concentration of [Ni²⁺] and [S²⁻] is 3.27x10⁻¹¹M for both.
Replacing:
Ksp = [3.27x10⁻¹¹M] [3.27x10⁻¹¹M]
<em>Ksp = 1.07x10⁻²¹</em>
<em></em>
Answer:
1.209g of MgO participates
Explanation:
In this problem, we have 0.030 moles of MgO that participates in a particular reaction.
And we are asked to solve for the mass of MgO that participates, that means, we need to convert moles to grams.
To convert moles to grams we need to use molar mass of the compound:
<em>1 atom of Mg has a molar mass of 24.3g/mol</em>
<em>1 atom of O has a molar mass of 16g/mol</em>
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That means molar mass of MgO is 24.3g/mol + 16g/mol = 40.3g/mol
And mass of 0.030 moles of MgO is:
0.030 moles MgO * (40.3g/mol) =
<h3>1.209g of MgO participates</h3>
Explanation:
pls, refer to the above picture, i hope you will find it helpful.
