Answer:
MgO- magnesium oxide
Cu(NO3)2- copper(11)nitrate
Li2CO3- lithium carbonate
Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
When an acid is neutralized by a base, that means moles of H+ = moles of OH-
moles of H+ = 0.5 M * 0.025 L HCl = 0.0125 moles H+
moles of OH- should be equal to 0.0125 moles, so
0.0125 moles = (x) * 0.025 L NaOH
x is the concentration of NaOH, which we want to find.
x = 0.5 M
The correct answer is C) 0.5 M.
Answer: How many moles of HCl was produced?
⇒ 0.261 moles of HCl
How many moles of MgCl2 reacted?
⇒ 0.131 moles of MgCl2
What mass of MgCl2 reacted?
⇒ 12.4 g MgCl2
Explanation:
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