Answer:
Explanation:
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In this case, since this acid-base neutralization is performed in a 1:2 mole ratio of acid to base as the former is a diprotic acid (two hydrogen ions in the molecule), we can write the following equation:
In such a way, we can solve for the molarity of the acid, given the molarity and concentration of the NaOH base and the volume of the acid:
Thus, we plug in the given data to obtain:
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Answer:
0.0006 mole
Explanation:
Considering:
Or,
Given :
For HCl :
Molarity = 0.100 M
Volume = 6.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 6.0×10⁻³ L
Thus, moles of potassium iodide :
Moles of HCl = 0.0006 moles
From the reaction shown below:-
1 mole of HCl reacts with 1 mole of sodium carbonate.
So,
0.0006 mole of HCl reacts with 0.0006 mole of sodium carbonate.
<u>Moles of sodium carbonate = 0.0006 moles</u>
Answer:
1.930 * 10⁻⁹ mg of Mn⁺² are left unprecipitated.
Explanation:
The reaction that takes place is:
Mn⁺² + S⁻² ⇄ MnS(s)
ksp = [Mn⁺²] [S⁻²]
If the pksp of MnS is 13.500, then the ksp is:
From the problem we know that [S⁻²] = 0.0900 M
We use the ksp to calculate [Mn⁺²]:
3.1623*10⁻¹⁴= [Mn⁺²] * 0.0900 M
[Mn⁺²] = 3.514 * 10⁻¹³ M.
Now we can calculate the mass of Mn⁺², using the volume, concentration and atomic weight. Thus the mass of Mn⁺² left unprecipitated is:
3.514 * 10⁻¹³ M * 0.1 L * 54.94 g/mol = 1.930 * 10⁻¹² g = 1.930 * 10⁻⁹ mg.