Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
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.
I believe the answer is 3
though alchemists were often superstitious, they left a rich legacy of modern chemists. what was their main contribution-
Explanation:
they were the first to preform experiments.
No se ha dicho nada de nada y que ha dicho la policía
