NaOH is a strong base and complete dissociation into Na⁺ and OH⁻ ions.
Therefore [NaOH] = [OH⁻]
To calculate the [OH⁻], we can first find the pOH as NaOH is a basic solution.
pH + pOH = 14
Since pH = 11.50
pOH = 14 - 11.50
pOH = 2.50
We can calculate [OH⁻] by knowing pOH
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 3.2 x 10⁻³ M
therefore [NaOH] = 3.2 x 10⁻³ M
Answer:
isotope 2
Explanation:
it has the highest percentage abundance
1,000 mL is the same as 10 dL.
Answer is: 4.02 grams of water are required.
Chemical reaction: BaH₂ + 2H₂O → Ba(OH)₂ + 2H₂.
Ideal gas law: p·V = n·R·T.
p = 755 mm Hg ÷ 760.0 mmHg / atm = 0.993 atm.
T = 25 + 273.15 = 298.15 K.
V(H₂) = 5.50 L.
R = 0,08206 L·atm/mol·K.
n(H₂) = 0.993 atm · 5.5 L ÷ 0,08206 L·atm/mol·K · 298.15 K.
n(H₂) = 0.223 mol.
From chemical reaction: n(H₂O) : n(H₂) = 1 : 1.
n(H₂O) = 0.223 mol.
m(H₂O) = 0.223 mol · 18 g/mol.
m(H₂O) = 4.02 g.
