Chemical reaction: SO₄²⁻ + Ba²⁺ → BaSO₄.
m(sample) = 1,543 g.
m(BaSO₄) = 0,2243 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,2243 g ÷ 233,4 g/mol.
n(BaSO₄) = 0,00096 mol.
n(BaSO₄) = n(SO₄²⁻).
ω(SO₄²⁻) = m(SO₄²⁻) ÷ m(sample).
ω(SO₄²⁻) = 0,00096 mol · 96 g/mol ÷ 1,543 g.
ω(SO₄²⁻) = 0,059 = 5,9%.
Is there any answer so I can answer it
Answer:
C
Explanation:
In the presence of a catalyst, both the forward and reverse reaction rates will speed up equally. :)
Answer:
pH of the buffer is 10.10
Explanation:
trimethylamine is a weak base that, in presence with its conjugate base, trimethylammonium ion, produce a buffer.
To determine the pH of the buffer we use H-H equation for weak bases:
pOH = pKb + log [Conjugate acid] / [Weak base]
<em>pKb is -log Kb = 4.20</em>
<em />
pOH = 4.20 + log [N(CH₃)₃] / [NH(CH₃)₃]
Replacing the concentrations of the problem:
pOH = 4.20 + log [0.20M] / [0.40M]
pOH = 3.90
As pH = 14 -pOH
<h3>pH of the buffer is 10.10</h3>
<em />
