4 g of ag2so4 will dissolve in 1l of water. calculate the solubility product (ksp) for silver (i) sulfate.
1 answer:
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.
You might be interested in
It produces nitrogen gas and water
4NH3+3O2+heat 2N2+6H 2O
A catalyst
A catalyst can be in many forms
Answer:
C15 H31 O4 S
Explanation:
molecular formula is also the same because the value of "n" is 1
Answer ( 3 ) :
<span>conversion of matter to energy .
</span>
hope this helps!
They become more stable because they achieve a full outer shell of valence electrons with the magic number of 8.