The low melting point of the wire is utilized in making electric fuse.
27 km = 27000m. Time = distance/speed = 27000/3x10^8 = 9x10^-5 s.
there are seven valence electrons<span> available of bonding in bromine(Br)</span>
Answer:
3.0x10⁻²M
Explanation:
Silver sulfate, Ag₂SO₄, has a product constant solubility equilbrium of:
Ag₂SO₄(s) ⇄ 2Ag⁺ + SO₄²⁻
When an excess of silver sulfate is added, some Ag₂SO₄ will react producing Ag⁺ and SO₄²⁻ until reach the equilbrium determined for the formula:
ksp = 1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
Assuming the Ag₂SO₄ that react until reach equilibrium is X, we can replace in Ksp expression:
1.4x10⁻⁵ = [Ag⁺]² [SO₄²⁻]
1.4x10⁻⁵ = [2X]² [X]
1.4x10⁻⁵ = 4X³
3.5x10⁻⁶ = X³
0.015 = X
As [Ag⁺] is 2X:
[Ag⁺] = 0.030 = 3.0x10⁻²M
The answer is:
<h3>3.0x10⁻²M</h3>