The number of moles of Al₂(SO₄)₃ is 1.2 mol
<u>Explanation:</u>
Given:
Volume of the solution, V = 501 mL
V = 0.501L
Molarity of the solution, M = 2.37M
Moles, n = ?
We know:
Molarity = 
On substituting the values we get:

Therefore, the number of moles of Al₂(SO₄)₃ is 1.2 mol
There are many differences between ideal gas and real gas; some of the main differences are as following:
- An ideal gas follows the formula PV=nRT but a real gas does not always follow this formula.
- There is no attraction between the molecules of an ideal gas. A real gas has significant particle attractions.
- The particles of an ideal gas lose no energy to its container. A real gas conducts and radiates heat, thereby losing energy.
- An ideal gas is infinitely compressible, a real gas will condense to a liquid at some pressure.
- Real gas particles have a volume and ideal gas particles do not.
- Real gas particles collide in-elastically (loses energy with collisions) and ideal gas particles collide elastically.
Answer:
[Cu²⁺] = 2.01x10⁻²⁶
Explanation:
The equilibrium of Cu(CN)₄²⁻ is:
Cu²⁺ + 4CN⁻ ⇄ Cu(CN)₄²⁻
And Kf is defined as:
Kf = 1.0x10²⁵ = [Cu(CN)₄²⁻] / [Cu²⁺] [CN⁻]⁴
As Kf is too high you can assume all Cu²⁺ is converted in Cu(CN)₄²⁻ -Cu²⁺ is limiting reactant-, the new concentrations will be:
[Cu²⁺] = 0
[CN⁻] = 0.33M - 4×2.2x10⁻³ = 0.3212M
[Cu(CN)₄²⁻] = 2.2x10⁻³
Some [Cu²⁺] will be formed and equilibrium concentrations will be:
[Cu²⁺] = X
[CN⁻] = 0.3212M + 4X
[Cu(CN)₄²⁻] = 2.2x10⁻³ - X
<em>Where X is reaction coordinate</em>
<em />
Replacing in Kf equation:
1.0x10²⁵ = [2.2x10⁻³ - X] / [X] [0.3212M +4X]⁴
1.0x10²⁵ = [2.2x10⁻³ - X] / 0.0104858X + 0.524288 X² + 9.8304 X³ + 81.92 X⁴ + 256 X⁵
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ = 2.2x10⁻³ - X
1.04858x10²³X + 5.24288x10²⁴ X² + 9.8304x10²⁵ X³ + 8.192x10²⁶ X⁴ + 2.56x10²⁷ X⁵ - 2.2x10⁻³ = 0
Solving for X:
X = 2.01x10⁻²⁶
As
[Cu²⁺] = X
<h3>[Cu²⁺] = 2.01x10⁻²⁶</h3>
B.)<span>A doctor would use technetium-99m because a short half-life indicates a quick measurement, and a geologist would use rubidium-87 because a longer half-life means that there is a longer time to measure old rock. </span>
Answer:
3Fe2O3 + H2 --> 2Fe3O4 + H2O