Explanation:
Moles of phosphorus pentachloride present initially = 2.5 mol
Moles of phosphorus trichloride at equilibrium = 0.338 mol
Initially
2.5 mol 0 0
At equilibrium:
(2.5 - x) mol x x
So, from above, the moles of phosphorus trichloride at equilibrium , x= 0.338 mol
Mass of 0.338 moles of phosphorus trichloride at equilibrium:
= 0.338 mol × 137.5 g/mol = 46.475 g
Moles of phosphorus pentachloride present at equilibrium :
= (2.5 - 0.338) mol = 2.162 mol
Mass of 2.162 moles of phosphorus pentachloride at equilibrium:
= 2.162 mol × 208.5 g/mol = 450.777 g
Moles of chloride gas present at equilibrium : 0.338 mol
Mass of 0.338 moles of chloride gas at equilibrium:
= 0.338 mol × 71 g/mol = 23.998 g
Aobt 1,200 halp you at all
Answer:
Specific gravity is the density of asubstance divided by the density of water. Since (at standard temperature and pressure) water has a density of 1 gram/cm3, and since all of the units cancel, specific gravity is usually very close to the same value as density(but without any units).
Answer:
S = 7.9 × 10⁻⁵ M
S' = 2.6 × 10⁻⁷ M
Explanation:
To calculate the solubility of CuBr in pure water (S) we will use an ICE Chart. We identify 3 stages (Initial-Change-Equilibrium) and complete each row with the concentration or change in concentration. Let's consider the solution of CuBr.
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0
C +S +S
E S S
The solubility product (Ksp) is:
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S²
S = 7.9 × 10⁻⁵ M
<u>Solubility in 0.0120 M CoBr₂ (S')</u>
First, we will consider the ionization of CoBr₂, a strong electrolyte.
CoBr₂(aq) → Co²⁺(aq) + 2 Br⁻(aq)
1 mole of CoBr₂ produces 2 moles of Br⁻. Then, the concentration of Br⁻ will be 2 × 0.0120 M = 0.0240 M.
Then,
CuBr(s) ⇄ Cu⁺(aq) + Br⁻(aq)
I 0 0.0240
C +S' +S'
E S' 0.0240 + S'
Ksp = 6.27 × 10⁻⁹ = [Cu⁺].[Br⁻] = S' . (0.0240 + S')
In the term (0.0240 + S'), S' is very small so we can neglect it to simplify the calculations.
S' = 2.6 × 10⁻⁷ M
I’m pretty sure the answer is D :)
