Answer:
0.0977M is the concentration of the final solution
Explanation:
In a dilution process, to the original solution (Stock solution) you add more solvent in order to decrease its original concentration. To know how many times the solution was diluted you must find the dilution factor (Ratio between the initial and final volume). That is:
Intial Volume: 5.73mL
Final volume: 5.73mL + 20.26mL = 25.99mL
Dilution factor: 25.99mL / 5.73mL = 4.536 → 4.536 times the solution is diluted. The concentration of the diluted solution is:
0.443M / 4.536 =
<h3>0.0977M is the concentration of the final solution</h3>
Answer:
Final concentration = 0.019 M
Explanation:
Initial Concentration [A]o = 0.27M
Rate constant, k = 0.75 s^-1
Final concentration [A] = ?
Time, t = 1.5s
The relationship between the variables is given by the equation;
ln[A] = ln[A]o - kt
ln[A] = ln(0.27) - (0.75)(1.5)
ln[A] = - 1.309 - 1.125
ln[A] = - 2.434
[A] = 0.019 M
The chemical reaction between the reactants:
3 AgNO₃ (aq) + FeCl₃ (aq) → 3 AgCl (s) + Fe(NO₃)₃ (aq)
Explanation:
We have the following chemical reaction:
3 AgNO₃ (aq) + FeCl₃ (aq) → 3 AgCl (s) + Fe(NO₃)₃ (aq)
Complete ionic equation:
3 Ag⁺ (aq) + 3 NO₃⁻ (aq) + Fe³⁺ (aq) + 3 Cl⁻ (aq) → 3 AgCl (s) + Fe³⁺ (aq) + 3 NO₃⁻ (aq)
We remove the spectator ions and we get the net ionic equation:
Ag⁺ (aq) + Cl⁻ (aq) → AgCl (s)
where:
(aq) - aqueous
(s) - solid
Learn more about:
net ionic equation
brainly.com/question/7018960
#learnwithBrainly
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
