Answer is: n<span>o, because the ion product is less than the Ksp of lead iodide. </span>
Chemical dissociation 1: KI(s) → K⁺(aq) + I⁻(aq).
Chemical dissociation 2: Pb(NO₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction: Pb²⁺(aq) + 2I⁻(aq) → PbI₂(s).
Ksp(PbI₂) = 7.1·10⁻⁹.
V = 500 mL ÷ 1000 mL/L = 0.5 L.
c(KI) = c(I⁻) = 0.0025 mol ÷ 0.5 L.
c(I⁻) = 0.005 M.
c(Pb(NO₃)₂) = c(Pb²⁺) = 0.00004 mol ÷ 0.5 L.
c(Pb²⁺) = 0.00008 M.
Q = c(Pb²⁺) · c(I⁻)².
Q = 8·10⁻⁵ M · (5·10⁻³ M)².
Q = 2·10⁻⁹; <span> the ion product.</span>
The total pressure is calculated as below
Total pressure = partial pressure of (nitrogen +oxygen + argon gas)
= 587 mm hg + 158 mm hg + 7 mm hg = 752 mmhg is the total pressure
Answer:
Explanation:
attached here is the diagram representing the structure
First step is to calculate the mass of Ag in each compound separately:
From the periodic table:
molar mass of Ag is 107.87 gm
molar mass of Cl is 35.45 gm
molar mass of Br is 79.9 gm
For AgCl, mass % of Ag = [107.87/143.32] x 100 = 75.26%
For AgBr, mass % of Ag = [107.87/187.77] x 100 = 57.45 %
Second step is to calculate the mass % of each compound in the mixture:
Assume mass % of AgCl is y and that of AgBr is (1-y) as the total percentage is 100% or 1
0.6094 = 0.7526 y + 0.5745 (1-y)
y = 0.8716
This means that the mixture is almost 87% AgCl and 13% AgBr
The mass % of chlorine and bromine together is (100%-60.94%) which is 39.06%
mass % of chlorine = (1-0.6094)(0.8716) x 100 = 34.044%
mass % of bromine = 39.04 - 34.044 = 5.056%
