Ksp of AgCl= 1.6×10⁻¹⁰
AgCl=Ag⁺ +Cl⁻
Ksp=[Ag⁺][Cl⁻]
Assume [Ag⁺]=[Cl⁻]=x
Ksp=x²
1.6×10⁻¹⁰=x²
x=0.000012
In FeCl₃:
FeCl₃------>Fe⁺³+ 3Cl⁻
as there is 0.010 M FeCl₃
So there will be ,
[Cl⁻]= 0.030
So
[Ag⁺]=Ksp/[Cl⁻]
=1.6×10⁻¹⁰/0.030
=5.3×10⁻⁹
so solubility of AgCl in FeCl₃ will be 5.3×10⁻⁹.
Answer:
B is sedimentary and A is metermorphic
Answer:
P= 7.01 atm
P(CO₂)= 2.34 atm
Explanation:
Step 1: Convert the temperature to Kelvin
We will use the following expression.
K = °C + 273.15
K = 32.0°C + 273.15 = 305.2 K
Step 2: Calculate the total number of moles of the mixture
We will use the following expression.
n = nCO₂ + nN₂ = 2.33 mol + 4.66 mol = 6.99 mol
Step 3: Calculate the total pressure of the mixture
We will use the ideal gas equation.
P × V = n × R × T
P = n × R × T / V
P = 6.99 mol × 0.0821 atm.L/mol.K × 305.2 K / 25.0 L
P= 7.01 atm
Step 4: Calculate the partial pressure of carbon dioxide
We will use the ideal gas equation.
P(CO₂) × V = nCO₂ × R × T
P(CO₂) = nCO₂ × R × T / V
P(CO₂) = 2.33 mol × 0.0821 atm.L/mol.K × 305.2 K / 25.0 L
P(CO₂)= 2.34 atm
<span>c. it will be four times less. </span>
