To measure the solubility product of lead (II) iodide (PbI2) at 25°C, you constructed a galvanic cell that is similar to what yo
u used in the lab. The cell contains a 0.5 M solution of a lead (II) nitrate in one compartment that connects by a salt bridge to a 1.0 M solution of potassium iodide saturated with PbI2 in the other compartment. Then you inserted two lead electrodes into each half-cell compartment and closed the circuit with wires. What is the expected voltage generated by this concentration cell? Ksp for PbI2 is 1.4 x 10-8. Show all calculations for a credit.
1 answer:
Answer:
0.2320V
Explanation:
Voltage can be defined as the amount of potential energy available (work to be done) per unit charge, to move charges through a conductor.
Voltage can be generated by means other than rubbing certain types of materials against each other.
Please look at attached file for solution to the problem.
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Answer:
Therefore the theoretical density of iron is 7.877 g/cm³ .
Therefore the number of vacancy per cm³ is
Explanation:
BCC structure contains 2 atoms per cell.
BCC : Body centered cubic.
Atomic mass of iron = 55.845 gram/mole.
Atomic mass of a chemical element is the mass of 1 mole of the chemical element.
Density: Density of a matter is the ratio of mass of the matter and volume of the matter.
Avogrdro Number is number of atoms per 1 mole.
Avogrdro Number= 6.023×10²³
Lattice parameter of iron = 2.866×10⁻⁸ cm
The theoretical density:
g/cm³ [x= 2,Since it is BCC structure]
=7.877 g/cm³
Therefore the theoretical density of iron is 7.877 g/cm³ .
Now we have to find out the number of unit cell of iron crystal having density 7.874 g/cm³.
⇒ x =1.9923 atom/cell
Therefore the vacancy of atom per cell = (2- 1.9923)=0.0077
Therefore the number of vacancy per cm³ is