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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441 g CaCO₃ would have to be decomposed to produce 247 g of CaO
<h3>Further explanation</h3>Reaction
Decomposition of CaCO₃
CaCO₃ ⇒ CaO + CO₂
mass CaO = 247 g
mol of CaO(MW=56 g/mol) :
From equation, mol ratio CaCO₃ : CaO = 1 : 1, so mol CaO :
mass CaCO₃(MW=100 g/mol) :
<u>Answer:</u> The average atomic mass of X is 28.09 amu
<u>Explanation:</u>
Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.
Formula used to calculate average atomic mass follows:
.....(1)
- <u>For isotope 1:</u>
Mass of isotope 1 = 27.979 amu
Percentage abundance of isotope 1 = 92.21 %
Fractional abundance of isotope 1 = 0.9212
- <u>For isotope 2:</u>
Mass of isotope 2 = 28.976 amu
Percentage abundance of isotope 2 = 4.70 %
Fractional abundance of isotope 2 = 0.0470
- <u>For isotope 3:</u>
Mass of isotope 3 = 29.974 amu
Percentage abundance of isotope 3 = 3.09 %
Fractional abundance of isotope 3 = 0.0309
Putting values in equation 1, we get:
Hence, the average atomic mass of X is 28.09 amu