Answer:
The molar solubility of lead bromide at 298K is 0.010 mol/L.
Explanation:
In order to solve this problem, we need to use the Nernst Equaiton:
E is the cell potential at a certain instant, E⁰ is the cell potential, n is the number of electrons involved in the redox reaction, [ox] is the concentration of the oxidated specie and [red] is the concentration of the reduced specie.
At equilibrium, E = 0, therefore:
[red] = 0.010 M
The reduction will happen in the anode, therefore, the concentration of the reduced specie is equivalent to the molar solubility of lead bromide.
Answer:
Saturated solution = 180 gram
Explanation:
Given:
Solubility of Z = 60 g / 100 g water
Given temperature = 20°C
Amount of water = 300 grams
Find:
Saturated solution
Computation:
Saturated solution = [Solubility of Z] × Amount of water
Saturated solution = [60 g / 100 g] × 300 grams
Saturated solution = [0.6] × 300 grams
Saturated solution = 180 gram
The cells that result from the reproductive division of one cell during mitosis or meiosis
Answer: Protons because they have a positive charge.
Explanation: