Answer:
E°= E°cathode- E° anode= 0.271-0.330= -0.59V
Explanation:
NB: the stoichiometry does not affect E°values,
And the more positive the E° values , the greater it's tendency to become spontaneous and hence irreversible, and the more negative the E° values the more likely to become less spontaneous and reversible, hence the above reaction is reversible
Answer:
0.00840
Explanation:
The computation of the mole fraction is as follow:
As we know that
Molar mass = Number of grams ÷ number of moles
Or
number of moles = Number of grams ÷ molar mass
Given that
Number of moles of CaI2 = 0.400
And, Molar mass of water = 18.0 g/mol
Now Number of moles of water is
= 850.0 g ÷ 18.0 g/mol
= 47.22 mol
And, Total number of moles is
= 0.400 + 47.22
= 47.62
So, Molar fraction of CaI2 is
= 0.400 ÷ 47.62
= 0.00840
The 3 and 2 to the right of the components are subscriptions.
Answer:
The study of the human body as a machine for the performance of work has its foundations in three major areas of study—namely, mechanics, anatomy, and physiology; more specifically, biomechanics, musculoskeletal anatomy, and neuromuscular physiology. Explanation:
Answer:
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M
Explanation:
Step 1: Data given
Molarity of Na2CrO4 = 0.010 M
Molarity of NaBr = 2.5 M
Ksp(PbCrO4) = 1.8 * 10^–14
Ksp(PbBr2) = 6.3 * 10^–6
Step 2: The balanced equation
PbCrO4 →Pb^2+ + CrO4^2-
PbBr2 → Pb^2+ + 2Br-
Step 3: Define Ksp
Ksp PbCrO4 = [Pb^2+]*[CrO4^2-]
1.8*10^-14 = [Pb^2+] * 0.010 M
[Pb^2+] = 1.8*10^-14 /0.010
[Pb^2+] = 1.8*10^-12 M
The minimum [Pb^2+] needed to precipitate PbCrO4 is 1.8*10^-12 M
Ksp PbBr2 = [Pb^2+][Br-]²
6.3 * 10^–6 = [Pb^2+] (2.5)²
[Pb^2+] = 1*10^-6 M
The minimum [Pb^2+] needed to precipitate PbBr2 is 1*10^-6 M
This means the amount of PbCrO4 will precipitate first, with a [Pb^2+] concentration of 1.8*10^-12 M