Answer:
a. Minimum 1.70 V
b. There is no maximum.
Explanation:
We can solve this question by remembering that the cell potential is given by the formula
ε⁰ cell = ε⁰ reduction - ε⁰ oxidation
Now the problem states the cell must provide at least 0.9 V and that the reduction potential of the oxidized species 0.80 V, thus
ε⁰ reduction - ε⁰ oxidation ≥ ε⁰ cell
Since ε⁰ oxidation is by definition the negative of ε⁰ reduction , we have
ε⁰ reduction - ( 0.80 V ) ≥ 0.90 V
⇒ ε⁰ reduction ≥ 1.70 V
Therefore,
(a) The minimum standard reduction potential is 1.70 V
(b) There is no maximum standard reduction potential since it is stated in the question that we want to have a cell that provides at leat 0.9 V
Answer:
Black holes are merely the most exotic example of the general principle that ... quantum computer stores bits on protons and uses magnetic fields to flip them. ... Powered by Standard Model software, the universe computes
Explanation:because they are mostly exotic
Make sure the equation is always balanced first. (It is balanced for this question already) 6.022 x 10^23 is Avogadro’s number. In one mole of anything there is always 6.022 x 10^23 molecules, formula units, atoms. For one mol of an element/ compound use molar mass (grams).
Multiply everything on the top = 8.61x10^47
Multiple everything on bottom= 1.20x10^24
Divide top and bottom = 7.15x10^23
Answer: 7.15x10^23 mol SO2
Answer:
- <u>You need to convert the number of atoms of Ca into mass in grams, using Avogadro's number and the atomic mass of Ca.</u>
Explanation:
The amount of matter is measured in grams. Thus, you need to convert the number of atoms of Ca (calcium) into mass to compare with 2.45 grams of Mg.
To convert the atoms of calcium into mass, you divide by Avogadro's number, to obtain the number of moles of atoms, and then divide by the atomic mass of calcium.
<u />
<u>1. Number of moles, n</u>
<u />
<u>2. Mass</u>
- mass = number of moles × atomic mass
- mass = 0.053969mol × 40.078g/mol = 2.16g
Then, 2.45 g of Mg represent a greaer mass than the 3.25 × 10²² atoms of Ca.
