43.8 kJ
<h3>
Explanation</h3>
There are two electrodes in a voltaic cell. Which one is the anode?
The lithium atom used to have no oxygen atoms when it was on the reactant side. It gains two oxygen atoms after the reaction. It has gained more oxygen atoms than the manganese atom. Gaining oxygen is oxidation. As a result, lithium is being oxidized.
Oxidation takes place at the anode of a cell. Therefore, the anode of this cell is made of lithium.
Lithium has an atomic mass of 6.94. Each gram of Li would contain 1/6.94 = 0.144 moles of Li atoms. Each Li atom loses one electron in this cell. Therefore, the number of electron transferred, <em>n</em>, equals 0.144 moles for each gram of the anode.
Let represents the electrical energy produced.
, where
- <em>n</em> is the <em>number of moles</em> electrons transferred,
- <em>F</em> is the Faraday's constant,
- <em>E</em> is the cell potential,
<em>n </em>= 0.144 mol, as shown above, and
<em>F </em>= 96.486 kJ / ().
Therefore,
.
The answer for this issue is:
The chemical equation is: HBz + H2O <- - > H3O+ + Bz-
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz]
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x.
Accept that x is little contrasted with 0.5 M. At that point,
Ka = 6.4X10^-5 = x^2/0.5
x = [H3O+] = 5.6X10^-3 M
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make)
They are the outer layer of the electron layers.
<h3>
Answer:</h3>
1 x 10^13 stadiums
<h3>
Explanation:</h3>
From the question;
1 x 10^5 people can fill 1 stadium
We are given, 1 x 10^18 atoms of iron
We are required to determine the number of stadiums that 1 x 10^18 atoms of iron would occupy.
We are going to assume that a stadium would occupy a number of atoms equivalent to the number of people.
Therefore;
One stadium = 1 x 10^5 atoms
Then, to find the number of stadiums that will be occupied by 1 x 10^18 atoms;
No. of stadiums = Total number of atoms ÷ Atoms in a single stadium
= 1 x 10^18 atoms ÷ 1 x 10^5 atoms
= 1 x 10^13 stadiums
Therefore, 1 x 10^18 atoms of iron would occupy 1 x 10^13 stadiums