Atoms of elements that are nonmetals tend to gain electrons and atoms of metallic elements tend to lose electrons. Metals have few electrons in their valence shells.
By losing those electrons, these metals achieve noble gas configuration and satisfy the octet rule.
Nonmetals that have close to 8 electrons in their valence shells readily accept electrons to achieve noble gas configuration.
An example is the reaction between calcium and oxygen. Calcium is a metal and has 2 valence electrons. Oxygen is a nonmetal and has 6 valence electrons.
Calcium gives up its two valence electrons and oxygen accepts them and an ionic bond is established resulting in the formation of anew compound namely calcium oxide.
The reaction N2O4 (g) <--> 2NO2 (g) is endothermic, meaning that it consumes heat to move towards formation of the products.
According to Le Chatelier's Principle, therefore, if heat is added, more product (NO2) will be produced, and equilibrium would shift towards the right side. This is choice 3.
The Bohr Model, which was proposed by Niels Bohr in 1913
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
Explanation:
Reaction equation is as follows.
Here, 1 mole of 
produces 2 moles of cations.
![[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58](https://tex.z-dn.net/?f=%5BNa%5E%7B%2B%7D%5D%20%3D%202%5BNa_%7B2%7DSO_%7B3%7D%5D%20%3D%202%20%5Ctimes%200.58)
= 1.16 M
![[SO^{2-}_{3}] = [Na_{2}SO_{3}]](https://tex.z-dn.net/?f=%5BSO%5E%7B2-%7D_%7B3%7D%5D%20%3D%20%5BNa_%7B2%7DSO_%7B3%7D%5D)
= 0.58 M
The sulphite anion will act as a base and react with 
to form 
and 
.
As, 
= 
= 
According to the ICE table for the given reaction,
Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
![K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHSO%5E%7B-%7D_%7B3%7D%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BSO%5E%7B2-%7D_%7B3%7D%5D%7D)
x = 0.0003 M
So, x = ![[HSO^{-}_{3}] = [OH^{-}]](https://tex.z-dn.net/?f=%5BHSO%5E%7B-%7D_%7B3%7D%5D%20%3D%20%5BOH%5E%7B-%7D%5D)
= 0.0003 M
![[SO^{2-}_{3}]](https://tex.z-dn.net/?f=%5BSO%5E%7B2-%7D_%7B3%7D%5D)
= 0.58 - 0.0003
= 0.579 M
Now, we will use ![[HSO^{-}_{3}]](https://tex.z-dn.net/?f=%5BHSO%5E%7B-%7D_%7B3%7D%5D)
= 0.0003 M
The reaction will be as follows.
Initial: 0.0003
Equilibrium: 0.0003 - x x x
= 
= 
Therefore, 
As, x <<<< 0.0003. So, we can neglect x.
Therefore, 
= 
x = 
x = ![[OH^{-}] = [H_{2}SO_{3}]](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%20%3D%20%5BH_%7B2%7DSO_%7B3%7D%5D)
= 
![[H^{+}] = \frac{10^{-14}}{[OH^{-}]}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B%5BOH%5E%7B-%7D%5D%7D)
= 
= 
M
Thus, we can conclude that the concentration of spectator ion is M.
