1. Ca(HCO3)2
2.Ca(HCOO)2
3. Ca(OH)2
4.NaOH
5.KCI
6.MgSO4
7.PbO
8.HCl
9.HNO3
10.H2SO4
11.NH3
12.(NH4)3PO4
13.NaOH
:)
Answer:
yeah,The oxidation state of an atom does not represent the "real" charge on that atom, or any other actual atomic property.Hydrogen has OS = +1, but adopts −1 when bonded as a hydride to metals or metalloids. Oxygen in compounds has OS = −2. This set of postulates covers .
Explanation:
This is a missing part of your question:
The equilibrium system between sulfur dioxide gas, oxygen gas, and sulfur trioxide gas is given.
So you need the equilibrium balanced equation of SO2, O2, SO3 reaction:
First, we will start with the original equation which is not balanced yet (to understand how we get it):
SO2 + O2 ↔ SO3
Here the number of O atom is not equal at the to sides
So we will start to balance our equation by make the number of O atom equal each other on both sides:
So we will start to put 2SO3 instead of SO3
and put 2SO2 instead of SO2 to balance also the S atom on both sides
So we will get this:
2SO2(g) + O2(g) ↔ 2SO3(g) (This is our equilibrium balanced equation)
know we have a number of O atom equals on each side = 6
and the sulfur equals on each side = 2
