Burning Mg in the air and reacting with O2 forming a white powder of MnO
So the equation is going to be:
Mn + O2 ⇒ MnO (this equation is not conserved)
to make it equilibrium:
1- First we should put 2Mno to equal the O2 on both sides.
So it will be:
Mg + O2⇒ 2MgO
2- Second we should put 2Mn to equal the Mn on both sides.
2Mg + O2⇒ 2MgO (this equation is conserved)
After putting the physical states the final equilibrium equation is going to be:
Δ
2Mg(s) + O2(g)⇒ 2MgO(s)
<h3>
Answer:</h3>
56.11 g/mol
<h3>
General Formulas and Concepts:</h3>
<u>Math</u>
<u>Pre-Algebra</u>
Order of Operations: BPEMDAS
- Brackets
- Parenthesis
- Exponents
- Multiplication
- Division
- Addition
- Subtraction
<u>Chemistry</u>
<u>Atomic Structure</u>
<h3>
Explanation:</h3>
<u>Step 1: Define</u>
[Compound] KOH
<u>Step 2: Identify</u>
[PT] Molar Mass of K - 39.10 g/mol
[PT] Molar Mass of O - 16.00 g/mol
[PT] Molar Mass of H - 1.01 g/mol
<u>Step 3: Find</u>
39.10 + 16.00 + 1.01 = 56.11 g/mol
Answer:
First, precipitate of AgCl is formed. Second, a soluble complex of silver and ammonia is formed. Third, AgCl is reproduced due to disappearance of ammonia complex in presence of
.
Explanation:
In presence of NaCl,
forms an insoluble precipitate of AgCl.
Reaction: 
In presence of
, AgCl gets dissolved into solution due to formation of soluble
complex.
Reaction: ![AgCl(s)+2NH_{3}(aq.)\rightarrow [Ag(NH_{3})_{2}]^{+}(aq.)+ Cl^{-}(aq.)](https://tex.z-dn.net/?f=AgCl%28s%29%2B2NH_%7B3%7D%28aq.%29%5Crightarrow%20%5BAg%28NH_%7B3%7D%29_%7B2%7D%5D%5E%7B%2B%7D%28aq.%29%2B%20Cl%5E%7B-%7D%28aq.%29)
In presence of
,
complex gets destroyed and free
again reacts with free
to produce insoluble AgCl
Reaction: ![[Ag(NH_{3})_{2}]^{+}(aq.)+2H^{+}(aq.)+Cl^{-}(aq.)\rightarrow AgCl(s)+2NH_{4}^{+}(aq.)](https://tex.z-dn.net/?f=%5BAg%28NH_%7B3%7D%29_%7B2%7D%5D%5E%7B%2B%7D%28aq.%29%2B2H%5E%7B%2B%7D%28aq.%29%2BCl%5E%7B-%7D%28aq.%29%5Crightarrow%20AgCl%28s%29%2B2NH_%7B4%7D%5E%7B%2B%7D%28aq.%29)