Answer is "0.05 mol".
<em>Explanation;</em>
We can do calculation by using a simple formula as
n = m/M
Where, n is the number of moles of the substance (mol), m is the mass of the substance (g) and M is the molar mass of the substance (g/mol).
Here,
n = ?
m = 2.80 g
M = 56.08 g/mol
By substitution,
n = 2.80 g /56.08 g/mol
n = 0.0499 mol ≈ 0.05 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 ![[Ag(NH_{3})_{2}]^{+}](https://tex.z-dn.net/?f=%5BAg%28NH_%7B3%7D%29_%7B2%7D%5D%5E%7B%2B%7D)
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)
Answer:
NaCl is the one of the example of ionic compound because na conducts positive charge and CL conduct negative charge
Independent- controlled in a scientific experiment to test the effects a dependent is being tested and measured in a scientific experimrent
