<h3>
Answer:</h3>
Anion present- Iodide ion (I⁻)
Net ionic equation- Ag⁺(aq) + I⁻(aq) → AgI(s)
<h3>
Explanation:</h3>
In order to answer the question, we need to have an understanding of insoluble salts or precipitates formed by silver metal.
Additionally we need to know the color of the precipitates.
Some of insoluble salts of silver and their color include;
- Silver chloride (AgCl) - white color
- Silver bromide (AgBr)- Pale cream color
- Silver Iodide (AgI) - Yellow color
- Silver hydroxide (Ag(OH)- Brown color
With that information we can identify the precipitate of silver formed and identify the anion present in the sample.
- The color of the precipitate formed upon addition of AgNO₃ is yellow, this means the precipitate formed was AgI.
- Therefore, the anion that was present in the sample was iodide ion (I⁻).
- Thus, the corresponding net ionic equation will be;
Ag⁺(aq) + I⁻(aq) → AgI(s)
Answer:
The answer to your question is 2.21 %
Explanation:
Data
mass of Na₂SO₄ = 10.9 g
mass of H₂O = 482 g
Formula
Percent by mass = 
x 100
Substitution
x 100
Simplification and result
x 100
Percent by mass = 0.022 x 100
Percent by mass = 2.21
In my opinion the answer is identical
The SAME number of molecules are in ANY “mole” of a compound or element. So, you only need to ... 24 g116 g/mol=0.207 moles of FeF3.
