Correct Question:
A spectator ion is (Select all that apply.)
- a piece of french fry contaminating the reaction mixture
- an ionic component of a reactant that is unchanged by the reaction
-in this experiment, nitrate ion
- your eye, carefully watching the progress of the reaction
Answer:
- an ionic component of a reactant that is unchanged by the reaction
Explanation:
A spectator ion is an ion that exists as a reactant and a product in a chemical equation. A spectator ion is one that exists in the same form on both the reactant and product sides of a chemical reaction.
Spectator ions are ions that are present in a solution but don't take part in the reaction. When reactants dissociate into ions, some of the ions may combine to form a new compound. The other ions don't take part in this chemical reaction and are therefore called spectator ions.
The correct option is therefore the option;
- an ionic component of a reactant that is unchanged by the reaction
Answer:
The process where substance react with oxygen is called combustion.
Explanation:
When substance react with oxygen combustion is occur. The substance which burned is called fuel and in this process large amount of heat is released to the surrounding. It is exothermic process.
For example:
4Li + O₂ → 2Li₂O
2Mg + O₂ → 2MgO
S + O₂ → SO₂
The product which is formed as a result of combustion reaction are called oxides.
In given examples we can see that lithium, magnesium and sulfur react with oxygen and product formed is oxides of respective elements such as lithium oxide ( Li₂O), magnesium oxide (MgO) and sulfur oxide ( SO₂ ).
Answer: Equilibrium constant for this reaction is 
.
Explanation:
Chemical reaction equation for the formation of nickel cyanide complex is as follows.
 + 2OH^{-}(aq)](https://tex.z-dn.net/?f=Ni%28OH%29_%7B2%7D%28s%29%20%2B%204CN%5E%7B-%7D%28aq%29%20%5Crightleftharpoons%20%5BNi%28CN%29_%7B4%7D%5E%7B2-%7D%5D%28aq%29%20%2B%202OH%5E%7B-%7D%28aq%29)
We know that,
K = 
We are given that, 
and, 
Hence, we will calculate the value of K as follows.
K =
K = 
=
Thus, we can conclude that equilibrium constant for this reaction is .
