<h3>
Answer:</h3>
Ag⁺(aq) +Cl⁻(aq) → AgCl(s)
<h3>
Explanation:</h3>
The questions requires we write the net ionic equation for the reaction between aqueous potassium chloride and aqueous silver nitrate.
<h3>Step 1: Writing a balanced equation for the reaction.</h3>
- The balanced equation for the reaction between aqueous potassium chloride and aqueous silver nitrate will be given by;
KCl(aq) + AgNO₃(aq) → KNO₃(aq) +AgCl(s)
- AgCl is the precipitate formed by the reaction.
<h3>Step 2: Write the complete ionic equation.</h3>
- The complete ionic equation for the reaction is given by showing all the ions involved in the reaction.
K⁺(aq)Cl⁻(aq) + Ag⁺(aq)NO₃⁻(aq) → K⁺(aq)NO₃⁻(aq) +AgCl(s)
- Only ionic compounds are split into ions.
<h3>Step 3: Write the net ionic equation for the reaction.</h3>
- The net ionic equation for a reactions only the ions that fully participated in the reaction and omits the ions that did not participate in the reaction.
- The ions that are not involved directly in the reaction are known as spectator ions and are not included while writing net ionic equation.
Ag⁺(aq) +Cl⁻(aq) → AgCl(s)
The transfer of energy that occurs when a force is applied over a distance is WORK.
Hope this helps!
Answer:
Explanation:
C decays by a process called beta decay. During this process, an atom of 14C decays into an atom of 14N, during which one of the neutrons in the carbon atom becomes a proton. This increases the number of protons in the atom by one, creating a nitrogen atom rather than a carbon atom.
Answer:
The equation is already balanced. There's an equal number of materials on each side of the equation.
Answer:
Boiling point of the solution is 100.78°C
Explanation:
This is about colligative properties.
First of all, we need to calculate molality from the freezing point depression.
ΔT = Kf . m . i
As the solute is nonelectrolyte, i = 1
0°C - (-2.79°C) = 1.86 °C/m . m . 1
2.79°C / 1.86 m/°C = 1.5 m
Now, we go to the boiling point elevation
ΔT = Kb . m . i
Final T° - 100°C = 0.52 °C/m . 1.5m . 1
Final T° = 0.52 °C/m . 1.5m . 1 + 100°C → 100.78°C
