Write the molecular and net ionic equations, including phases, for the reaction that occurs between aqueous sodium chloride, NaCl , and aqueous silver nitrate, AgNO 3 .

Molecular equation: NaCl(aq) + AgNO₃(aq)  →  NaNO₃(aq) + AgCl(s)↓

Net ionic equation:  Cl⁻(aq)  +  Ag⁺(aq)   → AgCl (s)

Explanation:

We determine the reactants for the reaction:

NaCl and AgNO₃

In aqueous solutions, salts from nitrate are soluble (all soluble)

In aqueous solutions, salts from chloride can make precipitate (with Ag⁺, Pb²⁺ Hg²⁺, Cu²⁺)

We dissociate the reactants:

NaCl  →  Na⁺  + Cl⁻

AgNO₃  →  Ag⁺  +  NO₃⁻

Molecular equation: NaCl(aq) + AgNO₃(aq)  →  NaNO₃(aq) + AgCl(s)↓

To build the net ionic equation we pick up the ions:

Na⁺(aq) + Cl⁻(aq)  + Ag⁺(aq)  + NO₃⁻(aq)  → NO₃⁻(aq)  + Na⁺(aq) + AgCl (s)

Those who are solid must be written as a compound, the ions are only from aq solutions. As we find NO₃⁻ and Na⁺ in both sides of the reactions we named them, spectators ions because they do not react. We can cancel them, so the net ionic equation is:  Cl⁻(aq)  +  Ag⁺(aq)   → AgCl (s)

Molecular equation: NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

net ionic equation: Ag+(aq) + Cl-(aq) → AgCl(s)

Explanation:

Step 1: Data given

aqueous sodium chloride = NaCl(aq)

aqueous silver nitrate = AgNO3(aq)

Step 2: The (un)balanced equation

NaCl(aq) + AgNO3(aq) → AgCl(s) + NaNO3(aq)

This equation is already balanced

Step 3: The net ionic equation

The net ionic equation, for which spectator ions are omitted - remember that spectator ions are those ions located on both sides of the equation - will , after canceling those spectator ions in both side, look like this:

Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) → AgCl(s) + Na+(aq) + NO3-(aq)

Ag+(aq) + Cl-(aq) → AgCl(s)

58 km per hour : d

answer: option (b) is the correct answer.

explanation:

a covalent bond is formed by sharing of electrons. whereas polarity arises due to difference in electronegativity of two atoms.

therefore, more is the electronegativity of an atom which combines with the titanium, more readily it will induce a partial positive and negative charge.

thus, out of the given options fluorine is most electronegative and has an incomplete octet. therefore, it will readily combine with titanium to share the electrons and complete its octet.