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4 years ago

Complete the balanced overall ionic equation for lead(ii) nitrate dissolving in water. $$pb(no3)2(s)

Chemistry

2 answers:

Vladimir79 [104]4 years ago

6 0

The balanced overall ionic equation for leads (ii) nitrate dissolving in water:

Pb (NO₃) ₂ (s) → Pb²⁺ (aq) + 2NO₃⁻ (aq)

<h3>Further explanation </h3>

The electrolyte in the solution produces ions.

The equation of a chemical reaction can be expressed in the equation of the ions

For strong electrolytes (the ionization rate = 1) is written in the form of separate ions, while the weak electrolyte (degree of ionization <1) is still written as an un-ionized molecule

In the ion equation, there is an ion spectator that is the ion which does not react because it is present before and after the reaction

When these ions are removed, the ionic equation is called the net ionic equation

For gases and solids including water (H₂O) can be written as an ionized molecule

So only the dissolved compound is ionized ((expressed in symbol aq)

Ionic compounds consist of cations (ions +) and anions (ions -)

Lead (ii) nitrate, Pb (NO₃) ₂ is a solution (ion compound) which, if dissolved in water, ionizes into Pb2 + cation and NO3- anion

Pb (NO₃) ₂ (s) → Pb²⁺ (aq) + 2NO₃⁻ (aq)

<h3>Learn more </h3>

The net ionic equation

brainly.com/question/8885824

ion stability

brainly.com/question/1781595

ionic bonding

brainly.com/question/1603987

Arisa [49]4 years ago

4 0

Missing question: <span>What will happen if we combine the solution of sodium iodide and the solution of lead(II) nitrate?
Chemical reaction 1: Pb(NO</span>₃)₂(s) → Pb²⁺(aq) + 2NO₃⁻(aq).
Chemical reaction 2: 2NaI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2NaNO₃(aq).
Ionic reaction:
2Na⁺(aq) + 2I⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → PbI₂(s) + 2Na⁺(aq) + 2NO₃⁻(aq).
Net ionic reaction: 2I⁻(aq) + Pb²⁺(aq) → PbI₂(s).

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azamat

Answer:

$K=1.12x10^9$

Explanation:

Hello there!

Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

$K=\frac{[NO_2]^2}{[NO]^2[O_2]}$

Then, we plug in the concentrations at equilibrium to obtain the equilibrium constant as follows:

$K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9$

In addition, we can infer this is a reaction that predominantly tends to the product (NO2) as K>>>>1.

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4 years ago

Complete the balanced overall ionic equation for lead(ii) nitrate dissolving in water. $$pb(no3​)2​(s)

Complete the balanced overall ionic equation for lead(ii) nitrate dissolving in water. $$pb(no3)2(s)