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Chapter 1: Problem 3

If equimolar solutions of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) and \(\mathrm{NaCl}\) are mixed, which ion will not be present in significant amounts in the resulting solution after equilibrium is established? (A) \(\mathrm{Pb}^{2+}\) (B) \(\mathrm{NO}_{3}^{-}\) (C) \(\mathrm{Na}^{+}\) (D) \(\mathrm{Cl}^{-}\)

Short Answer

Expert verified
The ion not present in significant amounts in the solution after equilibrium is established is (A) \(Pb^{2+}\).

Step by step solution

01

Identify the chemical equation

Firstly, write down the chemical equation of the reaction. When Pb(NO3)2 and NaCl react together, they undergo a double replacement reaction forming PbCl2 and NaNO3. Hence, the chemical equation becomes: Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3.
02

Understand the solubility

According to solubility rules, most nitrate salts are soluble, and most chloride salts are soluble, except for lead(II) chloride (PbCl2). Therefore, in this chemical reaction, Pb(NO3)2 and NaCl are soluble and they dissociate completely. However, PbCl2 is not soluble, and it precipitates out of the solution.
03

Analyze the ions at equilibrium

After reaching the equilibrium, the ions present in the solution will be: Cl-, Na+, NO3-. These ions are from soluble compounds, NaNO3 and remaining NaCl and Pb(NO3)2. Pb2+, from PbCl2, will not be present in significant amounts because PbCl2 is insoluble, and it remains as a precipitate, not an ion in the solution.

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Most popular questions from this chapter

A proposed mechanism for a reaction is as follows: $\begin{array}{cc}{\mathrm{NO}_{2}+\mathrm{F}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{F}+\mathrm{F}} & {\text { Slow step }} \\ {\mathrm{F}+\mathrm{NO}_{2} \rightarrow \mathrm{NO}_{2} \mathrm{F}} & {\text { Fast step }}\end{array}$ What is the order of the overall reaction? (A) Zero order (B) First order (C) Second order (D) Third order

\(\begin{array}{ll}{\mathrm{C}(s)+2 \mathrm{H}_{2}(g) \rightarrow \mathrm{CH}_{4}(g)} & {\Delta H^{\circ}=x} \\\ {\mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)} & {\Delta H^{\circ}=y} \\ {\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)} & {\Delta H^{\circ}=\mathrm{z}}\end{array}\) Based on the information given above, what is \(\Delta H^{\circ}\) for the following reaction? \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\) (A) \(x+y+z\) (B) \(x+y-z\) (C) \(z+y-2 x\) (D) \(2 z+y-x\)

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) Based on the information given in the table below, what is \(\Delta H^{\circ}\) for the above reaction? \(\begin{array}{cc}{\text { Bond }} & {\text { Average bond energy }(\mathrm{kJ} / \mathrm{mol})} \\ {\mathrm{H}-\mathrm{H}} & {500} \\\ {\mathrm{O}=\mathrm{O}} & {500} \\ {\mathrm{O}-\mathrm{H}} & {500}\end{array}\) (A) \(-2,000 \mathrm{kJ}\) (B) \(-500 \mathrm{kJ}\) (C) \(+1,000 \mathrm{kJ}\) (D) \(+2,000 \mathrm{kJ}\)

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. During this reaction, heat transfers from (A) the reactants to the products (B) the reactants to the system (C) the system to the surroundings (D) the products to the surroundings

Use the following information to answer questions 14-16 The radius of atoms and ions is typically measured in Angstroms \((A),\) which is equivalent to \(1 * 10^{-10} \mathrm{m} .\) Below is a table of information for three different elements. TABLE NOT AVAILABLE Neon has a smaller atomic radius than phosphorus because: (A) Unlike neon, phosphorus has electrons present in its third energy level. (B) Phosphorus has more protons than neon, which increases the repulsive forces in the atom. (C) The electrons in a neon atom are all found in a single energy level. (D) Phosphorus can form anions, while neon is unable to form any ions.
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