Question 1

Hydrogen is oxidized in the following chemical reaction.H₂ + Cl₂ $\rarr$ 2HCl

Accepted Answer

Hydrogen is losing electrons (being oxidized) in the reaction, while chlorine is gaining electrons (being reduced).

Question 2

The following reaction is a redox reaction. CaC₂(s) + H₂O(l) $\rarr$ HCCH(g) + CaO(s)

Accepted Answer

In this reaction, there is no transfer of electrons between the reactants and products, which is a requirement for a redox reaction. It is a hydrolysis reaction where water reacts with calcium carbide to produce acetylene and calcium oxide.

Question 3

The oxidation number of N in NO is +7.

Accepted Answer

The oxidation number of N in NO is +2, not +7.

Question 4

Silver chloride (AgCl) has an extremely low solubility in water; therefore, it is a weak electrolyte.

Accepted Answer

Silver chloride (AgCl) has an extremely low solubility in water, which means it dissociates into ions to a very small extent. However, being a salt, if it does dissolve, it dissociates completely into its ions (Ag+ and Cl-), making it a strong electrolyte. The term "weak electrolyte" refers to a substance that only partially dissociates into ions in solution, which is not the case for AgCl.

Question 5

Most compounds containing chlorides, bromides, and iodides are soluble except those containing Ag⁺, Hg₂²⁺, and Pb²⁺.

Accepted Answer

This statement is true.

Question 6

The following reaction is an acid-base neutralization reaction. H₂SO₄(aq) + CaBr₂(aq) $\rarr$ CaSO₄(s) + 2HBr(g)

Accepted Answer

This reaction is a double displacement reaction, not an acid-base neutralization, as it does not produce water.

Question 7

A weak acid or a weak base ionizes completely.

Accepted Answer

A weak acid or a weak base does not ionize completely. They only partially ionize in water.

Question 8

The following equation is an example of a net ionic equation. Na⁺(aq) + Br^-(aq) + Ag⁺(aq) + NO₃^-(aq) $\rarr$ AgBr(s) + Na⁺(aq) + NO₃^-(aq)

Accepted Answer

A net ionic equation only includes the ions and molecules directly involved in the reaction, excluding spectator ions like Na⁺ and NO₃⁻.

Question 9

If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, how many moles of Pb²⁺ were originally in the solution A) 7.25 $\rarr$ 10^-3 moles of Pb²⁺ B) 7.35 $\rarr$ 10^-3 moles of Pb²⁺ C) 7.45 $\rarr$ 10^-3 moles of Pb²⁺ D) 7.55 $\rarr$ 10^-3 moles of Pb²⁺ E) None of the above

Accepted Answer

First, we need to use the balanced chemical equation for the reaction between lead(II) nitrate and potassium iodide: Pb(NO3)2(aq) + 2KI(aq) → PbI2(s) + 2KNO3(aq) From the equation, we can see that 1 mole of Pb(NO3)2 reacts with 2 moles of KI to form 1 mole of PbI2. We know that 73.5 mL of 0.200 M KI was required to precipitate all of the lead(II) ion, which means that 0.200 moles of KI was used (since Molarity = moles/volume). Since every 2 moles of KI is used to form 1 mole of PbI2, then 0.100 moles of PbI2 was formed. Finally, from the balanced equation, we see that 1 mole of PbI2 contains 1 mole of Pb²⁺. So, the original amount of Pb²⁺ in the solution is also 0.100 moles. Therefore, the answer is 7.35 x 10^−10 moles of Pb²⁺ (Choice B).

Question 10

Which, if any of the following correctly shows the balanced molecular equation for the reaction that occurs when aqueous solutions of potassium iodide and lead(II) nitrate are combined. A)2KI(aq) + Pb(NO₃)₂(aq) $\rarr$ PbI₂(s) + KNO₃(aq) B)KI(aq) + Pb(NO₃)₂(aq) $\rarr$ PbI₂(s) + 2KNO₃(aq) C)2KI(aq) + Pb(NO₃)₂(aq) $\rarr$ PbI₂(s) + 2KNO₃(aq) D)2KI(aq) + 2Pb(NO₃)₂(aq) $\rarr$ PbI₂(s) + 2KNO₃(aq) E) None of the above

Accepted Answer

The balanced molecular equation for the reaction between potassium iodide and lead(II) nitrate can be written as follows:
2KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2KNO3(aq)
This equation shows that two moles of potassium iodide (2KI) react with one mole of lead(II) nitrate (Pb(NO3)2) to form one mole of lead(II) iodide (PbI2) and two moles of potassium nitrate (2KNO3).
Choice C is the only option that correctly shows the reactants and the products with the correct coefficients.

Question 11

Sugar dissolves in water, therefore it is a strong electrolyte.

Accepted Answer

Sugar, also known as sucrose, is a non-electrolyte as it does not ionize in water and therefore does not conduct electricity. Only substances that ionize in water, such as salts and acids, are considered strong electrolytes.

Question 12

The oxidation number of iodine increases by 6 in the following reaction.2MnO₄^- + I^- + H₂O $\rarr$ 2MnO₂ + IO₃^- + 2OH^-

Accepted Answer

In the reaction, iodine (I) changes from an oxidation state of -1 to +5, which is an increase of 6.

Quiz 4: Reactions in Aqueous Solutions

Hydrogen is oxidized in the following chemical reaction.H2 + Cl2 →\rarr→ 2HCl

The following reaction is a redox reaction. CaC2(s) + H2O(l) →\rarr→ HCCH(g) + CaO(s)

The oxidation number of N in NO is +7.

Silver chloride (AgCl) has an extremely low solubility in water; therefore, it is a weak electrolyte.

Most compounds containing chlorides, bromides, and iodides are soluble except those containing Ag+, Hg22+, and Pb2+.

The following reaction is an acid-base neutralization reaction. H2SO4(aq) + CaBr2(aq) →\rarr→ CaSO4(s) + 2HBr(g)

A weak acid or a weak base ionizes completely.

The following equation is an example of a net ionic equation. Na+(aq) + Br-(aq) + Ag+(aq) + NO3-(aq) →\rarr→ AgBr(s) + Na+(aq) + NO3-(aq)

If 73.5 mL of 0.200 M KI(aq) was required to precipitate all of the lead(II) ion from an aqueous solution of lead(II) nitrate, how many moles of Pb2+ were originally in the solution

Which, if any of the following correctly shows the balanced molecular equation for the reaction that occurs when aqueous solutions of potassium iodide and lead(II) nitrate are combined.