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Chapter 1: Problem 11

A 0.1-molar solution of which of the following acids will be the best conductor of electricity? (A) \(\mathrm{H}_{2} \mathrm{CO}_{3}\) (B) \(\mathrm{H}_{2} \mathrm{S}\) (C) \(\mathrm{HF}\) (D) \(\mathrm{HNO}_{3}\)

Short Answer

Expert verified
The 0.1-molar solution of \(\mathrm{HNO}_{3}\) will be the best conductor of electricity.

Step by step solution

01

Understanding and Establishing the Conductivity of Acids

The conductivity of an acid in an aqueous solution is determined by the number of ions it forms when dissolved. The more ions an acid forms, the better it will conduct electricity. Strong acids completely ionize in solution and will therefore be the best conductors.
02

Identifying the Strong Acid

The list of acids includes: \(\mathrm{H}_{2} \mathrm{CO}_{3}\), \(\mathrm{H}_{2} \mathrm{S}\), \(\mathrm{HF}\), and \(\mathrm{HNO}_{3}\). Among these, \(\mathrm{HNO}_{3}\) is a strong acid. This is because Nitric Acid, \(\mathrm{HNO}_{3}\), is a strong acid which means it will completely ionize in water to produce a large quantity of ions and hence the conductivity will be the highest.
03

Summarize the Findings

Based on the ionization properties of the acids, the 0.1-molar solution of \(\mathrm{HNO}_{3}\) will be the best conductor of electricity as it completely ionizes in water, providing the most ions to carry charge and hence conduct electricity.

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Most popular questions from this chapter

During a chemical reaction, NO(g) gets reduced and no nitrogen- containing compound is oxidized. Which of the following is a possible product of this reaction? (A) \(\mathrm{NO}_{2}(g)\) (B) \(\mathrm{N}_{2}(g)\) (C) \(\mathrm{NO}_{3}^{-}(a q)\) (D) \(\mathrm{NO}_{2}^{-}(a q)\)

Which expression below should be used to calculate the mass of copper that can be plated out of a 1.0 \(\mathrm{M} \mathrm{Cu}\left(\mathrm{NO}_{3}\right)_{2}\) , solution using a current of 0.75 A for 5.0 minutes? (A) \(\frac{(5.0)(60)(0.75)(63.55)}{(96500)(2)}\) (B) \(\frac{(5.0)(60)(63.55)(2)}{(0.75)(96500)}\) (C) \(\frac{(5.0)(60)(96500)(0.75)}{(63.55)(2)}\) (D) \(\frac{(5.0)(60)(96500)(63.55)}{(0.75)(2)}\)

Use the following information to answer questions 1-5. \(\begin{array}{ll}{\text { Reaction } 1 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=?} \\ {\text { Reaction } 2 : \mathrm{N}_{2} \mathrm{H}_{4}(l)+\mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g)} & {\Delta H=-37 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 3 : \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightarrow 2 \mathrm{NH}_{3}(g)} & {\Delta H=-46 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}} \\ {\text { Reaction } 4 : \mathrm{CH}_{4} \mathrm{O}(l) \rightarrow \mathrm{CH}_{2} \mathrm{O}(g)+\mathrm{H}_{2}(g)} & {\Delta H=-65 \mathrm{kJ} / \mathrm{mol}_{\mathrm{rxn}}}\end{array}\) If reaction 2 were repeated at a higher temperature, how would the reaction's value for \(\Delta G\) be affected? (A) It would become more negative because entropy is a driving force behind this reaction. (B) It would become more positive because the reactant molecules would collide more often. (C) It would become more negative because the gases will be at a higher (D) It will stay the same; temperature does not affect the value for \(\Delta G\) .

\(2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)\) 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) \(+2,000 \mathrm{kJ}\)

A solution of \(\mathrm{Co}^{2+}\) ions appears red when viewed under white light. Which of the following statements is true about the solution? (A) A spectrophotometer set to the wavelength of red light would read a high absorbance. (B) If the solution is diluted, the amount of light reflected by the solution will decrease. (C) All light with a frequency that is lower than that of red light will be absorbed by it. (D) Electronic transmissions within the solution match the wavelength of red light.
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