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Chapter 4: Problem 15

What is the difference between an ionic equation and a molecular equation?

Short Answer

Expert verified
An ionic equation represents compounds as dissociated ions, showing all the particles as they realistically exist in the solution. A molecular equation, on the other hand, shows the reaction occurring between the neutral compounds as a whole and not as separated ions.

Step by step solution

01

Understanding Molecular Equation

A molecular equation is a balanced chemical equation where the ionic compounds are expressed as molecules instead of component ions.
02

Understanding Ionic Equation

An ionic equation is a chemical equation where the electrolytes in aqueous solution are written as dissociated ions.
03

Key Differences

The key difference between these two lies in the fact that the ionic equation shows all the particles involved in the reaction as they realistically exist in the solution, while the molecular equation shows the neutral compounds as whole rather than as separated ions.

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Most popular questions from this chapter

What is the difference between a nonelectrolyte and an electrolyte? Between a weak electrolyte and a strong electrolyte?

This "cycle of copper" experiment is performed in some general chemistry laboratories. The series of reactions starts with copper and ends with metallic copper. The steps are: (1) A piece of copper wire of known mass is allowed to react with concentrated nitric acid [the products are copper(II) nitrate, nitrogen dioxide, and water]. (2) The copper(II) nitrate is treated with a sodium hydroxide solution to form copper(II) hydroxide precipitate. (3) On heating, copper(II) hydroxide decomposes to yield copper(II) oxide. (4) The copper(II) oxide is reacted with concentrated sulfuric acid to yield copper(II) sulfate. (5) Copper(II) sulfate is treated with an excess of zinc metal to form metallic copper. (6) The remaining zinc metal is removed by treatment with hydrochloric acid, and metallic copper is filtered, dried, and weighed. (a) Write a balanced equation for each step and classify the reactions. (b) Assuming that a student started with \(65.6 \mathrm{~g}\) of copper, calculate the theoretical yield at each step. (c) Considering the nature of the steps, comment on why it is possible to recover most of the copper used at the start.

Hydrogen halides (HF, HCl, HBr, HI) are highly reactive compounds that have many industrial and laboratory uses. (a) In the laboratory, HF and HCl can be generated by reacting \(\mathrm{CaF}_{2}\) and \(\mathrm{NaCl}\) with concentrated sulfuric acid. Write appropriate equations for the reactions. (Hint: These are not redox reactions.) (b) Why is it that HBr and HI cannot be prepared similarly, that is, by reacting \(\mathrm{NaBr}\) and \(\mathrm{NaI}\) with concentrated sulfuric acid? (Hint: \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is a stronger oxidizing agent than both \(\mathrm{Br}_{2}\) and \(\mathrm{I}_{2} .\) ) (c) \(\mathrm{HBr}\) can be prepared by reacting phosphorus tribromide \(\left(\mathrm{PBr}_{3}\right)\) with water. Write an equation for this reaction.

A 325-mL sample of solution contains \(25.3 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\). (a) Calculate the molar concentration of \(\mathrm{Cl}^{-}\) in this solution. (b) How many grams of \(\mathrm{Cl}^{-}\) are in \(0.100 \mathrm{~L}\) of this solution?

What factors qualify a compound as a salt? Specify which of the following compounds are salts: \(\mathrm{CH}_{4}\) \(\mathrm{NaF}, \mathrm{NaOH}, \mathrm{CaO}, \mathrm{BaSO}_{4}, \mathrm{HNO}_{3}, \mathrm{NH}_{3}, \mathrm{KBr} ?\)
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