Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 28

Commercial cold packs and hot packs are available for treating athletic injuries. Both types contain a pouch of water and a dry chemical. When the pack is struck, the pouch of water breaks, dissolving the chemical, and the solution becomes either hot or cold. Many hot packs use magnesium sulfate, and many cold packs use ammonium nitrate. Write reactions to show how these strong electrolytes break apart when they dissolve in water.

Short Answer

Expert verified
The dissociation reactions of magnesium sulfate (MgSO₄) and ammonium nitrate (NH₄NO₃) in water are as follows: For magnesium sulfate: \[MgSO_4(s) \rightarrow Mg^{2+}(aq) + SO_4^{2-}(aq)\] For ammonium nitrate: \[NH_4NO_3(s) \rightarrow NH_4^+(aq) + NO_3^-(aq)\]

Step by step solution

01

Write the chemical formulas for magnesium sulfate and ammonium nitrate

Magnesium sulfate has the chemical formula MgSO₄, while ammonium nitrate has the chemical formula NH₄NO₃.
02

Write the dissociation reaction for magnesium sulfate in water

When magnesium sulfate (MgSO₄) dissolves in water, it separates into magnesium ions (Mg²⁺) and sulfate ions (SO₄²⁻). The balanced dissociation reaction is: \[MgSO_4(s) \rightarrow Mg^{2+}(aq) + SO_4^{2-}(aq)\]
03

Write the dissociation reaction for ammonium nitrate in water

When ammonium nitrate (NH₄NO₃) dissolves in water, it separates into ammonium ions (NH₄⁺) and nitrate ions (NO₃⁻). The balanced dissociation reaction is: \[NH_4NO_3(s) \rightarrow NH_4^+(aq) + NO_3^-(aq)\] This completes the step-by-step solution for the dissociation reactions of magnesium sulfate and ammonium nitrate in water.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the concentration of all ions present in each of the following solutions of strong electrolytes. a. 0.100 mole of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) in \(100.0 \mathrm{mL}\) of solution b. 2.5 moles of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) in 1.25 L of solution c. \(5.00 \mathrm{g}\) of \(\mathrm{NH}_{4} \mathrm{Cl}\) in \(500.0 \mathrm{mL}\) of solution d. \(1.00 \mathrm{g} \mathrm{K}_{3} \mathrm{PO}_{4}\) in \(250.0 \mathrm{mL}\) of solution

What mass of barium sulfate can be produced when \(100.0 \mathrm{mL}\) of a 0.100-M solution of barium chloride is mixed with 100.0 mL of a 0.100-M solution of iron(III) sulfate?

Which of the following solutions of strong electrolytes contains the largest number of moles of chloride ions: \(100.0 \mathrm{mL}\) of \(0.30 \mathrm{M} \mathrm{AlCl}_{3}, 50.0 \mathrm{mL}\) of \(0.60 \mathrm{M} \mathrm{MgCl}_{2},\) or \(200.0 \mathrm{mL}\) of \(0.40 M\) NaCl?

In most of its ionic compounds, cobalt is either Co(II) or Co(III). One such compound, containing chloride ion and waters of hydration, was analyzed, and the following results were obtained. A \(0.256-\mathrm{g}\) sample of the compound was dissolved in water, and excess silver nitrate was added. The silver chloride was filtered, dried, and weighed, and it had a mass of 0.308 g. A second sample of 0.416 g of the compound was dissolved in water, and an excess of sodium hydroxide was added. The hydroxide salt was filtered and heated in a flame, forming cobalt(III) oxide. The mass of cobalt(III) oxide formed was 0.145 g. a. What is the percent composition, by mass, of the compound? b. Assuming the compound contains one cobalt ion per formula unit, what is the formula? c. Write balanced equations for the three reactions described.

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are 50.0 mL of 0.100 \(M\) hydrochloric acid, \(100.0 \mathrm{mL}\) of \(0.200 \mathrm{M}\) of nitric acid, \(500.0 \mathrm{mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and 200.0 mL. of 0.100 \(M\) rubidium hydroxide. Did the acids and bases exactly neutralize each other? If not, calculate the concentration of excess \(\mathrm{H}^{+}\) or \(\mathrm{OH}^{-}\) ions left in solution.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free