/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 When the following solutions are... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 46

When the following solutions are mixed together, what precipitate (if any) will form? a. \(\mathrm{Hg}_{2}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CuSO}_{4}(a q)\) b. \(\mathrm{Ni}\left(\mathrm{NO}_{3}\right)_{2}(a q)+\mathrm{CaCl}_{2}(a q)\) c. \(\mathrm{K}_{2} \mathrm{CO}_{3}(a q)+\mathrm{MgI}_{2}(a q)\) d. \(\mathrm{Na}_{2} \mathrm{CrO}_{4}(a q)+\mathrm{AlBr}_{3}(a q)\)

Short Answer

Expert verified
When the following solutions are mixed together, these precipitates will form: a. \(Hg_{2}SO_{4}\) b. No precipitate c. \(MgCO_{3}\) d. \(Al_{2}(CrO_{4})_{3}\)

Step by step solution

01

Write Double Replacement Reactions

Write the balanced double replacement reactions for each pair of solutions: a. \(Hg_{2}(NO_{3})_{2}(aq) + CuSO_{4}(aq) \rightarrow Hg_{2}SO_{4}(s?) + Cu(NO_{3})_{2}(aq) \) b. \(Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq) \rightarrow NiCl_{2}(s?) + Ca(NO_{3})_{2}(aq) \) c. \(K_{2}CO_{3}(aq) + MgI_{2}(aq) \rightarrow MgCO_{3}(s?) + 2KI(aq) \) d. \(Na_{2}CrO_{4}(aq) + AlBr_{3}(aq) \rightarrow Al_{2}(CrO_{4})_{3}(s?) + 6NaBr(aq) \)
02

Apply Solubility Rules

Check the solubility rules for each of the potential precipitates: a. \( Hg_{2}SO_{4} \) - Mercury(I) sulfate is insoluble, so it will form a precipitate. b. \( NiCl_{2} \) - Nickel(II) chloride is soluble, so it will not form a precipitate. c. \( MgCO_{3} \) - Magnesium carbonate is insoluble, so it will form a precipitate. d. \( Al_{2}(CrO_{4})_{3} \) - Aluminum chromate is insoluble, so it will form a precipitate.
03

Identify Precipitates

Identify the precipitate for each reaction: a. Precipitate: \( Hg_{2}SO_{4} \) b. Precipitate: None c. Precipitate: \( MgCO_{3} \) d. Precipitate: \( Al_{2}(CrO_{4})_{3} \) Now, we know the result when each of the mentioned solutions is mixed: a. \(Hg_{2}(NO_{3})_{2}(aq) + CuSO_{4}(aq)\) forms the precipitate \(Hg_{2}SO_{4}\). b. \(Ni(NO_{3})_{2}(aq) + CaCl_{2}(aq)\) forms no precipitate. c. \(K_{2}CO_{3}(aq) + MgI_{2}(aq)\) forms the precipitate \(MgCO_{3}\). d. \(Na_{2}CrO_{4}(aq) + AlBr_{3}(aq)\) forms the precipitate \(Al_{2}(CrO_{4})_{3}\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

The blood alcohol \(\left(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{OH}\right)\) level can be determined by titrating a sample of blood plasma with an acidic potassium dichromate solution, resulting in the production of \(\mathrm{Cr}^{3+}(a q)\) and carbon dioxide. The reaction can be monitored because the dichromate ion \(\left(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\right)\) is orange in solution, and the \(\mathrm{Cr}^{3+}\) ion is green. The balanced equation is \(16 \mathrm{H}^{+}(a q)+2 \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(a q) \longrightarrow\) $$ 4 \mathrm{Cr}^{3+}(a q)+2 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) $$ This reaction is an oxidation-reduction reaction. What species is reduced, and what species is oxidized? How many electrons are transferred in the balanced equation above?

How would you prepare \(1.00 \mathrm{~L}\) of a \(0.50 M\) solution of each of the following? a. \(\mathrm{H}_{2} \mathrm{SO}_{4}\) from "concentrated" \((18 M)\) sulfuric acid b. \(\mathrm{HCl}\) from "concentrated" \((12 \mathrm{M})\) reagent c. \(\mathrm{NiCl}_{2}\) from the salt \(\mathrm{NiCl}_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O}\) d. HNO \(_{3}\) from "concentrated" (16 M) reagent e. Sodium carbonate from the pure solid

Which of the following solutions of strong electrolytes contains the largest number of ions: \(100.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}, 50.0 \mathrm{~mL}\) of \(0.200 \mathrm{M} \mathrm{BaCl}_{2}\), or \(75.0 \mathrm{~mL}\) of \(0.150 \mathrm{M} \mathrm{Na}_{4} \mathrm{PO}_{4}\) ?

A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are \(50.0 \mathrm{~mL}\) of \(0.100 M\) hydrochloric acid, \(100.0 \mathrm{~mL}\) of \(0.200 M\) of nitric acid, \(500.0 \mathrm{~mL}\) of \(0.0100 \mathrm{M}\) calcium hydroxide, and \(200.0 \mathrm{~mL}\) 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.

A 6.50-g sample of a diprotic acid requires \(137.5 \mathrm{~mL}\) of a \(0.750\) \(M\) NaOH solution for complete neutralization. Determine the molar mass of the acid.
See all solutions

Recommended explanations on Chemistry Textbooks

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

The Earths Atmosphere

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.