Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Q34 E (page 829)

Nitric acid reacts with insoluble copper (II) oxide to form soluble copper (II) nitrate,Cu (NO3)2, a compound that has been used to prevent the growth of algae in swimming pools. Write the balanced chemical equation for the reaction of an aqueous solution of HNO3 with CuO.

Short Answer

Expert verified

The balanced chemical reaction for an aqueous solution is

\[2HN{O_3}(aq) + CuO(s) \to Cu{\left( {N{O_3}} \right)_2}(aq) + {H_2}O(l)\]

Step by step solution

01

Definition of the aqueous solution

Aqueousis a term used to describe a water-based system. The term aqueous can also refer to a solution or mixture in which water serves as the solvent

02

Writing balanced equation for aqueous solution

Nitric acid reacts with copper oxide and the product is Cu (NO3)2. The balanced equation is

\[2HN{O_3}(aq) + CuO(s) \to Cu{\left( {N{O_3}} \right)_2}(aq) + {H_2}O(l)\]

Cu (NO3)2also receives water and the equation is of the ionic form. The dissociate CuO is in solid state and not dissolved in water.

\[2{H^ + } + 2NO_3^ - + C{u^{2 + }} + {O^{2 - }} \to C{u^{2 + }} + 2NO_3^ - + {H_2}O\]

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 solute species in each of the following solutions of acids or bases. Assume that the ionization of water can be neglected, and show that the change in the initial concentrations can be neglected. Ionization constants can be found in Appendix H and Appendix I.

\((a) 0.0092M HClO\), a weak acid

\((b) 0.0784M {C_6}{H_5}N{H_2}\), a weak base

\((c) 0.0810{\rm{ }}M HCN\), a weak acid

\((d) 0.11M{\left( {C{H_3}} \right)_3}\;N\), a weak base

\((e) 0.120MFe\left( {{H_2}O} \right)_6^{2 + }\), a weak acid \({K_a} = 1.6 \times 1{0^{ - 7}}\)

Calculate \(pH\;and the\;pOH\) of each of the following solutions at\(2{5^o}C\)for which the substances ionize completely:

(a)\(0.200M HCl\)

(b)\(0.0143M NaOH\)

(c)\(3.0M HN{O_3}\)

(d) \(0.0031M Ca{(OH)_2}\)

Novocaine, \({C_{13}}{H_{21}}{O_2}\;{N_2}Cl\), is the salt of the base procaine and hydrochloric acid. The ionization constant for procaine is \(7 \times 1{0^{ - 6}}\). Is a solution of novocaine acidic or basic? What are \(\left( {{H_3}{O^ + }} \right),\left( {O{H^ - }} \right)\), and \(pH\) of a \(2.0\% \) solution by mass of novocaine, assuming that the density of the solution is \(1.0\;g/mL\).

Explain why the ionization constant, \({K_a}\), for\({H_2}S{O_4}\) is larger than the ionization constant for \({H_2}S{O_3}\).

The active ingredient formed by aspirin in the body is salicylic acid, \({C_6}{H_4}OH\left( {C{O_2}H} \right)\). The carboxyl group \(\left( { - C{O_2}H} \right)\)acts as a weak acid. The phenol group (an OH group bonded to an aromatic ring) also acts as an acid but a much weaker acid. List, in order of descending concentration, all of the ionic and molecular species present in a \(0.001M\) aqueous solution of \({C_6}{H_4}OH\left( {C{O_2}H} \right)\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

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