Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Give the formula of a compound of nitrogen that is (a) a weak base. (b) a strong acid. (c) a weak acid. (d) capable of oxidizing copper.

Short Answer

Expert verified
Question: Identify a compound of nitrogen for each of the following categories: (a) weak base, (b) strong acid, (c) weak acid, and (d) capable of oxidizing copper. Answer: (a) Ammonia (NH3), (b) Nitric acid (HNO3), (c) Hydrazoic acid (HN3), and (d) Nitric acid (HNO3).

Step by step solution

01

a) Weak base

A weak base is a base that does not fully dissociate into its ions in solution. Ammonia (NH3) is a weak base. It does not completely ionize in water, only partially reacting with water to form ammonium (NH4+) and hydroxide (OH-) ions.
02

b) Strong acid

For a nitrogen compound that is a strong acid, we can consider nitric acid (HNO3). It is a strong acid because it fully dissociates in water, ionizing to form hydrogen ions (H+) and nitrate ions (NO3-).
03

c) Weak acid

A weak acid refers to an acid that does not fully dissociate in water. For a nitrogen compound, this would be hydrazoic acid (HN3). In a solution, it only partially ionizes to produce hydrogen ions (H+) and azide ions (N3-).
04

d) Capable of oxidizing copper

For a nitrogen compound that can oxidize copper, we can refer to nitric acid (HNO3) again. In this reaction, nitric acid oxidizes copper into copper(II) ions while the nitrogen in nitric acid is reduced, ultimately producing nitrogen monoxide (NO) and water (H2O) as products. The overall reaction can be written as: 3Cu(s) + 8HNO3(aq) -> 3Cu(NO3)2(aq) + 2NO(g) + 4H2O(l)

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 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

Write a balanced equation for the reaction of ammonia with (a) \(\mathrm{Cu}^{2+}\) (b) \(\mathrm{H}^{+}\) (c) \(\mathrm{Al}^{3+}\)

When a solution of hydrogen bromide is prepared, \(1.283 \mathrm{~L}\) of HBr gas at \(25^{\circ} \mathrm{C}\) and 0.974 atm is bubbled into \(250.0 \mathrm{~mL}\) of water. Assuming all the HBr dissolves with no volume change, what is the molarity of the hydrobromic acid solution produced?

Write a balanced equation for the preparation of (a) \(\mathrm{N}_{2}\) from \(\mathrm{Pb}\left(\mathrm{N}_{3}\right)_{2}\). (b) \(\mathrm{O}_{2}\) from \(\mathrm{O}_{3}\) (c) \(\mathrm{S}\) from \(\mathrm{H}_{2} \mathrm{~S}\).

The amount of sodium hypochlorite in a bleach solution can be determined by using a given volume of bleach to oxidize excess iodide ion to iodine; \(\mathrm{ClO}^{-}\) is reduced to \(\mathrm{Cl}^{-}\). The amount of iodine produced by the redox reaction is determined by titration with sodium thiosulfate, \(\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ; \mathrm{I}_{2}\) is reduced to \(1^{-}\). The sodium thiosulfate is oxidized to sodium tetrathionate, \(\mathrm{Na}_{2} \mathrm{~S}_{4} \mathrm{O}_{6} .\) In this analysis, potassium iodide was added in excess to \(5.00 \mathrm{~mL}\) of bleach \(\left(d=1.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\). If \(25.00 \mathrm{~mL}\) of \(0.0700 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}\) was required to reduce all the iodine produced by the bleach back to iodide, what is the mass percent of \(\mathrm{NaClO}\) in the bleach?

Name the following compounds. (a) \(\mathrm{HBrO}_{3}\) (b) \(\mathrm{KIO}\) (c) \(\mathrm{NaClO}_{2}\) (d) \(\mathrm{NaBrO}_{4}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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