/*! 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 56 The two most important iron oxid... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The two most important iron oxide minerals are magnetite, \(\mathrm{Fe}_{3} \mathrm{O}_{4}\), and hematite, \(\mathrm{Fe}_{2} \mathrm{O}_{3}\). One is a ferrimagnetic material, while the other is an antiferromagnetic material. (a) Based on the oxidation states of iron, which one is more likely to be ferrimagnetic? (b) Would it be possible to use magnetic fields to separate these minerals?

Short Answer

Expert verified
(a) Magnetite (Fe3O4) is more likely to be ferrimagnetic due to its mixed oxidation states of iron, resulting in unequal magnetic moments. (b) Yes, it is possible to use magnetic fields to separate magnetite and hematite due to their differing magnetic properties.

Step by step solution

01

Determine the oxidation states of iron in the minerals

Magnetite has the chemical formula Fe3O4, which means it consists of three iron atoms (Fe) and four oxygen atoms (O). The oxidation state of oxygen is -2. To determine the overall charge of the compound, we use the formula nFe + mO = 0, where n is the oxidation state of iron and m is the count of each element. For Fe3O4, the oxidation state of iron (n) can be calculated as \(3n + 4(-2) = 0\). Similarly, hematite has the chemical formula Fe2O3, which means it contains two iron atoms and three oxygen atoms. The oxidation state of iron (n) can be calculated as \(2n + 3(-2) = 0\).
02

Calculate the oxidation states of iron

Using the equations above, we solve for the oxidation states of iron in magnetite and hematite: For magnetite (Fe3O4): \(3n + 4(-2) = 0 \Rightarrow 3n - 8 = 0 \Rightarrow n = \frac{8}{3}\). For hematite (Fe2O3): \(2n + 3(-2) = 0 \Rightarrow 2n - 6 = 0 \Rightarrow n = 3\). The oxidation states of iron in magnetite and hematite are \(\frac{8}{3}\) (approximately 2 and 3) and 3, respectively.
03

Identify the ferrimagnetic mineral

Ferrimagnetic materials exhibit magnetic properties in which the magnetic moments of atoms or ions on different sublattices are aligned in opposite directions and their magnitudes are unequal. Antiferromagnetic materials have magnetic moments that are aligned oppositely but are equal in magnitude, resulting in a net magnetic moment of zero. Magnetite, with the mixed oxidation states of approximately 2 and 3, is more likely to be ferrimagnetic. This is because the difference in the oxidation states results in unequal magnetic moments, leading to a net magnetic moment in magnetite. Hematite, with the same oxidation state of iron (3) across its lattice, is more likely to be antiferromagnetic due to the equal magnetic moments leading to a net magnetic moment of zero.
04

Assess the possibility of using magnetic fields for separation

Since magnetite is ferrimagnetic and hematite is antiferromagnetic, it is possible to use magnetic fields for their separation. The magnetite with its net magnetic moment will be attracted or repelled by magnetic fields, whereas the hematite with its net magnetic moment of zero will not respond to magnetic fields significantly. Thus, the difference in the magnetic properties of these iron oxides can be exploited to separate them by applying magnetic fields. To summarize: (a) Magnetite (Fe3O4) is more likely to be ferrimagnetic due to its mixed oxidation states of iron, resulting in unequal magnetic moments. (b) Yes, it is possible to use magnetic fields to separate magnetite and hematite due to their differing magnetic properties.

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 91Ó°ÊÓ!

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

Associated with every ferromagnetic solid is a temperature known as its Curie temperature. When heated above its Curie temperature, the substance no longer exhibits ferromagnetism but rather becomes paramagnetic. Use the kinetic- molecular theory of solids to explain this observation.

Write balanced chemical equations for each of the following verbal descriptions: (a) Vanadium oxytrichloride \(\left(\mathrm{VOCl}_{3}\right)\) is formed by the reaction of vanadium(III) chloride with oxygen. (b) Niobium(V) oxide is reduced to the metal with hydrogen gas. (c) Iron(III) ion in aqueous solution is reduced to iron(II) ion in the presence of zinc dust. (d) Niobium(V) chloride reacts with water to yield crystals of niobic acid \(\left(\mathrm{HNbO}_{3}\right)\).

In an electrolytic process nickel sulfide is oxidized in ? two-step reaction: $$ \begin{array}{r} \mathrm{Ni}_{3} \mathrm{~S}_{2}(s) \longrightarrow \mathrm{Ni}^{2+}(a q)+2 \mathrm{NiS}(s)+2 \mathrm{e}^{-} \\ \mathrm{NiS}(s) \longrightarrow \mathrm{Ni}^{2+}(a q)+\mathrm{S}(s)+2 \mathrm{e}^{-} \end{array} $$ What mass of \(\mathrm{Ni}^{2+}\) is produced in solution by passing a current of 67 A for a period of \(11.0 \mathrm{hr}\), assuming the cell is \(90 \%\) efficient?

For each of the following alloy compositions indicate whether you would expect it to be a substitutional alloy, an interstitial alloy, or an intermetallic compound: (a) \(\mathrm{Cu}_{0.66} \mathrm{Zn}_{0.34}\) (b) \(\mathrm{Ag}_{3} \mathrm{Sn},(\mathrm{c}) \mathrm{Ti}_{0.99} \mathrm{O}_{001}\).

Indicate whether each of the following compounds is expected to be diamagnetic or paramagnetic, and give a reason for your answer in each case: (a) \(\mathrm{NbCl}_{5}\), (b) \(\mathrm{CrCl}_{2}\), (c) \(\mathrm{CuCl}\), (d) \(\mathrm{RuO}_{4}\), (e) \(\mathrm{NiCl}_{2}\).

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.