Musical instruments like trumpets and trombones are made from an alloy called brass. Brass is composed of copper and zinc atoms and appears homogeneous under an optical mi-croscope. The approximate composition of most brass objects is a \(2 : 1\) ratio of copper to zinc atoms, but the exact ratio varies somewhat from one piece of brass to another.(a) Would you classify brass as an element, a compound, a homogeneous mixture, or a heterogeneous mixture? (b) Would it be correct to say that brass is a solution? [ Section 1.2\(]\)

Short Answer

Expert verified
(a) Brass is a homogeneous mixture. (b) Yes, it would be correct to say that brass is a solution.

Step by step solution

01

Understanding Elements, Compounds, and Mixtures

An element is a pure substance made of only one type of atom. A compound is a pure substance made of two or more elements chemically bound together in a fixed proportion. A mixture is a combination of two or more substances that are not chemically bound together. A homogeneous mixture has a uniform composition throughout, while a heterogeneous mixture does not have a uniform composition and its components are easily distinguishable.
02

Classifying brass

Brass is composed of copper and zinc atoms, and its composition varies somewhat from one piece of brass to another. The approximate ratio of copper to zinc atoms is \(2:1\). Since brass is composed of two different types of atoms (copper and zinc) and its composition is not fixed, it cannot be considered an element or a compound. Brass appears homogeneous under an optical microscope, so it can be classified as a homogeneous mixture.
03

Determining if brass is a solution

A solution is a type of homogeneous mixture where one substance is uniformly distributed in another substance at the molecular or ionic level. In brass, copper and zinc atoms are uniformly distributed, making it a homogeneous mixture. Thus, it would be correct to say that brass is a solution. Answer: (a) Brass is a homogeneous mixture. (b) Yes, it would be correct to say that brass is a solution.

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

The distance from Earth to the Moon is approximately \(240,000 \mathrm{mi.}\) (a) What is this distance in meters? (b) The peregrine falcon has been measured as traveling up to 350 \(\mathrm{km} /\) hr in a dive. If this falcon could fly to the Moon at this speed, how many seconds would it take? (c) The speed of light is \(3.00 \times 10^{8} \mathrm{m} / \mathrm{s} .\) How long does it take for light to travel from Earth to the Moon and back again? (\boldsymbol{d} ) ~ E a r t h ~ t r a v e l s ~ a r o u n d ~the Sun at an average speed of 29.783 \(\mathrm{km} / \mathrm{s} .\) Convert this speed to miles per hour.

Two positively charged particles are first brought close together and then released. Once released, the repulsion between particles causes them to move away from each other. (a) This is an example of potential energy being converted into what form of energy? (b) Does the potential energy of the two particles prior to release increase or decrease as the distance between them is increased.

(a) The diameter of Earth at the equator is 7926.381 \(\mathrm{mi}\) . Round this number to three significant figures and express it in standard exponential notation. (b) The circumference of Earth through the poles is \(40,008 \mathrm{km}\) . Round this number to four significant figures and express it in standard exponential notation.

Give the derived SI units for each of the following quantities in base Sl units: $$\begin{array}{l}{\text { (a) acceleration }=\text { distance } / \text { time }^{2}} \\ {\text { (b) force }=\text { mass } \times \text { acceleration }} \\ {\text { (c) work }=\text { force } \times \text { distance }} \\ {\text { (d) pressure }=\text { force/area }}\end{array}$$$$ \begin{array}{l}{\text { (e) power }=\text { work/time }} \\ {\text { (f) velocity }=\text { distance } / \text { time }} \\ {\text { (g) energy }=\text { mass } \times \text { (velocity) }^{2}}\end{array} $$

Suppose you decide to define your own temperature scale with units of \(\mathrm{O},\) using the freezing point \(\left(13^{\circ} \mathrm{C}\right)\) and boiling point \(\left(360^{\circ} \mathrm{C}\right)\) of oleic acid, the main component of olive oil. If you set the freezing point of oleic acid as \(0^{\circ} \mathrm{O}\) and the boiling point as \(100^{\circ} \mathrm{O},\) what is the freezing point of water on this new scale?

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