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

Explain the difference between a pure substance and a mixture.

Short Answer

Expert verified
A pure substance has a constant composition and properties throughout the sample and consists of a single type of particle, while a mixture is a combination of two or more substances that retain their individual properties and can be separated by physical means.

Step by step solution

01

Define a Pure Substance

A pure substance is a form of matter that has a constant composition and properties that are constant throughout the sample. It consists of a single type of particle, which can be an element or a compound. An element is a substance that cannot be separated into simpler substances by chemical means, while a compound is a substance formed when two or more chemical elements are chemically bonded together. Examples include water, gold, and oxygen.
02

Define a Mixture

A mixture is a combination of two or more substances that are not chemically bonded together and can be separated by physical means. The components of a mixture maintain their own properties and can be present in variable proportions. Mixtures can be homogeneous, where the composition is uniform throughout the sample, or heterogeneous, where the composition varies within the sample. Examples include salt water, air, and salad.
03

Contrast Pure Substances and Mixtures

The main difference between a pure substance and a mixture is in their composition and properties. A pure substance has a uniform and definite composition, while a mixture is composed of different substances that can vary in proportion. The components of a mixture retain their individual properties and can be physically separated, unlike the elements in a compound which are chemically bonded.

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!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chemical Composition
Understanding the chemical composition of substances is crucial when differentiating between pure substances and mixtures. A pure substance has a specific chemical composition, which means it contains only one type of particle, either atoms of a single element or molecules of a single compound. These particles are consistent throughout the substance, giving it a definite proportion of elements that constitute its makeup.

For instance, water (H2O) always consists of two hydrogen atoms bonded to one oxygen atom, and table salt (NaCl) is always comprised of sodium and chloride ions in a 1:1 ratio. In contrast, mixtures have a variable chemical composition. Air, for example, is a mixture containing nitrogen, oxygen, carbon dioxide, and other gases in varying amounts. Since mixtures are not chemically bonded, the proportion of their components can differ from one sample to another. This fundamental difference in chemical composition is what separates pure substances from mixtures.
Physical Separation
The concept of physical separation is pivotal in understanding mixtures. Unlike pure substances, the components of a mixture can be separated using physical methods because they do not entail changing the chemical identities of the substances involved. Separation techniques rely on physical properties like boiling point, density, solubility, or magnetic properties. For example, salt can be recovered from saltwater through the process of evaporation, which is possible because the water has a lower boiling point than salt.

Filtration and Distillation

These are two common methods of physical separation. Filtration is used when particles of different sizes need to be separated, while distillation separates substances based on differences in their boiling points. Both processes exemplify how mixtures can be taken apart to yield their original components, which is a distinct characteristic that differentiates them from pure substances.
Homogeneous and Heterogeneous Mixtures
Mixtures can be classified into two broad categories: homogeneous and heterogeneous. Homogeneous mixtures have a uniform composition throughout, meaning that their components are evenly distributed on a molecular level. Solutions like saltwater or vinegar are prime examples of homogeneous mixtures where the solute is completely dissolved in the solvent, creating a single-phase system that appears the same throughout.

Conversely, heterogeneous mixtures have a non-uniform composition where the individual substances remain distinct and can often be seen as separate phases. Examples include oil and water, which do not mix evenly, or a salad where the different ingredients can be visually and physically separated. Understanding these two types of mixtures helps in comprehending the diversity of matter and provides insight into the processes necessary for their separation and identification, reinforcing the distinction between mixtures and pure substances.

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

Identify each statement as being most like an observation, a law, or a theory. a. All coastal areas experience two high tides and two low tides each day. b. The tides in Earth's oceans are caused mainly by the gravitational attraction of the moon. c. Yesterday, high tide in San Francisco Bay occurred at 2: 43 A.M. and 3: 07 P.M. d. Tides are higher at the full moon and new moon than at other times of the month.

Nanotechnology, the field of building ultrasmall structures one atom at a time, has progressed in recent years. One potential application of nanotechnology is the construction of artificial cells. The simplest cells would probably mimic red blood cells, the body's oxygen transporters. Nanocontainers, perhaps constructed of carbon, could be pumped full of oxygen and injected into a person's bloodstream. If the person needed additional oxygen - due to a heart attack perhaps, or for the purpose of space travel- these containers could slowly release oxygen into the blood, allowing tissues that would otherwise die to remain alive. Suppose that the nanocontainers were cubic and had an edge length of \(25 \mathrm{nm}\). a. What is the volume of one nanocontainer? (Ignore the thickness of the nanocontainer's wall.) b. Suppose that each nanocontainer could contain pure oxygen pressurized to a density of \(85 \mathrm{~g} / \mathrm{L}\). How many grams of oxygen could each nanocontainer contain? c. Air typically contains about 0.28 g of oxygen per liter. An average human inhales about \(0.50 \mathrm{~L}\) of air per breath and takes about 20 breaths per minute. How many grams of oxygen does a human inhale per hour? (Assume two significant figures.) d. What is the minimum number of nanocontainers that a person would need in his or her bloodstream to provide 1 hour's worth of oxygen? e. What is the minimum volume occupied by the number of nanocontainers calculated in part d? Is such a volume feasible, given that total blood volume in an adult is about \(5 \mathrm{~L} ?\)

How many \(1-\mathrm{cm}\) squares would it take to construct a square that is \(1 \mathrm{~m}\) on each side? MISSED THIS? Read Section 1.6

The diameter of a hydrogen atom is \(212 \mathrm{pm}\). Find the length in kilometers of a row of \(6.02 \times 10^{23}\) hydrogen atoms. The diameter of a ping pong ball is \(4.0 \mathrm{~cm} .\) Find the length in kilometers of a row of \(6.02 \times 10^{23}\) ping pong balls.

A thief uses a can of sand to replace a solid gold cylinder that sits on a weight-sensitive, alarmed pedestal. The can of sand and the gold cylinder have exactly the same dimensions (length \(=22\) and radius \(=3.8 \mathrm{~cm}\) ). a. Calculate the mass of each cylinder (ignore the mass of 1 the can itself). (density of gold \(=19.3 \mathrm{~g} / \mathrm{cm}^{3},\) density of sand \(\left.=3.00 \mathrm{~g} / \mathrm{cm}^{3}\right)\) b. Does the thief set off the alarm? Explain.

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