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Chemistry Super Review - 2nd Ed.
Chemistry Super Review - 2nd Ed.
Chemistry Super Review - 2nd Ed.
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Chemistry Super Review - 2nd Ed.

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REA's Chemistry Super Review
Get all you need to know with Super Reviews!
2nd Edition


REA's Chemistry Super Review contains an in-depth review that explains everything high school and college students need to know about the subject. Written in an easy-to-read format, this study guide is an excellent refresher and helps students grasp the important elements quickly and effectively.

Our Chemistry Super Review can be used as a companion to high school and college textbooks, or as a handy resource for anyone who wants to improve their chemistry skills and needs a fast review of the subject.

Presented in a straightforward style, our review covers the material taught in a beginning-level chemistry course, including: atomic structure, bonding, chemical reactions, liquids, solids, gases, properties of solutions, chemical thermodynamics, and more.
The book contains questions and answers to help reinforce what students learned from the review. Quizzes on each topic help students increase their knowledge and understanding and target areas where they need extra review and practice.

LanguageEnglish
Release dateDec 15, 2012
ISBN9780738683867
Chemistry Super Review - 2nd Ed.

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Chemistry Super Review - 2nd Ed. - Editors of REA

Table

CHAPTER 1

Introduction


1.1 Matter and Its Properties

1.1.1 Definition of Matter

Matter occupies space and possesses mass. Mass is an intrinsic property of matter.

Weight is the force, due to gravity, with which an object is attracted to the earth.

Force and mass are related to each other by Newton’s equation (Newton’s Law), F = ma, where F = force, m = mass, and a = acceleration. Weight and mass are related by the equation w = mg, where w = weight, m = mass, and g = acceleration due to gravity.

Note that the terms mass and weight are often (incorrectly) used interchangeably in most literature.

1.1.2 States of Matter

Matter occurs in three states or phases: solid, liquid, and gas. A solid has both a definite size and shape. A liquid has a definite volume but takes the shape of the container, and a gas has neither definite shape nor definite volume.

1.1.3 Composition of Matter

Matter is divided into two categories: distinct substances and mixtures. Distinct substances are either elements or compounds. An element is made up of only one kind of atom. A compound is composed of two or more kinds of atoms joined together in a definite composition.

Mixtures contain two or more distinct substances more or less intimately jumbled together. A mixture has no unique set of properties: it possesses the properties of the substances of which it is composed.

In a homogeneous mixture, the composition and physical properties are uniform throughout. Only a single phase is present. A homogeneous mixture can be gaseous, liquid, or solid. A heterogeneous mixture, such as oil and water, is not uniform and consists of two or more phases.

1.1.4 Properties of Matter

Extensive properties, such as mass and volume, depend on the size of the sample. Intensive properties, such as melting point, boiling point, and density, are independent of sample size.

Physical properties of matter are those properties that can be observed, usually with our senses. Examples of physical properties are physical state, color, and melting point.

Chemical properties of a substance are observed only in chemical reactions involving that substance.

Reactivity is a chemical property that refers to the tendency of a substance to undergo a particular chemical reaction.

Chemical changes are those that involve the breaking and/or forming of chemical bonds, as in a chemical reaction.

Physical changes do not result in the formation of new substances. Changes in state are physical changes.

1.2 Conservation of Matter

1.2.1 Law of Conservation of Matter

In a chemical change, matter is neither created nor destroyed, but only changed from one form to another. This law requires that material balance be maintained in chemical equations.

1.3 Laws of Definite and Multiple Proportions

1.3.1 Law of Definite Proportions

A pure compound is always composed of the same elements combined in a definite proportion by mass.

Problem Solving Example:

It has been determined experimentally that two elements, A and B, react chemically to produce a compound or compounds. Experimental data obtained on combining proportions of the elements are:

(a) Which two laws of chemical change are illustrated by the above data? (b) If 80 g of element B combines with 355 g of a third element C, what weight of A will combine with 71 g of element C? (c) If element B is oxygen, what is the equivalent weight of element C?

(a) If one adds the weight of A to the weight of B and obtains the weight of the compound formed, the law of conservation of matter is illustrated. This law states that there is no detectable gain or loss of matter in a chemical change. Using the data from the experiments described, you find the following:

From these calculations one can see that the law of conservation of matter is shown.

Another important law of chemistry is the law of definite proportions. This law is stated: when elements combine to form a given compound, they do so in a fixed and invariable ratio by mass. One can check to see if this law is adhered to by calculating the ratio of the weight of A to the weight of B in the three experiments. If all of these ratios are equal, the law of definite proportions is shown.

The law of definite proportions is illustrated here.

(b) From the law of definite proportions, one can find the number of grams of B that will combine with 71 g of C. After this weight is found, one can find the number of grams of A that will react with 71 g of C by finding the amount of A that reacts with that amount of B. It is assumed that the amount of A that reacts with 71 g of C is equal to the amount of A that will react with the amount of B that reacts with 71 g of C. The amount of A that will react with this amount of B can be found by remembering, from the previous section of this problem, that A reacts with B in a ratio of 1:1.5.

(1) Finding the amount of B that would react with 71 g of C.

One is told that 80 g of B reacts with 355 g of C. By the law of definite proportions, a ratio can be set up to calculate the number of grams of B that will react with 71 g of C.

Let x = the number of grams of B that will react with 71 g of C.

16 g of B will react with 71 g of C.

(2) It is assumed that the same amount of A that will react with 16 g of B will react with 71 g of C. Therefore, using the fact that the ratio of the amount of A that reacts to the amount of B is equal to 1.5 (this fact was obtained in part (1)), one can calculate the amount of A that will react with 71 g of C.

Let x = the number of grams of A that will react with 16 g of B.

24 g of A will react with 16 g of B or 71 g of C.

(c) In finding the equivalent weight of C when B is taken to be oxygen, the law of definite proportions is used again. The equivalent weight of oxygen is 8 g. Knowing that 16 g of B reacts with 71 g of C, one can set up the following ratio:

Let x = weight of C if the weight of B is taken to be 8 g.

The equivalent weight of C when B is taken to be oxygen is 35.5 g.

1.3.2 Law of Multiple Proportions

When two elements combine to form more than one compound, different masses of one element combine with a fixed mass of the other element such that those different masses of the first element are in small whole-number ratios to each other.

Problem Solving Example:

Two different compounds of elements A and B were found to have the following composition: first compound, 1.188 g of A combined with 0.711 g of B; second compound, 0.396 g of A combined with 0.479 g of B. Show that these data are in accordance with the law of multiple proportions.

The law of multiple proportions can be stated: when two elements combine to form more than one compound, the different masses of one that combine with a fixed mass of the other are in the ratio of small whole numbers. This means that if one solves for the expected amount of B that is used in forming the second compound from the ratio of A:B in experiment 1, the experimental amount should be a multiple of the calculated value. This is seen more clearly after looking at the data.

In experiment 1, A combines with B in a ratio of 1.188 g A:0.711g B, or 1:0.598. In experiment 2, A combines with B in a ratio of 0.396 A:0.479 B, or 1:1.21. The law of multiple proportions states that 0.598 should be a small multiple of 1.21. Thus, 1.21/0.598 = 2 and the law is supported.

1.4 Energy and Conservation of Energy

1.4.1 Definition of Energy

Energy is usually defined as the ability to do work or transfer heat.

1.4.2 Forms of Energy

Energy appears in a variety of forms, such as light, sound, heat, mechanical energy, electrical energy, and chemical energy. Energy can be converted from one form to another.

Two general classifications of energy are potential energy and kinetic energy. Potential energy is due to position in a field. Kinetic energy is the energy of motion and is equal to one-half of an object’s mass, m, multiplied by its speed, s.

1.4.3 Law of Conservation of Energy

Energy can be neither created nor destroyed, but only changed from one form to another.

1.5 Measurement

Numbers that arise as the result of measurement may contain zero or more significant figures (significant digits). The general rule for multiplication or division is that the product or quotient should not possess any more significant figures than does the least precisely known factor in the calculation.

For addition and subtraction, the rule is that the absolute uncertainty in a sum or difference cannot be smaller than the largest absolute uncertainty in any of the terms in the calculation, i.e., the number of significant figures is limited by the number of digits to the right of the decimal point in the term that is known to the fewest decimal places.

Some examples of assigning significant figures (SF):

Examples of arithmetic using the rules for significant figures:

1.5.1 Length

1 meter = 39.37 inches, 12 inches = 1 foot

1 meter = 100 cm, 1 meter = 1,000 mm, 1 inch = 2.54 cm

1.5.2 Volume

Volume = height × length × width [for

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