Math Review - Southwest DeKalb High Schoolswdekalbhs.dekalb.k12.ga.us/Downloads/Math Review.pdfA number in scientific notation has two parts. The number in front of the “x 10”

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Math ReviewTable of Contents

Scientific Notation Page 2

Rounding Page 6

Conversion Factors Page 7

Fractions, Decimals, and Percents Page 11

Geometric Figures Page 17

Balancing Equations Page 21

Signed Numbers Page 27

Basic Algebra Page 31

Cross Multiplication Page 34

Significant Figures in Logarithms Page 40

Logarithms and Inverse Logarithms Page 42

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Scientific Notation

Any quantity can be expressed using a power of ten. As you move the decimal point, youmultiply by 10 as many times as necessary to make the numbers equal. Consider thefollowing examples:

325 = 32.5 x 10 = 3.25 x 101

325 = 3.25 x 10 x 10 = 3.25 x 102

325 = 0.325 x 10 x 10 x 10 = 3.25 x 103

Because 100 = 1 we can also express 325 as 325 x 100.

A number in scientific notation has two parts. The number in front of the “x 10” is calledthe coefficient. The power to which 10 is raised is called the exponent.

3.25 x 103

The coefficient must have one and only one digit in front of the decimal point. Theexponent will always be a whole number in chemistry, although the exponent can be afractional number in other disciplines.

There are three rules for using scientific notation:

Rule 1: To express a number in scientific notation, you move the decimal point tothe position such that there is one nonzero digit to the left of the decimal point.

Rule 2: If the decimal point is moved to the left, the exponent is positive.

Rule 3: If the decimal point is moved to the right, the exponent is negative.

A mnemonic that my help you remember how to keep the signs straight is “RegisteredNurses Love Patients,” or RN, LP, which stands for right--negative, left--positive. If youmove the decimal point to the right, the exponent is negative. If you move the decimalpoint to the left, the exponent is positive.

Coefficient

Exponent

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Scientific Notation Example 1:

Express 0.0003821 in scientific notation.

Solution:

0.0003821 = 3.821 x 10-4

The decimal point was moved to the right four spaces. If the decimal point movesto the right, the exponent is negative. Since it moved four spaces, the exponent is a–4. Notice that the decimal point was moved until one and only one nonzero digitwas in front of the decimal point.


Express 56,873 in scientific notation.

Solution:

5.6873 x 104

The decimal point was moved to the left four spaces. If the decimal point moves tothe left, the exponent is positive. Since it moved four spaces, the exponent is a +4.Notice that the decimal point was moved until one and only one nonzero digit was infront of the decimal point.

Sometimes it is necessary to change a number from one power of ten to another power often. When changing to a new power of ten, we must consider the sizes of both thecoefficient and the power of ten. In the number 3.2 x 105, the coefficient is 3.2 and thepower of ten is 5. If we wish to express this as another power of 10 that is equal in valueto 3.2 x 105, we must increase the power of ten if the coefficient decreases, and decreasethe power of ten if the coefficient increases This is shown in the illustration below.


Fill in the blank: 6.79 x 104 = ________ x 106

Solution:

The exponent has gotten larger. This means that we have multiplied the coefficientby 10 two more times. Because the exponent has gotten larger by 2 places, the

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coefficient must get smaller by two places. We must change the 6.79 to 0.0679,which is the value that goes in the blank.

6.79 x 104 = 0.0679 x 106

Although you will probably use your calculator to perform arithmetic operations onnumbers written in scientific notation, you should know the rules that govern theseprocesses. The rules are presented below.

Addition and Subtraction1. Express the numbers as the same power of 10.2. Perform the mathematical operation on the coefficients (add or subtract).3. Bring down the power of 10 unchanged.4. Express the answer in proper scientific notation.


Subtract 4.72 x 102 from 5.32 x 103

Solution:

Make the exponents equal before we add the numbers. We can make both of theexponents 102

or we can make them both 103 The result will be the same, nomatter which common exponent we choose. We’ll make them both 102. Therefore5.32 x 103 = 53.2 x 1024. Now we can subtract the two numbers. Since the finalanswer has more than one nonzero digit in the coefficient, we need to rewrite thenumber in proper scientific notation.

53.20 x 102

- 4.72 x 102

48.48 x 102 or 4.848 x 103

Multiplication and Division

1. To multiply two numbers expressed in scientific notation, multiply the coefficientsand add the exponents on the powers of 10.

2. To divide two numbers expressed in scientific notation, divide the coefficients andsubtract the exponents on the powers of 10.

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What is the product of 5.1 x 10-4 and 3.2 x 10-2?

Solution:

Multiply 5.1 by 3.2 to get 16. Add the exponents to get –6. The answer is 16 x 10-6.Change this number to proper scientific notation. The answer is 1.6 x 10-5.

Hints for Using Scientific Notation:

Is it necessary to use scientific notation?

Not every calculation requires scientific notation, but it is a very convenient system forhandling the very large and very small numbers commonly encountered in chemistry.Scientific notation shows up very frequently in chemistry, so mastering it now will pay off inthe future.

Can I use a calculator to solve problems with numbers in scientific notation?

Yes. You will need to find out how to input scientific notation numbers into your calculator.Check your calculator manual for directions. The most common mistake in the earlypart of this course is entering a number in scientific notation incorrectly into acalculator!

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Rounding

It is often necessary to round numbers after performing calculations. This is usuallynecessary when the calculated answer has more significant figures than are allowed. Therules for rounding numbers are presented below.

1. If the last digit to be kept is followed by a number less than 5, round by droppingall digits to the right of the kept number.

2. If the last digit to be kept is followed by a number equal to or greater than five,round up by adding 1 to the last digit to be kept and drop the digits that follow thekept number.

Rounding Example 1:

Round the following number to the indicated number of significant digits.

54.674873 rounded to four digits0.8765 rounded to two digits5.875 rounded to three digits

Solution:

The last digit to be kept in 54.674873 is a seven. The seven is followed by anumber less than five, so drop all of the digits after the seven in blue. The answeris 54.67.

The last digit to be kept in 0.8765 is a seven. The seven is followed by a numbergreater than or equal to 5, so round up. The answer is 0.88.

The last digit to be kept in 5.875 is a seven. The seven is followed by a numbergreater than or equal to 5, so round up. The answer is 5.88.

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Conversion Factors

Conversion factors are ratios of one object to another object. A ratio is a way of comparingtwo quantities. The quantities can be compared in three different ways: a to b, a:b, or a/b.At the local grocery store, a case of soda contains 24 cans. We can express the ratio inthree forms:

(a to b) 24 cans of soda to each case of soda (a:b) 24 cans of soda: 1 case of soda (a/b) 24 cans of soda/1 case of soda

This course uses conversion factors frequently to solve problems. Conversion factors areratios written in the fraction form (a/b).

Some of the conversion factors you will see in this course include metric-to-metricconversion factors and molar conversion factors such as:

1 gram of silver is 1000 mg of silver

1 liter of water is 1 x 106 µL of water

1 mole of CuCl2 contains 6.02 x 1023 molecules of CuCl2

The most common way that chemists represent ratios like these is in a fraction form. Theconversion factors would look like this in the fraction form.

1 gram silver 1 liter water 1 mole of CuCl21000 mg silver 1 x 106 µL water 6.02 x 1023 molecules of CuCl2

Each of these conversion factors can be written in the inverse form The first ratio weexamined involved cans of soda. We expressed that ratio as 24 cans of soda in 1 case ofsoda. The inverse form of this ratio is 1 case of soda contains 24 cans of soda. The keyidea here is that both the original ratio and its inverse form are true. It is true that onecase of soda contains 24 cans and it is true that there are 24 cans of soda in one case.

Conversion factors in chemistry can be written as shown above or in the inverse form.Regardless of which form is used, the comparison the conversion factor makes is true.For example, one gram of silver weighs 1000 mg and 1000 mg of silver weighs one gramare both true.

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Here are the conversion factors written in the inverse form.

1000 mg silver 1 x 106 µL water 6.02 x 1023 molecules of CuCl21 gram silver 1 liter water 1 mole of CuCl2

The obvious question at this point is: Which form of the conversion factor should you usein a problem? You use the form that allows the units to cancel. Units cancel when theunit that appears on the top also appears on the bottom of a fraction somewhere in thesolution. This style of problem solving, where you arrange the conversion factors so thatthe units cancel, is called unit analysis.

Conversion Factor Example 1

Which is the correct solution for the question: How many cases of soda can be made from1200 cans of soda?

1200 cans of soda x 24 cans of soda = 28800 cans of soda2/case of soda1 case of soda

1200 cans of soda x 1 case of soda = 50 cases of soda24 cans of soda

Solution:

In which solution do the units cancel? In the second solution, the unit “cans ofsoda” cancels. The correct solution will use conversion factors in a way that unitswill cancel. Blue lines are drawn through units that cancel.


How many milligrams of silver are in 2.76 grams of silver?

Solution:

This question uses the metric-to-metric conversion factor: 1 gram = 1000 mg. Setup the solution so that the units cancel. Blue lines are drawn through units thatcancel.

2.76 grams silver x 1000 mg silver = 2760 mg silver 1 gram silver

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If 8.35 x 109 µL of water were collected in a flask, how many liters of water were collected?

Solution:

You don’t need to concern yourself with whether the conversion factor is inverted ornot. Just make sure that the units cancel. Blue lines are drawn through units thatcancel.

8.35 x 109 µL water x 1 liter water = 8.35 x 103 liters water1 x 106 µL


2.5 moles of CuCl2 contain how many molecules of CuCl2?

Solution:

This question uses the conversion factor of moles to molecules. Set up the solutionso that the units cancel. Blue lines are drawn through units that cancel.

2.5 moles of CuCl2 x 6.02 x 1023 molecules CuCl2 = 1.5 x 1024 molecules CuCl21 mole CuCl2


The formula for sugar is C6H12O6. How many moles of sugar are present when 20.0 molesof oxygen are present?

Solution:

This question requires the chemical formula conversion factor in the solution. Youdon’t need to concern yourself with whether the conversion factor is inverted or not.Just make sure that the units cancel. Blue lines are drawn through units that cancel.

20.0 moles of oxygen x 1 mole C6H12O6 = 3.33 moles C6H12O6

6 moles oxygen

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Hints for using conversion factors:

How do you know what quantity to start with when solving problems?

Start with the quantity given in the problem and not with the conversion factor. In general,the first quantity in your unit analysis will not have a denominator. In example 3, westarted with the number of microliters of water:

8.35 x 109 µL water x 1 liter water = 8.35 x 103 liters water1 x 106 µL

Notice that the first quantity is a whole number and not a fraction. If you start with theconversion factor, it is much harder to decide whether to use the original or inverse form ofthe conversion factor.

Should I enter all of the numerator numbers into the calculator before I enter thedenominator numbers, or should I enter each conversion factor into the calculatorbefore I move on to the next conversion factor?

Either style of entering the numerical quantities will work. You should make a decision toalways use one style of entering numbers into a calculator. This helps you avoid skippinga number when you are calculating an answer.

In the example calculation below, you could enter the numbers in either of these ordersand obtain the same answer.

16 x 2.54 x 50 = 75.2 3 9

Enter the numbers into your calculator as:

16 x 2.54 ÷ 3 x 50 ÷ 9 =

or

16 x 2.54 x 50 ÷ 3 ÷ 9 =

You don’t need to press the ENTER key after each mathematical operation.

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Fractions, Decimals, and Percents

Fractions, decimals, and percentages are all related to each other. Each is used toindicate some part of a whole quantity. For example, 3/4, 0.75, and 75% are all used toexpress a quantity representing three parts out of four.

The fraction represents the number of parts of the whole (3 parts of the whole, which isdivided into 4 equal sized parts).

The decimal represents the part of one whole (0.75 of the whole).

The percentage represents the number of parts of a whole that has been divided into 100parts (75 parts of the whole which has been divided into 100 equal parts).

The fraction, 3/4, is converted to a decimal by dividing the numerator (the number on top ofthe fraction) by the denominator (the number on bottom of the fraction). Three divided byfour is 0.75.

The decimal is converted to the percentage by moving the decimal point two places to theright. This is equivalent to multiplying the decimal by 100 in order to convert it to thepercentage. In this conversion, 0.75 becomes 75%

To convert a percent to a decimal, we move the decimal point two places to the left. Thedecimal point in 75% is understood to be after the five in the percentage. In thisconversion, 75% becomes 0.75.

Divide pie up into 100segments

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In this course, we use percentages more often than decimals or fractions. The termpercent means “per cent” or “per 100.” When we say that 10% of people prefer Brand A,we mean that 10 people out of 100 people prefer Brand A. When someone has 71% ofthe votes in an election, it means that 71 people out of 100 people voted for that person. Achemist might say that a substance is 64% nitrogen. The chemist means that 64 atomsout of every 100 atoms are nitrogen.

A percentage is another conversion factor. It can be written as a fraction and used inproblem solving (also called unit analysis). Here are examples of the percentages weused above written as conversion factors.

10 people prefer Brand A100 people expressed an opinion

71 people voted for candidate100 people voted

64 atoms of nitrogen100 atoms present

The secret to using percentages correctly is to write the units correctly. It is veryimportant that the units for the “per 100” part indicate what substance is being measured orcounted. The numerator or top of the fraction should clearly indicate what small portion ofthe whole quantity is being measured or counted. In other words, a percentage should bewritten as:

The PartThe Whole

Percentage Example 1:

How would you present each percentage as a fraction?

47% of applicants were women15% of atoms were sulfur

Solution:

47 female applicants (the Part)100 total applicants (the Whole)

15 atoms sulfur (the Part)100 atoms total (the Whole)

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Sometimes we are asked to calculate percentages from actual measurements. In this typeof problem, a percent is not given in the problem.


What percent of people in the class are women if 30 of the 50 students are female?

Solution:

Remember to divide the part by the whole. What is the “whole” class number? 50.We divide the “part” (which is 30 female students) by the “whole” (which is 50 maleand female students). Then, we multiply by 100 to create the percentage.

30 women x 100 = 60%50 students

Chemists frequently use percentages in calculations. Remember, it is important that youinclude units in your calculations to avoid mistakes.


A total of 476 people were surveyed. When asked to choose the best tasting pizza, 34.2%of them chose Brand B. How many people selected Brand B as the tastiest pizza?

Solution:

First, convert the percentage into a conversion factor. Remember that thedenominator is always 100 (from the “per cent” part of the question).

The conversion factor we would write would be:

34.2 of people selected Brand B100 people surveyed

The total number of people we surveyed was MORE than 100. It was 476. How dowe include this information in our calculations?

476 people surveyed x 34.2 of people selected Brand B100 people surveyed

= 163 people selected Brand B

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Notice how the units cancel! We can check the answer to see if it is reasonable. If 476people were surveyed, is it reasonable that less than 476 chose Brand B? Yes. Thepercentage of people selecting Brand B was less than 50%. Is 163 less than half of 476?Yes.


If 54% of the mass of a mixture of iron and copper is iron, how much iron is there in asample that has a mass of 25 grams?

Solution:

What is the conversion factor we would write for the percentage?

54 grams of iron100 grams mixture

How did we choose the units “grams?” The unit was given in the problem. Now,let’s use the conversion factor that we obtained from the percentage in the solution.

25 grams mixture x 54 g iron = 14 grams iron100 gram mixture

Again, check to see if the answer is reasonable. If about half of the mixture wasiron and the mixture weighed 25 grams, we would expect an answer of about 12.5grams. Our estimate of the answer agrees with the calculated answer.


If a mixture of iron and copper was 54% iron, how much of the mixture would you need touse to have 200. grams of iron (plus the copper impurity in the mixture)?

Solution:

This question uses the same conversion factor as the previous example. But howshould you arrange the conversion factor so that the units cancel?

200. grams iron x 100 grams mixture = 370 g mixture 54 grams iron

Notice that we used the inverse form of the percentage. We know to do this so thatthe units cancel. If you set up the problem so that the units cancel, you will be onthe right track to get the correct answer!

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If a solution is 25.3 % (m/v), how many grams of solute are present in 250 mL of solution?

Solution:

What is the conversion factor we would write for the percentage?

25.3 grams of solute100 milliliters solution

How did we choose the units “grams” or “milliliters?” The unit of the solution wasm/v% or mass/volume percent. We could choose any unit for mass (grams,milligrams, or kilograms) or any unit of volume (milliliters or liter), but the mostcommonly chosen unit for a solution is gram and milliliter.

Now, let’s use the conversion factor that we obtained from the percentage in thesolution.

250 milliliters solution x 25.3 grams solute = 63.3 grams solute100 milliliters solution

Again, check to see if the answer is reasonable. If each 100 grams of solutioncontain about 25 grams of solute, then 200 grams of solution should contain 50grams of solute. We have a little more than 200 grams of solution, so we shouldhave more than 50 grams of solute.


If a solution is 16.4 % (m/v) glucose, how many mL of solution are needed to deliver 25.0grams of glucose to a patient?

Solution:

The conversion factor is 16.4 grams solute/100 mL of solution. Notice that we havea unit of mass in the numerator and a unit of volume in the denominator. We chosethe unit “milliliter” because we need a volume in the denominator. The “100” part ofthe denominator is understood.

25.0 grams glucose x 100 mL solution = 152 mL solution 16.4 grams glucose

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Notice that we used the inverse form of the percentage. We know to do this so thatthe units cancel. If you set up the problem so that the units cancel, you will be onthe right track to get the correct answer!

Hints for using percentages:

What units do you choose when turning the percentage into a conversion factor?

Look at the question to see what unit of measurement is being used or what substancewas being counted. If the question describes centimeters of metal tubing, then bothnumerator and the denominator will be in centimeters. If the question involved milliliters ofsolution, then both the numerator and the denominator will be in milliliters. Add otherwords to the units to help you distinguish “The Part” from “The Whole.”

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Geometric Figures

Geometric figures and the angles associated with them form an important foundation fordiscussing the shapes of molecules and the bond angles in molecules. You shouldbecome comfortable with the geometric names and angles used in this chapter.

There are 360° as you go around a circle.

The angles change as you cut the circle into different numbers of pieces. The angle of astraight line is 180°.

Molecules whose atoms have a 180° bond angle are said to have linear geometry.

The bond between the small blue atom and the large gray atom is in the same line as thebond between the two large gray atoms. The angle between these two bonds is 180° Ifyou had to describe the orientation of the blue atom to the two gray atoms, you would saythat they are all on the same line or that they have a linear geometry.

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The angles at the intersection of three lines equally spaced around a circle are120°.

Molecules whose bonds are 120° apart have trigonal planar geometry.

The angle between the single bond and the double bond is 120° The blue, black, and redatoms are all in the same plane. If you traced a line from the blue atom to the red atom tothe other blue atom and then back to the first blue atom, you would have a triangularshape. Thus, we describe the geometry as trigonal planar.

The angles at the intersection of two lines perpendicular and crossing each other are 90°.

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There are no 90° bond angles in common biological molecules. However, there aremany atoms that have four bonds. What angle do the four bonds make? The make 109°bond angles. Look at the molecules below. Are the bond angles 90° or 109°?

If you look carefully, you will see that the bond angle is more than 90° You can also seethat the four blue atoms are not in the same plane. Two of the blue atoms are muchsmaller than the other two blue atoms because, in space, they are behind the larger blueatoms. This kind of molecule is NOT planar. If we traced a line around the moleculesgoing from blue atom to blue atom, we would have a tetrahedral shape. The geometry ofthis molecule is tetrahedral.

Geometry Example 1:

The BF3 molecule, shown below, has a shape that is called trigonal planar. What are thebond angles in the molecule?

Solution:

Since the angles are formed by the intersection of three lines that come together atthe center, the bond angles are 120°.

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Geometry Example 2:

The BeCl2 molecule below has 180° bond angles. What is the geometry of this molecule?

Solution:

180° bond angles give the molecule linear geometry.

Hints for Using Geometries:

What is the most common mistake in molecular geometry?

Students often think of molecules as flat, planar structures. Once an atom makes 4 bonds,the geometry becomes very three-dimensional. There are NO 90° bond angles in commonbiological molecules. Carbon makes four bonds, but the angle between the bonds is 109°and the geometry is tetrahedral.

The second most common mistake is not memorizing the possible geometries of biologicalmolecules. You must have the names and angles memorized in order to understand thechemical and physical properties of organic and biological molecules.

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Balancing Equations

A mathematical equation is simply a sentence that states that two expressions are equal.One or both of the expressions will contain a variable whose value must be determined bysolving the equation. Linear equations are equations in which the variable being solved foris raised to a power no higher than 1. Equations in which the variable being solved for israised to a power of 2 are called quadratic equations. The problems in this text do notrequire you to solve quadratic equations, so that topic will not be covered here. Considerthe following linear equation.

5x - 3 = 2x + 9

To solve the equation for x, the first step is to get all of the terms involving x on one side ofthe equation, and all other terms on the other side. We are allowed to perform thefollowing operations on any equation without changing the equality that the equationrepresents.

1. Adding the same quantity to both sides of the equation.

2. Subtracting the same quantity from both sides of the equation.

3. Multiplying both sides of the equation by the same quantity.

4. Dividing both sides of the equation by the same quantity.

The basic rule is: Whatever you do to one side, do to the other side also.

In the equation 5x - 3 = 2x + 9, we will collect all of the x’s on the left side of theequation and all of the numbers on the right side. We must add 3 to each side to get rid ofthe numbers on the left side, and subtract 2x from each side to get rid of the x’s on theright side.

Original equation: 5x - 3 = 2x + 9(1) add three to both sides: + 3 + 3

5x = 2x + 12(2) subtract 2x from both sides: -2x -2x

3x = 12

Now to solve for x, we must divide both sides of the equation by 3.

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(3) divide both sides by 3: 3x = 123 3

Final Answer: x = 4

Balancing Equations Example 1

Solve for the value of x in the equation below.

6x + 7 = 4x + 23

Solution:

First, subtract 7 from both sides. Then, subtract 4x from both sides. Finally, divideboth sides by 2 to get the final answer: x = 8. You will be on track to get the rightanswer if you remember one simple rule: Whatever you do to one side, do to theother side also.

Now what does this have to do with balancing equations? You can use basic algebrarearrangements like we did in the example above to help you balance equations. Look atthis nuclear equation.

84218

Po � 24 He + ?

A balanced nuclear equation must have the same atomic number (the subscriptednumber) on each side of the equation. A balanced nuclear equation must have the sametotal mass number (the superscripted number) on each side of the equation.

We can set up two algebraic equations to describe this chemical reaction. For the massnumber, we can say:

Mass number: 218 = 4 + x

For the atomic number, we can say:

Atomic number: 84 = 2 + y

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Isolate the x or the y to solve the equation. In the mass number equation, subtract fourfrom both sides of the equation. In the atomic number equation, subtract two from bothsides. The equations become:

Mass Number: 214 = xAtomic Number: 82 = y

Finally, use the periodic table to identify the element that has 82 protons. The element isPb. The complete nuclear equation is:

84218

Po � 24

He + 82214

Pb


Balance this nuclear equation:

53131

I � 54131

Xe + ?

Solution:

Set up two algebraic equations, one for the mass number and one for the atomicnumber.

Mass Number: 131 = 131 + xAtomic Number: 53 = 54 + y

Isolate the variable, x or y. The equations become:

Mass Number: 0 = xAtomic Number: -1 = y

The balanced nuclear equation is:

53131 I �

54131 Xe +

10−

e

A beta particle was emitted.

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Let’s look at a simple application of algebra to the problem of balancing a chemicalequation.

Fe(s) + O2(g) � Fe2O3(s)

A balanced equation must have the same number of each kind of atom on both sides ofthe equation. The same number of iron atoms must appear on both sides of the equation.The same number of oxygen atoms must appear on both sides of the equation.

Let’s balance the iron atoms first. There is one iron atom on the left side of the equationand two iron atoms on the right side of the equation. What do you need to multiply the leftside of the equation by in order to make the two sides of the equation have the samenumber of iron atoms? Let this unknown multiplication factor be “x.” We can set up analgebraic equation for this question that looks like:

x(1) = 2

Solving for x by dividing both sides by one gives us the final answer of: x = 2. We need toput a coefficient of two in front of Fe(s) on the left to balance the iron atoms.

2 Fe(s) + O2(g) � Fe2O3(s)

There are two oxygen atoms on the left side of the equation and three oxygen atoms onthe right side of the equation. The algebra expression that represents this situation wouldneed to say that some unknown multiplication factor, y, times the two oxygen atoms on theleft will equal the three oxygen atoms on the right.

y(2) = 3

Solve for y by dividing both sides by two. The final answer is y = 3/2. We need to put acoefficient of 3/2 in front of the O2 on the left to balance the oxygen atoms. It is easier toleave this answer in the fractional form rather than convert it to a decimal number. You’llsee why soon.

2 Fe(s) + 3/2 O2(g) � Fe2O3(s)

The equation is now balanced. However, as often happens, a fraction appears in thebalanced equation. Chemical reactions, except in a few limited circumstances, should notbe balanced with fractions. In order to eliminate the fraction, we need to multiply all of thecoefficients by the number in the denominator. Remember the basic rule: Whatever youdo to one side, do to the other side also.

2 . 2Fe(s) + 2 .

3/2 O2(g) � 2 Fe2O3(s)

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The equation becomes:

4 Fe(s) + 3 O2(g) � 2 Fe2O3(s)


Balance the equation

C2H6(g) + O2(g) � CO2(g) + H2O(l)

Solution:

Begin by balancing the carbons. The algebra expression for carbons could be

2 = x(1).Solving for x gives the final answer: 2 = x. You need a coefficient of two in front ofCO2(g).

The algebra expression for hydrogen could be

6 = y(2).

Solving for y gives the final answer: 3 = y. You need a coefficient of 3 in front ofH2O(l).

C2H6(g) + O2(g) � 2 CO2(g) + 3 H2O(l)

The algebra expression for oxygen is more complex. There are two oxygen atomson the left side of the equation and 4 + 3 oxygen atoms on the right side of theequation. The algebra expression could be:

z(2) = 7

Solving for z gives the final answer: z = 7/2. The balanced equation is:

C2H6(g) + 7/2 O2(g) � 2 CO2(g) + 3 H2O(l)

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Although the equation is balanced, we need to eliminate the fraction from thechemical equation. We do this by multiplying coefficients on both sides of theequation by whatever value is in the denominator. In this case, we would multiplyby two.

2 C2H6(g) + 2 . 7/2 O2(g) � 2 . 2 CO2(g) + 2 . 3 H2O(l)

The equation becomes

2 C2H6(g) + 7 O2(g) � 4 CO2(g) + 6 H2O(l)

Hints for Balancing Equations:

Do I need to use algebra to balance an equation?

No, but it is the most reliable problem solving method. You are much less likely to make amistake if you write out an algebraic expression than if you attempt to solve the problem inyour head.

I got a negative numbers or a zero for my answer. What did I do wrong?

You did nothing wrong at all if you are balancing a nuclear equation. Beta and gammaradiation involves particles with a negative number or a zero. You should not get anegative number or a zero when balancing a chemical equation.

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Signed Numbers

Both positive and negative numbers are used to describe the values of quantities andproperties. Certain properties can have only positive values, such as the number of atomsin a molecule. Some variables, however, can have negative or positive values, such as thetemperature in degrees Fahrenheit or Celsius.

Positive numbers will generally be written with no sign in front of them, but negativenumbers will always have a minus sign in front of them. For example, a temperature of 20degrees below zero Fahrenheit will be expressed as -20° F, but a temperature of 20degrees above zero Fahrenheit will be expressed as 20° F.

In chemistry, you will have to perform arithmetic operations on both positive and negativenumbers. The rules for performing arithmetic operations on signed numbers are presentedbelow. Although you will usually perform these operations with your calculator, it is helpfulto know the rules so you will not be dependent on your calculator.

Adding Signed Numbers

1. When the signs of the numbers are the same, add the values and assign to theanswer the sign of the numbers being added.

2. When the signs of the numbers are different, subtract the smaller valuefrom the larger value. The sign of the answer will be the sign of the large number.

3. When adding more than two signed numbers, add all the positive numbers to getone positive number, and add all the negative numbers to get one negative number.Then add the one positive number to the one negative number, using the rulefor adding numbers with different signs.

Signed Numbers Example 1:

Add: - 12 + - 7

Solution:

The two numbers have the same sign. Add the numbers to get 19. Assign anegative sign to the answer since both values were negative.

The final answer is –19. Remember to use the +/- key to make a number negative.Using the subtraction key makes your calculator perform a mathematical operationrather than change the sign of a number.

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Add: – 8 + 9 + (-10) + 5

Solution:

Add the positive numbers to get 14, and add the negative numbers to get –18.Then, subtract 14 from 18 to get 4. The larger number was 18 and it was negative,so we give the answer a negative sign. The final answer is –4. Remember to usethe +/- key to make a number negative.

.

Subtracting Signed Numbers

To subtract signed numbers, we change the sign of the number being subtracted and usethe rules of addition.


Subtract: -35 – (-42)

Solution:

Change the sign of the number being subtracted. The equation becomes: -35 +42. Since the numbers have different signs, subtract the smaller number, 35, fromthe larger number, 42. This gives 7. The sign of the larger number is positive, sothe final answer is 7.

Multiplying and Dividing Signed Numbers

When multiplying and dividing signed numbers, we perform the mathematical operation onthe numbers. If both signs are the same, the answer is positive; if the two signs aredifferent, the answer is negative.

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Perform the following operation: (-8)(-6).

Solution:

Since the signs are the same, the product will be positive. The final answer is 48.


Perform the following operation: -16 / 4.

Solution:

Since the signs are different, the answer will be negative. The final answer is –4.


What is the temperature in Kelvin when the temperature of a gas is - 47 oC?

Solution:

When the temperature is a negative number, you add 273 to the number to obtainthe temperature in the Kelvin scale just like you did when the temperature waspositive. Subtract the smaller number from the larger number and assign the signof the larger number to the answer.

Temperature in Kelvin = Temperature in Celsius + 273

Temperature in Kelvin = - 47oC + 273

Temperature in Kelvin = 226 K

The temperature in Kelvin must be less than 273 since the Celsius temperature wasnegative. Use the “plus/minus” key on your calculator to enter the negativesign, not the “subtract” key.

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Hints for Using Signed Numbers:

I can’t remember all of the rules. What should I do?

If you can’t remember the rules, then you have to use common sense. If you add a largenegative number and a small positive number, the answer will be negative. Whensubtracting, rewrite the calculation with the sign change and then add the numbers. Again,use common sense to give the answer the same sign as the larger of the numbers youadded.

For multiplication and division, remember that “two wrongs make a right.” Two negativenumbers multiplied together make a positive number. The negative signs cancel eachother out. It’s the same idea as saying you are not not going to pay taxes. The doublenegative makes the statement a positive statement: you are going to pay taxes.

How do I enter negative numbers in my calculator?

The most common error when using negative numbers is that a student uses the “subtract”key on their calculator instead of the “plus/minus” key. In example two, we needed tocalculate the temperature in Kelvin of a gas that is below zero in the Celsius scale.

Temperature in Kelvin = - 47oC + 273

You could enter this into your calculator as:

47 plus/minus key addition key 273 equal or enter key

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Basic Algebra

Algebra encompasses the mathematical manipulations needed to solve for an unknown.The unknown is often called the variable. It is most often symbolized with an “x.” Inchemistry, the unknown or variable often is labeled with a different letter. The letter ischosen to remind the scientist what quantity he is solving for. If the quantity is pressure,the variable is “P.” If the quantity is temperature, the variable is “T.”

Many different kinds of equations can be solved using algebra. The most common algebraequation you will have to solve in chemistry has the form:

6x = 24

where a variable appears on one side of the equation with a coefficient. You can solve thisbasic algebra equation by dividing BOTH sides of the equation by the coefficient in front ofthe variable (in this case, the six).

6x = 246 6

On the side with the variable, the six cancels out. That is because 6 divided by 6 is one.On the other side of the equation (the side without the variable), 24 divided by 6 is four.

6x = 24 6 6

The equation now is: x = 4.

You will see this basic algebra equation frequently in this course. Boyle’s Law, wherepressure is compared to volume, is the good example of this algebraic manipulation.

Basic Algebra Example 1:

A gas occupies 3.5 L at 0.80 atm. If the volume is reduced to 1.7 L, what pressure will thegas experience?

Solution:

Boyle’s Law uses the equation P1V1 = P2V2 One of the four variables in the equationis unknown.

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Initial Conditions Final ConditionsP1 = 0.80 atm P2 = ?V1 = 3.5 L V2 = 1.7 L

When we fill in the numbers, the equation becomes:

(0.80 atm)(3.5 L) = (P2)(1.7 L)

(0.80 atm)(3.5 L) = (P2)(1.7 L)1.7 L 1.7 L

Divide both sides by the coefficient on the side of the equation with the variable P2

For this question, we divide both sides by 1.7 L. The unit “liter” cancels out on bothsides. The value “1.7” on the right side is canceled out by dividing by “1.7.”

1.6 atm = P2

Hints for solving basic algebra equations:

What does it mean when the question says ”solve for P2”?

This is just a mathematician’s way to tell you to do what we did in example 1. You shouldget P2 alone on one side of the equation.

Can I rearrange the equation so that both pressures are on the same side?

No. The equation P1V1 = P2V2 is NOT the same as P1P2 = V1V2. The units wouldn’tcancel and you would get a different answer. Although this alternative equation could besolved in the same manner as the first equation using basic algebra, you cannot changethe equation (or formula) and get the same answer.

I didn’t get the answer in the book, but I know I’m solving the equation using correctalgebra. What’s the problem?

The most common mistake in gas chemistry is not assigning the initial conditions and finalconditions correctly. If we take the information in the first example, but reverse the initialand final volumes, the table would look like:

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Initial Conditions Final ConditionsP1 = 0.80 atm P2 = ?V1 = 1.7 L V2 = 3.5 L

Instead of the volume getting smaller, the question now has the volume getting larger. Theequation becomes

(0.80 atm)(1.7 L) = (P2)(3.5 L)

0.39 atm = P2

We got a totally different answer because we are asking a different question. Read eachquestion carefully to determine what quantity is present initially and what quantity ispresent at the end.

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Cross Multiplication

Cross multiplication is a valuable mathematical solving technique because it allows you toconvert a fraction, which can be tricky to use, into a simpler equation. If, for example,there were 5 members on a team and you had 7 teams, you could set up an equality usingfractions to find the number of people on 7 teams.

5 team members = x 1 team 7 teams

To solve this problem with the two fractions more easily, you cross multiply. The term“cross multiply” was coined because students first drew a big “X,” or “cross,” to connect thetwo numbers that need to be multiplied together to solve the problem.

5 team members = x 1 team 7 teams

The arrows tell us to multiply 5 team members by 7 teams and to multiply 1 team by theunknown. This trick helps us turn a tricky fraction into a common algebra problem.

(5 team members)(7 teams) = (1 team)(x)

In the previous section of this review, we learned how to solve this basic algebra problem:divide both sides by the term “1 team.” Notice how the unit “team” cancels on both sidesof the equation.

(5 team members)(7 teams) = (1 team)(x)1 team 1 team

We are left with:

35 team members = x

The total number of team members (unknown x) is 35.

Cross multiplication is used frequently in gas chemistry to solve for one of the fourquantities that define a gas: pressure, temperature, volume, or moles of gas.

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Cross Multiplication Example 1

If a gas at 250 K has a pressure of 1.30 atm, what pressure will the gas experience if thegas is heated to 300 K?

Solution:

This is Gay-Lussac’s Law. The equation to compare pressure and temperature is:

P1 = P2

T1 T2

Initial Conditions Final ConditionsP1 = 1.30 atm P2 = ?T1 = 250 K T2 = 300 K

1.30 atm = P2 Insert values into equation250 K 300 K

(1.30 atm)(300 K) = (P2)(250 K) Convert to multiplication form

(1.30 atm)(300 K) = (P2)(250 K) Isolate the unknown P2

250 K 250 K

1.56 atm = P2 Solve for unknown P2

With a bit of practice, you will be able to solve this problem with fewer intermediatesteps.


When the volume of a gas at 325 K is decreased from 5.00 L to 3.00 L, what will the newtemperature of the gas be?

Solution:

This is Charles’s Law. It is no harder to solve for an unknown when it is in thedenominator of the fraction than it is to solve for an unknown when it is in thenumerator of the fraction. Why? Because we convert the fraction to multiplicationduring the process of cross multiplication. The equation to compare volume andtemperature is:

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V1 = V2

T1 T2

Initial Conditions Final ConditionsV1 = 5.00 L V2 = 3.00 L

T1 = 325 K T2 = ?

5.00 L = 3.00 L Insert values into equation325 K T2

(5.00 L)(T2) = (3.00 L)(325 K) Convert to multiplication form

T2 = 195 K Solve for unknown T2

The step in which both sides were divided by 5.00 L is not commonly written out bystudents.


When a 0.75 L sample of a gas at 100 K and 2.0 atm changes to 200 K and 2.5 atm, whatwill the new volume of the gas be?

Solution:

Volume, temperature, and pressure are changing in this question. Because morethan two quantities are changing, we should use the combined gas law to solve theproblem.

P1V1 = P2V2

T1 T2

Initial Conditions Final ConditionsP1 = 2.0 atm P2 = 2.5 atm

V1 = 0.75 L V2 = ?

T1 = 100 K T2 = 200 K

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(2.0 atm)(0.75 L) = (2.5 atm)(V2)100K 200K

We still cross multiply even though there is more than one quantity in thenumerator. The fraction is converted to:

(2.0 atm)(0.75 L)(200K) = (2.5 atm)(V2)(100K)

Divide both sides by 2.5 atm and by 100 K. The units “atm” and “K” will cancel.

1.2 L = V2


How many moles of NaCl are present in 0.285 L of a 2.50 M NaCl solution?

Solution:

The equation for molarity is:

Molarity = moles of solute liters of solution

At first, this doesn’t look like cross multiplication. What is the denominator for theterm “molarity” in the equation? The denominator is understood to be “one.” Thislets us rewrite the equation as:

Molarity = moles of solute1 liters of solution

Insert the values from the question.

2.50 M = moles NaCl1 0.285 L


(2.50 M)(0.285 L) = (1)(moles NaCl)

0.713 moles = moles NaCl

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How many liters of a 0.81 M HCl solution are required to provide 2.0 moles of HCl?

Solution:

Insert the values from the question and cross multiply.

0.81 M = 2.0 moles HCl1 liters of solution


(0.81 M)(liters of solution) = (1)(2.0 moles HCl)

Divide both sides by 0.81 M to get

liters of solution = 2.5 L

It is no harder to solve for an unknown that is in the denominator than it is to solvefor an unknown that is in the numerator. Cross multiplication makes manipulatingfractions quite easy.

Hints for using Cross Multiplication:

How do the units for molarity cancel?

The units for molarity are moles of solute/liters of solution. If we look at the molarityformula with units instead if numbers we see.

Moles of solute/liters of solution = moles of solute1 liters of solution

When we cross multiply, the equation becomes:

(Moles of solute/liters of solution)(liters of solution) = (1)(moles of solute)

The unit “liters of solution” cancels. It is just not very obvious how the units are cancelingunless you change the unit “molarity” to “moles of solute/liters of solution.”

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Does it matter which pressure is designated P1 and which is designate P2?

YES! It matters very much. Your answer will be different if you reverse the pressures.Read the question very carefully to determine what the initial pressure is and what the finalpressure is. Reversing initial and final conditions is one of the most common mistakes ingas chemistry.

Is it harder to solve for a denominator unknown than for a numerator unknown?

No. Cross multiplication is a wonderful mathematical trick to turn a fraction into a simplermultiplication. Therefore, it doesn’t matter where the unknown appears in the equation. Itwill always be simple to solve for because of cross multiplication.

If the unknown in the denominator really bothers you, you can reverse the initial equation.In the second example, we compared volume and temperature using the equation. Theunknown was T2

V1 = V2

T1 T2

We could have reversed the equation and then solved for T2 (now in the numerator).

T1 = T2

V1 V2

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Significant Figures in Logarithms

The logarithm of a number has two parts. The part in front of the decimal point is calledthe characteristic. These digits are not significant.

log (28.29) = 1.4516

It represents the power to which ten must be raised in order to get the number 28.29. Ifyou wrote 28.29 in scientific notation, the number would be 2.829 x 101 Ten is raised tothe power of one so the characteristic is one.

The part of the logarithm after the decimal point is called the mantissa.

log (28.29) = 1.4516

The numbers in the mantissa are all significant. In this example, the number 28.29 hasfour significant figures. In order for the logarithm of 28.29 to have four significant figures, itmust have four decimal places.

The number of decimal places in the logarithm of a number is equal to the numbersignificant figures in the number.

.Significant Figures in Logarithms Example 1:

How many decimal places would the logarithm of each number have?

0.05087-3.87285.02.0

Solution:

0.05087 has 4 significant figures, so the logarithm would have 4 decimal places3.87 has 3 significant figures, so the logarithm would have 3 decimal places285.0 has 4 significant figures, so the logarithm would have 4 decimal places2.0 has 2 significant figures, so the logarithm would have 2 decimal places

characteristic

mantissa

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Significant Figures in Logarithms Example 2:

How many significant figures are in the number if the logarithm of the number is:

-2.800.08836.3

Solution:

-2.80 has 2 decimal places, so the number would have 2 significant figures0.0883 has 4 decimal places, so the number would have 4 significant figures6.3 has 1 decimal place, so the number would have 1 significant figure

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Logarithms and Inverse Logarithms

The common logarithm (abbreviated as “log”) of a number is the power to which you mustraise 10 in order to obtain the number. It sounds more complicated than it is.

If the log of 2 is 0.3, we write the expression:

log 2 = 0.3

This means that if we raised 10 to the 0.3 power, the answer would be 2.

100.3 = 2

Here are a two more examples.

Logarithmic Equation Base 10 Equationlog 4 = 0.6 100.6 = 4log 236 = 2.373 102.373 = 236

Raising ten to various powers looks odd because so far in our study of chemistry, we onlyhave seen ten raised to whole number powers. Another reason raising ten to variouspowers looks odd is that we usually have a coefficient in front of the ten. We usually seenumbers written like 3.65 x 107, where ten is raised to a whole number power (7) and acoefficient (3.65) is written in front.

Let's look at the logarithms of some numbers written in scientific notation. You shouldnotice an important pattern. When working with a number in scientific notation, if thecoefficient is one, then the logarithm is just the power to which ten is raised.

LogarithmExpression

Log Expression inScientific Notation

Answer

log 1000 log 1 x 103 3log 100 log 1 x 102 2log 10 log 1 x 101 1log 0.01 log 1 x 10-2 -2log 0.00001 log 1 x 10-5 -5

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What happens when the coefficient is not one? We use a calculator!

The common logarithm of a number can be obtained by pressing the LOG button on yourcalculator. On some calculators, you enter the number and then press the LOG button.On other calculators, you press the LOG button and then enter the number.

The logarithms of several numbers are shown below. See if you get the same answers onyour calculator.

Logarithmslog 4.05 = 0.607log 0.054 = -1.27log 12.90 = 1.1106log 2.33 x 10-4 = -3.633log 8.7 x 107 = 7.9

Three important things to remember when working with logarithms:

� The logarithm of any number less than one is negative number.� The logarithm of any number greater than 1 is a positive number.� The log of one is zero: log 1 = 0 since 100 = 1

Logarithm Example 1:

What are the logarithms of 3.71 and 0.00739?

Solution:

log 3.71 = 0.569

log 0.00739 = -2.131

Remember that the logarithm of a number greater than one is positive and thelogarithm of a number less than one is negative. Watch the significant figures.Remember that the number of decimal places in the logarithm should equal thenumber of significant figures in the number.

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When a logarithm of a number is known, you must take the inverse log, or antilog, todetermine the number. The manner in which you do this depends on the calculator you areusing. Try one of these procedures to obtain an inverse logarithm.

Procedure 1: For calculators with an INV button, enter the number for which youwant to obtain the inverse log. Press the INV button (the inverse button) and thenthe LOG button.

Procedure 2: For calculators with a 2nd button, enter the number for which you wantto obtain the inverse log. Press the 2nd button. The 2nd button is usually a differentcolor from the rest of the buttons on the calculator. The 2nd button allows you to usethe mathematical operations on the calculator that are the same color as the 2nd

button. After pressing the 2nd button, press the button under the 10x symbol. The10x symbol is usually written above the LOG button.

Procedure 3: Use either procedure 1 or 2, but enter the number after you use theINV or 2nd button.

Find out how to use your calculator to get the inverse logarithms shown below.

Inverse logarithmsinvlog 3.0 = 1 x 103

invlog -4.60 = 2.5 x 10-5

invlog 1.61 = 41

Logarithm Example 2:

What is the inverse logarithm of –8.97

Solution:

invlog –8.97 = 1.1 x 10-9

For the logarithm of a number to be negative, as it is in this example, the numbermust be less than one. The logarithm –8.97 has two decimal places, so the inverselogarithm should have two significant figures.

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Hints for using logarithms:

Does it matter if I use the LOG button or the LN button on my calculator?

Yes! Common log (the LOG button) is based on a base-10 system. The natural logarithm(the LN button) is based on a base-e system. The two types of logarithms can beconnected using the equality:

ln x = 2.303 log x

They are related by multiplying the base-10 log value by 2.303.

I’m not getting the right answers but I’m entering the numbers into my calculatorcorrectly.

You probably are making a small mistake while calculating the logarithm or inverselogarithm on your calculator. The most common mistake is using the LN button instead ofthe LOG button. The next most common mistake is taking the inverse logarithm when youmean to take the logarithm, or visa versa. Check how you are using your calculator andconsult the manual to your calculator if you continue to have problems.

Documents

Math Review - Southwest DeKalb High Schoolswdekalbhs.dekalb.k12.ga.us/Downloads/Math Review.pdfA number in scientific notation has two parts. The number in front of the “x 10”