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Module 1 Basic Concepts Matter, Mathematical

Manipulation, Dimensional Analysis, Density,

Specific Gravity, Temperature, and Heat Transfer

- By Alyssa Jean-Mary
- Source Modular Study Guide for First Semester

Chemistry by Anthony J. Papaps and Marta E.

Goicoechea-Pappas

Basic Ideas What is Chemistry?

- Chemistry is the science that studies the

composition, characterization, and transformation

of matter - It started out as alchemy.

Basic Ideas What is the Scientific Method?

- The Scientific Method is a systematic approach to

research that involves collecting and analyzing

data. - The first step is to formulate a problem.
- The second step is to make observations and

conduct experiments on that problem. - The third step is to interpret the observations

and experiments from the second step. - Here, a hypothesis is first formed. A hypothesis

is a tentative explanation for a set of

observations that is made after the collection of

enough information. - After the formation of a hypothesis, a more

definite explanation is developed a law or a

theory is made. A law is a concise verbal or

mathematical statement of a relationship between

phenomena that is always the same under the same

conditions that is made after the collection of a

large amount of information. A theory is a step

further than a law. It is a unifying principle

that explains a body of facts and those laws that

are based on them. - The fourth step is testing the interpretation,

making sure additional observations and

experiments support the interpretation made is

the thrid step.

Basic Ideas What is Matter and how is it

classified?

Basic Ideas States of Matter

- The physical state that a sample of matter exists

in depends on the temperature - There are three physical states of matter
- Solid
- Solids have a definite shape and volume, a high

density, and are virtually incompressible. - Liquid
- Liquids have a definite volume, but assume the

shape of the container they are in. They also

have a high density like solids, but are slightly

compressible unlike solids. - Gas
- Gases assume the volume and shape of the

container they are in (i.e. no definite volume or

shape), have a low density unlike solids and

liquids, and are very compressible. - Solids have stronger forces of attraction and are

more ordered than liquids, and liquids, in turn,

have stronger forces of attraction and are more

ordered than gases.

Basic Ideas Conversion between States of Matter

- Matter can convert between the three physical

states by changes in temperature and pressure - Gas to Liquid cool or increase pressure

Condensation - Gas to Solid cool Deposition
- Liquid to Gas heat or decrease pressure

Vaporization (Evaporation) (Boiling) - Liquid to Solid cool Freezing

(Solidification) (Crystallization) - Solid to Gas heat Sublimation
- Solid to Liquid heat Melting (Fusion)

Basic Ideas Common Elements to Know

Basic Ideas Types of Solutions (i.e. -

Homogeneous Mixtures)

- Every solution has two components a solute and a

solvent. - The solute(s) are the component(s) that are

present in the lesser amount. - The solvent is the component that is present in

the greatest amount. - Liquid solutions are the most common, but there

are also gas solutions and solid solutions. For

example, an alloy is a solid solution. It is a

homogeneous mixture of metals.

Basic Ideas Characteristics of Solutions

- Solutions
- have uniform distribution.
- have components that do not separate upon

standing. - have components that cannot be separated by

filtration. - only vary their compositions within certain

limits. - are almost always transparent (i.e. you cannot

see through it).

Basic Ideas Energy

- Energy is the capacity to do work or transfer

heat. Two forms of energy are potential energy

(PE) and kinetic energy (KE). - Potential energy is the energy an object

possesses because of its position or composition.

This is the kind of energy that is found in

chemicals. For example, natural gas and gasoline

contain PE. - Kinetic energy is the energy of motion.

Basic Ideas Properties and Changes

- Properties
- A physical property is a property that can be

observed in the absence of any change in

composition. Some examples are color, odor,

taste, melting point, boiling point, freezing

point, density, length, and specific heat. - A chemical property is a property that matter

exhibits as it undergoes changes in composition.

Some examples are that coal and gasoline burn in

air to form carbon dioxide and water, iron reacts

with oxygen in the air to form rust, and bleach

turns hair blonde. - Changes
- A physical change is a change that is observed

without a change in composition. Some examples

are cutting wood, melting a solid, and boiling a

liquid. - A chemical change is a change that is observed

only when a change in composition is occurring.

Some examples are the burning of wood, the

rusting of iron, and the dying of hair.

Basic Ideas Additional Properties

- An extensive property is a property that depends

on the amount of material present. Some examples

are volume and mass. - An intensive property is a property that does not

depend on the amount of material present. Some

examples are melting point, boiling point,

freezing point, color, and density.

Basic Ideas Laws

- Law of Conservation of Mass There is no

observable change in the quantity of matter

during an ordinary chemical reaction. For

example, 58.7g of Ni and 12.0g of C make 70.7g

NiC when reacted together. - Law of Conservation of Energy Energy cannot be

created or destroyed, but it can be converted

from one form to another. For example, potential

energy can be converted to kinetic energy, but if

you have 10kJ of potential energy, then you are

going to get 10kJ of kinetic energy. - Law of Definite Proportions Different samples of

any pure compound contain the same elements, each

in the same proportion by mass. For example, all

samples of water (H2O) contain 11.1 hydrogen (H)

by mass and 88.9 oxygen (O) by mass.

Basic Ideas Accuracy vs. Precision

- Accuracy tells us how close a measurement is to

the true value. - Precision tells us how closely two or more

measurements agree with one another. - For example, Say you measured the density of a

substance in two separate trials to be 1.05g/mL

and 0.998g/mL, and the density of the substance

is actually 1.00g/mL. The accuracy of the

measurement is obtained by comparing it to

1.00g/mL, the actual density, whereas the

precision of the measurement is obtained by

comparing the 1.05g/mL to the 0.998g/mL.

Rounding Off Numbers

- The rounding off of a number depends on the

identity of the digit after the cut-off point

(i.e. the next digit). The digit before the

cut-off point is known as the previous digit.

The rules for rounding off are - Rule 1 If the next digit is less than 5, then

the previous digit remains the same. - Rule 2 If the next digit is greater than 5 or

5 followed by nonzeros, then the previous digit

is increased by one. - Rule 3 If the next digit is 5 or 5 followed by

all zeros, then the previous digit - A. remains the same if it is even OR
- B. is increased by one if it is odd.

- Steps to Round Off a Number
- Step 1 Identify the next digit.
- Step 2 Based on the next digit, identify the

rule that should be followed for rounding off. - Step 3 Based on the rule, either keep the

previous digit the same or increase it by one.

Examples of Rounding Off Numbers

- Example 1 Round off 2.637 to the second decimal

point. - Answer
- Step 1 The next digit is 7.
- Step 2 Since 7 is greater than 5, rule 2

applies. - Step 3 Since rule 2 applies, the previous

digit, 3, is increased by one, making it 4. So,

the number rounded off is 2.64. - Example 2 Round off 4.45120 to the second

decimal point. - Answer
- Step 1 The next digit is 1.
- Step 2 Since 1 is less than 5, rule 1 applies.
- Step 3 Since rule 1 applies, the previous

digit remains the same. So, the number rounded

off is 4.45. - Example 3 Round off 1.67500 to the second

decimal point. - Answer
- Step 1 The next digit is 5 followed by zeros.
- Step 2 Since the next digit is a 5 followed by

zeros, rule 3 applies. And since the previous

digit, 7, is odd, part B of rule 3 applies. - Step 3 Since part B of rule 3 applies, the

previous digit, 7, is increased by one, making

it 8. So, the number rounded off is 1.68.

Scientific Notation

- The general from of scientific notation is
- N x 10e
- N is a number that is between /- 1 and /- 9. e

is an exponent (i.e. a power of ten), and thus is

always a whole number. - Some examples are 4.5 x 1011 or 8.7 x 10-6.
- Steps to convert a number into scientific

notation - Step 1 If your number is not in scientific

notation, add 100 to it. - Step 2 Convert your number so that it is between

/- 1 and /- 9. - If you need to move the decimal place to the left

to do this, then the exponent needs to be

increased by the same amount. - If you need to move the decimal place to the

right to do this, then the exponent needs to be

decreased by the same amount.

Examples of Scientific Notation

- Example 1 What is 425630.5 in scientific

notation? - Answer
- Step 1 Since the number is not in scientific

notation, it becomes 425630.5 x 100. - Step 2 To get the number between /- 1 and /-

9, I have to move the decimal point to the left 5

places. Since I moved the decimal point to the

left, I have to increase my exponent by the same

amount, 5. So, my number in scientific notation

is 4.256305 x 105. - Example 2 What is -0.0000586 in scientific

notation? - Answer
- Step 1 Since the number is not in scientific

notation, it becomes 0.00586 x 100. - Step 2 To get the number between /- 1 and /-

9, I have to move the decimal point to the right

3 places. Since I moved the decimal point to the

right, I have to decrease my exponent by the same

amount, 3. So, my number in scientific notation

is 5.86 x 10-3. - Example 3 What is 625.366 x 105 in correct

scientific notation? - Answer
- Step 1 Since the number is already in scientific

notation, nothing is added to it. - Step 2 To get the number between /- 1 and /-

9, I have to move the decimal point to the left 2

places. Since I moved the decimal point to the

left, I have to increase my exponent by the same

amount, 2. So, my number in scientific notation

is 6.25366 x 107.

Significant Figures

- The amount of significant figures indicates how

accurate a measurement is. - Exact numbers have an infinite number of

significant figures, which means that there is an

infinite number of zeros after the number. These

zeros are not shown for convenience. For example,

the amount in 1 dozen, 12, is an exact number,

but we dont write 12.00000 every time. - Steps to identify the amount of significant

figures a number has - Step 1 Going from left to right, locate the

first nonzero digit. - Step 2 Again going from left to right, count the

amount of digits present in the number, starting

with the first nonzero digit. - Note If there are zeros at the end of a number

without a decimal point, they may or may not be

significant. Thus, if the zeros are significant,

a decimal point should be at the end of the

number. For example, 300 could have 1 significant

figure, 2 significant figures, or 3 significant

figures, but 300. has 3 significant figures.

- Example 1 How many significant figures are

present in the number 564.32? - Answer
- Step 1 The first nonzero digit is 5.
- Step 2 Starting with the 5, there are 5

significant figures. - Example 2 How many significant figures are

present in the number 0.00042? - Answer
- Step 1 The first nonzero digit is 4.
- Step 2 Starting with the 4, there are 2

significant figures.

Manipulating Powers of Ten

- Three ways to manipulate powers of ten
- If powers of ten are multiplied, the exponents

are added. - If powers of ten are divided, the exponents are

subtracted. - If powers of ten are raised to an exponent, the

exponents are multiplied.

- Example 1 What is 106 x 104?
- Answer Since it is multiplication, the exponents

are added together 6 4 10, so, it is 1010. - Example 2 What is 1013/10-3?
- Answer Since it is division, the exponents are

subtracted from each other 13 (-3) 16, so,

it is 1016. - Example 3 What is (105)3?
- Answer Since it is raised to an exponent, the

exponents are multiplied together 5 x 3 15,

so, it is 1015.

Numbers with Powers of Ten Multiplying and

Dividing

- Steps when multiplying and dividing numbers with

powers of ten - Step 1 Place the numbers (also called

coefficients) together and the powers of ten

together. - Step 2 Multiply or divide the numbers and add or

subtract the powers of ten, depending on which

operation is present. - Step 3 The final answer should
- have the same amount of significant figures as

the number with the least amount of significant

figures. - have been rounded off correctly.
- preferably be reported in scientific notation.

Examples of Multiplying and Dividing Numbers with

Powers of Ten

- Example 1 Multiplication What is (3.443 x 106)

x (0.43 x 104)? - Answer
- Step 1 (3.443 x 0.43) x (106 x 104)
- Step 2 1.48049 x 1010
- Step 3 3.443 has 4 significant figures and 0.43

has 2 significant figures, so the answer should

have 2 significant figures. To round off the

number to 2 significant figures, the previous

digit, 4, should become 5 since the next digit

is 8. So, it is 1.5 x 1010.

- Example 1 Division What is (4.25 x 104)/(3.21 x

102)? - Answer
- Step 1 (4.25/3.21) x (104/102)
- Step 2 1.32399 x 102
- Step 3 4.25 has 3 significant figures and 3.21

also has 3 significant figures, so the answer

should also have 3 significant figures. To round

off the number to 3 significant figures, the

previous digit, 2, should stay the same since

the next digit is 3. So, it is 1.32 x 102.

Numbers with Powers of Ten Adding and Subtracting

- Steps when adding and subtracting numbers with

powers of ten - Step 1 Make all of the powers of ten the same.
- Step 2 Once all of the powers of ten are the

same, add or subtract the numbers, depending on

which operation is present, but keep the powers

of ten the same - Step 3 The final answer should
- have the same amount of decimal places as the

number with the fewest decimal places at the time

of the addition or subtraction. - have been rounded off correctly.
- preferably be reported in scientific notation.

Examples of Adding and Subtracting Numbers with

Powers of Ten

- Example 1 Addition What is 5.34 x 1043 4.2 x

1044? - Answer
- Step 1 To make the powers the same, either

change 43 to 44 or 44 to 43. If 43 is turned into

44, since the value is increased by one, the

decimal point is moved to the left by one, making

it 0.534 x 1044. - Step 2 (0.534 4.2) x 1044 4.734 x 1044.
- Step 3 At the time of addition, 0.534 has 3

decimal places and 4.2 has 1 decimal place, so

the answer should have 1 decimal place. To round

off the number to 1 decimal point, the previous

digit, 7, should remain the same because the

next digit is 3. So, it is 4.7 x 1044.

- Example 1 Subtraction What is 0.98 x 1056

3.12 x 1055? - Answer
- Step 1 To make the powers the same, either

change 56 to 55 or 55 to 56. If 56 is turned into

55, since the value is decreased by one, the

decimal point is moved to the right by one,

making it 9.8 x 1055. - Step 2 (9.8 3.12) x 1055 6.68 x 1055.
- Step 3 At the time of addition, 9.8 has 1

decimal place and 3.12 has 2 decimal places, so

the answer should have 1 decimal place. To round

off the number to 1 decimal point, the previous

digit, 6, should be increased to 7, because the

next digit is 8. So, it is 6.7 x 1055.

Numbers with Powers of Ten Mixing

Multiplying/Dividing with Adding/Subtracting

- When all operations are combined, make sure to

carry out those in parenthesis first.

- Example 1 What is
- (2.63 x 105 x (3.64 x 10-4 1.22 x 10-3))/(7.36

x 106)2? - Answer
- (3.64 x 10-4 1.22 x 10-3) (0.364 x 10-3 -

1.22 x 10-3) (0.364 1.22) x 10-3 -0.856 x

10-3 - (7.36 x 106)2 (7.36 x 106) x (7.36 x 106)

(7.36 x 7.36) x (106 x 106) 54.1696 x 1012 - (2.63 x 105 x -0.856 x 10-3) (2.63 x -0.856) x

(105 x 10-3) -2.2513 x 102 - -2.2513 x 102/54.1696 x 1012 (-2.2513/54.1696)

x (102/1012) 0.04156 x 10-10 4.16 x 10-12

Units of Measurement Equalities in the English

System

- Volume
- 1 pint (pt) 16 fl oz (fluid ounce)
- 1 quart (qt) 2 pt
- 1 gallon (gal) 4 qt
- Mass
- 1 pound (lb) 16 ounces (oz)
- 1 ton 2000 lb
- Length
- 1 foot (ft) 12 inches (in)
- 1 yard (yd) 3 feet (ft)
- 1 mile (mi) 5280 ft
- Time
- 1 minute (min) 60 seconds (sec)
- 1 hour (hr) 60 minutes (min)
- 1 day 24 hours (hr)

Units of Measurement The Metric System and the

International System of Units (SI Units)

- These two systems are decimal systems, where the

units are related to each other by powers of ten.

Prefixes are used to indicate fractions and

multiples of ten. - The basic units for these systems are
- Volume liter (L)
- Mass gram (g)
- Length meter (m)
- Time second (s)
- The relation between length units and volume

units 1 mL 1 cm3 1 cc - The table to the left illustrates the value of

each prefix. The number in the table always gets

placed in front of the base unit and a 1 always

gets placed in from of the prefixed base unit.

For example, 1 millimeter 10-3 meters or 1 mm

10-3 m.

Units of Measurement Conversion between the

Metric and the English Systems

- Volume
- 1 qt 0.946 L
- Mass
- 1 lb 454 g
- Length
- 1 in 2.54 cm

Use of Conversion Factors in Calculations

- Obtained from equalities, which show the

relationship between two quantities that are

measuring the same quantity (i.e. mass, volume,

length, etc.) - From the equality, two different conversion

factors can be obtained - When using a conversion factor in a problem, you

use the one that cancels out the given units and

leaves the answer units the given units need to

be on the bottom of the conversion factor, with

the answer units on the top of the conversion

factor

- Equality
- x y,

where x and y are different units and the s

arent equal - Two conversion factors from the equality x/

y and y / x - If you want to convert y to x, you need to use

x / y - y ( x / y) x
- If you want to convert x to y, you need to use

y / x - x ( y / x) y

Problem Solving with Conversion Factors Factor

Label Method

- Step 1 Write down the given quantity and unit(s)

and the answer unit(s), with space between them

for the conversion factors. - Step 2 Identify the equalities needed to change

the given unit(s) to the answer unit(s) more

than one equality may be needed. - Step 3 Write down the two conversion factors

possible from each of the identified equalities. - Step 4 Arrange the conversion factors between

the given quantity and unit and the answer unit,

so that the given unit cancels, leaving the

answer unit the given unit needs to be on the

bottom of the first conversion factor and the

answer unit needs to be on the top of the last

conversion factor. - If the given quantity has two units, convert the

unit on top, and then convert the unit on the

bottom. Unlike when converting the unit on

top, which is described above, when converting

the unit on the bottom, the given unit needs to

be on the top of the first conversion factor for

the bottom and the answer unit needs to be on

the bottom of the last conversion factor for the

bottom. - Hint A quantity with two units, x/y, can be

written as x / 1 y, since dividing by one

doesnt change the value of the number. This

will help to choose the correct conversion factor

for the problem.

Examples of Using the Factor Label Method 1

- Example 1 English to English Conversion How

many feet are in 3 yards? - Answer
- Step 1 3 yd.

____ ft. - Step 2 1 yd. 3 ft. (this covers all the units

above, so no more equalities are needed) - Step 3 Conversion factors 1 yd. / 3 ft. OR 3

ft. / 1 yd. - Step 4 3 yd. (3 ft. / 1 yd.) 9 ft. this

conversion factor is needed to cancel yd. (given

unit is on bottom) and get ft. (answer unit is on

top) - Example 2 Metric to Metric Conversion How many

kilograms are in 90 decigrams? - Answer
- Step 1 90 dg.

____ kg. - Step 2 1 dg. 0.1 g. and 1 kg. 1000 g. (two

equalities are needed here since one does not

cover all of the units above ) - Step 3 Conversion factors 1 dg. / 0.1 g. OR 0.1

g. / 1 dg. and - 1 kg. / 1000 g. OR 1000 g. / 1 kg.
- Step 4 90 dg. (0.1 g. / 1 dg.) (1 kg. / 1000

g.) 0.0090 kg. the first conversion factor is

needed to cancel dg (given unit is on bottom) and

get g (unit is on top), which leads to the second

conversion factor, which cancels g (unit is on

bottom) to get kg (answer unit is on top)

Examples of Using the Factor Label Method 2

- Example 3 Metric to English Conversion How many

feet are in 5.85 centimeters? - Answer
- Step 1 5.85 cm.

____ ft. - Step 2 2.54 cm. 1 in. and 12 in. 1 ft. (two

equalities are needed here since one does not

cover all of the units above the first equality

converts between the english and the metric

system and the second one converts within the

english system) - Step 3 Conversion factors 2.54 cm. / 1 in. OR 1

in. / 2.54 cm. and 12 in. / 1 ft. OR 1 ft. / 12

in. - Step 4 5.85 cm. (1 in. / 2.54 cm.) (1 ft. /

12 in.) 0.192 ft. the first conversion factor

is needed to cancel cm. (given unit is on bottom)

and get in. (unit is on top), which leads to the

second conversion factor, which cancels in. (unit

is on bottom) to get ft. (answer unit is on top) - Example 4 Using Square Units How many m2 are in

105 in2? - Answer
- Step 1 105 in2

____ m2 - Step 2 2.54 cm. 1 in. and 1 cm. 0.01 m. (two

equalities are needed here since one does not

cover all of the units above the first equality

converts between the english and the metric

system and the second one converts within the

metric system) - Step 3 Conversion factors 2.54 cm. / 1 in. OR 1

in. / 2.54 cm. and 1 cm. / 0.01 m. OR 0.01 m. /

1 cm. - Step 4 105 in.2 (2.54 cm. / 1 in.)2 (0.01 m.

/ 1 cm.)2 0.0677 m.2 the first conversion

factor is needed to cancel in. (given unit is on

bottom) and get cm. (unit is on top), which leads

to the second conversion factor, which cancels

cm. (unit is on bottom) to get m. (answer unit is

on top) each conversion factor needs to be

squared so that the given units are cancelled

completely to give the answer units

Examples of Using the Factor Label Method 3

- Example 5 Using Cubed Units How many m3 are in

105 in3? - Answer
- Step 1 105 in3

____ m3 - Step 2 2.54 cm. 1 in. and 1 cm. 0.01 m. (two

equalities are needed here since one does not

cover all of the units above the first equality

converts between the english and the metric

system and the second one converts within the

metric system) - Step 3 Conversion factors 2.54 cm. / 1 in. OR 1

in. / 2.54 cm. and 1 cm. / 0.01 m. OR 0.01 m. /

1 cm. - Step 4 105 in.3 (2.54 cm. / 1 in.)3 (0.01 m.

/ 1 cm.)3 0.00172 m.3 the first conversion

factor is needed to cancel in. (given unit is on

bottom) and get cm. (unit is on top), which leads

to the second conversion factor, which cancels

cm. (unit is on bottom) to get m. (answer unit is

on top) each conversion factor needs to be

cubed so that the given units are cancelled

completely to give the answer units - Example 6 Using Double Units How many ft./sec.

are in 731 in./min.? - Answer
- Step 1 731 in./1 min.

____ ft./sec. - Step 2 For the top units, 12 in. 1 ft. (only

one equality is needed) for the bottom units,

1 min. 60 sec. (only one equality is needed) - Step 3 For the top units, 12 in. / 1 ft. OR 1

ft. / 12 in. for the bottom units, 1 min. / 60

sec. OR 60 sec. / 1 min. - Step 4 731 in./1 min. (1 ft. / 12 in.) (1

min. / 60 sec.) 1.02 ft./sec. the first

conversion factor (for the top units) is needed

to cancel in. (given unit is on bottom) and get

ft. (answer unit is on top), and the second one

(for the bottom units) is needed to cancel min.

(given unit is on top) and get sec. (answer unit

is on bottom)

Density (d) and Specific Gravity (sp.gr.)

- Density mass/volume OR d m/V
- The common units of density are
- g/mL for liquids and solids
- g/L for gases
- The density of a substance shows the relationship

between its mass and its volume. For example, if

the density of a substance is 3.45 g/mL, the

relationship between the mass and volume can be

written as the equality 3.45 g 1 mL. - Specific Gravity (sp.gr.) (density of

substance)/(density of water at 4?C) (density

of substance in g/mL)/(1.00g/mL) - Since the density of a substance is divided by

1.00 to obtain the specific gravity, the density

and specific gravity are numerically equivalent

(i.e., they are the same number since dividing

any number by 1.00 gives the same number).

Although they are numerically equivalent, they

are not equal. Since both the density of a

substance and the density of water have the same

units, they are cancelled out when the specific

gravity is obtained, making the specific gravity

a unitless quantity, unlike density, which always

has units.

To Obtain a Substances Density

- Since density is mass divided by volume, you need

to obtain both a substances mass and its volume

to obtain its density. - To obtain the mass, measure the substance on a

balance. - To obtain the volume
- If it is a liquid, measure it with a graduated

cylinder. - If it is a solid, one of two methods can be used
- One is by water displacement. Here, a certain

volume of water is placed in a graduated

cylinder, and the value is recorded (Vwater). The

solid is then dropped into the graduated cylinder

with the water. The new volume of water is then

recorded (Vwatersolid). To obtain the volume of

the solid (Vsolid), the initial volume of the

water is subtracted from the new volume of water

(Vsolid Vwatersolid - Vwater). - The other is through mathematical equations,

using one of the following equations - V(cubic solid) l x w x h (length (l) width

(w) height (h)) - V(rectangular solid) l x w x h
- V(sphere) (4/3) pr3 (radius (r) diameter

(d)/2) - V(cylinder) (pd2h)/4

Steps to Solving a Problem

- Step 1 Identify the information that is given in

the problem. - Step 2 Identify what information the problem is

looking for. - Step 3 Identify the equation that is needed to

obtain the information you are looking for, while

using the information you were given. - Step 4 Put the given information into the

equation found in Step 3 and solve for the

information you are looking for, making sure that

all of the given information is expressed in the

proper units.

Examples of Calculations of Density 1

- Example 1 What is the density of a liquid that

has a mass of 5.67 grams and a volume of 10.3 mL? - Step 1 5.67g m 10.3mL V
- Step 2 density (d)
- Step 3 d m/V
- Step 4 d 5.67 g / 10.3 mL 0.550 g/mL
- Example 2 What is the mass of a liquid if it

has a density of 1.50 g/mL and a volume of 14 mL? - Step 1 1.50 g/mL d 14mL V
- Step 2 mass (m)
- Step 3 d m/V
- Step 4 d m / V - m d V 1.50 g/mL 14

mL 21 g - Example 3 What is the volume of a liquid if it

has a mass of 14.6 grams and a density of 3.45

g/mL? - Step 1 14.6g m 3.45 g/mL d
- Step 2 volume (V)
- Step 3 d m/V
- Step 4 d m / V - V m / d 14.6 g / 3.45

g/mL 4.23 mL

Examples of Calculations of Density 2

- Example 4 If a 24g solid was placed in a

graduated cylinder containing 25.0mL of water and

the water raised to 34.2mL, what is the density

of the substance? - Answer
- Step 1 24g m 25.0mL Vwater 34.2mL

Vwatersolid - Step 2 density (d)
- Step 3 d m/V, where V Vsolid Vwatersolid

- Vwater - Step 4 V Vwatersolid Vwater 34.2mL-25.0mL

9.2mL d m/V 24g/9.2mL 2.6g/mL - Example 5 What is the density of a wooden block

that has a mass of 75g and measures 2.5cm by

3.5cm by 6.1cm? - Answer
- Step 1 75g m 2.5cm length (l) 3.5cm

width (w) 6.1cm height (h) - Step 2 density (d)
- Step 3 d m/V, where V V(rectangular solid)

l x w x h - Step 4 V 2.5cm x 3.5cm x 6.1cm 53.375cm3

53.375mL d m/V 75g/53.375mL 1.4g/mL

Examples of Calculations of Specific Gravity

- Example 1 What is the specific gravity of a

substance with a mass of 5.021 g and a density of

1.21 g/mL? - Step 1 5.021 g m 1.21 g/mL d
- Step 2 specific gravity (sp.gr.)
- Step 3 sp.gr. (density of substance in

g/mL)/(1.00g/mL) - Step 4 sp.gr. 1.21 g/mL / 1.00 g/mL 1.21
- Example 2 What is the specific gravity of a

substance with a mass of 5.021 g and a volume of

7.99 mL? - Step 1 5.021g m 7.99mL V
- Step 2 specific gravity (sp.gr.)
- Step 3 sp.gr. (density of substance in

g/mL)/(1.00g/mL), where d m/V - Step 4 d m/V 5.021 g / 7.99 mL 0.628 g/mL

sp.gr. (density of substance in

g/mL)/(1.00g/mL) 0.628 g/mL / 1.00 g/mL 0.628

Temperature Conversions

- There are three temperature units degrees

Fahrenheit (F), degrees Celsius (C), and Kelvin

(K). (Note K does not have a sign.) - To convert C to F F (1.8 C) 32
- To convert F to C C (F 32) / 1.8
- To convert C to K K C 273
- To convert K to C C K 273
- Note For the Celsius scale and the Kelvin scale,

there is a 100 difference between the freezing

point and the boiling point of water, but on the

Fahrenheit scale, the difference is 180.

Examples of Temperature Conversions

- Example 1 How many degrees Fahrenheit are in 73

degrees Celsius? - Answer F (1.8 C) 32 (1.8 73 C) 32

163.4 F - Example 2 How many degrees Celsius are in 200

degrees Fahrenheit? - Answer C (F 32) / 1.8 (200 F 32) /

1.8 93.3 C - Example 3 How many Kelvin are there in 650

degrees Celsius? - Answer K C 273 650 C 273 923 K
- Example 4 How many degrees Celsius are there in

543 Kelvin? - Answer C K 273 543 K 273 270 C
- Example 5 How many degrees Fahrenheit are there

in 321 Kelvin? - Answer
- C ? K 273 321 K -273 48 C
- F (1.8 C) 32 (1.8 48 C) 32 118.4

F

Heat Transfer

- During chemical changes and physical changes,

heat is transferred - If there is an evolution of heat (i.e. heat is

lost or released), it is an exothermic process. - If there is an absorption of heat (i.e. heat is

gained or absorbed), it is an endothermic

process. - The units to express the amount of heat

transferred are calories (cal) or joules (J),

where 1 cal 4.184J. - Specific heat is the amount of heat necessary to

raise the temperature of 1 g of substance by 1C.

Specific heat is a physical intensive property,

like density, melting point, and boiling point.

Each substance has its own specific heat. - Using the specific heat of a substance, the heat

that is absorbed or released in a given process

can be obtained by - q m x s x ?T
- where q is the heat energy (cal, kcal, J, or

kJ), m is mass (g), s is specific heat

(cal/gC), and ?T is the change in temperature

(?T T2 T1)(C).

Examples of Calculations of Heat Transfer

- Example 1 How much heat (in cal) is needed to

heat 54g of a substance (specific heat 0.212

cal/gC) from 72C to 98C? - Answer
- Step 1 Given 54g m 0.212 cal/gC s 72C

T1 98C T2 - Step 2 Looking for heat (in cal) q
- Step 3 Equation q m x S x ?T, where ?T T2

T1 - Step 4 ?T T2 T1 98C - 72C 26C q

(54g) x (0.212 cal/gC) x (26C) 297.648 cal

3.0 cal. - Example 2 How much of a substance is present if

56J of heat is added when the substance (specific

heat 0.532 cal/gC) is heated from 23C to

54C? - Answer
- Step 1 Given 56J q 0.532 cal/gC s 23C

T1 54C T2 - Step 2 Looking for mass m
- Step 3 Equation q m x S x ?T, where ?T T2

T1 - Step 4 ?T T2 T1 54C - 23C 31C q m

x S x ?T ? 56J (m) x (0.532 cal/gC) x

(31C), but since there is J on one side and cal

on the other side, the J have to be converted to

cal before solving for m 56J x (1 cal /4.184J)

13.384 cal 13.384 cal (m) x (0.532

cal/gC) x (31C), so m 0.81156 g 0.81 g. - Example 3 If 86cal of heat is added to 10g of a

substance (specific heat 0.411 cal/gC), and

the final temperature was measured to be 152C,

what is the initial temperature of the substance? - Answer
- Step 1 Given 86cal q 10g m 0.411 cal/gC

s 152C T2 - Step 2 Looking for initial temperature T1
- Step 3 Equation q m x S x ?T, where ?T T2

T1 - Step 4 86cal (10g) x (0.411 cal/gC) x (?T),

so ?T 20.92C ?T T2 T1 ? 20.92C 152C

T1, so T1 131.08C 1.3 x 102C

Some Specific Heats

- The larger the specific heat of the substance,

the more heat it needs to raise its temperature. - The specific heat of water is 1.00 cal/gC,

which is one of the largest specific heats. Since

approx. 60 of our body weight is water, this

large specific heat allows our body to maintain a

constant body temperature (around 37C) much

easier. So, our body can absorb or release a

large amount of energy with little change in body

temperature. - Examples of other specific heats
- wood 0.421 cal/gC
- gold 0.0306 cal/gC
- graphite 0.172 cal/gC

THE END