SUPPLEMENTAL LEARNING ACTIVITIES FOR GENERAL CHEMISTRY

Rebecca A. EikeyCollege of the Canyons

Guided Learning Activities

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TABLE OF CONTENTS

1: Chemistry Basics

1.1: Unit Conversions1.2: Dimensional Analysis1.3: Chemical Formulas for Ionic Compounds1.4: Nomenclature for Chemical Compounds

2: Chemical Reactions

2.1: Balancing Chemical Reactions2.2: Predicting Products of Double Displacement Reactions2.3: Writing Complete and Net Ionic Reactions2.4: Balancing Oxidation Reduction Reactions by the Half-Reaction Method

3: Stoichiometry

3.1: Mole Conversions3.2: Introduction to Stoichiometry3.3: Units of Concentration3.4: Solution Stoichiometry and Titrations3.5: Stoichiometric Calculations with Excess Reactants3.6: Determining Chemical Formulas by Combustion Analysis

4: Molecular Structure and Bonding

4.1: Electron Configurations4.2: Introduction to Lewis Dot Structures4.3: Predicting Molecular Shape with VSEPR Theory4.4: Formal Charge and VSEPR for Expanded Octets4.5: Assigning Hybrid Atomic Orbitals4.6: Introduction to Molecular Orbital Theory

5: Energy

5.1: Calorimetry5.2: Hess's Law and Enthalpy of Formation5.3: Using the Born Haber Cycle

6: Phases

6.1: Using the Ideal Gas Law6.2: Determining Vapor Pressure with Raoult's Law

7: Chemical Kinetics

8: Chemical Equilibrium

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9: Answers to Exercises

9.1.1: Unit Conversions

Index

Glossary

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CHAPTER OVERVIEW

1: Chemistry Basics1.1: Unit Conversions1.2: Dimensional Analysis1.3: Chemical Formulas for Ionic Compounds1.4: Nomenclature for Chemical Compounds

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1.1: Unit Conversions

Writing Conversion Factors

Converting between units is an important part of any science. Below is a table with measurements that are common in chemistry. Inthe table, useful equalities are given.

Measure Base Unit Abbreviation Conversion to Know

Length meter m 1 inch = 2.54 centimeter

Mass gram g 1 pound = 453.6 g

Volume liter L 1.057 quart = 1 liter1 cubic centimeter = 1 milliliter

Temperature Celsius C Celsius = ( Fahrenheit - 32)/1.8Kelvin = Celsius + 273.15

Energy joule J 1 calorie = 4.184 joule

To convert between different systems of measure, conversion factors can be applied. A conversion factor is written based on theequality between the two units. For example, a conversion factor for inches and centimeters can be written two ways:

Which of these two conversion factors is used depends on the desired conversion.

Write two conversion factors for the unit equality 12 inches = 1 foot.

Solution

The equality can be written as either or .

Write two conversion factors for the following unit equalities:

a. 1 mile = 1.61 kilometer

b. 4 cups = 1 quart

Answer

a. The equality can be written as either or .

b. The equality can be written as either or .

Another set of conversion factors that you’ll need to know are the metric prefixes. The metric system includes a set of prefixes thatare based on factors of 10. These prefixes are extremely useful because they are applied to many different types of measurements(e.g. length measurements, mass measurements, etc.) The following table includes a shortened list of unit prefixes and equalitiesthat Introductory Chemistry students should know.

Prefix Abbreviation Equality in General Notation* Equality in Scientific Notation*

nano- n- 1 m = 1,000,000,000 nm 1 m = 10 nm

micro- μ- 1 m = 1,000,000 μm 1 m = 10 μm

milli- m- 1 m = 1,000 mm 1 m = 10 mm

centi- c- 1 m = 100 cm 1 m = 10 cm

(base units: g, m, L, J, M, etc.)

o oo o

o

Example 1.1.1

12 in

1 ft

1 ft

12 in

Exercise 1.1.1

1 mi

1.61 km

1.61 km

1 mi

4 c

1 qt

1 qt

4 c

9

6

3

2

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Prefix Abbreviation Equality in General Notation* Equality in Scientific Notation*

kilo- k- 1,000 m = 1 km 10 m = 1 km

mega- M- 1,000,000 m = 1 Mm 10 m = 1 Mm

giga- G- 1,000,000,000 m = 1 Gm 10 m = 1 Gm

*These equalities use meters as an example. The same equalities can be written with an one of the base units.

Write two conversion factors that can be used to convert between meters and kilometers.

Solution

From the table above, 1000 m = 1 km. This can be written as or .

Write two conversion factors that can be used to convert between the following units:

a. microliters and litersb. milligrams and kilograms

Answer

a. Add texts here. Do not delete this text first. The conversion can be written as or .

b. For this equality, we must use two relationships from the table: 1000 mg = 1 g and 1000 g = 1 kg. We can divide thelatter by 1000 to get 1 g = 0.001 kg. This means that

The conversion can be written as or .

Often, conversion factors are not unit conversion factors per say, but are equivalencies that can be derived from physical orchemical properties of substances or systems. For example, the density of a substance is often used to relate its volume to its mass.Other common examples are given below.

Property Common Units

density g/mL, g/cm , lb/ft

speed m/s, mi/hr

concentration mol/L, g/mL

percent composition g species/g substance

Write a conversion factor for the density of gold: 19.32 g/cm .

Solution

The density can be written as or as .

3

6

9

Example 1.1.2

1000 m

1 km

1 km

1000 m

Exercise 1.1.2

 µL106

1 L

1 L

 µL106

1000 mg = 1 g = 0.001 kg (1.1.1)

1000 mg

0.001 kg

0.001 kg

1000 mg

3 3

Example 1.1.3

3

19.32 g

1 cm3

1 cm3

19.32 g

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Write a conversion factor for each of the following relationships.

a. 1 tablet contains 250 milligrams of acetaminophen.b. A water molecule has two hydrogen atoms.c. Sodium chloride is 39.33% sodium.

Answer

Add texts here. Do not delete this text first. a. The relationship can be written as or .

b. The relationship can be written as or .

c. The relationship can be written as or .

Performing Simple ConversionsTo perform single step conversions, a conversion factor that relates the given value to the desired value is identified. The givenvalue is multiplied by the conversion factor so that the given units are divided from the quantity.

When using this approach, it is important to include all units in calculations, treating them as algebraic quantities.

Perform the following conversion: 36 in. = ____ cm

Solution

We know that 1 in. = 2.54 cm. To convert 36 in. to centimeters,

Notice that the units of inches 'cancel'. The answer is in the remaining unit, centimeters.

Perform the following conversions.

a. 50.0 cm = _____ in.b. 4.57 x 10 nm = _____ m

Answer

a.

b.

Sometimes the conversion factors are based on chemical or physical properties.

Exercise 1.1.3

1 tablet

250 mg

250 mg

1 tablet

1  O moleculeH2

2 H atoms

2 H atoms

1  O moleculeH2

39.33 g Na

100 g NaCl

100 g NaCl

39.33 g Na

given units∗ = desired unitsdesired units

given units

Example 1.1.4

36    ∗ = 91.4 cmin.2.54 cm

1  in.

Exercise 1.1.4

4

50.0  ∗ = 19.7 cmcm1 in.

2.54  cm

4.57 ∗   ∗ = 4.57 ∗  m104 nm1 m

 109 nm10−5

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The density of iron is 7.87 g/cm . How much volume does 24.5 g of iron occupy?

Solution

The iron will occupy 3.11 cm .

Magnesium chloride is 25.5% magnesium. How many grams of magnesium are present in 5.24 x10 g of magnesium chloride(MgCl )?

Answer

The sample contains 1.34 x 10 g magnesium.

Special Cases in Unit Conversions – Temperature and Units with PowersTemperature conversions between the Celsius scale and the Kelvin scale require only subtraction, since the size of a degree in theCelsius and Kelvin scales is identical. When converting to and from the Fahrenheit scale, both the size of the degree and theoccurrence of 0° must be accounted for. The conversion equations are given in the first table.

What is 350 °F in the Kelvin scale?

Solution

The first step is to convert the temperature to Celsius.

Then the temperature in Celsius can be converted to the Kelvin scale.

Finally, to convert between units with powers, we must account for the power in the unit conversion. To do so, the entireconversion equality is raised to the desired power. For example, the conversion factor between cubic inches (in. ) and cubiccentimeters (cm ) can be determined from the inches to centimeters conversion factor. We know . To find therelationship between in. and cm , we can cube the equality.

How many square meters does an 80. ft rug occupy? (1 ft = 0.305 m)

Answer

First, we must determine the relationship between ft and m .

Example 1.1.5

3

24.5  ∗ = 3.11 cg1 cm3

7.87  gm3

3

Exercise 1.1.5

4

2

4

5.24x   ∗ = 1.34x  g Mg104 g MgCl225.5 g Mg

100  g MgCl2

104

Example 1.1.6

C = = 177 Co 350 F −32 o

1.8 o

K = 177 C +273.15 = 450. K o

3

3 1 in. = 2.54 cm3 3

(1 in. = 2.54 cm)3

 in =  c13 .3 2.543 m3

1 in = 16.39 c.3 m3

Exercise 1.1.7

2

2 2

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Then we can use this conversion factor to determine the area of the rug.

The rug occupies 7.4 square meters.

Extra Practice1. Perform the following unit conversions.

a. 12 mg = ______________ lb. h. 12 L = ______________ cm

b. 67.3 km = ______________ mm i. 0.00245 m = ______________ in.

c. 39.3 in. = ______________ cm j. 1.00 g = ______________ oz. (1 lb. = 16 oz.)

d. 60.0 cal = ______________ J k. 170 g/cm = ______________ lb/in

e. 1.2 x 10 mg = ______________ μg l. 2.5 x 10 km = ______________ in.

f. 5.2 m = ______________ ft m. 45 °F = ______________ °C

g. 62 °C = ______________ K n. -264 °F = ______________ K

2. How many liters of air are in a room that is 1200 m ?

3. What is the volume of a 59.5 g silver spoon? (density = 10.5 g/cm ).

4. The combustion of one gallon of gasoline will produce approximately 8.39 kg of carbon dioxide (CO ). What volume will theCO occupy? (density of CO = 1.96 g/L).

5. The density of silver is 10.5 g/cm . Express this density in terms of lb/ft .

6. The USDA recommends that a person’s sodium intake be limited to 2,400 mg. Table salt is 39.33% sodium. How many grams oftable salt can Jenny consume without surpassing this limit? (Assume there are no other sources of sodium in her diet.)

7. If one cup of coffee contains 95 milligrams of caffeine, how many cups of coffee will contain 5.00 grams of caffeine?

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(1 ft = 0.305 m)2

1 f =   = 0.0930 t2 0.3052 m2 m2

80.  ∗ = 7.4 ft2 0.0930 m2

1  ft2m2

3

2 2

-9 7

3 3

3

Ag3

2

2 2

3 3

Chemistry 151 - 1 Unit Conversions PartChemistry 151 - 1 Unit Conversions Part……

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1.2: Dimensional AnalysisPart A – Dimensional Analysis Strategy

In chemistry, many calculations rely on the units used to express the quantities. Many of the problems you see will utilize a processtermed dimensional analysis, or the factor-based method. Dimensional analysis involves adding, subtracting, dividing, andmultiplying units just as you would real numbers. When using dimensional analysis, we ‘cancel’ units to guide the problem solvingstrategy.

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1.3: Chemical Formulas for Ionic Compounds

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1.4: Nomenclature for Chemical Compounds

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CHAPTER OVERVIEW

2: Chemical Reactions2.1: Balancing Chemical Reactions2.2: Predicting Products of Double Displacement Reactions2.3: Writing Complete and Net Ionic Reactions2.4: Balancing Oxidation Reduction Reactions by the Half-Reaction Method

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2.1: Balancing Chemical ReactionsPart A – The Basics

According to the Law of Conservation of Matter, matter can neither be created nor destroyed. This means that in any chemicalreaction, any atoms that are present in the reactants should also be present in the products. Subscripts in a chemical formulaindicate how many of each atom is present. If there is a subscript outside of a parenthesis, it should be multiplied by all thesubscripts within the parentheses:

For example, in the compound above there are 2 aluminum atoms, 3 sulfur atoms and 12 oxygen atoms.

Identify and count the atoms of both the reactants and products in the following unbalanced chemical equation.

C H + O → CO + H O

Solution

Atom Number in Reactants Number in Products

C 2 (in C H ) 1 (in CO )

H 6 (in C H ) 2 (in H O)

O 2 (in O ) 3 (2 in CO + 1 in H O)

Identify and count the atoms of both the reactants and products in the following unbalanced reaction.

Al + H SO → Al (SO ) + H

Solution

Atom Number in Reactants Number in Products

Al 1 (in Al) 2 (in Al SO )

H 2 (in H SO ) 2 (in H )

S 1 (in H SO ) 3 (in Al (SO ) )

O 4 (in H SO ) 12 (in Al (SO ) )

Identify and count the atoms in both the reactants and products in the following reactions.

a) Al + HCl → AlCl + H

b) Pb(NO ) + NaCl → PbCl + NaNO

Answer a)

Atom Number in Reactants Number in Products

Al 1 (in Al) 1 (in AlCl )

H 1 (in HCl) 2 (in H )

Cl 1 (in HCl) 1 (in AlCl )

Example 2.1.1

2 6 2 2 2

2 6 2

2 6 2

2 2 2

Example 2.1.2

2 4 2 4 3 2

2 4

2 4 2

2 4 2 4 3

2 4 2 4 3

Exercise 2.1.1

3 2

3 2 2 3

3

2

3

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Answer b)

Atom Number in Reactants Number in Products

Pb 1 (in Pb(NO ) 1 (in PbCl )

N 2 (in Pb(NO ) ) 2 (in NaNO )

O 6 (in Pb(NO ) ) 3 (in NaNO )

Na 1 (in NaCl) 1 (in NaNO )

Cl 1 (in NaCl) 2 (in PbCl )

Part B – Adding coefficients to balance reactions

To balance a reaction means to make the number of atoms the same on both the reactants and products side. To do so, coefficientsneed to be added to the chemical equation. Coefficients are whole numbers that are placed in front of the element or compound inthe equation to indicate how many units of each substance participate in the chemical reaction.

When counting atoms, coefficients in front of a molecule should be multiplied by the subscripts of all atoms in the molecule:

In the reaction below, a ‘2’ was placed in front of both the O and the H O to balance the reaction. This means that there are two Omolecules and two H O molecules, as shown in the diagram below. This balances the atoms, making the number of atoms the sameon both the reactants and products side. When balancing chemical reactions, you can only change the coefficients – never changethe subscripts. Changing the subscripts changes the identity of the molecule. Coefficients of ‘1’ are omitted.

Figure : One methane molecule reacts with two oxygen molecules. Source: Openstax Chemistry

Balance each of the following chemical reactions by adding coefficients. It is best to add coefficients to substances with more thanone type of atom first and then add coefficients to substances with only one type of atom last.

3 2 2

3 2 3

3 2 3

3

2

2 2 2

2

2.1.1

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Part D – Extra Practice

Balance each of the following chemical reactions:

a) NO (g) + H (g) ® NH (g) + H O (g)

b) N H (l) + N O (l) ® N (g) + H O (g)

c) C H (g) + O (g) ® CO (g) + H O (g)

d) C H (l) + O (g) ® CO (g) + H O (g)

e) Fe O (s) + CO (g) ® Fe (s) + CO (g)

f) NO (g) + H O (l) ® HNO (aq) + NO (aq)

g) Hg (C H O ) (aq) + KCl (aq) ® Hg Cl (s) + KC H O (aq)

h) H PO (aq) + Ba(OH) (aq) ® H O (l) + Ba (PO ) (s)

i) Co(NO ) (aq) + (NH ) S (aq) ® Co S (s) + NH NO (aq)

j) CO (g) + CaSiO (s) + H O (l) ® SiO (s) + Ca(HCO ) (aq)

Balancing Chemical Eq

X + Y XY

Introduction Game

2 2 3 2

2 4 2 4 2 2

3 8 2 2 2

6 14 2 2 2

2 3 2

2 2 3

2 2 3 2 2 2 2 2 3 2

3 4 2 2 3 4 2

3 3 4 2 2 3 4 3

2 3 2 2 3 2

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2.2: Predicting Products of Double Displacement Reactions

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2.3: Writing Complete and Net Ionic Reactions

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2.4: Balancing Oxidation Reduction Reactions by the Half-Reaction Method

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CHAPTER OVERVIEW

3: Stoichiometry3.1: Mole Conversions3.2: Introduction to Stoichiometry3.3: Units of Concentration3.4: Solution Stoichiometry and Titrations3.5: Stoichiometric Calculations with Excess Reactants3.6: Determining Chemical Formulas by Combustion Analysis

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3.1: Mole Conversions

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3.2: Introduction to Stoichiometry

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3.3: Units of Concentration

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3.4: Solution Stoichiometry and Titrations

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3.5: Stoichiometric Calculations with Excess Reactants

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3.6: Determining Chemical Formulas by Combustion Analysis

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CHAPTER OVERVIEW

4: Molecular Structure and Bonding4.1: Electron Configurations4.2: Introduction to Lewis Dot Structures4.3: Predicting Molecular Shape with VSEPR Theory4.4: Formal Charge and VSEPR for Expanded Octets4.5: Assigning Hybrid Atomic Orbitals4.6: Introduction to Molecular Orbital Theory

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4.1: Electron Configurations

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4.2: Introduction to Lewis Dot Structures

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4.3: Predicting Molecular Shape with VSEPR Theory

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4.4: Formal Charge and VSEPR for Expanded Octets

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4.5: Assigning Hybrid Atomic Orbitals

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4.6: Introduction to Molecular Orbital Theory

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CHAPTER OVERVIEW

5: Energy5.1: Calorimetry5.2: Hess's Law and Enthalpy of Formation5.3: Using the Born Haber Cycle

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5.1: Calorimetry

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5.2: Hess's Law and Enthalpy of Formation

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5.3: Using the Born Haber Cycle

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CHAPTER OVERVIEW

6: Phases

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6.1: Using the Ideal Gas Law6.2: Determining Vapor Pressure with Raoult's Law

Topic hierarchy

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6.1: Using the Ideal Gas Law

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6.2: Determining Vapor Pressure with Raoult's Law

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7: Chemical Kinetics

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8: Chemical Equilibrium

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CHAPTER OVERVIEW

9: Answers to Exercises9.1.1: Unit Conversions

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9.1.1: Unit Conversions

Perform the following unit conversion.

12 g = _________ lb.

Answer

Perform the following unit conversion.

67.3 km = _________ mm

Answer

Perform the following unit conversion.

39.3 in. = _________ cm

Answer

Perform the following unit conversion.

60.0 cal = __________ J

Answer

Perform the following unit conversion.

1.2 x 10 mg = ________ μg

Answer

Exercise 9.1.1.1a

12 g x    =  0.026 lb.1 lb.

453.6 g

Exercise 9.1.1.1b

67.3 km x  x   =  6.73x  mm1000 m

1 km

1000 mm

1 m107

Exercise 9.1.1.1c

39.3 in.  x    = 99.8 cm2.54 cm

1 in.

Exercise 9.1.1.1d

60.0 cal x    =  251 J4.184 J

1 cal

Exercise 9.1.1.1e

-9

11.2 x   mg x   x    = 1.2x  μg10−91 g

1000 mg

 μg106

1 g10−6

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Perform the following unit conversion.

5.2 m = __________ ft

Answer

Add texts here.\(5.2 \ m^{3} \ x \ \dfrac {(

Perform the following unit conversion.

62°C = __________ K

Answer

9.1.1: Unit Conversions is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

Exercise 9.1.1.1f

3 3

Exercise 9.1.1.1g

C   +  273  =  335 K620

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IndexUunit conversions

1.1: Unit Conversions

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GlossarySample Word 1 | Sample Definition 1