APPLIED CHEMISTRY SEMESTER REVIEW - Wikispaceskmkunz.wikispaces.com/file/view/Applied+Chemistry+-+Semester... · applied chemistry semester review chapter 7 – chemical reactions

APPLIED

CHEMISTRY

SEMESTER

REVIEW CHAPTER 7 – CHEMICAL REACTIONS

CHAPTER 8 – CHEMICAL QUANTITIES IN REACTIONS

CHAPTER 9 – ELECTRONIC STRUCTURE

CHAPTER 7

CHEMICAL REACTIONS

Classify each of the following changes as physical or chemical:

a.water freezing into an icicle b. burning a match

c. breaking a chocolate bar d. digesting a chocolate bar

Problem 7.1 Classifying Chemical and Physical Change

Classify the following changes as physical or chemical:

a. chopping a carrot

b. developing a Polaroid picture

c. inflating a balloon

Study Check

Solution

a. Physical. Freezing water involves only a change from liquid water to ice. No change has occurred in the

substance water.

b. Chemical. Burning a match causes the formation of new substances that have different properties.

c. Physical. Breaking a chocolate bar does not affect its composition.

d. Chemical. The digestion of the chocolate bar converts it into new substances.

Identify each of the following as a physical change or a chemical reaction. If it is a chemical reaction, what is

the evidence?

a. propane burning in a barbecue b. chopping an onion c. using peroxide to bleach hair

Problem 7.2 Evidence of a Chemical Reaction

Striking a match is an example of a chemical reaction. What evidence would you see that indicates a chemical

reaction?

Study Check

Solution

a. The production of heat during burning is evidence of a chemical reaction.

b. Chopping an onion into smaller pieces is a physical change.

c. The change in hair color is evidence of a chemical reaction.

Hydrogen and nitrogen react to form ammonia, NH3-

3H2(g) + N2(g) →2NH3(g).

Ammonia

a. What are the coefficients in the equation?

b. How many atoms of each element are in the reactants and products of the equation?

Problem 7.3 Chemical Equations

When ethane (C2H6) burns in oxygen, the products are carbon dioxide and water. The balanced equation is as follows.

2C2H6(g) + 7O2(g) →4CO2(g) + 6H2O(g)

State the total number of atoms of each element on each side of the equation.

Study Check

Solution

a. The coefficients are three (3) in front of H2; one (1), which is understood, in front of N2; and two (2) in

front of NH3-.

b. In the reactants there are six hydrogen atoms and two nitrogen atoms. In the product there are also six

hydrogen atoms and two nitrogen atoms.

Balance the following equation:

Na3PO4(aq) + MgCl2(aq) → Mg3(PO4)2(s) + NaCl(aq)

Problem 7.4 Balancing Equations

Solution

In the equation, the correct formulas are written.


STEP 1

STEP 2 When we compare the number of ions on the reactant and product sides, we find that the equation is not

balanced. In this equation, it is more convenient to balance the phosphate polyatomic ion as a group

instead of its individual atoms.



Problem 7.4 (Cont.) Balancing Equations

Solution

We begin with the formula of Mg3(PO4)2, which is the most complex. A 3 in front of MgCl2

balances magnesium and a 2 in front of Na3PO4 balances the phosphate ion. Looking again at each

of the ions in the reactants and products, we see that the sodium and chloride ions are not yet

equal. A 6 in front of the NaCl balances the equation.

STEP 3



Problem 7.4 (Cont.) Balancing Equations


Fe(s) + O2(g) → Fe3O4(s)

Study Check

Solution

A check of the atoms indicates the equation is balanced. STEP 4

Problem 7.5 Identifying Reactions and Predicting Products

Solution

In the reaction of 1 mol solid carbon with oxygen gas, the energy of the carbon dioxide produced is 393 kJ

lower than the energy of the reactants.

a. Is the reaction exothermic or endothermic?

b. Write the balanced equation for the reaction, including the heat of the reaction.

c. Write the value of ∆H for this reaction.

Problem 7.6 Exothermic and Endothermic Reactions

The reaction of 1 mol hydrogen gas (H2) with iodine gas (I2) to form hydrogen iodide (HI) is endothermic and requires

50. kJ of heat.

a. Write a balanced equation for the reaction, including the heat of reaction.

b. Write the value of ∆H for this reaction.

Study Check

Solution

a. When the products have a lower energy than the reactants, the reaction is exothermic.

b. C(s) + O2(g) → CO2(g) + 393 kJ

c. ∆H = -393 kJ

Problem 7.7 Calculating Heat in a Reaction

The formation of ammonia from hydrogen and nitrogen has a H = -92.2 kJ.

How much heat in kilojoules is released when 50.0 g of ammonia forms?

Solution

List given and needed.

Given 50.0 g NH3 Need Heat in kilojoules (kJ) to form NH3

STEP 1

STEP 2 Plan Use conversion factors that relate the heat released to the moles of NH3 in the balanced equation.

Equalities/Conversion Factors STEP 3

Problem 7.7 (Cont.) Calculating Heat in a Reaction

The formation of ammonia from hydrogen and nitrogen has a H = -92.2 kJ.

How much heat in kilojoules is released when 50.0 g of ammonia forms?

Mercury(II) oxide decomposes to mercury and oxygen:

a. Is the reaction exothermic or endothermic?

b. How many kJ are needed to react 25.0 g mercury(II) oxide?

Study Check

Set Up Problem STEP 4

Solution

Copyright ©2008 by Pearson Education, Inc.

Upper Saddle River, New Jersey 07458

All rights reserved.

Basic Chemistry, Second Edition

By Karen C. Timberlake and William Timberlake

CHAPTER 8

CHEMICAL QUANTITIES IN REACTIONS

Calculate the total mass of reactants and products for the following equation when 1 mol CH4 reacts:

CH4(g) + 2O2 (g) → CO2(g) + 2H2O(g)

Problem 8.1 Conservation of Mass

For the following reaction, calculate the total mass of the reactants and

products when 4 mol NH3 react:

4NH3(g) + 3O2(g) → 2N2(g) + 6H2O(g)

Study Check

Solution

Interpreting the coefficients in the equation as the number of moles of each substance and multiplying by the

respective molar masses gives the total mass of reactants and products.

Consider the following balanced equation:

4Na(s) + O2(s) → 2Na2O(s)

Write the mole–mole factors for

a. Na and O2 b. Na and Na2O

Problem 8.2 Writing Mole–Mole Factors

Using the equation in Problem 8.2, write the mole–mole factors for O2 and Na2O.

Study Check

Solution

In the reaction of iron and sulfur, how many moles of sulfur are needed to react with 1.42 mol iron?

2Fe(s) + 3S(s) → Fe2S3(s).

Problem 8.3 Calculating Moles of a Reactant

Solution

Write the given and needed number of moles. In this problem, we need to find the number of

moles of S that react with 1.42 mol Fe.

Given 1.42 mol Fe Need moles of S

STEP 1

STEP 2 Write the plan to convert the given to the needed.

Use coefficients to write relationships and mole-mole factors. Use coefficients to write the

mole-mole factors for the given and needed substances.

STEP 3

In the reaction of iron and sulfur, how many moles of sulfur are needed to react with 1.42 mol iron?

2Fe(s) + 3S(s) → Fe2S3(s).

Problem 8.3 (Cont.) Calculating Moles of a Reactant

Using the equation in Sample Problem 8.3, calculate the number of moles of iron needed to completely react

with 2.75 mol sulfur.

Study Check

Solution

Set up problem using the mole-mole factor that cancels given moles. Use a mole-mole factor

to cancel the given moles and provide needed moles.

STEP 4

The answer is given with three significant figures because the given quantity 1.42 mol Fe has 3 SFs.

The values in the mole-mole factor are exact.

In a combustion reaction, propane (C3H8) reacts with oxygen. How many moles of CO2 can be produced when

2.25 mol C3H8 reacts?

C3H8(g) + 5O2(s) → 3CO2(g) + 4H2O(g)

Propane

Problem 8.4 Calculating Moles of a Product

Solution

Write the given and needed moles. In this problem, we need to find the number of moles of CO2

that can be produced when 2.25 mol C3H8 react.

Given 2.25 mol C3H8 Need moles of CO2

STEP 1

STEP 2 Write the plan to convert the given to the needed.

Use coefficients to write relationships and mole-mole factors. Use coefficients to write the

mole-mole factors for the given and needed substances.

STEP 3

In a combustion reaction, propane (C3H8) reacts with oxygen. How many moles of CO2 can be produced when

2.25 mol C3H8 reacts?

C3H8(g) + 5O2(s) → 3CO2(g) + 4H2O(g)

Propane

Problem 8.4 (Cont.) Calculating Moles of a Product

Using the equation in Problem 8.4, calculate the number of moles of oxygen that must react to

produce 0.756 mol water

Study Check

Solution

Set up problem using the mole-mole factor that cancels given moles. Use a mole-mole factor

to cancel the given moles and provide needed moles.

The answer is given with three significant figures because the given quantity 2.25 mol C3H8 has

3 SFs. The values in the mole-mole factor are exact.

STEP 4

In the formation of smog, nitrogen reacts with oxygen to produce nitrogen oxide. Calculate the grams of NO

produced when 2.15 mol O2 react.

N2(g) + O2(g) → 2NO(g)

Problem 8.5 Mass of Products from Moles of Reactant

Solution

Given 2.15 mol O2 Need grams of NO STEP 1

STEP 2 Plan

Equalities/Conversion Factors The mole-mole factor that converts moles of O2 to moles of NO is

derived from the coefficients in the balanced equation.

STEP 3

In the formation of smog, nitrogen reacts with oxygen to produce nitrogen oxide. Calculate the grams of NO

produced when 2.15 mol O2 react.

N2(g) + O2(g) → 2NO(g)

Problem 8.5 (Cont.) Mass of Products from Moles of Reactant

Using the equation in Sample Problem 8.5, calculate the grams of NO that can be produced when 0.734 mol

N2 reacts.

Study Check

Solution

Set Up Problem First, we can change the given, 2.15 mol O2, to moles of NO. STEP 4

Problem 8.6 Mass of Products from Mass of Reactant

Solution

Given 54.6 g C2H2 Need grams of CO2 STEP 1

STEP 2 Plan Once we convert grams of C2H2 to moles of C2H2 using its molar mass, we can use a mole-mole

factor to find the moles of CO2. Then the molar mass of CO2will give us the grams of CO2.

In a combustion reaction, acetylene (C2H2) burns with oxygen.

How many grams of carbon dioxide are produced when 54.6 g C2H2 is burned?

Problem 8.6 (Cont.) Mass of Products from Mass of Reactant

Solution

Equalities/Conversion Factors We need the molar mass of C2H2 and CO2. The mole-mole factor

that converts moles of C2H2 to moles of CO2 is derived from the coefficients in the balanced

equation.

STEP 3






Solution

Set Up Problem Using our plan, we first convert grams of C2H2 to moles of C2H2. STEP 4




Using the equation in Problem 8.6, calculate the grams of CO2 that can be produced when 25.0 g O2

reacts.

Study Check

Solution

Set Up Problem Using our plan, we first convert grams of C2H2 to moles of C2H2. STEP 4

Problem 8.7 Calculating the Mass of Reactant

Propane gas, (C3H8), a fuel for camp stoves and some specially equipped automobiles, reacts with oxygen to

produce carbon dioxide and water. How many grams of O2 are required to react with 22.0 g C3H8?

Solution

Given 22.0 g C3H8 Need grams of O2 STEP 1

STEP 2 Plan The 22.0 g C3H8 is first changed to moles of C3H8 using its molar mass. Then the moles of C3H8 are

changed to moles of O2, which can be converted to grams of O2.


Problem 8.7 (Cont.) Calculating the Mass of Reactant

Using the equation in Problem 8.7, calculate the grams of C3H8 that are needed to produce 15.0 g H2O.

Study Check

Propane gas, (C3H8), a fuel for camp stoves and some specially equipped automobiles, reacts with oxygen to

produce carbon dioxide and water. How many grams of O2 are required to react with 22.0 g C3H8?

Solution

Set Up Problem Using a sequence of three factors converts grams of C3H8 to grams of O2. STEP 4

You are going to have a dinner party. In your silverware drawer, there are 10 spoons, 8 forks, and 6 knives. If

each place setting requires 1 spoon, 1 fork, and 1 knife, how many people can you serve?

Problem 8.8 Limiting Reactants

You have a bicycle parts store. If there are 45 wheels and 18 bicycle seats, how many bicycles can you put together?

Study Check

Solution

Problem 8.9 Moles of Product from Limiting Reactant

Consider the reaction for the synthesis of methanol (CH3OH).

In the laboratory, 3.00 mol CO and 5.00 mol H2 are combined. Calculate the number of moles of CH3OH that

can form and identify the limiting reactant.

Solution

Given 3.00 mol CO and 5.00 mol H2

Need moles of CH3OH

STEP 1

STEP 2 Plan We can determine the limiting reactant and the excess reactant by calculating the moles of methanol

that each reactant would produce if it were all used up. The actual number of moles of CH3OH produced

is from the reactant that produces the smaller number of moles.

Problem 8.9 (Cont.) Moles of Product from Limiting Reactant




Solution

Equalities/Conversion Factors The mole-mole factors needed are obtained from the equation. STEP 3

Set Up Problem The moles of CH3OH from each reactant are determined in separate

calculations.

STEP 4





Solution

Set Up Problem The smaller amount (2.5 mol CH3OH) is all the methanol that can be produced.

Thus, H2 is the limiting reagent and CO is in excess.

STEP 4





Consider the reaction for the formation of copper(II) oxide from copper(I) oxide.

Complete the following table for the quantities of reactants and products if 5.0 mol of copper(I) oxide and 2.0 mol of

oxygen are allowed to react.

Study Check

Carbon monoxide and hydrogen gas react to form methanol, CH3OH.

If 48.0 g CO and 10.0 g H2 react, how many grams of methanol can be produced?

Problem 8.10 Mass of Product from a Limiting Reactant

Solution

Given 48.0 g CO and 10.0 g H2 Need grams of CH3OH STEP 1

STEP 2 Plan Convert the grams of each reactant to moles and calculate the moles of CH3OH that each reactant

can produce. Then convert the number of moles of CH3OH from the limiting reactant to grams of

CH3OH using molar mass.

Problem 8.10 (Cont.) Mass of Product from a Limiting Reactant



Solution

Equalities and Conversion Factors STEP 3




Solution

Set Up Problem The moles of CH3OH from each reactant can now be determined in separate

calculations.

Moles CH3OH produced from CO

Moles CH3OH produced from H2

STEP 4




Solution

Because CO produces the smaller number of moles of CH3OH, CO is the limiting reactant. Now

the grams of product CH3OH from these reactants is calculated by converting the moles of

CH3OH obtained from CO to grams using molar mass.

STEP 4

Hydrogen sulfide burns with oxygen to give sulfur dioxide and water. How many moles of sulfur dioxide are formed

from the reaction of 0.250 mol H2S and 0.300 mol O2?

Study Check

Problem 8.11 Mass of Product from a Limiting Reactant

Ammonia and fluorine react to form dinitrogen tetrafluoride, N2F4, and hydrogen fluoride.

If 5.00 g NH3 and 20.0 g F2 react, how many grams of hydrogen fluoride are produced?

Solution

Given 5.00 g NH3 and 20.0 g F2 Need grams of HF STEP 1

STEP 2 Plan We determine the limiting reactant and the excess reactant by calculating the possible number of

moles of HF that each reactant can produce. We are looking for the reactant that produces the smaller

number of moles of HF. In this calculation, the grams of each reactant is converted to moles and the

number of moles of HF each will produce is calculated. Then the number of moles of HF from the

limiting reactant, the smaller amount, is converted to the mass in grams of HF using its molar mass.





Solution




Solution

Set Up Problem The moles of HF from each reactant are determined in separate calculations.

The limiting reactant is F2 because it produces the smaller number of moles of HF product. Next

the quantity of the product, HF, is converted from moles to grams using its molar mass.

STEP 4




When silicon dioxide (sand) and carbon are heated, the ceramic material silicon carbide, SiC, and carbon monoxide are

produced. How many grams of SiC are formed from 20.0 g SiO2 and 50.0 g C?

Study Check

Problem 8.12 Calculating Percent Yield

On a spaceship, LiOH is used to absorb exhaled CO2 from breathing air to form LiHCO3.

What is the percent yield of the reaction if 50.0 g LiOH gives 72.8 g LiHCO3?

Solution

Given 50.0 g LiOH and 72.8 g LiHCO3 (actually produced)

Need % yield LiHCO3

STEP 1

STEP 2 Plan

Calculation of theoretical yield:

Problem 8.12 (Cont.) Calculating Percent Yield



Solution


Problem 8.12 (Cont.) Calculating Percent Yield

For the reaction in Problem 8.12, what is the percent yield if 8.00 g CO2 produces 10.5 g LiHCO3?

Study Check



Solution


CHAPTER 9

ELECTRONIC STRUCTURE AND PERIODIC TRENDS

A student uses a microwave oven to make popcorn. If the radiation has a frequency of 2500 megahertz (MHz),

what is the wavelength in meters of the microwaves?

Problem 9.1 Wavelength and Frequency

Solution

Given v = 2500 MHz Need m STEP 1

STEP 2 Plan


A student uses a microwave oven to make popcorn. If the radiation has a frequency of 2500 megahertz (MHz),

what is the wavelength in meters of the microwaves?

Problem 9.1 (Cont.) Wavelength and Frequency

If your dentist uses X rays with a wavelength of 2.2 × 10-10 m, what is the frequency of the X rays?

Study Check

Solution


Visible light contains colors from red to violet.

a. What color has the shortest wavelength?

b. What color has the lowest frequency?

Problem 9.2 The Electromagnetic Spectrum

Arrange the following in order of increasing frequencies: X rays, ultraviolet light, FM radio waves, and microwaves.

Study Check

Solution

a. Violet light has the shortest wavelength.

b. Red light has the lowest frequency

a. How does an electron move to a higher energy level?

b. When an electron drops to a lower energy level, how is energy lost?

Problem 9.3 Electron Energy Levels

Why did scientists propose that electrons occupied specific energy levels in the atom?

Study Check

Solution

a. An electron moves to a higher energy level when it absorbs an amount of energy equal to the difference in levels.

b. Energy equal to the difference in levels is emitted when an electron drops to a lower level.

Indicate the type and number of orbitals in each of the following energy levels or sublevels:

a. 3p sublevel b. n = 2

c. n = 3 d. 4d sublevel

Problem 9.4 Sublevels and Orbitals

Solution

a. The 3p sublevel contains three 3p orbitals.

b. The n = 2 principal energy level consists of 2s (one) and 2p (three) orbitals.

Indicate the type and number of orbitals in each of the following energy levels or sublevels:

a. 3p sublevel b. n = 2

c. n = 3 d. 4d sublevel

Problem 9.4 (Cont.) Sublevels and Orbitals

What is similar and what is different for 1s, 2s, and 3s orbitals?

Study Check

Solution

Write the orbital diagram, electron configuration, and the abbreviated electron configuration for nitrogen.

Problem 9.5 Writing Electron Configurations

Write the orbital diagram, electron configuration, and the abbreviated electron configuration for fluorine.

Study Check

Solution

Nitrogen with atomic number 7 has 7 electrons. To write the orbital diagram, draw boxes for the 1s, 2s, and 2p

orbitals.

Add 7 electrons, starting with the 1s orbital. Two electrons with opposite spins are added to the 1s and 2s orbitals.

For the 2p orbitals, the 3 remaining electrons are placed in separate orbitals with parallel spins.

The electron configuration for nitrogen shows the electrons filling the sublevels in order of increasing energy.

The abbreviated electron configuration for nitrogen is written by substituting the symbol [He] for the 1s2.

For each of the following elements, write the stated type of electron notation:

a. orbital diagram for silicon

b. electron configuration for phosphorus

c. abbreviated electron configuration for chlorine

Problem 9.6 Orbital Diagrams and Electron Configurations

Solution

a. Silicon in Period 3 has atomic number 14, which tells us that it has 14 electrons. To write the orbital diagram,we

draw boxes for the orbitals up to 3p.

Add 14 electrons, starting with the 1s orbital. Show paired electrons in the same orbital with opposite spins and place

the last 2 electrons in different 3p orbitals.

For each of the following elements, write the stated type of electron notation:

a. orbital diagram for silicon

b. electron configuration for phosphorus

c. abbreviated electron configuration for chlorine

Problem 9.6 (Cont.) Orbital Diagrams and Electron Configurations

Write the complete and abbreviated electron configuration for sulfur.

Study Check

Solution

Use the sublevel blocks on the periodic table to write the electron configuration for

a. bromine b. cesium

Problem 9.7 Using Sublevel Blocks to Write Electron

Configurations

Solution

a. Bromine is in the p block and in Period 4. STEP 1

STEP 2 Beginning with 1s2, go across the periodic table writing each filled sublevel block as follows:

STEP 3

Use the sublevel blocks on the periodic table to write the electron configuration for

a. bromine b. cesium

Problem 9.7 (Cont.) Using Sublevel Blocks to Write Electron

Configurations

Write the electron configuration for tin.

Study Check

Solution

b. Cesium is in the s block and in Period 6. STEP 1

STEP 2 Going across the periodic table starting with Period 1, the sublevel blocks fill as follows:

STEP 3

Using the periodic table, write the group number and the electron configuration of the valence electrons for the

following:

a. calcium b. selenium c. lead

Problem 9.8 Using Group Numbers

What are the group numbers and valence electron arrangements for sulfur and strontium?

Study Check

Solution

a. Calcium is in Group 2A (2). It has a valence electron configuration of 4s2.

b. Selenium is in Group 6A (16). It has a valence electron configuration of 4s24p4.

c. Lead is in Group 4A (14). It has a valence electron configuration of 6s26p2. The electrons in the 5d and 4f

sublevels are not valence electrons.

Why is the radius of a phosphorus atom larger than the radius of a nitrogen atom but smaller than the radius of

a silicon atom?

Problem 9.9 Atomic Radius

Arrange atoms of the following elements in order of decreasing atomic radius: Mg, S, and Na.

Study Check

Solution

The radius of a phosphorus atom is larger than the radius of a nitrogen atom because phosphorus has valence

electrons in a higher energy level, which is further from the nucleus. A phosphorus atom has one more proton than a

silicon atom; therefore, the nucleus in phosphorus has a stronger attraction for the valence electrons, which decreases

its radius compared to a silicon atom.

Indicate the element in each that has the higher ionization energy and

explain your choice.

a. K or Na b. Mg or Cl c. F or N

Problem 9.10 Ionization Energy

Arrange Sn, Sr, and I in order of increasing ionization energy.

Study Check

Solution

a. Na. In Na, an electron is removed from an orbital closer to the nucleus.

b. Cl. Nuclear charge increases the attraction for the outermost electrons for elements in the same period.

c. F. The increased nuclear charge of F requires a higher ionization energy compared to N.

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APPLIED CHEMISTRY SEMESTER REVIEW - Wikispaceskmkunz.wikispaces.com/file/view/Applied+Chemistry+-+Semester... · applied chemistry semester review chapter 7 – chemical reactions