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Tommy Smith Dr. Snyder 10th Grade 2012-2013

The above image was created by Tommy Smith


Table Of Contents Quarter 1 1.Periodic Table of Elements Handout

6. Translating Sentences into Equations

2. Conversion Sheet

7. Balancing Chemical Equations

3. Significant Figures Handout

8. Synthesis Reactions

4. Constructing A Model

9. Decomposition Reactions

5. The Mole Concept Handout

10. Single-Replacement Reactions

6. The Mole Concept

11. Double-Replacement Reactions

7. Atomic Models

12. Stoichiometry

Quarter 2 1. Atomic Theories (Cont.)

Quarter 4 1. Charles's Law

2. Atomic Orbitals Drawing

2. Gay-Lussac's Law

3. Electron Configuration

3. Elasticity of Gases

4. Designing Your Own Periodic Table

4. Crystal Systems Handout

5. Atomic Modelling handout

5.Constructing Crystal Systems

6. Lewis Structures

6. The Changing of Equilibrium

7. Molecular Geometry Handout

7. Analysis of Simple Mixtures

Quarter 3 1. Chemical Formulas Handout

8. Phase Diagram of H2O 9. Distinguishing Acids and Bases

2. Naming Ionic Compounds worksheet 10. Identifying Acids and Bases 3. Naming Chemical Compounds worksheet 11.Organic Compounds 4. Naming Covalent Compounds worksheet 12. Reflection Essay 5. Translating Equations into Sentences


1. Periodic Table of Elements Handout 2. Conversion Sheet 3. Significant Figures Handout 4. Constructing A Model 5. The Mole Concept Handout 6. The Mole Concept 7. Atomic Models


A. 2.5 cm/1 x 10mm/1 cm B. 3.4dm/1 x 1dkm/1dm C. 6.834hm/1 x 100 000mm/1hm D. 109.87dkm/1 x 1km/100dkm E. 1900.07cm/1 x 1hm/10 000cm F. 1km/1 x 1 000m/1km G. 2km/1 x 1 000 000mm/1km H. 3m/1 x 1km/1 000m I. 1dm/1 x 1hm/1 000dm J. 2cm/1 x 1hm/10 000cm K. 3hm/1 x 1 000dm/1hm L. 1hm/1 x 100m/1hm M. 4m/1 x 10dm/1m N. 3m/1 x 100cm/1m O. 0.5km/1 x 1 000m/1km P. 14m/1 x 1km/1 000m Q. 23.5mm/1 x 1m/1 000mm R. 14.2mm/1 x 1cm/10mm S. 43.9km/1 x 10 000dm/1km T. 5678.901mm/1 x 1m/1 000mm U. 0.234km/1 x 1 000 000mm/1km V. 34.25dkm/1 x 1km/100dkm W. 99.9999hm/1 x 10 000cm/1hm X. 8787.87mm/1 x 1dkm/10 000mm Y. 5555.555cm/1 x 1hm/10 000cm Z. 0.000302km/1 x 10 000mm/1km


1. a. .0120 m has three significant figures. b. 100.5 mL has four signific80ant figures c. 101 g has three significant figures d. 350 cm2 has two significant figures e. 0.97 km has two significant figures f. 1000 kg has one significant figures g. 180. mm has three significant figures h. 0.4936 L has four significant figures i .020700 s has five significant figures 2. a. 5 487 129 m rounded to 3 significant figures is 5490000m b. 0.013 479 265 mL rounded to 6 significant figures is 0.0134793 mL c. 31 947.972 cm2 rounded to 4 significant figures is 31950 cm2 d. 192.673 9 m2 rounded to 5 significant figures is 192.67 m2 e. 786.916 4 cm rounded to 2 significant figures is 790 cm f. 389 277 600 J rounded to 6 significant figures is 389 278 000 J g. 225 834.762 cm3 rounded to 7 significant figures is 225834.8 cm3 8. a. 13.75 mm x 10.1 mm x .91 mm = 126.376 25 mm3 = 130 mm3 b. 89.4 cm2 x 4.8 cm = 429.12 cm3 = 430 cm3 c. 14.9 m3/3.0 m2 = 4.96 m = 5.0 m d. 6.975 m x 30 m x 21.5 m = 4 498.875 m3 = 4 000 m3 9. 752 m x 319 m x 110 m = 26 387 680 m3 = 26 000 000 m3 10. a. 7.38 g + 1.21 g + 4.792 3 g = 13.384 3 g = 13.4 g b. 51.3 mg + 83 mg - 34.2 mg = 100.1 mg = 1.0 x 102 mg c. .007 L – 0.003 7 L + 0.012 L = 0.015 3 L =0.02 L d. 253.05 cm2 + 33.9 cm2 + 28 cm2 = 314.95 cm2 = 320 cm2 e. 14.77 kg + .086 kg - .391 kg = 14.465 kg = 14 kg f. 319 mL + 13.75 mL +20. mL = 352.75 mL = 350 mL


Constructing a Model I. Purpose: To understand how scientists make inferences about atoms II. Materials: closed container, various objects, balance III. Procedure First place your items in a closed container. Then take another student's container and, keeping it closed, shake it and see if you can determine the contents. Measure the mass of the container with the contents inside and estimate the mass of the contents. Then open the container without looking and feel the contents. Make adjustments to your estimation of the mass if needed. lastly remove the contents and observe them. Measure their mass on the balance. V. Data Part A: Closed Container Closed Container

# of Objects

Mass of Objects (g)

Kind of Objects

1

2

71.5

metallic

2

8

28

Hard, wood?

3

3

89

Metallic, hair clips

Mass of Objects (g)

Kind of Objects

Part B: Open Container without Looking Open Container # of Objects 1

3

78.6

Eraser, Duplo Lego, Magnet

2

15

29

Hairbands, Tag, Pieces of Paper

3

4

70

Bracelet, Bow

Mass of Objects (g)

Kind of Objects

Part C: Open Container with looking Open Contianer # of Objects 1

3

50.7

Eraser, Duplo Lego, Magnet

2

3

15.8

Earbuds, Ribbon, Paperclip

3

5

8.25

Bow, Charm Bracelet, Safety Pin, Sequin, Charm

VI. Analyze and Conclude 1. Scientists often use more then one method to gather data. How was this illustrated in the investigation? We used three different methods to gather data, shaking the box, feeling the object and then finally looking at the contents.


2. Of the observations made, which were quantitative and which were qualitative? The measurements of mass and the number of items inside were quantitative data. The types of objects were qualitative data. 3. Using the data you gathered, draw a model of the unknown objects and write a brief summary of your conclusions. In this one, number two, I felt several different textures, giving the illusion of many different objects. Also, the objects, in particular the ear-buds, had several components, which gave the illusion of several objects when shaken.

The above image was made by the author, Tommy Smith.


Problem Type 1 mol to g 1. Analyze: Given: 1.002 mol of chromium Unknown: mass in grams Plan: mol x g/1mol Calculate: 1.002 mol Cr x 52.00 g Cr/1 mol Cr = 52.10 g Cr Evaluate: Units, correct number of significant figures, correct There is a little more than 1 mol of substance present. So there would be a little more than 1 molar mass of chromium 2. Analyze: Given: 4.08 x 10-8 mol of sodium Unknown: mass in grams Plan: mol x g/1mol Calculate: 4.08 x 10-8 mol Ne x 20.18 g Ne/1 mol Ne= 8.23 x 10-7 g Ne Evaluate: Units, correct number of significant figures, correct There is very little substance present. Therefore there would be very little mass. 3.Analyze: Given: 7 mol of titanium Unknown: mass in grams Plan: mol x g/1mol Calculate: 7 mol Ti x 47.88 g Ti/1 mol Ti = 300 g Ti Evaluate: Units, correct number of significant figures, correct There are seven moles present. So seven times the molar mass of titanium would be the answer and 300 is about seven times the molar mass. Problem Type 2 grams to mol 1. Analyze: Given: 11.9 g of aluminum Unknown: number of moles of aluminum Plan: g x 1 mol/g Calculate: 11.9 g Al x 1 mol Al/26.98 g Al = 0.441 mol Al Evaluate: Units, correct

number of significant figures, correct


There is less than half the molar mass present. Ergo there would be less than half a mole. 2. Analyze: Given: 3.22 g of copper Unknown: number of moles of copper Plan: g x 1 mol/g Calculate: 3.22 g Cu x 1 mol Cu/63.55 g Cu = 0.0507 mol Cu Evaluate: Units, correct number of significant figures, correct There is roughly one twentieth of the molar mass present, so the answer is about one twentieth. 3. Analyze: Given: 2.72 x 10-7 g of lithium Unknown: number of moles of lithium Plan: g x 1 mol/g Calculate: 2.72 x 10-4 g Li x 1 mol Li/6.94 g Li = 3.92 x 10-5 mol Li Evaluate: Units, correct number of significant figures, correct The already small number is being made smaller, this makes the answer reasonable. Problem Type 3 number of atoms to moles 1. Analyze: Given: 3.01 x 1023 atoms of silver Unknown: number of moles of silver Plan: number of atoms x 1 mol/Avogadro's Number Calculate: 3.01 x 1023 atoms Ag x 1 mol Ag/(6.022 x 1023)atoms Ag = 0.500 mol Ag Evaluate: Units, correct number of significant figures, correct The number is about half of Avogadro's number, so it makes sense that the answer would be about half. 2. Analyze: Given: 2.25 x 1022 atoms of carbon Unknown: number of moles of carbon Plan: number of atoms x 1 mol/Avogadro's Number Calculate: 2.25 x 1022 atoms C x 1 mol C/(6.022 x 1023)atoms C = .0374 mol C Evaluate: Units, correct number of significant figures, correct The number is less than Avogadro's number by about one ten's place. So the answer should reasonably be within the hundred's place.


3. Analyze: Given: 2 x 106 atoms of oxygen Unknown: number of moles of oxygen Plan: number of atoms x 1 mol/Avogadro's Number Calculate: 2 x 106 atoms O x 1 mol O/(6.022 x 1023)atoms O = 3 x 10-18 mol O Evaluate: Units, correct number of significant figures, correct The number of atoms is so few that it would be so little of a mole. Problem Type 4 number of atoms to grams 1. Analyze: Given: 5.0 x 109 atoms of neon Unknown: number of grams of neon Plan: number of atoms x molar mass/Avogadro's Number Calculate: 5.0 x 109 atoms Ne x 20.18g Ne/(6.022 x 1023)atoms Ne = 1.7 x 10-13 g Ne Evaluate: Units, correct number of significant figures, correct The amount of atoms is so much less than Avogadro's number, so the mass would be much smaller than the molar mass. 2. Analyze: Given: 20 x 108 atoms of copper Unknown: number of grams of copper Plan: number of atoms x molar mass/Avogadro's Number Calculate: 20 x 108 atoms Cu x 63.55g Cu/(6.022 x 1023)atoms Cu = 2 x 10-13 g Cu Evaluate: Units, correct number of significant figures, correct The amount of atoms is so much less than Avogadro's number, so the mass would be much smaller than the molar mass. 3. Analyze: Given: 2.0 x 1023 atoms of gold Unknown: number of grams of gold Plan: number of atoms x molar mass/Avogadro's Number Calculate: 2.0 x 1023 atoms Au x 196.97g Au/(6.022 x 1023)atoms Au = 66 g Au Evaluate: Units, correct number of significant figures, correct The number of atoms is about one-third the mass of Avogadro's number, 66 is about one third of 196.97.


The Mole Concept I. Purpose: to identify each sample based on observation and numerical data II. Materials: sample set, balance, periodic table III. Procedure Make observations about sample A. Measure the mass on a balance. Use a periodic table of elements to identify the substance. Repeat for samples B, C, and D. IV. Data A. Observations Sample

Observations

A

Metallic, silver color; streaking of the metal

B

Silver color; silver color

C

Tarnished; dark color

D

Metallic, coppery color

B. Numerical Data Sample Mass (g) Number of moles A B

65 26.9

Number of atoms

Identity of the element

1

6.022 x 1023

Zinc

1

23

Aluminum

23

6.022 x 10

C

55.5

1

6.022 x 10

Iron

D

63.2

1

6.022 x 1023

Copper

V. Analyze and Conclude 1. What led you to the determination of each element? The molar mass of each substance is unique, so our masses would be similar to the average atomic mass. 2. If each element was exactly one mole, why were the masses different? The mass of each individual atom is different. 3. Determine the actual mass of a single atom of each sample. Sample A: 1.1 x 10-22 Sample B: 4.67 x 10-23 Sample C: 9.22 x 10-23 Sample D: 1.05 x 10-22 4. Write a one-two paragraph summary of the lab. Draw some general conclusions. This lab was helpful in going beyond book knowledge. I knew the basic concepts; this went and applied them. In this lab, the ones with the highest measured mass had the atoms with the highest mass. Each object was one mole, so they were all different shapes and sizes. This illustrated the differences in molar masses. Overall it was a really fun lab and I lokk forward to others like it.


Atomic Models The Billiard Ball Model Early 1800s Proposed by Dalton ~Explains many chemical properties and confirms the laws of conservation of mass and definite proportions ~Lacks a nucleus and all subatomic particles

The Raisin Bun Model Mid 1800s Proposed by Thomson ~Adds subatomic particle to the model and introduces the nucleus ~Does not explain the existence of electrons outside of the nucleus ~Does not include neutrons


The Planetary Model Late 1800s Proposed by Rutherford ~First real modern view of the atom ~Explains why the electron spins around the nucleus. Proposes that an atom is made mostly of empty space. ~Does not place electrons in definite levels around the nucleus. Does not relate the valence electrons atomic charge. ~Does not include neutrons in the nucleus

All images in this lab were created by Tommy Smith


1. Atomic Theories (Cont.) 2. Atomic Orbitals Drawing 3. Electron Configuration 4. Designing Your Own Periodic Table 5. Atomic Modelling handout 6. Lewis Structures 7. Molecular Geometry Handout

The above image of two bonded hydrogen atoms was made by the author, Tommy Smith.


Bohr's Model Early 1900s proposed by Bohr ~Explains the role of electrons in bonding ~Fully explains ionic and covalent bonding ~Places electrons in definite energy levels ~Relegates number of valence electrons to specific periods of the periodic table ~Does not explain the shapes of molecules or other abnormalities that result from unevenly shared electrons

Quantum Model Proposed by Einstein, Schroedinger, de Broglie, Planck, Hertz, Maxwell, and Ferni ~Explains shapes of molecules and other abnormalities that result from unevenly shared electrons


Atomic Orbitals

S Orbital

D orbital

P Orbital


Electron Configuration 1. Sulfur: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓ 1s 2s 2p 3s 3p 1s22s22p63s23p4 [Ne]3s23p4 2. Arsenic: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑ ↑ ↑ 1s 2s 2p 3s 3p 4s 3d 4p 1s22s22p63s23p64s23d104p3 [Ar]4s23d104p3 3. Cobalt: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑ ↑ ↑ 1s 2s 2p 3s 3p 4s 3d 1s22s22p63s23p64s23d7 [Ar]4s23d7 4. Phosphorus: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑ ↑ ↑ 1s 2s 2p 3s 3p 1s22s22p63s23p3 [Ne]3s23p3 5. Krypton: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ 1s 2s 2p 3s 3p 4s 3d 4p 1s22s22p63s23p64s23d104p6 [Ar]4s23d104p6 6. Iodine: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 1s22s22p63s23p64s23d104p65s24d10 [Kr]5s24d10 7. Potassium: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑ 1s 2s 2p 3s 3p 4s 1s22s22p63s23p64s1 [Ar]4s1


8. Scandium: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑ 1s 2s 2p 3s 3p 4s 3d 1s22s22p63s23p64s23d1 [Ar]4s23d1 9. Silicon: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑ ↑ 1s 2s 2p 3s 3p 1s22s22p63s23p2 [Ne]3s23p2 10. Fluorine: ↑↓ ↑↓ ↑↓↑↓↑ 1s 2s 2p 1s22s22p5 [He]2s22p5 11. Zinc: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ 1s 2s 2p 3s 3p 4s 3d 1s22s22p63s23p64s23d10 [Ar]4s23d10 12. Cesium: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑ 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s 1s22s22p63s23p64s23d104p65s24d105p66s1 [Xe]6s1 13. Zirconium: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑↓ ↑ ↑ 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 1s22s22p63s23p64s23d104p65s24d2 [Kr]5s24d2 14. Argon: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ 1s 2s 2p 3s 3p 1s22s22p63s23p6 [Ne]3s23p6


15. Xenon: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 1s22s22p63s23p64s23d104p65s24d105p6 [Kr]5s24d105p6 16. Tellurium: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑ ↑ 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 1s22s22p63s23p64s23d104p65s24d105p4 [Kr]5s24d105p4 17. Strontium:↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑↓ 1s 2s 2p 3s 3p 4s 3d 4p 5s 1s22s22p63s23p64s23d104p65s2 [Kr]5s2 18. Manganese: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑ ↑ ↑ ↑ ↑ 1s 2s 2p 3s 3p 4s 3d 1s22s22p63s23p64s23d5 [Ar]4s23d5 19. Indium:↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑ 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 1s22s22p63s23p64s23d104p65s24d105p1 [Kr]5s24d105p1 20. Germanium: ↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓ ↑↓ ↑↓↑↓↑↓↑↓↑↓ ↑ ↑ 1s 2s 2p 3s 3p 4s 3d 4p 1s22s22p63s23p64s23d104p2 [Ar]4s23d104p2


Designing Your Own Periodic Table I. Purpose: To design your own periodic table using information similar to that available to Mendeleev. II. Materials: periodic table, index cards IV. Procedure Fill out data on your cards. Make sure the data is comprehensive enough so someone could figure out the position of each element on the periodic table. Using elements from only periods 1, 2, and 13-17, label each card A-T. Make a key as you label them so that you do not have to solve your own puzzle. Swap cards with another student. Arrange his or her cards as you would expect them to be using the periodic table. When the other student is done check his or her arrangement and have them check your arrangement. Chart your first attempt at arranging the cards and then chart the correct version. V. Data Part A: Index Cards and Key See separate sheet for index cards Group 1

2

13

14

15

16

17

S(N)

P(O)

F(F)

C(P)

K(S) G(Cl)

1 2 Period

3

A(Mg)

4

N(Ga) I(Ge) B(As) Q(Se)

5

L(In) E(Sn) R(Sb) H(Te)

6

M(Tl)

7

T(Bi) O(Po) G(At)

D(Fr)

Part B: Initial Attempt at Organizing Elements Group 1

2

13

14

15

16

17

S

R

1 2 Period

D

3

K

4

G

5

Q

6

N

7

J

B

I

O C

E

F

A P

T

L H

M


Part C: Actual Order of Elements Group 1

2

13

2

D

K

J

3

G

4

Q

14

15

16

17

S

R

1 Period

I B

5 6 7

N

O C

E

F

A

L H

M

P T

VI. Analyze and Conclude 1. Keeping in mind that the information you have is similar to that available to Mendeleev in 1869, answer the following questions. a. Why are atomic masses used instead of atomic numbers? The atomic numbers were unknown at the time. b. Can you identify each element by name? Yes, element A is antimony. Element B is calcium. Element C is germanium. Element D is lithium. Element E is arsenic. Element F is tin. Element G is sodium. Element H is tellurium. Element I is aluminum. Element J is boron. Element K is beryllium. Element L is bromine. Element M is iodine. Element N is cesium. Element O is silicon. Element P is bismuth. Element Q is potassium. Element R is chlorine. Element S is sulfur. Element T is radium. 2. How many groups or families are there in your periodic table? There are seven. 3. Predict the characteristics of any missing elements. When you have finished, check your work using a periodic table. The element in period five, group one will have an atomic mass of about 84-85. The element in period 7, group 1 will have an atomic mass of about 220. The element in group 2, period five will have an atomic mass of about 86. The element in group fourteen period six will have an atomic mass of about 200-210. The element in group 16, period four will have an atomic mass of about eighty.


Lewis Structures 1.

2. 3.

4.

5. 6. 7. 8. 9. 10.

11. 12. 13. 14. 15. 16. 17.


18. 19. 20.


1. Chemical Formulas Handout 2. Naming Ionic Compounds worksheet 3. Naming Chemical Compounds worksheet 4. Naming Covalent Compounds worksheet 5. Translating Equations into Sentences 6. Translating Sentences into Equations 7. Balancing Chemical Equations 8. Synthesis Reactions 9. Decomposition Reactions 10. Single-Replacement Reactions 11. Double-Replacement Reactions 12. Stoichiometry 13. Boyle's Law


Translating Equations into Sentences 1. NO2(g) + HNO3(l) → HNO3(aq) + NO(g) Nitrogen dioxide gas combines with liquid hydrogen nitrate to produce nitric acid and nitrogen dioxide gas. 2. KClO3(s) → HCl(s) + O2(g) Solid potassium chlorate decomposes into solid hydrogen chloride and oxygen gas. 3. Zn(s) + HCl(aq) → ZnCl2(aq) + H2(g) Solid zinc combines with hydrochloric acid to produce zinc chloride and hydrogen gas. 4. Al2(SO4)3(s) + Ca(OH)2(aq) → Al(OH)3(aq) + CaSO4 (aq) Solid aluminum sulfate combines with aqueous calcium hydroxide to form aqueous aluminum hydroxide and aqueous calcium sulfate. 5. Na(s) + Cl2(g) → NaCl(s) Solid sodium combines with chlorine gas to form solid sodium chloride.


Translating Sentences into Equations 1. Aqueous dihydrogen dioxide decomposes to produce oxygen and water. H2O2(aq) → O2(g) + H2O(l) 2. Solid zinc sulfide and oxygen yield solid zinc oxide and sulfur dioxide gas. ZnS(s) + O2(g) → ZnO(s) + SO2(g) 3. Aqueous hydrogen chloride and aqueous magnesium hydroxide react to produce aqueous magnesium chloride and water. HCl(aq) + Mg(OH)2(aq) → MgCl2(aq) + H2O(l) 4. Aqueous hydrogen nitrate and aqueous calcium hydroxide react to produce aqueous calcium nitrate and water. HNO3(aq) + Ca(OH)2(aq) → CaNO3(aq) + H2O(l) 5. Solid calcium hydroxide and liquid water react to produce aqueous calcium hydroxide and hydrogen gas. Ca(OH)2(s) + H2O(l) → Ca(OH)2(aq) + H2(g)


Balancing Chemical Equations 1. Solid calcium metal reacts with water to form aqueous calcium hydroxide and hydrogen gas. Ca(s) + 2H2O(ℓ) → Ca(OH)2(aq) + H2 (g) 2. Nitrogen dioxide gas reacts with water to form aqueous hydrogen nitrate (nitric acid) and nitrogen monoxide gas. 3NO2(g)+ H2O(ℓ) → 2HNO3(aq) + NO(g) 3. Solid potassium chlorate decomposes to form solid potassium chlorate and oxygen gas. 2KClO3 → 2KCl(s) + 3O2(g) 4. Solid zinc oxide is made by reacting solid zinc sulfide with oxygen gas. Sulfur dioxide gas is also produced. 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g) 5. Solid calcium phosphate is reacted with aqueous sulfuric acid (hydrogen sulfate) to form phosphoric acid (hydrogen phosphate) and solid calcium sulfate. Ca3(PO4)2 (s) + 3H2SO4(aq) → 2H3PO4(aq) + 3CaSO4(s) 6. Solid sodium combines with chlorine gas to produce solid sodium chloride. 2Na(s) + Cl2(g) → 2NaCl(s) 7. When solid copper reacts with aqueous silver nitrate the products are aqueous copper(II) nitrate and solid silver. Cu(s) + 2AgNO3 → Cu(NO3)2(aq) + 2Ag(s) 8. In a blast furnace, the reaction between solid iron(III) oxide and carbon monoxide gas produces solid iron and carbon dioxide gas. Fe2O3(s) +3CO(g)

2Fe(s) + 3CO2(g)

9. When solid lithium reacts with aqueous aluminum sulfate the products are aqueous lithium sulfate and solid aluminum. 6Li(s) + Al2(SO4)3(aq) → 3Li2SO4(aq) + 2Al(s) 10. Hydrogen sulfide gas and oxygen gas produce sulfur dioxide gas and water vapor. 2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)


Synthesis Reactions 1. calcium + oxygen → calcium oxide 2Ca + O2 → 2CaO 2. iron (III) + sulfur → iron(III) sulfide 16Fe + 3S8 → 8Fe2S3 3. hydrogen + oxygen → water 2H2 + O2 → 2H2O 4. aluminum + bromine → aluminum bromide 2Al + 3Br2 → 2AlBr3 5. sodium + iodine → sodium iodide 2Na + I2 → 2NaI 6. calcium oxide + water → calcium hydroxide CaO + H2O → Ca(OH)2 7. chromium(III) + oxygen → chromium (III) oxide 4Cr + 3O2 → 2Cr2O3 8. silver + sulfur → silver sulfide 16Ag + S8 → 8Ag2S 9. magnesium + oxygen → magnesium oxide 2Mg + O2 → 2MgO 10. sodium + oxygen → sodium oxide 4Na +O2 → 2Na2O


Decomposition Reactions 1. Iron(III) oxide → iron(III) + oxygen Fe3O2 → 3Fe + O2 2. Barium hydroxide → barium oxide + water Ba(OH)2 → BaO + H2O 3. Magnesium chlorate → magnesium chloride + oxygen Mg(ClO3)2 → MgCl2 + 3O2 4. Potassium carbonate → potassium oxide + carbon dioxide K2CO3 → K2O + CO2 5. Magnesium hydroxide → magnesium oxide + water Mg(OH)2 → MgO + H2O 6. Silver chloride → silver + chlorine 7. strontium chlorate → strontium chloride + oxygen Sr(ClO3)2 → SrCl2 + 3O2 8. Magnesium carbonate → magnesium oxide + carbon dioxide MgCO3 → MgO + CO2 9. Aluminum chlorate → aluminum chloride + oxygen 2Al(ClO3)3 → 2AlCl3 + 9O2 10. beryllium hydroxide → beryllium oxide + water Be(OH)2 → BeO + H2O


Singe-Replacement Reactions 1. silver nitrate + nickel(II) → nickel(II) nitrate + silver 2AgNO3 + Ni → Ni(NO3)2 + 2Ag 2. aluminum bromide + chlorine → aluminum chloride + bromine 2AlBr3 + 3Cl2 → 2AlCl3 + 3Br3 3. Sodium iodide + bromine → sodium bromide + iodine 2NaI + Br2 → 2NaBr + I2 4. Calcium + hydrochloric acid → hydrogen + calcium chloride Ca + 2HCl → CaCl2 + H2 5. Magnesium + nitric acid → magnesium nitrate + hydrogen Mg + 2HNO3 → Mg(NO3)2 + H2 6. Potassium + water → potassium hydroxide + hydrogen 2K + 2H2O → 2KOH + H2 7. Zinc + magnesium chloride → no reaction Zn + MgCl2 → no reaction 8. Aluminum + water → aluminum oxide + hydrogen 2Al + 3H2O → Al2O3 + 3H2 9. Potassium iodide + bromine → potassium bromide + iodine 2KI + Br2 → 2KBr + I2 10. Magnesium + cobalt(II) nitrate → cobalt(II) + magnesium nitrate Mg + Co(NO3)2 → Co + Mg(NO3)2


Double-Replacement Reactions 1. Silver nitrate + hydrochloric acid → silver chloride + nitric acid AgNO3 + HCl → HNO3 + AgCl 2. Copper(II) chloride + sodium sulfide → copper(II) sulfide + sodium chloride CuCl2 + Na2S → CuS + 2NaCl 3. Iron(II) sulfide + hydrochloric acid → iron(II) chloride + hydrogen sulfide FeS + 2HCl → FeCl2 + H2S 4. Sulfuric acid + potassium hydroxide → potassium sulfate + water H2SO4 + 2KOH → K2SO4 + 2H2O 5. Nitric acid + calcium hydroxide → calcium nitrate + water 2HNO3 + Ca(OH)2 → Ca(NO3)2 + 2H2O 6. Lithium hydroxide + iron(III) nitrate → lithium nitrate + iron(III) hydroxide 3LiOH + Fe(NO3)3 → 3LiNO3 + Fe(OH)3 7. Lead(II) acetate + hydrogen sulfide → lead sulfide + hydrogen acetate Pb(CH3COO)2 + H2S → PbS +2HCH3COO 8. Aluminum iodide + mercury(II) chloride → aluminum chloride + mercury(II) iodide 2AlI3 + 3HgCl2 → 2AlCl3 + 3HgI2 9. Calcium acetate + sodium carbonate → calcium carbonate + sodium acetate Ca(CH3COO)2 + Na2CO3 → CaCO3 + 2NaCH3COO 10. ammonium chloride + mercury(I) acetate → ammonium acetate + mercury(I) chloride NH4Cl + HgCH3COO → NH4CH3COO + HgCl


Stoichiometry Problem Type 1: mol→mol 1. N2 + 3H2 → 2NH3 Analyze: Given: 6 mol H2 Unknown: mol NH3 Plan: mol H2 → mol NH3 mol H2 x mol NH3 = mol NH3 mol H2 Compute: 6 mol H2 x 2 mol NH3 = 4 mol NH3 3 mol H2 Evaluate: Significant figures: correct Units: correct Reasonable: The answer is two-thirds of six. 2. 2KClO3 → 2KCl + 3O2 Analyze: Given: 15 mol O2 Unknown: mol KClO3 Plan: mol O2 → mol KClO3 mol O2 x mol KClO3 = mol KClO3 mol O2 Compute: 15 mol O2 x 2 mol KClO3 = 10 mol KClO3 3 mol O2 Evaluate: Significant figures: correct Units: correct Reasonable: The answer is two-thirds of fifteen. 3. 2H2 + O2 → 2H2O Analyze: Given: 5.0 mol H2O Unknown: mol O2 Plan: mol H2O → mol O2 mol H2O x mol O2 = mol O2 mol H2O Compute: 5.0 mol H2O x 1 mol O2 = 2.5 mol O2 2 mol H2O


Evaluate: Significant figures: correct Units: correct Reasonable: The answer is half of five. 4. 4NH3 + 3O2 → 2N2 + 6H2O Analyze: Given: 4 mol N2 Unknown: mol NH3 Plan: mol N2 → mol NH3 mol N2 x mol NH3 = mol NH3 mol N2 Compute: 4 mol N2 x 4 mol NH3 = 8 mol NH3 2 mol N2 Evaluate: Significant figures: correct Units: correct Reasonable: The answer is the double of four. 5. Mg + 2HCl → MgCl2 + H2 Analyze: Given: 2.50 mol MgCl2 Unknown: mol HCl Plan: mol MgCl2 → mol HCl mol MgCl2 x mol HCl = mol HCl mol MgCl2 Compute: 2.50 mol MgCl2 x 2 mol HCl = 5.00 mol HCl 1 mol MgCl2 Evaluate: Significant figures: correct Units: correct Reasonable: The answer is twice two and a half. Problem Type 2: mol → mass 1. 6CO2 + 6H2O → C6H12O6 + 6O2 Analyze: Given: 3.00 mol H2O Unknown: mass(g) C6H12O6 Plan: mol H2O→ g C6H12O6 mol H2O x mol C6H12O6 x g C6H12O6 = g C6H12O6 mol H2O mol C6H12O6 Compute: 3.00 mol H2O x 1 mol C6H12O6 x 180.18 g C6H12O6 ≈ 90.1 g C6H12O6 6 mol H2O 1 mol C6H12O6


Evaluate: Significant figures: correct Units: correct Reasonable: The answer is about one-half of the molar mass of glucose. 2. 2NaN3 → 2Na + 3N2 Analyze: Given: .500 mol NaN3 Unknown: mass(g) N2 Plan: mol NaN3→ g N2 mol NaN3 x mol N2 x g N2 = g N2 mol NaN3 mol N2 Compute: .500 mol NaN3 x 3 mol N2 x 28.02 g N2 ≈ 21.0 g N2 2 mol NaN3 1 mol N2 Evaluate: Significant figures: correct Units: correct Reasonable: The answer is about three-fourths the mass of dinitrogen. 3. 2Mg + O2 → 2MgO Analyze: Given: 2.00 mol O2 Unknown: mass(g) MgO Plan: mol O2 → g MgO mol O2 x mol MgO x g MgO = g MgO mol O2 mol MgO Compute: 2.00 mol O2 x 2 mol MgO x 40.31 g MgO ≈ 161 g MgO 1 mol O2 1 mol MgO Evaluate: Significant figures: correct Units: correct Reasonable: The answer is about four times the molar mass of magnesium oxide. 4. SiO2 + 3C → SiC + 2CO Analyze: Given: 2.00 mol C Unknown: mass(g) SiC Plan: mol C→ g SiC mol C x mol SiC x g SiC = g SiC mol C mol SiC Compute: 2.00 mol C x 1 mol SiC x 40.1 g SiC ≈ 26.7 g SiC 3 mol C 1 mol SiC


Evaluate: Significant figures: correct Units: correct Reasonable: The answer is about two-thirds the mass of silicon carbide. 5. 2ZnO + C → 2Zn + CO2 Analyze: Given: 5.00 mol C Unknown: mass(g) ZnO Plan: mol C→ g ZnO mol C x mol ZnO x g ZnO = g ZnO mol C mol ZnO Compute: 5.00 mol C x 2 mol ZnO x 81.39 g ZnO ≈ 813.9 g ZnO 1 mol C 1 mol ZnO Evaluate: Significant figures: correct Units: correct Reasonable: The answer is ten times the mass of zinc oxide. Problem Type 3: mass → mol 1. 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) Analyze: Given: 824 g NH3 Unknown: mol NO Plan: g NH3 → mol NO g NH3 x mol NH3 x mol NO = mol NO g NH3 mol NH3 Compute: 824 g NH3 x 1 mol NH3 x 4 mol NO ≈ 48.4 mol NO 17.04 g NH3 4 mol NH3 Evaluate: Significant figures: correct Units: correct Reasonable: The answer is more than twenty times the molar mass, so this large answer makes sense. 2. 2H2O + 2NaCl Cl2 + 2NaOH + H2 Analyze: Given: 250. g NaCl Unknown: mol Cl2 Plan: g NaCl → mol Cl2 g NaCl x mol NaCl x mol Cl2 = mol Cl2 g NaCl mol NaCl


Compute: 250. g NaCl x 1 mol NaCl x 1 mol Cl2 ≈ 2.14 mol Cl2 58.44 g NaCl 2 mol NaCl Evaluate: Significant figures: correct Units: correct Reasonable: There are about five moles of sodium chloride. The answer is about half of that so it is reasonable. 3. Fe2O3(s) + 3CO(g) 2Fe(s) + 3CO2(g) Analyze: Given: 4000 g Fe2O3 Unknown: mol CO Plan: g Fe2O3 → mol CO g Fe2O3 x mol Fe2O3 x mol CO = mol CO g Fe2O3 mol Fe2O3 Compute: 4000 g Fe2O3 x 1 mol Fe2O3 x 3 mol CO ≈ 75.1 mol CO 159.7 g Fe2O3 1 mol Fe2O3 Evaluate: Significant figures: correct Units: correct Reasonable: There are about twenty moles of iron(III) oxide present and then three times that would be the answer. 4. H2SO4 + 2NaHCO3 → 2CO2 + Na2SO4 + 2H2O Analyze: Given: 150.0 g H2SO4 Unknown: mol NaHCO3 Plan: g H2SO4 → mol NaHCO3 g H2SO4 x mol H2SO4 x mol NaHCO3 = mol NaHCO3 g H2SO4 mol H2SO4 Compute: 150.0 g H2SO4 x 1 mol H2SO4 x 2 mol NaHCO3 ≈ 3.058 mol NaHCO3 98.09 g H2SO4 1 mol H2SO4 Evaluate: Significant figures: correct Units: correct Reasonable: There are a little less than two mole of sulfuric acid present, so that doubled would be a little less than four. Therefore this answer is reasonable. 5. Fe2O3 + 2Al → 2Fe + Al2O3 Analyze: Given: 99.0 g Al Unknown: mol Fe2O3 Plan: g Al → mol Fe2O3 g Al x mol Al x mol Fe2O3 = mol Fe2O3 g Al mol Al


Compute: 99.0 g Al x 1 mol Al x 1 mol Fe2O3 ≈ 1.83 mol Fe2O3 26.98 g Al 2 mol Al Evaluate: Significant figures: correct Units: correct Reasonable: There are a little less than four moles of aluminum. That halved would be less than two so the answer is reasonable. Problem Type 4: mass → mass 1. Sn(s) + 2HF(g) → SnF2(s) + H2 Analyze: Given: 30.00 g HF Unknown: g SnF2 Plan: g HF → g SnF2 g HF x mol HF x mol SnF2 x g SnF2= g SnF2 g HF mol HF mol SnF2 Compute: 30.00 g HF x 1 mol HF x 1 mol SnF2 x 156 g SnF2 ≈ 116.94 g SnF2 20.01 g HF 2 mol HF 1 mol SnF2 Evaluate: Significant figures: correct Units: correct 2. 2Na2O2 + 2H2O → 4NaOH + O2 Analyze: Given: 50.0 g Na2O2 Unknown: g O2 Plan: g Na2O2 → g O2 g Na2O2 x mol Na2O2 x mol O2 = mol SnF2 g Na2O2 mol Na2O2 Compute: 50.0 g Na2O2 x 1 mol Na2O2 x 1 mol O2 x 32 g O2 ≈ 116.94 g SnF2 77.98 g Na2O2 2 mol Na2O2 1 mol O2 Evaluate: Significant figures: correct Units: correct 3. H3PO4 + 2NH3 → (NH4)2HPO4 Analyze: Given: 2800000 g H3PO4 Unknown: g NH3 Plan: g H3PO4 → g NH3 g H3PO4 x mol H3PO4 x mol NH3 x g NH3 = mol NH3 g H3PO4 mol H3PO4 mol NH3


Compute: 2800000 g H3PO4 x 1 mol H3PO4 x 2 mol NH3 x 17.04 g NH3 ≈ 970000 g NH3 98 g H3PO4 1 mol H3PO4 1 mol NH3 Evaluate: Significant figures: correct Units: correct 4. 2NH4NO3 → 2N2 + O2 + 4H2O Analyze: Given: 36.0 g NH4NO3 Unknown: g N2 Plan: g NH4NO3 → g N2 g NH4NO3 x mol NH4NO3 g NH4NO3

x mol N2 x g N2 = g N2 mol NH4NO3 mol N2

Compute: 36.0 g NH4NO3 x 1 mol NH4NO3 x 2 mol N2 x 28.02 g N2 ≈ 12.6 g N2 80.06 g NH4NO3 2 mol NH4NO3 1 mol N2 Evaluate: Significant figures: correct Units: correct 5. 2H2O(l) + O2(g) + 2SO2(g) → 2H2SO4(aq) Analyze: Given: 50.0 g SO2 Unknown: g H2SO4 Plan: g SO2 → g H2SO4 g SO2 x mol SO2 x mol H2SO4 x g H2SO4= g H2SO4 g SO2 mol SO2 mol H2SO4 Compute: 50.0 g SO2 x 1 mol SO2 x 2 mol H2SO4 x 98.09 g H2SO4 ≈ 76.5 g H2SO4 64.07 g SO2 2 mol SO2 1 mol H2SO4 Evaluate: Significant figures: correct Units: correct


Boyle's Law 1. Analyze Given: P1= 700. torr V1= 200. mL P2= 350. torr Unknown: V2 Plan V1 = V2 P1 P2 Compute 200 mL = V2 700. torr 350 torr V2 = 100. mL Evaluate significant figures: correct units: correct reasonable: the pressure is doubled, so the volume is halved. 2. Analyze Given: P1= 0.48 atm V1= 435 mL P2= 0.75 atm Unknown: V2 Plan V1 = V2 P1 P2 Compute 435 mL = V2 0.48 torr 0.75 torr V2 = 278 mL Evaluate significant figures: correct units: correct reasonable: the pressure is almost doubled, so the volume is almost halved. 3. Analyze Given: P2= 180 mmHg V1= 2.4 x 105 mL V2= 1.8 x 103 mL


Unknown: P1 Plan V1 = V2 P1 P2 Compute 2.4 x 105 mL = P1

1.8 x 103 mL 180 mmHg

P1 = 1.35 mmHg Evaluate significant figures: correct units: correct reasonable: the volume is decreased very greatly, so the pressure is increased very greatly 4. Analyze Given: P1= 0.428 atm V1= 240. mL P2= 0.724 atm Unknown: V2 Plan V1 = V2 P1 P2 Compute 240. mL = V2 0.724 atm 0.428 atm V2 = 405 mL Evaluate significant figures: correct units: correct reasonable: the pressure is almost halved, so the volume is almost doubled. 5. Analyze Given: P1= 0.960 atm V1= 200.0 mL V2= 50.0 mL Unknown: P2 Plan V1 = V2 P1 P2 Compute


200.0 mL = 0.960 atm

50.0 mL P2

V2 = 3.84 atm Evaluate significant figures: correct units: correct reasonable: the volume is divide by four so the pressure must be multiplied by four.


1. Charles's Law 2. Gay-Lussac's Law 3. Elasticity of Gases 4. Crystal Systems Handout 5.Constructing Crystal Systems 6. The Changing of Equilibrium 7. Analysis of Simple Mixtures 8. Phase Diagram of H2O 9. Distinguishing Acids and Bases 10. Identifying Acids and Bases 11.Organic Compounds 12. Reflection Essay


Charles' Law 1. Analyze: Given: V1 = 2.75 mL V2 = 2.46 mL T1= 20º C = 293º K Unknown: T2 Plan: V1 = V2 T1 T2 Compute: 2.75 mL = 2.46 mL 293 K T2 T2 = 262º K = -11º C Evaluate: significant figures correct units correct reasonable because as volume goes down, the temperature also goes down. 2. Analyze: Given: V1 = 4.22 mL V2 = 3.87 mL T1= 65º C = 338º K Unknown: T2 Plan: V1 = V2 T1 T2 Compute: 4.22 mL = 3.87 mL 338º K T2 T2 = 310º K = 37º C Evaluate: significant figures correct units correct reasonable because as volume goes down, the temperature also goes down. 3. Analyze: Given: V1 = 80.0 mL T1 = 27º C = 300º K T2= 77º C = 350º K Unknown: T2 Plan: V1 = V2 T1 T2 Compute: 80.0 mL = V2 300º K 350º K


V2 = 93.3 mL Evaluate: significant figures correct units correct reasonable because as temperature goes up, the volume also goes up. 4. Analyze: Given: V1 = 125 mL V2 = 85.0 mL T2= 127º C = 400º K Unknown: T1 Plan: V1 = V2 T1 T2 Compute: 125 mL = T1

85.0 mL 400º K

T1 = 588º K = 315º C Evaluate: significant figures correct units correct reasonable because as volume goes down, the temperature also goes down. 5. Analyze: Given: T1 = -33º C = 240º K V2 = 54.0 mL T2= 160º C = 433º K Unknown: V1 Plan: V1 = V2 T1 T2 Compute: V1 = 54.0 mL 240º K 433º K V1 = 29.9 mL Evaluate: significant figures correct units correct reasonable because as temperature goes up, the volume also goes up.


Gay-Lussac's Law 1. Analyze: Given: P1 = 3.00 atm T1 = 25º C = 298º K T2= 52º C = 325º K Unknown: P2 Plan: P1 = P2 T1 T2 Compute: 3.00 atm = P2 298º K 325º K P2 = 3.27 atm Evaluate: significant figures correct units correct reasonable because as temperature goes up, the pressure also goes up. 2. Analyze: Given: P1 = 1.8 atm T1 = 20º C = 293º K P2= 1.9 atm Unknown: T2 Plan: P1 = P2 T1 T2 Compute: 1.8 atm = 293º K

1.9 atm T2

T2 = 309oK = 36oC Evaluate: significant figures correct units correct reasonable because as pressure goes up, the temperature also goes up. 3. Analyze: Given: P1 = 1.07 atm T1 = 120º C = 393º K T2= 205o C = 478oK Unknown: P2 Plan: P1 = P2 T1 T2 Compute: 1.07 atm = P2 393º K 478oK P2 = 1.30 atm


Evaluate: significant figures correct units correct reasonable because as the temperature goes up, the pressure also goes up. 4. Analyze: Given: P1 = 1.20 atm T1 = 22º C = 295º K P2= 2.00 atm Unknown: T2 Plan: P1 = P2 T1 T2 Compute: 1.20 atm = 295º K

2.00 atm T2

T2 = 491o K = 218oC Evaluate: significant figures correct units correct reasonable because as pressure goes up, the temperature also goes up. 5. Analyze: Given: P1 = 140.0 atm T1 = 67º C = 340º K P2= 50.0 atm Unknown: T2 Plan: P1 = P2 T1 T2 Compute: 140.0 atm = 340º K

50.0 atm T2

T2 = 121o K = -152oC Evaluate: significant figures correct units correct reasonable because as pressure goes up, the temperature also goes up.


1. In iron, iron oxide can be removed by heating the metal to 1250 oC which makes the oxygen break its bonds with the iron. Gold can be removed from ores by dissolving it in potassium cyanide or sodium cyanide. Then carbon is added to the mixture to remove the gold and is removed by heating the solution. 3. The mixture could be heated due to different melting and evaporation temperatures, such as in salt water. The mixture could also undergo chemical separation or more of the substance could be added to bring some out of solution. Conclusion: The mixtures involved in the experiment were all simple mixtures which could be separated. It was interesting using the magnet to remove the iron filings from mixture B. It was also interesting how mixture C could be separated by both floatation and solubility. The experiment helped in learning how to see the parts and predict how they could be separated.


The Dissolution of Salt I. Purpose: to determine the rate of dissolution of sodium chloride in 60mL of water II. Materials: salt, sea salt or coarse-grained salt, water, beaker, hot plate III. Procedure Place 1g of salt in 60mL of water at room temperature. Estimate the amount of salt dissolved at three minutes maximum or if it is all dissolved. Add 1g of sea salt to 60mL of water and estimate the amount of salt dissolved at three minutes maximum or if it is all dissolved. Add 1g of regular salt and stir the water for a maximum of three minutes estimate the amount dissolved. Heat 60mL of water to fifty degrees on the hot plate and add 1g of salt. Estimate the amount of salt dissolved. Heat the water to eighty degrees and repeat. IV. Data Variable

Control

Coarse Salt

Agitation

50째

80째

Time Elapsed

3 min.

3 min.

3 min.

3 min.

3 min.

Percent Dissolved

40.00%

20.00%

99.00%

60.00%

80.00%

V. Analyze and Conclude 1. What were the manipulated and responding variables? The manipulated variables are the surface area of the solute, the stirring of the solvent, and the temperature. The responding variable was the amount of solute that dissolved. 2. What variables were controlled for? The variables controlled for were surface area of the solute, stirring of the solvent, and temperature. 3. What did you learn about how different factors affect the rate of dissolution? Stirring has the most dramatic effect. Ninety-nine percent of the solvent dissolved in the water. Raising the temperature also had a large effect. When the temperature was raised to eighty degrees, the amount of solvent that dissolved was doubled. Reducing the surface area halved the amount of solute that was dissolved.


Identifying Acids and Bases 1. NH4+ (aq) + CN- (aq) ↔ HCN (aq) + NH3 (aq) Acid NH4+ Conjugate Base NH3 Base CN- Conjugate Acid HCN Arrhenius Acids: none Bases: none

Bronstead-Lowry Acids: NH4+, HCN Bases: CN-, NH3

Lewis Acids: NH4+, HCN Bases: CN-, NH3

2. (CH3)3N (aq) + H2O (l) ↔ (CH3)3NH+ (aq) + OH- (aq) Acid H2O Conjugate Base OH- Base (CH3)3N Conjugate Acid (CH3)3NH+ Arrhenius Acids: none Bronstead-Lowry Acids: H2O, (CH3)3NH+ Lewis Acids: H2O, (CH3)3NH+ Bases: (CH3)3N Bases: (CH3)3N, OHBases: (CH3)3N, OH3. HNO3 (aq) + H2O (l) ↔ H3O+ (aq) + NO3- (aq) Acid HNO3 Conjugate Base NO3- Base H2O Conjugate Acid H3O+ Arrhenius Acids: HNO3 Bases: none

Bronstead-Lowry Acids: HNO3, H3O+ Lewis Acids: HNO3, H3O+ Bases: H2O, OHBases: H2O, OH-

4. CH3COOH (aq) + H2O (l) ↔ H3O+ (aq) + CH3COO- (aq) Acid CH3COOH Conjugate Base CH3COO- Base H2O

Conjugate Acid H3O+

Arrhenius Acids: CH3COOH Bronstead-Lowry Acids:CH3COOH, H3O+ Lewis Acids:CH3COOH, H3O+ Bases: none Bases: H2O, H3O+ Bases: H2O, H3O+ 5. CN- (aq) + H2O (l) ↔ HCN (aq) + OH- (aq) Acid H2O Conjugate Base OH- Base CN- Conjugate Acid HCN Arrhenius Acids: none Bases: CN-

Bronstead-Lowry Acids:H2O, HCN Bases: OH-, CN-

Lewis Acids: H2O, HCN Bases: OH-, CN-

6. NH3 (aq) + H2O (l) ↔ NH4+ (aq) + OH- (aq) Acid H2O

Conjugate Base OH-

Arrhenius Acids: none Bases: NH3

Base NH3

Conjugate Acid NH4+

Bronstead-Lowry Acids:H2O, NH4+ Bases: OH-, NH3

Lewis Acids: H2O, NH4+ Bases: OH-, NH3

7. HBrO (aq) + H2O (l) ↔ H3O+ (aq) + BrO- (aq) Acid HBrO Conjugate Base BrO- Base H2O Conjugate Acid H3O+


Arrhenius Acids: HBrO Bases: none

Bronstead-Lowry Acids:HBrO, H3O+ Bases: H2O, BrO-

Lewis Acids: HBrO, H3O+ Bases: H2O, BrO-

8. CO3- (aq) + H2O (l) ↔ HCO3 (aq) + OH- (aq) Acid H2O Conjugate Base OH- Base HCO3- Conjugate Acid HCO3 Arrhenius Acids: none Bases: CO3-

Bronstead-Lowry Acids:HCO3, H2O Bases: OH, HCO3-

Lewis Acids: HCO3, H2O Bases: OH, HCO3-

9. HNO3 (aq) + SO4- (l) ↔ HSO4 (aq) + NO3- (aq) Acid HNO3 Conjugate Base NO3- Base SO4- Conjugate Acid HSO4 Arrhenius Acids: none Bases: none

Bronstead-Lowry Acids:HNO3, HSO4 Bases: SO4-, NO3-

Lewis Acids: HNO3, HSO4 Bases: SO4-, NO3-

10. H2SO4 (aq) + H2O (l) ↔ H3O+ (aq) + HSO4- (aq) Acid H2SO4 Conjugate Base OH- Base HSO4- Conjugate Acid H3O+ Arrhenius Acids: H2SO4 Bases: none

Bronstead-Lowry Acids:H2SO4, H3O+ Bases: H2O, HSO4-

Lewis Acids: H2SO4, H3O+ Bases: H2O, HSO4-


Organic Compounds


Reflection Essay The recurrent theme for this year of chemistry was the construction of matter on the molecular level. We started with some of the basic building blocks and then built towers on them. This meant going from atoms to their compounds and how they reacted.

We began with the basics, the atom. We first made a periodic table that color-coded the various types, such as the metals, nonmetals, and noble gases. Then, the measurement of these atoms became important so that we could share the quantities used in the experiments. Numerous calculations were made with the new unit of the mole and Avogadro’s number, which is 6.022 x 1023. We ended first quarter with the atomic theories which led up to the modern “beehive� theory.

In second quarter, after the theories had been completed, the electron orbitals and their configuration were discussed at length. Next Mendeleev was taught and how he managed to design his own periodic table. Mendeleev even managed to predict the properties of undiscovered elements and was astoundingly accurate. We then tried to take a leaf out of his book and tried to design our own periodic table with unknown elements. Last in the second quarter, we finally expanded our view and began looking at chemical compounds. Their structure was represented by the Lewis structures and the exact angles between the molecules were measured.

In the third quarter, the knowledge of the atoms was applied and the chemical formulas of the atoms were examined. The formulas were separated out into two categories, ionic and covalent. Ionic bonding was contingent on the number of valence electrons and the number of those electrons that it gave up. Covalent bonding also depended on the number of valence electrons, but this type depended on the number of bonds it could create. For instance, carbon can create four bonds, while sulfur can


create eight bonds. Then came the fun part, how those compounds would react with each other. First chemical equations have to be balanced. Then the type of reaction had to be identified. A reaction was either a synthesis reaction, a decomposition reaction, a single- or double-replacement reaction. As we started with measurement in the first quarter so we measured the products which in a unit called stoichiometry.

We began the fourth and final quarter with the laws of the volume, pressure, and temperature of gases. After the gases came the solids and how they are arranged in the crystal structure. Then how the phase of a material might be changed was learned. We discovered that these changes in phase could be graphed by a phase diagram. Equilibrium became an important concept as did Le Châtelier’s principle. Then solutions were studied. The rate of dissolution for salt in water was measured. Since acids are generally in aqueous solution, they were also studied. First acids have to be named if they are to be reported and it becomes necessary to distinguish between acids and bases. We then ended the year on the study of organic compounds.

My favorite part of this year was the experiments we did. In particular when the magnesium caught fire and when we did the experiments with acids and bases. I was fascinated by how the water underwent electrolysis and split into its most basic components, hydrogen and oxygen. Chemistry is the root and origin of all the sciences, because without chemistry, there could be no science.


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