Chemistry portfolio 2012 2013 by Elena Doman

Table of Contents

Quarter 1 -

Periodic Table Metric Conversions Worksheet Conversions Making Sense of Density Lab Significant Figures Constructing a Model Mole Concept Mole Types Mole Concept Lab Atomic Theories

Quarter 2 -

Atomic Theories p. 2 Atomic Orbitals Electron Configurations Design Your Own Periodic Table Atomic Modeling Schema Drawing Lewis Structures Molecular Geometry

Quarter 3 -

Chemical Formulas Naming Ionic Compounds Naming Chemical Compounds Naming Covalent Compounds Translating Equations into Sentences Translating Sentences into Equations Balancing Equations Synthesis Reactions Decomposition Reactions Single Replacement Reactions Double Replacement Reactions Stoichiometry Problem Types Boyle’s Law Problems

Quarter 4 - Charles’ Law Problems - Gay Lussac’s Law Problems - Elasticity of Gases - 6 Crystal Systems - Constructing Crystal Models - Changing of Equilibrium: Le Chatelier’s Principle - Analysis of Simple Mixtures - Phase Diagram of Water - Rate of Dissolution Lab - Naming Acids Worksheet - Distinguishing Characteristics of Acids and Bases - Identifying Acids and Bases - Organic Compounds

Reflection Essay

Portfolio Assignment #2:

Making Sense of Density

I. Purpose: To use density to identify eight unknown substances II. Materials: 1.) triple-beam balance; 2.) ruler; 3.) beaker with water; 4.) calculator; 5.) eight density blocks III.

Procedure: 1) 2) 3) 4) 5)

Select each block and determine the mass in grams Figure out the volume of each (15.625) Based upon the mass and volume, determine the density of each block (mass/volume) Place each block in the beaker with water and observe if the block floats or sinks Determine the substance of the blocks by looking up the densities of pure substances

IV. Data: Block # 1 2

Mass

Volume

Density

(g)

(cm3)

(g/cm3)

15.5 g 7.1 g

15.625 15.625

.992 .4544

Floats Floats

Polyprolene Pine

127.6 g

15.625

8.1664

Sinks

Steel

20 g

15.625

1.28

Sinks

Acrylic

44.4 g

15.625

2.8416

Sinks

Aluminum

144.2 g

15.625

9.2288

Sinks

Copper

13 g

15.625

.832

Floats

PUC

138.4 g

15.625

8.8576

Sinks

Brass

Sink/Float Substance

V. Conclusion 1. Convert each density measurement to kg/m3. Block #

g/cm3 1 2 3 4 5 6 7 8

Kg/m3 .992 .4544 8.1664 1.28 2.8416 9.2288 .832 8.8576

992 454 8166 1280 2842 9229 832 8858

2. Every pure substance has a specific density. You can use density to identify what a substance is by comparing the density you calculated to the known density of certain pure substances. The data above shows that we calculated the densities of blocks with unknown substances and then compared them to known densities we looked up. 3. If you were to examine a smaller piece of the same blocks, the density would not be different. There is a specific density for every pure substance. 4. It's important to dry off the blocks after dipping them in the water, because the water may soak into the block, therefore endangering the validity of the outcomes when you try to weigh the substance. 5. To find the density of olive oil, one would put the oil into a small container, weigh it, subtract the weight of the container, and then divide it by the volume of the oil. The correct density of olive oil is 800-920 kg/m3.

Portfolio Assignment

Constructing a Model

I. Purpose: To understand how scientists make inferences about atoms II. Materials: 1.) Closed container; 2.) various objects; 3.) balance III.

Procedure: 1) Pick another student’s closed container and observe it (e.g. try to guess how many objects are inside and what kind) 2) Weigh the box on the balance to determine the weight altogether 3) Now open the container and feel inside to produce a better guess at what is inside. 4) Then look inside to see if your guessed were correct about the contents 5) Weigh each object in grams

IV. Data: Part A: Closed Container Container # of Objects Mass of Objects (w/ box) Kind of Objects Block, small sphere, dice 1: Tommy 3 364 g Paper clips 2: Claire 3-5 90 g A figure, game piece 3: Philip 2 99 g

Part B: Open container without looking Container # of Objects Mass of Objects (w/ box) Kind of Objects Shark figure, bone, dice 1: Tommy 3 364 g Bracelet, bow, charm 2: Claire 3 90 g Rubber duck, spoon 3: Philip 2 99 g

Part C: Open container with looking

Container 1: Tommy 2: Claire 3: Philip

# of Objects 3 6

Mass of Objects 50 g 8.25 g

93 g

Kind of Objects Shark figure, bone, dice Flower charm, pin, 2 plastic gems, bracelet, bow Rubber duck, spoon

V. Analyze and Conclude: 1.) Scientists often use more than one method to gather data. This was illustrated in this experiment by first trying to figure out the information of the objects without opening the container, then opening it, then looking at the objects inside. Three different tables of data were used in this experiment. 2.) Out of the observations that were made, the number of objects and the mass of the objects were quantitative and the kind of objects were qualitative.

3.)

Tommy used a packing box as his container and placed a shark figure, a bone, and a dice. By shaking the box, I was able to presume that there were three objects and that it weighed 364 g all together. By reaching inside in the next method, I was able to confirm that there were three objects inside the box and ended up guessing what the objects were inside. The last method revealed that I was correct in my assumptions about the objects inside, which all weighed out to be 50 g.

Lab #3

The Mole Concept I. Purpose: To identify each sample based on the observation and numerical data; II. Materials: 1) mole sample set; 2) balance III. Procedure: 1) Observe each sample by its qualitative properties (i.e. silver, metallic…) 2) Weigh each sample on the balance and record the masses 3) Find out how many moles are in each sample 4) Record the number of atoms in each mole (Avogadro’s #) 5) Identify each sample by comparing your masses to the molar masses for each element on the Periodic Table IV. Data: A. Observations Sample Observations A Silver, metallic, heavy B Light silver, metallic, lighter C Darker silver, metallic. heavier D Bronze, metallic, heavier

B. Numerical Data Sample Mass (g) # of moles A 65 1 B 26.9 1 C 55.5 1 D 63.2 1

# of atoms 6.022x10^23 6.022x10^23 6.022x10^23 6.022x10^23

Identity of Element Zinc Aluminum Iron Copper

V. Analyze and Conclude 1) The mass is what led me to the determination of each element. 2) The masses were different even though each element was exactly one mole because different atoms of different elements weigh a different amount, which led us to determine the elements by their masses. 3) Determine the actual mass of a single atom of each sample: Sample Mass of Atom A 1.1 x 10^ -‐22 g B 4.47 x 10^ -‐23 g C 9.22 x 10^ -‐ 23 g D 1.05 x 10^ -‐22 g

4) In this lab, we were able to figure out the identities of each sample of moles by finding out their masses. To begin with, we observed their physical properties (i.e. how heavy, what color …). After doing so, we weighed them on the balance to find out their masses. Once weighed, we were able to figure out what element they were composed of by looking at the Periodic Table of Elements. By observing their physical properties and figuring out their masses, we were able to determine what element all the samples were. By completing this experiment, I learned that there is an easy way to determine the element of a mole sample. It also led me to understand more about what a mole is. Not only were we able to find out what each sample was, but was also able to find out how many atoms were in each mole sample by dividing the mass by Avogadro’s Number. By matching up the masses we weighed to the correct masses for each element of the Periodic Table, we defined each sample with ease.

Chemistry Lab:

Designing Your Own Periodic Table I.

Purpose: To design your won periodic table using information similar to that available to Mendeleev

II.

Materials: 1.) Periodic Table; 2.) index cards

III.

Procedure: 1.) Pick 20 elements from a periodic table and 20 index cards; 2.) On each index card, code each element as a letter and then write the elementâ&#x20AC;&#x2122;s atomic mass, melting point, boiling point, density, and its chemical properties on the card 3.) When finished, exchange cards with another person and try to arrange their elements in the right places by looking at their chemical properties and atomic mass like Mendeleev 4.) Look at the personâ&#x20AC;&#x2122;s key to see if you were correct

IV. Data: Part A: Index Cards and Key (At the Bottom) Part B: Initial attempt at arranging elements 1 2 3 4 5 6 7

1 Q A

16 D P

B L M

F T C

O G

H S

N I

Part C: Actual Order of Elements 1 2 3 4 5 6 7 V.

1 Q A

E B L

D J O

F T

R I

P K S H

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.) Mendeleev did not have information about each elementâ&#x20AC;&#x2122;s atomic number (it hadnâ&#x20AC;&#x2122;t been known in that time period yet) and it is much easier to find and sort elements by increasing atomic mass. b.) Some of them, yes. 2. There are 8 groups in my periodic table. 3. I can predict missing elements by looking at where the missing spots lie. For instance, in group one, it is most likely that the missing elements will have the same properties as the other elements in that group.

Synthesis Reactions 1.) Calcium + oxygen à Calcium Oxide 2Ca + O2 à 2CaO

2.) Iron (III) + sulfur à Iron (III) Sulfide 16Fe + 3S8 à 8Fe2S3

3.) Hydrogen + oxygen à Hydrogen Oxide 2H2 + O2 à 2H2O

4.) Aluminum + Bromine à Aluminum Bromide 2Al + 3Br2 à 2AlBr3

5.) Sodium + Iodine à Sodium Iodide 2Na + I2 à 2NaI

6.) Calcium Oxide + water à Calcium Hydroxide CaO + H2O à Ca(OH)2

7.) Chromium (III) + Oxygen à Chromium (III) Oxide 4Cr + 3O2 à 2Cr2O3

8.) Silver + sulfur à Silver Sulfide 16Ag + S8 à 8Ag2S

9.) Magnesium + oxygen à Magnesium Oxide 2Mg + O2 à 2MgO

10.) Sodium + Oxygen à Sodium Oxide 4Na + O2 à 2Na2O

Decomposition Reactions 1.) Iron (III) Oxide à Iron (III) + oxygen 2Fe2O3 à 4Fe + 3O2

2.) Barium Hydroxide à Barium Oxide + water Ba(OH)2 à BaO + H2O

3.) Magnesium Chlorate à Magnesium Chloride + oxygen Mg3(ClO)2 à MgCl2 + 3O2

4.) Potassium Carbonate à Potassium Oxide + Carbon Dioxide K2(CO3) à K2O + CO2

5.) Magnesium Hydroxide à Magnesium Oxide + water Mg(OH)2 à MgO + H2O

6.) Silver Chloride à Silver + Chlorine 2AgCl à 2Ag + Cl2

7.) Strontium Chlorate à Strontium Chloride + oxygen Sr(ClO3)2 à SrCl2 + O2

8.) Magnesium Carbonate à Magnesium oxide + carbon dioxide MgCO3 à MgO + CO2

9.) Aluminum Chlorate à Aluminum Chloride + oxygen 2Al(ClO3)3 à 2AlCl3 + 9O2

10.) Beryllium Hydroxide à Beryllium oxide + water Be(OH)2 à BeO + H2O

Single Replacement Reactions 1.) Silver Nitrate + Nickel à Nickel (II) Nitrate + silver 2AgNO3 + Ni à Ni (NO3)2 + 2Ag

2.) Aluminum Bromide + Chlorine à Aluminum Chloride + Bromine 2AlBr3 + 3Cl2 à 2AlCl2 + 3Br2

3.) Sodium Iodide + Bromine à Sodium Bromide + Iodine 2Na I + Br2 à 2NaBr + I2

4.) Calcium + Hydrochloric acid à Calcium Chloride + Hydrogen Ca + 2HCl à CaCl2 + H2

5.) Magnesium + Nitric Acid à Magnesium Nitrate + Hydrogen Mg + 2HNO3 à Mg(NO3)2 + H2

6.) Potassium + water à Potassium Hydroxide + Hydrogen 2K + 2H2O à 2KOH + H2

7.) Zinc + Magnesium Chloride à No Reaction Zn + MgCl2 à No Reaction

8.) Aluminum + water à Aluminum Oxide + Hydrogen 2Al + 3H2O à Al2O3 + 3H2

9.) Potassium Iodide + Bromine à Potassium Bromide + Iodine 2K I + Br2 à 2KBr + I2

10.) Magnesium + Cobalt (II) Nitrate à Magnesium Nitrate + Cobalt (III) Mg + Co(NO3)2 à Mg(NO3)2 + Co

Double Replacement Reactions 1.)

Silver Nitrate + Hydrochloric acid à Silver Chloride + Nitric acid AgNO3 + HCl à HNO3 + AgCl

2.) Copper (II) Chloride + Sodium Sulfide à Copper (II) Sulfide + Sodium Chloride CuCl2 + Na2S à CuS + 2NaCl

3.) Iron (II) Sulfide + Hydrochloric acid à Iron (II) Chloride + Hydrogen Sulfide FeS + 2HCl à FeCl2 + H2S

4.) Sulfuric acid + Potassium Hydroxide à Hydrogen Hydroxide + Hydrogen Sulfide H2SO4 + 2KOH à 2H(OH) + K2SO4

5.) Nitric acid + Calcium Hydroxide à Calcium Nitrate + water 2HNO3 + Ca(OH)2 à Ca(NO3)2 + 2H2O

6.) Lithium Hydroxide + Iron (III) Nitrate à Lithium Nitrate + Iron (III) Hydroxide 7.)

3LiOH + Fe(NO3)3 à LiNO3 + Fe(OH)3

Lead (II) Acetate + Hydrogen Sulfide à Lead (II) Sulfide + Hydrogen Acetate Pb(CH3COO)2 + H2S à PbS + 2HCH3COO

8.) Aluminum Iodide + Mercury (II) Chloride à Mercury (II) + Aluminum Chloride Iodide 9.)

2Al I3 + 3HgCl2 à 3HgI2 + 2AlCl3

Calcium Acetate + Sodium Carbonate à Calcium Carbonate + Sodium Acetate Ca(CH3OO)2 +NaCO3 à Ca(CO3)2 + 2NaCH3COO

10.) Ammonium Chloride + Mercury (I) Acetate à Ammonium Acetate + Mercury Chloride NH4Cl2 + HgCH3COO à NH4CH3COO + HgCl

Stoichiometry Problem Types Problem Type #1: Mol à Mol 1.) 2Na + 2H2O  2NaOH + H2 Analyze Given: 4 mol H2 Unknown: mol Na Plan Mol H2 x mol Na = mol Na mol H2 Compute 4.0 mol H2 x 2 mol Na = 8.0 mol Na 1 mol H2 Evaluate Units √ Sig Figs √ Reasonable: Because answer is 2x the given √ 2.) 2LiBr + Cl2  2LiCl + Br2 Analyze Given: 0.046 mol LiBr Unknown: mol LiCl Plan Mol LiBr x mol LiCl = mol LiCl mol LiBr Compute 0.046 mol LiBr x 2 mol LiCl = 0.046 mol LiCl 2 mol LiBr

Evaluate Units √ Sig Figs √ Reasonable: Because answer is multiplied by 1 which gives us the same number as given √ 3.) 4Li + O2  2Li2O Analyze Given: 2 mol Li Unknown: mol Li2O Plan Mol Li x mol Li2O = mol Li2O mol Li Compute 2 mol Li x 2 mol Li2O = 1 mol Li2O 4 mol Li Evaluate Units √ Sig Figs √ Reasonable: Because answer is one half the given √ 4.) 2H2O2  2H2O + O2 Analyze Given: 5.0 mol H2O2 Unknown: mol O2 Plan Mol H2O2 x mol O2 = mol O2 mol H2O2 Compute 5.0 mol H2O2 x 1 mol O2 = 2.5 mol O2 2 mol H2O2

Evaluate Units √ Sig Figs √ Reasonable: Because answer is one half the given √ 5.) 2NH3 + H2SO4  (NH4)2SO4 Analyze Given: 30.0 mol NH3 Unknown: mol (NH4)2SO4 Plan Mol NH3 x mol (NH4)2SO4 = mol (NH4)2SO4 mol NH3 Compute 30.0 mol NH3 x 1 mol (NH4)2SO4 = 15.0 mol (NH4)2SO4 2 mol NH3 Evaluate Units √ Sig Figs √ Reasonable: Because answer is one half the given √ Problem Type #2: Mol à Mass 1.) 2NaN3  3N2 + 2Na Analyze Given: .500 mol NaN3 Unknown: mass (g) N2 Plan Mol NaN3 x mol N2 x mass (g) N2 = mass (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 Units √ Sig Figs √ Reasonable: √ 2.) SiO2 + 3C  SiC + 2CO Analyze Given: 2.00 mol C Unknown: mass (g) SiC Plan Mol C x mol SiC mol C

x mass (g) SiC = mass (g) SiC mol SiC

Compute 2.00 mol C x 1 mol SiC 3 mol C

x 40.1 g SiC = 26.7 g SiC 1 mol SiC

Evaluate Units √ Sig Figs √ Reasonable: √ 3.) 2ZnO + C  2Zn + CO2 Analyze Given: 5.00 mol C Unknown: mass (g) ZnO Plan Mol C x mol ZnO mol C

x mass (g) ZnO = mass (g) ZnO mol ZnO

Compute 5.00 mol C x 2 mol ZnO 1 mol C

x 81.38 g ZnO = 813.0 g ZnO 1 mol ZnO

Evaluate Units √ Sig Figs √ Reasonable: Because answer is about 10x the mass of ZnO √ 4.) Fe2O3 + 2Al  2Fe + Al2O3 Analyze Given: 0.905 mol Al2O3 Unknown: mass (g) Fe Plan Mol Al2O3 x mol Fe x mass (g) Fe = mass (g) Fe mol Al2O3 1 mol Fe Compute 0.905 mol Al2O3 x 2 mol Fe x 55.8 g Fe = 101 g Fe 1 mol Al2O3 1 mol Fe Evaluate Units √ Sig Figs √ Reasonable: √ 5.) 2N2 + O2 + 4H2O  2NH4NO3 Analyze Given: 7.35 mol H2O Unknown: mass (g) NH4NO3 Plan Mol H2O x mol NH4NO3 mol H2O

x mass (g) NH4NO3 = mass (g) NH4NO3 1 mol NH4NO3

Compute 7.35 mol H2O x 2 mol NH4NO3 4 mol H2O

x 77.36 g NH4NO3 = 294 g NH4NO3 1 mol NH4NO3

Evaluate Units √ Sig Figs √ Reasonable: √ Problem Type #3: Mass à Mol 1.) 4NH3 + 5O2  4NO + 6H2O Analyze Given: 824 NH3 Unknown: mol NO Plan 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 Units √ Sig Figs √ Reasonable: As mass NH3 increases the mols of NO should also increase√ 2.) 2NaCl + 2H2O  2NaOH + Cl2 + H2 Analyze Given: 250 NaCl Unknown: mol Cl2

Plan 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 Units √ Sig Figs √ Reasonable: √ 3.) Fe2O3 + 3CO  2Fe + 3CO2 Analyze Given: 4.00 kg Fe2O3 Unknown: mol CO Plan kg Fe2O3 x 1000 g x mol Fe2O3 x mol CO = mol CO 1 kg g Fe2O3 mol Fe2O3 Compute 4.00 kg Fe2O3 x 1000 g x 1 mol Fe2O3 x 3 mol CO = 75.1 mol CO 1 kg 159.7 g Fe2O3 1 mol Fe2O3 Evaluate Units √ Sig Figs √ Reasonable: √ 4.) H2SO3 + 2NaHCO3  2CO2 + Na2SO4 + 2H2O Analyze Given: 150.0 g H2SO4 Unknown: mol NaHCO3

Plan 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 48.08 g H2SO4 1 mol H2SO4 Evaluate Units √ Sig Figs √ Reasonable: √ 5.) Fe2O3 + 2Al  2Fe + Al2O3 Analyze Given: 99.0 g Al Unknown: mol Fe2O3 Plan 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 Units √ Sig Figs √ Reasonable: √ Problem Type #4: Mass à Mass 1.) Sn + 2 HF  SnF2 + H2

Analyze Given: 30.00 g HF Unknown: g SnF2 Plan 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 156.71 g SnF2 = 117.5 g SnF2 20.01 g HF 2 mol HF 1 mol SnF2 Evaluate Units √ Sig Figs √ Reasonable: √ 2.) 2Na2O2 + 2H2O  4NaOH + O2 Analyze Given: 50.0 g Na2O2 Unknown: g O2 Plan g Na2O2 x mol Na2O2 x mol O2 x g O2 = g O2 g Na2O2 mol Na2O2 mol O2 Compute 50.0 g Na2O2 x 1 mol Na2O2 x 1 mol O2 x 31.98 g O2 = 10.3 g O2 77.96 g Na2O2 2 mol Na2O2 1 mol O2 Evaluate Units √ Sig Figs √ Reasonable: √

3.) N2 + 3H2  2NH3 Analyze Given: 1.40 g N2 Unknown: g H2 Plan g N2 x mol N2 x mol H2 x g H2 = g H2 g N2 mol N2 mol H2 Compute 1.40 g N2 x 1 mol N2 x 3 mol H2 x 2.02 g H2 = .303 g H2 28.02 g N2 1 mol N2 1 mol H2 Evaluate Units √ Sig Figs √ Reasonable: √ 4.) C3H7COOH + CH3OH  C3H7COOCH3 + H2O Analyze Given: 52.5 g C3H7COOH Unknown: g C3H7COOCH3 Plan g C3H7COOH x mol C3H7COOH x mol C3H7COOCH3 x g C3H7COOCH3 = g C3H7COOCH3 g C3H7COOH mol C3H7COOH mol C3H7COOCH3 Compute 52.5 g C3H7COOH x 1 mol C3H7COOH x 1 mol C3H7COOCH3 x 103. 14 g C3H7COOCH3 = 88.1 g C3H7COOH 1 mol C3H7COOH 1 mol C3H7COOCH3 60.9 g C3H7COOCH3 Evaluate Units √ Sig Figs √ Reasonable: √

5.) LiOH + HBr  LiBr + H2O Analyze Given: 10.0 g LiOH Unknown: g LiBr Plan g LiOH x mol LiOH x mol LiBr x g LiBr = g LiBr g LiOH mol LiOH mol LiBr Compute 10.0 g LiOH x 1 mol LiOH x 1 mol LiBr x 86.84 g LiBr = 36.3 g LiBr 23.94 g LiOH 1 mol LiOH 1 mol LiBr Evaluate Units √ Sig Figs √ Reasonable: √

Evaluate

Charles’ Law Problems 1.) v1 = 2.75 L, T1 = 20°C = 293°K, V 2 = 2.46 L T1V2 = T2 V1 (293° k)(2.46 L) 2.75 L

= 262° K

Units ✓ Sig Figs ✓ Reasonable: Temperature decreases as volume does 2.) v1 = 4.22 L, T1 = 65°C = 338°K, V 2 = 3.87 L T1V2 = T2 V1 (338° k)(3.87 L) 4.22 L

= 310° K

Units ✓ Sig Figs ✓ Reasonable: Temperature decreases as volume does 3.) v1 = 5.5 L, T1 = 25°C = 298°K, T 2 = 100°C = 373°K T2V1 = V2 T1 (373° k)(5.5 L) 298°K

= 6.9 L

Units ✓ Sig Figs ✓ Reasonable: Volume increases as temperature does

4.) T1 = 0.0°C = 273°K , V1 = 375 mL, V 2 = 500. mL

T1V2 = T2 V1 (273° k)(500. mL) 375 mL

= 364° k

Units ✓ Sig Figs ✓ Reasonable: Temperature increases as volume does 5.) V1 = 80.0 mL, T1 = 27°C = 300°K, T 2 = 77°C = 350°K T2V1 = V2 T1 (350° k)(80.0 mL) 300° k

= 99.3 mL

Units ✓ Sig Figs ✓ Reasonable: Volume increases as temperature does

Gay Lussac’s Law Problems 1.) P1 = 1.8 atm, T1 = 20°C = 293°K, P 2 = 1.9 atm P2T1 = T2 P1 (1.9 atm)(293° k) 1.8 atm

= 309° k

Units ✓ Sig Figs ✓ Reasonable: Temperature increases as pressure does 2.) P1 = 1.0 atm, T1 = 20°C = 293°K, T 2 = 500°C = 773°K

P1T2 = P2 T1 (773°K )(1.0 atm) 293°K

= 2.6 atm

Units ✓ Sig Figs ✓ Reasonable: Temperature increases as pressure does 3.) P1 = 3.0 atm, T1 = 100°C = 373°K, T 2 = 300°C = 573°K

P1T2 = P2 T1 (573°K )(3.0 atm) 373°K

= 4.6 atm

Units ✓ Sig Figs ✓ Reasonable: Pressure increases as temperature does

4.) P1 = 0.329 atm, T1 = 47°C = 320°K, T 2 = 77°C = 350°K

P1T2 = P2 T1 (350°K )(0.329 atm) 320°K

= .360 atm

Units ✓ Sig Figs ✓ Reasonable: Temperature increases as pressure does 5.) T1 = 27°C = 300°K , P1 = 0.625 atm, P 2 = 1.125 P2T1 = T2 P1 (1.125 atm)(300°K ) 0.625 atm

= 540°K

Units ✓ Sig Figs ✓ Reasonable: Temperature increases as pressure does

Q5.

1.) Some separation techniques used in manufacturing processes include comminution (which reduces size of particles), floatation process (separating minerals through floatation), and solvent extraction (dissolving one substance in another). A more specific one would be Parkes process, in which you can remove silver from lead by adding liquefied zinc. By adding the liquefied zinc, you are able separate the layers because of the immiscibility of zinc in the lead (thus, easily removed). Then, the zinc-‐silver solution is heated until the zinc vaporizes and leaves behind pure silver. If gold is present, you use the same process to separate the two metals. 2.) Milk looks like a single substance, but is actually a homogeneous mixture. You can separate it by breaking down the fat (cream) molecules in the milk and pressurizing it out through special homogenization machines, thus making the milk less fatty. Alcohol can be separated from water by heating.

A.) Describe all the phase changes a sample of solid water would undergo when heated to its critical temperature at a pressure of 1.00 atm. Solid à melt into a liquid à boil à evaporate into gas B.) Describe all the phase changes a sample of water vapor would undergo when cooled to 5°C at a pressure of 1.00 atm. It will condense into a liquid and continue as liquid cools. C.) At approximately what pressure will water be a vapor at 0°C? Approximately 0.006 atm. D.) Within what range of pressures will water be a liquid at temperatures above its normal boiling point? 1.00 – 217.75 atm.

Title: Rate of Dissolution Lab

II.

Purpose: to observe the rate of dissolution of salt in water in different procedures/factors.

III.

Materials: sea salt, table salt, beakers, water, balance, thermometer, glass mixing rod, hot plate

IV.

Procedure: 1.) Effect of Temperature: Pour a fixed amount of water into a beaker, pour a fixed amount of table salt, 2.) Effect of Surface Area: 3.) Effect of Agitation

V. Manipulated Variable Dissolving Time Temperature 20°C (room temp) 70°C 3 min 3 min 40% 80% Surface Area 1g salt, 60 mL water 1g salt, 60 mL water sea salt table salt 3 min 3 min 30% 40% Agitation Without With 3 min 1 min 4 sec 40% 100%

Analyze and Conclude: 1.) What were the manipulated and responding variables in the experiment? 2.) What variables were controlled for? 3.) What did you learn about how different factors affect the rate of dissolution? I learned that the agitation variable is the most effective in getting the salt to dissolve quickly. Raising the temperature will create more kinetic energy, therefore, aiding the speed of dissolution. Decreasing the surface area of the salt will give more space for the water to dissolve it.

Elena Doman Chemistry 2013 Reflection Essay

Looking back on all the assignments and projects I have done throughout the school year in Chemistry truly amazes me. I was never a real fan for Chemistry to begin with (and honestly, I canâ&#x20AC;&#x2122;t say that I am now), but after this class, I think Iâ&#x20AC;&#x2122;ve come to learn a lot more than I thought possible. It amazes me how much I have accomplished in this class and how much I ended up understanding. Besides, acquiring new knowledge, I have come to acquire a new respect for all the chemists in the world who took classes for years to become professionals in this area. It was not easy!

The first quarter was surprisingly the easiest quarter for me. This quarter mainly consisted of studying the basics of chemistry: the periodic table, the mathematics involved with it, and the atom. Looking into the specifics of the Periodic Table, we were introduced to all its elements, the labels and groups in which those elements were categorized, and some of the characteristics of those elements. We were taught how to distinguish which elements were metals, metalloids, or nonmetals and how they were arranged accordingly into the Periodic Table. The mathematics section consisted mainly of converting between metric units, measuring density, using significant figures in our calculations, and converting between moles and mass units. In learning the mathematics, a clearer view was presented to us in terms of the atom and how it has a definite mass and can be measured even though it is the smallest unit of matter.

Knowing this made it possible to classify certain elements according to their densities. In the end of the quarter, we were presented with the different atomic theories of scientists throughout the years. Each theory brought man a step closer to understanding the true characteristics and form of the atom, all the way up to todayâ&#x20AC;&#x2122;s theory. The first quarter was the most important in the curriculum and was the first step into understanding even more about the atomic world.

The second quarter consisted mainly of looking into the core of the atom and its characteristics. We were taught about the different orbital shapes and how the electrons surrounded each atom. It was observed that each element has its own unique electron configuration as well as its own special properties. By observing the special properties of these elements, we were able to blindly sort them into their rightful groups on the Periodic Table according to their behavior and atomic masses. Apart from distinguishing characteristics of the elements, we were taught about the different bonds between elements and how the electrons of those elements share and replace one other in the orbitals. This brought us into the concept of molecules and their different geometries.

The third quarter was a branch from the second, in which we were basically brought into writing and distinguishing different chemical formulas by using the elements from the Periodic Table. It was important to note how the electrons and different charges affected writing the elements into formulas. We were able to count the number of elements in each compound as well as name the different compounds that were introduced, including ionic, covalent, and chemical compounds. Alongside this, we

learned the mechanics in translating chemical equations into sentences (and vise versa) and as well as balancing the equations. The different type of reactions (synthesis, decomposition, single replacement, etc.) were introduced and classified. Lastly, we went through the process of doing Stoichiometry, which introduced more quantitative data.

The fourth quarter was focused on observing the characteristics of solids, liquids, and gases as well as the characteristics of acids and bases. One of the characteristics which was especially observed, was the elasticity of gases and how there is an inverse relationship between volume and pressure. A subject that proceeded from the gases was the definition of equilibrium and its principles. The reflection of solids led into the realm of crystal forms and the different crystal structures. The characteristics of water were also closely observed and soon branched into the topic of mixtures and solutions. Solubility and dissolution were both defined and made clearer in an experiment with salt and water. After looking into the properties of the states of matter, we looked into the characteristics of acids and bases. It was taught to identify the acids and bases in equations with their conjugate acids and bases as well as the properties of different acids/bases. Last, but not least, we looked upon a small bit of organic chemistry and how its deals with all carbon-based life (basically all living things).

It amazes me how much knowledge I have gained in just one Chemistry class and I do believe I have gained more than just knowledge. My will to persevere and to learn have also increased and I am proud of it. I can thank Dr. Snyder for that!