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) Select each block and determine the mass in grams 2) Figure out the volume of each (15.625) 3) Based upon the mass and volume, determine the density of each block (mass/volume) 4) Place each block in the beaker with water and observe if the block floats or sinks 5) Determine the substance of the blocks by looking up the densities of pure substances

IV. Data:

Block #

Mass (g)

Volume (cm3)

Density (g/cm3)

Sink/Float Substance

1 15.5 g 15.625 .992 Floats Polyprolene 2 7.1 g 15.625   .4544 Floats Pine

3 127.6 g 15.625   8.1664 Sinks Steel

4 20 g 15.625   1.28 Sinks Acrylic

5 44.4 g 15.625   2.8416 Sinks Aluminum

6 144.2 g 15.625   9.2288 Sinks Copper

7 13 g 15.625   .832 Floats PUC

8 138.4 g 15.625   8.8576 Sinks Brass

 

 

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

Block # g/cm3 Kg/m3

1 .992 992 2 .4544 454 3 8.1664 8166 4 1.28 1280 5 2.8416 2842 6 9.2288 9229 7 .832 832 8 8.8576 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 1: Tommy 3 364 g Block, small sphere, dice

2: Claire 3-5 90 g Paper clips

3: Philip 2 99 g A figure, game piece

Part B: Open container without looking

Container # of Objects Mass of Objects (w/ box) Kind of Objects 1: Tommy 3 364 g Shark figure, bone, dice

2: Claire 3 90 g Bracelet, bow, charm

3: Philip 2 99 g Rubber duck, spoon

 

 

 

Part C: Open container with looking Container # of

Objects Mass of Objects

Kind of Objects

1: Tommy 3 50 g Shark figure, bone, dice

2: Claire 6 8.25 g Flower charm, pin, 2 plastic gems, bracelet, bow

3: Philip 2 93 g 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  

#  of  atoms   Identity  of  Element  

A   65     1   6.022x10^23   Zinc  B   26.9   1   6.022x10^23   Aluminum  C   55.5   1   6.022x10^23   Iron  D   63.2   1   6.022x10^23   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 element’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 person’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 13 14 15 16 17 18 1 Q D 2 A K E J P 3 B 4 O H 5 L F G R S 6 T N 7 M C I

Part C: Actual Order of Elements

1 2 13 14 15 16 17 18 1 Q 2 A E D 3 J P 4 B O K 5 L F R S 6 T N G I H 7 M C

V. 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 element’s atomic number (it hadn’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 3LiOH + Fe(NO3)3 à LiNO3 + Fe(OH)3

7.) 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 2Al I3 + 3HgCl2 à 3HgI2 + 2AlCl3

9.) 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 x mass (g) SiC = mass (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 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 x mass (g) ZnO = mass (g) ZnO mol C mol ZnO

Compute 5.00 mol C x 2 mol ZnO x 81.38 g ZnO = 813.0 g ZnO 1 mol C 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 x mass (g) NH4NO3 = mass (g) NH4NO3 mol H2O 1 mol NH4NO3

Compute 7.35 mol H2O x 2 mol NH4NO3 x 77.36 g NH4NO3 = 294 g NH4NO3 4 mol H2O 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

Evaluate

 

Evaluate

Evaluate

 

Evaluate

 

Charles’ Law Problems

1.) v1 = 2.75 L, T1 = 20°C = 293°K, V2 = 2.46 L T1V2 = T2 V1 (293°k)(2.46 L) = 262°K 2.75 L Units ✓ Sig Figs ✓ Reasonable: Temperature decreases as volume does

2.) v1 = 4.22 L, T1 = 65°C = 338°K, V2 = 3.87 L

T1V2 = T2 V1 (338°k)(3.87 L) = 310°K 4.22 L Units ✓ Sig Figs ✓ Reasonable: Temperature decreases as volume does

3.) v1 = 5.5 L, T1 = 25°C = 298°K, T2 = 100°C = 373°K

T2V1 = V2 T1

(373°k)(5.5 L) = 6.9 L 298°K Units ✓ Sig Figs ✓ Reasonable: Volume increases as temperature does

4.) T1 = 0.0°C = 273°K, V1 = 375 mL, V2 = 500. mL T1V2 = T2 V1

(273°k)(500. mL) = 364°k 375 mL Units ✓ Sig Figs ✓ Reasonable: Temperature increases as volume does

5.) V1 = 80.0 mL, T1 = 27°C = 300°K, T2 = 77°C = 350°K T2V1 = V2 T1

(350°k)(80.0 mL) = 99.3 mL 300°k Units ✓ Sig Figs ✓ Reasonable: Volume increases as temperature does

Gay Lussac’s Law Problems

1.) P1 = 1.8 atm, T1 = 20°C = 293°K, P2 = 1.9 atm P2T1 = T2 P1

(1.9 atm)(293°k) = 309°k 1.8 atm Units ✓ Sig Figs ✓ Reasonable: Temperature increases as pressure does

2.) P1 = 1.0 atm, T1 = 20°C = 293°K, T2 = 500°C =

773°K P1T2 = P2 T1

(773°K)(1.0 atm) = 2.6 atm 293°K Units ✓ Sig Figs ✓ Reasonable: Temperature increases as pressure does

3.) P1 = 3.0 atm, T1 = 100°C = 373°K, T2 = 300°C =

573°K P1T2 = P2 T1

(573°K)(3.0 atm) = 4.6 atm 373°K Units ✓ Sig Figs ✓ Reasonable: Pressure increases as temperature does

4.) P1 = 0.329 atm, T1 = 47°C = 320°K, T2 = 77°C = 350°K P1T2 = P2 T1

(350°K)(0.329 atm) = .360 atm 320°K Units ✓ Sig Figs ✓ Reasonable: Temperature increases as pressure does

5.) T1 = 27°C = 300°K, P1 = 0.625 atm, P2 = 1.125

P2T1 = T2 P1

(1.125 atm)(300°K) = 540°K 0.625 atm 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.

I. 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)

3 min 40%  

70°C 3 min 80%  

Surface  Area   1g salt, 60 mL water sea salt 3 min 30%

1g salt, 60 mL water table salt

3 min 40%

Agitation   Without 3 min 40%

With 1 min 4 sec

100%  

V. 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 can’t say that I am now), but after this class, I think I’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 today’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!