Formatting Blackline Masters - Caddo Science€¦ · Web viewMuddy Water Na Solution As salt water Cl Metal NaCl nonmetal Metalloid Elements, ... Copper wire and silver nitrate

Unit 1, Activity 2, Specific Assessment Rubric

Blackline Masters, Chemistry Page 1Louisiana Comprehensive Curriculum, Revised 2008

Chemistry

Unit 1, Activity 2, Specific Assessment Rubric

3 2 0Measurements are to the correct number of significant figures

All measurements 2 or 3 measurements Less than 2 measurements

Units included All measurements 2 or 3 measurements Less than 2 measurements

Answers are within the range of acceptable error

All measurements 2 or 3 measurements Less than 2 measurements

Measurements finished within the prescribed time limit

All measurements 2 or 3 measurementsLess than 2 measurements

All safety rules followed

Questions Answered

Answered correctly

Answered incorrectly but supported by evidence

Answered incorrectly. No supporting evidence.

Blackline Masters, Chemistry Page Louisiana Comprehensive Curriculum, Revised 2008

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Unit 1, Activity 3, Accuracy and Precision Worksheet

Figure 1 Figure 2 Figure 3

1. Determine the accuracy and precision represented by each group of darts in the figures above. Explain your choices using complete sentences.

Figure 1 Figure 2 Figure 3

Precision?

Accuracy?

2. A basketball player throws 100 free-throws; 95 of these balls go through the goal; 5 miss the goal entirely. Describe the precision and accuracy of the free-throws.

3. The same player is having an off day; 5 balls go through the goal; the other 95 balls bounce off of the rim. Describe the precision and accuracy of the throws.

Blackline Masters, Chemistry Page Louisiana Comprehensive Curriculum, Revised 2008

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Unit 1, Activity 3, Accuracy and Precision Worksheet Answers

Figure 1 Figure 2 Figure 3 1. Determine the accuracy and precision represented by each group of darts in the figures above. Explain your choices using complete sentences.

Figure 1 Picture 2 Picture 3

Precision?

Good All of the darts are grouped in the same area.

Poor None of the darts are grouped in the same area.

Good All of the darts are grouped in the same area.

Accuracy?

PoorNone of the darts are grouped in the bull’s-eye.

PoorFew of the darts are grouped in the bull’s-eye.

GoodAll of the darts are grouped in the bull’s-eye.

2. A basketball player throws 100 free-throws; 95 of these balls go through the goal; 5 miss the goal entirely. Describe the precision and accuracy of the free-throws.

The player has good precision and good accuracy because so many of the balls go through the goal.

3. The same player is having an off day; 5 balls go through the goal; the other 95 balls bounce off of the rim. Describe the precision and accuracy of the throws.

The player has good precision because so many balls bounce off the rim but poor accuracy because so few balls make it through the goal.


Unit 2, Activity 1, Card Sort Template 1

Matter Homogeneous

Pure Substance Heterogeneous

Element Mixture

Compound


Unit 2, Activity 1, Card Sort Template 2

Muddy Water Na

Solution As

salt water Cl

Metal NaCl

nonmetal Metalloid


MATTER

PURESUBSTANCE

MIXTURE

HOMOGENEOUS

HETEROGENEOUSELEMENT

COMPOUND

METAL METALLOID NONMETAL

Na As Cl

MUDDY WATERSOLUTIONNaCl

SALT WATER

Can be either

Is

Chemically combine to form

Is either

Is

Is calledExample Example

Can be

ExampleExample Example Example

Unit 2, Activity 1, Sample Concept Map

Elements, Compounds and Mixtures Concept Map


Unit 2, Activity 2, Sample Word Grid

Sample: Homogeneous HeterogeneousCan be separated into individual components

The properties of the individual components are the same as properties of the sample

Salt

Water

Copper

Salt and water

Copper and water


Unit 2, Activity 2, Sample Word Grid Answers

Sample: Homogeneous HeterogeneousCan be separated into individual components

The properties of the individual components are the same as properties of the sample

Salt X X

Water X X

Copper X X

Salt and water X X X

Copper and water X X X

Conclusions:1. Salt (NaCl) is a homogeneous material that can be decomposed into individual elements

(sodium and chlorine). The properties of the salt differ from the properties of the elements. Salt is a compound.

2. Water (H2O) is a homogeneous material that can be decomposed into elements (hydrogen and oxygen). Water is a compound.

3. Copper is a homogeneous material that cannot be separated into components. Copper is an element.

4. Salt and water combine to form a homogeneous material that can be separated into parts. When the salt and water are mixed, their properties do not change. Salt water is a homogeneous mixture called a solution.

5. Copper shot and water not homogeneous because the copper and water are easily seen as individual parts. These parts can be separated easily. When the copper and water are mixed, their individual properties do not change. This is a heterogeneous mixture.


Unit 2, Activity 4, Three Worlds of Chemistry


Unit 2, Activity 5, Density

Each box has the same volume. If each

ball has the same mass, which box would

weigh more? Why?


Unit 2, Activity 5, Density


Unit 2, Activity 6, Split-Page Notes

Physical and Chemical Changes

Effervescent tablet in waterObservations

1. numerous bubbles formed as soon as tablet touched the water 2. bubbles rose to top of water and burst3. tablet disappeared4. bubbles stopped forming5. looks like nothing else is happening

Conclusion The bubbles contained a gas that escaped into the air. The

tablet was a solid that underwent a chemical change with the water to produce the gas bubbles. Once the tablet (reactant) was used up, no more gas bubbles (products) were formed, and the reaction stopped. There has been a change in the identity of the material. It is no longer an effervescent tablet. The production of a gas is evidence of a chemical change (reaction) taking place.

Cutting a piece of paper Observations

1. smaller pieces of paper are formed Conclusion

The smaller pieces of paper are exactly like the original piece of paper (reactant). There has been no change in the identity of the material. It is still paper (product).


Unit 3, Activity 3, Exploring the Periodic Table

1.01 H

4.00 He

6.94 Li

9.01 Be

10.81 B

12.01 C

14.01 N

16.00 O

19.00 F

20.18 Ne

22.99 Na

24.30 Mg

26.98 Al

28.08 Si

30.97 P

32.07 S

35.45 Cl

39.95 Ar

39.10 K

40.08 Ca

69.72 Ga

72.61 Ge

74.92 As

78.96 Se

79.90 Br

83.80 Kr

85.47 Rb

87.62 Sr

114.82 In

118.71 Sn

121.75 Sb

127.60 Te

126.90 I

131.29 Xe

132.90 Cs

137.33 Ba

204.38 Tl

207.2 Pb

208.98 Bi

(209) Po

(210) At

(222) Rn


Unit 3, Activity 4, GISTing

GISTing

The individual Gists are limited to 15 words.

Sample paragraph from notes:

Atomic radii

The atomic radius is ½ the distance between the centers of neighboring atoms. It is the size of the atom due to the size of the electron cloud.

Group trends

The atomic radii of the main group elements (s & p sublevels) generally increases down a group. The outermost electrons occupy energy levels that are farther from the nucleus.

Period trends

Atomic radius generally decreases across a period. This is caused by the increasing nuclear charge of the nucleus as you go across a period. More protons are in the nucleus and more electrons are in the same energy level. The increasing nuclear charge attracts the electrons and pulls them closer to the nucleus.

Class gist statements for each sentence of the paragraphs

1. Atomic radius means how big an atom is. _____ _____ _____ _____ _____ _____ _____

2. Atoms get bigger down a group because there are more energy levels. _____ _____ _____

3. Atoms get smaller across a period because more protons attract the electrons pulling them

closer.

Summary: Atomic radius (size of the atom) increases down a group because of more energy

levels and across a period because of a greater attraction between the larger number of protons

and the outer electrons.

After several gisting activities, you will be able to construct summaries. Gisting is a mental process and not necessarily a written one.

Blackline Masters, Chemistry

Page 14Louisiana Comprehensive Curriculum, Revised 2008

Unit 4, Activity 1, Vocabulary Self-Awareness

Term + - Definition Example

Chemical bond

Ionic bond

Covalent bond

Metallic bond

Electronegativity

Polar covalent bond

Nonpolar covalent bond

Formula unit

Molecule

Molecular formula


Unit 4, Activity 2, Ion Cards






Unit 4, Activity 3, Chemical Formulas and Nomenclature I

Write formulas for the following compounds:

1. copper (I) oxide _______________

2. aluminum hydroxide _______________

3. triphosphorus decasulfide _______________

4. zinc nitrate _______________

5. hydrobromic acid _______________

6. mercury (I) bromide _______________

7. boron tribromide _______________

8. sodium hydride _______________

9. barium perchlorate _______________

10. tetraphosphorus hexasulfide _______________

11. sulfuric acid _______________

12. calcium hypochlorite _______________

13. ammonium phosphite _______________

14. chromium (III) acetate _______________

15. hydrosulfic acid _______________

16. carbonic acid _______________

17. phosphorus pentafluoride _______________

18. cobalt (II) nitrate _______________

19. magnesium sulfate _______________

20. strontium phosphate _______________

21. dichlorine monoxide _______________

22. phosphorous acid _______________

23. disulfur dichloride _______________

24. iron (III) carbonate _______________

25. perchloric acid _______________


Unit 4, Activity 3, Chemical Formulas and Nomenclature I Answers

Write formulas for the following compounds:

1. copper (I) oxide __Cu2O_______

2. aluminum hydroxide __Al(OH)3_____

3. triphosphorus decasulfide __P3S10_______

4. zinc nitrate __Zn(NO3)2____

5. hydrobromic acid __HBr(aq)_____

6. mercury (II) bromide __HgBr2_______

7. boron tribromide __BBr3________

8. sodium hydride __NaH________

9. barium perchlorate __Ba(ClO4)2____

10. tetraphosphorus hexasulfide __P4S6________

11. sulfuric acid __H2SO4(aq)___

12. calcium hypochlorite __Ca(ClO)2____

13. ammonium phosphite __(NH4)3PO3___

14. chromium (III) acetate __Cr(C2H3O2)3_

15. hydrosulfic acid __H2S(aq)_____

16. carbonic acid __H2CO3(aq)___

17. phosphorus pentafluoride __PF5_________

18. cobalt (II) nitrate __Co(NO3)2____

19. magnesium sulfate __MgSO4______

20. strontium phosphate __Sr3(PO4)2____

21. dichlorine monoxide __Cl2O________

22. phosphorous acid __H3PO4(aq)___

23. disulfur dichloride __S2Cl2_______

24. iron (III) carbonate __Fe2(CO3)3___

25. perchloric acid __HClO4(aq)___


Unit 4, Activity 3, Chemical Formulas and Nomenclature II

Name the following compounds.

1. K2SO4 ______________________________

2. N2O4 ______________________________

3. BaClO4 ______________________________

4. HNO2(aq) ______________________________

5. FE2(SO4)3 ______________________________

6. NH4F ______________________________

7. BaI2 ______________________________

8. CrO3 ______________________________

9. Cu(C2H3O2)2 ______________________________

10. Ag2CO3 ______________________________

11. NaOH ______________________________

12. Ca3(PO4)2 ______________________________

13. ClF3 ______________________________

14. K2SO3 ______________________________

15. AlBr3 ______________________________

16. MgCl2 ______________________________

17. HC2H3O2(aq) ______________________________

18. P2O5 ______________________________

19. FePO4 ______________________________

20. SrBr2 ______________________________

21. Al2S3 ______________________________

22. LiBr ______________________________

23. NH3 ______________________________

24. PbO2 ______________________________

25. MgO ______________________________


Unit 4, Activity 3, Chemical Formulas and Nomenclature II Answers

Name the following compounds.

1. K2SO4 __potassium sulfate______________

2. N2O4 __dinitrogen tetroxide____________

3. BaClO4 __barium perchlorate_____________

4. HNO2(aq) __nitrous acid__________________

5. Fe2(SO4)3 __iron (III) sulfate_______________

6. NH4F __ammonium fluoride____________

7. BaI2 __barium iodide_________________

8. CrO3 __chromium (IV) oxide___________

9. Cu(C2H3O2)2 __copper (II) acetate_____________

10. Ag2CO3 __silver carbonate_______________

11. NaOH __sodium hydroxide______________

12. Ca3(PO4)2 __calcium phosphate_____________

13. ClF3 __chlorine trifluoride_____________

14. K2SO3 __potassium sulfite_______________

15. AlBr3 __aluminum bromide_____________

16. MgCl2 __magnesium chloride____________

17. HC2H3O2(aq) __acetic acid___________________

18. P2O5 __diphosphorous pentoxide________

19. FePO4 __iron (III) phosphate____________

20. SrBr2 __strontium bromide_____________

21. Al2S3 __aluminum sulfide______________

22. LiBr __lithium bromide_______________

23. NH3 __ammonia____________________

24. PbO2 __lead (IV) oxide_______________

25. MgO __magnesium oxide_____________


Unit 4, Activity 3, Molecular Geometry of Simple Molecules Student Sheet 1

Note: A represents the central atom in the molecule. B represents atoms bonded to the central atom. B can be identical atoms or different atoms. Directions:

1. Find the other students who have the same color balloons as you. Have someone inflate a balloon as much as possible without popping it. Inflate your balloon(s) to the same size.

2. Using the patterns below, tie the appropriate number and color balloons together. For example, for the AB2E model, tie 2 blue balloons and a white balloon together. For groups of 4 balloons, it is easier to tie 2 balloons together and then the other 2 balloons together, then twist the two groups together. For five-balloon groups, make sets of 2 and 3 balloons and twist. For six balloons, use 3 sets of 2 balloons twisted together.

3. Attach a piece of string to hang the finished model from the ceiling.


Type of Molecule

Number of Atoms

Attached to the Central

Atom

Balloons Needed for

Model

One colored balloon models

for electron pair geometries

AB2 2 2 pink *

AB3 3 3 blue *

AB2E 3 2 blue, 1 white

AB4 4 4 red *

AB3E 3 3 red, 1 white

AB2E2 3 2 red, 2 white

AB5 5 5 green *

AB4E 4 4 green, 1 white

AB3E2 3 3 green, 2 white

AB2E3 2 2 green, 3 white

AB6 6 6 yellow *

AB5E 5 5 yellow, 1

white

AB4E2 4 4 yellow, 2

white

Unit 4, Activity 3, Molecular Geometry of Simple Molecules Student Sheet 2

Number of lone pairs around the

Central Atom

Number of atoms attached

to the Central Atom

Electron Pair

Geometry

Bond angle of electron

pairs

Type of Molecule

Molecular Geometry

(Shape of the molecule)

Example


Unit 4, Activity 3, Molecular Geometry of Simple Molecules Answer Sheet

Number of lone

pairs on the

Central Atom

Number of atoms attached

to the Central Atom

Electron Pair Geometry

Bond angle of Electron

pairs

Type of Molecule

Molecular Shape

Example of

Molecule

0 2 Linear 180 AB2 Linear CO2

0 3Trigonal

planar120 AB3

Trigonal

planarBF3

0 4 Tetrahedral 109.5 AB4 Tetrahedral CH4

1 3 Tetrahedral <109.5 AB3ETrigonal

PyramidalNH3

2 2 Tetrahedral <109.5 AB2E2 Bent H2O

0 5Trigonal

Bipyramidal

90,

120,180AB5

Trigonal

BipyramidalPCl5

1 4Trigonal

Bipyramidal

90,

120,180AB4E *See-Saw SF4

2 3Trigonal

Bipyramidal90, 180 AB3E2 *T- structure IBr3

3 2Trigonal

Bipyramidal180 AB2E3 * Linear XeF2

0 6 Octahedral 90, 180 AB6 Octahedral SCl6

1 5 Octahedral 90, 180 AB5E*Pyramidal

PlanarIF5

2 4 Octahedral 90, 180 AB4E2

*Square

PlanarXeF4

Note: Molecular Shapes marked * may be omitted if time is a factor.


Unit 4, Activity 4, Chemical Bond Type Lab

Purpose: To observe characteristics of ionic and covalent bonds and to classify compounds as ionic or covalent based on those observations.

Modified from http://www.hse.k12.in.us/staff/ebutzin/Documents/ICP/Bonding/bond%20types%20lab.doc

Safety: Wear goggles. Do not taste or touch any chemicals. Follow guidelines pertaining to an open flame.

Materials Test tubes Thin stem pipettes Iron ring and stand Candle with foil

holder

Small foil pie pan Calcium chloride Citric acid Phenyl salicylate Potassium iodide

Sodium chloride Sucrose Conductivity probe Safety goggles

Procedure:1. Place a few crystals of sucrose, sodium chloride, phenyl salicylate, calcium chloride,

citric acid and potassium iodide in separate locations around the pie pan as shown in Figure B. Make sure all of the samples are approximately the same size. Do not allow the crystals to touch.

Write a brief description of each of the 6 substances in a data table.2. Testing melting point

Place the pie pan on the iron ring. Position the ring so it is just above the tip of a candle flame, as shown in Figure A. Light the candle to check that you have the correct height.

Place the candle under the middle of the pan and heat. Record the order in which the substances melt. If a compound doesn’t melt record N/A.

3. Testing the solubility in water


http://www.hse.k12.in.us/staff/ebutzin/Documents/ICP/Bonding/bond%20types%20lab.doc

Unit 4, Activity 4, Chemical Bond Type Lab

Place a few crystals of each substance in separate test tubes. Add about 1 mL of distilled water and agitate each.

Record the solubility in the data table (Yes – if it dissolves, No – if it does not dissolve).

4. Testing the conductivity in water Use the conductivity probe for each of the substances that WERE SOLUBLE in water

to determine if they conduct electricity or not. If the compound didn’t dissolve, do NOT try to measure the conductivity.

Rinse and dry the probe after each test.

Cleanup Rinse all test tubes with water and scrub with a test tube brush. Rinse off the pie pan and scrub with a test tube brush. Dry with a clean cloth. Wash hands and put away goggles.

Data Table

Compound Description

Melting Point

(1, 2, 3, 4, N/A)

Solubility in Water (Y/N) Conductivity

Calcium chloride

Citric acid

Phenyl salicylatePotassium

iodideSodium chloride

Sucrose

Write and defend a conclusion based on a logical analysis of your experimental data.


Unit 4, Activity 6, RAFTing

R – Role (role of the writer)

A – Audience (to whom or what the RAFT is being written)

F – Form (the form the writing will take, as in letter, song, etc.)

T – Topic (the subject focus of the writing)

R – H2O

A – Oil

F – Letter

T – Intermolecular Forces between molecules

Dear Oil,I know you would really like for us to get together. Unfortunately, my intermolecular

forces are too strong and will always keep us apart. I am a polar molecule. I am attracted to other polar molecules much more than I am

attracted to your nonpolar structure. I also have hydrogen bonding which really makes me extremely attractive to other like molecules. I guess you could say that the only thing we really have in common is a really weak dispersion force. Unfortunately, this will not be strong enough for us to base any lasting relationship.

Please feel free to look for another molecule with whom to combine. Perhaps you should look for a nonpolar molecule with no tendency to hydrogen bond.

Sincerely, Water


Unit 5, Activity 1, How Large Is a Mole?

Materials:1. Samples of 5 different types of beans 2. Container for measuring the mass of the beans3. Balance4. Calculator

Procedure:1. Measure the mass of each type of bean.

2. Using a ratio, students are to calculate the relative masses of the other beans by dividing the mass of the beans by the mass of the smallest bean of the five types used.

3. Count how many whole beans are needed to get the mass in grams equal to the relative mass calculated in step 2 for each type of bean.

4. Using the data in the relative mass column, place the empty container on the balance and zero (tare) the balance. Add beans one at a time to count how many whole beans are needed to get a mass in grams equal to the relative mass for each type of bean. (Mass of container in this example is 25.6g.)

Name of bean

Mass of the container and the beans (g)

Mass of beans

(g)

Relative mass(g)

Number of beans

from step 3

Average of Last Column

Calculate the average number of whole beans in a container by adding the number of beans in 1 container for each type of bean and dividing by 5.

Note: Use the number from the average of the last column box for all calculations.

The following ratios can be derived from the data:


Unit 5, Activity 1, How Large Is a Mole?

Use the data and the ratios to solve the following problems:

1. Calculate the number of containers given 350 beans of each type of bean.

2. Calculate the number of beans given 5.5 containers of each type of bean.

3. Calculate the mass of 350 containers of each type of bean.

4. Calculate the number of containers given 400 g. of each type of bean.

5. Calculate the number of beans given 400 g of each type of bean.

Write and defend a conclusion based on logical analysis of the data obtained from this activity.


Unit 5, Activity 1, How Large Is a Mole? Answer Sheet

This activity is designed to help students understand the concept of the mole as a definite number of particles. Using five varieties of different type beans, students will determine the relative mass of each type of bean and express the relative masses in grams. Have students work in groups and provide each group with five sets of 40 beans, a container,

and a balance. Have students determine the total mass of each type of bean. Enter the data into the table

provided. Using a ratio, students are to calculate the relative masses of the other beans by dividing the

mass of the beans by the mass of the smallest bean of the five types used. Using the data in the relative mass column, place an empty container on the balance and zero.

Add beans one at a time to count how many whole beans are needed to get a mass in grams equal to the relative mass for each type of beans. (Mass of container in this example is 25.6g.)

Name of beanMass of the

container and the beans (g)

Mass of beans(g)

Relative mass(g)

Number of beansfrom step

3

Red beans 46.7 21.1 19

Large lima beans 77.9 52.3 19

Chick peas 44.5 18.9 9.0 20 Lentils 27.7 2.1 1.0 19 Black eyes peas 35.3 9.7 4.6 20

Average of last column 19

Calculate the average number of whole beans in a container.

Note: Use the number from the average of the last column box for all calculations.

The following ratios can be derived from the data:

Note: Use the average number of beans for the beans/container ratio and the relative mass of each type of bean for the mass/container ratio.

Use the data to solve the problems as you would solve mole problems.

1. Calculate the number of containers given 350. beans of each type of bean.

2. Calculate the number of beans given 5.5 containers of each type of bean.


Unit 5, Activity 1, How Large Is a Mole? Answer Sheet

3. Calculate the mass of 350. containers of each type of bean.

4. Calculate the number of containers given 400.g. of each type of bean.

5. Calculate the number of beans given 400.g of each type of bean.

Sample Calculations:

For these calculations, the container was a cup.

Write and defend a conclusion based on logical analysis of the data obtained from this activity. Regardless of the type of bean used, the number of beans per cup is consistent (within an acceptable margin of error). Although the mass of each bean is different, the average number of beans per cup is also consistent. The data supports the idea that a cup of beans contains 19 whole beans regardless of the type of bean used or its relative mass.

When using this activity as an introduction to mole problems, the container will be used as an analogy to a mole.

For example:

Calculate the number of containers when given 350 beans of each type of bean.

Calculate the number of moles when given 350 atoms of any element.


Unit 5, Activity 3, Observing Chemical Reactions

Lab Station 1: Copper wire and silver nitrate solution

Materials: small piece of copper wire, pipettes of 0.2 M silver nitrate solution, test tube, test tube rack, waste container

Procedure:

1. Record a description of all reactants and products.

2. Make a hook on one end of the copper wire.

3. Hang the wire by the hook in a test tube.

4. Pour enough silver nitrate solution into the tube to cover most of the wire.

5. Place the test tube into the test tube rack and observe for one minute.

6. Record observations.

7. Empty the test tube contents into the large beaker (waste container).



Station 2: zinc + hydrochloric acid

Materials: mossy zinc, 0.1M hydrochloric acid, microplate, pipette, waste container


2. Place a piece of mossy zinc a well of a microplate.

3. Add 10 drops of hydrochloric acid to the well.


5. Empty the contents of the microplate into the large beaker (waste container).

Station 3: sodium chloride + silver nitrate

Materials: sodium chloride solution, 0.2 M silver nitrate solution, small test tube, pipette, large beaker


2. Fill a small test tube halfway with sodium chloride solution.

3. Add 3 to 5 drops of the silver nitrate solution.


5. Empty the contents of the microplate into the large beaker (waste container).



Station 4: Acetic acid + sodium hydrogen carbonate

Materials: Acetic acid (vinegar), sodium hydrogen carbonate (baking soda), 250 ml beaker, waste container


2. Place one teaspoon of sodium hydrogen carbonate into a small beaker.

3. Add three teaspoons of Acetic acid.


5. Empty the beaker contents into the large beaker (waste

container).

Station 5: Acetic acid + sodium hydrogen carbonate + phenolphthalein

Materials: Acetic acid (vinegar), sodium hydrogen carbonate (baking soda), phenolphthalein, 250 ml beaker, waste container


2. Place one teaspoon of sodium hydrogen carbonate into a small beaker.

3. Add three teaspoons of Acetic acid.


5. Empty the beaker contents into the large beaker (waste

container).


Unit 5, Activity 4, Split-Page Notetaking

Patterns for the types of chemical reactions1. Composition

(Synthesis)

A + X AX

2 reactants form one product:element + element one compound 2Na + Cl2 2NaCl compound + compound one compound CO2 + H2O H2CO3

2. Decomposition AX A + X

one compound two or more products2NaCl 2Na + Cl2

H2CO3 CO2 + H2OSingle Replacement Reaction:

A + BX B + AX or

Y + BX X + BY

element + compound different element + different

compound

Use the activity series of the elements to predict the products. Generally elements will replace any element below it on the chart. metals replace less active metals or hydrogen from a

compound

Cu + AgNO3 Ag + Cu(NO3)2 (reverse reaction will

not occur)

nonmetals replace less active nonmetals from a compound

I2 + NaCl Cl2 + NaI (reverse reaction will not occur)

Double Replacement Reaction:

AX + BY AY + BX

Neutralization:

compound + compound different compound + different compound Use a solubility table to predict precipitates (solids)

NaCl + AgNO3 NaNO3 + AgCl(s) (reverse reaction will not occur)

acid + base → salt + water HCl + NaOH → NaCl + H2ONeutralization is a type of double replacement reaction. An ionic salt is formed from the cation of the base and the anion of the acid. Neutralization is the reaction of the hydronium ions and hydroxide ions to form water molecules.


Unit 5, Activity 6, Can You Make Two Grams?

Possible combinations that form precipitates: Reaction Number

1. MgSO4• 7H2O + Ca(C2H3O2)2 • H2O → CaSO4(s) + Mg(C2H3O2)2 + 7H2O

246.50 g/mol 176.19 g/mol 136.15 g/mol 142.38g/mol 18.02g

3.62 g 2.59 g 2.00 g 2.09 g 2.12 g

2. MgSO4• 7H2O + Na2CO3 → MgCO3(s) + Na2SO4 + 7H2O

246.50 g/mol 105.99 g/mol 84.31 g/mol 142.02 g/mol 18.02g/mol

5.85 g 2.51 g 2.00g 3.37 g 2.99 g

3. MgSO4• 7H2O + K2CO3 → MgCO3(s) + K2SO4 + 7H2O

246.50 g/mol 138.21 g/mol 84.31 g/mol 174.27 g/mol 18.02 g/mol

5.85 g 3.28 g 2.00g 4.13 2.99g

4. ZnSO4 • 7H2O + Ca(C2H3O2)2 • H2O → CaSO4(s) + Zn(C2H3O2)2 + 8 H2O


4.22 g 2.59g 2.00 g 2.70 g 2.12g

5. ZnSO4 • 7H2O + Na2CO3 → ZnCO3(s) + Na2SO4 + 7 H2O

287.56 g/mol 105.99 g/mol 125.38g/mol 142.02 g/mol 18.02g/mol

4.59 g 1.69g 2.00g 2.27g 2.01g

6. ZnSO4 • 7H2O + K2CO3 → ZnCO3(s) + K2SO4 + 7 H2O

287.56 g/mol 138.21 g/mol 125.38g/mol 174.27 g/mol 18.02 g/mol

4.59 g 2.20g 2.00g 2.78g 2.01g

7. Ca(C2H3O2)2 • H2O + Na2CO3 → CaCO3(s) + 2NaC2H3O2 + H2O


3.52g 2.12g 2.00g 3.28g 036g

8. Ca(C2H3O2)2 • H2O + K2CO3 → CaCO3(s) + 2KC2H3O2 + H2O

176.19 g/mol 138.21 g/mol 100.09 g/mol 98.15 g/mol 18.02g/mol

3.52g 2.76g 2.00g 3.92g 0.36g


Unit 5, Activity 7, Vocabulary Self-Awareness


Oxidation

Reduction

Redox reaction

Oxidizing agent

Reducing agent

Skeleton equation


Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions Half- fill each well in the column with the indicated solution.

Column 1: Zn(NO3)2

Column 2: Pb(NO3)2

Column 3: Cu(NO3)2

1. Place one piece of zinc shot in each filled well in row 1.

2. Place one piece of lead shot in each filled well in row 2.

3. Place one piece of copper shot in each filled well in row 3.

Column

1 2 3

Row

1

2

3

4. Watch for two minutes, then record observations.

5. Make a list of the elements in order of reactivity.

6. Write the redox equation for each chemical change.


Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions Answer Sheet

1. Half- fill each well in the column with the indicated solution.

Column 1: Zn(NO3)2

Column 2: Pb(NO3)2

Column 3: Cu(NO3)2

2. Place one piece of zinc shot in each filled well in row 1.

3. Place one piece of lead shot in each filled well in row 2.

4. Place one piece of copper shot in each filled well in row 3.

Column

1 2 3

Row

1

2

3

5. Watch for two minutes, then record observations.

6. Make a list of the elements in order of reactivity.

7. Write the redox equation for each chemical change.

8. Write a conclusion based on your observations.

Elements in order of reactivity:

Zn, Pb, Cu

LEO the lion says GER (loss of electrons is oxidation, gaining electrons is reduction)

Redox equations:

Zn + Pb(NO3)2 → Pb + Zn(NO3)2

Zn0 → Zn 2+ + 2e- (oxidation)

Pb 2+ + 2e- → Pb0 (reduction)



Zn + Cu(NO3)2 → Cu + Zn(NO3)2

Zn0 → + Zn 2+ 2e- (oxidation)

Cu2++ 2e- → Cu0 (reduction)

Pb + Zn(NO3)2 → NR (no reaction)

Pb + Cu(NO3)2 → Cu + Pb(NO3)2

Pb0 → Pb2+ +2e- (oxidation)

Cu2++2e- → Cu0 (reduction)

Cu + Zn(NO3)2 → NR

Cu + Pb(NO3)2 → NR

Conclusion: Oxidation is the process by which electrons are removed from atoms or ions.

Zn is oxidized by the other two ions. Pb is only oxidized by the Cu2+ ion. Cu is not

oxidized by either ion. Zn gives up its electrons more easily than the other ions.The

element that is oxidized is the reducing agent therefore Zn is the strongest reducing

agent, followed by Pb and lastly, by Cu.

Reduction is the process by which electrons are added to atoms or ions. The element that

is reduced is the oxidizing agent. Cu is the strongest oxidizing agent, followed by Pb and

then Zn.

Oxidation and reduction must take place at the same time because the number of

electrons lost must equal the number of electrons gained.



Answers to ionic equations:

Molecular equation: Zn + Pb(NO3)2 → Pb + Zn(NO3)2

Ionic Equation: Zn0 + Pb2+ + 2NO3-1 → Pb0 + Zn 2+ + 2NO3

-1

Net Ionic Equation: Zn0 + Pb 2+→ Pb0 + Zn 2+

Molecular equation: Zn + Cu(NO3)2 → Cu + Zn(NO3)2

Ionic equation: Zn0 + Cu2+ + 2NO3-1 → Cu0+ Zn 2+ + 2NO3

-1

Net Ionic equation: Zn0 + Cu2+ → Cu0+ Zn 2+

Molecular equation: Pb + Zn(NO3)2 → NR (no reaction)

Molecular equation: Pb + Cu(NO3)2 → Cu + Pb(NO3)2

Ionic equation: Pb0 + Cu2+ +2NO3-1 → Cu0 + Pb2+ + 2NO3

-1

Net Ionic equation: Pb0 + Cu2+ → Cu0 + Pb2+

Molecular equation: Cu + Zn(NO3)2 → NR

Molecular equation: Cu + Pb(NO3)2 → NR


Unit 6, Activity 4, Heating Curve

Heat is added to a substance in the solid state. The energy added will increase the temperature of the substance to its specific melting point. The amount of energy required to raise the temperature depends on the specific heat (Cp) and the state of the substance. Specific heat is the amount of energy needed to raise the temperature of one gram of a substance by one degree Celsius.

At the melting point, the temperature stops rising and the substance starts to melt. The energy supplied is used to weaken the intermolecular forces of attraction and the temperature remains constant. The amount of energy needed to melt a substance depends on its heat of fusion (Hf). The molar heat of fusion is the amount of energy required to melt one mole of a substance at its melting point.

After the phase change is complete, the temperature rise will follow a different rate than that of the solid because the liquid state has a different heat capacity.

At the boiling point, the temperature stops rising and the substance starts to boil. The energy supplied is used to break the intermolecular forces of attraction and the temperature remains constant. The amount of energy needed to boil a substance depends on its heat of vaporization (Hv). The molar heat of vaporization is the amount of energy required to boil one mole of a substance at its boiling point. The Hv is higher than the Hf because of breaking the forces of attraction.

After the phase change is complete, the temperature rise will follow a different rate than that of the liquid because the gaseous state has a different heat capacity.


Boiling Point (Hv)

Melting Point (Hf)

Solid State Cp(s)

Liquid State Cp (l)

Gas State Cp (g)

Unit 6, Activity 4, Phase Diagrams

A phase diagram is a graph of the conditions of temperature and pressure at which the solid, liquid, and gaseous phases of a substance exist. The lines separating the phases are called phase boundaries. Each point on the phase boundary show the conditions under which the two phases exist in dynamic equilibrium.

Each point along the solid/liquid phase boundary represents the temperature and pressure combinations where the rate of the solid melting is equal to the rate of the liquid freezing. Each point represents a melting point.

Each point along the liquid/ vapor phase boundary represents the temperature and pressure combinations where the rate of the liquid boiling is equal to the rate of the vapor condensing. Each point represents a boiling point.

Each point along the solid/gas phase boundary represents the temperature and pressure combinations where the rate of the solid subliming is equal to the rate of the vapor condensing to a solid (called deposition). Each point represents a sublimation point.

The triple point indicates the only temperature and pressure conditions where the solid, liquid, and vapor phases are all in equilibrium.

The critical point (Pc) is the point above which a substance will always be a gas regardless of the pressure and temperature. The critical temperature is the highest temperature a substance can exist as a liquid. The critical pressure is the lowest pressure required for the substance to be a liquid at the critical temperature.

Phase Diagram for H2O Phase diagram for CO2


Unit 6, Activity 6, Exothermic and Endothermic Energy Diagrams

Heat of reaction (H) is the amount of heat released or absorbed during a chemical reaction.H = H products - H reactants

The heat content of the reactants is higher than the heat content of the products. Energy in the form of heat will be released when the products form. The heat of reaction (H) is negative.

Example:

2 H2(g) + O2(g) → 2H2O(g) H = - 483.6 kJ

The heat content of the reactants is lower than the heat content of the products. Energy in the form of heat must be absorbed (added) to form the products. The heat of reaction (H) is positive.

Example:

2 H2O(g) → 2H2(g) + O2(g) H = + 483.6 kJ


Unit 6, Activity 6, Energy Diagram (with activation energy)

For a reversible reaction, the activated complex is the same. The activated complex occurs at the maximum-energy position along the reaction pathway. The activation energy of the forward reaction is lower than the activation energy of the reverse reaction in this energy diagram. The H is the same amount for both reactions but the sign of H is negative for the forward reaction and is positive for the reverse reaction.


Unit 7, Activity 1, Vocabulary Self- Awareness


Solution

Solute

Solvent

Soluble

Electrolyte

Nonelectrolyte

Colloid

Solubility

Saturated solution

Unsaturated solution

Supersaturated

Solution equilibrium

Miscible


Unit 7, Activity 1, Vocabulary Self- Awareness

Immiscible

chromatography

molarity

molality

Colligative property

Vapor pressure

Nonvolatile

Volatile

Freezing-point depression

Boiling-point elevation


Unit 7, Activity 3, Solution Concentrations

Sample Problems using factor-label method.

Molarity (M) =

Example 1:

Calculate the molarity of a 1500 ml solution that contains 45.0 g of MgCl2.

M=

Example 2:

Calculate the mass of solute in 750.0 mL of a 0.500 M H2SO4 solution.

Mass of solute =

Example 3:

Calculate the volume of solution that can be made using a 6.00 M solution using 45.0 g C6H12O6.

Volume =

Molality (m) =

Example 1:

Calculate the molality of a solution containing 50.0 g of HC2H3O2 dissolved in 500.0 g Hm=O.

Example 2:

Calculate the mass of solute needed to make a 0.450 m NaOH solution containing 750.0 g H2O.


Unit 7, Activity 3, Solution Concentrations

Example 3:

Calculate the mass of solvent needed to make a 2.50 m H2SO4 solution containing 150.0 g of the acid.

Example:

Determine the mole fraction of glucose, C6H12O6, in a solution containing 425 g glucose dissolved in 750.0 g H2O.

Moles of glucose:

Moles of water:

Total moles of solute + solvent:2.36 mol + 41.7 mol= 44.06 mol

Mole fraction of C6H12O6:


Unit 7, Activity 5, Ice Cream Recipe

Recipe for 40 pint size bags of ice cream (1 bag per student):

1 gal whole milk1 pint half & half6 cups sugar6 t vanilla

Additional materials needed: spoon, large pot, several boxes of ice cream salt; two rolls of duct tape, several large bags of ice, enough newspaper for each student to have a section, plastic bags from grocery or discount stores

Combine all ingredients in the pot to make the ice cream mixture and heat until the sugar is dissolved. Stir the mixture often to prevent the mixture from scorching. Pour into the empty milk container. There may some extra mixture, so have a smaller container or zip top bag handy also.

Per student:

1 pint size freezer zip top bags1 gallon size freezer bags2 plastic bags a section of newspaper ice cream salt ice duct tape

Directions:

Fill the gallon bag half full of ice. Add 1/2 inch layer of ice cream salt. Put 1/2 cup of ice cream mixture in small bag. Seal the small bag and place duct tape over the sealed end. Put the small bag inside of large one. Add enough ice to fill the gallon bag. Seal and duct tape the sealed end of the large bag. Wrap the large bag with several layers of newspaper. Place the wrapped bag in a couple of plastic bags. Tie the ends of the plastic bags. Shake or rotate the bags gently for about 15 min. Have a large container such as a dish pan handy to empty the water/ salt mixture into. The water can be evaporated and the salt reused, if desired.

Powdered drink mixes can be made according to the directions on the package and used in place of the ice cream. A “slush” will be formed.


Unit 7, Activity7, pH Lab Carousels

Lab Carousel 1: At each station place: a microplate, red litmus paper, blue litmus paper, universal indicator and phenolphthalein (and any other available indicators), and stirring rod. Pipettes containing:

Station 1: vinegar Station 4: household ammonia solutionStation 2: distilled water Station 5: colorless sodaStation 3: KOH solution Station 6: HCl solution

Instruct students to test each solution with the indicator papers and indicators at each station and identify each of the solutions as acids or bases. Data should be recorded in a student –generated data table.

Lab Carousel 2: At each station place a microplate, pH paper and/ or pH meter, and stirring rod. Pipettes containing:

Station 1: vinegar Station 4: household ammonia solutionStation 2: distilled water Station 5: colorless sodaStation 3: KOH solution Station 6: HCl solution

Instruct students to determine the pH of the solutions and rank them in order of increasing pH. Data should be recorded in a student –generated data table.


Unit 8, Activity 2, Alkanes

Alkanes are saturated hydrocarbons (compounds containing only carbon and hydrogen) with the formula CnH2n+2, where “n” represents the number of carbon atoms. “Saturated” means that all C-C bonds are single bonds.

Names of organic compounds follow the rules of IUPAC (International Union of Pure and Applied Chemistry). Notice that each compound differs from the previous one by a –CH2 group. A homologous series in one in which the compounds differ from each other by a specific unit. The pattern for the first 10 alkanes is shown below.

Stem name Alkane name Formula Number of isomers

meth- methane CH4 1

eth- ethane C2H6 1

prop- propane C3H8 1

but- butane C4H10 2

pent- pentane C5H12 3

hex- hexane C6H14 5

hept- heptane C7H16 9

oct- octane C8H18 18

non- nonane C9H20 35

dec- decane C10H22 75

Isomers are compounds with the same molecular formula but different structural formulas.

Draw the isomers for pentane and hexane.


Unit 8, Activity 2, Alkanes

Rules for naming alkanes:

1. Pick out the longest continuous chain of carbon atoms and name it.2. Number the carbon atoms from the end that will give the lowest numbers possible to the

branches.3. Name the branches by adding –yl to the stem name and adding a number to indicate the

carbon atom the branch is attached to. The number will be followed by a dash. All branches must have a number with it. Numbers are separated by commas.

*If branches are different groups, they appear alphabetically in the name.4. If more than one of an alkyl group appears, a number prefix is used to denote the total

number of groups. 5. Dashes between carbon atoms do not need to be shown.

Examples:

6 5 4 3 CH3 CH2 CH2 CH CH3

2 CH2 1 CH3

Name: 3-methylhexane

1 2 3 4 5 CH3 CH CH2 CH CH3 CH3 CH3

Name: 2,4-dimethylpentane

CH3 CH3 CH3CHCHCH2CHCH2 CH3 CH2 CH3

Name: 3-ethyl-2,5-dimethylheptane

CH3 CH2 CH3CHCHCH3 CHCH2CH3 CH3 Name: 3,4,5-trimethylheptane

Name the isomers for pentane and hexane that were drawn on the previous sheet.


Unit 8, Activity 2, Alkanes Answer Sheet

Isomers of Pentane (formula C5H12):

1. CH3CH2CH2CH2CH3 n-pentane ( n means normal straight chain)

2. CH3CHCH2CH3 2- methylbutane CH3

3. CH3 2,2-dimethylpropane

CH3CCH3

CH3

Isomers of hexane (formula C6H14)

1. CH3CH2CH2CH2CH2CH3 n-hexane

2. CH3CHCH2CH2CH3 2-methylpentane

CH3

3. CH3CH2CHCH2CH3 3-methylpentane

CH3

4. CH3 CH CH CH3 2,3-dimethylbutane CH3 CH3

CH3

5. CH3 C CH2 CH3 2.3-dimethylbutane

CH3
