Unit 1, Activity 2, Specific Assessment Rubric
Blackline Masters, Chemistry Page 1 Louisiana Comprehensive Curriculum, Revised 2008
Chemistry
Unit 1, Activity 2, Specific Assessment Rubric
Blackline Masters, Chemistry Page Louisiana Comprehensive Curriculum, Revised 2008
1
3 2 0 Measurements 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 measurements
Less than 2 measurements
All safety rules followed
Questions Answered
Answered correctly
Answered incorrectly but supported by evidence
Answered incorrectly. No supporting evidence.
Unit 1, Activity 3, Accuracy and Precision Worksheet
Blackline Masters, Chemistry Page Louisiana Comprehensive Curriculum, Revised 2008
2
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.
Unit 1, Activity 3, Accuracy and Precision Worksheet Answers
Blackline Masters, Chemistry Page 3 Louisiana Comprehensive Curriculum, Revised 2008
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?
Poor None of the darts are grouped in the bull’s-eye.
Poor Few of the darts are grouped in the bull’s-eye.
Good All 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
Blackline Masters, Chemistry Page 4 Louisiana Comprehensive Curriculum, Revised 2008
Matter Homogeneous
Pure Substance Heterogeneous
Element Mixture
Compound
Unit 2, Activity 1, Card Sort Template 2
Blackline Masters, Chemistry Page 5 Louisiana Comprehensive Curriculum, Revised 2008
Muddy Water Na
Solution As
salt water Cl
Metal NaCl
nonmetal Metalloid
Unit 2, Activity 1, Sample Concept Map
Blackline Masters, Chemistry Page 6 Louisiana Comprehensive Curriculum, Revised 2008
MATTER
PURE SUBSTANCE
MIXTURE
HOMOGENEOUS HETEROGENEOUSELEMENT
COMPOUND
METAL METALLOID NONMETAL
Na As Cl
MUDDY WATER SOLUTION NaCl
SALT WATER
Can be either
Is
Chemically combine to form
Is either
Is
Is called Example Example
Can be
Example Example Example Example
Elements, Compounds and Mixtures Concept Map
Unit 2, Activity 2, Sample Word Grid
Blackline Masters, Chemistry Page 7 Louisiana Comprehensive Curriculum, Revised 2008
Sample: Homogeneous Heterogeneous
Can 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
Blackline Masters, Chemistry Page 8 Louisiana Comprehensive Curriculum, Revised 2008
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.
Sample: Homogeneous Heterogeneous
Can 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
Unit 2, Activity 4, Three Worlds of Chemistry
Blackline Masters, Chemistry Page 9 Louisiana Comprehensive Curriculum, Revised 2008
Unit 2, Activity 5, Density
Blackline Masters, Chemistry Page 10 Louisiana Comprehensive Curriculum, Revised 2008
Each box has the same volume. If each
ball has the same mass, which box would
weigh more? Why?
Unit 2, Activity 6, Split-Page Notes
Blackline Masters, Chemistry Page 11 Louisiana Comprehensive Curriculum, Revised 2008
Physical and Chemical Changes
Effervescent tablet in water Observations
1. numerous bubbles formed as soon as tablet touched the water
2. bubbles rose to top of water and burst 3. tablet disappeared 4. bubbles stopped forming 5. 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
Blackline Masters, Chemistry Page 12 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Louisiana Comprehensive Curriculum, Revised 2008
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.
Unit 4, Activity 1, Vocabulary Self-Awareness
Blackline Masters, Chemistry Page 14 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 15 Louisiana Comprehensive Curriculum, Revised 2008
Unit 4, Activity 2, Ion Cards
Blackline Masters, Chemistry Page 16 Louisiana Comprehensive Curriculum, Revised 2008
Unit 4, Activity 2, Ion Cards
Blackline Masters, Chemistry Page 17 Louisiana Comprehensive Curriculum, Revised 2008
Unit 4, Activity 2, Ion Cards
Blackline Masters, Chemistry Page 18 Louisiana Comprehensive Curriculum, Revised 2008
Unit 4, Activity 3, Chemical Formulas and Nomenclature I
Blackline Masters, Chemistry Page 19 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 20 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 21 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 22 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 23 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 24 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 25 Louisiana Comprehensive Curriculum, Revised 2008
Note: Molecular Shapes marked * may be omitted if time is a factor.
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 3 Trigonal
planar 120° AB3
Trigonal
planar BF3
0 4 Tetrahedral 109.5° AB4 Tetrahedral CH4
1 3 Tetrahedral <109.5° AB3E Trigonal
Pyramidal NH3
2 2 Tetrahedral <109.5° AB2E2 Bent H2O
0 5 Trigonal
Bipyramidal
90°,
120°,180° AB5
Trigonal
Bipyramidal PCl5
1 4 Trigonal
Bipyramidal
90°,
120°,180° AB4E *See-Saw SF4
2 3 Trigonal
Bipyramidal 90°, 180° AB3E2 *T- structure IBr3
3 2 Trigonal
Bipyramidal 180° AB2E3 * Linear XeF2
0 6 Octahedral 90°, 180° AB6 Octahedral SCl6
1 5 Octahedral 90°, 180° AB5E *Pyramidal
Planar IF5
2 4 Octahedral 90°, 180° AB4E2 *Square
Planar XeF4
Unit 4, Activity 4, Chemical Bond Type Lab
Blackline Masters, Chemistry Page 26 Louisiana Comprehensive Curriculum, Revised 2008
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.
Unit 4, Activity 4, Chemical Bond Type Lab
Blackline Masters, Chemistry Page 27 Louisiana Comprehensive Curriculum, Revised 2008
3. Testing the solubility in water • 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 salicylate
Potassium iodide
Sodium chloride
Sucrose
Write and defend a conclusion based on a logical analysis of your experimental data.
Unit 4, Activity 6, RAFTing
Blackline Masters, Chemistry Page 28 Louisiana Comprehensive Curriculum, Revised 2008
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?
Blackline Masters, Chemistry Page 29 Louisiana Comprehensive Curriculum, Revised 2008
Materials: 1. Samples of 5 different types of beans 2. Container for measuring the mass of the beans 3. Balance 4. 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:
beanscontainer
relative mass of beanscontainer
Unit 5, Activity 1, How Large Is a Mole?
Blackline Masters, Chemistry Page 30 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 31 Louisiana Comprehensive Curriculum, Revised 2008
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 bean Mass of the
container and the beans (g)
Mass of beans (g)
Relative mass (g)
Number of beans from step
3
Red beans 46.7 21.1 21121
10 0..
.=
19
Large lima beans 77.9 52.3
52 321
24 9..
.=
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.
(19 19 20 19 20) 19.4 195
wholebeans+ + + += =
Note: Use the number from the average of the last column box for all calculations. The following ratios can be derived from the data:
beanscontainer
relative mass of beanscontainer
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
Blackline Masters, Chemistry Page 32 Louisiana Comprehensive Curriculum, Revised 2008
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.
1. 350. beans X 1 cup19 beans
cups= 18 4.
2. 5.5 cups X 19 beans1 cup
104.5 beans=
1cup 21.1g3. 350red beans x × =388.7g19 red beans 1cup
1cup 24.9g350 lima beans× × = 458.68g19 lima beans 1cup
4. 400. g X 1 cup21.1 g
= 19.0 cups
5. 400. g X 1 cup21.1 g
X 19 red beans1 cup
= 360.2 red beans
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
Blackline Masters, Chemistry Page 33 Louisiana Comprehensive Curriculum, Revised 2008
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).
Unit 5, Activity 3, Observing Chemical Reactions
Blackline Masters, Chemistry Page 34 Louisiana Comprehensive Curriculum, Revised 2008
Station 2: zinc + hydrochloric acid Materials: mossy zinc, 0.1M hydrochloric acid, microplate, pipette, waste container 1. Record a description of all reactants and products.
2. Place a piece of mossy zinc a well of a microplate.
3. Add 10 drops of hydrochloric acid to the well.
4. Record observations.
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
1. Record a description of all reactants and products.
2. Fill a small test tube halfway with sodium chloride solution. 3. Add 3 to 5 drops of the silver nitrate solution. 4. Record observations. 5. Empty the contents of the microplate into the large beaker (waste
container).
Unit 5, Activity 3, Observing Chemical Reactions
Blackline Masters, Chemistry Page 35 Louisiana Comprehensive Curriculum, Revised 2008
Station 4: Acetic acid + sodium hydrogen carbonate
Materials: Acetic acid (vinegar), sodium hydrogen carbonate (baking soda), 250 ml beaker, waste container
1. Record a description of all reactants and products.
2. Place one teaspoon of sodium hydrogen carbonate into a small beaker.
3. Add three teaspoons of Acetic acid.
4. Record observations.
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
1. Record a description of all reactants and products.
2. Place one teaspoon of sodium hydrogen carbonate into a small beaker.
3. Add three teaspoons of Acetic acid.
4. Record observations.
5. Empty the beaker contents into the large beaker (waste
container).
Unit 5, Activity 4, Split-Page Notetaking
Blackline Masters, Chemistry Page 36 Louisiana Comprehensive Curriculum, Revised 2008
Patterns for the types of chemical reactions 1. 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 products 2NaCl →2Na + Cl2 H2CO3→ CO2 + H2O
Single 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 + H2O Neutralization 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?
Blackline Masters, Chemistry Page 37 Louisiana Comprehensive Curriculum, Revised 2008
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
287.56 g/mol 176.19 g/mol 136.15 g/mol 183.48 g/mol 18.02 g/mol
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
176.19 g/mol 105.99 g/mol 100.09 g/mol 82.03 g/mol 18.02 g/mol
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
Blackline Masters, Chemistry Page 38 Louisiana Comprehensive Curriculum, Revised 2008
Term
+ - Definition Example
Oxidation
Reduction
Redox reaction
Oxidizing agent
Reducing agent
Skeleton equation
Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions
Blackline Masters, Chemistry Page 39 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 40 Louisiana Comprehensive Curriculum, Revised 2008
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)
Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions Answer Sheet
Blackline Masters, Chemistry Page 41 Louisiana Comprehensive Curriculum, Revised 2008
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.
Unit 5, Activity 7, Introduction to Oxidation-Reduction Reactions Answer Sheet
Blackline Masters, Chemistry Page 42 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 43 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 44 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 45 Louisiana Comprehensive Curriculum, Revised 2008
Heat of reaction (ΔH) is the amount of heat released or absorbed during a chemical reaction. ΔH = 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 (ΔH) is negative. Example: 2 H2(g) + O2(g) → 2H2O(g) ΔH = - 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 (ΔH) is positive. Example: 2 H2O(g) → 2H2(g) + O2(g) ΔH = + 483.6 kJ
Unit 6, Activity 6, Energy Diagram (with activation energy)
Blackline Masters, Chemistry Page 46 Louisiana Comprehensive Curriculum, Revised 2008
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 ΔH is the same amount for both reactions but the sign of ΔH is negative for the forward reaction and is positive for the reverse reaction.
Unit 7, Activity 1, Vocabulary Self- Awareness
Blackline Masters, Chemistry Page 47 Louisiana Comprehensive Curriculum, Revised 2008
Term
+ - Definition Example
Solution
Solute
Solvent
Soluble
Electrolyte
Nonelectrolyte
Colloid
Solubility
Saturated solution
Unsaturated solution
Supersaturated
Solution equilibrium
Miscible
Unit 7, Activity 1, Vocabulary Self- Awareness
Blackline Masters, Chemistry Page 48 Louisiana Comprehensive Curriculum, Revised 2008
Immiscible
chromatography
molarity
molality
Colligative property
Vapor pressure
Nonvolatile
Volatile
Freezing-point depression
Boiling-point elevation
Unit 7, Activity 3, Solution Concentrations
Blackline Masters, Chemistry Page 49 Louisiana Comprehensive Curriculum, Revised 2008
Sample Problems using factor-label method.
Molarity (M) = moles of soluteliter of solution
Example 1: Calculate the molarity of a 1500 ml solution that contains 45.0 g of MgCl2.
M= 45.0 g MgCl1500 mL of solution
X 1 mol MgCl95.3 g MgCl
X 1000 mL1 L
0.31 M2 2
2
=
Example 2: Calculate the mass of solute in 750.0 mL of a 0.500 M H2SO4 solution.
Mass of solute = 750mL X 1L1000mL
X 0.500mol H SO1L
X 98.1g H SO1mol H SO
36.8g H SO2 4 2 4
2 42 4=
Example 3: Calculate the volume of solution that can be made using a 6.00 M solution using 45.0 g C6H12O6.
Volume = 45.0g C H O X 1mol C H O180.0g C H O
X 1L6.00 mol C H O
0.400 L6 12 6 6 12 6
6 12 6 6 12 6
=
Molality (m) = moles of solutekg of solution
Example 1: Calculate the molality of a solution containing 50.0 g of HC2H3O2 dissolved in 500.0 g Hm=O.
2 3 2 2 3 2 2
2 2 3 2 2
50.0g HC H O 1mol HC H O 1000g H Om= X X =1.67 m500.0g H O 60.0gHC H O 1kg H O
Example 2: Calculate the mass of solute needed to make a 0.450 m NaOH solution containing 750.0 g H2O.
2750.0g H O 1kg 0.450mol NaOH 40.0g NaOHmass = X X X =13.5g NaOH1000g 1kg 1mol NaOHsolute
Unit 7, Activity 3, Solution Concentrations
Blackline Masters, Chemistry Page 50 Louisiana Comprehensive Curriculum, Revised 2008
Example 3: Calculate the mass of solvent needed to make a 2.50 m H2SO4 solution containing 150.0 g of the acid.
2
2 4 2 4 2H O 2
2 4 2 4
150.0g H SO 1mol H SO 1kg H O 1000gmass = X X X = 612g H O98.1g H SO 2.50 mol H SO 1kg
moles of solute (or solvent)Mole Fraction (X) =moles of solute + solvent
Example: Determine the mole fraction of glucose, C6H12O6, in a solution containing 425 g glucose dissolved in 750.0 g H2O. Moles of glucose:
6 12 6 6 12 66 12 6
6 12 6
425g C H O 1mol C H OX = 2.36molC H O180g C H O
Moles of water:
2 22
2
750.0g H O 1mol H OX = 41.7 mol H O18.0g H O
Total moles of solute + solvent: 2.36 mol + 41.7 mol= 44.06 mol
cosX molglu e= Mole fraction of C6H12O6:
6 12 6
6 12 6
6 12 6 2
C H OC H O
C H O +H O
mol 2.36 molX = = = 0.5mol 44.06 mol
Unit 7, Activity 5, Ice Cream Recipe
Blackline Masters, Chemistry Page 51 Louisiana Comprehensive Curriculum, Revised 2008
Recipe for 40 pint size bags of ice cream (1 bag per student):
1 gal whole milk 1 pint half & half 6 cups sugar 6 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 bags 1 gallon size freezer bags 2 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
Blackline Masters, Chemistry Page 52 Louisiana Comprehensive Curriculum, Revised 2008
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 solution Station 2: distilled water Station 5: colorless soda Station 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 solution Station 2: distilled water Station 5: colorless soda Station 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
Blackline Masters, Chemistry Page 53 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 54 Louisiana Comprehensive Curriculum, Revised 2008
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
Blackline Masters, Chemistry Page 55 Louisiana Comprehensive Curriculum, Revised 2008
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
Top Related