Name _____________________
Inquiry – Questions – Hypothesis – Independent – Dependent – Controls – Control – Data Collection
– Data Presentation - Data Analysis – Hypothesis supported, refuted, inconclusive-conclusions-new
questions
Lab Safety – Right to know – lab rules – MSDS – Bunsen burner
Chemistry – Matter – Sources of Matter – Separation Techniques
I. Chemiluminescence Glow Sticks –
II. A student ask the question, does a glow stick used on a Halloween night last longer on a cold night?
III. Materials
Luminol
5 % Hydrogen Peroxide
0.10 M Sodium Hydroxide
Thermometer
Water at different temperatures
Beakers
Stop Watches
IV. Procedures
1. a) Place 200 mL of room temperature water (approximately 20oC measured with an alcohol
thermometer) in a 400 mL beaker
b) Place 10mL of 5% hydrogen peroxide in test tube 1.
b) Place 5 ml of .10 M NaOH(aq) in test tube 2.
2. a) Place test tubes 1 and 2 into the 400ml beaker filled with the 200 mL water at room
temperature.
b) Wait 5 minutes so that the 5 % hydrogen peroxide and NaOH (aq) obtain the
temperature of the water.
3. a) Add 1.0 g of luminol to a test tube tube 3
b) Add 10 mL of 5 % hydrogen peroxide to test tube 3
c) Add 5.0 mL of .10 M NaOH(aq) to test tube 3, start timing with a stopwatch, and place in
the 20oC water bath
4. Stop timing with the stopwatch when the mixture does not release visible light.
5. Repeat steps 1,2, and 3 at approximately 10oC, 30oC and 40oC.
200mL
water
Temperature
20oC
10mL
5%
Hydrogen
Peroxide
5 mL
.10 M
NaOH(aq)
1.0 g
Luminol
10ml
hydrogen
peroxide
NaOH(aq)
V. Data
20o C – glow time = 63 seconds
11oC - glow time = 128seconds
29 oC - glow time = 31 seconds
42 oC - glow time = 14 seconds
Reproducible results were obtained for multiple trials at approximately the same temperature
ranges
1. What is the independent variable of the above experiment and how was it measured?
The temperature of the reaction between the luminol, 5% hydrogen peroxide, and 1.0 M
NaOH(aq) was the independent variable (it was manipulated in the Experiment) and was
measured by an alcohol thermometer.
2. What is the dependent variable of the above the experiment and how was it measured?
The dependent variable ( it responded to the change in temperature) was the glow time
measured with a stopwatch.
3. What controls were performed in the above experiment? Why are controls needed?
The volume of 5 % hydrogen peroxide, mass of luminol, volume of 1.0 M NaOH(aq),
Container sizes. They were controlled in order to ISOLATE THE RELATIONSHIP between
the DEPENDENT VARIABLE ( Glow time ) and the INDEPENDENT VARIABLE
( Temperature of reaction mixture )
4. Create a graph that illustrates the relationships between temperature and duration of glow
Title: 2pts Glow Time of the reaction between luminol, 5 % hydrogen peroxide as a function of
temperature of reaction mixture.
120
100
1pt
Glow Time
In Seconds
80
60
40
20
0
10
20
30
40
1pt Temperature of reaction mixture in oC
What can you conclude given the above data and graph?
5. a) Calcium carbonate is the active ingredient Tums Antacid tablets. What source of matter does
calcium carbonate most likely come from and why? _____ 1pt
a) plants
d) crude oil / petroleum
b) lakes and oceans
e) rocks, minerals, ores
c) air
ii) Why? 2pts Calcium is a metal containing compound and is therefore it’s source
is rocks, minerals, ores. Calcium carbonate is a mineral
b) Plants are excellent sources of fibers such as cotton. A new fiber can be made from the
polymer Poly Vinyl Alcohol can be used to replace cotton in hospital garments. It is useful
because it dissolves in water about 200 oC. What is the source of matter Poly Vinyl Alcohol
is made from? Why is this source usually used as a substitute for naturally occurring
substances?
ii)Why? 3pts Crude oil – Crude oil is formed from decaying plant and animal matter
that has been exposed to heat and pressure for millions of years. Crude oil contains
the same raw materials found in plants and can be used to synthesize substances that
mimic those produced by plants and animals.
Separation Techniques
6. Separates and saves substances based on
differences in boiling points. Lowest boiling point
substances are removed and collected first.
____
Distillation
7. Separates substances based on the relative affinity
a substance has for the medium it is traveling through
such as paper.
____
Chromatography
8. Separates the elements found in a compound
based on a chemical reaction that occurs at the
positive and negative poles of a power supply. ____
Electrolysis
9. Separates and saves substances based on
their solubility differences
_____
Filtration
10. Separates substances based on the differences
in the density of liquids that do not mix
____
Separating Funnel
A.
B.
C.
D.
E.
F.
G.
H.
Chromatography
Centrifuge
Electrolysis
Filtration
Separating Funnel
Evaporation
Magnet
Distillation
11.
12.
Which one of the following could be used to separate hydrogen from water? F 1pt
Which one of the following can be used to separate and save the oils from the aqueous solutions
found in citrus fruits because the oils have a lower density than the aqueous solutions. A 1pt
13. Which one of the following can be used to separate insoluble sediment from solution containing a
soluble mineral. E – filtration 1pt
14. Which one of the following can be used to remove water from a mixture to determine the amount
of dissolved substances found in the solution? B – evaporation 1pt
A
B
C
E
D
F
15. a) If your Bunsen burner produces a yellow sooty flame what can you do the change it to a blue
flame? 2pts Increase fuel by adjusting the burner valve and then increasing the air intake
by adjusting the burner valve.
b) Which valve do you use to adjust the height of a flame? (The table valve or the burner valve)
1pt. Burner valve.
16. HCl (aq) is a known poison! It is dangerous. It is corrosive. The liquid and mist cause
severe burns to all body tissues. HCl may be fatal if swallowed or inhaled. Inhalation may cause
lung damage. Where can you find information about the substances such as HCl(aq) used in this
class? 1pt. Material Safety Data Sheets
What law informs workers in industrial work places of the risks associated with substances they
Use or come in contact with?
Right to Know Law
17. You are given an acid base indicator, a base and an acid. You know that the indicator is yellow in
acid and blue in base. Explain how you can determine which solution is more concentrated and
how many times more concentrated that solution is compared to the other solution. (4pts)
1. Add 1 drop of an indicator into a test tube and note the color
2. Add 3 drops of the acid to that test tube and note the color.
3. Add 1 drop of an indicator into second test tube and note the color
4. Add 3 drops of an indicator into that second test tube
5. Add 5 drops of indicator to a third test tube
6. Add 10 drops of acid to the third test tube.
7. Add drops of base until the indicator changes from the base color to the acid color.
Record number of drops of base needed.
8. Add 5 drops of indicator to a fourth test tube
9. Add 10 drops of base to a fourth test tube.
9. Add drops of acid until the indicator changes from the base color to the acid color.
Record number of drops of acid needed.
10. Repeat steps 5 to 7 three times and 8 to 9 three times.
The following table can be used to determine the relative concentrations of the acid and base
Indicator
Drops acid and color of
Drops of base added to change
indicator acid mixture
color of indicator from acid
red to green base
5 drops of green indicator
10 – color red
12 – color green
Indicator
Drops base added – and Color Drops of acid added to change
color of indicator from base
green to acid red
5 drops of green indicator
10 – color – green
9 – color red
Calculations - Conclusions
If 12 drops of base are needed to change the indicator and 10 drops of acid from the acid
Color to the base color then the acid is between 11/10 ( 1.1) and 12/10 ( 1.2 ) times more
Concentrated. 11 drops is not quite enough base to neutralize the 10 acid and the 12
drops is more than enough base to neutralize the 10 drops acid
If 9 drops of acid are needed to change the indicator and 10 drops of base from the base
color to the acid color then the acid is between 10/9 ( 1.11) and 10/8 ( 1.25 ) times more
Concentrated. 8 drops is not quite enough base to neutralize the 10 base and the 9
drops is more than enough base to neutralize the 10 drops base
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A column cartridge is composed of a polar stationary phase.
Isopropyl alcohol is a relatively non-polar solvent.
Water is a polar liquid.
A maroon pigment is extracted using isopropyl alcohol.
The pigment is then injected into the column cartridge and the maroon pigment stays in the
cartridge and the eluent ( product) is clear.
A 5 % water 95 % isopropyl solution is then injected into the cartridge containing the maroon
extract. The column cartridge turns red and the eluent ( product ) is orange.
A 25 % water 75 % isopropyl solution is then injected into the cartridge containing the red
pigement. The column cartridge turns clear and the eluent ( product ) is red.
What conclusions can you make given the above information about the maroon pigment.
The maroon pigment is a mixture of orange and red pigments. The eluent
after adding the 5 % water and 95 % isopropyl alcohol to the maroon mixture in the
column’s polar stationary phase is orange and the eluent is red after adding the
25 % water 75 % isopropyl alcohol solution to the polar stationary phase after extracting
the orange pigment. The column is clear after the second extraction indicator that the
pigment contains two pigments.
The orange pigment is more non-polar than the red pigment. It was extracted from the
polar cartridge stationary phase with the more non polar mobile phase mixture of 5 %
water and 95 % isopropyl alcohol.
The red pigment is less non-polar or more polar than the orange pigment. It was
extracted from the polar cartridge stationary phase with the more polar mobile phase
mixture of 25 % water and 75 % isopropyl alcohol.
18. A solution of gold nitrate is obtained. Explain in detail using the following terms how the
Gold can be extracted from the solution.
Electrons, reduction, cathode, anode, oxidation, red (+) wire, black (-) wire, water, oxygen
Hydrogen ions.
Black lead – negative electrode – cathode - releases electrons
Negative electrons attract the positive gold ions of the gold nitrate solution
Gold metal forms as the electrons reduce the positive gold ions to metal atoms
The nitrate ions migrate to the positive electrode
At the positive electrode – red lead – anode water decomposes into oxygen, hydrogen
ions, and releases electrons through the red lead. The oxygen of the water has
Lost electrons and has undergone the process of oxidation.