Blue Team Science
Portfolio
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By: Taruni Singanamala
Scientific Method
1. Scientific Discoveries Presentation
2. Scientific Method Scavenger Hunt
Please visit the following websites, read carefully and respond to the questions.
Website 1: h ttp://www.biology4kids.com/files/studies_scimethod.html
Questions:
1. What is the scientific method?
The scientific method is a process used by scientists to study the world around them.
2. What sample questions are given that science can answer?
“Why do dogs and cats have hair?”, “What is that?”
3. How does science allow the world to “advance, evolve and grow?”
Science allows the world to advance, evolve, and grow through coming up with &
answering/finding answers. New discoveries can be made by investigating and finding proper
evidence.
4. What is the difference between inductive and deductive reasoning?
Inductive reasoning is reasoning where multiple “premises” that are found to generally be true,
are used to create a conclusion. Deductive reasoning is where the conclusion is based off of
multiple “premises” that are generally assumed to be true. Furthermore, inductive reasoning
works from the bottom-up, while deductive reasoning works from the top-down.
Website 2: http://phet.colorado.edu/sims/html/balancing-act/latest/balancing-act_en.html
Questions:
1. Make some changes to the Lever.
2. What are the variables that you can change?
Amount of items on each side, weight of the items, type of items/objects, placement of each item,
and how far each item is from each other.
3. Conduct a simple experiment and discuss your basic results.
The experiment I decided to conduct was to test if the distance each item was placed on the
lever would impact how easily everything would balance out. I started out by placing the two
heaviest weights, 80 kg, and 60 kg, each placed on the furthest opposite points on the scale,
which was about 2 meters. I added a 30 kg object about 1.5 meters on the left of the scale, but
then noticed left side was too heavy, so I added a 20 kg item .25 meters right of the lever, and
both of the items balanced out. To test out my theory about distance, I moved the 20 kg item to
the 1 meter mark on the right, and the lever instantly unbalanced. For some of the weights or
objects I placed on the lever, I had to calculate the distance specifically according to the meter
marks on it.
4. What were your observations?
Depending on how you distribute the weight, balancing items will be either easier/harder. For
example, when I moved the 20 kg object to the 1 meter mark, the lever unbalanced. Before moving
the item from .25 meters, the lever was perfectly balanced.
Website 3: h ttps://www.youtube.com/watch?v=OgS46ksAawk
Questions:
1. Describe her basic experiment.
The experiment tested what the best way was to keep apples from browning. Specifically, two
separate containers were removed of oxygen, each with a different method. One of the
containers was removed of oxygen using a burnt candle method, and the other container was
deoxidized using baking soda and vinegar. The apples were then placed inside of the containers,
and later on, were checked to see if the apples browned.
2. What are the variables?
Independent Variable: Method used to prevent oxidation in apples.
Dependent Variable: How much the apples brown.
3. Does she have Constants and a Control? Explain
Yes, there were Constants and a Control Group in the experiment. Some of the constants were
the size of the containers, around the same sized apple slice, aluminum coverings, and the amount
of time spent in the container. The Control Group was the apple slice in the container with no
deoxidation methods.
4. What were her results?
The Control Group had a little bit of browning near the outer edges. The apple slice placed near
the burnt candle was a little browner than the Control Group was but overall they looked pretty
similar. Again, the apple slice placed in baking soda and vinegar was about the same as the Control
Group, and may possibly have been slightly browner. In general, all 3 of the apple slices looked
about the same.
3. Scientific Method Practice
Fire Extinguishers
Problem: What is the most effective type of fire extinguishers? Or Which extinguisher puts out
the fire the fastest? (CO2, water, or dry chemical)
Hypothesis: If CO2 is used to put out fire, then the flames will be put out faster.
IV: The type of fire extinguisher used to put out the flame.
DV: How fast the fire is extinguished/put out.
Constants: Where the fire is, the climate of the area that the fire is in, what type of fire (forest
fire, oil fire, etc.), the elevation of the fire.
Control: Letting the fire die out on its own.
Paper Towel Absorbency Presentation
4. Scientific Method Quiz
Directions: Read the following description of an experiment and complete the components of the
scientific method.
Experiment:
Option #1: Patrick believed that fish would become smarter and complete a maze faster if they
ate food that was placed in a microwave first. He had 100 fish that he could use for the
experiment. He evaluated their intelligence based on their time to complete the maze.
Option #2: Mr. Smithers believed that Caffeine may make people more alert. Mr. Smithers tested
100 people by using their scores in the same video game. Devin had 3 different brands of drinks
with 10 g, 20 g, and 30 g of caffeine respectively. He measured their scores on a video game
that had a range of 0-1000 points. Some of the players were not given caffeine drinks. on the
game
*Help Mr. Smithers design an effective experiment and write a conclusion that analyzes your
results.
Problem Statement
Does coffee raise people’s scores when they play video games?
Hypothesis
If someone drinks coffee when playing video games, then they will get a higher score than
usual.
Independent Variable
Keurig Starbucks Folgers No Coffee
Dependent Variable
Whether or not the coffee helps the player’s score on the video game.
Constants (Pick 2)
The brand of coffee. Ingredients put in Video Game Amount of caffeine in
coffee and amount of coffee.
them (sugar, milk,
creamer, ice, etc.)
Control
The group of players that played video games with no coffee.
Basic Procedures:
(List 5-8 steps)
1. Give 1 brand of coffee to each group of the 4 groups: Group 1 gets Keurig, Group 2
gets Starbucks, Group 3 gets Folgers, and Group 4 gets no coffee.
2. Then, make 3 more groups for each brand of coffee: 3 groups for Keurig with group 1
with 10 g of caffeine, group 2 20g, group 3 30g. Same thing for Starbucks, and Folgers.
3. After the groups are all finished playing the video games, record the results.
4. Create a table.
5. Transfer data into a graph, and compare results.
6. Finally, create a conclusion about your experiment.
Data Table: (Place data table here)
Table #1:
Video Game Score
Brand of Coffee Total
Keurig 518
Starbucks 879
Folgers 634
No Coffee 692
Table #2:
Coffee Video Game Score
Keurig(10g) 200
Keurig(20g) 234
Keurig(30g) 658
Starbucks(10g) 302
Starbucks(20g) 491
Starbucks(30g) 799
Folgers(10g) 117
Folgers(20g) 230
Folgers(30g) 421
Graph: (Place graph here)
Graph #1
Graph #2
Conclusion (Purpose, Hypothesis, Description, Data or evidence, Improvements, Conclusion):
As you have noticed, the purpose of this experiment was to test whether or not caffeine (coffee)
helps people stay alert, specifically for higher results on video games. To test this, I based the
experiment off of my/Mr. Smithers’ hypothesis: If gamers drink a caffeinated drink (coffee), then
their video game score will be higher than usual. To conduct this experiment, there are a few
steps one must follow-
1. Give 1 brand of coffee to each group of 4 groups: Group 1 gets Keurig, Group 2 gets
Starbucks, Group 3 gets Folgers, and Group 4 gets no coffee.
2. Then, make 3 more groups for each brand of coffee: 3 groups for Keurig with group 1 with
10 g of caffeine, group 2 20g, group 3 30g. Same thing for Starbucks, and Folgers.
3. After the groups are all finished playing the video games, record the results.
4. Create a table.
5. Transfer data into a graph, and compare results.
6. Finally, create a conclusion about your experiment.
Moving on, my hypothesis was correct. Even if the improvement in the scores were low, all the
groups that drank coffee had a significant growth in their score. For instance, the Starbucks
group’s score improved by about 890 points, Folgers by about 620, and lowest out of all 3, Keurig
scored about 510 more. However, the group that drank no coffee gained about 690 points, which
means that this group was 2nd best after Starbucks.
Moreover, one improvement that could have been made to the experiment, is whether or not
each group was playing the same video game. This should have been in the constants, but since it
wasn’t included, the results are not accurate. Another constant that should have been included
was how often the gamers play video games.
All things considered, the hypothesis that if gamers drink a caffeinated drink (coffee), then their
video game score will be higher than usual, was overall accurate and a success.
5. Science Articles: Cassini Spacecraft
Article: h ttps://www.space.com/38153-cassini-huygens-saturn-mission-titan.html
“Habitable Titan? Cassini, Huygens Revealed Wonders of Saturn’s Biggest Moon” written by
Space.com contributor Ian O’Neill talks about the the Cassini spacecraft, which was on a mission
to discover “Titan” an Earth-like planet/moon. On October 15, 1997, the Cassini-Huygens mission
was launched by NASA, ESA, and the Italian Space Agency. Mainly, the point of the mission was to
land and explore Titan, and reveal important information about the planet. Many scientists have
been amazed and intrigued by Titan, therefore, the Cassini spacecraft was going to be an
important breakthrough in the scientific world. In fact, no other moon in the solar system has a
thick atmosphere like Titan. It seemed very Earth-like from a distance, due to it’s yellow/green
atmosphere. In order to explore this fascinating world, the Cassini spacecraft passed through the
thick atmosphere, while taking important photographs/visuals of Titan’s atmospheric layers and
chemical processes over a course of two and a half hours. As you can see, the Cassini Spacecraft
was important because it helped scientists and researchers learn more about Titan, a moon in the
solar system that seemed relatively similar to planet Earth.
Metric System and Density
1. Metric System Quiz
Lab Template
Mystery Objects
2. Density Lab Report
I. Investigation Design
A. Problem Statement:
How can density be used to identify unknown metals/objects?
B. Hypothesis:
If density is known, then unknown metals can be identified with further calculations.
C. Independent Variable: x
Levels of IV
Copper Bronze Zinc Tin Aluminum Brass
D. Dependent Variable:y
Density (g/cm3)
E. Constants:
Volume of Water Units Procedures
F. Control:
Water (the density of water is 1 g/mm3).
G. Materials: (List with numbers)
1. Graduated Cylinder
2. Dropper
3. Triple-Beam Balance
4. Measuring Cup
5. Water
6. Objects (8 metals)
7. Chromebook
8. Paper Towels
H. Procedures: (List with numbers and details)
1. First, place the metal on the scale to weigh it.
2. Measure out 50mL of water.
3. Then insert the metal in the water.
4. Find difference between 50 ml of water and the volume of the metal.
5. Divide the weight of the metal out by the volume to find density.
6. And lastly, put data into a table and create a graph.
II. Data Collection
A. Qualitative Observations:
https://docs.google.com/document/d/1EWq7aiq3hnGkQqbGkpn_PtoLGz_xgXlWwVfSYs5s66A/edit
B. Quantitative Observations: (Key data)
→ 1 . Data Table
Data Table 1:
Volume Before Volume After Volume Object
Object Mass (g) (mL) (mL) (cm3) Density (g/cm3)
A 68.3 50 58 8 8.5
B 267.3 N/A N/A 27 9.9
C 72.5 50 58 8 9.06
D 28.8 50 53 3 9.6
E 29 50 54 4 7.25
F 29 50 54 4 7.25
G 22 50 57 7 3.14
H 29.6 50 61 11 2.69
Data Table 2:
Unknown Objects Mass (g) Volume Before Volume After Volume Object Density (g/cm3)
1 28.7 50 53 3 9.6
2 29 50 54 4 7.25
3 267.2 N/A N/A 27 9.9
4 68.3 50 58 8 8.5
5 29.1 50 54 4 7.27
6 29.5 50 61 11 2.68
7 72.5 50 58 8 9.06
8 22 50 58 8 2.75
2. Graphs
Graph 1:
(g/cm3) she= Mass of Object (g)
Graph 2: she= Volume of Water
Before (mL)
Fgb= Density of Object (g/cm3)
she= Volume of Water
After (mL)
she= Volume of Object
(cm3)
Fgb= Density of Object
she= Mass of Object
(g)
she= Volume of Water
Before (mL)
she= Volume of Water
After (mL)
she= Volume of Object
(cm3)
3. Calculations
Show 3 Math Examples
Copper
D = m/v
D= 28.8 g
3 cm3
D = 9.6 g/cm3
Aluminum
D = m/v
D= 22 g
7 cm3
D = 3.14 g/cm3
Zinc
D = m/v
D= 29.1 g
4 cm3
D = 7.27 g/cm3
III. Data Analysis/Conclusion
As you have noticed, the purpose of this experiment was to match numbered metals with lettered
metals (labels) using density, and to eventually figure out what specific metals they are. As a
result, my group and I created a hypothesis: If the density is known, then unknown metals can be
identified with further calculations. To conduct the experiment, there were a few steps we had
to follow-
1. First, we took a labeled metal and put it on the triple-beam balance to weigh it.
2. We then measured out 50mL of water in a graduated cylinder.
3. After that, we slipped the metal into the water.
4. Next, we found the difference between 50 ml of water and the volume of the metal.
5. Then we divided the weight of the metal out by the volume to find density.
6. After doing this same procedure to all of the metals, we put the data into a table and
created a graph.
These mystery metals were first given to us with letter labels (A-H). When we had finished
putting in data with these metals, we were given the same metals, except they were labeled from
1-8, not in the same order as A-H. Our job was to then match A-H with the correct number using
the data we collected for both times, and then compare. According to our data, the density of
letter D came out as 9.6 g/cm3, as did #1. Hence, we concluded that D is the same as 1. However,
when we finished the experiment, we were also given the actual metal name of each of the ones
we got. For example, D and 1 turned out to be Copper. Other examples include A and 4, whose
density was 8.5 g/cm3. A and 4 turned out to be Brass (Alley).
All things considered, our hypothesis, if the density is known, then unknown metals can be
identified with further calculations, was overall accurate and the experiment was also an overall
success.
IV. Research and Applications
*How does Density relate to Plate Tectonics?
Density relates to plate tectonics because density plays a role in how the plates move and their
buoyancy (buoyancy: the ability or tendency to float in water, air, or any other fluid). Basically,
underneath the Earth’s crust is a thick layer of molten rock, and as the rock gets hotter in the
center, it moves up to the surface, and then eventually cools down and sinks to the middle again.
This is called convection and is how the plates on the crust move. The more dense (cool) the
molten rock is (the more buoyant), the more it sinks. Corresponding, the less dense (hot) it is, the
more it rises to the surface and moves the plates. These differences between the density due to
the temperature are only one example of how Plate Tectonics relate to Density.
V. References and Citations
● “Causes of Tectonic Plate Movement.” S tudy.com, Study.com,
http://study.com/academy/lesson/causes-of-tectonic-plate-movement.html
● “Introduction to Geological Sciences.” G eological Sciences 101,
www.geo.cornell.edu/geology/classes/Geo101/101week9_f05.html
3. Density Quiz
4. Scientific Method Test Review
1. If the density of water is 1 gram/cm3, this means that the mass of 100 cm3 of water
should be 1 00 grams
2. Wood floats in water. If you measured the mass of the same volume of wood and water
the wood would have a lower mass.
3. The density of hot and cold water are different mainly because the molecules in cold
water are packed together and move slowly, while the molecules in hot water are more
spread apart and move faster.
4. Calculate the density of sulfuric acid if 35.4 mL of the acid is 65.14 g.
D = m/v
D = 65.14 g/35.4 mL
D = 2305.956 g/cm3
The density of sulfuric acid is 2305.956 g/cm3.
5. The density of silver is 10.49 g/cm3. If a sample of pure silver has a volume of 27 cm3,
what would be its mass?
M = d * v
M = 10.49 g/cm3 * 27 cm3
M = 283.23 g
The mass of the silver is 283.23 grams.
6. A student finds a rock on the way to school. In the laboratory he determines that the
volume of the rock is 34.5 cm3, and the mass is 48.3 g. What is the density of the rock?
D = m/v
D = 48.3 g/34.5 cm3
D = 1.4 g/cm3
The density of the rock is 1.4 g/cm3.
7. A scientist conducted an experiment to determine how the amount of salt in a body of
water affects the number of plants that can live in the water. In this experiment the
dependent variable is t he number of plants that can survive
8. In an experiment, the one variable that is changed by the experimenter is called the
independent variable
9. A scientist who wants to study the effects of fertilizer on plants sets up an experiment.
Plant A gets no fertilizer, Plant B gets 5 mg. of fertilizer each day, and Plant C gets 10mg.
of fertilizer each day. Which plant is the control group? P lant A
10. Homer notices that his shower is covered in a strange green slime. Homer decides to
spray half of the shower with coconut juice thinking this will kill the slime. He sprays the
other half of the shower with water. After 3 days of "treatment" the green slime on the
coconut juice side of the shower dies. The dependent variable in his experiment is t he
amount of slime still left on the shower
11. A scientist plants two rows of corn for experimentation. She puts fertilizer on row 1 but
does not put fertilizer on row 2. Both rows receive the same amount of water and light
intensity. She checks the growth of the corn over the course of 5 months. What is a
constant in this experiment. Same amount of water and light intensity
12. A student hypothesized that the amount of sunlight a sunflower plant receives
determines the number of sunflower seeds the plant produces. In her experiment, the
number of seeds produces is the d ependent variable
13. Sarah wanted to find out if temperature has an effect on the growth of bread mold. She
grew the mold in nine Petri dishes containing the same amount and type of nutrients.
Three were kept at 0 C, three were kept at 90 C, and three were kept at room
temperature, 27 C. The containers were examined and the growth of the bread mold was
recorded each Friday for five weeks. Which of the following is her hypothesis? D epending
on the temperature, bread mold will grow differently.
14. 5.9 km = 5 900 m
15. 756.0 cg = 7 .56 g
16. 23,000 mm = 0 .023000 m
5. Scientific Method Test
Phase Changes
1. Phase Change of Water Activity
Directions:
● Melt the ice water and record the temperatures every 30 seconds until you reach the
boiling point of water.
● Record the temperatures on the following data table:
https://docs.google.com/a/cheshire.k12.ct.us/spreadsheets/d/1HtGu3B-HRWJxmGTVLMQY
gu1TbWx4N4uGlbo_A0Blfks/edit?usp=drive_web
C onstruct a graph of your results. *Use Link on Classroom
● Respond to the Critical Thinking Questions
Graph:
Critical Thinking Questions:
1. When did the temperatures stay the same on the graph? Why did the temperatures stay
the same at 2 points during the lab?
The temperature stayed the same during the first and last section of the graph, and this
is because it takes extra joules to convert one matter into a different one, which is why
we need to use formulas such as Heat of Fusion and Heat of Vaporization.
2. How would the graph be different if we tried this experiment with Gold? Explain:
The graph would be different because since the density of Gold would differ from
water, it takes a different amount of joules needed to change the states of gold.
3. Describe the motion of the molecules throughout the experiment. Find diagrams that show
the motion.
When a substance starts out solid, the molecules are tight, compact, and move only slightly.
But as more energy is added, the molecules start moving around more and this breaks the
solid structure, and thus turns into a liquid state, and as even more energy is increased, the
molecules separate completely and float around each other.
4. How does the Average Kinetic Energy change throughout the experiment? (Be specific)
The average kinetic energy slowly increases throughout each state of matter, changing
the substance’s arrange of molecules.
5. Suppose you had 200 mL of ice in one beaker and 400 mL of ice in another beaker.
Compare and explain the following in the beakers after they have reached the boiling point:
A.Heat Energy: The 400 mL of ice will contain more heat energy than the 200 mL of ice
because it takes more heat to warm up a larger amount of a substance.
B.Temperature: They will have the same temperature since the boiling point of ice is the same
no matter the amount.
C.Average Kinetic Energy: The average kinetic energy will be different because the mass of the
two beakers are different, and the 200mL will have less since it has less mass and therefore
takes less energy to reach the boiling point.
D.Specific Heat: The specific heat will also contrast because the two beakers have different
masses, with the 200 mL beaker taking less specific heat than 400mL because again, it has a
smaller mass
E. Latent Heat (Define it)
Latent heat is the energy released or absorbed during a constant heating or freezing
process
7. Why do we put water in a car’s engine? Explain.
W e put water in a car’s engine because it takes a lot of energy to heat up water, or more
specifically, it has lots of latent heat that is released while it is being heated up constantly,
so is useful for making a heavy object like a car move.
8. Crystallization of Moth Crystals
SH **Change in temperature is the difference
6. Phase Changes Quiz Review
Apply the following Equations:
Heat = Mass * Heat of Fusion
Heat = Mass * Change in Temperature *
between melting point and boiling point
*Always put answers in scientific notation
Heat = Mass * Heat of Vaporization
Data Table:
Metal Mass Heat of Melting Pt. Boiling Heat of Specific Heat
Heat Energy
Fusion (C) Pt. (C) Vaporization (cal/gC) (cal)
(cal/g) (cal/g) 1
Water 65 g 80 0 100 540 0.21
Aluminum 65 g 95 660 2467 2500 0.03
Gold 65 g 15 1063 2800 377
*SHOW ALL MATH STEPS
A. Aluminum
Heat = Mass * Hf usion
Heat = 65g * 95cal/g
Heat = 6175 cal
I melted the aluminum and now it’s a liquid
Heat = Mass * Change in Temperature * Specific Heat
Heat = 65g * Δ 1807 * 0.21cal/g
Heat = 24665.55 cal
Heat = Mass * Hvaporization
Heat = 65g * 2500cal/g
Heat = 162,500 cal
162,500 = 1.625 * 105 c al
B. Gold
Heat = Mass * Hfusion
Heat = 65g * 15cal/g
Heat = 975 cal
I melted gold and now it’s a liquid
Heat = Mass * Change in Temperature * Specific Heat
Heat = 65g * Δ 1737 * 0.03cal/g
Heat = 3387.15 cal
Heat = Mass * Hvaporization
Heat = 65g * 377cal/g
Heat = 24,505 cal
24,505 = 2.4505 * 104
C. Water
Heat = Mass * Hfusion
Heat = 65g * 80cal/g
Heat = 5200 cal
Heat = Mass * Change in temperature *Specific Heat
Heat = 65g * Δ 100 * 1cal/g
Heat = 6500 cal
Heat = Mass * Hvaporization
Heat = 65g * 540cal/g
Heat = 35,100 cal
35,100 cal = 3.51 * 104
Graph your Results:
Questions:
1. How are the substances different?
a. The substances are different because because they are made out of contrasting
molecular structures
2. What is the difference between Heat and Temperature?
a. Heat is the amount of energy that flows from one object to another, and
temperature is the degree or intensity of heat of an object
3. Place your Heat Energy results in Scientific Notation
a. Gold(2.4505 * 104)
b. Aluminum(1.625 * 105 c al)
c. Water(3.51 * 104 )
4. Why do metals have such low specific heats? How does this relate to Conductors?
a. Metals have low specific heats because it takes less energy for metals to
absorb heat and change the structure of its molecules, and this relates to
conductors because conductors are substances that absorb heat very well
without producing as much energy as substances that are not conductors.
5. How are Heat and Temperature different for the following pictures of boiling water?
Explain: (Hint: Use the Heat equation)
Heat and temperature are different for the following pictures of boiling water because
the picture with the boiling water ocean will have more heat and a higher temperature than the
boiling water contained in a beaker, because according to the heat equations, the more mass an
object has, the more heat it uses to reach it to the boiling point, and thus, the temperature and
heat must be higher/greater than something with less mass.
3. Phase Change Quiz
Calculate Heat Energy:
Apply the following Equations:
Heat = Mass * Heat of Fusion
Heat = Mass * Change in Temperature * SH
Heat = Mass * Heat of Vaporization
Data Table:
Metal Mass Heat of Melting Pt. Boiling Heat of Specific Heat
Heat Energy
Fusion (C) Pt. (C) Vaporization (cal/gC) (cal)
(cal/g) (cal/g)
Water 37 g 80 0 100 540 1 1.998 *
104 cal
Silver 37 g 26 961 2212 2356 0.057 8.7172 *
104 cal
Directions: Determine the Heat Energy required to completely evaporate the substances in the
data table.
*SHOW ALL MATH STEPS
A. Water
Heat = Mass * Hfusion
Heat = 37g * 80 cal/g
Heat = 2960 cal
Heat = Mass * Change in temperature *Specific Heat
Heat = 37g * Δ 100 * 1 cal/g
Heat = 3700 cal
Heat = Mass * Hvaporization
Heat = 37g * 540 cal/g
Heat = 19,980 cal
Scientific Notation: 19,980 cal = 1.998 * 104 cal
B. Silver
Heat = Mass * Hf usion
Heat = 37g * 26 cal/g
Heat = 962 cal
Heat = Mass * Change in temperature *Specific Heat
Heat = 37g * Δ 1251 * 0.057 cal/g
Heat = 2638.359 cal
Heat = Mass * Hv aporization
Heat = 37g * 2356 cal/g
Heat = 87,172 cal
Scientific Notation: 87,172 cal = 8.7172 * 104 cal
Graph your Results:
Questions:
1. How are Heat and Temperature different for the following pictures of boiling water? Explain:
(Hint: Use the Heat equation)
Heat and temperature are different for the following pictures of boiling water because the
picture with the boiling water ocean will have more heat and a higher temperature than the
boiling water contained in a beaker, because according to the heat equations, the more mass an
object has, the more heat it uses to reach it to the boiling point, and thus, the temperature and
heat must be higher/greater than something with less mass.
2. How can you use the unit (cal/gC) to explain the difference between Water and Silver?
The unit “cal/g” can be used to explain the difference between water and silver. For instance, the
specific heats of the two substances (measured in cal/g) show a difference in the amount of heat
needed in order to increase the temperature of the object/substance. The specific heat of water
is 1 cal/gC, whereas the specific heat of silver is 0.057 cal/gC. As seen in the data, this shows
that silver is a better conductor, and is easier to heat up than water. Generally, the specific
heats of water and silver help differentiate the two. Furthermore, each of the substance’s heat
of fusion and heat of vaporization (both measured in cal/g) are used to explain the differences
between silver and water. Water’s heat of fusion is 80 cal/g and its heat of vaporization is 540
cal/g. On the other hand, silver’s heat of fusion is 26 cal/g and its heat of vaporization is 2356
cal/g. Because of many factors such as specific heat, heat of fusion, and heat of vaporization, the
amount of total heat energy needed to evaporate the substances will come out as different
solutions, therefore, showing how water and silver are different. Given the facts above, the unit
“cal/g” (used in specific heat, heat of fusion, and heat of vaporization) can be used to explain the
difference between water and silver.
Classification of Matter
1. Classifying Matter Quiz Review
Research Heterogeneous and Homogeneous Mixtures and write down characteristics and
examples in the chart below:
Heterogeneous Mixtures Homogeneous Mixtures
Definition - any mixture that isn’t uniform in Definition - any mixture that is uniform in
composition (varies, visible differences) composition (non-visible differences,
consistent)
Characteristics - at least two or more
ingredients are mixed together but remain Characteristics - ingredients/substances are
physically separate mixed together and combine into one
undetectable phase
Examples - sand in water, oil and water
mixture, soil Examples - air, water, brass
Determine the Mass % of each component within the following Mixtures and Make Pie Charts:
25 grams of Large Rocks 36 grams of Fine Grained Sand
125 grams of Small Rocks 3 grams of Salt
75 grams of Coarse Grained Sand 19 grams of Copper (Cu)
Total Mass (g) = 283
Large Rocks = 25/283 = 0.08 * 100 = 8%
Small Rocks = 125/293 = 0.44 * 100 = 44%
Coarse Grained Sand = 75/283 = 0.27 * 100 = 27%
Fine Grained Sand = 36/283 = 0.13 * 100 = 13%
Salt = 3/283 = 0.01 * 100 = 1%
Copper = 19/283 = 0.07 * 100 = 7%
23 grams of Fine Grained Sand
175 grams of Large Rocks
35 grams of Small Rocks 11 grams of Salt
89 grams of Coarse Grained Sand 53 grams of Copper (Cu)
Total Mass (g) = 386
Large Rocks = 175/386 = 0.45 * 100 = 45%
Small Rocks = 35/386 = 0.09 * 100 = 9%
Coarse Grained Sand = 89/386 = 0.23 * 100 = 23%
Fine Grained Sand = 23/386 = 0.06 * 100 = 6%
Salt = 11/386 = 0.03 * 100 = 3%
Copper = 53/386 = 0.14 * 100 = 14%
Determine the Mass % of each element in the following compounds: (Choose 4 Compounds)
Positive Ions Negative Ions
Sodium +1 Phosphate PO4 -3
Calcium +2 Carbonate CO3-2
Potassium +1 Sulfate SO4 -2
Lithium +1 Nitrate NO3 -1
1. NaPO4
Na(1) = 23/118 = 0.2 * 100 = 20%
P(1) = 31/118 = 0.26 * 100 = 26%
O(4) = 64/118 = 0.54 * 100 = 54%
2. CaCO3
Ca(1) = 40/100 = 0.4 * 100 = 40%
C(1) = 12/100 = 0.12 * 100 = 12%
O(3) = 48/100 = 0.48 * 100 = 48%
Conclusion: *Explain the difference between Mixtures and Compounds using evidence (Data) from
your charts.
Mixtures and compounds are different classifications of matter. All components in a mixture don’t
chemically react and have properties that aren’t combined in a specific ratio. On the other hand, a
compound is a molecule made of 2 or more elements. Elements in compounds are chemically
combined in a specific ratio. A compound is a substance (constant composition), whereas a mixture
is either heterogeneous (particles unevenly distributed) or homogeneous (particles evenly
distributed). As the data shows, the heterogeneous mixture of rocks and sand were made of
different compounds. Copper (Cu) and Salt (NaCl) were two of the compounds that comprised of
the mixture. The mixture composed of various rocks, sand, salt, and copper. Copper and Salt (two
compounds) can only be separated chemically. As a whole, the entire mixture of rocks and sand
can be picked apart and physically separated easily, but individual components inside of the
mixture (such as copper and salt) can only be broken through a chemical process. All in all,
mixtures and compounds differ from each other.
*How did you separate the Salt from the Sand? Discuss the role of Solute and solvent as well as
Heat Energy. You should also discuss IONS.
In the experiment, the water played the role of the solvent, which is the liquid that the solute
dissolved in and the sodium chloride played the role of the solute, which is the substance being
dissolved. The mystery compound that was discovered at the end of the experiment was salt. The
positive ion was the sodium and the negative ion was the chloride. Once the water evaporated,
there was nothing left to separate the sodium and chloride, therefore they had to combine,
making the salt visible.
2. Classifying Matter Quiz
I. Directions: Identify the following as either a Heterogeneous Mixture, Homogeneous Mixture,
Element or Compound. Write the following letters in Column B for your choices:
A. Heterogeneous
B. Homogeneous
C. Element
D. Compound
Column A Column B
Salad A
Copper C
Lemonade B
Rocks, sand, gravel A
Salt Water B
Gold C
Sodium Chloride ( NaCl) D
Air (Oxygen, nitrogen, carbon monoxide…) B
K2S O4 D
Twix, snickers, pretzels, popcorn in a bag A
II. Directions: Determine the Mass % of each mixture and construct the appropriate graphs.
Mixture A Mass (g) %
Large Rocks 125 52%
Small Rocks 75 31%
Coarse Sand 32 13%
Iron 9 4% %
53%
Mixture B Mass (g)
205
Large Rocks
Small Rocks 58 15%
Coarse Sand 97 25%
Iron 29 7%
Calculation Examples ( Provide 2 Examples showing how you determined the Mass %)
125+75+32+9 =241
241- Total Mass (g)
↓
Large Rocks: 125/241= 0.518 * 100 = 52%
Iron: 9/241= 0.0373 * 100 = 4%
Graphs:
Mixture A
Mixture B
PartIII. Determine the Mass % of Elements in each Compound:
K2 SO4 - Potassium Sulfate
K(2)39 = 78/174 = 0.448 * 100 = 45%
S(1)32 = 32/174 = 0.183 * 100 = 18%
O(4)16 = 64/174 = 0.367 * 100 = 37%
Na3 P O4 - Sodium Phosphate
Na(3)23 = 69/164 = 0.420 * 100 = 42%
P(1)31 = 31/164 = 0.189 * 100 = 19%
O(4)16 = 64/164 = 0.390 * 100 = 39%
Graphs:
IV. Conclusion:
1. Explain the difference between Mixtures and Compounds using data. Compare the pie charts.
Mixtures and compounds are different classifications of matter. All components in a mixture don’t
chemically react and have properties that aren’t combined in a specific ratio. On the other hand, a
compound is a molecule made up of 2 or more elements. Elements in compounds are chemically
combined in a specific ratio. A compound is a substance (constant composition), whereas a mixture
is either heterogeneous (particles unevenly distributed) or homogeneous (particles evenly
distributed). As the data shows, due to the contrast in their total mass and mass percentages on
the pie charts, the rock lab and the compound lab are different because they are, as I said
before, two different classifications of matter; compounds and mixtures.
2. E xplain how you separated the Salt from the Sand. Use as much new vocabulary as you can.
We filtered the sand with water to get a mixture with a mystery substance. We then took this
mixture and boiled it on the hot plate. After the water had evaporated, the mystery substance
was revealed; Sodium Chloride(Salt). In the experiment, the water played the role of the solvent,
which is the liquid that the solute dissolved in and the sodium chloride played the role of the
solute, which is the substance being dissolved. As you know, the mystery compound that was
discovered at the end of the experiment was salt. The positive ion was the sodium and the
negative ion was the chloride. To be more specific about the process, once the water evaporated,
there was nothing left to separate the sodium and chloride, therefore they had to combine,
making the salt visible.
Solubility
1. Solubility Graph Practice
Directions: C onstruct a solubility graph that contains 5 substances from the chart. (Temperature
on X-axis and Solubility on Y-axis)
Salt Solubility Data*
Salt Name Chemical Tempe
Formula rature
(○C )
0 10 20 30 40 50 60 70 80 90 100
Ammonium NH4 Cl 29.4 33. 37.2 ? 45.8 ? 55.2 ? 65.6 ? 77.3
Chloride 3
Potassium KNO3 13.9 21. 31.6 45.3 61.4 83.5 106. ? ? ? ?
Nitrate 2 0
Sodium NaNO3 73 ? 87.6 ? 102 ? 122 ? 148 ? 180
Nitrate
Barium Ba(OH)2 1.67 ? 3.89 ? 8.22 ? 20.9 ? 101. ? ?
Hydroxide KCl 4 4
Potassium LiCl
Chloride K2 SO4 28.1 31. 34.2 ? 40. ? 45.8 ? 51.3 ? 56.3
Lithium 2 0 ? 98.4 ? 112 ? 128
Chloride
Potassium 69.2 ? 83.5 ? 89.
Sulfate 8
7.4 9.3 11.1 13.0 14.8 16.5 18.2 19. 21.4 22.9 24.1
8
Sodium NaCl 35.7 35. 36.0 36.2 36.5 36.8 37.3 37. 38.1 38.6 39.2
Chloride CuSO4 8 6
Copper (II) KI 14.3 17. 20.7 24.2 28.7 33.8 40. 47. 56.0 67.5 80.
Sulfate 4 0 0 0
(A nhydrous) 128 ? 144 ? 162 ? 176 ? 192 ? 206
Potassium * Solubility values are given in grams of salt per 100 grams of water
Iodide
Critical Thinking Questions:
1. How does the solubility of NaCl vary with the temperature of the water? E xplain using
your data and your graph.
As I’ve noticed through the data and graph, the solubility of NaCl (salt) varies depending on the
temperature. In the beginning, while the temperature of water is at 0°C, the solubility of the salt
is 35.7g. As the temperature increases to 10°C, the solubility is now 35.8g. In addition, the
solubility of salt in water heated to 50°C was 36.8g. Lastly, as the temperature increases to
100°C, the solubility changes to 39.2g. Although the difference between solubility was slight, it
still proved that the solubility was altered as water temperatures changed up/down.
2. What generalization can you make about the relationship between solubility and
temperature? P rovide Evidence (Data)
I’ve noticed that as temperature increases, so does the solubility in most substances. With this in
mind, substances such as ammonium chloride, potassium nitrate, sodium nitrate, barium hydroxide,
potassium chloride, lithium chloride, potassium sulfate, sodium chloride, copper sulfate, and
potassium iodide (all substances in the data table above) have solubilities that increase as the
temperature rises. Potassium sulfate becomes more soluble as temperatures increase.
Approximately 7.4g of potassium sulfate dissolve in 0°C water. At about 50°C, 16.5g of potassium
sulfate can be fully dissolved. As the data shows, the solubility of potassium sulfate increased as
the temperature was raised by 50°C.
3. Estimate the solubility of each salt at certain temperatures by filling in the following table.
Use your graph to determine the solubilities.
Temper
Salt Name ature
(○C)
5 15 25 35 45 55 65
Potassium Sulfate 8.5 10 11.5 13 15 16 18.2
Sodium Chloride 35.8 36 36.2 36.3 36.6 38 38.2
Copper (II) Sulfate 15 18 22.3 25.5 30 35 43.7
(Anhydrous)
Solubility Curve Practice Problems Worksheet 1
You'll notice that for most substances, solubility increases as temperature increases. As discussed
earlier in solutions involving liquids and solids typically more solute can be dissolved at higher
temperatures. Can you find any exceptions on the graph?
NH3 and Ce2 (SO4) 3 decrease as the temperature increases. Everytime time temperature increases,
the solubility of those substances curves down (decreases).
Here's an example of how to read the graph. Find the curve for KClO3 .
At 30°C approximately 10g of KClO3 will dissolve in 100g of water. If the temperature is
increased to 80°C, approximately 40g of the substance will dissolve in 100g (or 100mL) of water.
Directions: Use the graph to answer the following questions. REMEMBER UNITS!
1) What mass of solute will dissolve in 100mL of water at the following temperatures?
a. KNO3 a t 70°C = 1 30g
b. NaCl at 100°C= 4 0g
c. NH4C l at 90°C= 7 0g
d. Which of the above three substances is most soluble in water at 15°C. = NaCl
2) Types of Solutions
On a solubility curve, the lines indicate the concentration of a s aturated solution - the maximum
amount of solute that will dissolve at that specific temperature.
Values on the graph u nder a curve represent unsaturated solutions - more solute could be
dissolved at that temperature.
Label the following solutions as saturated or unsaturated. If unsaturated, write how much more
solute can be dissolved in the solution.
Solution Saturated or Unsaturated? If unsaturated: How much
more solute can dissolve in
a solution that contains 70g of Unsaturated the solution?
NaNO3 at 30°C (in 100 mL H2O)
26g
a solution that contains 50g of Unsaturated 2g
NH4 C l at 50°C (in 100 mL H2 O)
a solution that contains 20g of Unsaturated 1g
KClO3 at 50°C (in 100 mL H2 O)
a solution that contains 70g of Unsaturated 57g
KI at 0°C (in 100 mL H2 O )
Homework – Use the Solubility Graphs on Page 1
1. a. What is the solubility of K Cl at 5°C? 2 8g
b. What is the solubility of K Cl a t 25°C? 34g
c. What is the solubility of C e2( SO4 )3 at 10°C? 1 4g
d. What is the solubility of C e2 (SO4)3 at 50°C? 5g
2. a. At 90°C, you dissolved 10 g of KCl in 100. g of water. Is this solution saturated or
unsaturated? U nsaturated
b. How do you know?
I know that this solution is unsaturated because it is under the KCL curve.
3. A mass of 100 g of NaNO3 is dissolved in 100 g of water at 80ºC.
a) Is the solution saturated or unsaturated? Unsaturated
b) As the solution is cooled, at what temperature should solid first appear in the solution?
Explain.
The first solid should first appear in the solution at 35 degrees celsius. I know this because the
point the curve is at is 35°C.
4. Use the graph to answer the following two questions:
Which compound is most soluble at 20 ºC? KI
Which is the least soluble at 40 ºC? Ce2( SO4)3
5. Which substance on the graph is least soluble at 10°C? K ClO3
6. A mass of 80 g of KNO3 i s dissolved in 100 g of water at 50 ºC. The solution is heated to 70ºC.
How many more grams of potassium nitrate must be added to make the solution saturated?
Explain your reasoning (S ee question #2 on the other side for a hint)
50 more grams are needed to make the solution saturated at 70 degrees celsius (130g).
3. Solubility and Naming Compounds Quiz
Charge
Part I.
Directions: Write the symbol of the element with the charge.
Formula
1. Sodium Na +1
2. Neon Ne 0
3. N itrate Nitrogen N -3
4. Chlorine Cl -1
5. Magnesium Mg +2
6. Silver Ag +1
7. Sulfur S -2
8. Phosphorus P -3
9. Aluminum Al +3
10. Calcium Ca +2
Part II.
Directions: Write the name for the compounds:
11. Na3PO4 Sodium Phosphate
12. Li2 (SO4 ) Lithium Sulfate
13. (NH4 ) 2 CO3 Ammonium Carbonate
14. MgCl2 Magnesium Chloride
15. Ca(NO3)2 Calcium Nitrate
16. BeF2 Beryllium Fluoride
Part III.
Directions: Write the chemical formula for the following compounds (Use your ions):
17. Calcium carbonate
Ca+ 2 + CO3- 2 = CaCO3
18. Ammonium phosphate
NH4+ 1 + PO4-3 = (NH4 )3PO4
19. Magnesium hydroxide
Mg+ 2 + OH-1 = MgOH2
20. Potassium sulfate
K+ 1 + SO4 -2 = K2SO4
Part IV.
Directions: Determine the Mass % of Oxygen in Al2 ( SO4 ) 3 or A gNO3
Atomic Mass: A l (27) S (32) O (16) Ag (108) N (14)
Ag (1) = 108/170 = 0.64 * 100 = 64%
N (1) = 14/170 = 0.08 * 100 = 8%
O (3) = 48/170 = 0.28 * 100 = 28%
The mass percentage of oxygen is 28%.
Part V.
Directions: Write an essay about the graph below. U se data!
Vocabulary: Unsaturated, saturated, supersaturated, Ions, Heat, Temperature, grams, solubility,
chemical formula
I decided to pick barium nitrate, Ba(NO3)2, as my compound to help me analyze the graph given:
Barium Nitrate [Ba(NO3)2] is a salt powder composed of barium and the nitrate ion. As you
can see from the graph, a solution of 25 grams is saturated at 15℃, and to make this
supersaturated, the amount of solute in the solution can be increased, or the temperature level
can be decreased. In this case, the solute is barium nitrate, so an increase in the amount of 25
grams to 50 grams and a decrease in temperature like from 15℃ to 9℃ will make the solution
supersaturated. Another example of this is a solution of 70 grams of Ba(NO3)2 that is saturated
at 38℃. This specific amount is also supersaturated at a change of temperature to 15℃ or a
change in mass to 85 grams. Moving on, att 15℃, there will be 45 grams of Ba(NO3)2 at the
bottom of the beaker. Given the points above, there are many observations that can be made
from the solubility graph given.
Chemical Reactions
1. Law of Conservation of Mass Presentation