Green Science Portfolio 2016 -
2017
Evan Levine
Survey Graph Conclusion
1. Data Table
2. Graph
3. Write a short conclusion of 5 sentences
Sports Responses
Soccer
Basketball 4
Baseball 3
Other 2
9
The purpose of this graph and data collection was to find out what my peers
favorite sport was. The responses for the graph and the data where, soccer was 4 of my
peers favorite sport, basketball was 3 of my classmates favorite sport, baseball was 2 of
my classmates favorite sport, and 9 of my peers had other favorite sports that were not
in the graph. Most of my classmates had another sport that was their favorite. Baseball
was the least liked sport by all of students. Soccer and basketball were in between the
most and least liked sports. In conclusion, almost half of my classmates had another
favorite besides the ones listed on the graph.
Experiment Presentation
n
Quiz: Scientific Method
Directions: R ead the following description of an experiment and complete the
components of the scientific method.
Experiment: Mr. Smithers believes that a special compound could help his workers
produce more “widgets” in one week. The chemical supply store sent him 3 different
compounds to try on his 100 workers. The following are the chemicals:
A. Sodium chloride
B. Magnesium hydroxide
C. Calcium sulfate
D. Water
*Help Mr. Smithers design an effective experiment and write a conclusion that analyzes
your results.
Problem Statement
What compound will be the most effective?
Hypothesis
If sodium chloride is tested then it will make the workers produce more widgets in one week.
Independent Variable
Sodium chloride Magnesium Calcium sulfate Water
hydroxide
Dependent Variable
How many widgets are produced in one week
Constants (Pick 2) Same amount of time to make the widgets
Same amount of liquid used
Control
Water
Basic Procedures:
(List 5-8 steps)
1. First, he has to divide up his workers into four groups. One for each one of the
compounds in which will be 25 people per group
2. Next, he has to divide up the compounds to make each of them has the same amount
(50 ml of the compounds per day)
3. After that, they let the workers work to seed how many widgets they make in a week's
worth of time
4. Then, Smithers has to collect the data
5. Finally, Smithers will make a data table, graph, and make a conclusion
Data Table: (Place data table here)
Compounds Results
Sodium 130 Widgets
chloride
Magnesium
hydroxide 100 Widgets
Calcium 180 Widgets
sulfate
Water 80 Widgets
Graph: (Place graph here)
Conclusion:
In conclusion, my hypothesis was incorrect because Calcium sulfate was the most
effective instead of Sodium chloride. Calcium sulfate was the most effective with 180 widgets
made. Water was the least effective with only 80 widgets made. Finally, Magnesium hydroxide
and Sodium chloride with 130 widgets and 100 widgets made. In all, Mr.Smithers should use
Calcium sulfate to make more widgets.
Density Lab Report
Name - Evan Levine
Class - Science7
Teacher - Lopez
Date - 9/19/16
Investigation Title:
Identifying Unknown Metals
I. Investigation Design
A. Problem Statement:
How can density help me identify unknown metals?
B. Hypothesis:
If density is known then I can identify the unknown metals
C. Independent Variable: x
Levels of IV - Types of Metals
Aluminum Iron Brass Copper Tin Zinc Aluminum
AI AL
D. Dependent Variable:y
What is the density of the metals?
E. Constants: Same materials used to find Same materials used to find
Same amount of water used volume mass
F. Control:
Water
G. Materials: (List with numbers)
1. Graduated cylinder
2. Dropper
3. Triple Beam Balance
4. Beaker
5. Ruler
H. Procedures: (List with numbers and details)
1. First, my group and I got all of the materials we needed in order to identify the materials
2. Second, we got filled the graduated cylinder to 50 milliliter so we could find the volume of
the unknow metals
3. Next, we measured the metals on the triple beam balance to find the mass of them
4. Then, we used the mass and volume of the unknown metals to find the density
5. Once, we had the data we made a data table to organize the data
6. After that, we used our data to identify the metals
7. From there, we used the data we used to identify the data into a graph
8. Finally, we will make a conclusion about the data individually
II. Data Collection
A. Qualitative Observations:
1. All of the metals were different shapes and sizes
a. Aluminum AI - Long, gray, cylinder, heavy
b. Iron - Orange and brown, short, rectangular
c. Brass - Brownish gold, rectangular, medium sized
d. Copper - Brownish, rectangular, medium
e. Tin - Medium, gray
f. Zinc - Small, gray, rectangular
g. Aluminum AL - Big, cubed, heavy
B. Quantitative Observations: (Key data)
1. Data Table
Object Volume Volume Density
Aluminum AI Mass Before Volume Object (g/cm3)
Iron (g) (mL) After (mL) (cm3) 4 4.88
Brass 50 54 8 7.88
Copper 19.5 50 58 89
Tin 63 50 58 3 9.33
68 50 53 4 7.25
28 50 54
29
Zinc 28.5 50 54 4 7.13
Aluminum AL 29
50 61 11 2.64
2. Graph
3. Calculations
Show 3 Math Examples
Copper
D = m/v
D= 27 g
3 cm3
D = 9 g/cm3
1. Aluminum AI
a. D = m/v
b. 19.5 g
4cm3
c. D = 4.88 g/cm3
2. Iron
a. D = m/v
b. 63 g
8cm3
c. D = 7.88 g/cm3
3. Brass
a. D= m/v
b. 68 g
8cm3
c. D = 8.5 g/cm3
III. Data Analysis/Conclusion
In conclusion, my hypothesis was correct because density can help us identify unknown
metals. Copper’s density was the highest with 9.33g/cm3 . Aluminum AI’s density was the
smallest with 2.64g/cm3. All, of the other metals had densities that were in between 9.33 and
2.64g/cm3 . In conclusion, density will help you identify unknown metals.
IV. Research and Applications
Plate Tectonics
1. How do the plates move? 1
a. Convection
i. When hot rocks rise and cold rocks fall
2. How does Density relate to the motion of the plates above the Earth’s mantle? 2
a. This article says that there is more movement of plates that are less dense
3. How are the Continental and Oceanic Plates different?
a. The Continental crust is less dense than the Oceanic crust
4. How does Density relate to the collisions of the Earth’s plates? 3
a. The denser Oceanic Crust will sink and the Continental Crust will stay up
How does plate tectonics relate to density? Plate tectonics are the movement of
plates, Continental and Oceanic. The less dense plates, Continental, move more easily
than the more dense Oceanic. There are different results due to the different plates that
collide. When an Oceanic and Continental plate collide it creates a trench and a
volcano. This happens because when the Oceanic plate sinks it cuts into the ocean
floor which creates a the trench and then a subduction zone is built which causes a
volcanic mountain chain forms above the subduction zone which makes volcanic
mountains on land. When two Continental plates collide it creates mountains because
neither one of the plates can sink so they move upward. Finally, when two Oceanic
plates collide it creates a trench and a volcano like before but the volcano ends up on
land. In conclusion, density is very important to tectonic plates because it causes
different reactions that help shape Earth.
1 "How do plates move? - The Geological Society." 2013. 25 Sep. 2016
<https://www.geolsoc.org.uk/Plate-Tectonics/Chap2-What-is-a-Plate/Plate-Movement>
2 "Understanding plate motions [This Dynamic Earth, USGS]." 2006. 22 Sep. 2016
<h ttp://pubs.usgs.gov/gip/dynamic/understanding.html>
3 "Convergent Plate Boundaries - USGS Geology in the Parks." 2006. 25 Sep. 2016
<http://geomaps.wr.usgs.gov/parks/pltec/converge.html>
V. References and Citations
● 2 or 3 web links
Density QUIZ
1. The scientist collected an object with a density of 6.4 g/cm3 and a
volume of 79 cm3. What is the mass of this object?
M = DV
M = 6.4 g/cm3 (79cm3)
M = 505.6 g
2. An irregularly shaped stone was lowered into a graduated cylinder
holding a volume of water equal to 50.0mL. The height of the water rose
to 68 mL. If the mass of the stone was 125.0g, what was its density?
D= M
V
125g
D= 18ml
D = 6.4 g/cm3
3. A scientist had 350.0 grams of Gold (Au) and a 530.0 gram sample of
Silver on the lab table. Which metal would have a greater volume
(cm3 )? Explain. *Show all work.
Gold Silver
GOLD
V= M
D
350g
V= 19.32 g/cm3
V = 18.12 cm3
Silver
V = M
D
530g
V= 10.49 g/cm3
V = 50.52 cm3
Silver has a higher density than gold because the mass is higher and
the density is lower which makes the volume higher.
4. Explain why the Titanic sank after hitting the iceberg. Use data to
explain your answer.
The Titanic sank after hitting the iceberg because, since the mass
was so big the water started to fill up the mass increasing the volume which
made the density increase causing the Titanic to sink.
Scientific Method Presentation
Phase Change of Water Lab
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:
Construct a graph of your results. *U se Link on Classroom
● Respond to the Critical Thinking Questions
Graph:
Critical Thinking Questions:
1. Why did the temperatures stay the same at 2 points during the lab?
It stayed the same because of the phase change between ice to water and water
to vapor
2. How does this relate to the heat trapped in the atmosphere? Find a
diagram that illustrates this concept.
This relates to heat trapped in the atmosphere because when there is water in
atmosphere and it gets heated up it forms rain.
3. What is the role of energy during the phase changes?
The role of energy is important because it changes the temperature in order for
the phase change to happen.
4. Describe the motion of the molecules throughout the experiment.
The motion of the molecules increases which created the different forms of water.
5. How does the Average Kinetic Energy change throughout the experiment?
The Average Kinetic Energy changed throughout the experiment because when
the particles increased, it increased the heat. This changes throughout the
experiment because when the phase change happens the Average Kinetic Energy
slows down.
6. Suppose you had 200 mL of ice in one beaker and 400 mL of ice in another
beaker. Compare the following in the beakers after they have reached the
boiling point:
A. Heat Energy - The 400 mL of ice beaker will have a bigger Heat Energy
because it has more molecules that will move than the 200 mL of ice that
has less molecules.
B. Temperature -
C. Average Kinetic Energy
D. Specific Heat
E. Latent Heat
Phase Change Lab Correct
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:
Construct a graph of your results. *U se 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 temperatures stayed the same during the phase change because
during the phase change the temperature stays the when the water changes from
ice to water and water to vapor.
2. How would the graph be different if we tried this experiment with Gold?
Explain:
The graph would be different because gold would require much more heat
due to different melting points. Also, gold is more dense than water.
3. What is the role of energy during the phase changes?
Energy plays important because it changes the temperature in order for the
phase change to happen.
4. Describe the motion of the molecules throughout the experiment. Find
diagrams that show the motion.
The motion of the molecules increases which created the different forms of
water.
5. How does the Average Kinetic Energy change throughout the experiment?
The Average Kinetic Energy changed throughout the experiment because
when the particles increased, it increased the heat. This changes throughout the
experiment because when the phase change happens the Average Kinetic Energy
slows down.
6. 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 - T he 400 mL of ice beaker will have a bigger Heat Energy
because it has more molecules that will move than the 200 mL of ice that
has less molecules.
B. Temperature - The temperature will be the same in both because in order
for the water to evaporate the water was to be at 100 C°
C. Average Kinetic Energy - B oth of the beakers will require the same Average
Kinetic Energy because they both need the same for them to has a phase
change.
D. Specific Heat - The specific heat will be the same because both beaker are
filled with the same substance water
E. Latent Heat - 2 00 mL will have a higher Latent Heat because it has less
molecules to melt ice into water and water into vapor
Quiz: Phase Changes
Directions: A nalyze the following data table with data collected by a scientist that wanted to
study how Heat Energy affects the Phase Changes of 2 different metals. Respond to the
questions below and perform all necessary calculations.
Data Table:
Metal Mass Heat of Melting Boiling Heat of Specific Heat
Fusion Pt. (C) Pt. ( C) Vaporization Heat Energy
(cal/g) (cal/gC) (cal)
(cal/g)
Aluminum 65 g 95 660 2467 2500 0.21 193340.55
Gold 65 g
15 1063 2800 377 0.03 28867.15
Scientific Method (__4_ out of 4)
Independent Variable:
The Metals being tested; Aluminum and Gold
Dependent Variable:
Heat needed to complete the phase changes
Constant:
The Mass
Control:
Water
Calculate Heat Energy: * SH
Apply the following Equations: Boiling Heat of
Heat = Mass * Heat of Fusion Pt. ( C) Vaporization
Heat = Mass * Change in Temperature
Heat = Mass * Heat of Vaporization (cal/g)
Data Table:
Metal Mass Heat of Melting Specific Heat
Fusion Pt. (C) Heat Energy
(cal/g) (cal/gC) (cal)
Aluminum 65 g 95 660 2467 2500 0.21 193340.55
Gold 65 g 15 1063 2800 377 0.03 28867.15
*SHOW ALL MATH STEPS
Math Steps (__4__ out of 4)
A. Aluminum
Heat = Mass * Heat of Fusion
Heat = 65 g * 95 cal/g
Heat = 6175 calories
Heat = Mass * Change in Temperature * SH
Heat = 65 g * (2467 C - 660 C) * .21 cal/gC
Heat = 65g * 1807 C * .21 cal/gC
Heat = 24665.55
Heat = Mass * Heat of Vaporization
Heat = 65g * 2500 cal/g
Heat = 162500 calories
Total Heat Energy = 193340.55 Calories
B. Gold
Heat = Mass * Heat of Fusion
Heat = 65g * 15 cal/g
Heat = 975 calories
Heat = Mass * Change in Temperature * SH
Heat = 65g * (2800 C - 1063 C) * .03 cal/gC
Heat = 65g * 1737 C * .03 cal/gC
Heat = 3387.15 calories
Heat = Mass * Heat of Vaporization
Heat = 65g * 377 cal/g
Heat = 24505 calories
Total Heat Energy = 28867.15 Calories
Graph your results (__4__ out of 4):
Write a Conclusion (____ out of 4):
In conclusion, Aluminum will need more Heat Energy to complete the process of
going through the phase change. Aluminum needs 193340.55 calories of heat while gold
only needs 28867.15 calories. All in all, a lot more Heat Energy is required for Aluminum
to go through the phase change than gold.
Questions:
1. How are Heat and Temperature different for the following pictures of boiling w ater?
Explain: (Hint: Use the Heat equation)
The heat and temperature are different for the pictures because the picture of the
ocean has a bigger mass than the picture of the beaker full of water. The ocean will
require more Heat Energy than the water in the beaker because of the bigger mass.
2. Water has a Specific Heat of 1.0 cal/gC and Gold has a Specific Heat of 0.03 cal/gC.
Use the data to explain the difference between their numbers.
Boiling Point and Elevation Presentation
Mass% Practice
Activity: Mass % Practice with Mixtures and Compounds
1. A scientist recorded the following data about a sample of rocks and sand:
37 grams of Large Rocks 75 grams of Fine Grained Sand
59 grams of Small Rocks 5 grams of Salt
125 grams of Coarse Grained Sand 25 grams of Copper (Cu)
2. D etermine the % of each component in this Heterogeneous Mixture and construct a pie
chart showing your results.
3. Data Table:
Mixture Grams %
Component
Large Rocks 37g 11.3
Small Rocks 59g 18
Coarse 38.3
Grained Sand 125g
Fine Grained
Sand 75g 23
Salt 5g 1.5
Copper 25g 7.7
Total 326g 100
4. Pie Chart:
5. Math Examples
Large Rocks -
37/326
.1134… * 100
11.3%
Large Rocks -
59/326
.180… * 100
18%
Coarse Grained Sand
125/326 = .3834… * 100
38.3%
____________________________________________________________________________
1. A second scientist recorded the following data about a different sample of rocks and
sand:
48 grams of Large Rocks 175 grams of Fine Grained Sand
78 grams of Small Rocks 2 grams of Salt
56 grams of Coarse Grained Sand 17 grams of Copper (Cu)
2. Determine the % of each component in this Heterogeneous Mixture and construct a pie
chart showing your results.
3. Data Table:
Mixture Grams %
Component
Large Rocks 48g 12.7
Small Rocks 78g 20.7
Coarse 14.8
Grained Sand 56g
Fine Grained
Sand 175g 46.5
Salt 2g 0.5
Copper 17g 4
Total 376g 100
4. Pie Chart:
5. Math Examples
Large Rocks
48/376
.1276… * 100
12.8%
Small Rocks
78/376
.2074… * 100
27.7%
Coarse Grained Sand
56/376 = .1489… * 100
14.9%
1. A third scientist received a 250 gram sample of Silver Nitrate - A gNO3
2. Chart for Mass % of a Compound
Questions:
1. How are the samples from these scientists different?
The samples are different because they all have different masses.
2. How are Compounds different from Heterogeneous Mixtures? Provide evidence.
The Compounds are different from Heterogeneous Mixtures because they can’t be separated. A
Heterogeneous Mixture is a mixture that can be separated like ice in a soda glass, but a
homogeneous mixture, like the Compounds in the experiment, the Compounds are inseparable.
Quiz: Classifying Matter
QUIZ: Classifying Matter
I. Directions: I dentify 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 D
K2SO4 D
Twix, snickers, pretzels, popcorn 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
Total 241 100
Mixture B Mass (g) %
Large Rocks 205 53
Small Rocks 58 15
Coarse Sand 97 25
Iron 29 7
Total 389 100
Calculation Examples ( Provide 2 Examples showing how you determined the Mass %)
Mixture B Large Rocks
205/389 = .52699… * 100
53%
Mixture B Iron
29/389 = .0745… * 100
7%
Graphs:
Mixture A
Mixture B
Part III. Determine the Mass % of Elements in each Compound:
K2 SO4 - Potassium Sulfate
(Show Math Here)
Potassium (2)
39g (2) = 78g
78/174
.45 * 100
45%
Sulfur
32g/174
.18 * 100
18%
Oxygen (4)
16 (4) = 64
64/174
.37 * 100
37%
Na3P O4 - Sodium Phosphate
(Show Math Here)
Sodium ( 3) 23g = 69g
69/164 = .42 * 100
42%
Phosphorus 31g
31/164= .19
19%
Oxygen (4) 16 = 64g
64/164 = .39 * 100
39%
IV. Conclusion: E xplain the difference between Mixtures and Compounds using data. Compare
the pie charts.
In conclusion, mixtures and compounds are different for a one major reason, which is
that compounds are chemically fused while mixtures are not. This means that compounds can’t
be easily separated like a mixture can. An example of this in the data above is Mixture A. In
Mixture A, there are many substances that can be taken apart like sand, iron, and rocks. To find
the mass percent of a substance you have to take mass of a substance then divide by the total
mass (241g) then, multiply by 100 to get a percent. For example, the mass of iron was 29g
which you divide by the total mass which is 241g to get .07 then multiply by 100 to get 7%.
However in a compound, you have to have certain elements that can’t be taken apart that make
it a compound. An example of this is when I had to identify the total mass of Sodium Phosphate.
Sodium Phosphate is made up of 3 elements that are chemically combined, Sodium,
Phosphorus, and Oxygen. To determine the mass of each element you don’t have to weigh
them like in a mixture but you have to use the periodic table. An example of finding the percent
of an element in a compound is Oxygen in Sodium Phosphate. In the equation, Na3P O4 , Oxygen
has a 4 next to which means there is 4 molecules of Oxygen in Sodium Phosphate. So, you
have to find the mass of Oxygen (16g) and multiply it by 4 to get the mass of 64g. Next, you
have to take the mass of Oxygen and divide it by the total mass of Sodium Phosphate (164g) to
get .39 which you have to multiply by 100 to get 39%. Also, the percents of the substances in
the mixture will never be the same because each mixture has a random amount of each
substance, like in the mixtures above. In a compound, since, the elements are chemically
combined the amount of atoms in each element will always be the same. In conclusion, don’t
assume that a mixture and a compound are the same.
Bonus:
Explain how you separated the Salt from the Sand. Use as much new vocabulary as you can.
You can separate salt from by identifying a solute, a solution, and a solvent. The solute
was sand because it was the substance being dissolved by the solvent. The solvent is water
because it separates the salt from the sand and dissolves the salt. Finally, the solution is the
sand because it is being separated from the salt. After pour water into the sand you catch the
water into a beaker and put it on a hot plate to let it evaporate. When the water evaporates salt
should be the only thing left in the beaker.
Quiz: Solubility
Directions: Use the Solubility Graph to answer the following questions.
Graph
I. Solubility Graph
Questions:
1. What is the Solubility of KClO3 at 40 C?
About 15g
2. What is the Solubility of NH4 Cl at 70 C?
60g
3. What Temperature would 80 grams of KNO3 completely dissolve and become saturated?
50 C
4. Suppose you have 120 grams of NaNO3 a t 30 C. Is the solution Unsaturated, Saturated or
Supersaturated and how many grams can you add/or take away to make it Saturated?
It is supersaturated so you need to take away about 23g of solute to make it saturated at 97C.
5. Suppose you have 120 grams of NaNO3 at 30 C. What could you do to the Beaker to make
the solution Saturated? (Use Data from graph here)
Since it is supersaturated you would have to heat the hot plate up to about 25 more degrees
to make it saturated at 55C.
6. Suppose you have 70 grams of KNO3 at 60 C. Is the solution Unsaturated, Saturated or
SuperSaturated and how many grams can you add/or take away to make it Saturated?
Since it is unsaturated you would have to put a bout 40g more of solute to make it saturated at
100g.
7. Suppose you have 70 grams of KNO3 at 60 C. What could you do to the Beaker to make the
solution Saturated? (Use Data from graph here).
Since it is unsaturated you would have to take the beaker off the hotplate to cool it down 15C
to make it saturated at 45C.
II. Soluble vs. Insoluble
Directions: U se your Solubility Rules Chart to determine if the following compounds are Soluble
or Insoluble.
Compound Soluble or Insoluble Identify the Rule # Used
Sodium chloride Soluble 1
Silver nitrate Soluble 4 and 2
Ammonium nitrate Soluble 1 and 2
Calcium carbonate Insoluble 8
Zinc sulfide Insoluble 7
AgCl Insoluble 4 and 3
Soluble 1 and 5
Na2 SO4 Insoluble
Calcium phosphate Insoluble 10
3
PbBr2
III. Use your Solubility Rules to Determine how the beaker would look in the following chemical
reactions:
Reaction #1
Potassium Chloride + Silver Nitrate → Potassium Nitrate + Silver Chloride
Ions- K+Cl- + Ag+NO3 - → K+ N O3 - + Ag+Cl-
Reaction- + AgNO3 → KNO3 + AgCl
KCl Insoluble
Soluble Soluble
Soluble
Reaction #2
Lithium Phosphate + Calcium Sulfate → Lithium Sulfate + Calcium Phosphate
Ions- Li+PO4- 3 + Ca+ 2S O4 - 2 → Li+S O4-2 + Ca+2PO4- 3
Reaction + CaSO4 → Li2S O4 + Ca3( PO4 ) 2
Li3PO4 Soluble Soluble Insoluble
Soluble
IV. Conclusion:
Write a conclusion explaining the results of one of the reaction. You should focus on the
appearance of the final beaker. Your conclusion should also discuss the % of Oxygen between
2 of the compounds in the same reaction.
In Reaction Two we had to see if a reaction would occur if we put Lithium Phosphate and
Calcium Sulfate in the same beaker. First, we had to find out if both were soluble in order for the
reaction to take place, which they were because of rules 1 and 5. When we did this it created
two new compounds, Lithium Sulfate and Calcium Phosphate. Since Calcium Phosphate is
insoluble it stays at the bottom of the beaker while Lithium Sulfate, soluble, floats around the
beaker. Then, we were told to compare the percent of Oxygen in Calcium Phosphate and
Lithium Phosphate. To do this we had to find the total of amu of both compounds which was,
102 amu for Lithium Phosphate and 135 amu for Calcium Phosphate. Then we had to divide the
mass of Oxygen (64 for both) over the total amu of both to get the percent of Oxygen. In Lithium
Phosphate the total percent of Oxygen was 63% while in Calcium Phosphate it was 43%. Thus,
the percent of Oxygen in Calcium Phosphate was smaller than the percent of Oxygen in Lithium
Phosphate.
V. What is wrong with the following formula: (PO4)2N a
This is is incorrect because Sodium (Na+ ) has a positive charge and should be written
first while Phosphate has a PO4 -3 has a negative charge and should be written second.