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To find this we set up an experiment, with both bismuth and gold weighing 55 grams. The
density of both the solid and the liquid gold was 19.32 g/cm3 with the volume of both being
2.84679 cm3. The density of the bismuth was 9.78g/cm3 with a volume of 5.62372 cm3.

First, we found the heat energy at the heat of fusion with the equation of mass (55
grams) times the heat of fusion (12.4 calories/grams for bismuth, and 15 calories/grams for gold).
With this, gold had a greater heat energy of 825 calories, compared to the heat energy of bismuth
which was 682 calories.

Following this, we multiplied the mass by the change in temperature from the melting
point to the boiling point, which was multiplied by the specific heat. The formula for gold was
55 grams times 1292 degrees Celcius by 0.0294 calories/grams degrees Celsius.

After this, gold remained with a higher heat energy than bismuth as the total heat energy
for bismuth at this point was 2771 calories, while the heat energy of gold was 3691.05 calories.

After this, we took the heat of vaporization and multiplied it by the mass of the metal.
Due to gold having a low heat of vaporization, 337 calories, and bismuth having a staggering heat
of vaporization, 160,000 calories bismuth ended up with the higher total heat energy (8,002,771
calories) compared to gold (22,226.05 calories.) The furnace that we ended up buying had a heat
of 25,000 calories, meaning it could melt the gold but not the bismuth.

While finding out which metal we would be able to use, we saw different average kinetic
energies, that ended up with a dramatic change by the point that both metals had become a gas.
With bismuth becoming a liquid before a gold, bismuth ended up with the molecules spreading out
into the form of a liquid sooner than the gold, meaning that bismuth had a greater kinetic energy.
By the time the metals started to boil, bismuth still had a greater kinetic energy than gold.
However, by the time that the two metals had finished boiling, gold had the greater kinetic
energy as it had finished boiling way before bismuth did.

Overall, we found that gold has a lower heat energy than bismuth and that gold would be
a better choice than bismuth for our company, as with a lower heat energy it is easier to get a
furnace that has enough power for the metal than the 8 million heat energy that bismuth has.

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Quarter 2

Blue Portfolio 2018-19
Part 2

1. Cover Page: Classifying Matter
2. Classifying Matter Vocabulary
3. Activity: Classifying a Candy Mixture

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Classification of Matter

Textbook: Chapter 15 (448-473)

1. Vocabulary
Directions: Write the definition and master the words on Quizlet. Include a
screenshot on this document showing your mastery of the words. Include a
picture that represents each word.

Substance Heterogeneous Solubility Solvent
A type of matter component of a
with a fixed Mixture the maximum solution in the
composition. largest quantity
A mixture in amount of a and does the
dissolving
which different solute that can

materials can be dissolve in a

distinguished specific solvent

easily under specific

conditions, such

as temperature

and pressure

Element Homogeneous Atom Saturated
All of the atoms Mixture
in a substance Contains two or basic unit of a Containing the
have the same more gaseous,
identity liquid, or solid chemical largest possible
substances
blended evenly element amount of a

smallest particular solute

particle/unit of an (single bonds)

element that *still

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throughout has the
properties of that
element*
composed of 3
basic subatomic
particles:
protons
neutrons
electrons

Compound Solution Solute Unsaturated
A substance in
which the atoms A homogeneous component of a a term used to
of two or more
elements are mixture of solution in the describe an
combined in a
fixed proportion particles so smallest quantity organic molecule

small that they and is dissolved that contains at

cannot be seen least one pi bond

with a or one ring

microscope and

will never settle

to the bottom of

their container.

Mixture Concentration Suspension Supersaturated
A material made the amount of
up of two or solute dissolved a heterogeneous containing an
substances that in a measured
can be easily amount of mixture amount of a

containing a substantially

liquid in which greater than that

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separated by solvent visible particles required for
physical means settle saturation as a
result of having
been cooled
from a higher
temperature to a
temperature
below that at
which saturation
occurs

https://quizlet.com/_5duh4f

2. Classification of Matter
*Provide Examples of each form of matter. Include a picture.

Heterogeneous Homogeneous Element Compound

Mixture Mixture

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Critical Thinking:
● How are the examples for Heterogeneous and Homogeneous
MIxtures different?

You can take apart a heterogeneous mixture because it is a collection of
items that can be taken apart with human force at any time but a
homogeneous mixture cannot be taken apart with human force.

● How are Elements and Compounds similar and different?
Both elements and compounds are filled with a set number of atoms. For an
element, all of the atoms in the substance have the same identity but in
contrast, a compound contains two or more different atoms, with the
proportion being set and unchangeable.

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Activity: Candy Mixture

Directions: You will create a candy and snack mixture and determine the Mass% of each
component found within your mixture. Your mixture must have at least 4 different components.
Your group must sort and classify the mixture and find the mass of each component and then
create a pie chart.

Table: (Use this table to record your results)

Type of Candy Mass Percentage
Red Skittles 8.7 Grams
Yellow Skittles 2.2 Grams 8.7/60.9=0.1428
Orange Skittles 1.2 Grams 5714285*100=14
Purple Skittles 3.2 Grams .2857142857
Blue M&M’S 2.7 grams Percentage:
Red M&M’S 3.7 grams 14.29%

2.2/60.9=0.0361
2479474*100=3.
61247947455
Percentage:
3.61%

1.2/60.9=0.0197
0443349*100=1.
97044334975
Percentage:
1.97%

3.2/60.9=0.0525
4515599*100=5.
25451559934
Percentage:
5.25%

2.7/60.9=0.0443
3497536*100=4.
43349753695
Percentage:
4.43%

3.7/60.9=0.0607
5533661*100=6.
07553366174
Percentage:
6.08%

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Orange M&M’S 1.7 grams 1.7/60.9=0.0279
Green M&M’S 3.7 grams 1461412*100=2.
Yellow M&M’S 0.7 grams 79146141215
Brown M&M’s 0.7 grams Percentage:
Swedish Fish 15.7 Grams 2.79%
Twix 16.7 grams
Blue Tray=2.3 Grams 3.7/60.9=0.0607
Total mass=60.9 5533661*100=6.
Graph: 07553366174
Percentage:
6.08%

0.7/60.9=0.0114
9425287*100=1.
14942528736
Percentage:
1.15%

0.7/60.9=0.0114
9425287*100=1.
14942528736
Percentage:
1.15%

15.7/60.9=0.257
79967159*100=2
5.7799671593
Percentage:
25.78%

16.7=60.9=0.274
22003284*100=2
7.4220032841
Percentage:
27.42%

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Mixture A:

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Mixture B:

Math
Total Mass=
Total Mass= Mass Object 1+ Mass Object 2+ Mass Object 3+ Mass Object 4+ Mass Object 5+
Mass Object 6+ Mass Object 7+ Mass Object 8+ Mass Object 9+ Mass Object 10+ Mass Object
11+ Mass Object 12
Total Mass= 8.7 grams + 2.2 grams + 1.2 grams + 3.2 grams + 2.7 grams + 3.7 grams + 1.7
grams + 3.7 grams + 0.7 grams + 0.7 grams + 15.7 grams + 16.7 grams
Total Mass= 60.9 grams

Percentage of Red Skittles= Mass of Red Skittles/Total Mass
Percentage of Red Skittles= 8.7 grams/60.9 grams
Percentage of Red Skittles= .143
Percentage of Red Skittles= 14.3%

Percentage of Yellow Skittles= Mass of Yellow Skittles/Total Mass
Percentage of Yellow Skittles= 2.2 grams/60.9 grams
Percentage of Yellow Skittles= 0.036
Percentage of Yellow Skittles= 3.6%

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Percentage of Orange Skittles= Mass of Orange Skittles/Total Mass 61/310
Percentage of Orange Skittles= 1.2 Grams/60.9 Grams
Percentage of Orange Skittles= 0.0197
Percentage of Orange Skittles= 2.0%

Percentage of Purple Skittles= Mass or Purple Skittles/Total Mass
Percentage of Purple Skittles= 3.2 grams/60.9 grams
Percentage of Purple Skittles= 0.0525
Percentage of Purple Skittles= 5.3%

Percentage of Blue M&M’s= Mass of Blue M&M’s/Total Mass
Percentage of Blue M&M’s= 2.7 grams/60.9 grams
Percentage of Blue M&M’s= 0.044
Percentage of Blue M&M’s= 4.4%

Percentage of Red M&M’s= Mass of Red M&M’S/Total Mass
Percentage of Red M&M’s= 3.7 grams/60.9 grams
Percentage of Red M&M’s= 0.0607
Percentage of Red M&M’s= 6.1%

Percentage of Orange M&M’s= Mass of Orange M&M’s/Total Mass
Percentage of Orange M&M’s= 1.7 grams/60.9 grams
Percentage of Orange M&M’s= 0.0279
Percentage of Orange M&M’s= 2.8%

Percentage of Green M&M’s= Mass of Green M&M’s/Total Mass
Percentage of Green M&M’s= 3.7 grams/60.9 grams
Percentage of Green M&M’s= 0.0607
Percentage of Green M&M’s= 6.1%

Percentage of Yellow M&M’s= Mass of Yellow M&M’s/Total Mass
Percentage of Yellow M&M’s= 0.7 grams/60.9 grams
Percentage of Yellow M&M’s= 0.011
Percentage of Yellow M&M’s= 1.1%

Percentage of Brown M&M’s= Mass of Brown M&M’s/Total Mass
Percentage of Brown M&M’s= 0.7 grams/60.9 grams
Percentage of Brown M&M’s= 0.011
Percentage of Brown M&M’s= 1.1%

Percentage of Swedish Fish= Mass of Swedish Fish/Total Mass
Percentage of Swedish Fish= 15.7/60.9

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Percentage of Swedish Fish= 0.2577
Percentage of Swedish Fish= 25.8%
Percentage of Twix= Mass of Twix/Total Mass
Percentage of Twix= 16.7/60.9
Percentage of Twix=0.274
Percentage of Twix=27.4%

Candy Lab Day 2
Directions:

1. Complete the candy lab
2. Find information and data from 2 other groups and make pie charts showing their data.
3. Use the data to show how the candy was an example of a Heterogeneous Mixture.
4. How are the mixtures different from the following compounds:

a. H2O - Water
Hydrogen 1amu2 oxygen 16 amu1
Hydrogen= 2amu oxygen=16 amu
Mass of Hydrogen= 18 amu
11.1%= Hydrogen
88.9%= oxygen

b. K3PO4 - Potassium Phosphate 62/310
Potassium 39amu3 Phosphorus 31 amu1 oxygen 16 amu4

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Potassium= 117 amu, Phosphorus= 31 amu, Oxygen= 64 amu
Mass Of Potassium Phosphate= 212amu

c. Na2SO4 - Sodium sulfate
Sodium 23amu2 Sulfur 32amu1 oxygen 16amu4
Sodium=46amu Sulfur=32amu oxygen= 64amu

Mass of Sodium Sulfate= 142amu

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d. NaCl - Sodium chloride

Sodium 23amu1 Chloride 35amu1
Sodium= 23amu Chlorine= 35amu

Mass of Sodium Chloride= 58amu

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5. Read chapter and watch video playlist

Challenge: Elements contained in above compounds:
*You can use these masses to determine the mass of one molecule of each compound
H - Hydrogen (1amu)
O - Oxygen (16amu)
P - Phosphorus (31amu)
K - Potassium (39amu)
Na - Sodium (23amu)
Cl - Chloride (35amu)
S - Sulfur (32amu)

Conclusion: How is your sample an example of a Heterogeneous mixture? Explain how it is
different from the mixtures other groups. (Use data) How is this different from a compound? Use
an example from the past assignment or research some compounds.

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The example of random candies that we put together is an example of a heterogeneous
mixture. This is because it is a mixture of different objects (M&M’s, Sour Skittles, Swedish Fish,
Twix) that can be easily separated at any time. With our mixture, the final mass totaled to be 60.7
grams. The highest percentage of this mixture was the one Twix, which was 27.4% while weighing
16.7 grams. The lowest percentage of the mixture was both Yellow and Brown M&M’s, each being
0.7 grams by themselves and each being 1.1% of the mixture. This varied from the other two
mixtures that we looked at, Mixture A and Mixture B (Due to not having data from two other
groups, a spin the wheel website was used to see how many of a selected candy was in the
mixture, after finding the mass of that candy). Mixture A was three bags of M&M’s with each bag
having a total of 11 M&M’s in it. Mixture B was M&M’s already out of their bags, Almond Joy’s,
and Twix. Both mixtures shared similarities and differences with the original mixture which led to
different results. For Mixture A, due to it being all M&M’s all but one color, orange, showed an
increase in mass and each color of candy showed an increase of percent of the final mass. In that
experiment, due to the sample of candy only being M&M’s, the amount for each was able to rise.
For Mixture B, the sample was between three candies, Twix, Almond Joy, and M&M’s, and the
results were very different than both Mixture A and our Mixture. In our mixture, there was only one
Twix, but due to there only being 3 different candies and this being done with a spin the wheel,
the amount of Twix, and therefore the final mass and percentage of the pie chart were
substantially higher. In this mixture, M&M’s were not separated into colors, which led to a lower
total and therefore a lower mass as M&M’s in total only made up three percent of the mixture.
One thing that all three of these samples of candy have in common are each of these are
mixtures. A mixture is a sample of something that can be taken apart with human force,
something that you could do with all three of these mixtures, despite being very different from
each other. In contrast, a compound is like a mixture, except human force cannot take it apart. An
example of a compound is water, which is made up of 16amu of oxygen and 2amu of Hydrogen.
This leads to percentages of 88.9% oxygen and 11.1% hydrogen. There is no human force that
can take apart the Hydrogen and oxygen that makes up water, therefore making it a compound
unlike the mixtures of candy.

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Directions: Analyze the data tables below and use the data to answer the questions about the
rock samples. YOu will do a similar lab tomorrow in class.

Table 1: Rock Mixture A Mass % of Sample
Component 35 g
12 g 35/76=0.460526316*100
Large Rocks 7g 46.05%
Small Rocks 3g
Fine Grained Sand 17 g 12/76=0.157894737*100
Coarse Grained Sand 2g 15.79%
Elements 76 g
Table Salt 7/76=0.0921052632*100
Total 9.21%

3/76=0.0394736842*100
3.95%

17/76=0.223684211*100
22.37%

2/76= 0.0263157895*100
2.63%

100%

Table 2: Rock Mixture B Mass % of Sample
Component 154 g
41 g 154/235=0.655319149*100
Large Rocks 18 g 65.53%
12g
Small Rocks 41/235=0.174468085*100
7g 17.45%
Fine Grained Sand 3g
18/235=0.0765957447*100
Coarse Grained Sand 7.66%

Elements 12/235=0.0510638298*100
5.1%
Table Salt
7/235=0.029787234*100
2.98%

3/235=0.0127659574*100

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1.28%

Total 235 g 100%

Questions:
1. How are the mixtures different?

These two mixtures, while containing the same objects are very different. For the first mixture, the
total mass of all of the objects comes up to 76 grams, while with Mixture B large rocks in itself has
over double the mass. For each object in Mixture B, it has a greater mass than in Mixture A. For
both mixtures A and B, the Large Rocks had the greatest percent with a 46.05% for Mixture A
while in Mixture B it was a little over 65 at 65.53% of the Mixture.

2. Which group had a greater % of Fine-Grained Sand?
Mixture A had the greater percentage of Fine-Grained Sand. This was because while it had a
lower mass, the difference between the masses wasn’t too drastic while the difference between
the final mass was way higher. To find the percentages of these objects, I took the mass of the
object, divided it by the final mass and then multiplied it by 100 to get a percentage. With Mixture
A, the formula was 7/76 which led to 0.0921052632*100, giving a final answer of 9.21%. WIth
Mixture B, the formula was 18/235 which led to the equation of 0.0765957447*100 which gave the
Fine-Grained Sand of Mixture B a final percent of 7.66% meaning that the percentage was 1.55%
lower than Mixture A.

3. Why are these examples of Heterogeneous Mixture?
These are both examples of a Heterogeneous Mixture. This is because the rocks and sand aren’t
chemically combined. While with the compound of table salt cannot be taken apart, it can be
taken apart from the rest of the mixture meaning that even with a compound in it, this is still a
mixture.

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QUIZ: Classifying Matter

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 Heterogeneous

Copper Element

Lemonade Homogeneous

Rocks, sand, gravel Heterogeneous

Salt Water Homogeneous

Gold Element

Sodium Chloride (NaCl) Compound

Air (Oxygen, nitrogen, carbon monoxide…) Homogeneous

K2SO4 Compound
Twix, Snickers, pretzels, popcorn in a bag Heterogeneous

II. Directions: Determine the Mass % of each mixture and construct the appropriate graphs.

Mixture A Mass (g) %
Large Rocks 125
125/241=0.5186
Small Rocks 75 72199*100
51.86%
Coarse Sand 32
75/241=0.31120
Iron 9 332*100
31.11%

32/241=0.13278
0083*100
13.28%

9/241=0.037344
3983*100
3.73%

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Total Mass 241 100%

Mixture B Mass (g) %
Large Rocks 205
Small Rocks 58 205/389=0.5269
Coarse Sand 97 92288*100
Iron 29 52.7%
Total Mass 389
58/389=0.14910
0257*100
14.91%

97/389=0.24935
7326*100
24.93%

29/389=0.07455
01285*100
7.45%

100%

Calculation Examples (Provide 2 Examples showing how you determined the Mass %)
Mixture A:
Total Mass= Large rocks + Small rocks + Coarse-Grained sand + Iron
Total Mass= 125+75+32+9
Total Mass= 241 grams

Mass% of Large Rocks
Mass of Large Rocks/Mass of Mixture=*100
Mass%=125/241
Mass%=0.518672199*100
Mass%=51.8672199
Mass% of Large Rocks= 51.87%

Mixture B:
Total Mass= Large rocks + Small rocks + Coarse-Grained sand + Iron
Total Mass= 205+58+97+29
Total Mass= 389 grams

Mass% of Iron
Mass of Iron/Mass of Mixture=*100
Mass% of Iron= 29/389
Mass% of Iron= 0.0745501285*100

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Mass% of Iron= 7.45501285
Mass% of Iron=7.45%

Graphs:
Mixture A

Mixture B

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Part III. Determine the Mass % of Elements in each Compound: 72/310

K2SO4 - Potassium Sulfate
(Show Math Here)
K - 39 amu
S - 32 amu
O - 16 amu
Total Mass= K(2)+S+O(4)
39(2)+32+16(4)
78+32+64
Total Mass= 174

K= 78/174
K=0.448275862
K=44.83%

S=32/174

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S=0.183908046
S=18.39%

O=64/174
O=0.367816092
O=36.78%

Na3PO4 - Sodium Phosphate
(Show Math Here)
Na - 23 amu
P - 31 amu
O - 16 amu
Na(3)+P+O(4)
23(3)+31+16(4)
69+31+64
Total Mass=164
Na=69/164
Na=0.420731707
Na=42.07%

P=31/164
P=0.18902439
P=18.90%

O=64/164
O=0.390243902
O=39.02%

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Graphs: 67
A. Mixture 54
95
M&M
Skittles
Twix

B. Compound (pick one)
Na3PO4

Sodium 69
Phosphorous 31
Oxygen 64

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IV. Conclusion:

1. Explain the difference between Mixtures and Compounds using data. Compare the pie charts.
The difference between Mixtures and Compounds are that a mixture is a random grouping of
items that can be taken apart at any time,
is a controlled group of elements combined, each with a specific number. The first pie chart, which
is the mixture can be altered at any time. You can add more M&M’s or take away some Skittles or
even add a new candy but it will always be a mixture. With a compound, the elements in it can’t
be altered by adding or subtracting from the original total due to it being preset in a certain way.
An example of this is the compound of Na3PO4 (Sodium Phosphate). This is an equation stating
that in Sodium Phosphate, there is 3 NA (Sodium), 1 P (Phosphorous), and 4 O (Oxygen). If you
alter that in any way, it is no longer Sodium Phosphate. However, with a mixture of candy, there is
no set equation to the mixture (and nothing chemically combining it) meaning that you can change
what’s in the mixture at any point that you choose.

2. Explain how you separated the Salt from the Sand. Use as much new vocabulary as you can.

To separate the salt from the sand, my group dissolved the sand (which is a solute) in the water
(solvent) by putting all of the fine-grained sand in the coffee paper sheets, over the funnel.

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Following this, we poured water in through the coffee paper, which funneled the salt out of the
sand and into the water, making salt water. After Thanksgiving break, we checked and the water
had evaporated with the salt in its place.

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Quiz
Name: Martin Roberg
Class: Galileo

QUIZ: Solubility and Compounds

I. Write the formula for the following compounds:

1. Sodium phosphate Na+1 PO -3
4
Na3PO4

2. Magnesium carbonate Mg+2Co3-2
MgCo3

3. Ammonium carbonate NH4+1 CO3-2
(NH4)2CO3

4. Lithium sulfate Li+1So4-2
Li2So4

5. Aluminum hydroxide Al+3OH-1
Al(OH)3

II. Write the name of the following compounds:

6. CaCO3 Calcium Carbonate

7. Ag3PO4 Silver Phosphate

8. K2S Potassium Sulfide

9. Mg(ClO3)2 Magnesium Chloride

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10. ZnF2 Zinc Fluoride

Question: Explain how you used your Periodic Table to find the charges of 2 of the elements:

III. Solubility - Read the questions and analyze the solubility graphs. Explain the appearance of
the beaker with the following solutions:
1. Suppose you have 160 g of Potassium nitrate at 55 C.

This is supersaturated. This is because the temperature is too low to boil the Potassium
Nitrate. Because the temperature is too low to boil the Potassium Nitrate, it is at the bottom of the
beaker. The total amount of Potassium Nitrate at the bottom of the beaker is exactly 55% of the
original total because 88 g of Potassium Nitrate would dissolve at 55 degrees Celsius. To boil it
dissolve it at 160 g the temperature would have to be about 76 degrees Celsius. To boil To
dissolve at 55 degrees Celsius, you would only have a total of around 88 grams of Potassium
Nitrate.

2. Suppose you have 200 g of Potassium Iodide at 60 C.

IV. Mass % You are exploring the Mississippi River and collect water samples. You find the
following ions in the water:

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Sodium Na+1 86/310
Phosphate PO4-3
P(1)= 31 amu
O(4)= 64 amu
31+64=95 amu
P=31/95
P=0.326315789*100
P=32.63%
O=64/95
O=0.673684211*100
O=67.37%
Sulfate SO4-2
S(1)= 32 amu
O(4)= 64 amu
32+64=96 amu
S=32/96
S=0.333333333*100
S=33%
O=64/96=0.666666667*100
O=67%
Magnesium Mg+2
Mg(1)= 24 amu

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Mg=24/24

Mg=100%

Carbonate CO -2
3

C(1)=12 amu

O(3)=48 amu

12+48= 60 amu

C=12/48

C= 0.2

C= 20%

O=48/60

O=0.8

O=80%

Lithium Li+2

Li(1)=7 amu

Li=7/7

Li=100%

Na (1)= 23 amu

P(1)= 31 amu

O(11)= 176 amu

S(1)= 32 amu

Mg(1)= 24 amu

C(1)= 12 amu

Li(1)= 7 amu

23+31+176+32+24+12+7=305 amu

Na= 23/305

Na=0.0754098361*100

Na=7.54%

P=31/305

P=0.101639344*100

P=10.16%

O=176/305

O=0.57704918*100

O=57.70%

S=32/305

S=0.104918033

S=10.49%

Mg=24/305

Mg=0.0786885246*100

Mg=7.87%

C=12/305

C=0.0393442623*100

C=3.93%

Li=7/305

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Li=0.0229508197*100
Li=2.30%
Your company needs you to compare the Oxygen content in 2 of the compounds. Choose 2
compounds and report the compound with the greatest and least % of oxygen. Construct Pie
Charts of both compounds.

Lithium Sulfate

Lithium Carbonate 88/310

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The Compound Lithium Carbonate has a higher percentage of Oxygen, despite having a lower
total AMU of Oxygen. Lithium Carbonate had 16 amu lower for Oxygen, but the total mass was 36
amu lower than the mass of Lithium Sulfate. The reason that the total amu of Lithium Sulfate was
higher than Lithium Carbonate was that the atomic mass of Carbon is lower than the atomic mass
of Sulfur. Carbon has an atomic mass of 12, while the atomic mass of Sulfur is 32 amu. Because
of this, Lithium Carbonate had a greater mass percentage of Oxygen than Lithium Sulfate.

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12345 67 Unknow
n

1 NA Bubbly Pink Milky Very Milky Minty Very

silver before Goldenr Milky White Color Milky

stirred od White White

turned before

white stirred,

more

solid after

2 Bubbly NA Dark Dark Magenta Started Turned Magent

Silver Magent Blood Magen Teal, a

a Red ta, Milky

Turned

Milky

Pink

3 Pink Dark NA Dark Magenta Wine Purple Magent

before Magent Red like a

stirred a color

turned

white

4 Milky Dark Dark NA Milky Turquo Yellow Milky

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Goldenr Blood Red Yellow ise Yellow
od Red Clear
No
5 Very Magent Magent Milky NA White Murky Reactio
Silver n
Milky a a Yellow
Milky White
White Blue
Murky
before NA Silver
Murky NA
stirred, Silver

more

solid

after

6 Milky Started Wine Turquoi White NA

White Magent like se

a, color

Turned

Milky

Pink

7 Minty Turned Purple Yellow Murky Milky

Color Teal, Silver Blue

milky

Unknow Very Magent Magent Milky Clear No White
n Milky a a Yellow Reaction

White

The Undefined is Number 5: Calcium Chloride

Activity: Predicting Products

CHOOSE 1

Challenge #1: Your company needs to produce 250 grams of Potassium nitrate. You need to find
2 compounds that will react to help you obtain this important compound. You will need to:

1. Show all ions
2. Show criss-cross of ions
3. Balance the reaction
4. Name the 4 compounds
5. Provide the mass in grams of all 4 substances showing the Law of Conservation of Mass.

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Test:

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