Gas Laws: Self-Learning Kit in Science 10
Gas Laws: Self-Learning Kit in Science 10
Department of Education
Division of Iloilo
ILOILO NATIONAL HIGH SCHOOL
Luna, Street, La Paz, Iloilo City
Grade10
Gas laws
FOURTH QUARTER
SY 2021-2022
Quarter 4-Week 1
OTHER UNITS: mmHg, OTHER UNITS: OTHER UNITS: I mol= 6.02 x 1023
torr, Pa (Pascal), mL, cm3, m3 O
C, OF number of
psi (pound per square atoms/molecules
inch)
Concepts:
Robert Boyle- an English chemist and physicist that proposed the volume pressure
relationship among gases.
Boyle’s Law – states that “At constant temperature and amount of gas, the volume occupied
by a given mass of gas is inversely proportional to the applied pressure.”
Remember what an inverse relationship means? This means that increasing or decreasing one
variable will result to an opposite change to the other
variable.
For a gas, the relationship between volume and pressure (at constant mass and temperature)
can be expressed mathematically as follows
P ∝ (1/V)
Where P is the pressure exerted by the gas and V is the volume occupied by it. This
proportionality can be changed into an equation by adding a constant, k.
P = k*(1/V) ⇒ PV = k
The pressure v/s volume curve for a fixed amount of gas kept at constant temperature is
illustrated below.
As per Boyle’s law, any change in the volume occupied by a gas (at constant quantity and
temperature) will result in a change in the pressure exerted by it. In other words, the product of
the initial pressure and the initial volume of a gas is equal to the product of its final pressure and
final volume (at constant temperature and number of moles). This law can be expressed
mathematically as follows:
P1V1 = P2V2
Where,
This expression can be obtained from the pressure-volume relationship suggested by Boyle’s law.
For a fixed amount of gas kept at a constant temperature, PV = k. Therefore,
P1V1 = k (initial pressure * initial volume)
P2V2 = k (final pressure * final volume)
∴ P1V1 = P2V2
This equation can be used to predict the increase in the pressure exerted by a gas on the walls
of its container when the volume of its container is decreased (and when its quantity and
absolute temperature remain unchanged).
EXAMPLES:
1. A fixed amount of a gas occupies a volume of 1L and exerts a pressure of 400 kPa
on the walls of its container. What would be the pressure exerted by the gas if it is
completely transferred into a new container having a volume of 3 liters (assuming the
temperature and quantity of gas remains constant)?
Given:
Solution:
As per Boyle’s law,
P1V1 = P2V2
P2 = (P1V1)/V2
P2 = (1L * 400 kPa)/3L = 133.33 kPa
Therefore, increasing the volume of the gas three times will decrease its exerted pressure by 1/3.
2. A gas exerts a pressure of 3 kPa on the walls of container 1. When container 1 is
emptied into a 10-liter container, the pressure exerted by the gas increases to 6 kPa.
Find the volume of container 1. Assume that the temperature and quantity of the gas
remain constant.
Given:
INITIAL CONDITION FINAL CONDITION
Initial volume (V1) = ? Final volume (V2) = 10L
Initial pressure (P1) = 3kPa Final pressure (P2)= 6kPa
Solution:
P1V1 = P2V2
V1 = (P2V2)/P1
Activity 1
A. Solve the following problems using Boyle’s Law. Show your complete solution.
[APPLICATION]
1. If a syringe is filled with 100 mL of air at 1 atm, how much pressure must be applied
to reduce the volume to 60 mL?
2. If a gas at 25.0 °C occupies 3.60 liters at a pressure of 1.00 atm, what will be its
volume at a pressure of 2.50 atm?
3. A sample of Ne gas occupies 0.220L at 0.86 atm. What will be its volume at 29.4kPa?
B. Based on the given data of gas in a closed cylinder below, answer the following questions:
[ANALYSIS, EVALUATION]
3. What will happen to the volume of the gas if the pressure is increased four times that
of the initial reading? Why?
C. When we breathe, we actually inhale and exhale about 500ml of air twelve times per
minute. Do a simple research and in 3-5 sentences, explain how Boyle’s Law is applied in
this process. [COMPREHENSION]
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Quarter 4-Week 2
Charles's law (also known as the law of volumes) is an experimental gas law that
describes how gases tend to expand when heated. A modern statement of Charles's law is:
When the pressure and number of moles on a sample of a dry gas is held constant, the Kelvin
temperature and the volume will be in direct proportion.
This relationship of direct proportion can be written as: V T
So this means:
V = k or V = k T
T
where:
V is the volume of the gas,
T is the temperature of the gas (measured in kelvins),
and k is a non-zero constant.
Fig. A. Temperature-Volume Relationship (constant pressure and number of moles)
This law describes how a gas expands as the temperature increases; conversely, a
decrease in temperature will lead to a decrease in volume. For comparing the same
substance under two different sets of conditions, the law can be written as:
V = V or V = T
1 2 or V x T = V x T
2 2 1 2 2 1
T 1 T 2 V T 1 1
Further , V = V x T or V = V x T or T = V x T
1 2 1 2 1 2 1 1 2 or T = V x T
2 2 1
T 2 T V 1 2 V 1
EXERCICES:
1. Find the new volume of a flexible container if its original volume was 10 liters at 35
C if it is heated to 50 C. Pressure and the number of moles are held constant.
o o
V = V x T = 10 L x 323 K
2 1 2
0
T 308 K
1
0
= 3230 L
L 308
V = 10.5 L (take note that as the temperature was
2
2. At what temperature will the 50-Li flexible container at 33 C double its volume with
o
T =V xT
2 2 1 = 100 L x 273 K 0
V 1 50L
= 27300 K 0
50
T = 546 K (Take note that the temperature has to
2
0
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Activity 2
DIRECTIONS: Solve the following problems using the Charle’s Law. Show your given and do
the step-by-step processes and cancellation of common units. Put your final answer in a box.
1. Compute for the final volume of a flexible container at 45 Li volume and 45 oC temperature if
it is cooled down to 10 oC, with its pressure and number of moles held constant. Will the
container increase its volume or decrease? (APPLICATION)
2. What would be the original volume of a flexible container with original temperature of 22 oC
if it was heated to 44 oC with the final volume of 200 Li ? Would its original volume be lower or
higher than its final volume? (ANALYSIS)
3. At what temperature will you heat/cool a flexible container to a new volume of 500 Li with
the initial temperature of 44 oC and 400 Li original volume with its pressure and number of
moles held the same? Will you heat up or cool it down? (ANALYSIS)
4. What would be the original temperature of a flexible container at 500 liters original volume if
you heated/cooled it to 300 K and had a new volume of 450 liters at the same pressure and
number of moles? Did you heated or cooled the system? (ANALYSIS)
5. Give the general relationship of temperature and volume with the pressure and number of
moles being held constant. (EVALUATION)
6. Relate their relationship according to the Kinetic Molecular Theory. Why would its volume
change when it is heated or being cooled? (COMPREHENSION)
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Quarter 4 -Week 3
Concepts:
• Living systems require essential biomolecules to be able to function well and to
sustain growth and development, as well as, used for physiological
aspects.
• Biomolecules refer to any molecule that is produced by a living system. It
includes large macromolecules such as carbohydrates, the chief energy source of
all living organisms; lipids, high energy giving molecules; proteins, for the repair
and functions of body tissue; and nucleic acid, the genetic carrier of organisms.
• Carbohydrates are organic compounds consisting of carbon, hydrogen and
oxygen which are the main energy source and as essential structural
components of all organisms. The general formula C (H O) is commonly used to
x 2 y
Example(s):
As a child, you may have been told that sugar is bad for you. Well, that's not exactly
true. Essentially, carbohydrates are made of sugar, from a single sugar molecule to
thousands of sugar molecules all attached together. Why? One reason is to store
energy. But that does not mean you should eat it by the spoonful.
Carbohydrates are probably the most abundant and widespread organic substances in
nature, and they are essential constituents of all living things. Carbohydrates are
formed by green plants from carbon dioxide and water during the process
of photosynthesis. Carbohydrates are illustrated as:
This molecule of the sugar glucose consists of 6 carbon atoms bonded together
as a chain with additional atoms of oxygen and hydrogen. Fructose is
recognized by having a five member ring and having six carbons, a hexose.
Glucose is six membered ring, while fructose is a five-membered ring. Both
rings contain an oxygen atom.
Lipids are important fats that serve different roles in the human body. A common
misconception is that fat is simply fattening. However, fat is probably the reason we are
all here. Throughout history, there have been many instances when food was scarce.
Our ability to store excess caloric energy as fat for future usage allowed us to continue
as a species during these times of famine. So, normal fat reserves are a signal that
metabolic processes are efficient and a person is healthy.
Phospholipids make up only about 2 percent of dietary lipids. They are water-
soluble and are found in both plants and animals. Phospholipids are crucial for
building the protective barrier, or membrane, around your body’s cells. In fact,
phospholipids are synthesized in the body to form cell and organelle membranes.
In blood and body fluids, phospholipids form structures in which fat is enclosed
and transported throughout the bloodstream.
Sterols are the least common type of lipid. Cholesterol is perhaps the best well-
known sterol. Though cholesterol has a notorious reputation, the body gets only
a small amount of its cholesterol through food—the body produces most of it.
Cholesterol is an important component of the cell membrane and is required for
the synthesis of sex hormones, vitamin D, and bile salts.
8. Take a look at the illustration below, how are they different?
mjtingatinga
Monosaccharide like glucose is composed of one sugar unit, disaccharide like sucrose is
composed of two sugar units and polysaccharide is composed of many sugar or
saccharide units as in the case of starch.
9. Which food molecule would you eat if…
a. …you needed an instant energy?
b. …you have not eaten for a week?
a. Carbohydrates are needed if you needed for instant energy.
b. If you haven’t eaten for a week, your body will use stored lipids.
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Reminder: Please follow the format used in Activity 1.
ACTIVITY: 3
Our food directly affects our bodies. This activity encourages you to think critically
about what happens to the food you eat. Read and answer each question below.
2. The liver is one destination for digested monosaccharides. How does the liver
utilizes monosaccharides? (Applying/Analyzing)
4. Study the diagrams below, which show carbohydrate molecules. Beside each
molecule, write whether it is a monosaccharide, a disaccharide, or a polysaccharide.
(Evaluating/Rembering)
a.
b.
c.
Quarter 4-Week 4
Concepts:
Nucleic acids are the main information-carrying molecules of the cell, and, by
directing the process of protein synthesis, they determine the inherited characteristics of
every living thing. The two main classes of nucleic acids are deoxyribonucleic acid (DNA)
and ribonucleic acid (RNA). DNA is the master blueprint for life and constitutes the
genetic material in all free-living organisms and most viruses. RNA is the genetic material
of certain viruses, but it is also found in all living cells, where it plays an important role
in certain processes such as the making of proteins.
“Double Helix”
Chemical
structure of DNA
Summary:
Biomolecules Building blocks Functions Examples
Proteins Amino acids Provide cell Keratin (found in
structure, send hair and nails),
chemical signals, hormones,
speed up chemical enzymes,
reactions, etc antibodies
Nucleic acids Nucleotides Store and pass on DNA, RNA
genetic information
Drill:
I. Complete the structure of DNA by writing the unknown nitrogen bases (remember, A always
pairs with T and C always pairs with G).
T
A
G A
G
II. Using the DNA sequence, make a complimentary mRNA strand (remember to substitute U
for T in mRNA).
mRNA
Answers:
I. Complete the structure of DNA by writing the unknown nitrogen bases (remember, A always
pairs with T and C always pairs with G).
II. Using the DNA sequence, make a complimentary mRNA strand (remember to substitute U
for T in mRNA).
mRNA UGAGAUUCGCGGAUGCGUAAUUGC
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ACTIVITY 4
INSTRUCTION: In a separate sheet of paper, copy and answer ALL the questions, illustrations
and tables below.
II. Copy and complete the structure of DNA by writing the unknown nitrogen bases
(remember, A always pairs with T and C always pairs with G). (ANALYSIS)
C C
T T T
A G A G A G
III. Using the DNA sequence, make a complimentary mRNA strand for both the human and the
DNA Sequence C C A T A G C A C G T T A C A A C G T G A A G G T A A
mRNA
Table 2. Cow insulin protein sequence
DNA Sequence C C G T A G C A T G T T A C A A C G C G A A G G C A C
mRNA
Answers:
I. Free Response
1. Amino acids are organic compounds that combine to form proteins. Amino acids and
proteins are the building blocks of life. They're needed for vital processes like the building of
proteins and synthesis of hormones and neurotransmitters
2.
3. Our body needs protein to stay healthy and work the way it should. More than 10,000 types
are found in everything from our organs to our muscles and tissues to our bones, skin, and
hair. It helps repair and build our body's tissues, allows metabolic reactions to take place and
coordinates bodily functions.
II.
C T T C
A G G T
A A G
A
G A A
T G
C T C T C
DNA Sequence C C A T A G C A C G T T A C A A C G T G A A G G T A A
mRNA G GU A U C G U G C AA U G U U G C A C U U C C A U U
DNA Sequence C C G T A G C A T G T T A C A A C G C G A A G G C A C
mRNA G G C A U C G U A C AA U G U U G C G C U U C C G U G
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Quarter 4-Week 5
At the end of the session, I would be able to:
S10MT-IVe-g-23 Apply the principles of conservation of mass to chemical reactions.
Specific Objectives:
a. State the law of Conservation of mass.
b. Identify the components of chemical reactions.
c. Write a chemical equation.
Concepts:
• Law of Conservation of Mass states that during a chemical reaction, matter is neither
created nor destroyed. In any chemical reaction, the total mass of the reactants must
equal the total mass of the reactants.
According to the law of Conservation of mass, atoms are not created or destroyed in a
chemical reaction. However, they may be rearranged, forming new substances.
Example:
about the state or properties of products and reactants. Symbols to signify the physical
states of the substances are often used.
Examples:
1. (g) for gas or an arrow pointing upward( ), such as H 2(g) or H2 ( )
2. (l) for liquid, such as H2O (l)
3. (s) for solid substances or precipitate, or you may use the arrow pointing downward
( ) or underline the formula of the solid substance in the product, such as AgCl (s)
, AgCl , or AgCl.
4. (aq )which means aqueous refers to a substance in water solution, such as
H2SO4 (aq) ; the substance is dissolved in water.
5. Or heat indicates that heat is supplied to the reaction.
6. Pt a formula written above or below the yield sign indicates its use as a catalyst.
Catalysts are used to speed up chemical reactions without being consumed.
The word equation is changed into symbols and formulas to give the chemical
equation
C (s) + O2 (g) CO2 (g)
3. The law of conservation of mass should be satisfied. The number of atoms for each
element should be equal in the reactant side and the product side of the reaction.
C (s) + O2 (g) CO2 (g)
1 atom------ C ---- 1 atom
2 atoms ---- O ---- 2 atoms
Note: If the equation is not balanced, do not change the formulas. You can change
only the number of molecules of the substances in the equation by writing a
coefficient. Then, apply the distributive property.
Example:
2H2 + O2 2H2O
Coefficient Coefficient
4 atoms -------- H ----------4 atoms
2 atoms -------- O --------- 2 atoms
The count shows that the equation is now balanced. It reads: Two molecules of
hydrogen combining with one molecule of oxygen form two molecules of water.
Example(s):
A. Identify the components of the given chemical equation.
Answer:
Drill:
A. Write the chemical equation of the given chemical reactions.
1. When calcium carbonate is heated, a calcium oxide and carbon dioxide are
produced.
Answer:
Calcium carbonate heat Calcium oxide + carbon dioxide
2. Gaseous dinitrogen trioxide reacts with water to form aqueous nitrous acid.
Answer:
Dinitrogen trioxide + water Nitrous acid
N2O3 (g) + H2O (l) HNO2 (aq)
2 atoms -------- N ------- 1 atom
4 atoms -------- O -------2 atoms
2 atoms -------- H ------- 1 atom
The equation is not balanced. Therefore, write the coefficients.
Coefficient
Reactant Product
2.
NaHCO3 (s) + HCl (aq) NaCl (aq) + H2O (l) + CO2 (g)
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ACTIVITY 5
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Quarter 4 - Week 6
Concepts:
Chemical reaction – a process in which at least one new substance is produced as a result of
chemical change.
Chemical equation – a written statement that uses chemical symbols, and chemical formulas
instead of words to describe the changes that occur in a chemical reaction.
Example:
Balanced chemical equation – a chemical equation that has the same number of atoms of
each element involved in the reaction on each side of the equation. A balanced equation
assures that the law of conservation of mass is obeyed.
This balanced equation tells us that four NH3 molecules react with three O2 molecules to
produces two N2 molecules and six H2O molecules.
• The preceding notation denotes two molecules of H2O; it also denotes a total of four (4)
H atoms and two (2) O atoms.
Chemical reactions occur everywhere in the world around you, not just in a chemistry lab.
Digestion is a complex process that involves thousands of chemical reactions. When you put
food in your mouth, water and the enzyme amylase breaks down sugar and other
carbohydrates into simpler molecules. Hydrochloric acid and enzymes break down proteins in
your stomach. Sodium bicarbonate released into the small intestine neutralizes the acid and
protects the digestive tract from dissolving itself.
Example(s):
A. Balance the chemical equation FeI2 + Cl2 à FeCl3 + I2
1. Make a list of all of the elements on each side under the equation for both the reactants
and products.
Elements Reactants Products
Fe 1 1
I 2 2
Cl 2 3
2. Put the number of atoms that are in each of the elements as shown above.
3. Determine which atoms are not balanced.
4. Balance one atom at a time using coefficients.
• There are 3 chlorine atoms on the right side of the equation and two atoms of chlorine
on the left. For the chlorine atoms to balance, six will be needed on each side; 6 is the
lowest number that both 3 and 2 will divide into evenly. In order to obtain six atoms of
chlorine on each side of the equation, the coefficient 3 is placed in front of Cl2 and the
coefficient 2 in front of FeCl3.
FeI2 + 3Cl2 à 2FeCl3 + I2
There are now six chlorine atoms on each side of the equation.
Elements Reactants (left) Products (right)
Fe 1 1
I 2 2
Cl 2 x 3=6 3 x 2= 6
• The number of iron atom on the right side has already been set at 2 by the coefficient
previously place in front of FeCl3. Two iron atoms are needed on the reactant side of the
equation instead of the one iron atom now present. This is accomplish by placing the
coefficient 2 in front of FeI2.
2FeI2 + 3Cl2 à 2FeCl3 + I2
Elements Reactants (left) Products
(right)
Fe 1 x 2= 2 2
I 2 2
Cl 2 x 3= 6 3 x 2= 6
• Only one element is left to balance-iodine. The number of iodine atoms on the left side
of the equation is already set at four (2FeI2). In order to obtain the four iodine atoms in
the right side of the equation, we place the coefficient 2 in front of I2.
5. As a final check on the correctness of the balancing procedure, count the atoms on each side
of the equation.
2FeI2 + 3Cl2 à 2FeCl3 + 2I2
Elements Reactants (left) Products (right)
Fe 1 x 2=2 1 x 2=2
I 2 x 2=4 2 x 2=4
Cl 2 x 3 =6 3 x 2=6
Reactants Products
Elements
Ca 1 1
C 2 2
H 2 2+2 = 4
O 1 2
• Place coefficient 2 before H2O. We now have equal numbers of oxygen atoms. The
equation is now balanced.
CaC2 + 2 H2O à Ca(OH)2 + C2H2
• Place coefficient 2 before H2O. We now have equal numbers of oxygen atoms, but the
number of hydrogen is now unbalanced.
H2 + O2 à 2H2O
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Quarter 4-Week 7
Concepts: Knowledge of rates of chemical reactions can be a very useful tool in understanding
everyday activities. Most of the practices that we do had been laid down from generations and
we do not even understand how it works. Many of our activities involves chemical reactions
and speed of these chemical reactions are affected by several factors which includes
temperature, acidity (pH), concentrations of reactants and catalyst. The following are
applications of kinetics in our daily activities.
1. The preservation of food. Many factors are attributed to the
spoiling of food which includes bacterial action, enzyme activity (already
present in food or coming from contamination) and oxidation from
oxygen in the air. Delaying the chemical reaction in the spoiling of food
is the principle behind food preservation. How is food preservation done?
First is lowering the temperature that is why we have refrigerators. By
cooling food we lower enzyme action thus slowing down the spoiling
process. Second, we add vinegar to change the pH. Enzyme action is
maximum at corresponding pH. Outside of that pH, enzymes become
deactivated slowing down the chemical reaction of spoiling. Bacteria and
virus are also deactivated by change in pH of their environment.
4. Control of pollution also involves kinetics. Gas solid carbonation control is a process
that is used to control pollution. This involves a fast reaction between a highly reactive
magnesium, calcium and carbon dioxide to form stable carbonates, a process of removing
carbon dioxide from the environment.
contact of material with moisture, salt, extreme pH, heat. To avoid the untoward effect of
corrosion, measures are done to avoid the trigger factors and slow down the reaction to
happen. Some of these are coating metals with grease, paint, or a more stable metal.
1. Adding vinegar to food is making the food acidic to slow down the spoiling process
2. The purpose of painting metal surfaces is merely for aesthetic reason.
3. A small fire maybe contained by covering it with a blanket. The purpose of covering is to
deprive it of oxygen.
4. In the process of fermentation of wine, yeast is added to improve the flavor.
5. Planting more tress will absorb more carbon dioxide which could lower down green house
effect of carbon dioxide.
******************************************************************
Reminder: Please follow the format used in Activity 1.
Activity 7.a
Exercise 1
Materials:
2 pcs clean nail (2”), 1 teaspoon of salt, 25 mL water, small amount of oil
DIRECTIONS:
Wipe both nails with a piece of paper towels. Coat nail number 1 with oil. Meantime dissolve
one teaspoon of salt in 25mL of water. Divide the salt solution in 2 separate containers.
Submerge both nails in separate containers of the salt solution. Note which one is coated with
oil. Leave the set up overnight and check the next day for changes in both nails. Give an
observation of both nails.
2. What was the purpose of the oil coating in nail number 1? (ANALYSIS)
4. Site an example of a situation at home wherein grease, paint or oils is used to stop the
6. How will these factors the affect the rates of chemical processes affect our daily lives like the
ripening of fruits and spoiling of fish? (COMPREHENSION)
******************************************************************
Activity 7.B
Exercise 2
Directions: Answer with true or false and determine that factor affects its rate of
chemical reaction/process. (EVALUATION/RECALL)
Yield ____________________________________
_____6. Refrigeration is lowering down the temperature of food to slow down the
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REFERENCES:
Canyong, Patty Mendoza et.al, Science Grade 10, Vicarish Publication and Trading, Inc.,
c.2015, pp.256-272
Gilbert, Thomas R., et.al. (2015). Chemistry: The Science in Context. Fourth Edition,
W.W. Norton and Company, pp. 984-992
https://www.britannica.com/science/nucleic-acid/Deoxyribonucleic-acid-DNA
https://biology4isc.weebly.com/protein.html
http://www.chem4kids.com/files/bio_nucleicacids.html
Briones, Lagrimas, Templora, Vilma and Tibajares Jr., Ignacio. Chemistry Power. SMPC,
Sta. Cruz, Manila, 2010.
(Second Edition). Phoenix Publishing House, Inc.
Lim, A., et. al (2015). Science for the 21st Century Learner 8. Diwa Learning Systems,
Inc
Ferriols-Pavico, J., et. al (2017). Exploring Life Through Science Series: The New Grade
10 (Second Edition). Phoenix Publishing House, Inc.
Bayquen, A., (2012). Exploring Life Through Science: Chemistry (Laboratory Manual)
(Second Edition). Phoenix Publishing House, Inc.
Lim, A., et. al (2015). Science for the 21st Century Learner 8. Diwa Learning Systems,
Inc.
GRADE 10
GOAL:
TLE (OAP): Come up with compost.
(HAIRDRESSING) - Perform hair parting and identify the styling products that suit
your overall hair condition based on your assessment of the client’s hair type. Take photos of
the most important part of your work.
B. SCIENCE
For OAP takers, identify and explain the factors that hasten the decomposition rate of
chemical reactions in your material production (compost). Attach pictures of your compost
materials labelled Day 1 to 10.
For Hairdressing takers, explain how the principle on the rate of chemical
reactions influences the selection of materials for hair preparations and hair care.
For Dressmaking takers, identify the materials used in preparing the ladies’
trousers. Explain how the rate of chemical reaction influences the appropriateness of materials
selection.
The output The output shows Some parts of Most parts of the
clearly shows evidence of the output are output are not
Content evidence of in- knowledge of the not clearly evident of the
depth knowledge topic. evident of the knowledge of the
50% of the topic. knowledge on topic.
the topic.
Creativity The output is The output is The output is The output is not
very organized organized and minimally organized and show
20% and highly creative organized and no effort
creative creative
Promptness The output was The output was The output was The output was
in submission submitted on submitted within the submitted a day submitted 2 days
time day but not on the after the after the deadlines
15% specific time deadline.
Total= 100
%
(100 pts)
A. Photos
B. Tabular Data
Column A- Materials Used
Column B- Factors that hasten the rate of chemical reactions
C. Explanation(s)
A. Photos
B. Styling Products Used
C. Explanation(s)
A. Photos
B. Materials Used
C. Explanation(s)
Prepared by:
SCIENCE
JIM JAPITANA
NELRHEA GRACE B. PAREN
APPLE FUNTANILLA
TLE
JOSEFA CANTARA
ELNA MARQUEZ
REYME GALLARDA
APRIL JOY P. GARBINO