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Gas Laws: Self-Learning Kit in Science 10

The document is a self-learning kit for Grade 10 Science focusing on gas laws, specifically Boyle's Law and Charles's Law. It outlines the relationships between pressure, volume, and temperature of gases, providing mathematical expressions and practical applications. The kit includes activities and exercises for students to apply their understanding of these concepts.
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0% found this document useful (0 votes)
23 views38 pages

Gas Laws: Self-Learning Kit in Science 10

The document is a self-learning kit for Grade 10 Science focusing on gas laws, specifically Boyle's Law and Charles's Law. It outlines the relationships between pressure, volume, and temperature of gases, providing mathematical expressions and practical applications. The kit includes activities and exercises for students to apply their understanding of these concepts.
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PDF, TXT or read online on Scribd
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Republic of the Philippines

Department of Education
Division of Iloilo
ILOILO NATIONAL HIGH SCHOOL
Luna, Street, La Paz, Iloilo City

Junior High School

Grade10

SELF-LEARNING KIT IN SCIENCE 10

Gas laws
FOURTH QUARTER
SY 2021-2022

Development Team of Self-Learning Kit in Science 10

Writers: Nelrhea Grace B. Paren Apple D Funtanilla


Ma. Leonora J. Tingatinga Cristine Eve G. Casiple
Vina D. Caraso Jim V. Japitana

Editor: Jim V. Japitana


Master Teacher I

Content & Language Validator: Lilian C. Palmos


Department Head, Science

Evaluator: Dr. Sherry H. Tampani


Assistant Principal II for Academics, JHS

Implementer : Delorah Cecilia L. Fantillo


OIC- Principal IV, INHS
ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

Quarter 4-Week 1

At the end of the session, I would be able to:


S9MT-IIIj-20- Investigate the relationship between volume and pressure at constant
temperature of a gas.
a. Investigate the relationship between volume and pressure at constant
temperature of a gas;
b. explain the relationship between volume and pressure of a gas;
c. cite the importance of volume-pressure relationship in daily life.

FOUR VARIABLES THAT DESCRIBE GASES


PRESSURE (P) VOLUME (V) Temperature (T) Amount of Gas (n)
-force exerted by gas -space occupied -amount of kinetic -actual number of gas
particles against the by gas particles energy of gas particles
walls of its containers particles
STANDARD UNIT: STANDARD UNIT: STANDARD UNIT: STANDARD UNIT:
atmospheres (atm) Liters (L) Kelvin (K) moles (mol)

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.

Let us take the container with a fixed amount of gas


below as an example. What happens to its volume
when the applied pressure is increased? According to
Boyle’s Law, we should see a decrease in its volume
when the pressure is increased if the temperature and
amount of gas is held constant

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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,

• P1 is the initial pressure exerted by the gas


• V1 is the initial volume occupied by the gas
• P2 is the final pressure exerted by the gas
• V2 is the final volume occupied by the gas

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)

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

∴ 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:

INITIAL CONDITION FINAL CONDITION


Initial volume (V1) = 1L Final volume (V2) = 3L
Initial pressure (P1) = 400 kPa Final pressure (P2)= ?

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

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

V1 = (6 kPa * 10 L)/3 kPa = 20 L


Therefore, the volume of container 1 should be 20 L or twice that of container 2 since the initial
pressure was ½ of the final pressure.
Application of Boyle’s Law
If a scuba diver rapidly ascends from a deep zone towards the surface of the water, the
decrease in the pressure can cause the gas molecules in his/her body to expand. These gas bubbles
can go on to cause damage to the diver’s organs and can also result in death. This expansion of the
gas caused by the ascension of the scuba diver is another example of Boyle’s law. Another similar
example can be observed in the deep-sea fish that die after reaching the surface of the water (due to
the expansion of dissolved gasses in their blood).
*****************************************************************************
Reminders: Please follow the same format below in answering the activity sheets throughout
the quarter. Write your answers legibly. Use short bond paper.
Subject: ___________________________ Subject Teacher: _________________________________
Name: ____________________________ Grade & Section: ________________________________
Week No.__________________________ Activity No. ___________________________________
Adviser:____________________________ Score: _________________________________________

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]

Reading Pressure Volume Temperature


(kPa) (L) (K)
1 5 40 475
2 10 20 475
3 15 13.3 475
4 17 11.8 475
1. What happened to the volume of the gas when the pressure doubled from 5kPa to
10kPa?
2. How would you describe the values of Temperature as pressure in the cylinder is
increased?

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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]

****************************************************************************
Quarter 4-Week 2

At the end of the session, I would be able to:


S9MT-IIj-20: Investigate how gases behave based on the motion and relative distances
between gas particles.
Investigate the relationship:
b. Volume and Temperature at constant pressure
c. Explain these relationships using the Kinetic Molecular Theory
Concepts:

The kinetic molecular theory of gases is a simple, historically significant model of


the thermodynamic behavior of gases, with which many principal concepts of thermodynamics were
established. The model describes a gas as a large number of identical
submicroscopic particles (atoms or molecules), all of which are in constant, rapid, random motion.
Their size is assumed to be much smaller than the average distance between the particles. The
particles undergo random elastic collisions between themselves and with the enclosing walls of the
container. The basic version of the model describes the ideal gas and considers no other interactions
between the particles.

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:

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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

Answer: Given: V = 10 liters; T = 35 + 273 = 308 K T = 323 K


1 1
0
2
0

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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

increased, the volume also increased)

2. At what temperature will the 50-Li flexible container at 33 C double its volume with
o

pressure and number of moles being held constant?

Answer: Given : V = 50 L ; T = 33 + 273 = 306K ; V = 2x50 = 100 L


1 1 2

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

increase twice in order for the volume


double with the pressure and number of
moles being held constant)

****************************************************************************

Reminder: Please follow the format used in Activity 1.

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)

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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)

*****************************************************************************

Quarter 4 -Week 3

At the end of the session, I would be able to:


S10MT- IVc-d-22- Recognize the major categories of biomolecules such as carbohydrates,
lipids, proteins and nucleic acids.
1. Describe the composition, function and structure of
carbohydrates and lipids and the functions of these
biomolecules.

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

represent many carbohydrates, which means “watered carbon.”


• Carbohydrates (Sugars and Starches)
1. Functions - provide, store energy, build macromolecules, and
spare protein and fat for other uses .

a. Sugar - supplies glucose to the brain and provide energy to


cells around the body

b. Starch - provide the majority of fuel for your body, so it makes up a

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

significant part of your caloric intake.

2. Composition - made up of molecules of carbon (C), hydrogen (H), and


oxygen (O)

a. Monosaccharides - carbohydrate made up of one type of


sugar (ex. Glucose)
b. Disaccharides - carbohydrates made up of two sugars
bonded together. (ex. Glucose + Glucose = Maltose)
c. Polysaccharide - complex carbohydrate made up of chains of
monosaccharides
ex. Starch - food storage compound found in plants
Cellulose - makes up the cell wall of plants
Glycogen - main storage form of glucose in animals
and humans

• Lipids are biomolecules that serve as energy storage, signalling, lubricant,


cushion and fundamental part of the cell membrane.

Lipids (Fats, Oils, Waxes)


1. Function
a. Fat - stores energy (twice as much as carbohydrates)
b. Plant Wax
c. Cholesterol
2. Composition - Carbon, hydrogen and oxygen atoms, and in some
cases contain phosphorus, nitrogen, sulfur and other elements.

Triglyceride - consists of 3 fatty acids and one molecule of


glycerol (fatty acids are made from CH2 units)
3. Insoluble in Water
4. Fats that are in a liquid state at room temperature are OILS

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

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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.

Drill: How well do you know carbohydrates and lipids?

1. What atoms make up carbohydrates? Lipids?

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

Carbon, Hydrogen and Oxygen make up carbohydrates and lipids.


2. What are the functional groups for carbohydrates and lipids?
The functional group for carbohydrates are monosaccharides. Glycerol and fatty
acids are the functional group for lipids.
3. What are the three types of carbohydrates?
Carbohydrate is divided into different groups of sugar or saccharide units. These
are the (1) monosaccharide, consisting of 1 sugar unit (2) disaccharide,
consisting of two sugar units and (3) polysaccharide, consisting of large units of
sugar. The most common monosaccharides are glucose or blood sugar
(dextrose), fructose or fruit sugar and galactose or milk sugar. The
disaccharides are sucrose or table sugar (glucose and fructose), lactose or milk
sugar (glucose and galactose) and maltose or malt sugar (2 glucose units).
Examples of polysaccharides (long chain of glucose) are starch in plants,
glycogen (animal starch) and cellulose.

4. What is the function of carbohydrates in our body?


The main function of carbohydrates is store energy.

5. What will happen to excess glucose in our body?


The excess glucose in our body is converted to glycogen.

6. What function does lipids do in our body?


Lipids function as energy storage, signalling, lubricant, cushion and fundamental
part of the cell membrane.

7. What are the three main types of lipids?


The three main types of lipids are triglycerides(or triacylglycerols) ,
phospholipids, and sterols. Triglycerides make up more than 95 percent of lipids
in the diet and are commonly found in fried foods, vegetable oil, butter, whole
milk, cheese, cream cheese, and some meats and are naturally found in many
foods, including avocados, olives, corn, and nuts. We commonly call the
triglycerides in our food as “fats” and “oils.” Fats are lipids that are solid at room
temperature, whereas oils are liquid. As with most fats, triacylglycerols do not
dissolve in water.

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.

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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.

******************************************************************************
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.

1. Besides long-term energy storage, what other functions do lipids serve in


organisms and in cells? List down 2. (Remembering)

2. The liver is one destination for digested monosaccharides. How does the liver
utilizes monosaccharides? (Applying/Analyzing)

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

3. Which food molecule would you eat if… (Evaluating/Applying)


a. …you needed a quick boost of energy?
b. …you haven’t eaten in days?
c. …you had a race tomorrow afternoon?
d. …you were getting ready for hibernation?

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.

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

Quarter 4-Week 4

At the end of the session, I would be able to:


Recognize major categories of biomolecules such as carbohydrates, lipids, proteins and
nucleic acids.

Concepts:

The third biomolecule is protein. Proteins are a diverse class of biological


molecules, which play a crucial role in almost every biological process. Proteins are
made of amino acids. Amino acids contain Carbon, Hydrogen, Oxygen, Nitrogen, Sulfur.

Protein is one of the most


important nutrients in our food
because it is the chief constituent
of the body cells, of body tissues
and of body fluids.

Protein has multiple


functions:
1. builds and repairs
body tissue.
2. maintains cell
growth in the
formation of new
body tissue. This is
especially
important if the
body is growing
rapidly, injured, or
under stress.
3. aids in the
formation of
enzymes, some
hormones and
antibodies.
4. provides as energy
if sufficient
carbohydrates and
fats are not
supplied by the

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

The forth biomolecule is nucleic acid. A nucleic acid is a long chain of


nucleotide units. A nucleotide is a molecule made up of three parts: a sugar, a base,
and a phosphate group. Nucleotides of deoxyribonucleic acid, or DNA, contain the sugar
deoxyribose (see it ends in –ose). Nucleotides of ribonucleic acid, or RNA, contain the
sugar ribose.

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.

Each nucleic acid contains four of five possible nitrogen-


containing bases: adenine (A), guanine (G), cytosine (C), thymine (T), and uracil (U).
All nucleic acids contain the bases A, C, and G; T, however, is found only in DNA,
while U is found in RNA. A always pairs with T through two hydrogen bonds, and G always
pairs with C through three hydrogen bonds.

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

“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

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

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).

DNA Sequence ACTCTAAGCGCCTACGCATTAACG

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).

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

II. Using the DNA sequence, make a complimentary mRNA strand (remember to substitute U
for T in mRNA).

DNA Sequence ACTCTAAGCGCCTACGCATTAACG

mRNA UGAGAUUCGCGGAUGCGUAAUUGC

*************************************************************************

Reminder: Please follow the format used in Activity 1.

ACTIVITY 4
INSTRUCTION: In a separate sheet of paper, copy and answer ALL the questions, illustrations
and tables below.

I. Free Response (APPLICATION)


1. Define amino acid.
2. Draw a generalized protein.
3. Why do we eat protein foods? (What is the function?

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

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(for INHS classroom use only) Chemistry 10

III. Using the DNA sequence, make a complimentary mRNA strand for both the human and the

cow (remember to substitute U for T in mRNA). (APPLICATION AND ANALYSIS)

Table 1. Human insulin protein sequence

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

III. Table 1. Human insulin protein sequence

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

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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

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 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

****************************************************************************

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:

• A Chemical equation is the symbolic representation of a chemical reaction or the


chemist’s shorthand for describing the course of reaction. There is a chemical reaction
when you observe a gas being released, a solid substance being formed from two clear
liquid substances or a change in color, or a change in temperature. The reactants
( the starting substances) the elements or compounds that enter into a reaction are
written on the left . The products ( the ending materials)the elements or compounds
that are formed as a result of a chemical reaction are written on the right. An arrow
( ), a double arrow ( ) , usually separates the reactants from the products.
• The coefficients next to the symbols of entities indicate the number of moles of a
substance produced or used in the chemical reaction. It should contain information

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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.

Components of a Chemical Equation

• Steps in Writing Chemical Equations


1. The chemical equation represents facts. Write a word equation with the names for
all reactants on the left and all products on the right:
Example:
1. When solid carbon is burned with excess oxygen gas, carbon dioxide gas is
produced.
Carbon + Oxygen Carbon dioxide
Reactants Products
2. Convert the equation to a skeleton formula equation by writing the correct formulas
for all reactants and products. Be sure that your formulas correctly represent the
particles in the reaction. Indicate the symbols to signify the physical states of the
substances.

The word equation is changed into symbols and formulas to give the chemical

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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

The count shows that the equation is balanced.

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.

1. C3H 8 (g) + 5 O2 (g) 3 CO2 (g) + 4 H2O (l)


a. What are the products in the given reaction?
Answer: The products are 3 CO2 (g) and 4 H2O (l) .
b. How many oxygen atoms are there in the products?
Answer: 3(2 Oxygen atoms) = 6 atoms + 4(1 Oxygen atom) = 4 atoms
6 Oxygen atoms + 4 Oxygen atoms = 10 oxygen atoms
In the given reaction, the product can be found on the left side of the equation.
Identify and count the number of oxygen atoms present. The first product have
three (3) coefficients multiply by two (2) atoms of Oxygen will give us six(6).
Oxygen atoms Then, on the second product four have (4) coefficients times
one (1) atom of oxygen is equal to four(4). Add the sum of the oxygen atoms
will give us ten(10) atoms of Oxygen.
c. What is the physical state of O2 in the reaction?
Answer: The physical state of O2 in the reaction is gas.
B. Write the chemical equation of the given chemical reaction.
a. An aqueous solution of sulfurous acid is formed when water in the liquid form
combines with sulfur dioxide gas.

Answer:

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Sulfur dioxide + Water Sulfurous acid


SO2 (g) + H2O (l) H2SO3 (aq)
1 atom ---- S ----- 1 atom
3 atoms -- O ----- 3 atoms
2 atoms --- H ----- 2 atoms

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

CaCO3 (s) CaO (s) + CO2 (g)


1 atom ---- Ca ---- 1 atom
1 atom ---- C ----- 1 atom
3 atoms --- O ----- 3 atoms

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.

N2O3 (g) + H2O (l) 2 HNO2 (aq)

Coefficient

2 atoms -------- N ------- 2 atoms


4 atoms -------- O -------4 atoms
2 atoms -------- H ------- 2 atoms
The count shows that the equation is now balanced.

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B. Label the components of the given chemical equations.

Yield No. of molecules (coefficient)


1.
H2SO4 (aq) + BaCl2 (aq) BaSO4 (s) + 2HCl (aq)

Reactant Product

States of matter No. of atoms (subscript)

2.
NaHCO3 (s) + HCl (aq) NaCl (aq) + H2O (l) + CO2 (g)

Reactant yield Product

*****************************************************************************

Reminder: Please follow the format used in Activity 1.

ACTIVITY 5

A. Write the chemical equations of the given chemical reactions. Applying/Analyzing


1. Solid sulfur burns in oxygen gas to form sulfur dioxide gas.
2. Lime, which is the common name for calcium oxide, reacts with gaseous sulfur
dioxide to form calcium sulfite.
3. In photosynthesis, carbon dioxide and water react to form glucose and oxygen gas.
4. Lithium metal reacts with water to produce lithium hydroxide and hydrogen gas.
5. Heating sulfuric acid produces water, oxygen and sulfur dioxide.

B. Label the components of the given chemical equations. Remembering


1. Zn(s) + FeSO4 (aq) ZnSO4 (aq) + Fe (s)

2. Ca(NO3)2 (aq) + 2NaOH (aq) 2NaNO3 (aq) + Ca (OH) 2 (aq)

3. 2KCl (aq) + Pb(NO3) 2 (aq) PbCl 2(s) + 2KNO 3 (aq)

C. Guide Questions: Applying/Analyzing/Evaluating


1. In your own understanding, what is Law of Conservation of Mass?

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2. How will you determine that chemical reactions take place?


3. Give one (1) example of chemical equation. Identify and label its component.
4. What are the steps to keep in mind when writing a chemical equation?

*****************************************************************************

Quarter 4 - Week 6

At the end of the session, I would be able to know:


S10MT-IVe-g-23-Apply the principles of conservation of mass to chemical reactions.
• Balance chemical equations
• Give examples of chemical reactions in daily life

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.

Guidelines for Balancing Chemical Equations

An unbalanced chemical equation is brought into balance by adding coefficients to the


equation to adjust the number of reactant or product molecules present.

An equation coefficient is a number that is placed to the left of a chemical formula in a


chemical equation; it changes the amount, but not the identity, of the substance. In the
notation 2H2O, the 2 on the left side is a coefficient; 2H2O means two molecules of H2O, and
3H2O means three molecules of H2O. Thus equation coefficients tell how many molecules or
formula units of a given substance are present.

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The following is a balanced chemical equation, with the encircled coefficients.

This balanced equation tells us that four NH3 molecules react with three O2 molecules to
produces two N2 molecules and six H2O molecules.

A coefficient of 1 in a balanced equation is not explicitly written; it is considered to be


understood. Both Na2SO4 and Na2S have “understood coefficients” of 1 in the following
balanced equation: Na2SO4 + 2C à Na2S + 2CO2

• An equation coefficient placed in front of a formula applies to the whole formula. By


contrast, subscripts, which are also present in formulas, affect only parts of a formula.

• 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.

Steps in Balancing Chemical Equation:


1. Count the number of atoms of each element on both product and the reactant side.
2. Determine which atoms are not balanced.
3. Balance one atom at a time using coefficients.
• write the coefficients in their lowest possible ratio
4. Make sure that there are equal numbers of atoms for each element on
both sides of the equation.

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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.

2FeI2 + 3Cl2 à 2FeCl3 + 2 I2

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Elements Reactants (left) Products (right)


Fe 1 x 2= 2 2
I 4 2 x 2= 4
Cl 2 x 3= 6 3 x 2= 6

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

B. Balance the chemical equation CaC2 + H2O à Ca(OH)2 + C2H2

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

Elements Reactants Products


Ca 1 1
C 2 2
H 2 x 2=4 4
O 1 x 2=2 2

C. Balance the chemical equation H2 + O2 à H2O


Elements Reactants Products
H 2 2
O 2 1

• 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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Elements Reactants Products


H 2 2 x 2=4
O 2 1 x 2=2
• To balance the number of hydrogen atoms, place coefficient 2 before hydrogen. The
equation is now balanced.
2H2 + O2 à 2H2O
Elements Reactants Products
H 2x2=4 2 x 2=4
O 2 1 x 2=2
*****************************************************************************
Reminder: Please follow the format used in Activity 1.
ACTIVITY 6
A. Tell whether the following chemical equations are balanced or unbalanced. If unbalanced,
write the correct balanced chemical equation. (Analysis)
1. 2Fe+ 3Cl2 à 2FeCl3
2. Al + Br2 à 2AlBr3
3. SiCl4 + 2H2O à SiO2 + 4HCl
4. AgI + Na2S à Ag2S + NaI
5. 2Al + 6HCl à 2AlCl3 + 3H2
B. Balance the following equations. (Evaluation)
1. Fe + O2 à Fe2O3
2. N2 + H2 à NH3
3. CH4 + O2 à CO2 + H2O
4. Na3PO4 + HCl à NaCl + H3PO4
5. SnCl4 + (NH4)3PO4 à Sn3(PO4)4 + NH4Cl
6. As + O2 à As2O3
7. KClO4 à KCl + O2
8. Mg(OH)2 + HCl à MgCl2 + H2O
9. FeS + O2 à Fe2O3 + SO2
10. P + O2 à P4O10
C. Give 5 examples of chemical reactions in our daily lives. (Application)

****************************************************************************

Quarter 4-Week 7

At the end of the session, I would be able to:


S10MT-IVh-j-20: Explain how the factors affecting rates of chemical reactions are applied in
food preservation and materials production, control of fire, pollution, and corrosion.

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

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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.

2. Materials production may also benefit from controlling rates of reactions. In


manufacturing return of investment is at maximum when the reactions in the synthesis of
materials is fast. The higher the production rate the bigger the profit.

3. Burning follows a certain mechanism. Three factors should be present to create a


flame. Fuel, oxygen and ignition temperature. Fire is not produce in the absence of anyone of
these three. When fire is already present, removal of anyone of the three factors is to be
considered. If fire is present in a confined
space, fire extinguishers are used to put
off the flame. Many of these fire
extinguishers contain carbon dioxide.
Carbon dioxide when sprayed into the
flame envelops the flame depriving it of
oxygen. In this case the concentration of
one reactant is decreased and this slows
down the reaction thus the flame is
extinguished. In a larger scale, the
cheapest way of putting off fire is the use
of water. Water is effective in putting off fire by lowering down temperature and by oxygen
displacement.

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.

5. Corrosion is the gradual destruction of a material, commonly a metal through a


chemical reaction with substances present in the
environment forming a more stable compound, an oxide,
hydroxide, or a sulfide. This is an undesirable process
since it destroys functionality as well as aesthetic value
of a material so that much effort is spent to stop it from
happening. Several trigger factors of corrosion are

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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.

Answer with true or false

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.

Answers: 1. True, 2. False, 3. True, 4. False, 5, true

******************************************************************
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.

1. Which nail is rusted? _____ (IDENTIFICATION)

2. What was the purpose of the oil coating in nail number 1? (ANALYSIS)

3. How was the rusting reaction delayed? (COMPREHENSION)

4. Site an example of a situation at home wherein grease, paint or oils is used to stop the

rusting process. (ANALYSIS)

5. What factor affected the rate of rusting of the nail? (RECALL)

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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)

_____1. Adding vinegar increases pH of food. ______________________________

_____2. Enzymes are catalyst. _________________________________________

_____3. Increased speed of reaction in manufacturing will result to higher

Yield ____________________________________

_____4. Oxygen alone can produce fire. _________________________________

_____5. Corrosion is a redox reaction. __________________________________

_____6. Refrigeration is lowering down the temperature of food to slow down the

spoiling process. ___________________________

_____7. Heating of bananas increase ripening process. ______________________

****************************************************************************

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

Carbohydrate Structure, What are lipids? https://med.libretexts.org/ Retrieved March 1,


2021

https://www.britannica.com/science/carbohydrate, Retrieved March 1, 2021

http://www.ricediversity.org/outreach, Retrieved March 11, 2021

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

McMurry, John. Foundation of Chemistry. Cengage Learning. 5 Shenton Way,


Singapore, 2011
https://www.khanacademy.org/science/high-school-biology/hs-biology-foundations/hs-
biological-macromolecules/a/hs-biological-macromolecules-review

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

MAJOR PERFORMANCE TASK FOR QUARTER 4

(A collaboration between Science and TLE)


OBJECTIVES:

TLE (OAP): Prepare compost.


TLE ( Hairdressing): Analyze other Factors Affecting Hair Design
TLE ( Dressmaking): Make and cut materials for lady’s trousers.
SCIENCE: Explain how the factors affecting rates of chemical reactions are applied
in food preservation and materials production, control of fire, pollution and corrosion.

GOAL:
TLE (OAP): Come up with compost.

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TLE (Hairdressing): Analyse factors affecting hair design.


TLE (Dressmaking): Use appropriate materials for lady’s trousers.

ROLE: Student Researcher


AUDIENCE: Professional Hair Stylist, Fashion Designer and Agriculturist.
SITUATION: As a student researcher, doing your tasks of your chosen specialization in TLE
subject. You will identify the materials used and how these affect the rates of chemical
reactions in the material production.

PRODUCT/PERFORMANCE AND PURPOSE:

For Online and Modular Learners:


A. TLE

(OAP) – Prepare compost using identified materials available in your household or


area. Explain the purpose of the material. Take photos during the Preparation, Construction
and Material production.

(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.

(DRESSMAKING) - Make a Ladies' Trousers using appropriate materials for the


seasons and the occasions. Take photos of the most important part of your work.

(Follow the template given below)

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.

(Follow the template given below)

For Online Learners: Submit your output in the Google Classroom.

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

RUBRICS FOR SUCCESS


Rubric for SCIENCE and TLE

CRITERIA EXCELLENT SATISFACTORY FAIR NEED


IMPROVEMENT (1)
(4) (3) (2)

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

Neatness and Presentation is Presentation is Presentation Presentation


Organization organized in an mostly organized in organization is organization is
easy-to- an easy-to- not easy to difficult to follow
15% understand understand format. understand.
format

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)

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ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

TEMPLATE: OAP and Science

Subject: Subject Teacher:


Name: Grade & Sec:
Adviser: Score:

In short bond papers, do the following:

A. Photos
B. Tabular Data
Column A- Materials Used
Column B- Factors that hasten the rate of chemical reactions
C. Explanation(s)

TEMPLATE: Hairdressing and Science

Subject: Subject Teacher:


Name: Grade & Sec:
Adviser: Score:

In short bond papers, do the following:

A. Photos
B. Styling Products Used
C. Explanation(s)

TEMPLATE: Dressmaking and Science

Subject: Subject Teacher:


Name: Grade & Sec:
Adviser: Score:

In short bond papers, do the following:

A. Photos
B. Materials Used
C. Explanation(s)

adfuntanilla, mljtingatinga, ceugrsaniba, ngbparen,vdcaraso, jvjapitana Page 36


ILOILO NATIONAL HIGH SCHOOL
Luna Street, La Paz, Iloilo City
(for INHS classroom use only) Chemistry 10

Prepared by:

SCIENCE

JIM JAPITANA
NELRHEA GRACE B. PAREN
APPLE FUNTANILLA

TLE
JOSEFA CANTARA
ELNA MARQUEZ
REYME GALLARDA
APRIL JOY P. GARBINO

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