Downloaded 8,726 times
![D
EPED
C
O
PY
415
3. Place a small portion of steel wool in the test tube.
4. Add 10 mL CuSO4
solution.
5. Cover the mouth of the test tube with the rubber/ cork stopper .
6. Get the mass of the set-up using the same 100mL-beaker.
Record the mass in Table 10.
7. Heat the lower part of the test tube gently for 2 minutes while moving
it to and fro. Make sure that the rubber/ cork stopper covers the mouth of
the test tube and the test tube is held with a test tube holder in a slanted
position.
Q17. Describe the appearance of the steel wool.
Q18. What is the evidence that a chemical change happened?
8. Allow the test tube to cool completely in the 100-mL beaker.
9. Get the mass of the set-up again. Record your observation in Table 10.
Table 10. Law of Conservation of Mass
BEFORE HEATING Mass (g)
(a) Mass of the test tube, stopper,
and beaker
(b) Mass of the test tube, stopper,
and beaker and Mass of the Steel
wool + CuSO4
solution
(c) Mass of the Steel wool +
CuSO4
solution [(b)+(a)]
Total Mass of Reactants:
AFTER HEATING
(d) Mass of the test tube, stopper,
and beaker and Mass of the Steel
wool + CuSO4
(e) Mass of the Steel wool +
CuSO4
solution [(d)-(a)]
Q19. Why is it important for the test tube to be sealed?
Q20. How will you compare the total mass before and after the reaction ?
Part 2. Paper Clip Reaction Model
Materials:
• 1 box of different colored paper clips
• Periodic table
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.](https://image.slidesharecdn.com/sci10lmu4-150909105511-lva1-app6892/75/Science-10-Learner-s-Material-Unit-4-71-2048.jpg)
![D
EPED
C
O
PY
418
NOTE: If there is only one set , we do not write 1 anymore.
6. Get the molar mass of N2
, H2
, and NH3
, multiply their masses by their
coefficient, then get the total mass of the reactants and compare to the total
mass of the products. The first equation is done for you.
molar mass (g/mol) : H=1 O=16 N=14
2 H2
+ O2
→ 2 H2
O
2(2 x1g/mol) + (2 x 16g/mol) → 2 [ (2x 1g/mol) + 16g/mol ]
4 g/mol + 32 g/mol → 2 ( 18 g/mol)
36g → 36g
Q29. Do the same with the second equation
Q30. How will you compare the total mass of the reactants and the total
mass of the products?
This now follows the Law of Conservation of Mass.
KEY CONCEPTS:
Law of Conservation of Mass states that mass is conserved in a
chemical reaction. The total mass of the reactants is equal to the total
mass of the products. No new atoms are created or destroyed, there
was only grouping or regrouping (rearrangement) of atoms.
_____N2
+ _____H2
> _____NH3
Figure 8
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.](https://image.slidesharecdn.com/sci10lmu4-150909105511-lva1-app6892/75/Science-10-Learner-s-Material-Unit-4-74-2048.jpg)
![D
EPED
C
O
PY
415
3. Place a small portion of steel wool in the test tube.
4. Add 10 mL CuSO4
solution.
5. Cover the mouth of the test tube with the rubber/ cork stopper .
6. Get the mass of the set-up using the same 100mL-beaker.
Record the mass in Table 10.
7. Heat the lower part of the test tube gently for 2 minutes while moving
it to and fro. Make sure that the rubber/ cork stopper covers the mouth of
the test tube and the test tube is held with a test tube holder in a slanted
position.
Q17. Describe the appearance of the steel wool.
Q18. What is the evidence that a chemical change happened?
8. Allow the test tube to cool completely in the 100-mL beaker.
9. Get the mass of the set-up again. Record your observation in Table 10.
Table 10. Law of Conservation of Mass
BEFORE HEATING Mass (g)
(a) Mass of the test tube, stopper,
and beaker
(b) Mass of the test tube, stopper,
and beaker and Mass of the Steel
wool + CuSO4
solution
(c) Mass of the Steel wool +
CuSO4
solution [(b)+(a)]
Total Mass of Reactants:
AFTER HEATING
(d) Mass of the test tube, stopper,
and beaker and Mass of the Steel
wool + CuSO4
(e) Mass of the Steel wool +
CuSO4
solution [(d)-(a)]
Q19. Why is it important for the test tube to be sealed?
Q20. How will you compare the total mass before and after the reaction ?
Part 2. Paper Clip Reaction Model
Materials:
• 1 box of different colored paper clips
• Periodic table
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.](https://crownmelresort.com/image.slidesharecdn.com/sci10lmu4-150909105511-lva1-app6892/75/Science-10-Learner-s-Material-Unit-4-71-2048.jpg)
![D
EPED
C
O
PY
418
NOTE: If there is only one set , we do not write 1 anymore.
6. Get the molar mass of N2
, H2
, and NH3
, multiply their masses by their
coefficient, then get the total mass of the reactants and compare to the total
mass of the products. The first equation is done for you.
molar mass (g/mol) : H=1 O=16 N=14
2 H2
+ O2
→ 2 H2
O
2(2 x1g/mol) + (2 x 16g/mol) → 2 [ (2x 1g/mol) + 16g/mol ]
4 g/mol + 32 g/mol → 2 ( 18 g/mol)
36g → 36g
Q29. Do the same with the second equation
Q30. How will you compare the total mass of the reactants and the total
mass of the products?
This now follows the Law of Conservation of Mass.
KEY CONCEPTS:
Law of Conservation of Mass states that mass is conserved in a
chemical reaction. The total mass of the reactants is equal to the total
mass of the products. No new atoms are created or destroyed, there
was only grouping or regrouping (rearrangement) of atoms.
_____N2
+ _____H2
> _____NH3
Figure 8
All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means -
electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.](https://crownmelresort.com/image.slidesharecdn.com/sci10lmu4-150909105511-lva1-app6892/75/Science-10-Learner-s-Material-Unit-4-74-2048.jpg)
Uploaded byPRINTDESK by Dan
Science 10 Learner’s Material Unit 4
Measure 100 mL of water using a graduated cylinder and pour it into a 500-mL beaker or tin can. 2. You: Use a pipette or syringe to carefully add 20 mL of cooking oil on top of the water in the beaker/tin can. Observe what happens. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015. D EPED C O PY 357 3.
In this document
Powered by AI
Introduction of the Science Module developed by educators in the Philippines, encouraging feedback from stakeholders.
Overview of Unit 4 covering Matter and Its Interactions, including modules on Gases, Chemical Reactions, and Biomolecules.
Content about gases' properties, behavior, and fundamental laws including pre-assessment and activities on gas behavior.
Experiments demonstrating gas properties such as mass, volume, temperature, and pressure influences.
Introduction to Boyle's Law with experimental activities to investigate the relationship between volume and pressure.
Details on Boyle's Law followed by sample problems illustrating the relationship between pressure and volume.
Introduction to the module on chemical reactions, objectives, and pre-assessment regarding chemical changes.
Tests for identifying chemical reactions, including indicators for reactions between various substances.
Classification of chemical reactions into combinations, substitutions, and others, with formulas and examples.
Activities demonstrating the Law of Conservation of Mass through experiments and balancing equations.
Discussion on factors affecting rates of chemical reactions including temperature, concentration, and catalysts.
Introduction to biomolecules including carbohydrates, lipids, proteins, and nucleic acids and their functions.
Discussion on carbohydrates, their testing, and the identification of different types through chemical tests.
Insight into proteins, their structure, functions, and the importance of amino acids for bodily functions.
Focus on nucleic acids, their roles in genetics, structure differences between DNA and RNA.
Summary of key concepts learned in the module and a summative assessment to evaluate understanding.
More Related Content
Similar to Science 10 Learner’s Material Unit 4
Science 10 Learner’s Material Unit 4
- 1. D EPED C O PY i 10 Science Department of Education Republicof the Philippines This book was collaboratively developed and reviewed by educators from public and private schools, colleges, and/or universities. We encourage teachers and other education stakeholders to email their feedback, comments, and recommendations to the Department of Education at action@deped.gov.ph. We value your feedback and recommendations. Learner’s Material Unit 4 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 2. D EPED C O PY ii Science – Grade10 Learner’s Material First Edition 2015 Republic Act 8293, section 176 states that: No copyright shall subsist in any work of the Government of the Philippines. However, prior approval of the government agency or office wherein the work is created shall be necessary for exploitation of such work for profit. Such agency or office may, among other things, impose as a condition the payment of royalties. Borrowed materials (i.e., songs, stories, poems, pictures, photos, brand names, trademarks, etc.) included in this book are owned by their respective copyright holders. DepEd is represented by the Filipinas Copyright Licensing Society (FILCOLS), Inc. in seeking permission to use these materials from their respective copyright owners. All means have been exhausted in seeking permission to use these materials. The publisher and authors do not represent nor claim ownership over them. Only institutions and companies which have entered an agreement with FILCOLS and only within the agreed framework may copy from this Learner’s Material. Those who have not entered in an agreement with FILCOLS must, if they wish to copy, contact the publishers and authors directly. Authors and publishers may email or contact FILCOLS at filcols@gmail.com or (02) 439-2204, respectively. Published by the Department of Education Secretary: Br. Armin A. Luistro FSC Undersecretary: Dina S. Ocampo, PhD Printed in the Philippines by REX Book Store, Inc. Department of Education-Instructional Materials Council Secretariat (DepEd-IMCS) Office Address: 5th Floor Mabini Building, DepEd Complex Meralco Avenue, Pasig City Philippines 1600 Telefax: (02) 634-1054, 634-1072 E-mail Address: imcsetd@yahoo.com Development Team of the Learner’s Material Authors: Herma D. Acosta, Liza A. Alvarez, Dave G. Angeles, Ruby D. Arre, Ma. Pilar P. Carmona, Aurelia S. Garcia, Arlen Gatpo, Judith F. Marcaida, Ma. Regaele A. Olarte, Marivic S. Rosales, Nilo G. Salazar Reviewers: Eligio C. Obille Jr., Marlene B. Ferido, Ma. Helen DH Catalan, Vic Marie Camacho, Lilia M. Rabago, Cerilina M. Maramag Illustrators: Joseph V. Bales, Ramon C. Gatpo, Regaele A. Olarte, Marivic S. Rosales, Ruel C. Quindoy, Antonio I. Basilla, Jose Leo Vic O. Albaño DepEd Specialists: Joseph R. Jacob, Maria Amparo R. Ventura Photo Credits: Herma D. Acosta, Dave G. Angeles, Liza A. Alvarez, Ruby D. Arre, Aurelia S. Garcia, Judith F. Marcaida, Regaele A. Olarte, Jane Chavarria, Nilo G. Salazar Layout Artists: Matthew Daniel V. Leysa and Mary Grace Ann G. Cadisal All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 3. D EPED C O PY Unit 4: Matterand Its Interactions Overview Module 1: Behavior of Gases I. Introduction --------------------------------------------------------------------------------351 II. Learning Competencies/Objectives ------------------------------------------------351 III. Pre-Assessment ------------------------------------------------------------------------352 IV. Reading Resources and Instructional Activities -------------------------------- 355 Activity 1: Getting to Know Gases --------------------------------------------355 Activity 2: Boyle’s Law -----------------------------------------------------------362 Activity 3: Charles’ Law ----------------------------------------------------------369 Activity 4: Gay-Lussac’s Law ---------------------------------------------------375 Activity 5: Combined Gas Laws -----------------------------------------------380 Activity 6: Squashing the Bottle ------------------------------------------------388 Activity 7: A Gaseous Outlook --------------------------------------------------391 V. Summary/Synthesis/Feedback ------------------------------------------------------394 VI. Summative Assessment -------------------------------------------------------------396 References and Links ------------------------------------------------------------399 Module 2: Chemical Reactions I. Introduction --------------------------------------------------------------------------------400 II.LearningCompetencies/Objectives--------------------------------------------------401 III. Pre-Assessment ------------------------------------------------------------------------401 TABLE OF CONTENTS IV. Reading Resources and Instructional Activities --------------------------------403 Activity 1: Everything has changed -------------------------------------------403 Activity 2: What’s in a Reaction? ----------------------------------------------408 Activity 3: We simply click together -------------------------------------------411 Activity 4: How much can you take? ------------------------------------------414 Activity 5: Balancing Act ---------------------------------------------------------419 Activity 6: Race to the Finish Line --------------------------------------------422 Activity 7: Making Connections ------------------------------------------------430 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 4. D EPED C O PY V.Summary/Synthesis/Feedback-------------------------------------------------------437 VI. Summative Assessment-------------------------------------------------------------438 Glossary of Terms -----------------------------------------------------------------441 References and Links ------------------------------------------------------------442 Module 3: Biomolecules I. Introduction --------------------------------------------------------------------------------443 II. Learning Competencies/Objectives ------------------------------------------------444 III. Pre-Assessment ------------------------------------------------------------------------444 IV. Reading Resources and Instructional Activities --------------------------------446 Activity 1: Test for Carbohydrates and Lipids -------------------------------447 Activity 2: A. Test for Proteins --------------------------------------------------462 V. Summary/Synthesis/Feedback ------------------------------------------------------472 Glossary of Terms -----------------------------------------------------------------472 VI. Summative Assessment -------------------------------------------------------------473 References and Links-------------------------------------------------------------475 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 5. D EPED C O PY 349 UNIT 4 Matter andIts Interactions All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 6. D EPED C O PY 350 Unit 4: Matterand Its Interactions Overview In Grade 9, you have learned about chemical bonding and its various types. You have learned how chemical bonding occurs and how particles rearrange to form new substances. Basic mole concept was also introduced to you, relating mass and number of particles of substances. You were also able to analyze the bonding characteristics of carbon which results in the formation of large variety of compounds. In Grade 10, you will learn that the rearrangement of particles happen when substances undergo chemical reaction. You will get to know how Law of Conservation of Mass applies to chemical reaction by analyzing masses and number of atoms of substances before and after a chemical reaction. Moving up from bonding characteristics of carbon, you will study about biomolecules such as carbohydrates, lipids, proteins, and nucleic acids. Also in Grade 10 Chemistry, you will investigate how gases behave in different conditions based on knowledge of the motion of and distances between gas particles. You will be able to explain behaviour of gases using the assumptions in the Kinetic Molecular Theory. You will also learn the relationships between volume, temperature, and pressure using established gas laws. Unit 4 is composed of the following modules: Module 1: Behavior of Gases Module 2: Chemical Reactions Module 3: Biomolecules Each module is filled with interesting and fun activities that will guide you in your journey to achieving optimum learning. Let your journey begin….. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 7. D EPED C O PY 351 I. Introduction This moduleoffers interesting discussion about gases. You will have a chance to get to know important concepts that will make you appreciate the properties and the behavior of gases. Most gases are invisible. We can name as many solids and liquids that we see around us but not gases. It is only the very few colored ones like the black smoke produced by smoke belchers that can be seen. Unseen gases are present, to name a few, in a bottle that seems to be empty, in the production of food by the plant, and even in playing our favorite sports. Can you play your favorite sports like volleyball and basketball without the ball sufficiently filled with air or gas? Even our very own existence requires the presence of unseen gases. We take in oxygen and we exhale carbon dioxide. Can we survive here on earth without the desirable gases which support life? You learned in Grade 8 that like other solids and liquids, gases are also made up of molecules that behave differently. Most of the properties of gases can be attributed to the random and scattered arrangement of its molecules, which are located as far away as possible from each other because they have very weak intermolecular force of attraction. II. Learning Competencies/Objectives To keep you on track while you are studying this module, let’s have the following learning competencies/objectives in mind: BEHAVIOR OF GASES Unit 4 MODULE 1 • Investigate the relationship between: o volume and pressure at constant temperature of a gas; o volume and temperature at constant pressure of a gas. • Explain the above mentioned relationships using the Kinetic Molecular Theory. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 8. D EPED C O PY 352 Before you engageyourself in studying this module, please answer the pre- assessment. III. Pre-Assessment Direction: Write the letter of the correct answer. 1. Which example has particles that can be drawn closer to occupy smaller volume? a. fruit juice b. block of wood c. air inside the syringe d. ice cube 2. Which of the following phenomena does NOT involve the application of gas pressure? a. burning fuels b. falling leaves c. vulcanizing tire d. rising hot air balloons 3. Last summer vacation, the Cruz family decided to go to Pagudpod, Ilocos Norte to have a beach party. On their way to Ilocos, all of them were surprised when the tire suddenly exploded. What is the probable explanation for the blown out tire during a long summer drive? a. High temperature causes a decrease in volume. b. The amount of the gases inside the tire is increased. c. The mass of the gases inside the tire increases causing a blown up tire. d. The volume of gases increases as the temperature increases, causing a blown up tire. 4. How can you possibly prove that gases have negligible mass? a. put a balloon in a digital balance before and after you fill it with air b. feel the weight of the samples on both hands c. ask two persons to hold a box filled with air d. support your claim of through equation All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 9. D EPED C O PY 353 5. Each ofthe following containers is air tight and has the same number of gas molecules. Which container has the highest pressure? 6. Each of the following containers has the same size. Which of following containers has the most compressed gas molecules? 7. All the gas samples have the same temperature and mass. In which of the following conditions will the gas sample have the highest density? 8. What happens to the density of a gas as its volume decreases at constant pressure and temperature? a. decreases b. increases c. stays the same d. unpredictable For numbers 9 to11, the choices are: a. Boyle’s Law b. Charles’ Law c. Combined Gas Law d. Ideal Gas Law 9. What law explains the mechanism of gas compressor? 10. What gas law best explains the explosion of the heated aerosol container? 11. What gas law explains the relationship among the volume, pressure, temperature, and the number of moles of gases? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 10. D EPED C O PY 354 12. How willyou represent the molecules of carbon dioxide at 30°C? 13. What kind of movement is exhibited by gas molecules? a. vibrational movement b. rotational movement c. translational movement d. combination of a, b and c 14. How does the temperature affect the average kinetic energy of gas molecules? a. as the temperature decreases the average kinetic energy of gas molecules decreases b. as the temperature decreases the average kinetic energy of gas molecules increases c. as the temperature decreases the average kinetic energy of gas molecules remains the same d. as the temperature decreases the average kinetic energy of gas molecules fluctuates 15. What will happen to the gas pressure as the temperature increases, if the amount and volume of the gas are kept constant? a. the gas pressure remains the same b. the gas pressure decreases c. the gas pressure increases d. there is no significant effect Have your answers checked and keep the result. You will learn about the explanations in your right and wrong answers as you study this module. Are you familiar with the properties of gases? The first activity will give you ideas on the properties of gases. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 11. D EPED C O PY 355 IV. Reading Resourcesand Instructional Activities Activity 1 Getting to Know Gases Objective: Prove that gases have the following properties: mass, volume, temperature, and pressure. Materials: For Activity A: For Activity B: 3 rubber balloons of the same kind pipette and aspirator or syringe digital balance 100-mL graduated cylinder balloon pump (optional) 200 mL water 20 mL cooking oil For Activity C: For Activity D: thermometer (360°C) Erlenmeyer flask alcohol lamp alcohol lamp tripod tripod wire gauze wire gauze match match denatured alcohol denatured alcohol ice 500-mL beaker or any tin can Procedure: A. Gases and Its Mass 1. Measure the mass of the deflated balloon using a digital balance with a 0.01 precision (sensitive up to two decimal places). All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 12. D EPED C O PY 356 2. Inflate the balloon using a balloon pump and seal the opening by securely twisting/looping the end. 3. Measure the mass of the inflated balloon using a digital balance. 4.Do three trials and record your data. Note: Keep the inflated balloon to be used in procedure D. Table 1. Data for the Mass of Gas inside the Balloon Trial Mass of the deflated balloon (g) Mass of the inflated balloon (g) Difference in mass (Inflated-deflated) (g) 1 2 3 Average Q1. Is the mass of the deflated balloon different from the mass of the inflated balloon? Q2. Which is heavier, the inflated or the deflated balloon? Why? Q3. What can you infer in this activity? Discover more about gases as you proceed to the next activities. B. Gases and Its Volume 1. Put approximately 50.0 mL of water in the graduated cylinder. 2. Cover the water with cooking oil up to approximately 70.0 mL. Let the oil settle at the top of the water. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 13. D EPED C O PY 357 3. Dip thetip of the pipette in the water-oil mixture until it reaches the water portion of the mixture. Carefully press the aspirator at the other end of the pipette to introduce air in the mixture. A syringe can be used as a substitute for pipette. 4. Carefully remove the pipet from the water-oil mixture. Read the final volume after introducing air in the water-oil mixture. Note: If pipette and aspirator are not available, you may instead use syringe. 5. Perform three trials and write your data on Table 2. Table 2. Data for the Volume of Air Trapped in the Water-Oil Mixture Trial Volume of water plus oil (mL) Total volume when air was introduced (mL) Difference in mass (Inflated-deflated) (mL) 1 2 3 Average Q1. What happens to the volume reading of the water-oil mixture when air is introduced to it? Q2. What does it indicate? C. Gases and Its Temperature 1. Pour approximately 150 mL of water in a beaker or any tin can. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 14. D EPED C O PY 358 2. Measure theinitial temperature of the air just above the water level. 3. Fill the beaker with crushed ice up to the water level. After 5 minutes, measure the temperature of the air just above the water level. 4. Assemble the wire gauze, tripod, and alcohol lamp. Set aside the iced water. Replace the content of the beaker with tap water. Place the beaker with water on the wire gauze. 5. Heat the water until it boils and get the temperature of the air just above the water level. 6. Perform three trials and write your data on Table 3. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 15. D EPED C O PY 359 Table 3. Temperatureof Water Vapor Trial Temperature of the Air (°C) Initial (room temperature) Above the ice water Above the boiling water 1 2 3 Average Q1. Is there a difference in the temperature of air among the three set-ups? Q2. Explain the difference in temperature of air. Note: Use the boiling water for the next set-up. D. Gases and Its Pressure 1. Transfer the hot water into the Erlenmeyer flask. 2. Carefully place the inflated balloon on the mouth of the Erlenmeyer flask with hot water. Observe what happens. Q1. What happens to the inflated balloon? Q2. What causes this phenomenon? 3. Remove the inflated balloon from the Erlenmeyer flask. 4. Get a deflated balloon and place it at the mouth of the Erlenmeyer flask. 5. Assemble the wire gauze, tripod, and alcohol lamp. Heat the Erlenmeyer flask with a deflated balloon. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 16. D EPED C O PY 360 Q3. What happensto the shape of the balloon? Q4. What causes the balloon to change its shape and size? Draw what happens to the balloon. You have just observed that gases have volume, mass, temperature, and exert pressure. From your daily experiences, can you enumerate some instances where these properties are shown? The warm temperature we are experiencing is from the heat trapped by the greenhouse gases (carbon dioxide, methane and water vapor to name a few). The basketball is filled with air. So, it bounces while you are dribbling it. The same is true with the other kinds of ball. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 17. D EPED C O PY 361 Now that wehave proven that gases have mass, volume, temperature, and pressure, let us now be familiar with the units being used to express these properties of gases. Can you identify whether a unit represents volume or pressure or temperature? Below is the list of these units. Start familiarizing yourself with them. Table 4. Commonly Used Units for Volume and Pressure Variable SI Unit Metric Unit English Unit Volume cubic meter (m3 ) cubic decimeter (dm3 ) cubic centimeter (cm3 ) liter (L) milliliter (mL) quart (qt) gallon (gal) Pressure Pascal (Pa) atmosphere (atm) millimeters of mercury (mm Hg) centimeters of mercury (cm Hg) torr lb/in2 (psi) When you open a can or bottle of softdrinks, it fizzes because of the escaping dissolved carbon dioxide due to change of pressure. When the wind blows, it exerts pressure too. There are a lot of manifestations of gases though we cannot see them. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 18. D EPED C O PY 362 Remembering these equivalentswill also be of great help: Volume units and their equivalents: 1 mL = 1 cm3 1 L = 1 dm3 1 m3 = 1000 L Source: http://www.metric-conversions.org/volume/cubic-meters-to-liters.htm Pressure units and their equivalents: 1 atm = 760 mm Hg = 76 cm Hg = 760 torr = 101325 Pa = 14.6956 psi Temperature units and their equivalents: 0˚C = 273.15 K 0˚C = 32˚F You will encounter most of these units as we go along. For the meantime, let us investigate if there are interrelationships among the properties of gases. Let us start with the effect of pressure to the volume of gases at constant temperature. Perform the next activity. Activity 2 Boyle’s Law Objective: • Investigate the relationship between volume and pressure of gases at constant temperature. Materials: • 25 mL syringe • 5” by 3” illustration board • set of weights • 6” by 4” by 0.25” wood • ruler • candle or glue gun • glue stick • match (if you opted to use candle) Procedure: 1. Fill the syringe with air by pulling the plunger. See to it that the volume reading is at approximately 25.0 mL. 2. Seal the opening of the syringe with the melted glue stick. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 19. D EPED C O PY 363 3. Bore ahole that is very close to the size of the opening of the syringe in a 6” by 4” flat wood. Screw the wood on a stable object. Insert in an upright position the sealed part of the syringe in the hole of the wood, be sure it is sturdy. 4. Paste a 5” by 3” illustration board at the end of the plunger. This will serve as the holder of the weights. You have just prepared a Boyle’s Law Apparatus. 5. Carefully place a 200-gram weight on the holder and get the volume reading. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 20. D EPED C O PY 364 6. Place oneat a time different weights to the plunger. If you do not have set of weights, you may use books of the same kind. Be sure to get the mass of each book. 7. Record the mass and volume reading using Table 5. Table 5. Observation on Volume Changes Trial Volume (cm3 ) Mass (g) Pressure (N/m2 ) Initial Reading 1 2 3 4 5 Note: P = Force/Area Force = mass (kg) x acceleration due to gravity (9.8m/s2 ) πr2 = Surface Area of the syringe Q1. What happens to the volume of the syringe as the set of weights is added on top of it? Q2. What happens to the pressure on the syringe when the set of weights is added? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 21. D EPED C O PY 365 8. Plot agraph with the pressure at the y axis and volume at the x axis. Q3. Describe the graph. Q4. What is the relationship between volume and pressure of gases at constant temperature? The activity you have performed enables you to observe Boyle’s Law, which can be used to describe the relationship between the volume and pressure of gases at constant temperature. Based on the result of your activity, what can you infer? In your Grade 9 lesson on living things and their environment, you made use of the lung model to explain the respiratory system. Do you still have the model with you? Try to use it again. What do you notice as you pull the bigger balloon that represents the diaphragm? Yes, the lungs expand! Let’s try to explain it with the use of Boyle’s Law. Pulling the rubber balloon represents inhaling. As you inhale, the lung cavity expands, causing the pressure inside the lungs to decrease and become lower than the outside pressure. As a result, air flows from the higher pressure area, which is outside the body, into the lungs. Exhaling is the opposite process; when you release the rubber which represents the diaphragm, the balloon representing the lungs decreases in volume. This phenomenon happens during exhaling. When the diaphragm contracts as you exhale, it results to a decrease in the lung volume, increasing the pressure inside the chest cavity and causing air to flow out of the lungs. Try to breath in and breath out and mindfully observe what happens to your lung cavity. Interestingly, as you inhale and exhale, approximately 500 mL of air gets in and out of your lungs. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 22. D EPED C O PY 366 Here is anotherthing that can happen which can be explained through Boyle’s Law. Have you observed the air exhaled by the fishes in the aquarium? It gets bigger and bigger as it rises because the pressure at the bottom of the aquarium is higher than the pressure near the surface. Where else do you see applications of the relationship between pressure and volume of gases? Similarly, this is what you observed when you perform Activity 2. Gas particles have a very weak intermolecular force of attraction, hence they move as far as possible from each other. They have the tendency to occupy all the spaces they are contained in. If the pressure is increased, the volume will be decreased forcing the gas particles to move closer to one another. The observations in Activity 2 can be expressed in the Boyle’s Law equation: 1 V α at constant T and n P Where: V = volume, P = pressure, T = temperature and n = amount of the gas. How will you read the above sited equation? It is read as: The volume of a gas is inversely proportional to its pressure, if temperature and amount of a gas are held constant. It can also be read as: At constant temperature, the volume occupied by a fixed amount of gas is directly proportional to the reciprocal of pressure (1/P). The relationship between the volume and pressure of gases at constant temperature was first stated by Robert Boyle during the 16th century. He performed an experiment wherein he trapped a fixed amount of air in the J-tube, he changed the pressure and controlled the temperature and then, he observed its effect to the volume of the air inside the J-tube. He found out that as the pressure is increased, the volume decreases. He finally concluded that the volume of a fixed amount of gas is inversely proportional to its pressure at constant temperature. Robert Boyle (1627-1691) All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 23. D EPED C O PY 367 Let’s take alook at the equation again and try to change the proportionality sign (α ) with the equal sign (=). 1 V α at constant (k) P k V = Thus, k = VP P The latter equation is simply read as: The product of Pressure and Volume is constant. What is the value of Vx P in Table 6? Table 6. Data on Volume-Pressure Relationship Trial Volume (L) Pressure (atm) VxP 1 2.0 10.00 2 4.0 5.00 3 8.0 2.50 4 16.0 1.25 Were you able to verify the meaning of proportionality constant? Let us apply the equation you learned about Boyle’s Law. Since volume and pressure of the gas can be varied, let P1 and V1 be the initial pressure and volume respectively and P2 and V2 be the final pressure and volume respectively. According to Boyle’s Law, PV= k therefore: V1 P1 = k V2 P2 = k then V1 P1 = V2 P2 You are now equipped with the fundamental knowledge to cope with the problem solving activities related to Boyle’s Law. Let’s try to solve this problem: All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 24. D EPED C O PY 368 The inflated balloon that slipped from the hand of Renn has a volume of 0.50 Lat sea level (1.0 atm) and it reached a height of approximately 8 km where the atmospheric pressure is approximately 0.33 atm. Assuming that the temperature is constant, compute for the final volume of the balloon. Source: http://regentsprep.org/Regents/math/algtrig/ATP8b/exponentialResource.htm In analyzing the problem, it is important that you categorize the initial and final conditions of the variables: Initial Conditions Final Conditions V1 = 0.50 L V2 = ? P1 = 1.0 atm P2 = 0.33 atm By applying Boyle’s Law, can you predict what will happen to the final volume? Yes, you’re right! The final volume will increase. Let’s compute for the numerical value of the final volume by substituting the given values to this equation. V1 P1 = V2 P2 V2 = V1 P1 / P2 V2 = ( 0.50 L) ( 1.0 atm) = 1.5 L (0.33 atm) Did you notice the decrease in pressure and how it affects the final volume? The pressure decreased by 1/3. That is why, the volume increased by 3-folds. Try to multiply V1 by P1 and V2 by P2. Does it have the same product? Isn’t it amazing? Answer the following problems for a better grasp of the lesson: 1. Oxygen gas inside a 1.5 L gas tank has a pressure of 0.95 atm. Provided that the temperature remains constant, how much pressure is needed to reduce its volume by ½? 2. A scuba diver needs a diving tank in order to provide breathing gas while he is underwater. How much pressure is needed for 6.00 liters of gas at 1.01 atmospheric pressure to be compressed in a 3.00 liter cylinder ? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 25. D EPED C O PY 369 3. A sample of fluorine gas occupies a volume of 500 mL at 760 torr. Given that thetemperature remains the same, calculate the pressure required to reduce its volume by 1/3. You have shown good mastery of the concepts on Boyle’s Law, thus you can now proceed to the next activity. This time, we will find out if there is a relationship between volume and temperature at constant pressure. Activity 3 Charles’ Law Objective: Investigate the relationship between volume and temperature at constant pressure. Materials: • rubber balloon • thermometer • tap water • alcohol lamp • hot water • tape measure • ice Procedure: 1. Prepare 3 beakers (1 for ice water, 1 for tap water, and another one for hot water). 2. Inflate a balloon. 3. Measure the circumference of the balloon using a tape measure. 4. Get the temperature reading of the hot water. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 26. D EPED C O PY 370 5. Put theballoon in hot water for 2 minutes, then measure again its circumference. 6. Do three trials and get the average of the results. 7. Repeat procedures 3 to 6 using tap water. 8. Repeat procedures 3 to 6 . This time use ice water. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 27. D EPED C O PY 371 9. Record theresults in the Table 7. Table 7. Data on Determining the Size of the Balloon at Different Temperatures Set-up Average Temperature (°C ) Average Circumference of the Balloon (cm) before after difference Warm Water Tap Water Ice Water Q1. What happens to the size of the balloon as the temperature decreases? Q2. How does the change in the temperature relate to the volume of gas in the balloon? The learning experiences you have fromActivity 3 focuses on the volume- temperature relationship. Can you enumerate familiar events you observe in your community and household which are related with the volume-temperature relationship in gases? The sky lanterns we use in celebrating New Year, Christmas, weddings, and other important occasions operate on the concept of volume-temperature relationship. Have you tried releasing a sky lantern? It is like a mini-hot air balloon; as the temperature increases, the sky lantern obtains its full volume and rises in the atmosphere. It rises and rises as the temperature increases because the density of gases decreases as gases expand due to the increase in temperature. This explains that the increase in volume and decrease in density cause the sky lantern to float in the air! All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 28. D EPED C O PY 372 Mathematically, Charles’ Lawcan be expressed as: V α T at constant P Where: V = volume and T = temperature expressed in Kelvin Why is there a need to convert °C to K? Kelvin is the basic unit for measuring temperature in the International System (SI). “It denotes the absolute temperature scale whereby 0K or absolute zero is defined as the temperature when molecules will have the lowest energy.” Removing the proportionality symbol (α) and using the equality sign (=) the equation will be as follows: V = k T or k = V T Thus, in a direct proportion, the quotient of the variable is constant. If you are going to consider the initial and final conditions, you will arrive at the following equations: V1 = k and V2 = k T1 T2 Whereas, V1 is the initial volume and V2 is the final volume T1 is the initial temperature and T2 is the final temperature If the volume-temperature ratios are the same in the initial and final conditions, then we will arrive at this equation: The volume - temperature relationship in gases (k = V/T) was determined by and named after Jacques Charles. In his experiment, Jacques Charles trapped a sample of gas in a cylinder with a movable piston in water bath at different temperatures. Jacques Charles found out that different gases decreased their volume by factors 1/273 per °C of cooling. With this rate of reduction, if gas will be cooled up to -273°C, it will have zero volume! Interesting, isn’t it? Charles’ Law states that at constant pressure, the volume of a fixed amount of gas is directly proportional to the Kelvin (K) temperature.Jacques Charles (1746- 1823) V1 = V2 T1 T2 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 29. D EPED C O PY 373 To further illustratethe mathematical equations above, let us have the following: A gas cylinder was measured to have different volumes at different temperature as shown in Table 8. Complete the table with the necessary information. Table 8. Data on Volume-Temperature Relationship Trial Volume Reading (ml) Temperature (o C) Temperature (K) 1 25 2 2 30 57 3 35 102 4 40 152 Note: To convert °C to K, use this formula: K = °C + 273.15 Plot the data from Table 8 in a graph by placing the volume in the y axis and temperature at Kelvin scale in the x axis. How is this graph different from the graph you obtained in Activity 2? Let’s apply Charles’ Law in solving problems related to volume- temperature relationship in gases. Sample Problem: An inflated balloon with a volume of 0.75 L at 30°C was placed inside the freezer where the temperature is -10°C. Find out what will happen to the volume of the balloon if the pressure remains constant. Support your answer with computation. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 30. D EPED C O PY 374 Just like what we did before, let’s start with the given variables: Convert thetemperature to Kelvin. K = °C + 273 = 30 + 273 = 303K Solve for the final volume. V1 T2 (0.75L) (263K) 197.25L V2 = = = = 0.65L T1 303K 303 Were you able to predict it correctly? Try to divide V1 by T1 and V2 by T2. Did you obtain the same quotient? Amazing! The volume decreases because the temperature decreases too. In this case, the volume between the gas molecules decreases because the kinetic energy is also affected by temperature. Do you realize the relationship of Charles’ Law to Kinetic Molecular Theory? Gas molecules move slowly at low temperature, thus there is less collision and so it will occupy smaller space. Answer the following Charles’ Law problem to facilitate mastery of concepts on the volume-temperature relationship: 1. A cylinder with a movable piston contains 250 cm3 air at 10°C. If the pressure is kept constant, at what temperature would you expect the volume to be 150 cm3 ? 2. A tank ( not rigid) contains 2.3 L of helium gas at 25°C. What will be the volume of the tank after heating it and its content to 40°C temperature at constant pressure? 3. At 20°C, the volume of chlorine gas is 15 dm3 . Compute for the resulting volume if the temperature is adjusted to 318K provided that the pressure remains the same. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 31. D EPED C O PY 375 Aside from Boyle’sand Charles’ laws, there is another gas law that you need to be familiar with. Have you ever wondered how temperature affects the pressure of the gas at constant volume? The next activity will help you visualize the effect of increasing the pressure on the temperature of gases at constant volume. Activity 4 Gay-Lussac’s Law Objective: Investigate the relationship between temperature and pressure at constant volume. Materials: • 110°C thermometer • Erlenmeyer flask/bottle • cork or rubber stopper • denatured alcohol • Liquid dropper Procedure: 1. Insert the thermometer into the stopper. Precaution: Lubricate the thermometer with a small amount of grease before insertion. 2. Put 5 drops of denatured alcohol in the Erlenmeyer flask. 3. Cover the Erlenmeyer flask with the stopper that you prepared in Procedure 1. The size of the stopper should fit the mouth of the Erlenmeyer flask. Wait for 2 minutes before measuring the temperature. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 32. D EPED C O PY 376 4. Shake the Erlenmeyer flask for 2 minutes and take the temperature reading. CAUTION: Carefullyhold the thermometer to avoid breakage. 5. Perform three trials and record the data. Table 9. Data on Temperature of the Gas Before and After Shaking the Erlenmeyer flask Trial Temperature (Co ) Before Shaking After Shaking 1 2 3 Average Q1. What happens to the drops of denatured alcohol after 2 minutes? after another 2 minutes ? Q2. Compare the pressure exerted by the denatured alcohol molecules before and after shaking? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 33. D EPED C O PY 377 Q3. How isthe temperature of gas molecules affected by pressure or vice versa? Thepreviousactivityrevealedtousthetemperature-pressurerelationship at constant volume in gases. Can you think of some phenomena which can be explained by this relationship? Are you familiar with the pressure cooker? The pressure cooker is airtight, so pressure builds up inside the pressure cooker as the liquid inside comes to a boil. The resulting trapped steam causes the internal temperature to rise more than what it can normally do at normal atmospheric pressure. Thus, the cooking of hard meat and fibre is done at a short period of time. This means that when the temperature of gases increases its pressure also increases or vice versa. Hence, we can state the Gay-Lussac’s Law as: At constant volume, the pressure of a fixed mass of gas is directly proportional to the absolute temperature. Gay-Lussac’s Law can be expressed mathematically as P α T at constant Volume It is can be written as: P P = k T or k = T Since there is a direct proportionality between the pressure and temperature of gases at constant volume, it can be shown in this equation: P1 P2 = T1 T2 The person who is credited with the determination of the temperature-pressure relationship in gases at constant volume is Joseph Louis Gay-Lussac. He deduced that the pressure of the gas is directly proportional to its temperature. Joseph Louis Gay-Lussac (1746- 1823) All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 34. D EPED C O PY 378 Consider this table: Table10. Data on Temperature-Pressure Relationship of Gases Trial Pressure (atm) Temperature (K) P/T 1 1.0 100 2 2.0 200 3 3.0 300 4 4.0 400 Plot a Temperature-Pressure graph using the data in the Table 10. What kind of relationship is depicted in the graph? Let us apply Gay-Lussac’s Law in problem solving: Sample Problem: The pressure of a nitrogen gas inside a rigid tank is 1.5 atmosphere at 30°C. What will be the resulting pressure if the tank is cooled to 0°C? Identify the given: Initial Conditions Final Conditions P1 = 1.50 atm P2 = ? T1 = 30o C = 303 T2 = 0o C = 273K All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 35. D EPED C O PY 379 Convert the temperatureto Kelvin. K = °C + 273 K = 30° + 273 = 303K Then substitute the given values to this equation. P1 / T1 = P2 / T2 P2 = P1 T2 / T1 P2 = (1.50 atm) (273 K) / 303 K = 1.35 atm Were you able to determine correctly that there will be a decrease in the pressure of nitrogen gas? That’s the beauty of understanding the relationship between temperature and pressure of gases. Practice makes perfect! Answer the following problems on Gay-Lussac’s Law to ensure mastery of concepts on the temperature-pressure relationship: 1. A certain light bulb containing argon has a pressure of 1.20 atm at 18°C. If it will be heated to 85°C at constant volume, what will be the resulting pressure? Is it enough to cause sudden breakage of the bulb? 2. At 20°C a confined ammonia gas has a pressure of 2.50 atm. At what temperature would its pressure be equal to 760 mmHg? 3. The helium tank has a pressure of 650 torr at 25°C. What will be the pressure if the temperature is tripled? You have demonstrated pretty well your skills in problem solving. Good job! All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 36. D EPED C O PY 380 Let’s have areview: Table 11. Gas Laws’ Working Formula Gas Law Working Formula Boyle’s Law V1 P1 = V2 P2 Charles’ Law V1 = V2 T1 T2 Gay-Lussac’s Law P1 = P2 T1 T2 The above cited laws show the relationship of two variables in gases. In the next activity, you will observe the interrelationship among the three variables of gases as to volume, temperature, and pressure. Activity 5 Combined Gas Laws Objective: Determine the relationship among temperature, pressure, and volume of gases at constant number of moles. Materials: • liquid dropper • cylindrical container with cover • denatured alcohol • match/candle • ruler Procedure: 1. Get a cylindrical container made of hard carton and bore a hole near its bottom. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 37. D EPED C O PY 381 3. Cover andhold the cylindrical container in such a way that your thumb is covering the hole near the base. 4. Shake the container vigorously for 1 minute. 5. Place the container on the table or arm rest. As quickly as possible, place a lighted match/candle near the hole. Observe what will happen. Cautions: The container of the denatured alcohol should be placed as far as possible from your working area because it is flammable. Immediately wash your hands with plenty of water after this procedure. 2. Remove the cover of the cylindrical container and put 5 drops of denatured alcohol. Caution: Denatured alcohol is toxic or poisonous. It can cause blindness. BE CAREFUL! All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 38. D EPED C O PY 382 Q1. What happensto the cylindrical container when a source of heat is placed near the hole? Q2. Why do you need to shake the container after putting 5 drops of denatured alcohol? Q3. How is the volume of the gases related to its temperature and pressure? Can you think of applications involving combined gas law? The weather balloon which carries instruments upward to be able to send back information on atmospheric pressure, humidity, temperature, and wind speed through radiosonde also applies Combined Gas Law. As the weather balloon rises up from the ground, it responds to three variable changes in the surroundings; volume, pressure, and temperature. Have you ever notice the warning label in the aerosol container? What is the temperature requirement for its storage? Have you seen an explosion of a can of this kind? The explosion of this container is also an application of Combined Gas Law.“The exposure to high temperature increases the kinetic energy of the gases causing an increase in the pressure due to the increased collision of the gases on the walls. An increase in pressure would result in expansion of volume. But because the can is contained, thus the container explodes.” No one is credited for the Combined Gas Law. Putting together Boyle’s Law and Charles’ Law together will result to this statement. The pressure and volume of a gas are inversely proportional to each other, but are both directly proportional to the temperature of that gas. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 39. D EPED C O PY 383 Translating it tomathematical equation will give us the following: VP kT kT PV T = or V = or P = or k = K P V T The constant k in the equation above is known as the universal gas constant. It is the result of the combination of the proportionality constants in the three gas laws. Note that the formula is equal to a constant, thus it is possible to compute for the change in volume, temperature, or pressure using the following proportion: P1 V1 P2 V2 = T1 T2 Let’s use the Combined Gas Law in determining change in the final volume, temperature, or pressure of gases. Sample Problem: The oxygen tank manufacturer used to produce 5.0 L oxygen tanks at 2000 psi and 25°C . Statistics suggests that the 3.0 L oxygen tank at 1500 psi more marketable. What temperature requirement is needed to produce a 3 L oxygen tank at 1500 psi? The given values are: Initial Conditions Final Conditions V1 = 5.0 L V2 = 3.0 L T1 = 25o C = 298K T2 = ? P1 = 2000psi P2 = 1500 psi Computing for temperature requirement: P1 V1 P2 V2 = T1 T2 T1 P2 V2 T2 = P1 V1 (298K) ((1500psi) (3.0L) T2 = (2000psi) (5.0L) T2 = 134K ≈ 130 K All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 40. D EPED C O PY 384 Answer the followingproblems: 1. Helium gas has a volume of 250 mL at 0°C at 1.0 atm. What will be the final pressure if the volume is reduced to 100 mL at 45°C? 2. The volume of a gas at 27°C and 700.0 mmHg is 600.0 mL. What is the volume of the gas at -20.0°C and 500.0 mmHg? 3. A 2.5 L of nitrogen gas exerts a pressure of 760 mmHg at 473 K. What temperature is needed to reduce the volume to 1.75 L at 1140 torr? It is really important to know how the properties of gases affect us and our environment. There is a lot more as you move on to the next activities. Do you still remember the mole concept? Can you still recall what a mole means? The number of moles quantifies the amount of a substance. What could be the possible relationship of the amount of gas in a mole to its volume? Can you make a prediction about it? In 1811, Avogadro wrote in a paper that, “Equal volumes of all gases, kept at the same pressure and temperature, contain the same number of molecules.” Avogadro was the first to suggest that the volume of a gas is directly proportional to the number of moles of gas present at a given temperature and pressure. If the volume of gases is directly proportional to the number of mole whose symbol is n, what will be the mathematical equation for the volume-mole relationship? Can you still recall the way we represent the relationship in a mathematical equation? During the first half of the nineteenth century, Lorenzo Romano Amedeo Carlo Avogadro, Count of Quaregna and Cerreto, made important contributions in shedding light on reaction stoichiometry. He provided explanations as to why compounds reacted in definite ratios and on how the amount of gas affects its volume. Experimentally, the most convenient way of quantifying the amount of gas is through its mass.Avogadro played an important role in providing evidence of the existence of atoms. Eventually the number of molecules in a mole is named after him. Amedeo Avogadro (1776-1856) All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 41. D EPED C O PY 385 Using the proportionalitysymbol, we can express the proportionality between the volume and the number of mole of a gas as: V α n at constant T and P Mathematically, the Avogadro’s Hypothesis can be expressed as: V = k n where V is the volume of gas n is the amount of gas in moles and k is a proportionality constant This can also be expressed as: V1 V2 = or V1 n2 = V2 n1 n1 n2 Let’s have this table: Table 12. Data on Avogadro’s Hypothesis Volume (L) No. of moles (mol) V/n (L/mol) 2.50 0.50 5.00 1.0 7.50 1.5 10.00 2.0 12.50 2.5 Did you obtain a constant value for V/n ? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 42. D EPED C O PY 386 PredicthowtheVolume-Molegraphwouldlooklike.Verifyyourprediction, plot a graph. Let’sapply Avogadro’s Hypothesis in solving this problem. What will be the final volume of a 5.00 L He gas which contains 0.965 mole of at 30°C and 1.00 atmosphere, if the amount of this gas is increased to 1.80 moles provided that temperature and pressure remains unchanged? As we have done in the past lessons, let’s start analysing the problem by identifying the initial and final conditions: Initial Conditions Final Conditions V1 = 5.0 L V2 = ? n1 = 0.965 mol n2 = 1.80 mol P1 = 1.00atm P2 = 1.00atm T1 = 30o C T2 = 30o C All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 43. D EPED C O PY 387 Since the temperatureand pressure are held constant, we will use this formula: V1 n2 V2 = n1 (5.0L) (1.80mol) = = 9.3L 0.965 mol Let’s have more problem sets! 1. A 7.25 L sample of nitrogen gas is determined to contain 0.75 mole of nitrogen. How many moles of nitrogen gas would there be in a 20 L sample provided the temperature and pressure remains the same. 2. Consider the following chemical equation: 2NO2 (g) N2 O4 (g) If 50.0 mL of NO2 gas is completely converted to N2 O4 gas, under the same conditions, what volume will the N2 O4 occupy? Can we observe Avogadro’s Hypothesis in real life scenarios? Try to observe the baking of bread or cake at the nearest bakery in your place. How can you explain the phenomenon of having a bigger bread or cake compared with the dough? Can you also use this law to explain the production of balloons and the way vulcanizing shop deals with flat tires? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 44. D EPED C O PY 388 Activity 6 Squashing theBottle Adopted from Apex Objective: Show the relationship among volume, temperature, pressure and number of moles. Materials: • two empty, plastic, 1.5-litre bottles with cover • hot water • ice cubes • hammer • plastic bag Procedure for Activity A: 1. Fill one-third of the bottle with hot water. 2. After a few seconds, empty the bottle and put the cover at once. Q1. What happened when you covered the bottle? Q2. What caused it to happen? Procedure for Activity B: 1. Put some ice cubes in a plastic bag. Crush the cubes with a hammer. 2. Put the crushed ice cubes in the bottle. Put the cover on. 3. Shake the bottle so that the inner portion is thoroughly chilled. Observe the bottle. Q4. What happened to the bottle? Q5. Explain the phenomenon. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 45. D EPED C O PY 389 Let’s us nowrecall the previous gas laws that we have learned in this module. The different gas laws are: 1 Boyle’s Law: V α (n and T are constant) P Charles’ Law: V α T (n and P are constant) Avogadro’s Law: V α n (P and T are constant) Combining the three laws, you will get: nT V α = P Using the sign of equality will result to this equation: RnT V = or PV = nRT P where: V = volume in liters P = pressure in atmosphere n = moles T = temperature in Kelvin R = universal gas constant, 0.0821 L. atm mol. K Do you have an idea on how we arrived at the value of proportionality constant (R)? Based on the equation above, can you state the ideal gas law in your own words? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 46. D EPED C O PY 390 The Ideal GasEquation is useful in illustrating the relationship among the pressure, volume, temperature, and number of moles of a gas. This equation is used to describe gases that behave ideally. Do gases behave ideally? Discuss among your group members and prove your answer. Validate your answer by consulting Science Teachers, reading books, and internet search to name a few. Let’s apply the ideal gas law equation in this problem: What is the volume of a container that can hold 0.50 mole of gas at 25.0°C and 1.25atm? The given are: Pressure: 1.25 atm Temperature: 25.0°C + 273 = 298 K No. of moles: 0.50 mole We are asked to calculate for the volume so let’s substitute the given values to this equation: PV = nRT nRT V = P (0.50 mole) (0.0821 L atm/mol. K) (298K) = 1.25 atm = 9.8L Let’s use the ideal gas equation in the following problems: 1. Calculate the pressure exerted by a 0.25 mole sulfur hexafluoride in a steel vessel having a capacity of 1250 mL at 70.0°C. 2. Fermentation of glucose produce gas in the form of carbon dioxide, how many moles of carbon dioxide is produced if 0.78 L of carbon dioxide at 20.1°C and 1.00 atm was collected during the process? 3. A sample of liquid acetone is placed in a 25.0 mL flask and vaporized by the heating to 75°C at 1.02 atm. The vapor weighs 5.87 g. Calculate the number of moles of the acetone. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 47. D EPED C O PY 391 Having enough informationabout the behaviour of gases you are now ready to explain the Kinetic Molecular Theory. Activity 7 A Gaseous Outlook Adopted from Apex Objective: Determine the application of gas laws in daily occurrences. Materials: Activity A Activity B Activity C • string • sticky tape • medium-sized balloon • drinking straw • glass bottle • medium-sized balloon • sink with hot and • cold water • bowl • drinking glass • water A. Jet-Propelled Balloon 1. Thread a string through the straw and tie its ends tightly between two points at equal heights in a room (e.g., handles or hooks). 2. Inflate the balloon and keep the neck closed between your fingers. 3. Fix the balloon underneath the drinking straw with the sticky tape and pull the balloon along to one end of the string. 4. Pull your fingers against the mouth of the balloon then let go. Q1. Explain why the balloon shoots along the thread at a speed using the concept of the gas laws. Q2. What does this prove regarding the compressibility of gases? B. The Rising Water 1. Put the glass into the water upside down. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 48. D EPED C O PY 392 2. Lift theglass up, but without the rim going above the surface of the water. Observe what happens. Q1. What happened to the level of the water inside the glass? Q2. What caused this to happen? Q3. If the rim of the glass was raised above the surface of the water what might have happened? Let us try to make ourselves familiar with the Kinetic Molecular Theory and try to relate the above mentioned concepts with the said theory. Kinetic Molecular Theory states that: a. Gases are composed of molecules. The distances from one molecule to another molecule are far greater than the molecules’ dimensions. These molecules can be considered as spherical bodies which possess negligible mass and volume. Figure 1. Molecules of Gases b. Gas molecules are always in constant random motion and they frequently collide with one another and with the walls of the container. Collision among molecules are perfectly elastic, that is, energy may transfer from molecule to molecule as the result of collision but the total energy of all the molecules in the system remains the same/constant. Figure 2. Molecules of Gases in Random Motion All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 49. D EPED C O PY 393 c. There isa neither attractive nor repulsive force between or among gas molecules. d. Movement of gas molecules is affected by temperature. The average kinetic of the molecules is directly related to the temperature of gas. The Kinetic Molecular Theory (KMT) explains the properties of gases and describes the behavior of gases. You can relate the early discussions that we had with this theory. So far, you have learned that gases have mass, volume, temperature and it exerts pressure. The pressure exerted by gas molecules is due to collision among gas molecules and with the walls of the container. The frequency of collision is affected by temperature because gas molecules move faster at high temperature, on the other hand, they move slowly at low temperature. The faster the movement of the molecules, the more frequent the collision, causing an increase in pressure. Let’s check whether you understand the Kinetic Molecular Theory. Try to answer the following: Direction: Identify and underline the possible weakness or flaws in the postulates. Write TRUE if the postulate is accurate and FALSE if the postulate is flawed. Postulates 1. A gas consists of a collection of small particles traveling in straight line motion and obeying Newton’s Laws. 2. The molecules in a gas occupy negligible volume. 3. Collisions between molecules are perfectly elastic (that is, no energy is gained nor lost during the collision). 4. There are negligible, attractive, or repulsive forces between molecules. 5. The average kinetic energy of a molecule is constant. Lifted from “Applied Academics for Excellence” (APEX) www.chem.wisc.ed All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 50. D EPED C O PY 394 IV. Summary/Synthesis/Feedback • Gas is one of the phases of matter. It has no definite shape and size. It canbe compressed easily. • Properties of gases include mass, volume, temperature, and pressure. o The amount of a gas or its mass could be expressed in moles or grams. The mass of gases is negligible. o The volume of a gas is the amount of space occupied by the gases. Gases have the tendency to occupy all the spaces of the container that they are confined. They have weak intermolecular force of attraction; hence they are arranged as far away as possible from each other. The common units used in expressing the volume of a gas are liter (L) and milliliter (mL). o The temperature of a gas is the measure of the hotness or coldness of an object. It is proportional to the average kinetic energy of its molecules. It can be measured in Celsius or Kelvin. Kelvin is the absolute scale. o The pressure of a confined gas is the average effect of the forces of the colliding molecules. It can be measured in atmosphere, torr, psi, cmHg or mmHg. It can be quantified using this equation: P =F/ A Where: P= pressure, F = force, and A= area F = ma Where: F = force, m = mass and a = acceleration • The properties of gases can affect one another. They are related to each other. o The volume of a gas is directly related to its temperature at constant pressure. o The pressure of a gas is directly related to its temperature at constant temperature. o The volume of a gas is inversely related to its pressure at constant temperature. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 51. D EPED C O PY 395 o The amountof a gas in a mole is directly related to its volume at constant pressure and temperature. • The properties of gases can be varied. The relationships of these properties can be quantified experimentally with the aid of the different laboratory apparatus or by using the different gas laws as follows: o Boyle’s Law V1 P1 = V2 P2 o Charles’ Law V1 / T1 = V2 / T2 o Gay-Lussac’s Law P1 / T1 = P2 / T2 o Avogadro’s Law V1 n2 = V2 n1 o Combined Gas Law V1 P1 T2 = V2 P2 T1 o Ideal Gas Law PV=nRT Where V= volume, P = pressure, T = temperature, n = amount of gas in moles R = 0.0821 L.atm./mol.K 1= used as a subscript,means initial condition 2= used as a subscript,means final condition • Not all gases behave ideally. Most of the gases found in nature conform to the principles of Boyle’s Law, Charles’ Law, Gay-Lussac’s Law, Avogadro’s Law, and Combined Gas Law. • The following conversion factors are useful in solving gas law related problems: a. For volume 1 mL = 1 cm3 1 L = 1 dm3 1 m3 = 1000 L http://www.metric-conversions.org/volume/cubic-meters-to-liters.htm b. For pressure 1 atm = 760 mmHg = 76 cmHg = 760 torr = 101,325 Pa = 14.6956 psi c. For temperature 0˚C = 273.15K 0˚C = 32 ˚F All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 52. D EPED C O PY 396 • The behaviorof the gas molecules can be explained by the Kinetic Molecular Theory. It states that: a. Gases are composed of molecules. The distances from molecule to molecule are far greater than the molecules’ dimensions. These molecules can be considered as spherical bodies which possess negligible mass and volume b. Gas molecules are always in constant random motion and they frequently collide with each other and with the walls of the container. Collisions among molecules are perfectly elastic, that is, energy may transfer from molecule to molecule as the result of collision, but the total energy of all the molecules in the system remains the same/constant. c. There is a negligible attractive or repulsive force between or among gas molecules. d. Movement of gas molecules is affected by temperature. The average kinetic of the molecules is directly related to the temperature of gas. V. Summative Assessment 1. Jane can still pump air in the party balloon even though it is already inflated. What explains this phenomenon? a. balloons look better if its size is bigger b. balloons are made up of plastic c. the air inside the balloon is hot d. air molecules can be compressed 2. What is most likely to happen when an aerosol can is heated? a. the can will be deformed b. the can will stay the same c. the can will eventually explode d. the can will tarnish 3. Each container with varying volume has 1.0 mole of oxygen gas at 30.0°C. In which container will pressure be the lowest? 4. Which of the following phenomena best illustrates Charles’ Law? a. carbon dioxide being dissolved in water b. expansion of the balloon as it is being submerged in hot water c. breathing apparatus being used by a patient d. leavening agent causing the fluffiness of cake products All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 53. D EPED C O PY 397 5. Which ofthe following pair/s is/are correctly matched? a. I & II b. III & IV c. I, III, & IV d. I, II, III, & IV 6. Which of the following samples is highly compressible at high pressure and expandable at high temperature? a. oxygen gas b. aluminium sheet c. water d. ice 7. Records show that the incident of tire explosion is high during summer season. Which of the following gives the best explanation for this observation? a. there are more travellers during summer vacation b. high temperature during summer season causes the air inside the tire to expand c. vehicles’ tires are not well maintained d. there is too much air inside the tires 8. Which is most likely to happen when a closed vessel filled with gas is shaken for 2 minutes? a. the temperature inside the vessel increases b. the pressure inside the vessel increase c. the temperature and pressure inside the vessel increase d. both the temperature and pressure inside the vessel increase 9. Determine what will happen to the temperature of a confined gas as the pressure decreases. a. the gas temperature stays the same b. the gas temperature decreases c. the gas temperature increases d. there is no enough data 10. Gab wants to have a portable oxygen tank. A 5.00 liter oxygen gas exerts a pressure of 1.00 atmosphere. How much pressure is needed for this gas to be compressed in a 2.00 liter cylinder, provided there is no temperature change? a. 3.0 atm b. 2.5 atm c. 2.0 atm d. 1.5 atm All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 54. D EPED C O PY 398 11. The temperatureof nitrogen gas contained in a not rigid vessel is reduced from 100°C to 5.0°C? Which of the following describes the resulting behavior of nitrogen gas molecules? I. The average kinetic energy suddenly increases, thus the pressure increases II. The average kinetic energy suddenly decreases, thus the pressure decreases III. The volume occupied by the gas molecules suddenly increases, thus the container expand IV. The volume occupied by the gas molecules suddenly decreases, thus the container shrink a. I & III b. II & IV c. I & IV d. II & II 12. A balloon with a volume of 200 mL at 30°C is submerged in hot water to obtain a temperature of 50°C. Find out what will happen to the volume of the balloon, provided the pressure remains the same. a. the volume of the balloon will become higher than 200 mL b. the volume of the balloon will become lower than 200 mL c. the volume of the balloon will stay the same d. there is no enough data 13. A 2.0 g (approximately 0.045 mole) sample of dry ice (solid carbon dioxide) is placed in an evacuated 3.5 L vessel at 30°C. Compute for the pressure inside the vessel after all the dry ice has been converted to carbon dioxide gas. (R=0.0821 L. atm/mol.K) a. 0.32 atm b. 0.45 atm c. 0.67 atm d. 1.0 atm 14. What is the explanation to your answer in item number 13? a. the gaseous form of dry ice exerts the same pressure with its environment because it adopts the atmospheric pressure b. the gaseous form of dry ice exerts lower pressure due to the bigger volume that results to lesser collisions of the gas particles. c. the gaseous form of dry ice will have the same pressure because its composition remains the same d. the gaseous form of dry ice will either have high or low pressure 15. What do you expect to happen to the volume of a gas if its pressure is doubled and its temperature is reduced to half? a. its volume is increased b. its volume is doubled c. its volume remains unchanged d. its volume is decreased All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 55. D EPED C O PY 399 References and Links PrintedMaterials: Briones, L.L., Templora, V. F., Tibajares, I. S. Jr. (2010). Chemistry Power Science and Technology III, vol.2, Manila: St. Mary’s Publishing Corp. Chang, R. (1998). Chemistry, 6th ed.,Boston:Mc.Graw-Hill Davis, R. E., Sarquis, M., Frey, R., Sarquis, J. L., (2009). Modern Chemistry. Teacher’s Ed., Orlando: Holt , Rinehart and Winston LeMay, E.H. Jr., Robblee, K.M., Brower, H., Brower D.C., Beall H. (2000). Chemistry Connections to Our Changing World. 2nd ed., New Jersey: Prentice Hall, Inc. Electronic Sources: http://chemteacher.chemeddl.org/services/chemteacher/index. php?option=com_content&view=article&id=9 http://www.chm.davidson.edu/vce/GasLaws/AvogadrosLaw.html http://www.grc.nasa.gov/WWW/K-12/airplane/Animation/frglab2.html http://phet.colorado.edu/simulations/sims.php?sim=Gas_Properties http://www.mhhe.com/physsci/chemistry/essentialchemistry/flash/gasesv6.swf http://intro.chem.okstate.edu/1314F00/Laboratory/GLP.htm http://preparatorychemistry.com/Bishop_animations.htm http://www.chemistry.co.nz/avogadro.htm http://www.chemteam.info/GasLaw/Gas-Avogadro.html https://www.khanacademy.org/science/chemistry/ideal-gas-laws/v/ideal- gas-equation--pv-nrt http://www.articlesbase.com/k-12-education-articles/avogadros-law- problems-with-solutions-6621701.html http://www.chm.davidson.edu/vce/GasLaws/GasConstant.html http://www.britannica.com/EBchecked/topic/475388/pressure http://wps.prenhall.com/wps/media/objects/602/616516/Chapter_09.html http://www.aiche.org/sbe/conferences/international-conference-biololecular- engineering-icbe/2013/events/experimental-computational-tools- engineering-biomolecules https://encryptedtbn0.gstatic.com/images?q=tbn:ANd9GcTXN5x_t_ qB2uvF19bKgYvegm1_bLTRKOr9CShmeSb5LRWyuJiu Photos by: Francis E.Ansing and Ma. Victoria G. Senase All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 56. D EPED C O PY 400 I. Introduction From thetime we get up in the morning to the time that we sleep at night, chemical changes are taking place, within us and outside of us. Plants grow through photosynthesis, foods that we eat are digested by the body, metals corrode, raw materials are being converted to useful products, new medicines are being developed, more versatile and cost effective materials are being made. Various chemical changes that occur around us have significant effects to our environment and consequently to our health. Chemical changes occurring in industries result to products that are useful to us. The wastes we throw continue to undergo chemical changes and this has an impact on our well- being as well. The irresponsible use of fertilizers, herbicides and pesticides have negatively affected plants and aquatic life. We continue to pollute the atmosphere with vehicle and industrial gas emissions. In your lower grade levels, you were exposed to some chemical reactions, you’ve tested the reactivity of some metals and you’ve seen the color changes of an indicator when tested with acids and bases. You have also learned in chemical bonding, that atoms gain stability by losing or gaining electron/s. In this module , you will further understand how a chemical change proceeds, how bonds are broken and new bonds are formed, and how chemical reactions are translated into chemical equations, where rearrangements of atoms causes the formation of new substance/s. A lot of these chemical changes made the quality of our lives better. CHEMICAL REACTION Unit 4 MODULE 2 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 57. D EPED C O PY 401 This module containsthe following lessons and activities: 1. Identifying chemical change Evidences of chemical reactions Chemical equation 2. Types of chemical reactions 3. Law of conservation of mass 4. Factors affecting reaction rate II. Learning Competencies/Objectives The learner should be able to: 1. Write chemical equations; 2. Apply the principles of conservation of mass to chemical reactions; 3. Classify reactions according to the different types; 4. Identify the factors that affect reaction rates and explain them according to collision theory; and 5. Explain how the factors affecting rates of chemical reactions are applied in food preservation and materials production, fire control, pollution, and corrosion. III. Pre-Assessment 1 -5 Multiple Choice. Choose the correct answer. 1. During a chemical reaction, a. atoms are destroyed b. atoms are rearranged c. elements are destroyed d. new elements are produced How do chemical reactions take place? What is the significance of studying the rates of reaction? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 58. D EPED C O PY 402 2. A chemicalreaction is a process in which a. all reactants change state b. products change into reactants c. the law of conservation of mass applies d. all of these 3. What determines an atom’s ability to undergo chemical reactions? a. protons b. neutrons c. innermost electrons d. outermost electrons 4. How is a chemical equation is balanced? a. changing subscripts b. erasing elements as necessary c. adding coefficients d. adding elements as necessary 5. What are the products in the equation below? Zn + CuSO4 -----> ZnSO4 + Cu a. Zn and Cu b. Zn and CuSO4 c. ZnSO4 and Cu d. Zn only 6 -10 Write true if the statement is correct and false if incorrect, and change the underlined word/s to make the statement correct. 6. Generally, the higher the concentration of the reacting substances, the faster is the reaction. 7. At lower temperature, chemical reactions occur at slower rates. 8. The bigger the surface area of the reactants, the faster the rate of reaction. 9. Catalysts increase the rate of reaction by providing a reaction pathway with a higher activation energy. 10. The minimum energy required to start a reaction is called bond energy. 11 - 12 Balance the following chemical equations, then classify the reaction according to its type Chemical Equation Type of Reaction 11. CaCO3 = HCI ---> CaCI2 + H2 CO3 12. AqNO3 = Zn ---> Zn (NO3 )2 + Ag All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 59. D EPED C O PY 403 13-15 Explain inconcise and brief sentences. 13. What is the function of MnO2 in the production of oxygen from hydrogen peroxide in this reaction: H2 O2 H2 O + O2 MnO2 14. Why would iron fillings rust faster than an iron nail? 15. Enzymes are in molds and bacteria that spoil food. Explain, using your knowledge of factors affecting the rate of reaction, why food doesn’t spoil as fast when it is refrigerated as it would at room temperature. IV. Reading Resources and Instructional Activities How do you know if a certain change that has taken place involves a chemical reaction? What indicators/ evidences should be present to consider it a chemical reaction? Activity 1 will help you identify those indicators/evidences of chemical reactions. Activity 1 Everything has changed Objectives: • Perform a laboratory activity involving chemical reactions; • Distinguish evidences of chemical reactions. Materials: • Mg ribbon (Mg) • Iron nail (Fe) • 30 volumes Agua oxigenada • Hydrogen peroxide (H2 O2 ) • Manganese dioxide (MnO2 ) • 10% copper sulfate (CuSO4 ) solution • 10% sodium hydroxide (NaOH) solution • Denatured alcohol • Vinegar • Baking soda • Matches • Alcohol lamp • Tripod • Crucible tong • Beakers or small transparent bottles • Test tubes • Test tube rack • Thermometer • Forceps or crucible tong • Iron nail/shoe tack • spatula or small teaspoon All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 60. D EPED C O PY 404 Procedure A. IronNail-Copper Sulfate Reaction 1. Fill a test tube with 10 mL of copper sulfate solution. 2. Drop the nail gently into the solution. 3. Place the test tube in the test tube rack for a few minutes. (You may proceed to the next procedure while waiting for any change.) Table 1. Iron Nail-Copper Sulfate Reaction Materials Color Before Mixing Color After Mixing Copper solution Nail Q1. What happened to the color of the copper sulfate solution? Q2. What happened to the color of the nail? Procedure B. Magnesium Ribbon Reaction 1. Cut about 10 cm of magnesium ribbon. 2. Light the alcohol lamp. 3. Hold the magnesium ribbon with a crucible tong or forceps. 4. Place the magnesium ribbon over the flame. Q3. What happened to the magnesium ribbon when you directly burned it? Q4. What substance in the air could have reacted with magnesium during burning? Q5. Describe the appearance of the product formed. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 61. D EPED C O PY 405 Table 2. MagnesiumRibbon Reaction Materials Before Burning During Burning Color Appearance Color Appearance Magnesium Procedure C. Hydrogen Peroxide (Agua Oxigenada) Reaction This procedure should be done fast. 1. Pour 20 mL of agua oxigenada in a small beaker. 2. Using a spatula add a pinch of manganese dioxide (MnO2 ) to the beaker. Q6. What happened to the mixture? 3. Place a lighted match on top of the beaker near the bubbles. (Figure 1) 4. Observe what happens to the flame. Q7. Describe the change you observe in the flame. Table 3. Hydrogen Peroxide (Agua Oxigenada) Reaction Material Before Reaction With addition of MnO2 Agua oxigenada Procedure D. Vinegar and Baking Soda Reaction This procedure should be done swiftly 1. Pour 20 mLof vinegar in a small beaker. 2. Get the temperature of vinegar. 3. Add a tablespoon of baking soda to the beaker. Figure 1 Figure 2 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 62. D EPED C O PY 406 Q8. What doyou observe in the mixture? 1. Place a lighted match on top of the beaker near the bubbles. ( Figure 2) 2. Observe what happens to the flame. Q9. Describe what you observe in the flame. Table 4. Vinegar and Baking Soda Reaction OBSERVATION Material Before Reaction During Reaction vinegar baking soda Procedure E. Copper Sulfate-Sodium Hydroxide Reaction 1. Pour 5 mL of aqueous copper solution in one test tube. 2. Pour 5 mL of aqueous sodium hydroxide in another test tube. 3. Slowly combine the two solutions. 4. Observe what happens. Q10. What did you observe at the bottom of the test tube? 5. Shake the mixture. 6. Observe what happens. Q11. Compare the appearance before and after shaking Table 5. Copper Sulfate-Sodium Hydroxide Reaction APPEARANCE Materials Before Reaction After Reaction (copper sulfate + sodium hydroxide) Copper sulfate solution hydroxide solution All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 63. D EPED C O PY 407 You have learnedin your Grade 9 Chemistry that substances undergo chemical bonding so that atoms can become more stable. Chemical bonding results to breaking of bonds and formation of new bonds, thus new substances are formed. Formation of new substances means a chemical reaction has taken place. Oxygen is vital to life. One interesting reaction which involves oxygen is the production of fire. Fire has fascinated people for so long, that the ancient people even regarded it as one of the earliest elements. Fire was so important to them and they described it as an element that changes everything. The earliest theory about burning was the Phlogiston Theory. This theory by George Ernst Stahl in the 17th century stated that when a material burns, it releases a substance known as phlogiston, and this theory was accepted for a very long time. Antoine Lavoisier through his careful observations from his experiments, debunked the phlogiston theory as he discovered that instead of releasing a substance (phlogiston) a material accurately burns as it reacts (uses) with oxygen. This is now known as the Theory of Oxidation, and this is accepted up to this day. KEY CONCEPTS: When a physical change occurs there is no breaking and forming of bonds. There are certain things that will help us identify if a chemical reaction has taken place. We call these evidences of chemical reactions. 1. Production of light 2. Evolution of gas 3. Temperature change 4. Change in intrinsic properties (color, odor) 5. Formation of precipitate Figure 3 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 64. D EPED C O PY 408 For burning tooccur, 3 factors should be present in proper conditions and proportions. 1. Fuel 2. Oxygen 3. Heat In our country, we are reminded that March is a Fire Prevention month, as this month signals the start of summer, the season when countless fires break out all over the country, “An ounce of prevention is better than a pound of cure” is a motto we all need to remember. Various materials acts as fuel to sustain fire, so various fire prevention and control measures are Later as you progress in your lessons, you will get to learn more chemical reactions which may bring benefit or harm to life as well as to the environment. In the next activity you will learn how chemical reactions can be presented in a shorter way. It is through this presentation that chemical reactions will later be analyzed for classification. Activity 2 “What’s in a Reaction?” Objectives: • Distinguish between reactants and products. • Write a chemical equation. http://pslc.ws/fire/howwhy/triangle.htm (accessed: (Mar.4, 2014) Figure 4 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 65. D EPED C O PY 409 A. Reactants andProducts. Reactants are substances that are used up to form new substances in a chemical reaction. The following chemical reactions took place in Activity 1 procedure A to E. 1. Iron reacts with copper sulfate (CuSO4 ) and forms iron (II) sulfate (FeSO4 ) and copper. 2. Magnesium combines with oxygen gas (O2 ) to produce magnesium oxide 3. Hydrogen peroxide (H2 O2 ) in the presence of manganese dioxide (MnO2 ) produces water and oxygen gas. 4. Acetic acid (CH3 COOH) and sodium bicarbonate (NaHCO3 ) produce sodium acetate with the release of carbon dioxide (CO2 ) gas and water. 5. Copper sulfate (CuSO4 ) reacts with sodium hydroxide (NaOH) to produce insoluble copper (II) hydroxide Cu(OH)2 and sodium sulfate (Na2 SO4 ) solution. Fill in the table below with the Reactants and Products from the chemical reactions above. Below each number, write the symbol or formula of the reactant and product. Table 6. Reactants and Products Reaction Reactants Products 1 2 3 4 5 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 66. D EPED C O PY 410 B. Symbols usedin Chemical Equation There are other symbols used in writing a chemical equations: Table 7. Symbols and their Meanings Symbol Meaning + to show combination of reactants or products To produce; to form; to yield (s), (l), (g), (aq) (s)-solid (l)-liquid (g)-gas (aq)-aqueous (substance is dissolved in water) Reversible reaction Heat Indicates that heat is supplied to the reaction Pt A formula written above or below the yield sign indicates its use as a catalyst or solvent Using the symbols and formulas in Table 6 and the symbols in Table 7, write the chemical reaction using these symbols to complete chemical equation. Table 8 Chemical Equation Reaction Chemical Equation 1 2 3 4 5 In the next activity you will classify the chemical reactions you encountered in the laboratory activity “Everything has changed”. KEY CONCEPTS: A chemical equation is a chemist’s shorthand for a chemical reaction. The equation distinguishes between the reactants, which are the starting materials and the products which are the resulting substance/s. It shows the symbols or formulas of the reactants and products, the phases (solid, liquid, gas) of these substances, and the ratio of the substances as they react. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 67. D EPED C O PY 411 Activity 3 We SimplyClick Together Objectives: • Classify reactions according to their types, based on how atoms are grouped or regrouped. • Classify chemical reactions. Materials: • Activity Guide • Students tabulated data from activity 2 “What’s in a Reaction?” Procedure: 1. Bring out your filled up (answered) table from activity 2 ”What’s in a Reaction?” Guide Questions: Q12. In the second chemical reaction, how many reactants are used? How many product/s is/are formed? Q13. In the third chemical reaction, how many reactants are used? How many product/s is/are formed? Q14. In the first chemical reaction, what changes did copper and iron undergo during the reaction? What can you conclude about iron? Q15. In the 4th chemical reactions, how many reactants and products are involved? What kind of substance are they? Q16. In the fifth chemical reaction, both the reactants and products are compounds made up of positive and negative ions, what did you notice with the pairing of the positive and negative ions in the reactant and product side? 2. Refer to the guide card in classifying these six chemical reactions. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 68. D EPED C O PY 412 GUIDE CARD A. COMBINATION(Synthesis) REACTION: A reaction when 2 or more reactants combine to form a single product. The general formula for this reaction is : A + B AB B. DECOMPOSITION REACTION: In this reaction, a single reactant breaks down into simpler ones. (2 or more products). This is the reverse of combination reaction. The general formula for this reaction is: AB A + B C. SINGLE DISPLACEMENT (Replacement) REACTION: This is when one element replaces another element from a compound. The more active element takes the place of the less active element in a compound. The general formula for this reaction is: A + BC AC + B D. DOUBLE DISPLACEMENT REACTION (Metathesis): This is when the positive ions (cations) and negative ions (anions) of different compounds switch places, forming two entirely different compounds. The general formula for this reaction is: AB + CD AD + CB E. COMBUSTION (Burning) REACTION: This is when oxygen combines with a hydrocarbon (compound containing hydrogen and carbon) to form a water and carbon dioxide. Example of which is the burning of butane gas C4 H 10 + O2 CO2 + H2 O F. ACID-BASE REACTION: This is a speacial kind of double displacement reaction that takes place when an acid and base react with each other. The H+ of the acid reacts with the OH- of the base forming water. The other product is salt. Example of which is: HCI + NaOH NaCI = H2 O All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 69. D EPED C O PY 413 KEY CONCEPTS: Chemical reactions can be classified according to the following types: A. COMBINATIONREACTION: Reactants combine to form a single product. The general formula for this reaction is: A + B ------→ AB B. DECOMPOSITION REACTION: In this reaction, a single reactant breaks down into simpler ones. ( 2 or more products). This is the reverse of combination reaction. The general formula for this reaction is : AB ------→ A + B C. SINGLE DISPLACEMENT (Replacement) REACTION. This is when one element replaces another element from a compound. The more active element takes the place of the less active element in a compound. The general formula for this reaction is: A + BC ------→ AC + B D. DOUBLE DISPLACEMENT REACTION (Metathesis). This is when the positive ions (cations) and negative ions (anions) of different compounds switch places, froming two entirely different compounds. The general formula for this reaction is: AB + CD → AD + CB E. COMBUSTION (Burning) REACTION This when oxygen combines with a hydrocarbon to form water and carbon dioxide. F. ACID-BASE REACTION: This is a special kind of double displacement that takes place when an acid and base react with each other. HCl + NaOH ----→ NaCl + H2O Table 9. Types of Chemical Reactions Reaction Chemical Equation Type of Chemical Reaction 1 2 3 4 5 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 70. D EPED C O PY 414 In the nextactivity, you will see how mass is conserved during a chemical reaction and how this is explained by the Law of Conservation of Mass. Activity 4 How much can you take? Objective: • Perform an activity that illustrates Law of Conservation of Mass. Part 1. Laboratory Activity on Law of Conservation of Mass Materials: • Steel wool • 10% CuSO4 solution • Test tube • Rubber/cork stopper • Test tube holder • Beaker • Alcohol burner • Wire gauze • Tripod • Matches Procedure: 1. Place a dry and clean test tube and a rubber/ cork stopper in a dry and clean 100 mL-beaker. 2. Get the total mass of the dry and clean test tube and the stopper, and the 100 mL-beaker. Record it in Table 10. ENRICHMENT: Classify the following unbalanced chemical equations according to the six types of chemical reactions. A. Combination B. Decomposition C. Single displacement D. Double displacement E. Combustion F. Acid-base 1. NaOH + KNO3 NaNO3 + KOH 2. CH4 + O2 CO2 + 2 H2 O 3. Fe + NaBr FeBr3 + Na 4. CaSO4 + Mg(OH)2 Ca(OH)2 + MgSO4 5. NH4 OH + HBr H2 O + NH4 Br 6. P4 + O2 P2 O5 7. NaNO3 NaNO2 + O2 8. C18 H18 + O2 CO2 + H2 O 9. H2 SO4 + NaOH NaSO4 + H2 O 10.NiSO4 + Li3 PO4 Ni3 (PO4 )2 + Li2 SO4 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 71. D EPED C O PY 415 3. Place asmall portion of steel wool in the test tube. 4. Add 10 mL CuSO4 solution. 5. Cover the mouth of the test tube with the rubber/ cork stopper . 6. Get the mass of the set-up using the same 100mL-beaker. Record the mass in Table 10. 7. Heat the lower part of the test tube gently for 2 minutes while moving it to and fro. Make sure that the rubber/ cork stopper covers the mouth of the test tube and the test tube is held with a test tube holder in a slanted position. Q17. Describe the appearance of the steel wool. Q18. What is the evidence that a chemical change happened? 8. Allow the test tube to cool completely in the 100-mL beaker. 9. Get the mass of the set-up again. Record your observation in Table 10. Table 10. Law of Conservation of Mass BEFORE HEATING Mass (g) (a) Mass of the test tube, stopper, and beaker (b) Mass of the test tube, stopper, and beaker and Mass of the Steel wool + CuSO4 solution (c) Mass of the Steel wool + CuSO4 solution [(b)+(a)] Total Mass of Reactants: AFTER HEATING (d) Mass of the test tube, stopper, and beaker and Mass of the Steel wool + CuSO4 (e) Mass of the Steel wool + CuSO4 solution [(d)-(a)] Q19. Why is it important for the test tube to be sealed? Q20. How will you compare the total mass before and after the reaction ? Part 2. Paper Clip Reaction Model Materials: • 1 box of different colored paper clips • Periodic table All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 72. D EPED C O PY 416 Procedure: 1. Sort outyour paper clips according to color. Designate a color for each element. Element Color of paper clip Hydrogen (H) White Nitrogen (N) Blue Oxygen (O) Red 2. By connecting paper clips together (follow the color coding in number (1), make model representations for these molecules : a. O2 , H2 , H2 O Prepare at least 3 sets of each molecule as shown in the figure below. b. N2 , H2 , NH3 Prepare at least 4 sets of each molecule 3. You will be working on balancing 2 chemical equations. a. H2 + O2 → H2 O b. N2 + H2 → NH3 4. Starting with the first equation: a. Break up one set of O2 since H2 O has only 1 Oxygen. b. Connect this single O atom to the one set of H2 you have prepared to form 1 set of H2 O c. Get another set of H2 and connect to the single O atom left to form a new set of H2 O. Figure 5.1 Figure 5.2 Figure 6.1 Figure 6.2 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 73. D EPED C O PY 417 Guide Questions: Q21. Howmany set/s of H2 have you used? ______ Q22. How many set/s of O2 have you used? ______ Q23. How many set/s of H2 O have you created? ______ These number of set/s represent coefficient which is the whole number placed before the formula of the reactants and products. Q24. Write the corresponding coefficients in the chemical equation. NOTE: If there is only one set, we do not write 1 anymore. 5. Do the same with the second equation ____N2 + ____ H2 → ____ NH3 NOTE: You can use more than 2 sets. Guide Questions: Q25. How many set/s of N2 have you used? ______ Q26. How many set/s of H2 have you used? ______ Q27. How many set/s of NH3 have you created? ______ Q28. Write the corresponding coefficients in the chemical equation. ____N2 + ____ H2 → ____ NH3 _____ H2 + _____O2 > _____H2 O Figure 7 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 74. D EPED C O PY 418 NOTE: If thereis only one set , we do not write 1 anymore. 6. Get the molar mass of N2 , H2 , and NH3 , multiply their masses by their coefficient, then get the total mass of the reactants and compare to the total mass of the products. The first equation is done for you. molar mass (g/mol) : H=1 O=16 N=14 2 H2 + O2 → 2 H2 O 2(2 x1g/mol) + (2 x 16g/mol) → 2 [ (2x 1g/mol) + 16g/mol ] 4 g/mol + 32 g/mol → 2 ( 18 g/mol) 36g → 36g Q29. Do the same with the second equation Q30. How will you compare the total mass of the reactants and the total mass of the products? This now follows the Law of Conservation of Mass. KEY CONCEPTS: Law of Conservation of Mass states that mass is conserved in a chemical reaction. The total mass of the reactants is equal to the total mass of the products. No new atoms are created or destroyed, there was only grouping or regrouping (rearrangement) of atoms. _____N2 + _____H2 > _____NH3 Figure 8 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 75. D EPED C O PY 419 The next activityreinforces your knowledge of Law of Conservation of Mass by balancing the chemical equations, involving the chemical reactions in the previous activity you performed. Activity 5 Balancing Act Objectives: • Recognize that the number of atoms of each element is conserved in a chemical reaction as atoms in the reactants only rearrange themselves to form the products • Apply the concept of Law of Conservation of Mass in balancing chemical equations Material: Table 11. Types of Chemical Reactions Reaction Chemical Equation Type of Chemical Reaction 1 2 3 4 5 Procedure: 1. Analyze the informations that can be gathered in the chemical equation : 2 H2 + O2 → 2 H2 O 2 molecules of H2 + 1 molecule of O2 → 2 molecules of H2 O 2 moles of H2 + 1 mole of O2 → 2 moles of H2 O 4 atoms of H + 2 atoms of O → 4 atoms of H and 2 atoms of O Figure 9 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 76. D EPED C O PY 420 Note that thecoefficient (number) placed before the formulas indicate the number of molecules or moles. Determining the correct coefficients balances the number of atoms in the reactant and in the product side, allowing it to follow the Law of Conservation of Mass. 2. Bring out your data on Table 9 Types of Chemical Reactions, balance the chemical equations guided by the steps in balancing equations below this table. Table 11. Balanced Chemical Equations Reaction Chemical Equation Type of Chemical Reaction 1 2 3 4 5 Steps in Balancing Equations: Write the unbalanced chemical equation, make sure you have followed correctly the rules in writing formulas of compounds. • Take note of the elements present in the reactant and product side. • Count the number of atom/s of each element present in the reactant and product side. • Apply the Law of Conservation of Mass to get the same number of atoms of every element on each side of the equation.Balance chemical equations by placing the appropriate coefficients before the symbol or formula. Do not change the subscripts of the formula in an attempt to balance the equation as it will change the identity of the components. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 77. D EPED C O PY 421 ENRICHMENT: Balance the followingchemical equations, making sure to apply the principle of the Law of Conservation of Mass. 1. Zn + HCl ZnCl2 + H2 2. CH4 + O2 CO2 + H2 O 3. Fe + NaBr FeBr3 + Na 4. SiCl4 + H2 O SiO2 + HCl 5. N2 + O2 + H2 O HNO3 6. P4 + O2 P2 O5 7. NaNO3 NaNO2 + O2 8. C3 H8 + O2 CO2 + H2 O 9. Fe + H2 O H2 + Fe3 O4 10.Al + O2 Al2 O3 A burning vehicle and a puppy are undergoing a similar kind of chemical reaction. What reaction could this be? KEY CONCEPTS: • For a chemical equation to conform to the Law of Conservation of Mass, it has to be balanced. • Chemical equations are balanced by placing the appropriate coefficients before the symbols or formulas of reactants and products. • Certain steps are observed in balancing reactions. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 78. D EPED C O PY 422 In the nextactivity you will learn why chemical reactions occur and why they occur at different rates. Activity 6 Race to the Finish Line Objectives: • explain how the factors affecting rates of chemical reactions are applied in food preservation, control of fire, pollution, corrosion and materials production • recognize the importance of controlling rates of reactions in technology Part 1. Collision Theory Task: Analysis of molecular representation of collision theory. Source: http://wps.prenhall.com/wps/media/objects/3082/3156859/blb1404/bl14fg16.jpg (accessed: Oct.29, 2014) Figure 10 Source: http://i.ytimg.com/vi/OkGzaSOkyf4/maxresdefault.jpg (accessed: Oct.29, 2014) Figure 11 The illustrations above show the effective and ineffective collision of molecules to effect a chemical reaction. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 79. D EPED C O PY 423 Guide Questions: Q31. Whatcauses a chemical reaction? Q32. What must happen for a chemical reaction to take place? Q33. Describe fruitful / effective collision resulting to formation of products. In 1888 Svante Arrhenius suggested that particles must possess a certain minimum amount of kinetic energy in order to react. The energy diagram is shown below. Energy diagrams are used to analyze the changes in energy that occur during a chemical reaction. The energy of the reactants must be raised up over an energy barrier. Activation energy is the energy required to initiate a reaction and force the reactants to form an activated complex.The activated complex is located at the peak of the energy diagram for a reaction. Source: http://www.bing.com/images/search?q=Activation+enegy&go=&qs=n&form=QBIR&p q=activation+energy&sc=8-17&sp=- (1&sk=#view=detail&id=C4330FFCC22298D717 98C4462372111054F635D6&selectedIndex=96 (Accessed: July 4, 2014) Figure 11.1 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 80. D EPED C O PY 424 Source: https://www.chem.tamu.edu/class/majors/tutorialnotefiles/factors.htm (accessed: July 4,2014) Figure 11.2 Q34. What is the effect of a catalyst on the activation energy? COLLISION THEORY: Collision theory explains how collision between reactant molecules may or may not result in a successful chemical reaction. Based this theory, not all collisions between the molecules result in the formation of products. Effective collisions between molecules, which result in the formation of products, only occur when the following two conditions are met: (a) the colliding molecules should possess a minimum kinetic energy, known as activation energy, to start a chemical reaction. (b) the reactant molecules should be in correct orientation when they collide. Activation energy is needed to break the bond between reactant molecules to form new bonds leading to formation of the products. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 81. D EPED C O PY 425 PART 2: FactorsAffecting Reaction Rates In this experiment, students will study the effect that temperature, reactant concentration, particle size, catalysts and surface area have on chemical reaction rates. Equipments: • 7 clear plastic cups • mortar and pestle • 2 medium sized test tubes • 2 test tube holders Reagents: • 20 volume hydrogen peroxide (Agua oxigenada) • Manganese dioxide • water • 4 seltzer tablets or denture cleaner in tablet form • 3 5cm x 5cm colored crepe paper/ Japanese paper • 25% household bleach solution • 50% household bleach solution • 75% household bleach solution A. Effect of Particle Size or Surface Area on Reaction Rate Procedure: 1. Get 2 clear plastic cups, half fill each plastic cups with water. KEY CONCEPTS: COLLISION THEORY: Reactions can only happen when the reactant particles collide, but most collisions are NOT successful in forming product molecules despite the high rate of collisions. Reactants should have sufficient energy, and their molecules should be in proper orientation for a successful collision to happen. The minimum kinetic energy required for reaction is known as the activation energy. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 82. D EPED C O PY 426 2. Obtain twodenture cleaner tablets. Powderize one tablet using mortar and pestle. 3. Simultaneously drop the whole tablet and powderized tablet in the 2 separate plastic cups. 4. Observe the reactions for several minutes and record the time it takes for each tablet to stop fizzing . Table 12. Effect of Particle Size or Surface Area on Reaction Rate Effect of Particle Size of Surface Area on Reaction Rate Reaction Condition Reaction Rate Time (sec) denture cleaner (whole) in water denture cleaner (powderized) in water Guide Questions: Q35. a. Which tablet fizzed for a longer period of time? b. How might you explain any difference ? Q36. a. Describe in your own words the effect of particle size or surface area on the rate of a reaction. B. Effect of Temperature on Reaction Rate Procedure 5. Fill one glass with cold water and another glass with hot water. 6. Drop a denture cleaner tablet into each glass. 7. Observe the reactions that occur. Record the time it takes for each tablet to stop fizzing . Q37. Is there any noticeable difference between the two reactions? Q38. What is the effect of temperature on reaction rate? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 83. D EPED C O PY 427 Table 13. Effectof Temperature on Reaction Rate Effect of Particle Size of Surface Area on Reaction Rate Reaction Condition Reaction Rate Time (sec) denture tablet in cold water denture tablet in hot water C. Effect of a Catalyst on Reaction Rate 8. Place 10mL of hydrogen peroxide (H2 O2 ) in 2 separate test tubes. Place one test tube in a hot water bath. Note the rate bubbles form. 9. Add a pinch of manganese dioxide in the second test tube. Note the rate bubbles form. Q39. How will you compare the rate at which bubbles were produced? Q40. Study the chemical equation below. Chemical Equation: H2 O2 (l) → H2 O (l) + O2 (g) MnO2 Figure 12 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 84. D EPED C O PY 428 Notice the reactantsand resulting products. Q41. Where is the MnO2 written in the equation? Q42. Do you think the MnO2 reacted with H2 O2 ? Q43. MnO2 only acted as a catalyst. What role do you think a catalyst play in a chemical reaction? Going back to the diagram below, recall the effect of catalyst on activation energy . D. Effect of Concentration on Reaction Rate Caution: Wear a mask while performing this experiment. Procedure: 10. Prepare in separate plastic cups, different concentrations of household bleach solution 100% (no water added) 50% (half part bleach solution- half part water) 25% (1/4 part bleach solution – ¾ part water) . 11. Prepare 3 pieces of 5cm x 5cm sized brightly colored crepe paper or Japanese paper. 12. Drop the pieces of crepe paper into the 3 plastic cups simultaneously. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 85. D EPED C O PY 429 13. Compare therate of discolorization of the papers in the 3 beakers. Record your observation in the table below. Table 14. Effect of Concentration on Reaction Rate Concentration Reaction Rate 25% solution 50% solution 100% solution Q44. Did you get the same rate of reaction? Q45. Describe in your own words the effect of concentration on the rates of reaction. Q46. How will you explain using the Collision theory the factors affecting reaction : a. Surface area of reactants b. Temperature c. Catalyst d. Concentration ENRICHMENT: Write TRUE on the space provided if the statement is correct. Rewrite the statement, if the statement is false. 1. Catalysts speed up chemical reactions but are not changed by them. 2. Heat, light, or change in odor can indicate a physical change. 3. Activation energy is the minimum energy required for reactions to start. 4. Low temperature speeds up reaction rates. 5. A low concentration of chemical slows reaction rate. KEY CONCEPTS: The rate of chemical reaction is affected by the following factors: temperature, surface area of reactants, presence of catalyst, concentration of reactants. Every factor that affects reaction rate can be understood relative to collision theory. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 86. D EPED C O PY 430 The following activity, will deepen your understanding of the benefits and harmposed by some chemical reactions, and will guide you in exploring why rate of some chemical reactions need to be controlled. Activity 7 Making Connections Objectives: • Explain how factors affecting the rate of chemical reactions are applied in food preparation, control of fire, corrosion prevention, etc. • Analyze effect of chemical reactions on life and the environment through visual presentation. PART 1. 1. Analysis of set of pictures linking to acid rain : What effect does acid rain has on limestone/ marble statues? http://www.petersommer.com/blog/category/news/ exhibitions/ Figure 13. Effect of Acid Rain on Marble All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 87. D EPED C O PY 431 What effect does acidrain has on plant growth? http://www.connecticutvalleybiological.com/acid-rain-and-the- environment-acidity-and-plant-growth-p-15860.html Nitric oxide NO2 , a product of combustion of gasoline in automobiles is one of the culprits in the formation of acid rain. Referring to Fig. 15, analyze how NO2 is converted to nitric acid HNO3 http://www.elmhurst.edu/~chm/vchembook/193nox.html Figure 15. Reaction of Nitrogen Dioxide with Water From Figure 16, identify other problems posed by acid rain. Figure 14. Effect of Acid Ran oon Plant Growth All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 88. D EPED C O PY 432 http://envis.tropmet.res.in/menu/ENVIS_Acid_Rain/images/acidImage/Acid_Rain_Arriving. png Figure 16 2. Discussionon acid rain, a chemical reaction that has environmental issues. An example of a chemical reaction that has an environmental concern is the acid rain. Acid rain has been the leading significant cause of destruction in our environment. In infrastructure, it is the cause of corrosion of metals in alloys like steel in buildings, bridges, and transport vehicles. This is due to the displacement reaction of active metals with hydrogen in acids. Materials with historical and cultural values such as monuments and statues are also destroyed by acid rain. They are mostly made up of limestone and marble which like metals form a chemical reaction with acids, lead to their dissolution. Marine life is also affected by acid rain. It causes the pH of bodies of water to decrease; this change in pH will increase marine life mortality, retard fish growth, decrease egg production and embryo survival. Acid rain also tends to dissolve vital minerals in the soil. Crops grown in these depleted soils give poor yields, if they grow at all. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 89. D EPED C O PY 433 Analyzing the issue: Q47.What natural processes can contribute to acid rain? Q48. How is acid rain produced? Q49. What adverse effect can acid rain pose on living organisms and its environment? Q50. Who should be responsible for cleaning up the pollution problem? Q51. What measures are taken to address the problem? PART 2 : Visual presentation ( any form of media) of the effects of chemical reactions on life and the environment Group Activity 1. Using any form of media, prepare a visual presentation of a chemical reaction involved in: a. Food processing and preservation b. Fire control c. Corrosion Control d. Photochemical Smog e. Haber Process f. Catalytic Converter g. Car air bag h. formation of ozone layer in the stratosphere i. formation of acid rain In areas of high automobile traffic, such as in large cities, the amount of nitrogen oxides emitted into the atmosphere can be quite significant. In urban areas, the main source of acid rain is from automobiles. Other sources are thermal power plants and coal mining industries. Gas emissions like CO2, CO, SO2, NO2, and NO from these sources react with water vapor in the air producing acids. Rain contaminated with these acids are what we know now as acid rain. Removing the offending oxides from exhaust and using alternate energy sources are much preferred courses of action at the present time. One of the most important means of reducing sulfur emissions is the swith to low sulfur fuels. Another is the scrubbing of stack gases before they are released to the atmosphere. In this process, the stack gases percolate through a solution that absorbs the oxides of sulfur. The solution is renewed frequently, and waste sulfur can be recovered from the spent solution. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 90. D EPED C O PY 434 2. Research on how a specific chemical reaction poses useful or harmful effects to life andthe environment. 3. Present to class your visual presentation • During your planning session, be reminded to follow the GRASP Task Design Prompts to assist you in the organization of your activity. Goal Your task is to create a visual presentation of benefits/ harm posed by a particular chemical reaction using any form of media. Role You have been asked to gather/ collect researches on chemical reaction assigned to your group. Your job is to understand fully the concepts and issues involved. Audience The target audience is the whole class and a local public official (e.g., barangay chairman) or a member of your community who may be involved in your assigned topic. You need to encourage/ convince your audience to draw pledges/ policies that will help mitigate the problem/ promote the benefits in your topic. Situation The challenge has to do with preparation of the visual presentation: choosing and documenting appropriate resources, summarizing and making the research coherent. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 91. D EPED C O PY 435 Product, Performance andPurpose You will create a visual presentation supported by research in order to better understand and appreciate the principles involved in chemical reactions. Standards and Criteria for Success Your performance needs to meet the following criteria: • Creative (visual presentation is clear/visually appealing) • Meaningful (giving importance to the understanding of the benefits and harm posed) • Illustrative (discussing thoroughly how these reactions may cause harm or how we can benefit from them) Though this is a group task, you will individually assess your performance using the Critical Thinking Rubric below. Critical Thinking Rubric: 4 3 2 1 Identifying the important information I determine what concepts and relationships are important in a complex system of abstract and concrete information. I can usually tell what concepts and relationships are important in a system. Sometimes, I have trouble telling the difference between important and unimportant concepts and relationships in a system. I often get important and unimportant information mixed up. Making Inferences I use what I know about the subject along with my personal experiences and knowledge to make reasonable inferences. I use my inferences to draw conclusions about information. I analyze new information and make reasonable inferences. With help, I can make inferences, but sometimes my inferences are not based on good reasons. I usually cannot make inferences about what I am learning. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 92. D EPED C O PY 436 4 3 21 Evaluating Sources I use several strategies for evaluating the reliability of a variety of different kinds of sources. I use some strategies for evaluating sources. Sometimes, I am fooled by information that is not reliable. I often cannot tell the difference between reliable and false information. Learning Independently I do whatever I need to do to learn more about ideas and concepts that are new to me. I make an effort to learn more about ideas and concepts that are new to me. If someone reminds me, I learn more about ideas and concepts that are new to me. I am usually happy with what I already know about information, and I do not bother to find out more. Communicating I can clearly and thoroughly explain my opinions by giving good reasons for them, orally and in writing. I can explain my opinions by giving good reasons for them, orally and in writing. With prompting and guidance, I can explain my opinions orally and in writing. I cannot explain my opinions so that they make sense. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 93. D EPED C O PY 437 IV. Summary/Synthesis/Feedback • Reactantsare the substances that enter into a chemical reaction, and products are the resulting substances. Substances that undergo a chemical reaction experience a change in their physical and chemical properties. • When a physical change occurs there is no breaking and forming of bonds. There are certain things that will help us identify if a chemical reaction has taken place. We call these evidences of chemical reactions. These are: production of light, evolution of gas, temperature change, color change, and formation of precipitate. • A chemical equation is a chemist’s shorthand for a chemical reaction. The equation distinguishes between the reactants, which are the starting materials and the products which are the resulting substance/s. It shows the symbols or formulas of the reactants and products, the phases (solid, liquid, gas) of these substances, the ratio of the substances as they react. • Chemical reactions are classified in to the following types: combination: A + B → AB decomposition: AB → A + B single displacement: A + BC → AC + B double displacement: AB + CD → AD + CB combustion ( reaction with oxygen producing carbon dioxide and water), acid-base: reaction between acid and base • COLLISION THEORY: Reactions can only happen when the reactant particles collide. Reactants should have sufficient energy, and their molecules should be in proper orientation for a successful collision to happen. • Activation Energy, Ea, is the minimum amount of energy needed for a reaction to occur. • The rate of chemical reaction is affected by the following factors: temperature, surface area of reactants, presence of catalyst, concentration of reactants. • Every factor that affects reaction rate can be understood relative to Collision theory. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 94. D EPED C O PY 438 V. Summative Assessment I.Multiple Choice: Choose the correct answer. 1. Analyze the diagram on the left, what evidence shows that the reaction’s product is a gas? a. bubbles are forming and collected b. the gas is not soluble in water c. acids always produce gases when they react with a solid d. there is no filter funnel and paper to remove unreacted solid. 2-3 Refer to the illustration below: The following depicts the formation of methanol ( CH3 OH). 2. What would be the skeleton equation for this reaction? a. C + Cl2 + O2 → CH3 ClH b. C + H2 + O2 → CH3 OH c. C2 + H2 + O2 → CH3 OH d. C + H + O → CH3 OH 3. If the formula for methanol is CH3 OH, what would be the balanced chemical equation for this reaction? a. C3 + 2H2 + O2 → 2CH3 OH b. 2C + 4H2 + O2 → 2CH3 OH c. 2C + 2H2 + O2 → 2CH3 OH d. C + H + O → CH3 OH All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 95. D EPED C O PY 439 4. Which ofthe following is the correct balanced reaction? a. 2 C3 H8 + 10O2 → 6CO2 + 8H2 O b. C3 H8 + O2 → CO2 + H2 O c. C3 H8 + O2 → 3CO2 + 2H2 O d. C3 H8 + 5O2 → 3CO2 + 4H2 O 5. Quicklime ( CaO ) is used as a drying agent. When water is added to this, slaked lime Ca(OH)2 is formed. What type of reaction is this? a. combination b. single displacement c. decomposition d. double displacement 6. Fresh fish and meat that are not stored in a refrigerator show signs of spoilage in less than a day. What has caused this spoilage? a. temperature changes b. presence of microorganisms c. oxygen in air d. all of the above 7. The rate of reaction increases as the temperature increases. Which of the following statements provides the best explanation for this? a. At lower temperatures the particles do not collide with each other. b. At higher temperatures the particles have more energy, move faster, and collide more often. c. Higher temperature has higher activation energy. d. Increasing the temperature increases the number of particles, so they collide more often. 8. Which of the following statements about collisions is correct? a. Reaction will occur even without collision of molecules. b. All colliding particles have the same amount of energy. c. Only fast-moving particles collide with each other. d. Reactions can happen if the colliding particles have enough energy. 9. Reactions eventually stop. What is generally the reason for this? a. The catalyst has been used up. b. The particles have run out of energy. c. One or more of the reactants has been used up. d. Wrong catalyst was used. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 96. D EPED C O PY 440 10. In areaction with hydrochloric acid, why does powdered magnesium reacts faster than the same mass of magnesium ribbon? a. The powdered magnesium contains more atoms than the magnesium ribbon. b. The powdered magnesium is hotter than the magnesium ribbon. c. Thepowderedmagnesiumhasabiggersurfaceareathanthemagnesium ribbon. d. The powdered magnesium has a smaller surface area than the magnesium ribbon. 11. Marble reacts with hydrochloric acid to produce calcium chloride, water and carbon dioxide. In which of these mixtures is the rate of reaction likely to be the greatest? a. 1 g of marble chips in 100 cm3 of hydrochloric acid at 20°C. b. 1 g of powdered marble in 100 cm3 of hydrochloric acid at 30°C. c. 1 g of powdered marble in 100 cm3 of hydrochloric acid at 20°C. d. 1 g of marble chips in 100cm3 of hydrochloric acid at 30°C. 12. Manganese dioxide is a black powder that catalyzes the breakdown of hydrogen peroxide to water and oxygen. Which of the following statements is correct? a. The mass of manganese dioxide will stay the same during the reaction. b. The catalyzed reaction will produce more oxygen than the uncatalyzed reaction. c. The particles in the catalyzed reaction will have more energy than in the uncatalyzed reaction. d. Manganese dioxide will cause production of more water. 13-15 Explain briefly. 13-15 Based on your knowledge of factors affecting the rate of reaction, why is there a danger of explosion in places like coal mines where there are large quantities of powdered, combustible materials? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 97. D EPED C O PY 441 Glossary of Terms Acid-basereaction Reaction between an acid and a base producing salt and water. Activated complex The specie temporarily formed by the reactant molecules as a result of collision before they form the product. Activation energy The minimum amount of energy required to start a chemical reaction. Catalyst A substance that hastens a chemical reaction without itself being consumed. Chemical equation An equation that uses chemical symbols and formulas to represent a chemical reaction Chemical reaction A process in which a substance is changed into one or more new substance/s. Coefficient The number placed before the formulas, used to balance a chemical equation. Collision Theory Reactions can only happen when the reactant particles collide. Combination reaction Two or more reactants form a single product. Combustion reaction Reaction with oxygen producing water and carbon dioxide. Decomposition reaction A single reactant breaks down into 2 or more products. Double displacement reaction This is when the positive ions (cations) and negative ions (anions) of different compounds switch places, forming two entirely new compounds. Inhibitor Substance that slows down or retards a chemical reaction. Precipitate An insoluble solid that separates from the solution. Product The resulting substance after a chemical reaction. Reactant The substance entering a chemical reaction. Single displacement reaction A more active element replaces another element from a compound. Smog Coined from the words smoke and fog, an environmental pollutant. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 98. D EPED C O PY 442 References and Links PrintedMaterials: Chang (2006 )Chemistry 11th Edition McGraw-Hill IncNew York Le May, Beall, Robblee , Brower (2000) Chemistry Connections to Our Changing World Teachers edition Prentice Hall Upper Saddle River, NJ Padolina, Antero, Alumaga (2010) Conceptual and Functional Chemistry Modular Approach Vibal Publishing House, Quezon City Phil., Silverberg ( 2006) Chemistry 4th Edition Mc Graw-Hill Inc New York Wilbraham , Staley, Matta Waterman (2002) Chemistry Prentice Hall Inc, New Jersey Zumdahl (2000) Basic Chemistry 4th edition Houghton Mifflin Co, New York Dep Ed Project EASE Module 17 Lesson 1 Electronic Sources: http://www.elmhurst.edu/~chm/vchembook/193nox.html http://chemistry.mtu.edu/~pcharles/SCIHISTORY/PhlogistonTheory.html How to prevent fire http://www.ulm.edu/police/fire-extinguishers http://www.bbc.co.uk/schools/gcsebitesize/science/add_aqa_pre_2011/ chemreac/energychangesrev3.shtml http://www.bbc.co.uk/schools/gcsebitesize/science/add_edexcel/chemical_ reactions/rates/quiz/q63137499/ http://pslc.ws/fire/howwhy/triangle.htm http://www.bing.com/images search?q=Collision+Theory+of+Chemical+- Reactions& Form=IQFRDR#view=detail&id=F1991A8C 155EB- 0FABE1D598B0 507B71895F 5DE2A&selectedIndex=12 http://www.bing.com/images search?q=Activation+energy&go=&qs=n&- form=QBIR &pq=activation+energy&sc=8-17&sp=-1&sk=#view=de- tail&id=C4330F FCC22298D71798C4462372111054F635D6&selecte- dIndex=96 http://sun.menloschool.org/~dspence/arda/chem_project/web_wan/fertilizer2. htm http://wps.prenhall.com/wps/media/objects/3082/3156859/blb1404/bl14fg16. jpg( accessed: Oct. 29, 2014) http://iytimg.com/vi/OkGzaSOkyf4/maxresdefault.jpg (accessed: Oct. 29, 2014) https://www.chem.tamu.edu/class/majors/tutorialnotefiles/factors.htm (accessed: July 4, 2014) http://www.connecticutvalleybiological.com/acid-rain-and-the-environment- acidity-and-plant-growth-p-15860.html http://envis.tropmet.res.in/menu/ENVIS_Acid_Rain/images/acidImage/Acid_ Rain_Arriving.png All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 99. D EPED C O PY 443 I. Introduction Think aboutthe food you eat everyday. Different types of food give you different nutrients for energy, growth and repair. These were introduced to you when you were at the elementary grades. Also, in Grade 9, you have learned that the bonding characteristics of carbon result in the formation of larger variety of compounds. In this module, you will learn more about compounds which are essential to life. These compounds belong to four main classes of biomolecules: carbohydrates, lipids, proteins, and nucleic acids. Carbohydrates and lipids are generally made up of carbon, hydrogen and oxygen. Proteins and nucleic acids and some derivatives of carbohydrates and lipids also contain nitrogen. You will also have the opportunity to test food for the presence of biomolecules. At the end of Module 3, you will be able to answer the following key question. BIOMOLECULES Unit 4 MODULE 3 • What differentiates the biomolecules from each other? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 100. D EPED C O PY 444 II. Learning Competencies/Objectives Atthe end of this module, the learners are expected to: • Recognize the major categories of biomolecules such as carbohydrates, lipids, proteins and nucleic acids; • Differentiate the biomolecules from each other in terms of their structure and function. III. Pre-Assessment Direction: Analyze each question carefully then choose the letter of the correct answer. 1. Which of the following is NOT a major source of protein? A. fish B. egg C. milk D. vegetable 2. Which of the following contains the most lipids? A. banana B. champorado C. olive oil D. cheese All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 101. D EPED C O PY 445 3. Which ofthe following is a correct pair? A. glucose: disaccharide C. starch: polysaccharide B. sucrose: monosaccharide D. triglyceride: polysaccharide 4. Which is a correct pair of an example of protein and its function? A. enzymes: speed up reactions in the body and eventually used up in the process. B. collagen: provides strength and flexibility to connective tissues. C. actin and myosin: supplies amino acids to baby mammals D. hemoglobin: helps regulate blood sugar levels 5. Maria wanted to determine what types of biomolecules are present in the three unknown substances that her teacher gave her. The following table shows her results. Substance Iodine Test Biuret Test Benedict’s Test A Black solution (+) (-) (-) B (-) (+) (-) C (-) (-) (+) Which of the following statements is TRUE? A. Substances A and B are proteins while substance C is a lipid. B. Substance A contains starch and substance B and C contain nucleic acid C. Substances A and C are carbohydrates where A is an amylose in starch and B is a protein and C maybe a simple sugar D. Substance B is a carbohydrate and substances A and C are lipids For numbers 6 to 9 please refer to the structures below: All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 102. D EPED C O PY 446 6. Which ofthe given structures (A, B, C, or D) represents molecules that provide energy and are very soluble in water? 7. Which of the given structures (A, B, C, or D) represents hydrophobic molecule that is used as storage of energy? 8. Which of the given structures (A, B, C, or D) represent the molecules that store the hereditary traits of humans? 9. Which of the given structures (A, B, C, or D) represent the building blocks of bigger molecules necessary for structural integrity of organisms? 10. Which of the biomolecules contain other elements aside from carbon, hydrogen, and oxygen? A. carbohydrates, lipids C. nucleic acids, proteins B. proteins, lipids D. nucleic acids, lipids IV. Reading Resources and Instructional Activities Figure 1. These are foods rich in carbohydrates and lipids Lookatthepicturesabove.Whichfoodcanbeclassified ascarbohydrates or lipids? In order to find out between carbohydrates and lipids, you can perform Activity 1. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 103. D EPED C O PY 447 Activity 1 Test forCarbohydrates and Lipids Objective: • To detect the presence of carbohydrates and lipids in food samples using chemical tests. Materials: • Iodine solution or tincture of iodine • Benedict’s solution • food samples for testing carbohydrates ( cooked pasta, cracker, • cooked rice, corn syrup, table sugar, pineapple) • food samples for testing lipids (oil, peanut butter, egg, fried chicken, • butter, milk, burger) • 6 pcs.small test tubes or vials per group • 6 pcs. test tube holders per group • 2 pcs.droppers per group • mortar & pestle per group • 1 spot plate per group All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 104. D EPED C O PY 448 Procedure: A. Carbohydrates Iodine Testfor Starch 1. Place ½ teaspoon of each food sample on the well of a spot plate. Make sure that the food samples are far from each other. 2. Add 3 drops of Lugol’s Iodine solution or tincture of iodine on each food sample. 3. Note that Lugol’s iodine solution or tincture of iodine changes from yellow to blue or black in the presence of starch. 4. Write your observation in Table A. B. Benedict’s Test for Reducing Sugar 1. Place a pinch of the food samples to be tested into a test tube. 2. Add 1 full dropper of Benedict’s solution to each test tube. 3. Gently shake the test tube or vial. 4. Place the test tubes in the hot water bath for 2-3 minutes. After 2-3 minutes, return the test tubes to the test tube racks. If the substance in your test tube contains sugar, Benedict solution will change color. Positive Test: Benedict’s solution changes from blue to green (very small amount of reducing sugar), to yellow (higher amount of reducing sugar) to orange or brick red (highest amount of reducing sugar). The change in color is due to the formation of the brick red precipitate, Cu2 O. CAUTION: Always use a test tube holder to handle hot test tubes. 5. Observe your test tube (using white paper as a background). Record the amount of sugar present in Table 1. Amount of Sugar in Food 0 None + Trace ++ Little Sugar +++ Moderate Sugar ++++ Much Sugar Color Blue Blue green Green Yellow Orange/Red All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 105. D EPED C O PY 449 C. Ethanol EmulsionTest for Fats and Oils Adapted:http://brilliantbiologystudent.weebly.com/ethanol-emulsion-test-for-lipids.html (accessed: July 15, 2014) Solid sample: 1. Crush a pinch of food sample and place in a dry test tube. 2. Add ethanol to about 2 cm3 above the level of the sample and shake thoroughly. 3. Allow the solid to settle for about 3 minutes and decant the ethanol into another test tube. 4. Add 2 cm3 of distilled water to the test tube. 5. Write observations in Table 2. Liquid sample: 1. Add a few drops of the liquid food sample to a dry test tube. 2. Add 2 cm3 ethanol and shake it thoroughly 3. Add 2 cm3 of distilled water. 4. Write observations in Table 3. Test for Carbohydrates and Lipids Table 2. Results of Carbohydrate Test Food Sample Test for Simple/ Reducing Sugars/ Benedict’s Test Iodine Test Cooked pasta Cracker Cooked rice Corn syrup Table sugar Pineapple All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 106. D EPED C O PY 450 Table 3. Resultsof the Ethanol Emulsion Test for Lipids Food Sample Colorless Layer of Cloudy White Suspension Oil Peanut Butter Egg Fried Chicken Butter Milk Burger Mashed potato Q1. Which of the foods samples tested would your body use for a quick burst of energy? Which could be used for energy when no carbohydrates are available? Q2. Why it is that Benedict’s test gives a negative (-) result with sucrose or table sugar? Q3. What kind of foods rich in fats should be taken in moderation? Why? Carbohydrates Since food is always a part of our lives it is important that we know the nutrients found in the food we eat. The following discussions will give you a clearer avenue to understand carbohydrates. Figure 2. Foods rich in carbohydrates All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 107. D EPED C O PY 451 Figure 2 showssome foods that are rich in carbohydrates. Carbohydrates are the major source of energy for the body. These are simple sugar, starch and cellulose. All carbohydrates contain carbon, hydrogen, and oxygen. They may be classified into the following: Monosaccharides From the prefix “mono” which means one, monosaccharide is the simplest sugar and the basic subunit of a carbohydrate.These compounds are white solids at room temperature. Because they have polar, hydroxyl (-OH) groups in their molecular structures, they are very soluble in water. The most common monosaccharides are glucose (also called dextrose) and fructose. http://joelbergerdc.com/tag/glucose-vs-fructose/ Figure 3. Structure of Glucose and Fructose Although both of these monosaccharides have the formula C6 H12 O6 , their structural formulas differ. As figure 3 shows, glucose in water solution forms a ring made up of five carbon atoms and one oxygen atom, and fructose in a water solution forms a ring made up of four carbon atoms and one oxygen atom. Both compounds have five-OH groups in their structures. Compounds with the same molecular formulas are called isomers. So, glucose and fructose are isomers. Though they have the same molecular formula, these sugars cannot be used in the same way by cells in the body. The arrangement of the C, H, and O atoms determines the shape and properties of each sugar. In Grade 8, you have learned about how carbohydrates and proteins are broken down in digestion. For fats and lipids, their digestion is completed in the small intestine and is broken down primarily into fatty acids and glycerol. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 108. D EPED C O PY 452 During digestion, carbohydratesare broken down into monosaccharide which is absorbed into the blood and transported to the cells providing “instant” energy to perform our activities. Sometimes we eat too much, especially when we are tired, the excess glucose is stored in the liver as glycogen for later use. It is very important to have a steady supply of glucose in the blood to maintain body functions. As what they say, too much or too little of anything may lead to some diseases. When too much glucose is in the blood, the pancreas secrete a hormone called insulin which stimulates cells in the liver, muscles and fat to absorb glucose and transform it into glycogen or fats, which can be stored for a period of time. When blood glucose drops, the pancreas secretes glucagon, which causes the liver, muscles and fat to convert glycogen back to glucose. Fruits like grapes, apple or atis contain a monosaccharide called fructose or fruit sugar.It is considered the sweetest naturally occurring sugar.Due to its sweetness, fructose is sometimes used as a low calorie sweetener because less fructose is needed to produce the same sweetness that table sugar does. Starchy food that we eat is widely distributed in the plant world.Thus, its main constituent glucose is found in all plants and in the sap of trees.However, glucose is also found in glycogen that is produced in animal cells. Disaccharides In the morning, Aaron Jay’s mother prepares his coffee; he always adds half a teaspoon of table sugar. He remembered his TLE (Technology and Livelihood Education) teacher who mentioned one time in their class that the sugar we use to sweeten coffee is a disaccharide. It is also called sucrose with the molecular formula C12 H22 O11. He wondered how sucrose, which is disaccharide, is formed. In their chemistry class, their teacher explained that the formation and breakdown of sucrose to glucose involves two reactions. Condensation reaction is a reaction in which two molecules or parts of the same molecule combine. During the condensation of monosaccharides to form disaccharides, one molecule of water is lost.When two glucose molecules are combined, maltose is formed and water is lost during the process. A Hydrolysis reaction occurs when the bond between monosaccharides is broken with the addition of a water molecule. Q1. What is the name of the dissacharide found in cheese and other milk products? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 109. D EPED C O PY 453 After he finisheddoing his homework, Aaron Jay drinks his milk. When he is about to jump into his bed to have a good night sleep, he has this bloated feeling along with a build up of intestinal gas.He feels uneasy and cannot sleep. He swears he will never drink milk again! The following morning in his chemistry class, his teacher discussed another important disaccharide- Lactose or milk sugar. Lactose is made up of a sugar called galactose and glucose. In our body, a specific enzyme, lactase is necessary to help break the bond between the two monosaccharides when lactose is digested. People who cannot digest milk products are called “lactose intolerant” because they do not produce the enzyme (lactase) necessary to break the bond between glucose and galactose. Since lactose molecules are too large to be absorbed into the circulatory system, they continue through the digestive system, where they are eventually broken down by bacteria in the large intestine. These bacteria digest monosaccharides, producing carbon dioxide gas in the process. As a result, a common symptom of lactose intolerance is a build up of intestinal gas along with a bloated feeling, and more often the passing out of undigested lactose as diarrhea. After the discussion, he concluded that he maybe “lactose intolerant.” Figure 4. Structure of Disaccharides Figure 4 shows that when two monosaccharides join together by combination reaction, a glycosidic bond will be formed between the two monosaccharide molecules. The reaction produces water as a side product. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 110. D EPED C O PY 454 Polysaccharides In the evening,he did not drink milk anymore. Instead he ate fruits before going to bed. The following morning he had the same routine-ate his breakfast and went to school. As the bell rang, Aaron Jay rushed to the canteen to eat his lunch. It included local tubers like sweet potato or camote and green, leafy vegetables like malungay and kangkong. Again, he remembered the result of their activity no.1 wherein the food samples like sweet potato and ripe banana turned blue-black when stained with iodine solution. In their class discussion, these foods contain polysaccharides (the prefix poly means many) or complex carbohydrates. They are large molecules that are made up of many smaller units that are joined together.The reason why these foods turn blue-black is because they contain starchy components. After lunch, he returned to their classroom.Their discussion was about the three common polysaccharides- starch, glycogen, and cellulose. The breakdown of starch requires a water molecule to provide a hydrogen atom and a hydroxyl group to the site where the bond is broken. With the help of enzymes in the digestive system, the glucose units can be separated from one another. When a glucose molecule is separated from the rest of the starch polymer; it can be absorbed and used as fuel by your cells. Since it takes time for glucose to be separated from the polysaccharide, it is released to the cells gradually. Thus, the glucose from starch reaches muscle cells over a period of time providing energy as it is needed. For this reason, athletes often eat meals rich in complex carbohydrates before an athletic event. https://courses.ecampus.oregonstate.edu/ans312/one/carbs_story.htm Figure 5. Structure of Starch All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 111. D EPED C O PY 455 Starch is thechief storage form of carbohydrates in plants and the most important source of carbohydrate in human nutrition. A starch molecule is a polysaccharide assembled from the simple sugar glucose; it can contain anywhere from five hundred to several hundred thousand glucose molecules joined by covalent bonds into a single structure. Starch is made up of two types of polysaccharides: amylose, which is a coiled or helical structure, and amylopectin, which is branched. Plants make starch. All individuals whose intake of glucose is excessive will store the excess glucose as fat for long term storage and some are converted to another polysaccharide glycogen. Glycogen is a polysaccharide that is similar to starch because it is also composed of alpha glucose units. It differs from starch since glycogen shows a higher degree of branching and is a polysaccharide that is made by animal. On the other hand, starch contains both straight chain and branched polysaccharides with much less branching than that of glycogen, and is made only by plant. http://www.natuurlijkerwijs.com/english/Glycogen_metabolism.htm Figure 6. Structure of Glycogen Figure 6 shows the structure of glycogen which consists of long polymer chains of glucose units connected by an alpha glycosidic linkage.It is a multibranched polysaccharide of glucose that serves as a form of energy storage in animals. The polysaccharide structure represents the main storage form of glucose in the body Glycogen is the readily available energy stored in liver and muscles and the one that is easily metabolized. Fats are stored in adipose tissues but unlike glycogen, are not as readily metabolized. They are used during prolonged exercise or activity. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 112. D EPED C O PY 456 https://myorganicchemistry.wikispaces.com/ Cellulose?responseToken=1a9131f668de1a94603bbdfb79f69128 Figure 7. Structureof Cellulose The glucose molecules in cellulose chains (refer to Figure 7) are arranged in such a way that hydrogen bonds link hydroxyl groups of adjacent glucose molecules to form insoluble fibrous sheets. These sheets of cellulose are the basic component of plant. People cannot digest cellulose, but when we eat foods rich in fiber, which is cellulose, it speeds the movement of food through the digestive tracts. It is a food for herbivorous animals like cows, carabaos, goats, and horses. These animals have microorganisms in their digestive tracts that can digest cellulose. They have a special stomach chamber that holds the plants they eat for a long period of time, during which these microorganisms can break down the cellulose into glucose. The protozoans in the gut of insects such as termites also digest cellulose. Being of great economic importance, cellulose is processed to produce papers and fibres, and is chemically modified to yield substances used in the manufacture of items such as plastics, photographic films, and rayon. Other cellulose derivatives are used as adhesives, explosives, thickening agents for foods, and in moisture-proof coatings. Likewise, starch has many industrial applications in addition to its importance in human nutrition. It is used in the manufacture of paper, textiles, pharmaceuticals, and biodegradable polymers, and as an additive in foods. Formulative Assessment: Q2. Why do you think marathon runners eat a meal rich in carbohydrates the day before the race? All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 113. D EPED C O PY 457 After the discussion,Aaron Jay was amazed at how carbohydrates contribute to energy production and the manufacture of important products for human consumption. Lipids In the previous lesson, you have learned that carbohydrates are important in providing “instant” energy for cells. There is another class of biomolecules called lipids that have the “job” of storing energy for later use. Lipids are also found in hormones and cell membrane components. Foods rich in lipids Lipids have different structural types such as carboxylic acids or fatty acids, triglycerides or neutral fats, steroids, and waxes, to name a few. Naturally occuring esters are lipids that contain one or more long-chain carboxylic acids called fatty acids. These are insoluble in water but soluble in nonpolar solvents. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 114. D EPED C O PY 458 When Aaron Jayaccidentally mixed oil and water he observed that they do not mix. He was late in his Chemistry class the following morning but he was able to catch up the discussion of his teacher on lipids. His teacher explained that oil and water do not mix because they do not have the same polarity. Also, oils are composed primarily of long hydrocarbon chains. They are formed reaction between an alcohol and one or more long–chain carboxylic acids. The most abundant of the lipids are the fats and oils, also called triglycerides. Table 4 below shows the structures of common fatty acids. The presence of double bonds in the fatty acids lowers its melting point. At room temperature, lauric acid is solid while linoleic acid is liquid. Table 4. Structures of Some Common Fatty Acids Name Structural Formula Melting Point (o C) Lauric CH3 (CH2 )10 COOH 44 Myristic CH3 (CH2 )12 COOH 53 Palmitic CH3 (CH2 )14 COOH 63 Stearic CH3 (CH2 )16 COOH 70 Oleic CH3 (CH2 )7 CH=CH(CH2 )7 COOH 16 Linoleic* CH3 (CH2 )4 (CH=CHCH2 )2 (CH2 )6 COOH -5 Linolenic* CH3 CH2 (CH=CHCH2 )3 (CH2 )6 COOH -11 Arachidonic* CH3 (CH2 )4 (CH=CHCH2 )4 (CH2 )2 COOH -50 Source: Padolina, M.C.D., Antero, E.S., Alumaga, M.J.B & Estanilla, L.C. (2004). Conceptual and Functional Chemistry Fats are solids at room temperature and contain saturated fatty acids. Aaron Jay still remembered that all saturated hydrocarbons contain single bonds and they are produced only by animals. Examples of animal fats are lard and butter. Oils are liquids at room temperature and contain unsaturated fatty acids. Again, he recalled that unsaturated hydrocarbons contain one or more double bonds. Most oils, such as vegetable oil, corn oil, and olive oil are produced by plants. Table 5 gives the fatty acid content of some glycerides. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 115. D EPED C O PY 459 Table 5. FattyAcid Content of Some Triglycerides Source Saturated Unsaturated Others Myristic Palmitic Stearic Oleic Linoleic Animal Fat Butter 10 29 9 27 4 31 Lard 2 30 18 41 6 5 Beef 3 32 25 38 3 2 PlantOil Corn 1 10 4 34 48 4 Soybean - 7 3 25 56 9 Peanut - 7 5 60 21 7 Olive 1 6 4 83 7 - Sometimes we prefer to buy a product in solid form rather than in liquid. Which do you prefer? Spreading margarine on a pandesal or pouring oil on it? Of course, margarine is more acceptable to consumers when it is solid because it looks more like butter. However, margarine is made from vegetable oils that are liquid at room temperature. The oils can be processed to form solid margarine. How is this done? Can we consider fats good or bad? It depends. If you eat in moderation, fats are good sources of body fuel.They are considered good emergency food and are efficient energy storage system. However, an excess quantity of fats is not good for the heart. The reason why fats are not good for the heart is because they tend to clog arteries and overwork the heart.While carbohydrates are the main source of energy in your body, your system turns it to fat as a backup energy source when carbohydrates are not available. Vitamins A, D, E, and K cannot function without adequate daily fat intake since they are fat soluble vitamins. If you don’t meet your daily fat intake or follow a low fat diet, absorption of these vitamins may be limited resulting in impaired functioning. Steroids are another class of lipids whose molecules are composed of fused rings of atoms. The most important steroid is cholesterol. It is a sterol because of the presence of alcohol or the hydroxyl functional group.It is found mainly in animal cells although cell membranes of plants may contain small quantities of cholesterol as well as its major derivatives, sitosterol. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 116. D EPED C O PY 460 http://sphweb.bumc.bu.edu/otlt/MPH-Modules/PH/PH709_BasicCellBiology/PH709_ BasicCellBiology24.html Figure 11. Structureof Cholesterol Figure 11 shows the unique structure of cholesterol which consists of four linked hydrocarbon rings forming the bulky steroid structure. There is a hydrocarbon tail linked to one end of the steroid and a hydroxyl group linked to the other end. Cholesterol is known as a “sterol” because it contains an alcohol functional group-OH. Cholesterol is present in most animal membranes with varying amounts but is absent in prokaryotes. Cholesterol plays an important role in eukaryotes and especially abundant in cell membranes of animal cells. Small amount of cholesterol can also be found in the membrane of some organelles inside the cells, such as the mitochondrion and the endoplasmic reticulum. It is not only abundant in cell membrane, but also in brain tissues of the nervous system. An important nerve cell, myelin, covers nerve axons to help conduct the electrical impulses that make movement, sensation, thinking, learning, and remembering possible. Studies have shown that cholesterol was found to be the most important factor in the formation of synapses, which greatly affect our memory and learning ability. Animals are able to use cholesterol to synthesize other steroids like cortisone, testosterone, and estrogen. These hormones are already discussed in Grade 9. Although cholesterol is an essential lipid for humans, excessive levels of cholesterol in the blood can lead to deposits in the arteries of the heart. These arterial deposits are a leading cause of heart disease. (LeMay Jr, 2000) All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 117. D EPED C O PY 461 Aaron Jay’s journeyto the world of carbohydrates and lipids gave him a clearer view of the importance of these biomolecules in providing the body with energy. However, he still wants to know which type of molecule has the higher calorie content. His teacher explained that a calorie is actually a unit of heat energy. We think of calories as something that are present in food and all food have calories. However, your body sees calories as energy in the form of heat. Heat energy is what really fuels our body in the same way that gasoline fuels your car’s energy. Now all foods have calories and different foods have different amounts of calories. Calories are provided by fats, carbohydrates, and proteins. Fats have the highest concentration of calories.On the average, that’s nine calories per gram of pure fat. Proteins and carbohydrates each have four calories per gram of pure protein or pure carbohydrate on the average. So understanding the role of calories in your diet can help you balance your calories in with your calories out, and help you achieve weight management goals. On the sample Nutrition Facts label, the serving size of this food is 1 cup and there are 2 servings in this container. There are 260 calories per serving of this food. If you eat the entire container of this product, you will eat 2 servings. That means you double the calories (260 x 2= 520 calories) If you eat 2 servings, you will have eaten over 500 calories. Retrieved: http://www.health.gov/dietaryguidelines/dga2005/healthieryou/html/chapter5.html All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 118. D EPED C O PY 462 Wait! You stillneed to explore another activity to enhance your knowledge on the identification of protein present in foods. Q3. Carbohydrates and lipids are composed of the same chemical elements, but in diffeent proportions. Both are used primarily as energy sources for cell metabolism. Which type of molecule has the higher calorie content per gram? Explain the reasons for your answers. Activity 2 A. Test for Proteins Objectives: • Perform standard chemical test for proteins. • Relate indicator reactions to the presence of organic nutrients. Materials: • 0.5 M sodium hydroxide solution • 0.5 M copper (II) sulfate solution • droppers • test tubes • test tube racks • food samples (egg white, cooked fish, cooked meat, cooked legumes, taho) Procedure: Biuret Test 1. Place a pinch of food sample to be tested into a test tube. 2. Add 5 drops of NaOH and 5 drops of CuSO4 solution to the test tube. 3. Gently shake the test tube. 4. Observe the content of each test tubes (using white paper as background). If the food contains protein, it will turn pink or blue- violet. Record the amount (0, +, ++, +++, ++++) of proteins for each food substance in table C. Positive Test: Biuret is clear or light blue in the absence of protein and pink or blue-violet in the presence of protein. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 119. D EPED C O PY 463 Amount of Protein in Food 0 None + Trace ++ Little protein +++ Moderate protein ++++ Much protein ColorLight Blue Light pink Pink Blue-violet Dark blue- violet CAUTION: Biuret reagent can burn your skin. Wash off spills & splash immediately with plenty of water.Inform the teacher when this occur. Data/Results Table C Food Samples Biuret Test egg white cooked fish cooked meat cooked legumes taho Q4. Describe what you observed in each test tube. Q5. Which foods may be used for building body parts? B. The Denaturation of Proteins Adapted from Sourcebook on Practical Work for Teacher Trainers, High School Chemistry volume 2, UP-NISMED Objectives: • Identify the agents for the denaturation of proteins. • Relate the denaturation of proteins to home or ordinary activities. • Explain what happens to proteins upon denaturation. Materials: • dilute egg white solution • test tubes • 0.1 M copper (II) sulfate solution • dropper • conc. HCl • alcohol burner • ethanol All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 120. D EPED C O PY 464 Procedure: 1. Set upfour test tubes (labeled A, B, C, and D) in a test tube rack. Place about 2 cm3 of the egg white solution in each test tube. Add a few drops of each of the following reagent solutions to separate egg white samples in test tubes A, B, and C. A. 0.1 M Copper(II) sulfate solution B. conc. HCl C. ethanol Take Note: Preparation of egg white sample: Mix together one portion of egg white with five portions of water in a small beaker. Add a very small amount of sodium chloride. Observe what happens in each test tube. Q6. Describe what you observed in each test tube. Q7. Copper sulfate is used as a fungicide in the garden. Explain the relation of this application to what you have just observed. Q8. A 70% solution of ethanol in water is used as a disinfectant. Explain the basis for this application. 2. Get test tube D and apply heat. Observe any change. Q9. Describe what happens. Q10. Give other examples of ordinary activities at home that involve the denaturation of proteins. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 121. D EPED C O PY 465 Proteins Proteins are madeup of the elements carbon, hydrogen, oxygen, nitrogen and sulfur. Let’s continue the story of Aaron Jay on his journey this time to the world of proteins. From the result of his activity, he was able to know that egg white, fish, meat, and cheese are foods rich in proteins. He learned from their discussion that proteins are found in all living cells.They are the second most common molecules found in the human body (after water) and make up about 10% to 20% of the mass of a cell. So whenever Aaron Jay eats protein-rich foods, his digestive system breaks the long protein chains into simpler substances called amino acids. He learned from his Chemistry class that amino acids are the building blocks of proteins. Of the 20 amino acids found in human protein, only 11 can be synthesized by the body and 9 have to be supplied by the foods we eat. These 9 amino acids are also called essential amino acids. Adults only need to obtain eight of them: valine, isoleucine, leucine, lysine, methionine, phenylalanine, threonine and tryptophan. The ninth amino acid - histidine - is only essential for infants. Your body doesn’t store amino acids, so it needs a regular daily supply of these essential building blocks. Nonessential is a slightly misleading label because these amino acids actually fill essential roles, but since they’re synthesized by your body, they’re not an essential part of your diet. Of the 11 nonessential amino acids, eight are called conditional amino acids. When you’re sick or under significant stress, your body may not be able to produce enough of these amino acids to meet your needs. The list of conditional amino acids includes arginine, glutamine, tyrosine, cysteine, glycine, proline, serine, and ornithine. The remaining three - alanine, asparagine, and aspartate - are nonessential. Aaron Jay also learned that whenever he eats protein foods, he is supplied with amino acids for the rebuilding of his body system. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 122. D EPED C O PY 466 http://entrytest-preparation.blogspot.com/2014/01/amino-acids.html Figure 12. Structureof amino acid Figure 12 shows the structure of amino acids. Amino acids are organic molecules that contain two functional groups: a basic NH2 amino group and an acidic- COOH carboxylic acid group. Figure 13. Peptide Bond When two amino acids react with each other in an acid-base reaction, a peptide is formed. The basic amino group of one amino acid reacts with the acidic carboxylic group of another amino acid, forming the peptide, and a molecule of water is lost. This reaction shown above is classified as a condensation reaction because the two amino acid molecules join together and water is formed. The bond formed is called a peptide bond, and the product is a dipeptide because it is made up of two amino acid units. Longer chains are called polypeptides and chains of 50 or more amino acids are called proteins. After the discussion on essential & non essential amino acids, Aaron Jay’s teacher discussed the primary, secondary, tertiary, and quarternary structures of proteins. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 123. D EPED C O PY 467 Figure 14. Primary,Secondary ,Tertiary, and Quaternary Proteins Proteins are characterized by their primary, secondary, tertiary and quaternary structures.The kind of amino acids, which make up the chain, the sequence in which the amino acids are arranged and the length of the chain distinguishes the primary structure of proteins. The secondary structures of proteins refer to the coiling of the protein chain into a α–helix structure, formation of b sheets, or twisting into random structures. These structures are the results of interactions between R groups, H-bonding or formation of –S-S- bonds between chains. Protein molecules are so long that they automatically coil, fold or twist. The resulting shape is unique for each polypeptide in a particular medium, at a particular pH. The tertiary structure describes the shape of the coiled chain when it is folded or hydrated in its natural state. (Adapted: Practical Work for Teacher Trainers, High School Chemistry volume 2, UP-NISMED) All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 124. D EPED C O PY 468 The quaternary proteinstructure involves the clustering of several individual peptides into a final specific shape. A variety of bonding interactions including hydrogen bonding, salt bridges and disulfide bonds hold the various chains into a particular geometry. Proteins perform varied functions in the body. How they perform their functions depend on their composition and structures. The particular form and shape each protein molecule takes determines or dictates its function within the organism. Aaron Jay remembered the result of their activity on denaturation of protein. When denaturing agents change the secondary and tertiary structures of proteins, the protein functions are impaired. The protein molecules in egg white fold and aggregates, which dissolve in water. The long string of molecules unfolds once it is denatured by such agents as heat, salt, baking soda, rubbing alcohol, etc. From the results of the activity 2, Aaron Jay learned that denaturation finds many applications at home. An example is the extraction of oil from coconut milk emulsion (gata). Proteins act as the emulsifying agent. When the coconut milk emulsion is heated, oil separates from water and is then recovered. The tasty solid residue remaining (latek) after water evaporates is denatured protein. Also, the preservation of food by pickling and salting also involves denaturation of proteins. Vinegar and salts are agents for denaturation. Decay microorganisms are killed when their cell proteins are denatured. (Adapted: Practical Work for Teacher Trainers, High School Chemistry volume 2, UP- NISMED) Aaron Jay also learned that protein malnutrition, also known as Kwashiorkor, affects children in underdeveloped countries.Although protein malnutrition can be classified as a type of malnutrition; protein malnutrition usually goes hand in hand with calorie malnutrition and referred to as Protein- Energy Malnutrition (PEM). Another type of protein is the enzymes. It is known as biological catalysts. In Grade 8 biology, you have learned the amazing action of catalysts particularly during digestion process.These molecules speed up biochemical reactions without themselves being used up in the process. They are also highly specific. That is, they act only on certain molecules called substrates (reactants), while leaving the rest of the system unaffected. The role of an enzyme can be compared to a lock and a key. The lock will not open unless you use the right key. In the same manner an enzyme works for a specific substrate like the enzyme lactase. Its role is to breakdown the sugar lactose into glucose All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 125. D EPED C O PY 469 and galactose. Youmust appreciate the role of enzymes in the body. Without them, chemical reactions in the body may be too slow to occur at normal condition and may affect the normal functioning of the different systems of the body. After the discussion on proteins, Aaron Jay was amazed at how diverse this group is and the myriad of functions they possess that are very important to all living things. NUCLEIC ACIDS Mother and daughters Photo credit: Edwin Manalang Aaron Jay wonders why siblings resemble each other, or how a mother and her daughters look alike. He will discover the answer as he explores the next lesson. Nucleic acids are molecules that code for hereditary traits by controlling the production of protein. Like proteins, nucleic acids are long chain of polymers consisting of simpler units or monomers. There are two kinds of nucleic acids: DNA, or deoxyribonucleic acid; and RNA, or ribonucleic acid. DNA found mainly in the cell nuclei contains the genetic information that codes for the sequences of amino acids in proteins. RNA is found in many places in the cell and carries out the synthesis of proteins. The monomers of nucleic acids are nucleotides. They are made up of three parts: a five carbon sugar (pentose), a phosphate group, and a ring- shaped base containing nitrogen. In this model, the sphere represents a phosphate group, the pentagon represents a five–carbon sugar (pentose) and the rectangle represents a nitrogen-containing base. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 126. D EPED C O PY 470 Figure 15. Showa model of a nucleotide. The double-helix consists of two linear strands of polymerized nucleotides that bound about each other. The two strands are held together by hydrogen bonds that form between pairs of nucleotides. Adenine (A) forms hydrogen bonds with a thymine (T) of the other strand. Cytosine (C) forms hydrogen bonds with a guanine (G) of the other strand. Figure 16. A model of a double helix for DNA All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 127. D EPED C O PY 471 Below is asummary of the differences between the two kinds of nucleic acids: (http://www.diffen.com/difference/DNA_vs_RNA) DNA RNA Deoxyribonucleic Acid Ribonucleic Acid Description It contains the genetic instruction used in the development and functioning of all living organisms. It is responsible for the template in the synthesis of proteins which in turn control the operation & function of the cell Function Long-term storage and transmission of genetic information Transfer the genetic information for the creation of proteins from the nucleus to the ribosomes Sugar and Bases Deoxyribose sugar Phosphate backbone; Four Bases: adenine, guanine, cytosine, and thymine Ribose sugar Phosphate backbone; Four Bases: adenine, guanine, cytosine, and uracil Pairing of Bases A-T (Adenine-Thymine) G-C (Guanine-Cytosine) A-U (Adenine-Uracil) G-C (Guanine-Cytosine) The process by which an identical copy of the original DNA is formed is called DNA replication. An analogy of DNA replication is opening a zipper.As you open, each side of the zipper acts as a template for the synthesis of a new, complementary strand. The result is two new DNA molecules, which have the same base pair sequence as the original double helix. Proteins are the ones responsible for observable traits like curly hair, blue eyes, dark skin, etc. DNA and RNA molecules direct the synthesis of proteins in the cells. However, this is beyond the scope of this module. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 128. D EPED C O PY 472 V. Summary/Synthesis/Feedback (LeMayJr, 2000) Carboydrates • They are molecules made from aldehydes and ketones containing numerous hydroxyl groups. • Monosaccharides are composed of a single ring. • Disaccharides consist of two monosaccharides that are chemically combined. • Polysaccharides are polymers containing numerous monosaccharide monomers. Lipids • They are water insoluble molecules that are composed of carbon, hydrogen and oxygen. • Fats and oils are triglycerides that are combinations of glycerol and three fatty acids. Proteins • Proteins are polymers of amino acids. They are found as structural materials in hair, nails and connective tissues. • Enzymes are proteins that act as biological catalysts. Nucleic Acids • Deoxyribonucleic acid (DNA) and Ribonucleic acid (RNA) are nucleic acids.Both DNA and RNA are polymers that are made up of nucleotides. • Nucleotides are molecules that are composed of three parts: a five carbon sugar, a nitrogen-containing base, and a phosphate group. Glossary of Terms • Biomolecule is any molecule that is produced by a living organism, including large macromolecules such as proteins, polysachharides, lipids and nucleic acids. • Condensation reaction is a process by which two molecules form a bond with the removal of a molecule of water. • Hydrolysis is a reaction in which water is added to a reactant, breaking the reactant into two product molecules. • Monomer is a small molecule that joins with other similar molecules to make a polymer; repeating units of a polymer • Polymer is a large organic molecule consisting of small repeating units called monomers. All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 129. D EPED C O PY 473 VI. Summative Assessment Direction:Analyze each question carefully then choose the letter of the correct answer. 1. Nutritional chemists have found that burning 1 gram of fat releases twice the amount of heat energy as burning 1 gram of starch. Based on this information, which type of biomolecule would cause a person to gain more weight? a. carbohydrate c. proteins b. fat d. nucleic acid 2. Lipids are insoluble in water because lipid molecules are _________? a. hydrophilic b. neutral c. hydrophobic d. Zwitter ions 3. Which of the following groups are all classified as polysaccharide? a. sucrose, glucose and fructose c. glycogen, sucrose and maltose b. maltose, lactose and fructose d. glycogen, cellulose and starch 4. Amino acids are the building blocks of which group of biomolecules? a. proteins b. carbohydrates c. lipids d. nucleic acid 5. Which of the following is the major function of carbohydrates? 1. structural framework 2. storage 3. energy production a. 1 only b. 2 only c. 3 only d. 1 & 3 only 6. In which organs are glycogen stored in the body? A. liver and spleen C. liver and bile B. liver and muscle D. liver and adipose tissue 7. When digesting a complex carbohydrate, water is added and simple sugar is obtained through which process? a. Photosynthesis c. Hydrolysis b. Condensation d. Dehydration 8. What kind of molecule is represented by the structure below? CH3 CH2 CH2 CH2 CH=CHCH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 COOH a. monosaccharide c. saturated fatty acid b. unsaturated fatty acid d. phospholipid All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 130. D EPED C O PY 474 9. Disaccharide isformed by combining two monosaccharides. What do you call the process of combining 2 or more simple sugars? a. Hydrolysis c. Condensation b. Peptide bonding d. Saccharide bonding 10. Which of the following elements is NOT present in carbohydrates? a. carbon b. oxygen c. nitrogen d. hydrogen 11. Which of the following biomolecules contain only the elements carbon, hydrogen and oxygen? a. carbohydrates and lipids c. proteins and nucleic acids b. lipids and proteins d.nucleic acids and carbohydrates 12. Which of the following sugars are the components of lactose? a. glucose & galactose c. glucose & fructose b. fructose and galactose d. glucose and glucose 13. What type of chemical bond is illustrated by the arrows below? a. sugar-sugar bond c. peptide bond b. glycerol- fatty acid bond d. hydrogen bond 14. Which of the following sugars are the components of maltose? a. glucose & galactose c. glucose & fructose b. fructose and galactose d. glucose and glucose 15. The sugar in RNA is _____________, the sugar in DNA is ________. a. deoxyribose, ribose b. ribose, deoxyribose c. ribose, phosphate d. ribose, uracil All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 131. D EPED C O PY 475 References and Links PrintedMaterials: Davis, Raymond E., Frey, Regina, Sarquis, Mickey, Sarquis Jerry L. (2009). Modern Chemistry (Teacher’s edition) Holt, Rinehart and Winston, USA LeMay, Jr. Eugene H., et al. (2000). Chemistry Connections to Our Changing World (Teacher’s Edition) Prentice Hall, Inc. Upper Saddle River, NJ 07458 Chang, Raymond, (1998). Chemistry. 6th edition. Mc Graw-Hill Companies, Inc. Comparison between DNA and RNA. Retreived from http://www.diffen.com/ difference/ DNA vs RNA Electronic Sources: http://learningcenter.nsta.org/products/symposia_seminars/ACS/ webseminar11.aspx Images for chemical structure of glucose and fructose. Retrieved from http://www.nsta.org/publications/press/extras/morechemistry. Images for the hydrolysis of sucrose. Retrieved from http://www.mhhe.com/biosci/pae/botany/uno/graphics/uno01pob/vrl/ images/0019.gif Images for the chemical structures of starch and cellulose. Retrieved from https://www.google.com.ph/search?q=chemical+structure+of+starch&client =firefox-a&hs=Qgc&rls=org.mozilla: Images for the chemical structure of triglycerides. Retrieved from https://www.google.com.ph/search?q=chemical+structure+of+triglycerides& client=firefox-a&rls=org.mozilla:en-US:official&channel=sb&source= lnms&tbm=isch&sa=X&ei=f6USU5XkFYyXkgWFp4DoBA&ved=0CA cQ_AUoAQ&biw=1207&bih=518 Image for hydrogenation reaction. Retreived from http://www.chemguide.co.uk/organicprops/alkenes/hydrogenation.html Pictures of food samples-Carbohydrates, Lipids, And Proteins. Retreived from http://www.slideshare.net/gurustip/carbohydrathttp://edtech2.boisestate.edu/ jonfreer/502/jigsaw.htmles-lipids-and-proteins-presentation Identifying Biomolecules in Foods.Retreived from http://www.cteonline.org/portal/default/Curriculum/Viewer/ Curriculum?action=2&view=viewer&cmobjid=177679. March 6, 2014 Biomolecules Jigsaw Activity. Retreived from http://edtech2.boiestate.edu/jonfreer/502/jigsaw.html. March 6,2014 All rights reserved. No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.
- 132. D EPED C O PY 476 http://joelbergerdc.com/tag/glucose-vs-fructose/ https://courses.ecampus.oregonstate.edu/ans312/one/carbs_story.htm http://www.natuurlijkerwijs.com/english/Glycogen_metabolism.htm https://myorganicchemistry.wikispaces.com/ Cellulose?responseToken=1a9131f668de1a94603bbdfb79f69128 http://pixshark.com/phospholipid-bilayer-diagram.htm http://en.wikipedia.org/wiki/Micelle https://courses.ecampus.oregonstate.edu/ans312/one/lipids.htm http://sphweb.bumc.bu.edu/otlt/MPH-Modules/PH/PH709_BasicCellBiology/ PH709_BasicCellBiology24.html http://entrytest-preparation.blogspot.com/2014/01/amino-acids.html http://www.physicalgeography.net/fundamentals/10h.html All rights reserved.No part of this material may be reproduced or transmitted in any form or by any means - electronic or mechanical including photocopying – without written permission from the DepEd Central Office. First Edition, 2015.