Physical Science Module Q3 Chemistry 24-25 PDF

PHYSICAL SCIENCE Level: SENIOR HIGH SCHOOL Semester: SECOND Subject Group: CORE SUBJECT Quarter: THIRD Course Description: Evolution of our understanding of matter, motion, electricity, magnetism, light, and t...

PHYSICAL SCIENCE Level: SENIOR HIGH SCHOOL Semester: SECOND Subject Group: CORE SUBJECT Quarter: THIRD Course Description: Evolution of our understanding of matter, motion, electricity, magnetism, light, and the universe from ancient times to the present; applications of physics and chemistry concepts in contexts such as atmospheric phenomena, cosmology, astronomy, vision, medical instrumentation, space technology, drugs, sources of energy, pollution and recycling, fitness and health, and cosmetics. Course Requirements: Below is the list of activities that must be completed and submitted with their corresponding percentage. WEEK ACTIVITIES Date of Completion Final Grade 1 EAA 1 2 Mini PT 1 3 EAA 2 4 Mini PT 2 5 EAA 3 6 Mini PT 3 7 EAA 4 8 Final Performance Task TOTAL QUARTER 3 CULMINATING PERFORMANCE TASK GOAL: The goal is to examine and identify the active component in a specific beauty product ROLE: You are a sales representative and you are tasked to be familiar with the beauty product that you will be selling AUDIENCE: You need to convince your customer (that will be your PhySci teacher) SITUATION: You will be choosing a specific beauty or cosmetic product in the household. PRODUCT: You will create pamphlet or brochure about the product of your choice. This should describe the components and the modes of action (how does the product work?). It should also include instructions on how to use the product. It may be printed or handwritten. Use a piece of short bond paper which may be folded. Be creative. STANDARDS: Your output must meet the following criteria: Content: 40% Organization: 40% Creativity: 20%  Colegio de Los Baños – PHYSICAL SCIENCE 1 PRE-REQUISITE ASSESSMENT: What is chemistry? LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Periodic table of elements PRE-REQUISITE SKILL: Interpretation of Periodic Table of Elements TIME ALLOTMENT: 4 HRS CONSULTATION: For inquiries and clarifications regarding the lesson, you may contact your teacher thru his/her FB Messenger or thru email RUA: At the end of the lesson, you should be able to: Give evidence for and explain the formation of the light elements in the Big Bang theory Give evidence for and describe the formation of heavier elements during star formation and evolution Write the nuclear fusion reactions that take place in stars, which lead to the formation of new elements INSTITUTIONAL VALUES: Precision and Accuracy, Scientific Literacy, Excellence Students will be able to apply Precision and accuracy in explaining the formation of heavier elements Scientific literacy in describing the process of element formation Excellence in solving problems related to process like helium burning during element formation OVERVIEW OF THE LESSON This lesson is all about basic method of star formation and its life cycle as well as the processes involved in formation of heavier elements. STUDENT’S EXPERIENTIAL LEARNING Scientists believe that the Solar System was formed when a cloud of gas and dust in space, made up of elements formed deep within stars, started to collapse, forming a solar nebula. This began to spin as it collapsed, eventually giving birth over billions of years to the Solar System as we know it today. Apart from hydrogen and helium, formed just after the Big Bang, most of the elements we come into contact with have been made in the stars. Planet Earth and all of its life forms are made of elements formed billions of years ago deep within the cores of stars now long dead. How are light and heavy elements formed? The lightest elements (hydrogen, helium, deuterium, lithium) were produced in the Big Bang nucleosynthesis. According to the Big Bang theory, the temperatures in the early universe were so high that fusion reactions could take place. This resulted in the formation of light elements: hydrogen, deuterium, helium (two isotopes), lithium and trace amounts of beryllium. Nuclear fusion in stars converts hydrogen into helium in all stars. In stars less massive than the Sun, this is the only reaction that takes place. In stars more massive than the Sun (but less massive than about 8 solar masses), further reactions that convert helium to carbon and oxygen take place in successive stages of stellar evolution.  Colegio de Los Baños – PHYSICAL SCIENCE 2 UNIVERSAL ELEMENT FORMATION Stellar nebula Average star Red giant Planetary nebula White dwarf Moments after the An average or As medium sized A planetary nebula is a A white dwarf is a Big Bang, energy medium star is stars exhaust their huge shell of gas and dust small, very dense, begins to condense less than 3 times hydrogen content, ejected during the last hot star that is made into matter, protons the mass of the they expand up to stage (red giant) of the life mostly of carbon. and neutrons are Sun. Stars are 100 times their of a medium star. These faint stars are formed, and then powered by original size to Elements such as He, C, what remain after a the first element nuclear fusion in become red giants. O, N, Ne and smaller red giant star loses its (hydrogen) is their cores, mostly The nuclear fusion amounts of heavier outer layers. They formed. Hundreds converting reactions occurring elements are present. are about the size of of millions of years hydrogen into within a red giant Planetary nebulae allow the Earth and will later in stellar helium and are H→ He and these elements to be eventually lose their nebulae, the liberating He→ C. Our Sun will returned to the interstellar heat to become a hydrogen gas tremendous follow this path over medium. The remains of cold, dark black clouds coalesce amounts of the next 5 billion the carbon core of a red dwarf. The sun will and, under gravity, energy.(JPL- years. This red giant giant evolve into a white eventually turn into a form protostars. Caltech NASA) is Aldebaran in the dwarf star. Ex.Eskimo white dwarf and then Nuclear fusion constellation Nebula in Gemini a black dwarf.(NASA processes begin Taurus. (NASA) (Andrew Fruchter (STScI) and H.Richer converting et al., WFPC2, HST, (University of British hydrogen into NASA) Columbia) helium. One example of a stellar Massive star Super Red giant Supernova Neutron star nursery is the Supergiants are the From the cataclysmic Stars with a mass Eagle Nebula. Massive stars, element factories of explosion of between 1.5 and 3 ( T.A.Rector & more than 3 times our universe. The the supernova, the times the mass of the B.A.Wolpa, the mass of the nuclear fusion heavier elements form. sun will end up NOAO/AURA/NSF) Sun, mostly reactions occurring as neutron stars. A convert hydrogen are H→ He, He→ C, The supernova is the final neutron star is a very into helium. Rigel C→ Ne, Ne→ O, stage in the life of small, super-dense is the brightest star O→ Si and Si→ Fe. massive stars. The outer star that is composed in the constellation region of the star mostly of tightly called Orion and Betelgeuse in the collapses and it instantly packed neutrons. A one of the constellation Orion rebounds off the inner rapidly spinning brightest stars in is a super red giant. core in a cataclysmic neutron star is known the sky. It is a blue It is about 20 times explosion. The extremely as a pulsar. (NASA) (very hot) supergiant, over as massive as the high level of energy 100 times bigger Sun. The lifetime of allows further fusion Black hole than the Sun. this type of star is reactions to occur, Black holes are all (Glen Youman) relatively short by producing heavy that remain after comparison with the elements stars with masses Sun – millions of like gold, silver and over 3 times that of years as opposed to uranium. The supernova the sun supernova. billions of years. image shows Tycho’s A black hole is a (A.Dupree and Supernova Remnant – massive object (or R.Gillard (CfA), this expanding gas cloud region) in space that R.Gilliland (STScI)) is all that remains after a is so dense that, star went supernova. In within a certain radius 1572, Danish astronomer (the Schwarzschild Tycho Brahe noticed the radius determines presence of a ‘new’ bright the event horizon), its light in the night sky and gravitational field recorded its position and does not let anything intensity in his writings. It escape from it – not has been named in his even light. honour. ( APOD/NASA/CXC/SAO)  Colegio de Los Baños – PHYSICAL SCIENCE 3 Nucleosynthesis is the process that creates new atomic nuclei from pre-existing nucleons (protons and neutrons). The first nuclei were formed about three minutes after the Big Bang, through the process called Big Bang nucleosynthesis. 3 types of Nucleosynthesis: ✔ Big Bang Nucleosynthesis ✔ Stellar Fusion or Nucleosynthesis ✔ Supernova Nucleosynthesis Big Bang Nucleosynthesis- Due to the rapid cooling because of expansion, nucleosynthesis ground to a halt about three minutes after the Big Bang occurred. This left us with mostly H isotopes (p, D and T), He isotopes and a very tiny bit of other elements like Li. The relative abundance of He and H did not change much until today. Stellar Fusion or Nucleosynthesis takes place in a. Main sequence stars Processes of element formation: 1. Proton- proton chain reaction The proton–proton chain reaction is one of two known sets of nuclear fusion reactions by which stars convert hydrogen to helium. The first step in all the branches is the fusion of two protons into deuterium. As the protons fuse one of them undergoes beta decay, converting into a neutron by emitting a positron and an electron neutrino  Colegio de Los Baños – PHYSICAL SCIENCE 4 2. CNO cycle The CNO cycle (for carbon–nitrogen–oxygen) is one of the two known sets of fusion reactions by which stars convert hydrogen to helium, the other being the proton–proton chain reaction (pp-chain reaction). Unlike the latter, the CNO cycle is a catalytic cycle. In the CNO cycle, four protons fuse, using carbon, nitrogen, and oxygen isotopes as catalysts, to produce one alpha particle, two positrons and two electron neutrinos. b. Red giants (evolved main sequence stars Processes of element formation: 1. Alpha ladder The alpha process or alpha ladder, is one of two classes of nuclear fusion reactions by which stars convert helium into heavier elements, the other being the triple-alpha process. The triple-alpha process consumes only helium, and produces carbon. After enough carbon has accumulated, the reactions below take place, all consuming only helium and the product of the previous reaction. 2. Triple alpha process The triple-alpha process is a set of nuclear fusion reactions by which three helium-4 nuclei (alpha particles) are transformed into carbon. 3. Slow neutron capture (s) process  Colegio de Los Baños – PHYSICAL SCIENCE 5 Supernova nucleosynthesis occurs during supernova explosions which happens when massive (red) giants explodes. In a supernova explosion, neutron capture reactions take place (this is not fusion), leading to the formation of heavy elements. This is the reason why it is said that most of the stuff that we see around us come from stars and supernovae (the heavy elements part). Process of element formation: Rapid neutron capture (r) process Examples of element making (nucleogenesis) in helium burning reactions: 12 C + 4 He → 16 O 16 O + 4 He → 20 Ne 20 Ne + 4 He → 24 Mg 24 Mg + 4 He → 28 Si 24 Si + 4 He → 32 S Man-made elements Only 90 of the 118 known elements occur naturally, the rest were formed synthetically. By smashing atoms together in machines known as particle accelerators, it was discovered that new elements could be made. For example, bombarding atoms of the element curium with atoms of neon made element 106 – seaborgium Man-made elements Only 90 of the 118 known elements occur naturally, the rest were formed synthetically. By smashing atoms together in machines known as particle accelerators, it was discovered that new elements could be made. For example, bombarding atoms of the element curium with atoms of neon made element 106 – seaborgium.  Colegio de Los Baños – PHYSICAL SCIENCE 6 WEEK 1 ANSWER SHEET (Please submit only the answers. Do not return the entire module.) Name: ________________________________ Section: _______________________ LAST NAME, FIRST NAME MIDDLE INITIAL ENGAGEMENT Enabling Assessment Activity No.1. Nucleosynthesis A. Fill in the blanks for the following nuclear reactions. 1. 8 Be + 4 He → _____ 2. _____+ 4 He → 20 Ne 3. 24 Mg + 4 He → ______ 4. _____ + 4 He → 36 Ar 5. 36 Ar + 4 He → _____ B. Identify if the following equation is TRUE or FALSE. 1. 55Mn + 4He →29Co 2. 40Ar + 4He → 44 Sc 3. 31P + 4He → 35Cl 4. 39K +4He + 4He + 4He → 51V 5. 7Li + 4HE + 4He + 4He → 20 Ne C. Give your own examples of nucleogenesis reactions. 1. ____________________________________________________________ 2. ____________________________________________________________ 3. ____________________________________________________________ 4. ____________________________________________________________ 5. ____________________________________________________________ ASSIMILATION Answer in 3 sentences. How do the changes in the Sun from being average star to a red giant affect the planets and other celestial bodies within our solar system? ___________________________________________________________________ SIGNATURE OVER PRINTED NAME OF PARENT/GUARDIAN  Colegio de Los Baños – PHYSICAL SCIENCE 7 PRE-REQUISITE ASSESSMENT: LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Basics in Atoms PRE-REQUISITE SKILL: Analyzing elements int the Periodic Table TIME ALLOTMENT: 4 HRS CONSULTATION: For inquiries and clarifications regarding the lesson, you may contact your teacher thru his/her FB Messenger or thru email RUA: At the end of the lesson, you should be able to: Describe how elements heavier than iron are formed Describe the ideas of the Ancient Greeks on the atom Describe the ideas of the Ancient Greeks on the elements Describe the contributions of the alchemists to the science of chemistry Point out the main ideas in the discovery of the structure of the atom and its subatomic particles Cite the contributions of J.J. Thompson, Ernest Rutherford, Henry Moseley, and Niels Bohr to the understanding of structure of the atom Describe the nuclear model of the atom and the location of its major components (protons, neutrons, and electrons) Explain how the concept of atomic number led to the synthesis of new elements in the laboratory Write the nuclear reactions involved in the synthesis of new elements Cite the contribution of John Dalton toward the understanding of the concept of the chemical elements Explain how Dalton’s theory contributed to the discovery of other elements INSTITUTIONAL VALUES: Precision and Accuracy, Scientific Literacy, Excellence Students will be able to apply Precision and accuracy in identifying number of proton, neutron and electrons Scientific literacy in describing the relationship of atomic number to mass number Excellence in solving problems related to ionic charges of elements in the periodic table OVERVIEW OF THE LESSON This lesson is all about basic concept of atoms, including sub-atomic particles like proton, neutron and electron, as well as the relationship of mass number and atomic number to sub-atomic particles STUDENT’S EXPERIENTIAL LEARNING Atoms are the basic units of matter and the defining structure of elements. The term "atom" comes from the Greek word for indivisible, because it was once thought that atoms were the smallest things in the universe and could not be divided. We now know that atoms are made up of three particles: protons, neutrons and electrons — which are composed of even smaller particles, such as quarks.  Colegio de Los Baños – PHYSICAL SCIENCE 8 Fig. 9. Elements according to the Greeks Atoms were created after the Big Bang 13.7 billion years ago. As the hot, dense new universe cooled, conditions became suitable for quarks and electrons to form. Quarks came together to form protons and neutrons, and these particles combined into nuclei. This all took place within the first few minutes of the universe's existence. It took 380,000 years for the universe to cool enough to slow down the electrons so that the nuclei could capture them to form the first atoms. The earliest atoms were primarily hydrogen and helium, which are still the most abundant elements in the universe. Gravity eventually caused clouds of gas to coalesce and form stars, and heavier atoms were (and still are) created within the stars and sent throughout the universe when the star exploded (supernova). Atomic structure The atom has two distinct regions: a. the nucleus(very small, dense, and positively charged, and where the protons and neutrons reside) b. the region outside the nucleus where electrons are located.  Colegio de Los Baños – PHYSICAL SCIENCE 9 Subatomic particles a. Electron: negatively charged particle discovered by JJ Thompson b. Proton: positively charged particle discovered by Ernest Rutherford c. neutron: neutral particle discovered by James Chadwick Atomic Number is equal to the number of protons. Mass Number is equal to the sum of the number of protons and neutrons 𝒑+𝒏 𝒑−𝒆 Elements are represented by a nuclide symbol: 𝒑𝑿 Where 𝒑 + 𝒏 = mass number (no of protons + no of neutrons) 𝒑 = atomic number (no of protons)  Colegio de Los Baños – PHYSICAL SCIENCE 10 𝒑 − 𝒆 = charge (no of protons - no of electrons) Atomic structure of the first 10 elements  Colegio de Los Baños – PHYSICAL SCIENCE 11 WEEK 2 ANSWER SHEET (Please submit only the answers. Do not return the entire module.) Name: ________________________________ Section: _______________________ LAST NAME, FIRST NAME MIDDLE INITIAL Mini-Performance Task No.1. Sub-atomic particles Element Atomic Mass Proton Electron Neutron Charge Number Number Sodium 11 23 14 28 14 16 35 17 18 18 22 20 26 2+ S 18 16 2- 65 30 28 35 35 36 183 68 109 6+ Pb 207 82 78 U 90 146 2+ 40 + 19𝐾 What is the notation to be used if nitrogen gained 3 electrons and its mass number is 15? _________________________________________________________________ SIGNATURE OVER PRINTED NAME OF PARENT/GUARDIAN DATE: ______________  Colegio de Los Baños – PHYSICAL SCIENCE 12 PRE-REQUISITE ASSESSMENT: Why do elements become cations when electrons move out of the system? LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Electronegativity of elements PRE-REQUISITE SKILL: Electronegativity and molecular property analysis TIME ALLOTMENT: 4 HRS CONSULTATION: For inquiries and clarifications regarding the lesson, you may contact your teacher thru his/her FB Messenger or thru email RUA: At the end of the lesson, you should be able to: Determine if a molecule is polar or nonpolar given its structure Relate the polarity of a molecule to its properties INSTITUTIONAL VALUES: Precision and Accuracy, Scientific Literacy, Excellence Students will be able to apply Precision and accuracy in identifying if molecules are polar or non-polar Scientific literacy in describing the properties of molecules based on their polarity Excellence in solving problems related to eletronegativity and polarity OVERVIEW OF THE LESSON This lesson is all about basic concept of polarity, shape of molecules and how do these shapes affect polarity or physical properties of a molecule STUDENT’S EXPERIENTIAL LEARNING Polar molecules are those that possess regions of positive and negative charge. Water is an example of a polar material. The type of bonds it has, when coupled with its shape, gives one end of the molecule a slight positive charge (the hydrogen end) and the other a slight negative charge (the oxygen end). Intramolecular Forces of Attraction (Chemical Bonding) A chemical bond is the force that holds atoms together in a molecule or compound (i.e. ionic or covalent bonds), while intermolecular forces of attraction are the forces that hold molecules together in more condensed phases like solids and liquids. Types of Chemical Bonds 1. Ionic Bonds NaCl  Colegio de Los Baños – PHYSICAL SCIENCE 13 2. Covalent Bonds In molecule H2, the hydrogen atoms share the two electrons via covalent bonding. Bonds within the most organic compounds are described as covalent Metallic Bond Metallic bonds are present in samples of pure elemental metals, such as gold or aluminum, or alloys, like brass or bronze. Lewis Structures Lewis diagrams, also called electron-dot diagrams, are used to represent paired and unpaired valence (outer shell) electrons in an atom. This is proposed by Gilbert N. Lewis. 1. Lewis structures Atoms A Lewis symbol is a symbol in which the electrons in the valence shell of an atom or simple ion are represented by dots placed around the letter symbol of the element. Each dot represents one electron.  Colegio de Los Baños – PHYSICAL SCIENCE 14 Sample Lewis Structure 2. Lewis Structures for Ions of Elements The chemical symbol for the element is surrounded by the number of valence electrons present in the ion. The whole structure is then placed within square brackets, with a superscript to indicate the charge on the ion Lithium ion Fluorine ion 3. Lewis Structures for Ionic Compounds The overall charge on the compound must equal zero, that is, the number of electrons lost by one atom must equal the number of electrons gained by the other atom. Lithium fluoride, LiF Potassium Oxide, Li2O 4. Lewis Structures for Covalent Compounds In a covalent compound, electrons are shared between atoms to form a covalent bond in order that each atom in the compound has a share in the number of electrons required to provide a stable, Noble Gas, electronic configuration.  Colegio de Los Baños – PHYSICAL SCIENCE 15 5. Lewis Structures for Electron-rich Compounds Elements with atomic number greater than 13 often form compounds or polyatomic ions in which there are “extra” electrons. 6. Lewis Structures for Electron-poor Compounds There is another type of molecule or polyatomic ion in which there is an electron deficiency of one or more electrons needed to satisfy the octets of all the atoms. The boron atom is two electrons shy of its octet. You may ask about the formation of a double bond (and even resonance). But fluorine and boron are not in the list that can form double bonds (C, P, O, N, S) and so the compound is electron poor. Bond Polarity or simply polarity, describes how equally bonding electrons are shared between atoms. It is a physical property of compounds and affects other physical properties such as solubility, melting points and boiling points. 𝐸𝑁 = 𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑦 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑒𝑙𝑒𝑚𝑒𝑛𝑡 ∆𝐸𝑁 Bond type Example 𝐸𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑦 ∆𝐸𝑁 ∆𝐸𝑁=0 Completely nonpolar I2 I:2.66 2.66-2.66=0 ∆𝐸𝑁 ≤0.5 Nonpolar covalent CH4 C: 2.55 2.55-2.20=0.35 H: 2.20 0.5< ∆𝐸𝑁 ≤ Polar covalent H2O H: 2.20 3.4-2.20=1.24 1.6 O: 3.44 1.6< Ionic (metal CaBr2 Ca:1.00 2.96-1.00=1.96 ∆𝐸𝑁 ≤2.0 +nonmetal) Br:2.96 HF H: 2.20 3.98-2.20=1.78 Polar covalent F: 3.98 (nonmetals) 2.0< ∆𝐸𝑁 Completely NaCl Na: 0.93 3.16-0.93=2.23 polar Cl: 3.16  Colegio de Los Baños – PHYSICAL SCIENCE 16 A polar molecule may be polar as a result of polar bonds or as a result of an asymmetric arrangement of non-polar bonds and nonbonding pairs of electrons. Example 1. hydrogen fluoride, HF Example 2. ammonia, NH3 Geometric Considerations For ABn molecules, where A is the central atom and B are all the same types of atoms, there are certain molecular geometries which are symmetric. Therefore, they will have no dipole even if the bonds are polar. These geometries include linear, trigonal planar, tetrahedral, octahedral and trigonal bipyramid. Molecular geometries with exact cancelation of polar bonding to generate a non-polar molecule Predicting Molecular Polarity Tip-off – You are asked to predict whether a molecule is polar or nonpolar; or you are asked a question that cannot be answered unless you know whether a molecule is polar or nonpolar. (For example, you are asked to predict the type of attraction holding the particles together in a given liquid or solid.)  Colegio de Los Baños – PHYSICAL SCIENCE 17 General Steps - Step 1: Draw a reasonable Lewis structure for the substance. Step 2: Identify each bond as either polar or nonpolar. (If the difference in electronegativity for the atoms in a bond is greater than 0.4, we consider the bond polar. If the difference in electronegativity is less than 0.4, the bond is essentially nonpolar.) If there are no polar bonds, the molecule is nonpolar. If the molecule has polar bonds, move on to Step 3. Step 3: If there is only one central atom, examine the electron groups around it. If there are no lone pairs on the central atom, and if all the bonds to the central atom are the same, the molecule is nonpolar. (This shortcut is described more fully in the Example that follows.) If the central atom has at least one polar bond and if the groups bonded to the central atom are not all identical, the molecule is probably polar. Move on to Step 4. Step 4: Draw a geometric sketch of the molecule. Step 5: Determine the symmetry of the molecule using the following steps. Describe the polar bonds with arrows pointing toward the more electronegative element. Use the length of the arrow to show the relative polarities of the different bonds. (A greater difference in electronegativity suggests a more polar bond, which is described with a longer arrow.) Decide whether the arrangement of arrows is symmetrical or asymmetrical If the arrangement is symmetrical and the arrows are of equal length, the molecule is nonpolar. If the arrows are of different lengths, and if they do not balance each other, the molecule is polar. If the arrangement is asymmetrical, the molecule is polar. EXAMPLE – Predicting Molecular Polarity: Decide whether the molecules represented by the following formulas are polar or nonpolar. (You may need to draw Lewis structures and geometric sketches to do so.) a. CO2 b. OF2 c. CCl4 d. CH2Cl2 e. HCN Solution: a. The Lewis structure for CO2 is The electronegativities of carbon and oxygen are 2.55 and 3.44. The 0.89 difference in electronegativity indicates that the C-O bonds are polar, but the symmetrical arrangement of these bonds makes the molecule nonpolar. If we put arrows into the geometric sketch for CO2, we see that they exactly balance each other, in both direction and magnitude. This shows the symmetry of the bonds. b. The Lewis structure for OF2 is The electronegativities of oxygen and fluorine, 3.44 and 3.98, respectively, produce a 0.54 difference that leads us to predict that the O-F bonds are polar. The molecular geometry of OF2 is bent. Such an asymmetrical distribution of polar bonds would produce a polar molecule.  Colegio de Los Baños – PHYSICAL SCIENCE 18 c. The molecular geometry of CCl4 is tetrahedral. Even though the C-Cl bonds are polar, their symmetrical arrangement makes the molecule nonpolar. d. The Lewis structure for CH2Cl2 is The electronegativities of hydrogen, carbon, and chlorine are 2.20, 2.55, and 3.16. The 0.35 difference in electronegativity for the H-C bonds tells us that they are essentially nonpolar. The 0.61 difference in electronegativity for the C-Cl bonds shows that they are polar. The following geometric sketches show that the polar bonds are asymmetrically arranged, so the molecule is polar. (Notice that the Lewis structure above incorrectly suggests that the bonds are symmetrically arranged. Keep in mind that Lewis structures often give a false impression of the geometry of the molecules they represent.) e. The Lewis structure and geometric sketch for HCN are the same: The electronegativities of hydrogen, carbon, and nitrogen are 2.20, 2.55, and 3.04. The 0.35 difference in electronegativity for the H-C bond shows that it is essentially nonpolar. The 0.49 difference in electronegativity for the C-N bond tells us that it is polar. Molecules with one polar bond are always polar.  Colegio de Los Baños – PHYSICAL SCIENCE 19 WEEK 3 ANSWER SHEET (Please submit only the answers. Do not return the entire module.) Name: ________________________________ Section: _______________________ LAST NAME, FIRST NAME MIDDLE INITIAL ENGAGEMENT Enabling Assessment Activity No.2. Lewis Structure and Bond Polarity Determine the polarity of the following compounds. IONIC COMPOUNDS COVALENT COMPOUNDS 1. KI a. Br2 a. Lewis Structure a. Lewis Structure b. ∆𝐸𝑁 = b. ∆𝐸𝑁 = c. Polarity = c. Polarity = 2. NaBr b. SF5 a. Lewis Structure a. Lewis Structure b. ∆𝐸𝑁 = b. ∆𝐸𝑁 = c. Polarity = c. Polarity =  Colegio de Los Baños – PHYSICAL SCIENCE 20 ∆𝐸𝑁 Bond type Electronegativity ∆𝐸𝑁=0 Completely Na: 0.93 nonpolar Ca:1.00 P: 2.19 ∆𝐸𝑁 ≤0.5 Nonpolar H: 2.20 covalent C: 2.55 0.5< Polar Br:2.96 ∆𝐸𝑁 ≤ 1.6 covalent F: 3.98 1.6< Ionic (metal ∆𝐸𝑁 ≤2.0 +nonmetal) Polar covalent (nonmetals) 2.0< ∆𝐸𝑁 Completely polar ASSIMILATION Answer in 3-5 sentences. Why does alcohol dissolve in water but does not dissolve in oil? (10pts) _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ _____________________________________________________________________________________ __________________________________________________________________________ ___________________________________________________________________ SIGNATURE OVER PRINTED NAME OF PARENT/GUARDIAN  Colegio de Los Baños – PHYSICAL SCIENCE 21 PRE-REQUISITE ASSESSMENT: What is the connection of polarity to dissolution process LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Electronegativity of elements, ionic charges PRE-REQUISITE SKILL: Identifying the polarity of a molecule TIME ALLOTMENT: 4 HRS CONSULTATION: For inquiries and clarifications regarding the lesson, you may contact your teacher thru his/her FB Messenger or thru email RUA: At the end of the lesson, you should be able to: Describe the general types of intermolecular forces Explain the effect of intermolecular forces Describe the general types of intermolecular forces Give the type of intermolecular forces in the properties of substances Explain the effect of intermolecular forces on the properties of substances Explain how the uses of the following materials depend on their properties: Medical implants, prosthesis Sports equipment Electronic devices Construction supplies for buildings and furniture Household gadgets Explain how the properties of the above materials are determined by their structure INSTITUTIONAL VALUES: Precision and Accuracy, Scientific Literacy, Excellence Students will be able to apply Precision and accuracy in describing the type of inter-molecular forces of attraction Scientific literacy in describing the effects of IMFA Excellence in solving problems by classifying molecules based on IMFA present OVERVIEW OF THE LESSON This lesson is all about basic concept of polarity, shape of molecules and how do these shapes affect polarity or physical properties of a molecule STUDENT’S EXPERIENTIAL LEARNING The term “INTERmolecular forces” is used to describe the forces of attraction BETWEEN atoms, molecules, and ions when they are placed close to each other This is different from INTRAmolecular forces which is another word for the covalent bonds inside molecules. When two particles experience an intermolecular force, a positive (+) charge on one particle is attracted to the negative (-) on the other particles. When intermolecular forces are strong, the atoms, molecules or ions are strongly attracted to each other, and draw closer together. These are more likely to be found in condensed states such as liquid or solid. When intermolecular forces are weak, the atoms, molecules or ions do not have a strong attraction for each other and move far apart. Intermolecular forces are electrostatic in nature and include van der Waals forces and hydrogen bonds. Molecules in liquids are held to other molecules by intermolecular interactions, which are weaker than the intramolecular interactions that hold the atoms together within molecules and polyatomic ions.  Colegio de Los Baños – PHYSICAL SCIENCE 22 Transitions between the solid and liquid or the liquid and gas phases are due to changes in intermolecular interactions but do not affect intramolecular interactions. intermolecular forces weaker than ionic or covalent bonds many properties of liquids reflect strengths of intermolecular forces three types of intermolecular forces: dipole-dipole forces, London dispersion forces, and hydrogen- bonding forces o also called van der Waals forces o less than 15% as strong as covalent or ionic bonds o electrostatic in nature, involves attractions between positive and negative species Ion-Dipole Forces – exists between an ion and partial charge at one end of a polar molecule (dipoles) magnitude of attraction increases as either the charge of ion or magnitude of dipole moment increases Dipole-Dipole Forces – exists between neutral polar molecules which are very close together weaker than ion-dipole forces for molecules of approximately equal mass and size, the strengths of intermolecular attractions increase with increasing polarity increase dipole moment ® increase boiling point London Dispersion Forces – interparticle forces that exist between nonpolar atoms or molecules London dispersion forces increase with increasing molecular size & molecular mass. Hydrogen Bonding – special type of intermolecular attraction that exists between the hydrogen atom in a polar bond and an unshared electron pair on a nearby electronegative ion or atom hydrogen bond with F, N, and O is polar Page 22 of 57 Colegio de Los Baños – PHYSICAL SCIENCE 23 Page 23 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 24 WEEK 4 ANSWER SHEET (Please submit only the answers. Do not return the entire module.) Name: ________________________________ Section: _______________________ LAST NAME, FIRST NAME MIDDLE INITIAL Lesson 3: Intermolecular Forces of Attraction (IMFA) SY 2024-2025 NAME DATE ACCOMPLISHED: SECTION Intermolecular forces are attractive forces that act between molecules or particles in the solid or liquid states. Certain properties of matter like melting point, boiling point, or freezing point can be used to indicate the strength of the intermolecular forces of attraction present in the substances. Objectives: At the end of the activity, students shall be able to: Identify the intermolecular forces of attraction (IMFA) present in each of the substances Determine the effect(s) of the IMFA(s) on the properties of substances Materials needed: Quantity Material 1 small bottle Water 1 small bottle Ethanol (a.k.a. ethyl alcohol) or isopropyl alcohol (whichever is available) 1 small bottle Acetone 3 pieces Small vial (or any small bottle/container) 3 pieces Medicine dropper or straw 6 pieces One-peso coin *Safety Precautions: Proper handling of the substances you will be using. Avoid direct inhalation of the vapors of the substances. Use mask. 1. Prepare your camera to document your activity. Fill one vial with water and label it as water. Do same thing for another vial then label it as ethanol. Repeat the procedure for acetone. 2. Using the first three coins, drop each liquid on a 1-peso coin and count the number of drops the coin can hold. Record the number of drops on the table below. 3. Using the next set of three coins, put a drop of the liquid and determine how much time it takes one drop to evaporate. (One coin for water, another coin for alcohol, and another one for acetone.) Record the time on the table below. 4. Attach here the photos of you doing the activity. Page 24 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 25 5. Gather the results and record them in the table below. Answer the discussion questions. Results and Discussion I. A. Your photos: B. Your results Water Alcohol Acetone Number of Drops Time it took for one drop to evaporate (Min & Sec) C. Your classmates WATER Classmate 1 Classmate 2 Classmate 3 (NAME) (NAME) (NAME) Number of Drops Time it took for one drop to evaporate (Min & Sec) Page 25 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 26 ETHANOL / ISOPROPYL ALCOHOL Classmate 1 Classmate 2 Classmate 3 (NAME) (NAME) (NAME) Number of Drops Time it took for one drop to evaporate (Min & Sec) ACETONE Classmate 1 Classmate 2 Classmate 3 (NAME) (NAME) (NAME) Number of Drops Time it took for one drop to evaporate (Min & Sec) II. Answer the following questions: a. Which molecules can hold more drops on the coin? b. Which molecules took longer to evaporate? Why do you think so? c. Are the molecules that can hold the lesser number of drops the same as the molecules that took less time to evaporate? d. Based on the formula and geometries of the substances, are the molecules that can hold more drops on the coin polar or nonpolar? What about those that took longer to evaporate? Page 26 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 27 e. Compare and contrast your results with your classmates’. Explain the differences. f. Conclusion Reference(s): Source 1 Source 2 ___________________________________________________________________ SIGNATURE OVER PRINTED NAME OF PARENT/GUARDIAN DATE: ______________ Page 27 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 28 PRE-REQUISITE ASSESSMENT: What are the different factors that speed up a reaction? LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Electronegativity of elements, ionic charges, functional groups PRE-REQUISITE SKILL: Shape and polarity of molecules TIME ALLOTMENT: 4 HRS CONSULTATION: For inquiries and clarifications regarding the lesson, you may contact your teacher thru his/her FB Messenger or thru email RUA: At the end of the lesson, you should be able to: Explain how the structures of biological macromolecules such as carbohydrates, lipids, nucleic acid, and proteins determine their properties and functions INSTITUTIONAL VALUES: Precision and Accuracy, Scientific Literacy, Excellence Students will be able to apply Precision and accuracy of determining sturctural properties and functions of certain biomolecules Scientific literacy on importance of proteins, carbohydrates, lipids and nucleic acids Excellence in solving problems related to functional groups of biomolecules OVERVIEW OF THE LESSON This lesson is all about basic concepts of biological molecules and their functional groups; including the properties of each functional groups that constitutes the biomolecules. STUDENT’S EXPERIENTIAL LEARNING Large biological molecules perform a wide range of jobs in an organism. Some carbohydrates store fuel for future energy needs, and some lipids are key structural components of cell membranes. Nucleic acids store and transfer hereditary information, much of which provides instructions for making proteins. Proteins themselves have perhaps the broadest range of functions: some provide structural support, but many are like little machines that carry out specific jobs in a cell, such as catalyzing metabolic reactions or receiving and transmitting signals. Core Concepts of Macromolecular Structure and Function The four major classes of macromolecules are carbohydrates, lipids, proteins, and nucleic acids. Three of the four classes of macromolecules—carbohydrates, proteins, and nucleic acids—form chainlike molecules called polymers. Monomers are connected by covalent bonds that form through the loss of a water molecule. A polymer is a long molecule consisting of many similar or identical building blocks linked by covalent bonds. Page 28 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 29 The repeated units are small molecules called monomers. Some of the molecules that serve as monomers have other functions of their own. Carbohydrate The word carbohydrate may be broken down to carbon and hydrate. From the chemical formula of carbohydrate, notice that the ratio of C:H:O is 1:2:1, which can be rewritten as Cn(H2O)n. Carbohydrates can be seen as hydrates of carbon. This is a traditional but incorrect understanding of carbohydrates but it still presents a useful picture of the molecule. Another term for carbohydrate is saccharide. This term is derived from the Latin word saccharum referring to sugar--a common carbohydrate. Carbohydrates are classified either as simple or complex. Simple sugars are monosaccharides and disaccharides. Complex sugars are polysaccharides. Carbohydrates are the primary energy source of the human body. The different saccharides that humans eat are converted to glucose which can be readily used by the body. Around 4 kilocalories is derived from one gram of carbohydrate. Should there be an excessive consumption of carbohydrates, the excess is converted to glycogen which is stored in the liver and in muscles. Glycogen is a slow-releasing carbohydrate. Types of Carbohydrates a. Monosaccharide (one saccharide) Glucose Used in dextrose, blood sugar; the form utilized by the human body Galactose Found in milk and milk products Fructose Found in fruits and honey The above monosaccharides all have the same chemical formula of C6H12O6 and its structure is the one that made the difference in its properties. For example, galactose (163-169oC) has a higher melting point than glucose (148-155oC). Glucose is sweeter than galactose. Page 29 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 30 b. Disaccharide (two saccharides) Maltose (Glucose + Glucose) Found in malt Sucrose (Glucose + Fructose) Found in regular table sugar, sugarcane, and sugar beet Lactose (Glucose + Galactose) Found in milk and milk products Individual saccharides are connected via glycosidic bonds. A water molecule is released when two saccharides are combined. Glucose, galactose, and fructose are isomers or compounds that have the same molecular but different structural formulas. c. Polysaccharide (many saccharides) Starch / Amylose Composed of 250 - 400 glucose molecules connected via α-1-4-glycosidic bond. Storage form of glucose in plants. Amylopectin like amylose but has more branches attached via α-1-6 glycosidic bond Storage form of glucose in plants. Glycogen Composed of more glucose, more highly branched (same type of bond as amylopectin) Storage form of glucose in animals, stored in the liver and muscles Cellulose Composed of glucose units connected via β-1-4 glycosidic bond, linear chain arranged in a parallel manner Structural material in plants--cell wall in wood, wood fiber Cannot be digested by humans Protein Protein is a good energy source. Proteins, like carbohydrates gives 4 kilocalories of energy per gram but they are not all used up by the body as primary energy source. Protein is more than just a good energy source. Proteins have other structural and enzymatic functions that are important to the human body. They act as transport, storage and as antibodies. Examples of proteins with different functions to relate structure to function/properties. Proteins are composed of amino acids in the similar way that carbohydrates are composed of saccharides. Depending on the sequence of the different amino acids, proteins will acquire certain structure and functions. The word protein came from the Greek term proteios meaning first. One can think of protein as the beginning of life. From egg albumin being pure protein to sperm and egg cells, we all start from proteins. Proteins are composed of four elements, namely, carbon, hydrogen, oxygen and nitrogen. Sulfur and other metals are sometimes also found in proteins. If carbohydrates are made up of saccharides, proteins are made up of amino acids. Page 30 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 31 There are 20 amino acids. The combination of many amino acids creates protein. Amino acids are joined together with a peptide bond. Proteins are also called polypeptides. The diagram below shows that water is released in the formation of peptide bonds. This is similar to the formation of complex saccharides. Examples of proteins Keratin is a structural protein found in hair, skin, and nails. It is a highly cross-linked protein containing α-helix and β-pleated sheets. Sheep’s wool is made largely of keratin. Fibroin is found in silk. Silk has a smooth and soft texture. It is one of the strongest natural fibers that have high resistance to deformation. It is also a good insulation. Silk is primarily composed of β-pleated sheets. The long polypeptide chain doubles back on its own running parallel connected together by H- bonds. Collagen is a major insoluble fibrous protein found in connective tissues such as tendons, ligaments, skin, cartilage and the cornea of the eye. It comprises as much as 30% of proteins in animals. Its strength is attributed to its triple helix structure comprising of α-helices braided together. When several triple helices combine, they form the fibrils that make up connective tissues. Myoglobin is a polypeptide that stores oxygen in muscles. It is a globular protein comprised of 153 amino acids in a single polypeptide chain. It contains a heme group which has an iron (II) ion at its center. This is where the oxygen is stored Hemoglobin is a globular protein that carries oxygen from the lungs to the bloodstream. It is composed of four sub-units, each containing a heme group that enables it to transport four oxygen molecules at a time. Enzymes function to catalyze chemical reactions. They either speed up a reaction, lower the needed energy for a reaction to take place, or bind substances to their specific partners. Enzymes themselves are very specific as can be seen in their shape. Examples of enzymes are: Lipase - help in digestion of fats Pepsin - help in breaking down proteins into peptides (smaller units) Sucrase - also called invertase, help in the digestion of sugars and starches Page 31 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 32 Lipids Lipids provide more energy per gram than carbohydrates and proteins They provide 9 kilocalories of energy per gram! Aside from being an energy source or storage, lipids also serve other functions such as material for cell membrane, insulation to maintain body temperature, aid in digestion, and as signal molecules. Based on the structure of a triglyceride you will notice that - it has three chains connected to a backbone - it has long chains mainly composed of C and H - it is composed of C, H, O The word lipid comes from the Greek word “lipos” which means fat. Lipids are a family of biomolecules having varied structures. They are grouped together simply because of their hydrophilic property (water- fearing). They are soluble in non-polar solvents such as ether, acetone, and benzene. Lipids can be classified into four categories: a. Wax A wax is a simple lipid which is an ester of a long-chain alcohol and a saturated fatty acid. The alcohol may contain from 12-32 carbon atoms. Waxes are found in nature as coatings on leaves and stems. The wax prevents the plant from losing excessive amounts of water. It serves as coating of leaves and fruits in plants. It is solid at room temperature but melts easily. It is also found in animals. Animals also have it in feathers and skin as protection. Common examples of wax are beeswax from honeycomb, carnauba wax from palm trees, and spermaceti wax from whale fat (used by sailors for candles in olden times). b. Triglycerides Triglycerides are a type of fat (lipid) found in your blood. When you eat, your body converts any calories it doesn't need to use right away into triglycerides. The triglycerides are stored in your fat cells. Later, hormones release triglycerides for energy between meals. Having a high level of triglycerides in your blood can increase your risk of heart disease. But the same lifestyle choices that promote overall health can help lower your triglycerides, too. c. Phospholipids Phospholipids contains glycerol, two fatty acids, and a phosphate group. Unlike other lipids, phospholipids have a polar and non-polar end. This property allows it to transport molecules in the bloodstream. It is also a major component in the cell membrane. d. Steroids Steroid came from the Greek word stereos meaning solid. It is composed of fused-ring structures, namely, three six-membered ring and a five-membered ring. The steroid groups is very wide, ranging from sterol and bile acid to sex hormones to adrenal hormones to hormones regulating the molting of insects and many more. Page 32 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 33 Cholesterol is one of the most common steroids. It is amphiphilic, meaning that it has a polar head and an extensive non-polar tail. Nucleic Acid Nucleic acids play an essential role in the storage, transfer, and expression of genetic information. Nucleic acid was discovered by a twenty-four-year-old Swiss physician named Friedrich Miescher in 1868. He was puzzled that an unknown substance in white blood cells did not resemble carbohydrates, proteins, or lipids. He was able to isolate the substance from the nucleus and initially called it nuclein. He eventually was able to break down nuclein into protein and nucleic acids. He found out that nucleic acids contain carbon, hydrogen, oxygen, nitrogen, and phosphorus. If carbohydrates are composed of saccharide units, proteins of amino acids, and lipids of fatty acids, nucleic acids are composed of nucleotides. Nucleic acids are also known as polynucleotides. A nucleotide has three parts: a. Nitrogenous base b. Five-carbon carbohydrate or sugar c. Phosphate group DNA DNA is responsible for storing and coding genetic information in the body. The structure of DNA allows for genetic information to be inherited by children from their parents. As the nucleotides adenine, thymine, guanine, and cytosine in DNA will only pair in a certain sequence (adenine with thymine, and guanine with cytosine), every time a cell duplicates the strand of DNA can specify the sequence in which the nucleotides should be copied. As such, accurate copies of DNA can be made and passed down from generation to generation. Inside DNA, instructions for all the proteins an organism will make are stored. RNA RNA plays an important role in protein synthesis and regulates the expression of the information stored in DNA to make these proteins. It is also how genetic information is carried in certain viruses. The various functions of RNA include: Creating new cells in the body Translating DNA into proteins Acting as a messenger between DNA and ribosomes Helps ribosomes choose the correct amino acids to create new proteins in the body. These functions are carried out by RNA with different names. These names include: Transfer RNA (tRNA) Page 33 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 34 Ribosomal RNA (rRNA) Messenger RNA (mRNA). ATP The nucleic acid adenosine triphosphate (ATP), made up of an adenine nitrogenous base, a 5-carbon ribose sugar, and three phosphate groups, is involved in generating energy for cellular processes. Page 34 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 35 WEEK 5 ANSWER SHEET (Please submit only the answers. Do not return the entire module.) Name: ________________________________ Section: _______________________ LAST NAME, FIRST NAME MIDDLE INITIAL ENGAGEMENT Enabling Assessment Activity No.3. Biomolecules Choose the BEST answer and write the CAPITAL letter of the answer before each number. ____ 1. What elements make up a carbohydrate? A. hydrogen, calcium, oxygen ____ 6. One function of a carbohydrate is B. hydrogen, carbon, oxygen _______________. C. carbon, potassium, oxygen A. To provide the body with immediate energy D. carbon, magnesium, hydrogen B. To keep the heart functioning smoothly E. nitrogen, carbon, oxygen C. To store and transport genetic material D. To control the rate of the biochemical reactions ____ 2. _________ is known as “animal starch”. E. To regulate the body’s metabolism A. Glucose B. Cellulose C. Fructose ____ 7. The carbohydrate that provides support D. Glycogen in plants is called __________. E. Lactose A. Chitin ____ 3. ____________ is the monosaccharide B. Dextrose found in starch. C. Lipids A. Glucose D. Cellulose B. Fructose E. Amylose C. Maltose D. Lactose ____ 8. Glucose, galactose and fructose are E. Amylose __________. A. Disaccharides ____ 4. __________ is a carbohydrate that B. Isotopes cannot be digested by C. Polymers humans. D. Isomers A. Cellulose E. Amines B. Starch C. Glucose ____ 9. The small repeating units that make up D. Fructose proteins are called __________. E. Maltose A. Fatty acids B. Amino acids ____ 5. Long chains of sugars are called C. Monosaccharides _________. D. Ethylene A. Polypeptides E. Styrene B. Polysaccharides C. Polynucleotides D. Polyunsaturated E. Dipeptide Page 35 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 36 ____ 10. The sequence of amino acids in a ____ 15. What kind of molecule is represented by polypeptide is called the protein’s __________. the structure below? A. primary structure B. secondary structure CH3CH2CH2CH2CHCHCH2CH2CH2CH2CH2C C. tertiary structure H2CH2COOH D. ouarternary structure A. A sugar E. crystal structure B. A disaccharide C. A dipeptide ____ 11. Biopolymers formed from the linkage of D. A saturated fatty acid monomers in the form of nucleotides are called E. An unsaturated fatty acid A. nucleic acids B. carbohydrates ____ 16. Table sugar is a form of C. rubber A. protein D. lipids B. lipid E. proteins C. carbohydrate D. nucleic acid ____ 12. Enzymes are _______. E. steroid A. monosaccharides B. lipids ____ 17. When a protein is boiled, it loses all C. proteins levels of organization except the primary level. D. nucleic Acids When this happens, the protein is said to be: E. polysaccharides A. hydrolyzed B. dehydrated ____ 13. Fats and oils are composed of what two C. denatured groups of molecules? D. plasmolyzed A. glucose and fructose E. folded B. starch and sugar C. water and cellulose ____18. The group of biologically important D. glycerol and fatty acids organic compounds responsible for storage and E. RNA and DNA transfer of information is A. carbohydrates ____ 14. Which of the following is a polymer of B. phospholipids glucose? C. polypeptides A. starch D. nucleic acids B. glycogen E. polysaccharides C. cellulose D. A nd B E. A, B, and C Page 36 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 37 ____ 19. The structure on the left is a(n) B. Carbohydrate, lipid __________ and the structure on C. Carbohydrate, amino acid the right is a(n) ________. D. Nucleotide, amino acid E. Nucleotide, carbohydrate ____ 20. DNA is a ________ A. Carbohydrate B. Lipid C. Fatty Acid D. Nucleic Acid A. Lipid, polypeptide E. Protein ASSIMILATION Answer in 3-5 sentences. Are all cholesterols bad? Justify your answer (10 pts) ___________________________________________________________________ SIGNATURE OVER PRINTED NAME OF PARENT/GUARDIAN Page 37 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 38 PRE-REQUISITE ASSESSMENT: Are steroids considered as biomolecules? Justify your answer LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Chemical changes, ionic charges, chemical reactions PRE-REQUISITE SKILL: Basic laboratory and measurement skills TIME ALLOTMENT: 4 HRS CONSULTATION: For inquiries and clarifications regarding the lesson, you may contact your teacher thru his/her FB Messenger or thru email RUA: At the end of the lesson, you should be able to: Use simple collision theory to explain the effects of concentration, temperature, and particle size on the rate of reaction Define catalyst and describe how it affects reaction rate INSTITUTIONAL VALUES: Precision and Accuracy, Scientific Literacy, Excellence Students will be able to apply Precision and accuracy on describing effects of different factors on reaction rate Scientific literacy and excellence on understanding and explaining the collision and kinetic molecular theory STUDENT EXPERIENTIAL LEARNING Chemical kinetics is the branch of chemistry which addresses the question: “how fast do reactions go? Studies the rate at which a chemical process occurs. Kinetics also sheds light on the reaction mechanism (exactly how the reaction occurs). Reactions can be slow like rusting of iron and food spoilage and fast or instantaneous like combustion. Factors that affect Reaction Rate Constant 1. Temperature: ✓ At higher temperatures, reactant molecules have more kinetic energy, move faster, and collide more often and with greater energy. a. Collision Theory: When two chemicals react, their molecules have to collide with each other with sufficient energy for the reaction to take place. b. Kinetic Theory: ✓ Increasing temperature means the molecules move faster. 2. Concentrations of reactants Page 38 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 39 ✓ As the concentration of reactants increases, so does the likelihood that reactant molecules will collide. 3. Catalysts ✓ Speed up or slow down reactions by lowering activation energy ✓ Types of catalyst: a. Heterogeneous Catalyst: catalyst in a different phase from the reactants b. Homogeneous Catalyst: catalyst in the same phase as the reactants c. Transition metals as catalysts: provide an alternative route for reactants and products that have a lower activation enthalpy. 4. Surface area of a solid reactant ✓ Bread and butter theory: More area for reactants to be in contact 5. Pressure of gaseous reactants or products ✓ Increased number of collisions promotes faster rate of reaction. Collision Theory of Reaction Kinetics States that the rate of a chemical reaction is proportional to the number of collisions between reactant molecules The more often reactant molecules collide, the more often they react with one another, and the faster the reaction rete. Only a small fraction of the collisions are effective collisions resulting to chemical reactions. The number of particles possessing enough energy is dependent on the temperature of the reactants. Requirements for an effective collision (for a chemical reaction to occur): 1. The reactants must collide with each other. 2. The molecules must have sufficient energy to initiate the reaction 3. The molecules must have the proper orientation. Page 39 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 40 WEEK 6 ANSWER SHEET (Please submit only the answers. Do not return the entire module.) Name: ________________________________ Section: _______________________ LAST NAME, FIRST NAME MIDDLE INITIAL Mini-Performance Task No.3. Catalysts in a Reaction Conduct the activities. Write your observation and identify which factor affect the reaction. Procedure Observation Which factor affected the process? 1. Dissolve 1/2 teaspoon of instant coffee in hot water. Dissolve ½ teaspoon instant coffee in cold water. 2. Dilute vinegar in water to get three different concentrations and place in plastic transparent cups or glass. a. 3 tablespoons of vinegar b. 2 tablespoons of vinegar + 1 tablespoon of water c. 1 tablespoon of vinegar + 2 tablespoons of water Add 1 teaspoon of baking soda to each. 3. Shake an unopened bottled of carbonated soda. 4. Imagine cooking pork liempo sinigang. Why do you think you is it necessary to cut the pork liempo into smaller pieces? ___________________________________________________________________ SIGNATURE OVER PRINTED NAME OF PARENT/GUARDIAN DATE: ______________ Page 40 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 41 PRE-REQUISITE ASSESSMENT: Why do some people use papaya in cooking tinola? LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Mole concept, balancing chemical reactions PRE-REQUISITE SKILL: Calculating molar mass and moles present in a reaction TIME ALLOTMENT: 4 HRS CONSULTATION: For inquiries and clarifications regarding the lesson, you may contact your teacher thru his/her FB Messenger or thru email RUA: At the end of the lesson, you should be able to: Calculate the amount of substances used or produced in a chemical reaction Calculate percent yield of a reaction Determine the limiting reactant in a reaction and calculate the amount of product formed Recognize that energy is released or absorbed during a chemical reaction INSTITUTIONAL VALUES: Precision and Accuracy, Scientific Literacy, Excellence Students will be able to apply Precision and accuracy on determining limiting and excess reactants Scientific literacy on balancing chemical equations Excellence on calculating amount of product formed during a reaction STUDENT EXPERIENTIAL LEARNING A chemical reaction is the process by which substances bond together (or break bonds) and, in doing so, either release or consumes energy. Indications that a chemical reaction has occurred may be water is produced, formation of solids, formation of bubbles, color change, change in temperature, change in taste of consumable goods and change in odor. A chemical equation is the shorthand that scientist use to describe a chemical reaction. Reactants → Products H2 + O2 → H2O 2H2 + O2 → 2H2O Balancing Chemical Equations Chemical reactions are represented by chemical equations. The Law of Conservation of Mass is the principle followed in balancing the coefficients of chemical equations. Steps in balancing a chemical equation: Page 41 of 57  Colegio de Los Baños – PHYSICAL SCIENCE 42 The simplest and most generally useful method for balancing chemical equations is “inspection,” better known as trial and error. The following is an efficient approach to balancing a chemical equation using this method. Steps in Balancing a Chemical Equation 1. Identify the most complex substance. 2. Beginning with that substance, choose an element that appears in only one reactant and one product, if possible. Adjust the coefficients to obtain the same number of atoms of this element on both sides. 3. Balance polyatomic ions (if present) as a unit. 4. Balance the remaining atoms, usually ending with the least complex substance and using fractional coefficients if necessary. If a fractional coefficient has been used, multiply both sides of the equation by the denominator to obtain whole numbers for the coefficients. 5. Check your work by counting the numbers of atoms of each kind on both sides of the equation to be sure that the chemical equation is balanced. Example: To demonstrate this approach, let’s use the combustion of n-heptane as an example. 1. Identify the most complex substance. The most complex substance is the one with the largest number of different atoms, which is C7H16C7H16. We will assume initially that the final balanced chemical equation contains 1 molecule or formula unit of this substance. 2. Adjust the coefficients. Try to adjust the coefficients of the molecules on the other side of the equation to obtain the same numbers of atoms on both sides. Because one molecule of n-heptane contains 7 carbon atoms, we need 7 CO2 molecules, each of which contains 1 carbon atom, on the right side: C7H16+O2→7CO2+H2O 3. Balance polyatomic ions as a unit. There are no polyatomic ions to be considered in this reaction. 4. Balance the remaining atoms. Because one molecule of n-heptane contains 16 hydrogen atoms, we need 8 H2O molecules, each of which contains 2 hydrogen atoms, on the right side: C7H16+O2→7CO2+8H2O The carbon and hydrogen atoms are now balanced, but we have 22 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the oxygen atoms by adjusting the coefficient in front of the least complex substance, O2, on the reactant side: C7H16(l)+11O2(g)→7CO2(g)+8H2O(g) 5. Check your work. The equation is now balanced, and there are no fractional coefficients: there are 7 carbon atoms, 16 hydrogen atoms, and 22 oxygen atoms on each side. Always check to be sure that a chemical equation is balanced.The assumption that the final balanced chemical equation contains only one molecule or formula unit of the most complex substance is not always valid, but it is a good place to start. Consider, for example, a similar reaction, the combustion of isooctane (C8H18C8H18). Because the combustion of any hydrocarbon with oxygen produces carbon dioxide and water, the unbalanced chemical equation is as follows: C8H18(l)+O2(g)→CO2(g)+H2O(g) 1. Identify the most complex substance. Begin the balancing process by assuming that the final balanced chemical equation contains a single molecule of isooctane. 2. Adjust the coefficients. The first element that appears only once in the reactants is carbon: 8 carbon atoms in isooctane means that there must be 8 CO2 molecules in the products: C8H18+O2→8CO2+H2O  Colegio de Los Baños – PHYSICAL SCIENCE 43 3. Balance polyatomic ions as a unit. This step does not apply to this equation. 4. Balance the remaining atoms. Eighteen hydrogen atoms in isooctane means that there must be 9 H2O molecules in the products: C8H18+O2→8CO2+9H2O The carbon and hydrogen atoms are now balanced, but we have 25 oxygen atoms on the right side and only 2 oxygen atoms on the left. We can balance the least complex substance, O2, but because there are 2 oxygen atoms per O2 molecule, we must use a fractional coefficient (25/2) to balance the oxygen atoms: C8H18+25/2O2→8CO2+9H2O Equation 11 is now balanced, but we usually write equations with whole-number coefficients. We can eliminate the fractional coefficient by multiplying all coefficients on both sides of the chemical equation by 2: 2C8H18(l)+25O2(g)→16CO2(g)+18H2O(g) 5. Check your work. The balanced chemical equation has 16 carbon atoms, 36 hydrogen atoms, and 50 oxygen atoms on each side. Balancing equations requires some practice on your part as well as some common sense. If you find yourself using very large coefficients or if you have spent several minutes without success, go back and make sure that you have written the formulas of the reactants and products correctly. Mass relationships in chemical reactions A properly balanced chemical equation shows the molar ratios of each of the species present, whether they are reactants or products EX: the combustion of propane as an example: C3H8 +5O2 → 3CO2 + 4H2O The ratios found in this equation are as follows: 1 mol propane : 5 mol oxygen 1 mol propane : 3 mol carbon dioxide 1 mol propane : 4 mol water 5 mol oxygen : 3 mol carbon dioxide 5 mol oxygen : 4 mol water 3 mol carbon dioxide : 4 mol water Theoretical and Actual Yield The theoretical yield is the amount of a given product one would expect would be produced based solely on the molar ratios and the amount of each starting material. The actual yield is a measured quantity and will either be given in problems or it will be the amount you need to find. Mass to Mass Calculations IF given the reaction: A → B This is the typical path to determine the mass of B, when mass of A is given grams A → moles A → moles B → grams B BCA Method Before Change After Note: This method isn’t about being “easier” or more simple. Rather, it is about having a better conceptual understanding of the process of stoichiometry calculations. BCA Tables: The Process Step 1: Write the balanced equation.  Colegio de Los Baños – PHYSICAL SCIENCE 44 Step 2: Make sure you have moles for your starting value. (Convert from grams to moles using the molar mass if needed.) Step 3: Insert the starting moles into the BCA Table and complete the “B” Row. If no starting amount of reactant is specified in the question, students write “XS” to show there is more than enough reactant for the reaction to proceed to completion. Before the reaction has started, no products have formed yet so students write in zero. Step 4: Calculate the changes necessary based on the mole ratio of the balanced equation. (Complete the “C” Row.) The “change” line is where the molar ratio from the balanced chemical reaction comes in. Use the coefficients from the balanced equation to fill in the rest of the change line. Reactants are consumed that is why the sign is (-). Products are formed that is why the sign is (+) Step 5: Calculate the “A” Row. To fill in the “after” line, the student performs some simple arithmetic. Step 6: Convert any values from the “A” Row into grams (if needed) by using the molar mass. Examples: 1. Dihydrogen sulfide gas, which smells like rotten eggs, burns in air to produce sulfur dioxide and water. How many moles of oxygen gas would be needed to completely burn 2.4 moles of hydrogen sulfide? How many moles of each product would be produced? Step 1: Write the balanced equation. 2 H2S (g) + 3 O2 (g) → 2 SO2 (g) + 2 H2O (g) Step 2: Make sure you have moles for your starting value. (Convert from grams to moles if needed.) The given is already in moles so no need to convert. Step 3: Insert the starting moles into the BCA Table and complete the “B” Row. Step 4: Calculate the changes necessary based on the mole ratio of the balanced equation. (Complete the “C” Row.) Step 5: Calculate the “A” Row. 2 H2S 3 O2 2 SO2 2 H2O B 2.4 𝑚𝑜𝑙 XS 0 0 C −2.4 𝑚𝑜𝑙 3 𝑚𝑜𝑙𝑒𝑠 𝑂2 2 𝑚𝑜𝑙𝑒𝑠 𝑆𝑂2 2 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑂 −2.4 𝑚𝑜𝑙 𝐻2 𝑆 ( ) 2.4 𝑚𝑜𝑙 𝐻2 𝑆 ( ) 2.4 𝑚𝑜𝑙 𝐻2 𝑆 ( ) 2 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆 2 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆 2 𝑚𝑜𝑙𝑒𝑠 𝐻2 𝑆 = −3.6 𝑚𝑜𝑙 A 0 XS 2.4 𝑚𝑜𝑙 2.4 𝑚𝑜𝑙 How many moles of each product would be produced? Answer: 2.4 𝑚𝑜𝑙 SO2, 2.4 𝑚𝑜𝑙 H2o 2. A chemist has 23.5 g of copper (II) chloride and lots of aluminum foil. How many grams of each product can the chemist produce by reacting the copper (II) chloride with the aluminum foil? Step 1: Write the balanced equation. 2 Al + 3 CuCl2 → 2 AlCl3 + 3 Cu Step 2: Make sure you have moles for your starting value. (Convert from grams to moles using the molar mass.) 23.5 grams CuCl2 → moles Molar Mass of CuCl2 Cu: (1)(63.55)=63.55 1 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 Cl: (2)(35.45)=70.90 23.5𝑔 𝐶𝑢𝐶𝑙2 ( ) = 0.1748 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 134.45𝑔 𝐶𝑢𝐶𝑙2 134.45 g/mol Step 3: Insert the starting moles into the BCA Table and complete the “B” Row. Step 4: Calculate the changes necessary based on the mole ratio of the balanced equation. (Complete the “C” Row.) Step 5: Calculate the “A” Row. 2 Al 3 𝐶𝑢𝐶𝑙2 2 AlCl3 3Cu B XS 0.1748 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 0 0 C 2 𝑚𝑜𝑙 𝐴𝑙 -0.1748 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 2 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3 3𝑚𝑜𝑙 𝐶𝑢 −0.1748𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 ( ) +0.1748 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 ( ) +0.1748 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 ( ) 3 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 3 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 3 𝑚𝑜𝑙 𝐶𝑢𝐶𝑙2 = −0.1165 A XS 0 0.1165𝑚𝑜𝑙 0.1748 𝑚𝑜𝑙 Step 6: Convert any values from the “A” Row into grams (if needed) by using the molar mass. 0.1165 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3 → 𝑔 𝐴𝑙𝐶𝑙3 Molar Mass of CuCl2 133.33𝑔 𝐴𝑙𝐶𝑙3 0.1165 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3 ( ) 1 𝑚𝑜𝑙 𝐴𝑙𝐶𝑙3 Al: (1)(26.98)=26.98 = 𝟏𝟓. 𝟓𝟑𝒈 𝑨𝒍𝑪𝒍𝟑 Cl: (3)(35.45)=106.35 133.33 g/mol 0.1748 𝑚𝑜𝑙 𝐶𝑢→ 𝑔𝐶𝑢 Atomic Mass of Cu: 63.55 g/mol 63.55𝑔 𝐶𝑢 0.1748𝑚𝑜𝑙 𝐶𝑢 ( ) = 𝟏𝟏. 𝟏𝟏𝒈 𝑪𝒖 1 𝑚𝑜𝑙 𝐶𝑢 Limiting and Excess Reagents The limiting reactant is the one that will be used up before any of the others An excess reactant is one that is not completely used up when the reaction is complete For limiting excess reactant problems, the amount of the two reactants are given initially. Because of this, the moles of the limiting reactant are what determine the theoretical yield of a reaction To find out which reactant is the limiting reactant you use the molar ratios given in the balanced chemical equation and examine the lesser amount of reactant in moles. Stoichiometric Calculations - Limiting Reactants and Percentage Yield "If one reactant is entirely used up before any of the other reactants, then that reactant limits the maximum yield of the product." Problems of this type are done in exactly the same way as the previous examples, except that a decision is made before the ratio comparison is done. The decision that is made is "What reactant is there the least of?" Example Problem #1 Methane, CH4, burns in oxygen to give carbon dioxide and water according to the following equation: CH4 + 2 O2 ------> CO2 + 2 H2O In one experiment, a mixture of 0.250 mol of methane was burned in 1.25 mol of oxygen in a sealed steel vessel. Find the limiting reactant, if any, and calculate the theoretical yield, (in moles) of water.  1 Colegio de Los Baños – PHYSICAL SCIENCE Solution: In any limiting reactant question, the decision can be stated in two ways. Do it once to get an answer, then do it again the second way to get a confirmation. According to the equation: 1 mol CH4 = 2 mol O2 If we use up all the methane then: 1 mol CH4 = 2 mol O2 0.25 mol x x = 0.50 mol of O2 would be needed. We have 1.25 mol of O2 on hand. Therefore we have 0.75 mol of O2 in excess of what we need. If the oxygen in is excess, then the methane is the limiting reactant. Confirmation: If we use up all the oxygen then 1 mol CH4 = 2 mol O2 x 1.25 mol x = 0.625 mol of methane. We don't have 0.625 moles of methane. We have only 0.25 moles. Therefore the methane will be used up before all the oxygen is. Again the methane is the limiting reactant. We now use the limiting reactant to make the mole comparison across the bridge to find the amount of water produced. 1 mol CH4 = 2 H2O 0.25 mol x x = 0.50 mol of H2O would be produced. Finish off with a statement: When 0.25 mole of methane and 1.25 mole of oxygen are mixed and reacted according to the equation, the methane is the limiting reactant and the maximum yield of water will be 0.50 moles. Example Problem #2 Chloroform, CHCl3, reacts with chlorine, Cl2, to form carbon tetrachloride, CCl4, and hydrogen chloride, HCl. In an experiment 25 grams of chloroform and 25 grams of chlorine were mixed. Which is the limiting reactant? What is the maximum yield of CCl4 in moles and in grams? Solution: Start with the equation: CHCl3 + Cl2 -------> CCl4 + HCl Did you check to see if it was balanced? Calculate the molecular masses of the species needed in the problem. CHCl3 = 1 C = 1(12.01) = 12.01 Cl2 = 2 (35.45) = 70.90 g/mol 1 H = 1(1.01) = 1.01 H2O = 2 H = 2 (1.01) = 2.02 3 Cl = 3(35.45) = 106.35 1 O = 1 (16.00) = 16.00 119.37 g/mol 18.02 g/mol Then calculate the moles of each of the reactants to be used. moles of CHCl3 = g = 25.00 g = 0.21 moles of CHCl3 are present. mm 119.37 g/mol  2 Colegio de Los Baños – PHYSICAL SCIENCE moles of Cl2 = g = 25.00 g = 0.35 moles of chlorine are present. mm 70.90 g/mol Decision time. Which of the two reactants do you have the least of? From the balanced equation you can see that the chloroform and chlorine reactant in a one to one ratio. If we use all the chloroform then we get the following equation. 1 CHCl3 = 1 Cl2 0.21 mol x x = 0.21 moles of chlorine are needed. We need 0.21 moles of chlorine. We have 0.35 moles of chlorine. Therefore chlorine is in excess. The chloroform must be the limiting reactant. Confirmation: IF we use all the chlorine then: 1 CHCl3 = 1 Cl2 x 0.35 mol x = 0.35 moles of chloroform are needed. If we use all the chlorine then we need 0.35 moles of chloroform. We have only 0.21 moles of chloroform. It is the reactant that we will run out of first. Therefore it is the limiting reactant. Use the limiting reactant to cross the ratio bridge and find the number of moles of water made. 1 CHCl3 = 2 H2O 0.21 mol x x = 0.42 moles of H2O will be made. Calculate the grams of water produced. grams = moles * molecular mass = 0.42 mol * 18.02 g/mol = 7.57 grams of water  3 Colegio de Los Baños – PHYSICAL SCIENCE WEEK 7 ANSWER SHEET (Please submit only the answers. Do not return the entire module.) Name: ________________________________ Section: _______________________ LAST NAME, FIRST NAME MIDDLE INITIAL ENGAGEMENT Enabling Assessment Activity No.4. Limiting Reactants Solve the following problems while following the steps done in the previous example 1. KOH + HNO3 → KNO3 + H2O Mass of reactants: 10g 10g Molar Mass Number of Moles Mass of Products Which is the limiting reactant? Which is the excess reactant? How much of KNO3 is formed in grams? How much of excess reactant remained in grams? 2. CH4 + 2 O2 → CO2 + 2H2O Given: 5g 10g Molar Mass Number of Moles Mass of Products Which is the limiting reactant? Which is the excess reactant? How much of CO2 is formed in grams? How much of excess reactant remained in grams? ASSIMILATION Answer in 3-5 sentences. What is the importance of balancing chemical equations? (10pts) ___________________________________________________________________ SIGNATURE OVER PRINTED NAME OF PARENT/GUARDIAN  4 Colegio de Los Baños – PHYSICAL SCIENCE PRE-REQUISITE ASSESSMENT: Why is it important to balance the equation first before calculation? LEARNING MATERIALS: Module, pen, paper, old chemistry books, scientific calculator, internet (if applicable) PRE-REQUISITE CONTENT KNOWLEDGE: Mole concept, balancing chemical reactions PRE-REQUISITE SKILL: Calculating molar mass and moles present