Science Reviewer PDF AY 2024-2025

SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 B2 INORGANIC CHEMISTRY T1.1 INTRODUCTION TO CHEMISTRY Deals with the study of inorganic...

SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 B2 INORGANIC CHEMISTRY T1.1 INTRODUCTION TO CHEMISTRY Deals with the study of inorganic compounds. Inorganic compounds are TOPIC OVERVIEW compounds which do NOT contain carbon-hydrogen bond. A. INTRODUCTION TO CHEMISTRY a. Chemistry b. Branches of Chemistry B3 BIOCHEMISTRY c. Chemistry as the Central Science Deals with the study of chemical processes inside the biological system. A CHEMISTRY B4 ANALYTICAL CHEMISTRY Chemistry deals with properties, composition, structure and changes Deals with qualitative and quantitative that material undergo. methods to analyze properties of matter B5 PHYSICAL CHEMISTRY Is concerned with the application of the techniques and theories of physics to the study of chemical systems. C CHEMISTRY AS THE CENTRAL SCIENCE B BRANCHES OF CHEMISTRY - Organic Chemistry - Inorganic Chemistry - Biochemistry - Analytical Chemistry - Physical Chemistry B1 ORGANIC CHEMISTRY T1.2 PROPERTIES OF MATTER A branch of chemistry which deals with the study of carbon compounds. Carbon is capable of TOPIC OVERVIEW forming long C - C chains (called A. PROPERTIES OF MATTER catenation) a. Physical Properties PAGE 1 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 b. Chemical Properties B. CHANGES IN MATTER a. Physical Changes b. Chemical Changes ○ A PROPERTIES OF MATTER A3 CHEMICAL PROPERTIES Characteristics that allow us to Descriptions of the ability of a substance recognize matter and distinguish to undergo to a chemical change them from the other substances. ○ EXAMPLES: Flammability, Corrosiveness, Acidity, Toxicity, Chemical Activity, Tarnish Matter Anything that occupies space and has mass. B CHANGES IN MATTER Mass represents the amount of matter. Forms of Energy are NOT Matter Fire B1 PHYSICAL CHANGES Sun Sound A physical change is a process that changes the physical properties of a A1 PHYSICAL PROPERTIES substance without changing its chemical composition. Changes in physical state are physical changes. Physical Property A physical property is a characteristic that we can observe and measure without changing the composition of a substance KINDS OF PHYSICAL PROPERTIES INTENSIVE PROPERTIES - Does NOT depend on the amount of substance being examined B2 CHEMICAL CHANGES A chemical change is a process where one or more substances are converted into one or more new substances (Also called a chemical reaction) ○ EXTENSIVE PROPERTIES - DEPENDS on the amount of substance. PAGE 2 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 PROPERTIES OF THE THREE STATES Fixed Shape Shape SOLID Shape not set by remains container rigid ○ No fixed Takes Can be shape shape of poured LIQUID filled portion of the container ○ No fixed Takes Fills shape shape of container T1.3 STATES OF MATTER GAS the container TOPIC OVERVIEW B IN MOLECULAR LEVEL A. MATTER MATTERS a. Observable Properties of the Three States B. IN MOLECULAR LEVEL B1 PARTICLES a. Kinetic Molecular Theory (KMT) b. Observable Properties of the A particle is an extremely tiny piece of Three States in Molecular Level matter, and scientists believe that C. TRANSFORMATION OF THE everything in the universe is made up of DIFFERENT STATES particles a. Melting Particles can range in size, from atoms or b. Freezing molecules, or to much smaller c. Evaporation submicroscopic particles like electrons. d. Condensation e. Sublimation f. Deposition D. LESSON SYNTHESIS A MATTER MATTERS MATTER MATTERS Can be classified based on physical state and composition B2 KINETIC MOLECULAR THEORY (KMT) A1 OBSERVABLE PROPERTIES OF THE Helps us interpret the microscopic THREE STATES properties of solids, liquids and gases. According to the theory, all matter consists of extremely tiny particles, which PAGE 3 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 are in constant motion. The energy of the moving object is called kinetic energy (KE) ○ If the speed of the object increases, its KE increases. ○ If the speed of the object decreases, its KE decreases. B3 Observable Properties of the Three States in Molecular Level C TRANSFORMATION OF THE DIFFERENT STATES PROPERTIES OF THE THREE STATES IN MOLECULAR LEVEL WATER EXISTS IN THREE DIFFERENT STATES Little Particles Little or The amount of energy affects the space fixed in no arrangement and rate of movement of between place but volume atoms or molecules in matter, which SOLID particles vibrate change leads to the different phase changes. around a under fixed moderat position e pressure Some Particles Can be free move compress LIQUID space past one ed lightly between another by particles moderat e pressure C1 MELTING (FUSION) - SOLID TO LIQUID Particles Particles Compres When solid substances such as ice cubes, are move sed metals, or wax are supplied with enough GAS widely through under amount of heat, they reach their melting separate space moderat point and are changed into liquids. d with e much pressure free C2 FREEZING - LIQUID TO SOLID space PAGE 4 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 When liquid releases heat, the molecules begin to move at a lower pace. The space between the molecules decreases, causing the molecules to become compact. C3 EVAPORATION - LIQUID TO GAS When a liquid absorbs enough heat, it reaches its boiling point and changes into a gas. C4 CONDENSATION - GAS TO LIQUID T1.4 CLASSIFICATION OF MATTER BASED ON COMPOSITION TOPIC OVERVIEW A. CLASSIFICATION OF MATTER a. Mattter is composed of Atoms C5 SUBLIMATION - SOLID TO GAS b. Atoms can combine to Molecules c. Classifying matter according to The process in which a substance is composition transformed from the solid phase directly d. Pure Substance VS Mixture to the gas state without passing the B. PURE SUBSTANCES liquid phase. a. Pure substances can be Classified into Elements and Compounds C6 DEPOSITION - GAS TO SOLID b. Elements i. Elements are organized The opposite of sublimation where a in the periodic table substance is transformed from the gas ii. How to represent the phase directly to the solid state without particulate nature of passing the liquid phase. Elements c. Compounds i. How to represent the D LESSON SYNTHESIS particulate nature of compounds C. MIXTURES a. Heterogeneous Mixtures b. Homogeneous Mixtures c. Representations of the particulate nature of mixture PAGE 5 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 Chemical and physical properties may D. LESSON SYNTHESIS vary.. ○ Examples: oil and water, sugar solution. A CLASSIFICATION OF MATTER B PURE SUBSTANCES SCANNING TUNNEL MICROSCOPE STM is a powerful instrument that The aluminum in soda can is pure if it is allows one to image the sample not combined with any other substances surface at the atomic level. such as the plastic and paint coating. The sand is NOT a pure substance but if A1 MATTER IS COMPOSED OF ATOMS we removed all the dirt, minerals, and other contaminants, it would be the pure An atom is the smallest unit of matter substance, silica. that retains its chemical properties B1 PURE SUBSTANCES CAN BE A2 ATOMS CAN COMBINE TO FORM CLASSIFIED INTO ELEMENTS AND MOLECULES COMPOUNDS Molecules are composed of two or more Element atoms bound together in a discrete It is composed of only one type of atom. arrangement. It is a substance that cannot be broken The atoms bound together in a down into simpler substances even by molecule can be from the same element chemical reaction. or from different elements. Compound A compound is a substance composed of two or more elements combinds in A3 CLASSIFYING MATTER ACCORDING TO definite proportions. COMPOSITION It can be broken down into smaller compounds, or their basic elements, by Pure Substances Have uniform (the chemical change. same) chemical composition throughout and from sample to sample B2 ELEMENTS A4 PURE SUBSTANCE VS MIXTURE ELEMENTS ARE ORGANIZED IN THE PERIODIC TABLE… Pure substances Chemical symbol identifies an Made of only one matter; thus the element. composition is the same throughout. A chemical symbol is one or two-letter Chemical and physical properties are designation of an element. The fire constant letter of a symbol is always capitalized. ○ Examples: pure water, hydrogen If the symbol contains two letters, the gas, gold. second letter is lowercase. Mixture All elements (118) are organized in the Made up of several substances that are periodic table. not chemically bonded. PAGE 6 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 C MIXTURES HOW DOES THE COMPOSITION OF A MIXTURE DIFFER FROM A SUBSTANCE? While pure substances have the uniform (the same) chemical composition throughout, Mixtures are How to represent the particulate nature of composed of two or more substances elements? and may or may not have the same uniform composition Atoms of molecules are represented by single spheres. Molecules of elements are represented by two or more spheres of the same size and color joined together. Mixture A mixture is a combination of two or more elements or compounds Mixtures differ from pure compounds in B3 COMPOUNDS that their components can be separated by physical processes. Chemical formula describes the Type of Mixtures composition of a compound. Heterogeneous Mixtures Homogeneous Mixtures C1 HETEROGENEOUS MIXTURES How to represent the particulate nature of Compounds? Molecules of compounds are represented by two or more spheres of different sizes and colors joined together. C2 HOMOGENEOUS MIXTURES PAGE 7 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 h. Fractional Distillation i. Chromatography j. Centrifugation A TECHNIQUES IN SEPARATING COMPONENTS C3 REPRESENTATION OF THE PARTICULATE NATURE OF MIXTURE A1 SIFTING OR SIEVING Used to separate a dry mixture which contains substances of different sizes by passing it through a sieve, a device containing tiny holes. A2 USING MAGNETS It is a process which uses magnets to D LESSON SYNTHESIS obtain the substances with magnetic properties from a mixture A3 DECANTATION It is a process of separating a solid from a liquid, by pouring off the liquid, after allowing the solid to settle in the liquid. A4 FILTRATION It is a process of separating a solid from a T1.5 SEPARATION OF MIXTURES liquid by pouring the mixture through a porous material such as filter paper TOPIC OVERVIEW A5 EVAPORATION A. TECHNIQUES IN SEPARATING COMPONENTS OF MIXTURES a. Sifting or Sieving It is a process in which a liquid is b. Using Magnets converted to vapor leaving the insoluble c. Decantation solids in the dish d. Filtration e. Evaporation A6 SUBLIMATION f. Sublimation g. Simple Distillation PAGE 8 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 It is a process of purifying solids by turning into vapor its volatile components PHASE CHANGE DIAGRAM A phase change diagram is a graphical A7 SIMPLE DISTILLATION representation of the correlation between heat input and the temperature of a substance. It can be It is a process of separating two miscible used to determine the melting point liquids because of the differences in the and the boiling point of a substance. boiling point. A8 FRACTIONAL DISTILLATION A1 HEATING CURVE It is a process of separating the component of a liquid one by one while it boils. A more complex process of distillation. A9 CHROMATOGRAPHY It is a process of separating the component colors of a mixture by dipping a medium into a filter paper soaked in a solvent. Point A to B = Solid Point B to C = Equilibrium of a Solid and A10 CENTRIFUGATION Liquid (Melting) Point C to D = Liquid It is a technique that, by rotating the Point D to E = Equilibrium of a Liquid sample at high speeds, separate solids and Gas (Vaporizing) suspended in a liquid (or liquids of Point E to F = Gas different densities) A2 COOLING CURVE T1.5.1 PHASE CHANGE DIAGRAMS TOPIC OVERVIEW A. PHASE CHANGE DIAGRAM a. Heating Curve b. Cooling Curve c. Phase Change Diagram Changed A PHASE CHANGE DIAGRAM PAGE 9 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 Point A to B = Gas A1 TYPES OF SOLUTION Point B to C = Equilibrium of a Gas and Liquid (Condensation) Point C to D = Liquid Point D to E = Equilibrium of a Liquid and Solid (Freezing) Point E to F = Solid A3 PHASE CHANGE DIAGRAM COMBINED A2 SOLUBLE/INSOLUBLE We say soluble if the solid solute completely dissolved in the liquid solvent and insoluble if it is not. A2 MISCIBLE/IMMISCIBLE T1.6 SOLUTIONS If the components are mixed, they are miscible. If they cannot be mixed, they TOPIC OVERVIEW are Immiscible A. SOLUTION a. Types of Solutions A2 ELECTROLYTE AND b. Soluble/Insoluble NONELECTROLYTE c. Miscible/Immiscible d. Electrolyte and Nonelectrolyte e. Concentration A solute that dissolves completely (into f. Solubility ions) in an aqueous solution is a strong B. FACTORS THAT AFFECT SOLUBILITY electrolyte. A nonelectrolyte does not dissociate into ions. A SOLUTION Homogeneous mixture. A2 CONCENTRATION Solute - Substance being dissolved, usually present in the larger amount. The relative amounts of solute and Solvent - Substance doing the dissolving, solvent. usually present in the larger amount. Aqueous Solution - A solution where the solvent is water. A2 SOLUBILITY PAGE 10 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 It is defined as the amount of solute that can be dissolved in a given amount of Saturated Solution solvent at a given temperature. Contains the maximum amount of solute in a given amount of solvent (no more solute dissolves) A FACTORS THAT AFFECT SOLUBILITY Supersaturated Solution Structure Contains more solute than a saturated “Like dissolves like” When the polarity of solution (becomes unstable, crystals the solvent is similar to the polarity of the form) solute, the solvent will dissolve the solute. Temperature A USING THE SOLUBILITY CURVE OF The solubility of most solids increases in SALT AND SUGAR water as the temperature of the solution increases. But gasses are less soluble as temperature increases. SOLUBILITY CURVE Pressure It is a graphic representation of the The solubility of gasses (but not of liquids variation with changing temperature or solids) is directly proportional to the of the solubility of a given substance in pressure of the gas above the liquid. a given solvent. T1.7 TYPES OF SOLUTIONS TOPIC OVERVIEW A. TYPES OF SOLUTION BASED ON THE AMOUNT OF SOLUTE B. USING THE SOLUBILITY CURVE OF SALT AND SUGAR a. Sample 1 & Answer b. Sample 2 & Answer c. Sample 3 & Answer C. USING THE SOLUBILITY CURVE TO IDENTIFY THE TYPE OF SOLUTION a. Sample 1 & Answer b. Sample 2 & Answer c. Sample 3 & Answer d. Sample 4 & Answer e. Sample 5 & Answer A TYPES OF SOLUTION BASED ON THE AMOUNT OF SOLUTE Unsaturated Solution Contains less solute than a saturated solution (more solute dissolves) PAGE 11 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 A1 SAMPLE 1 & ANSWER What is the mass of the sugar that can be dissolved in 100 mL of water at 20C? ○ 200g / 100mL A2 SAMPLE 2 & ANSWER What is the mass of sugar that can be dissolved in 200 mL of water at 90C ○ 410 / 100 = x / 200 ○ 410 x 200 / 100 B1 SAMPLE 1 & ANSWER ○ 820g / 200 CLASSIFY BASED ON THE TYPE OF SOLUTION. A3 SAMPLE 3 & ANSWER A solution that contains 60 g of 𝑁𝐻4𝐶𝐼 at 50C (100mL of H2O) What is the mass of sugar that can be dissolved in 50mL of water at 50C ○ 260 / 100 = x / 50 ○ 260 x 50 / 100 ○ 130g / 50ml B USING THE SOLUBILITY CURVE TO IDENTIFY THE TYPE OF SOLUTION USING THE SOLUBILITY CURVE TO IDENTIFY THE TYPE OF SOLUTION If the coordinate falls exactly in the saturation curve, the solution is ○ saturated. ○ Supersaturated Solution If the coordinate falls above the saturation curve, the solution is said to be supersaturated. Meaning, some B2 SAMPLE 2 & ANSWER solutes will be undissolved. If the coordinate falls below the CLASSIFY BASED ON THE TYPE OF SOLUTION. saturation curve, the solution is said to A solution that contains 70g of 𝐾𝑁𝑂3 at be unsaturated. Meaning, more solute can still be dissolved. 30o C (100mL of 𝐻2𝑂). PAGE 12 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 A solution that contains 35g of 𝐾𝐶𝐼𝑂3 at 30oC (200mL of 𝐻2𝑂). Supersaturated Solution Supersaturated Solution B3 SAMPLE 3 & ANSWER B5 SAMPLE 5 & ANSWER CLASSIFY BASED ON THE TYPE OF SOLUTION. A solution that contains 90g of 𝑁𝐴𝑁𝑂3 at CLASSIFY BASED ON THE TYPE OF SOLUTION. 40oC (100mL of 𝐻2𝑂). A solution that contains 2.5g of 𝑁𝐻3 at (0oC (25mL of 𝐻2𝑂). Unsaturated Solution B4 SAMPLE 4 & ANSWER Unsaturated Solution CLASSIFY BASED ON THE TYPE OF SOLUTION. T1.8 UNITS OF CONCENTRATION PAGE 13 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 a. What is the percent by mass TOPIC OVERVIEW concentration of NaCl in this solution? b. What mass of NaCl is present in 10.0g of UNITS OF CONCENTRATION this solution? D. PERCENT BY MASS Solution for A: a. Sample 1 & Answer 25.0g (mass solute) + 125.0g (mass b. Sample 2 & Answer solvent) = 150g (mass solution) E. PERCENT BY VOLUME % NaCl = 25.0/125.0g x 100 a. Sample 1 & Answer % NaCl = 16.67% b. Sample 2 & Answer ➔ Note: Always ROUND OFF to 2 decimal F. MASS/VOLUME PERCENT places, even if whole number. a. Sample 1 & Answer Solution for B: b. Sample 2 & Answer Mass solute = x G. PARTS PER MILLION 16.67% = x/10.0g x 100 a. Sample 1 & Answer (16.67)(10.0g)/100 = x (cross multiply) b. Sample 2 & Answer x = 1.67g H. PARTS PER BILLION a. Sample 1 & Answer b. Sample 2 & Answer A2 SAMPLE 2 & ANSWER CONCENTRATION Problem: What is the mass percentage of Fe in Relative amounts of solute and solvent a piece of metal with 87.9g of Fe in a 113g that make up the solution sample? UNITS OF CONCENTRATION ➔ Note: If given says “SAMPLE,” we assume Gives the precise quantitative it is the SOLUTION. measurements of the solution which Solution: depends on the purpose of the % Fe = 87.9g/113g x 100 solution and the methods of % Fe=77.79% preparation. B PERCENT BY VOLUME A PERCENT BY MASS IS most often used when liquids are dissolved in liquids. It is calculated by dividing the mass It is calculated by dividing the solute solute to the mass of the solution and volume to the volume of the solution and multiplying by 100%. multiplying by 100%. Mass of solution can be determined by adding the mass of the solute and the mass of the solvent. B1 SAMPLE 1 & ANSWER A1 SAMPLE 1 & ANSWER To prepare a solution, we dissolve 12.5 mL ethanol in sufficient water to give a total volume of 85.4 mL. What is the perfect Problem: Consider a solution prepared by by volume concentration of the ethanol? adding 25.0g NaCl to 125.0g of water Solution: PAGE 14 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 5 ethanol = 12.5 / 85.4 x 100 1.7% = x / 2000mL x 100 % ethanol: 14.64 (1.7%)(2000mL) / 100 = 100x / 100 x = 34.00 B2 SAMPLE 2 & ANSWER D PARTS PER MILLION (ppm) Rubbing alcohol is commonly used as an antiseptic for small cuts. It is sold as 70% Use for very dilute solutions in which the (v/v) solution off isopropyl alcohol in concentration is too small to express water. What volume of isopropyl alcohol conveniently as percentage. is used to make 320 mL of rubbing Parts per million is calculated by dividing alcohol? the mass of solute by the mass of 6 Solution: solution and multiplying by 1 million (10 ) 70% = x / 320mL x 100 (70%)(320mL) / 100 = 100x / 100 X = 224.00 mL C MASS / VOLUME PERCENT D1 SAMPLE 1 & ANSWER It is calculated by dividing the soute If there is 0.0551 g of As in 346 of solution, mass by the volume of solution and what is the As concentration in ppm? multiplying it by 100% Solution 6 ppm As = 0.0551g / 348g x 10 ppm As = 158.33 D2 SAMPLE 2 & ANSWER C1 SAMPLE 1 & ANSWER If there is 1.5 mg of 𝑆𝑂2 in 0.25 L of What is the concentration in percent mass/volume of 150 mL of solution solution, what is the sulfur dioxide containing 25 g dissolved medicine concentration in ppm? What is the tablet? potential effect of this concentration of Solution: one’s health? % tablet = 25.0g / 150mL x 100 Solution % tablet = 16.67 ppm 𝑆𝑂2 = 1.5 mg / 0.25L ppm 𝑆𝑂2 = 6.00 C2 SAMPLE 2 & ANSWER E PARTS PER BILLION (ppb) Many people use a solution of 𝑁𝐴3 𝑃𝑂4 to Use for very dilute solutions in which the clean walls before putting up wallpaper. concentration is too small to express The recommended concentration is 1.7% conveniently as percentage. (m/v). What mass of 𝑁𝐴3 𝑃𝑂4 is needed to Perts per billion is calculated by dividing the mass of solute by the mass of make 2.0 L of solution? Solution: 2.0L x 1000 = 2000mL PAGE 15 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK  SCIENCE AY 2024 - 2025 1ST QUARTER – SIR MANUBAY– DAY 1, 2, 3 9 solution and multiplying by 1 billion (10 ) E1 SAMPLE 1 & ANSWER A sample of water river was analyzed, and it was determined that there was 0.000012 g of Pb in 1000.0 g river water. What is the concentration of Pb, reported in units of ppb? Solution 9 ppb Pb = 0.000012 / 1000.00 x 10 ppb Pb = 12.00 ppb E2 SAMPLE 2 & ANSWER A fungus that grows on peanuts produces a deadly toxin. When ingested in large amounts, this toxin destroys the liver and can cause cancer. Any shipments of peanuts that contains more that 25 ppb of this dangerous fungus is ejected. A company receives 20 tons of peanuts to make peanut butter. What is the maximum amount (in g) of fungus that is allowed? Solution: 7 20 tons to grams = 1.814𝑥10 g 7 9 25 ppb = x / 1.814𝑥10 g x 10 7 9 (25 ppb)(1.814𝑥10 ) / 10 = x x = 0.45g PAGE 16 JULIAN PORTO – 9 - ST. CLAUDIA – JULIAN AUGUSTINE PORTO ON FACEBOOK