EoT3 Chemistry Grade 10 Advanced Revision 2023-2024 PDF

EoT_Term 3_ 2023-2024 Chemistry ‫الكيمياء‬ EoT3-Coverage Grade 10 Advanced 3‫مراجعة الهيكل الفصل‬ ‫ متقدم‬10 ‫الصف‬ By Mr. Mouad 0557903129 ...

EoT_Term 3_ 2023-2024 Chemistry ‫الكيمياء‬ EoT3-Coverage Grade 10 Advanced 3‫مراجعة الهيكل الفصل‬ ‫ متقدم‬10 ‫الصف‬ By Mr. Mouad 0557903129  Metallic Bonds In a metallic lattice, the outer energy levels of the metal atoms overlap. The electron sea model proposes that all metal atoms in a metallic solid contribute their valence electrons to form a “sea” of electrons that surrounds the metal cations in the lattice.  Metallic Bonds The electrons in the outer energy levels of bonding metallic atoms are called delocalized electrons. They can move from one atom to the next. When an atom’s outer electrons move freely throughout the solid, a metallic cation is formed. A metallic bond is the attraction of a metallic cation for delocalized electrons.  The physical properties of metals at the bulk scale can be explained by metallic bonding. These properties provide evidence of the strength of metallic bonds. Examples of physical properties of metals: — melting and boiling points — thermal and electrical conductivity — malleability, ductility, and durability — hardness and strength  Metallic Bonds In general, metals have moderately high melting points and high boiling points, as shown below. The melting points of metals are less extreme than the boiling points because the cations and electrons are mobile in a metal. It does not take much energy for them to move past each other. During boiling, atoms must completely separate from the lattice. This requires more energy.  Metallic Bonds The movement of mobile electrons around positive metallic cations makes metals good conductors. Mobile electrons move heat from one place to another quickly. Mobile electrons easily move as part of an electric current when an electric potential is applied to a metal.  Metallic Bonds Metals are malleable (can be hammered into sheets). Metals are ductile (can be drawn into wire). The figure below shows how an applied force causes metal ions to move through delocalized electrons, making metals malleable and ductile.  Metallic Bonds Metals are durable. Metallic cations are strongly attracted to electrons around them and not easily removed from the metal. As the number of delocalized electrons increases, so does hardness and strength.  Metal Alloys An alloy is a mixture of elements that has metallic properties. Examples: stainless steel, brass, cast iron The properties of alloys generally differ from the properties of the elements that they contain. Example: Steel is iron mixed with at least one other element. Some properties of iron are present, but steel is stronger than iron. Metal Alloys  Metal Alloys Alloys are classified into two basic types. Substitutional alloy: Some of the atoms in the original metal are replaced by other metals of similar atomic size. Example: sterling silver Interstitial alloy: This alloy is formed when the small holes in a metallic crystal are filled with smaller atoms. Example: carbon steel Covalent Bond a bond that is formed when atoms share their electrons. Electronegativity (EN) is a measure of the ability of an atom to attract electrons in a covalent bond. Bond Character: refers to the type of the bond: Ionic, polar covalent, or non-polar covalent. Polar: Has two different sides (example: “N” pole and - “S” pole in a magnet. Or “+” & “ ” in a battery) Non-Polar: Has no side (No different charges in the sides of the molecule) Electronegativity and Bond Character  REVISION 1. What happens to electronegativity across the periodic table: 1. “Up to Down” Decreases 2. “Left to Right” Increases 2. Which one has more electronegativity: C or F? what is the electronegativity C: 2.55 difference? F: 3.98 Electronegativity difference: 3.98-2.55 = 1.43 15  Electronegativity and Bond Character Bond Character The table lists the character and type of chemical bond that forms with differences in electronegativity. EN= Electronegativity Greater than 1.7 Transfer of electrons Between 0.4 & 1.7 Electrons unequally shared Below 0.4 Electrons equally shared  Quiz What type of bond will form between identical atoms with an electronegativity balance or difference of zero? A mostly ionic C nonpolar covalent B polar covalent D mostly covalent  Quiz If two atoms have an electronegativity difference greater than 1.7, what type of bond are they likely to form? A polar covalent B mostly covalent C mostly ionic D nonpolar covalent  Electronegativity and Bond Character Unequal sharing of electrons results in a polar covalent bond. Bonding is often not clearly ionic or covalent. The graph summarizes the range of chemical bonds between two atoms. This graph shows that the difference in electronegativity between bonding atoms determines the percent ionic character of the bond. Above 50% ionic character, bonds are mostly ionic.  Polar Covalent Bonds When a polar bond forms, the shared electron pair or pairs are pulled toward one of the atoms. The electrons spend more time around that atom than the other atom. This results in partial charges at the ends of the bond (we call them dipoles). Dipoles (‫ )األقطاب‬means 2 sides with different charges. Partial Charges ‫شحنات جزئية‬ Check Point Dipole: A bond or molecule whose ends have Partial Charges opposite charges (+ and -) not ions. 22  Polar Covalent Bonds Molecular Polarity Covalently bonded molecules are either polar or non-polar. Non-polar molecules are not attracted by an electric field. Polar molecules align with an electric field.  Polar Covalent Bonds Polarity and Molecular Shape Compare water, H2O, and CCl4: Both bonds are polar. The molecular shapes, determined by VSEPR, are bent and tetrahedral, respectively. O–H bonds are asymmetric in water. There is a definite positive end and a definite negative end. Thus, polar.  Polar Covalent Bonds C – Cl bonds are symmetrical in CCl4. Electric charge measured at any distance from the center is identical on all sides. Partial charges are balanced, and are thus nonpolar.  Properties of Covalent Compounds Intermolecular Forces (attraction forces ‫)قوى تجاذب‬ Covalent bonds between atoms are strong, but attraction forces between molecules are weak. The weak attraction forces are known as intermolecular forces, or van der Waals forces. The forces vary in strength but are weaker than the bonds in a molecule or ions in an ionic compound. -Intermolecular forces happen between covalent compounds or covalent molecules not atoms. -Intermolecular forces are weaker than ionic bonds and covalent bonds  Properties of Covalent Compounds Intermolecular Forces Non-polar molecules exhibit a weak dispersion force, or induced dipole. The force between two oppositely charged ends of two polar molecules is a dipole-dipole force. The more polar the molecule, the stronger the dipole- dipole force. A hydrogen bond is an especially strong dipole-dipole force between a hydrogen end of one dipole and a fluorine, oxygen, or nitrogen atom on another dipole. Check the table in the next slide  The Kinetic-Molecular Theory Assumption = ‫فرضية‬ Collision =‫تصادم‬ ‫النظرية الحركية الجزيئية‬ Elastic = ‫مرن‬ Consist = ‫تتكون‬ 5 assumptions of the theory Particle Size (‫)حجم الجسيمات‬ Gases consist of small particles separated from one another by empty space. Because gas particles are far apart, they experience no significant attractive or repulsive forces. Particle Motion (‫)حركة الجسيمات‬ Gas particles are in constant, random motion. Collisions between gas particles are elastic. An elastic collision is one in which no kinetic energy is lost.  The Kinetic-Molecular Theory Particle Energy Kinetic energy “KE” of a particle depends on mass and velocity. KE is kinetic energy, m is the mass of the particle, and ⱱ is its velocity. Temperature is a measure of the average kinetic energy of the particles in a sample of matter.  Quiz Which of the following is NOT an assumption of the kinetic-molecular theory? A Gases consist of small particles separated from each other by empty space. B Gas particles are in constant, random motion. C There are significant attractive forces between particles in gases. CORRECT D Mass and velocity determine kinetic energy.  Quiz Which of the following is an assumption of the kinetic-molecular theory? 1- Gases consist of big particles separated from each other by empty space. 2- Gas particles are in constant, random motion. 3- There are significant attractive forces between particles in gases. 4- The Kinetic Energy doesn’t depend on the mass and velocity. Between collisions, the particles move in straight lines  Explaining the Behavior of Gases Gases have low density. The density of solid gold (Au) is The density of Chlorine gas (Cl2) is 19.3 g/ml 2.95 x10-3 g/ml The density of Gold is around 6000 times bigger. Why? Density = Mass/Volume As the kinetic-molecular theory states, a great deal of space exists between gas particles. Thus, there are fewer chlorine molecules than gold atoms in the same volume.  Continue: Explaining the Behavior of Gases Compression and Expansion (‫)االنكماش والتمدد‬ In a closed container, compression and expansion change the volume occupied by a constant mass of particles. Relate the change in volume to the density of the gas particles in each cylinder.  Explaining the Behavior of Gases Diffusion and Effusion Gases easily flow past each other because there are no significant forces of attraction. Diffusion is the movement of one material through another. Effusion is a gas escaping through a tiny opening.  Explaining the Behavior of Gases Graham’s law of effusion states that the rate of effusion for a gas is inversely proportional to the square root of its molar mass. Graham’s law also applies to diffusion. A and B refer to Gas (A) and Gas (B)  Page 244: Example 1 GRAHAM’S LAW SOLVE FOR THE UNKNOWN Use with Example Problem 1. State the ratio derived from Graham’s law. Problem RateNH3 molar massHCl Ammonia has a molar mass of 17.0 = g/mol; hydrogen chloride has a molar RateHCl molar massNH3 mass of 36.5 g/mol. What is the ratio of Substitute molar massHCl = 36.5 g/mol and molar their diffusion rates? massNH3 = 17.0 g/mol. 36.5 g/mol = 1.47 17.0 g/mol KNOWN UNKNOWN The ratio of diffusion rates is 1.47. molar massHCl = 36.5 g/mol ratio of diffusion Ammonia (NH3) gas is 1.47 rates = ? times faster than HCl gas. molar massNH3 = 17.0 g/mol N = 14 g/mol Ne = 20.2 g/mol C = 12 g/mol O = 16 g/mol N = 14 g/mol Ne = 20.2 g/mol C = 12 g/mol O = 16 g/mol 1 atm = 760 mmHg = 760 torr = 101.3 KPa = 14.7 psi = 1.01 bar 44  Gas Pressure Dalton’s Law of Partial Pressures Dalton’s law of partial pressures states that the total pressure of a mixture of gases is equal to the sum of the pressures of all the gases of the mixture. The partial pressure of a gas depends on the number of moles, the size of the container, and temperature. The partial pressure of a gas is independent of the type of gas. At a given temperature and pressure, the partial pressure of 1 mol of any gas is the same.  Gas Pressure Partial pressure can be used to calculate the amount of gas produced in a chemical reaction. Ptotal = P1 + P2 + P3 +...Pn Dalton’s Law 47 Dalton’s Law: Application 48  Plenary Practice Problems (Page 248)  Intramolecular Forces ❑ Intramolecular forces: Attractive forces that hold particles together in ionic bonds. Covalent bonds. metallic bonds. ❑ The prefix intra- means within (between atoms). Intramolecular Forces Force Model Basis of Example Attraction Ionic cations and NaCl anions Covalent positive and nuclei and shared H2 electrons Metallic metal cations and Fe mobile electrons The 3 main types of Intermolecular Forces are: 1. Dipole-Dipole Forces 2. “Hydrogen Bonding” Forces 3. Dispersion Forces (Also called London Forces)  Types of Intermolecular Forces 1. Dipole-Dipole Forces Dipole-dipole forces are attractions between oppositely charged regions of polar molecules. Neighboring polar molecules orient themselves so that oppositely charged regions align. Permeant Identify the types of forces that are “dipole-dipole” represented in this figure. attraction  Types of Intermolecular Forces 2. Hydrogen Bonds N: Nitrogen O: Oxygen F: Fluorine Hydrogen bonds are special dipole-dipole attractions. Occurs between molecules containing a hydrogen (H) atom bonded to a small, highly electronegative atom (Like: N, O or F) with at least one lone electron pair. The hydrogen bonds between water molecules are stronger than typical dipole-dipole attractions because the bond between hydrogen and oxygen is highly polar.  Reminder: The 3 types of Intermolecular forces are: 1. Dipole-Dipole Forces. Dipole-Dipole Forces and Hydrogen bonding happens in Polar Molecules 2. Hydrogen Bonding Forces. 3. Dispersion Forces “London Forces”.  Types of Intermolecular Forces 3. Dispersion Forces Dispersion forces are weak forces that result from temporary shifts in the density of electrons in electron clouds. Figure 12.9 When two molecules are close together, the electron clouds repel each other, creating temporary dipoles. The δ sign represents an area of partial charge on the molecule. Explain what the δ+ and δ- signs on a temporary dipole represent.  Permanent Forces Vs. Temporary dipoles Permanent: ‫دائم‬ Temporary: ‫مؤقت‬ In all Molecules ( Only in Polar Molecules Nonpolar and polar molecules) 59  Types of Intermolecular Forces 3. Dispersion Forces: Dispersion forces are weak forces that result from temporary shifts in the density of electrons in electron clouds. Weaker than Dipole-Dipole and Hydrogen Bonding. Depend on the number of electrons and the size of molecules: ❖The size of molecules increases, the Dispersion force becomes bigger. ❖The number of electron in a molecule gets high, the Dispersion force becomes bigger. Hydrogen bonds explain why water is a liquid at room temperature, while compounds of comparable mass are gases. Look at the data in Table 12.3. The difference between methane and water is easy to explain. Because methane molecules are nonpolar, the only forces holding the molecules together are relatively weak dispersion forces. The difference between ammonia and water is not as obvious. Molecules of both compounds can form hydrogen bonds. Yet, ammonia is a gas at room temperature, which indicates that the attractive forces between ammonia molecules are not as strong. Because oxygen atoms are more electronegative than nitrogen atoms, the O–H bonds in water are more polar than the N–H bonds in ammonia. As a result, the hydrogen bonds between water molecules are stronger than those between ammonia molecules.  Boyle’s Law: There is an inverse relationship between Pressure and Volume when the temperature is constant. If the pressure goes down, the volume goes up, and vice versa  Boyle’s Law Both the amount of gas and its Temperature must be constant.  Page 279 BOYLE’S LAW Use with Example Problem 1. Problem SOLVE FOR THE UNKNOWN A diver blows a 0.75-L air bubble 10 m Use Boyle’s law. Solve for V2 , and calculate the new volume. under water. As it rises to the surface, the State Boyle’s law. pressure goes from 2.25 atm to 1.03 atm. P1V1 = P2V2 What will be the volume of air in the Solve for V2. bubble at the surface? P V2 = V1 1 P2 Substitute V1 = 0.75 L, P1 = 2.25 atm, and P2 KNOWN UNKNOWN = 1.03 atm. V1 = 0.75 L V2 = ? L 2.25 atm V2 = 0.75 L 1.03 atm P1 = 2.25 atm Multiply and divide numbers and units. P2 = 1.03 atm 2.25 atm V2 = 0.75 L =1.6 L 1.03 atm Page 280  There is an inverse relationship between Pressure and Volume when the temperature is constant. If the pressure is constant, what will be the relationship between temperature and volume?  Charles’s Law How Temperature and Volume are Related As temperature increases, so does the volume of a gas sample when the amount of gas and pressure remain constant.  Charles’s Law Graphing Temperature and Volume Note: the Temperature A temperature of 0 K corresponds to 0 is in Kelvin “K” not mL. Doubling temperature doubles Celsius °C volume. Absolute zero is zero on the Kelvin scale.  Using Charles’s law Both the amount of gas and its pressure must be constant. Note: the Temperature is in Kelvin “K” not Celsius °C  Page 282 CHARLES’S LAW Use with Example Problem 2. Problem SOLVE FOR THE UNKNOWN A helium balloon in a closed car occupies a Convert degrees Celsius to kelvins. volume of 2.32 L at 40.0°C. If the car is Apply the conversion factor. parked on a hot day and the temperature TK = 273 + TC inside rises to 75.0°C, what is the new volume of the balloon, assuming the Substitute T1 = 40.0°C. pressure remains constant? T 1 = 273 + 40.0°C = 313.0 K Substitute T2 = 75.0°C. KNOWN UNKNOWN T2 = 273 + 75.0°C = 348.0 K T1 = 40.0°C V2 = ? L V1 = 2.32 L T2 = 75.0°C  CHARLES’S LAW Use with Example Problem 2. SOLVE FOR THE UNKNOWN (continued) Use Charles’s law. Solve for V2, and substitute the known values into the rearranged equation. State Charles’s law. V1 V2 = T1 T2 Solve for V2 T V2 = V1 2 T1 Substitute V1 = 2.32 L, T 1 = 313.0 K, and T2 = 348.0 K. 348.0 K V2 = 2.32 L 313.0 K Multiply and divide numbers and units. 348.0 K V2 = 2.32 L = 2.58 L 313.0 K  Gay-Lussac’s Law How Temperature and Pressure of a Gas are Related Gay-Lussac’s law states that the pressure of a fixed amount of gas varies directly with the Kelvin temperature when the volume remains constant. Note: The volume must be fixed “Not changing”  Gay-Lussac’s Law When the volume is fixed, the pressure and temperature of a fixed amount of a gas will have a direct relationship.  Page 284 GAY-LUSSAC’S LAW KNOWN UNKNOWN Use with Example Problem 3. P1 = 5.00 atm P2 = ? atm Problem T1 = 25.0°C The pressure of the oxygen gas inside a canister is 5.00 atm at T2 = -10.0°C 25.0°C. The canister is located at a camp high on Mount Everest. If the SOLVE FOR THE UNKNOWN temperature there falls to -10.0°C, what is the new pressure inside the Convert degrees Celsius to kelvins. canister? Apply the conversion factor. TK = 273 + TC Substitute T1 = 25.0°C. T1 = 273 + 25.0°C = 298.0 K Substitute T2 = -10.0°C. T2 = 273 + (-10.0°C) = 263.0 K  GAY-LUSSAC’S LAW Page 284 Use with Example Problem 3. SOLVE FOR THE UNKNOWN (continued) Use Gay-Lussac’s law. Solve for P2, and substitute the known values into the rearranged equation. State Gay-Lussac’s law. P1 P2 = T1 T2 Solve for P2 T P2 = P1 2 T1 Substitute P1 =5.00 atm, T1 = 298.0 K, and T2 = 263.0 K. 263.0 K P2 = 5.00 atm 298.0 K Multiply and divide numbers and units. 263.0 K P2 = 5.00 atm = 4.41 atm 298.0 K  The Combined Gas Law The combined gas law states the relationship among pressure, temperature, and volume of a fixed amount of gas. For a given amount of gas, the product of pressure and volume, divided by the Kelvin temperature, is a constant. The gas amount must be fixed (doesn’t change) THE COMBINED GAS LAW Page 286 Use with Example Problem 4. Problem A gas at 110 kPa and 30.0°C fills a flexible SOLVE FOR THE UNKNOWN container with an initial volume of 2.00 L. Convert degrees Celsius to kelvins. If the temperature is raised to 80.0°C and Apply the conversion factor. the pressure increases to 440 kPa, what is TK = 273 + TC the new volume? Substitute T1 = 30.0°C. KNOWN UNKNOWN T1 = 273 + 30.0°C = 303.0 K P1 = 110 kPa V2 = ? L Substitute T2 = 80.0°C. P2 = 440 kPa T2 = 273 + 80.0°C = 353.0 K T1 = 30.0ºC T2 = 80.0ºC V1 = 2.00 L THE COMBINED GAS LAW Use with Example Problem 4. SOLVE FOR THE UNKNOWN (continued) Use the combined gas law. Solve for V2, and substitute the known values into the rearranged equation. State the combined gas law. Multiply and divide numbers and units. P1V1 P2 V2 110 kPa 353.0 K = V2 = 2.00 L = 0.58 L T1 T2 440 kPa 303.0 K Solve for V2. P T2 V2 = V1 1 P2 T1 Substitute V1 = 2.00 L, P1 = 110 kPa, P2 = 440 kPa, T 2 = 353.0 K, and T1 = 303.0 K. 110 kPa 353.0 K V2 = 2.00 L 440 kPa 303.0 K Summary 86  Avogadro’s Principle Avogadro’s principle: It states that equal volumes of gases at the same temperature and pressure contain equal numbers of particles or equal number of moles. 1) A Container that has a volume of 10 Liters contains 5 moles of H2.Another container at the same temperature and pressure and of a volume of 10 Liters contains N2 (Nitrogen). How many moles of Nitrogen are in the second container? 5 moles 90  The Molar Volume of gases The molar volume of a gas is: the volume a 1 mol of a gas occupies (Fills) at 0.00°C temperature and 1.00 atm pressure. The conditions of the temperature equaling 0.00°C and the pressure being 1.00 atm are known as standard temperature and pressure (STP). Avogadro (The scientist ‫ )العالم افوجادرو‬showed experimentally that 1 mol of any gas occupies a volume of 22.4 L at STP. Because the volume of 1 mol of a gas at STP is 22.4 L, you can use 22.4 L/mol as a conversion factor 91 whenever a gas is at STP. MOLAR VOLUME Page 289 MM C = 12.01 Use with Example Problem 5. g/mol MM H = 1.01 g/mol Problem The main component of natural SOLVE FOR THE UNKNOWN gas used for home heating and Determine the molar mass for methane. cooking is methane (CH4). Determine the molar mass mass. Calculate the volume that 2.00 kg of methane gas will occupy at MM = (1 x 12.01) + (4 x 1.01) = 16.05 g/mol STP. Express the molecular mass as g/mol to arrive at the molar mass. KNOWN UNKNOW = 16.05 g/mol N m = 2.00 kg V = ? L T = 0.00ºC P = 1.00 MOLAR VOLUME Page 289 SOLVE FOR THE UNKNOWN (continued) Determine the number of moles of methane. Convert the mass from kg to g. 1000 g 2.00 kg = 2.00 × 103 g 1 kg Divide mass by molar mass to determine the number of moles. m 2.00 × 103 g = = 125 mol M 16.05 g/mol Use the molar volume to determine the volume of methane at STP. Use the molar volume, 22.4 L/mol, to convert from moles to the volume. 22.4 L V = 125 mol × = 2.80 × 103 L 1 mol  Practice: Page 290 Molar Mass “N”= 14 *Remember: at STP conditions, 1 g/mol mol of any gas has a volume of 22.4 L.  The Ideal Gas Law Pressure Volume Temperature Number of moles atm L K0 mol L. atm The ideal gas constant is represented by R and is 0.0821 mol. K Note that the units for R are simply the combined units for each of the four variables. 97 97  The Ideal Gas Constant “R” R can have different Values depending on pressure unit 1) The Value of R when pressure is in KPa 8.314 is……………… 2) The Value of R when pressure is in mmHg is……………… 62.4 3) The Value of R when pressure is in atm is……………… 0.0821 98 98  Example: Page 291 Pressure is in atm so R = 0.0821 99 99  The Ideal Gas Law Page 292 100  The “Molar Mass” from M → Molar Mass the Ideal Gas Law m → mass n → moles D → Density Deriving the Molar Mass (M) 𝑚 𝑊𝑒 𝑘𝑛𝑜𝑤 𝑡ℎ𝑎𝑡: 𝑀 = and n 𝑚 Rearrange the first equation: n = 𝑀 Replace the n in the ideal gas law equation by m/M  The “Density” from the Ideal Gas Law 𝑚 If you know the following: 𝐷= 𝑉 If you manipulate both equations 𝑹𝑻 above. What would be the correct 𝟏) 𝑫 = equation for the density “D”? 𝑴𝑷 𝑴𝑷 𝟐) 𝑫 = 𝑹𝑻 𝑹𝑷 𝟑) 𝑫 = 𝑴𝑻  Calculating the density of a gas using the Ideal Gas Law Question (1) What is the density “D” of a sample of Hydrogen gas (H2) that has a pressure (P) of 1.106 atm, a Volume (V) of 10.0 L and its measured temperature (T) is 300 K? M of “H2” = 2.00 g/mol Known P=1.106 atm V = 10.0 L T = 300 K M =2.00 g/mol R = 0.0821 L.atm/mol.K 2.00 𝑥 1.106 𝑫= = 𝟎. 𝟎𝟖𝟗𝟖 𝒈/𝑳 0.0821 𝑥 300 Unknown D=??  Ideal Gases Vs. Real Gases Remember! What does it mean to be an “Ideal Gas”? An ideal gas follows the gas laws under all conditions of temperature and pressure. PV= nRT Boyle’s Law, Charles’s Law, Gay-Lussac’s Law and Avogadro's Principle Real Vs. Ideal Gases In Real Gases (Not Ideal): The gas has a volume (Volume is not zero) (when the pressure is very high) Attraction and repulsion forces (Intermolecular forces) exists between gas Particles (When the Temperature is very low) If the size of the gas particles is very big (Volume not zero) If the gas molecules are very polar (Intermolecular forces) The nature of the particles making up a gas also affects how ideally the gas behaves. For example, polar gas molecules, such as water vapor, generally have larger attractive forces between their particles than nonpolar gases, such as helium. The oppositely charged ends of polar molecules are pulled together through electrostatic forces, as shown in Figure 13.9. Therefore, polar gases do not behave as ideal gases. Also, the particles of gases composed of larger nonpolar molecules, such as butane ( C 4 H 10 ), occupy more actual volume than an equal number of smaller gas particles in gases such as helium (He). Therefore, larger gas particles tend to exhibit a greater departure from ideal behavior than do smaller gas particles.  When would a gas behave like a real gas? Very low Very high Temperature Pressure Size of molecules Very polar gas is very big molecules volume-volume problems in gas Stoichiometry (Page: 297) 114 Practice Problems: Page 297 Page 297  “Volume-Mass” problems For gases in Chemical Reactions Stoichiometry and Volume–Mass Problems All masses given must be converted to moles or volumes before being used as part of a ratio. Also, remember that the temperature units used must be kelvins. 117  Page: 298 Now let’s apply Avogadro’s law to calculate the mass of a gas in a chemical reaction: Remember mole = mass To find the mass you need to: 1. Calculate the volume using Avogadro’s law 2. Use the Ideal gas law (PV = nRT) to convert the volume to moles, and then the moles to mass (m= M x n) Molar Mass (M) of Ammonia = 17.04 g/mol 118 119 Mass = Moles x Molar Mass m=nxM 120  Another Solution Known: Unknown VNH3 = 10 L mNH3 = ?? 𝑀𝑃𝑉 P = 3 atm, 𝒎= 𝑅𝑇 T = 298 K M = 17.04 g/mol 17.04 𝑥 3 𝑥 10 R = 0.0821 𝑚= = 𝟐𝟎. 𝟖𝟗 = 𝟐𝟏 𝒈 0.0821 𝑥 298 121  To solve the problems in the book (Page 299), you must be familiar with the following problems from term 2: Mole-to-Mole Mole-to Mass or Mass-to-Mole Mass to Mass At STP: 1 mol of any gas has a volume of 22.4 L  Practice: Page 299 Known: unknown: M of NH4NO3 (Ammonium VNO2= 0.1 L Mass nitrate) = 80 g/mol (NH4NO3) Step 1: Find the moles of N2O Step 2: Mole-to-Mole conversion (N2O → NH4NO3) Step 3: Mass = Moles x Molar Mass 123  Practice: Page 299 Known: Unknown: Molar Mass (Fe) = 55.85 g/mol Iron “Fe” VO2 ?? In Liters “L” mFe=52 g Step 1: 52 𝑔 = 0.931 𝑚𝑜𝑙 𝐹𝑒 Moles = Mass/Molar Mass 55.85 𝑔/𝑚𝑜𝑙 Step 2: 3 𝑚𝑜𝑙 𝑂2 0.931 𝑚𝑜𝑙 𝐹𝑒 𝑥 = 0.698 𝑚𝑜𝑙 𝑂2 Mole-to-Mole conversion 4 𝑚𝑜𝑙 𝐹𝑒 (Fe → O2) Step 3: 22.4 𝐿 0.698 𝑚𝑜𝑙 𝑂2 𝑥 = 𝟏𝟓. 𝟔 𝑳 𝑶𝟐 Find the moles of O2 gas at STP. 1 𝑚𝑜𝑙 124 Practice: Page 299  The End With my best wishes Dear Students ☺