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c324cec6-92f1-42f7-a5f0-79fbe6caa106 medium difficulty easy examples for Given the average atomic mass of an element, find the % abundance of its isotopes Summary Example 1: Average atomic mass: 32.7 Isotope A mass: 31 Isotope B mass: 34 To find the percentage abundance: Let's assume the abundance of Isotope A is x, and the abundance of Isotope B is y. Equation 1: (x * 31) + (y * 34) = 32.7 Equation 2: x + y = 100 Solving the equations, we find that x = 70 and y = 30. Answer: Isotope A: 70% abundance Isotope B: 30% abundance Example 2: Average atomic mass: 42.9 Isotope A mass: 42 Isotope B mass: 44 To find the percentage abundance: Let's assume the abundance of Isotope A is x, and the abundance of Isotope B is y. Equation 1: (x * 42) + (y * 44) = 42.9 Equation 2: x + y = 100 Solving the equations, we find that x = 60 and y = 40. Answer: Isotope A: 60% abundance Isotope B: 40% abundance Example 3: Average atomic mass: 56.4 Isotope A mass: 55 Isotope B mass: 58 To find the percentage abundance: Let's assume the abundance of Isotope A is x, and the abundance of Isotope B is y. Equation 1: (x * 55) + (y * 58) = 56.4 Equation 2: x + y = 100 Solving the equations, we find that x ≈ 62.15 and y ≈ 37.85 (rounded to two decimal places). Answer: Isotope A: Approximately 62.15% abundance Isotope B: Approximately 37.85% abundance

Lesson 41 More Families of Elements & Periodic Trends Chapter 7 SABIS Grade 10 Part 3 Lesson 41 More Families of Elements & Periodic Trends Points explained Meaning of the word “stable” Reactions of the alkali metals with chlorine Reactions of the alkali metals with water Reactions of the alkali metals with hydrogen Flame test for Li+ , Na+ and K+ Summary of chemistry of the alkali metals 7.6 The Halogens 7.6.1 Physical properties of the halogens 7.6.2 Covalent bonding in the halogens Defining a covalent bond Differences between covalent and ionic bonding 7.6.3 Boiling points and melting points of the halogens 7.6.4 Atomic radii and volumes 7.6.5 Chemistry of the halogens Reactions with the alkali metals Summary 7.6.6 Chemistry of the halide ions The halides are stable Test for the halide ions Relative reactivity of the halogens 7.7 Hydrogen - A Family by Itself 7.7.1 Physical properties 7.7.2 Chemistry of hydrogen Reaction with the alkali metals 7.8 The Third-row Elements 7.8.1 Physical properties of the third-row elements 7.8.2 Compounds of the third-row elements The hydrides The chlorides The oxides Summary 7.9 The periodic table: chemical reactivity 🔬Understanding Stability, Alkali Metals & Halogens 📚Pre-Requisite Questions: What does it mean for an element to be "stable"? 🤔 What happens when alkali metals react with chlorine? 🧪 Can you describe the flame test results for Li+, Na+, and K+? 🔥 Break for Reflection 🤔✍️ (Answers: 1. A stable element has a full outer electron shell and doesn't tend to react. 2. When alkali metals react with chlorine, they form ionic salts. 3. Li+ burns with a crimson flame, Na+ with a yellow flame, and K+ with a lilac flame.) 🚀 Lesson Begins! 🧱 Meaning of the Word “Stable” In the chemistry world, "stable" doesn't mean standing still! It means an atom has a full outer shell of electrons and is not looking to react. They're like that chill friend who's content with what they have! 😌 💥Reactions of the Alkali Metals 💦With Water Splash alert! Alkali metals react violently with water, producing heat, hydrogen gas, and an alkali metal hydroxide. Think of it as a bath bomb that's too explosive for the tub! 🛀💣 🎈With Hydrogen Pairing up! Alkali metals can combine with hydrogen to form metal hydrides, releasing energy in the process. It's like an energetic dance duo! 💃🕺 🔥Flame Test for Li+, Na+, and K+ Ready for some fireworks? 🎆 In the flame test, Li+ produces a red/crimson flame, Na+ gives a yellow flame, and K+ presents a lilac flame. It's like a mini festival of lights in the lab! 🎇 7.6 The Halogens The Halogens, just like the Alkali Metals, are an interesting bunch. They're like the goths of the periodic table, always looking to gain an electron to achieve stability. 🕶️💀 🧪Covalent Bonding in the Halogens Halogens form covalent bonds by sharing electrons. Imagine sharing your favorite pizza with a friend—that's how halogens share electrons to become stable. 🍕❤️ 🔥Boiling Points and Melting Points of the Halogens Halogens have higher boiling and melting points as we move down the group, thanks to the increasing number of electrons which cause stronger intermolecular forces. It's like adding more logs to the fire—the more you have, the higher the flame! 🏕️🔥 ⚖️Atomic Radii and Volumes Atomic radii also increase as we go down the group. It's kind of like siblings—the older ones tend to be bigger! 🧑🤝🧑 🔥Chemistry of the Halogens Halogens are pretty reactive. Their reactions with alkali metals form ionic salts, and they're not shy about displacing less reactive halogens. It's like a game of musical chairs! 🎶🪑 7.7 Hydrogen - A Family by Itself Hydrogen is unique. Despite being the lightest and simplest element, its properties don't quite fit into any group. So, it charts its own path—just like a lone wolf. 🐺⛰️ 7.8 The Third-row Elements The third-row elements are like the middle kids of the periodic table. They have their quirks and surprises! So, let's dive deeper into their physical properties and compounds. 🏊♂️🌊 7.9 The Periodic Table: Chemical Reactivity The periodic table is not just a chart; it's a tale of reactivity, trends, and atomic friendships. Keep exploring, keep learning! 🚀 Review Questions: What is meant by a stable element? a. It has a full outer electron shell b. It has no protons c. It is radioactive d. None of the above What happens when alkali metals react with water? a. Nothing b. They dissolve c. They produce heat and hydrogen gas d. They turn into halogens Which element doesn't fit well into any group in the periodic table? a. Hydrogen b. Helium c. Oxygen d. Nitrogen Which of the following is NOT a property of the halogens? a. They form ionic bonds b. They form covalent bonds c. They have high boiling and melting points d. They are very reactive (Answers: 1. a, 2. c, 3. a, 4. a) End of Lesson 3 ⭐Keep studying, keep learning!⭐

e813ddb8-83c8-4092-a77b-068374b615c1 Mass of a Nucleus Summary The mass of a nucleus refers to the total mass of protons and neutrons present within the nucleus of an atom. It is a fundamental property that determines the overall mass of an atom. To understand the mass of a nucleus, let's consider an everyday example: a fruit bowl. Imagine each fruit in the bowl represents a proton or a neutron. The combined mass of all the fruits in the bowl would be analogous to the mass of the nucleus, which is composed of protons and neutrons. The mass of a nucleus is measured in atomic mass units (amu), with 1 amu being approximately equal to the mass of a proton or a neutron. The number of protons, known as the atomic number, determines the element, while the sum of protons and neutrons gives the mass number of an atom. For instance, let's take the element carbon. A carbon nucleus contains 6 protons and usually 6 neutrons, resulting in a total mass of approximately 12 atomic mass units. In a similar manner, let's consider a bag of marbles. Each marble can represent a proton or a neutron. The total weight of all the marbles in the bag would correspond to the mass of the nucleus, which is determined by the combined mass of protons and neutrons. The mass of a nucleus is crucial in understanding the stability and behavior of atoms. Isotopes, which are atoms of the same element with different numbers of neutrons, have different masses due to the varying number of neutrons in their nuclei. Mass defects are also observed in nuclei. The mass of a nucleus is slightly less than the combined mass of its individual protons and neutrons. This difference in mass is known as the mass defect and is a consequence of Einstein's famous equation, E=mc². To illustrate, think of a jar filled with marbles representing protons and neutrons. If you were to calculate the combined mass of all the marbles, it would be slightly greater than the actual mass of the filled jar due to the mass defect. The mass defect occurs because some of the mass of the nucleus is converted into binding energy, which holds the nucleus together. This binding energy is released during nuclear reactions, such as fusion or fission, where the total mass of the products differs from the mass of the original nucleus. An everyday example of mass defect and binding energy can be observed in the energy released from a nuclear power plant. The difference in mass between the reactant nuclei (such as uranium) and the product nuclei (after fission) is converted into a large amount of energy. In summary, the mass of a nucleus refers to the combined mass of protons and neutrons within an atom's nucleus. It is a fundamental property that influences the stability and behavior of atoms. Examples like a fruit bowl or a bag of marbles help illustrate the concept of the mass of a nucleus, as well as mass defects and binding energy associated with nuclear reactions. Understanding the mass of a nucleus is essential for comprehending atomic structure, isotopes, and the energy transformations that occur in nuclear processes.

b567f9ed-b3f8-4466-a97e-8b7ebfa59331 Law of Conservation of Matter Summary Matter can never be created or destroyed. It follows that in a chemical reaction mass and atoms are conserved. As a chemical reaction involves a rearrangement of atoms number of molecules is not conserved

dd1eaaf6-c348-4254-ad95-50705fa7a898 Heating a piece of paper until it turns black, then cooling it Summary Chemical

e27ae662-ec77-4658-83d4-ed508d10c1bd < Back Previous Next ecessary to know the bonds present in both the reactants and products How to complete bond energy calculations Write a balanced equation if none is present already Optional - draw the displayed formula in order to identify the type and number of bonds more easily Add together all the bond energies for all the bonds in the reactants – this is the ‘energy in’ Add together the bond energies for all the bonds in the products – this is the ‘energy out’ Calculate the enthalpy change: Enthalpy change (Δ H ) = Energy taken in - Energy given out Worked Example Hydrogen and chlorine react to form hydrogen chloride gas: H2 + Cl2 ⟶ 2HCl The bond energies are given in the table below. Bond Energy (kJ) H–H 436 Cl–Cl 242 H–Cl 431 Calculate the overall energy change for this reaction and use this value to explain whether the reaction is exothermic or endothermic. Answer: Calculate the energy in 436 + 242 = 678 (kJ) Calculate the energy out 2 x 431 = 862 (kJ) Calculate the energy change 678 - 862 = –184 (kJ) Since the energy change is a negative number, energy is being released (to the surroundings) Therefore, the reaction is exothermic Examiner Tips and Tricks When calculating enthalpy change using bond energies, it is helpful to write down a displayed formula equation for the reaction before identifying the type and number of bonds, to avoid making mistakes. So, the reaction for the above worked example is: H-H + Cl-Cl → H-Cl + H-Cl Worked Example Hydrogen reacts with iodine to form hydrogen iodide. H2 + I2 ⟶ 2HI The relevant bond energies are shown in the table below. Bond Energy (kJ) H–I 295 H–H 436 I–I 151 Calculate the overall energy change for this reaction and use this value to explain why the reaction is exothermic. Answer: Calculate the energy in 436 + 151 = 587 (kJ) Calculate the energy out 2 x 295 = 590 (kJ) Calculate the energy change 587 - 590 = -3 (kJ) The reaction is exothermic because: More energy is released than taken in Worked Example Hydrogen bromide decomposes to form hydrogen and bromine: 2HBr ⟶ H2 + Br2 The overall energy change for this reaction is +103 kJ. The relevant bond energies are shown in the table below. Bond Energy (kJ) H–Br 366 Br–Br H–H 436 Calculate the bond energy of the Br–Br bond. Answer: Calculate the energy in 2 x 366 = 732 (kJ) State the energy out 436 + Br–Br Overall energy change = energy in - energy out +103 = 732 - (436 + Br–Br) +103 = 732 - 436 - Br–Br Calculate the bond energy of the Br–Br bond Br–Br = 732 - 436 - 103 Br–Br = + 193 (kJ) Chemical energetics Next Topic

709b1fea-aad5-46ea-96bb-eccb9b2eda5f Measurements and Calculations Calculations with Significant Figures Next topic k-chemistry.com/concepts-definition/calculations-with-significant-figures-examples Summary Significant figures (or "sig figs") are the digits in a measurement that carry meaning regarding its precision. This includes all nonzero digits, any zeros between nonzero digits, and trailing zeros that appear after a decimal point. They reflect how accurately a quantity is known and are essential in scientific calculations to avoid overstating precision. The more significant figures present, the more exact the measurement is considered to be.

ee94519c-b2d4-4c87-8548-7520326f3762 Law of Conservation of Mass Summary Same as Conservation of Mass.

All Content Chapters Overview All Content Grade 10 SABIS Curriculum Topics ● Chapter 1 Topics. Lesson 1 1️⃣ Safety in the Lab 🧪👩🔬 This chapter kicks off with safety procedures that are key to ensuring safe laboratory practices. 2️⃣ Lab Apparatus 🧰🔍 Next, we explore different laboratory tools such as pipettes, burettes, test tubes, and more, along with their specific uses. 3️⃣ Drawing Equipment 🎨🖌️ Here, we'll learn how to accurately draw and label scientific apparatus. 4️⃣ Bunsen Burner Use 🔥👨🔬 We then delve into the steps for safely lighting and using a Bunsen burner. Remember, safety is a top priority! 🚀 Lesson 2 1️⃣ Lab Apparatus more details 🔬✨ Discover the Enchanting World of Lab Apparatus! 🧪🌈 Immerse yourself in a vibrant symphony of scientific tools! From the versatile beaker 🥼 to the precise pipette 🧪, each apparatus unveils a kaleidoscope of visual wonders. Let's embark on an awe-inspiring journey of experimentation and unleash the magic of chemistry! 🌟 Lesson 3 1️⃣ Crystallization: Lesson 4 1️⃣ Filtration : ● Chapter 2 Topics Chapter 2 Part 1 : The Three States of Matter and Changes of State Chapter 2 Part 2 : Cooling Curve and Physical Constants Chapter 2 Part 3: Boyle's Law and Mathematical Representations Chapter 2 Problems and Questions ● Chapter 3 Topics Chapter Three Part One: Introduction to Chemistry: Qualitative Properties and Atomic Theories Chapter Three Part Two: Identifying Substances and Mixtures Chapter Three Part Three : Atomic Symbols, Chemical Formulas, and Molecular Models Chapter Three Part Four: Avogadro's Number and Molar Mass ● Chapter 4 Topics 🧠 Exothermic and Endothermic Processes 🧠Physical and Chemical Changes 🧠Conservation of Matter and Balancing Chemical Equations ● Chapter 5 Topics ● Chapter 6 Topics ● Chapter 7 Topics ● Chapter 8 Topics ● Chapter 9 Topics Chapter 1,2,3 Summary Part 2 of 4 SABIS Grade 10 Final Revision .pdf Download PDF • 7.89MB Chapter 3 Summary Part 2 Part 1 of 4 SABIS Grade 10 Final Revision .pdf Download PDF • 1.97MB Chapter 4 Presentation Chapter 8 Notes Notes Ch 8 SABIS CHEMISTRY Grade 10 .pdf Download PDF • 4.07MB Chapter 9 Notes Part 1 Notes Ch 9 SABIS CHEMISTRY Grade 10 Part 1 .pdf Download PDF • 10.22MB

2c19d6fd-a7e9-4491-9399-7be2b14b1f46 Conservation of molecules? Summary Molecules are not necessarily conserved in chemical reactions.

9c31eef2-81e0-4428-b3e5-8b1df166529a easy examples for Given the average atomic mass of an element, find the % abundance of its isotopes Summary Example 1: Average atomic mass: 15.8 Isotope A mass: 14 Isotope B mass: 16 To find the percentage abundance: Let's assume the abundance of Isotope A is x, and the abundance of Isotope B is y. Equation 1: (x * 14) + (y * 16) = 15.8 Equation 2: x + y = 100 Solving the equations, we find that x = 40 and y = 60. Answer: Isotope A: 40% abundance Isotope B: 60% abundance Example 2: Average atomic mass: 18.9 Isotope A mass: 17 Isotope B mass: 20 To find the percentage abundance: Let's assume the abundance of Isotope A is x, and the abundance of Isotope B is y. Equation 1: (x * 17) + (y * 20) = 18.9 Equation 2: x + y = 100 Solving the equations, we find that x = 30 and y = 70. Answer: Isotope A: 30% abundance Isotope B: 70% abundance Example 3: Average atomic mass: 27.5 Isotope A mass: 26 Isotope B mass: 28 To find the percentage abundance: Let's assume the abundance of Isotope A is x, and the abundance of Isotope B is y. Equation 1: (x * 26) + (y * 28) = 27.5 Equation 2: x + y = 100 Solving the equations, we find that x = 60 and y = 40. Answer: Isotope A: 60% abundance Isotope B: 40% abundance

1cedca35-cc88-447f-bbba-186888504cbe AP Chemistry Cheat Sheets . . . . Summary