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The burning of a magnesium ribbon in air Grade 10 SABIS SABIS Exothermic

Reaction Mechanism: reactions rarely proceed in a single step as written. They take place in a series of smaller steps called a reaction mechanism. Grade 10 SABIS ​

Combustion Reactions Grade 10 SABIS SABIS Reaction when a substance reacts rapidly with a gas producing heat and light, for eg., burning a substance in the presence of air

Physical properties of metals: shiny, ductile (pulled into wires), malleable (hammered into thin sheets), conduct electricity. Grade 10 SABIS ​

Microscopic changes that take place when a liquid is warmed Grade 10 SABIS ​ When a liquid is warmed in thermochemistry, several microscopic changes occur at the molecular level. These changes involve the increased kinetic energy of the liquid molecules and their interactions, leading to observable macroscopic effects such as expansion and changes in physical properties. As the liquid is heated, the temperature of the system rises, and this increase in temperature corresponds to an increase in the average kinetic energy of the liquid molecules. The molecules gain energy and move more rapidly, exhibiting increased vibrational, rotational, and translational motion. The increased kinetic energy causes the intermolecular forces between the liquid molecules to weaken. In the liquid state, these forces, such as hydrogen bonding or London dispersion forces, hold the molecules together in a cohesive arrangement. As the molecules gain energy, the forces become less effective at maintaining this cohesion. The weakened intermolecular forces result in an expansion of the liquid. The increased molecular motion and reduced intermolecular forces allow the molecules to move farther apart, leading to an increase in volume. This expansion is commonly observed in liquids when they are heated. Additionally, the increased kinetic energy can lead to changes in the physical properties of the liquid. For example, the viscosity of the liquid may decrease as the molecules move more freely and with less resistance. The surface tension may also decrease as the cohesive forces weaken, affecting the behavior of the liquid at interfaces. Furthermore, in some cases, when a liquid is heated sufficiently, it may undergo a phase change and transform into a gas. This transition occurs at the boiling point, where the vapor pressure of the liquid becomes equal to the external pressure. The heated liquid absorbs energy to overcome intermolecular forces and transition into a gas phase. It's important to note that the microscopic changes in a liquid being warmed are reversible. When the liquid is cooled, the molecules lose kinetic energy, and the intermolecular forces regain their effectiveness, leading to a decrease in volume and a return to the initial state. Understanding the microscopic changes that occur when a liquid is warmed is crucial in thermochemistry and various applications. It allows us to analyze energy transfers, phase transitions, and the behavior of substances under different temperature conditions. In summary, when a liquid is warmed in thermochemistry, microscopic changes take place at the molecular level. The increased kinetic energy of the molecules weakens the intermolecular forces, resulting in expansion, changes in physical properties, and, in some cases, phase transitions. Recognizing and studying these microscopic changes enhances our understanding of energy transfer and the behavior of liquids at different temperatures.

Mathematical Representation ​ ​ P1V1 = P2V2, which signifies that the product of initial pressure and volume equals the product of final pressure and volume.

7. The decomposition of water into H2 and O2 gas. Endothermic Grade 10 SABIS SABIS

Recognizing the Reverse Reaction in Thermochemistry Grade 10 SABIS ​ Recognizing the reverse of an equation Write the reverse of the equation 2C(s) + 3H2 (g) → C2H6 (g) ΔH = − 84.5 kJ C2H6 (g) → 2C(s) + 3H2 (g) ΔH = + 84.5 kJ In thermochemistry, it is important to understand that chemical reactions can proceed in both the forward and reverse directions. The reverse reaction is simply the opposite of the forward reaction, where the products become the reactants, and the reactants become the products. To recognize the reverse of an equation, we look at the reactants and products and interchange their positions. In this case, the given equation is: 2C(s) + 3H2(g) → C2H6(g) ΔH = -84.5 kJ To write the reverse equation, we switch the positions of the reactants and products: C2H6(g) → 2C(s) + 3H2(g) ΔH = +84.5 kJ By reversing the equation, we also reverse the sign of the heat of reaction (∆H). In the original equation, the heat of reaction is -84.5 kJ, indicating that the reaction releases 84.5 kJ of heat energy. In the reverse equation, the heat of reaction becomes +84.5 kJ, indicating that the reaction now absorbs 84.5 kJ of heat energy. It's important to note that the reverse reaction occurs under different conditions compared to the forward reaction. While the forward reaction may be exothermic (releasing heat), the reverse reaction becomes endothermic (absorbing heat) due to the change in the sign of the heat of reaction. Understanding the reverse of an equation is crucial in thermochemistry, as it allows us to recognize that a reaction can proceed in both directions depending on the prevailing conditions. The reverse reaction is often observed when the products of a reaction have a higher concentration or are removed from the system, causing the reaction to shift towards the reactants. In summary, recognizing the reverse of an equation involves interchanging the positions of the reactants and products and changing the sign of the heat of reaction (∆H). In the given example, the reverse of the equation 2C(s) + 3H2(g) → C2H6(g) with a heat of reaction of -84.5 kJ is C2H6(g) → 2C(s) + 3H2(g) with a heat of reaction of +84.5 kJ. Understanding the reverse reaction is essential in thermochemistry to comprehend the bidirectional nature of chemical reactions and the associated heat changes.

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Volume Ratio at STP Grade 10 SABIS SABIS Write the volume ratio at STP conditions for a given reaction equation

Given the % abundance of isotopes, find the average atomic mass Grade 10 SABIS ​ Given the percentage abundance of isotopes: It's like knowing the proportion of different ingredients in a recipe. Isotopes: Imagine them as different types of toppings on a pizza. Each topping represents a specific isotope, and the percentage abundance tells us how much of each topping is used. The average atomic mass is like the overall flavor profile of the pizza, combining the tastes of all the different toppings. To find the average atomic mass, we'll multiply the mass of each isotope by its percentage abundance and then sum up the results. For example, let's consider an element with two isotopes: Isotope A and Isotope B. Let's assume Isotope A has a mass of 10 and an abundance of 40%, while Isotope B has a mass of 12 and an abundance of 60%. To find the average atomic mass, we'll calculate (10 * 0.40) + (12 * 0.60), which gives us the weighted sum of the masses. This calculation represents the weighted contribution of each isotope to the overall average atomic mass. In our everyday lives, we can relate this concept to calculating the average grade in a class, where each student's grade contributes differently based on their percentage weight in the final calculation. Let's consider another example with three isotopes: Isotope X, Isotope Y, and Isotope Z. Assuming Isotope X has a mass of 8 and an abundance of 20%, Isotope Y has a mass of 10 and an abundance of 30%, and Isotope Z has a mass of 12 and an abundance of 50%. To find the average atomic mass, we'll calculate (8 * 0.20) + (10 * 0.30) + (12 * 0.50). This calculation takes into account the masses and the respective percentage abundances of each isotope. In a practical context, we encounter similar situations when determining an average score in a game, where each player's score contributes differently based on their playing time or performance. The average atomic mass reflects the overall tendency of the element's isotopes, just as the average temperature in a region represents the general climate conditions over time. By knowing the percentage abundance of isotopes, scientists can gain insights into the natural distribution of elements and how they vary in different samples or locations. Analyzing the average atomic mass is vital in fields such as analytical chemistry, geology, and environmental science, where precise knowledge of isotopic compositions helps unravel natural processes and environmental changes. To summarize the process, we calculate the weighted sum of the masses of each isotope, taking into account their respective percentage abundances. By finding the average atomic mass, we obtain a representative value that encompasses the contributions of different isotopes, much like obtaining an average rating for a product based on customer reviews. In essence, by understanding the percentage abundance of isotopes and their respective masses, we can determine the average atomic mass, providing valuable information about the element's composition and its significance in various scientific disciplines.