11 17 How Are Equations Balanced
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The document provides an overview of a chemistry lesson on balancing chemical equations, including examples of balancing the equations H2 + O2 → H2O and H2 + Cl2 → HCl. It outlines the 3-step process for balancing equations: counting atoms on both sides, changing coefficients and multiplying, and checking the tally. Practice questions are assigned as homework.
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1. The document discusses balancing chemical equations through a three step process: counting atoms on both sides, changing coefficients and multiplying, and checking the tally.
2. An example reaction of N2 + H2 → NH3 is used to demonstrate the balancing process. After changing the coefficient of H2 to 3, the equation is balanced with N2 + 3H2 → NH3.
3. Students complete a balancing worksheet for homework by writing questions and showing the work to balance each equation in three steps.
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This study guide covers various stoichiometry calculations students will need to know for an upcoming test, including:
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2) Determining the mass and volume of products from a balanced chemical equation.
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This document contains 7 questions regarding chemical equilibrium calculations. The questions involve writing equilibrium constants, calculating equilibrium constants based on given partial pressures or concentrations at equilibrium, and determining unknown values at equilibrium given some known values. Specifically, it asks the student to:
1) Write the equilibrium constant expression for a dissolution reaction.
2) Write the equilibrium constant expression for a complex ion formation reaction.
3) Calculate the equilibrium constant for a decomposition reaction given partial pressures at equilibrium.
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1. The document discusses balancing chemical equations through a three step process: counting atoms on both sides, changing coefficients and multiplying, and checking the tally.
2. An example reaction of N2 + H2 → NH3 is used to demonstrate the balancing process. After changing the coefficient of H2 to 3, the equation is balanced with N2 + 3H2 → NH3.
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This document contains 7 questions regarding chemical equilibrium calculations. The questions involve writing equilibrium constants, calculating equilibrium constants based on given partial pressures or concentrations at equilibrium, and determining unknown values at equilibrium given some known values. Specifically, it asks the student to:
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2) Write the equilibrium constant expression for a complex ion formation reaction.
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1) The document solves the logarithmic equations log2(log3(log4(a))) = 0, log3(log4(log2(b))) = 0, and log4(log2(log3(c))) = 0.
2) It finds that a = 64, b = 16, and c = 9 by taking the bases of the logarithms to the power of the entire expression.
3) When a, b, and c are summed according to the given equation, the answer is 13.
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This document discusses solubility equilibria and the formation of precipitates. It defines key terms like solubility product constant (Ksp), explains how to calculate Ksp values from molar solubility and vice versa, and shows examples of using Ksp to determine whether precipitates will form when solutions are mixed. The key points are that solubility is dependent on equilibrium, saturated solutions have concentrations where Ksp = Q, and precipitates form when mixing produces Q > Ksp (supersaturation).
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Here is a stoichiometry question I wrote for the given reaction:
If 3.25 grams of barium chloride are available to react, how many grams of barium sulfate can be produced?
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Here are the steps to solve this problem:
1) Balance the equation: Zn + I2 → ZnI2
2) Convert given mass of Zn to moles:
125.0 g Zn → 2.25 mol Zn (molar mass = 65.38 g/mol)
3) Use mole ratio from balanced equation:
2.25 mol Zn → 2.25 mol ZnI2
4) Convert moles of ZnI2 to grams:
2.25 mol ZnI2 x 143.32 g/mol = 322.5 g ZnI2
So the mass of ZnI2 produced from 125.0 g of Zn is 322.5 g.
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Here are the steps to solve this problem:
1) Given: 12.04 X 1023 molecules of H2O
2) Conversion factors:
1 mol H2O / 6.02 X 1023 molecules H2O
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3) Calculations:
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The document provides information and instructions for a class. It includes:
- Yesterday's exit slip data showed 95-99% of students understanding a topic.
- Today's objective is to review for an upcoming exam on Thursday through working on a study guide and jeopardy game.
- The teacher emphasizes studying purposefully and asks students to focus on following class rules around interfering with learning and leaving seats without permission.
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The law of conservation of mass requires that both sides of a chemical reaction equation have the same number and types of atoms. This ensures that the total mass is conserved before and after the reaction.
The key is that both sides must have the same number of each type of atom, not just the same substances or same number of molecules. The correct answer is C.
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1) The document solves the logarithmic equations log2(log3(log4(a))) = 0, log3(log4(log2(b))) = 0, and log4(log2(log3(c))) = 0.
2) It finds that a = 64, b = 16, and c = 9 by taking the bases of the logarithms to the power of the entire expression.
3) When a, b, and c are summed according to the given equation, the answer is 13.
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This document discusses solubility equilibria and the formation of precipitates. It defines key terms like solubility product constant (Ksp), explains how to calculate Ksp values from molar solubility and vice versa, and shows examples of using Ksp to determine whether precipitates will form when solutions are mixed. The key points are that solubility is dependent on equilibrium, saturated solutions have concentrations where Ksp = Q, and precipitates form when mixing produces Q > Ksp (supersaturation).
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Here is a stoichiometry question I wrote for the given reaction:
If 3.25 grams of barium chloride are available to react, how many grams of barium sulfate can be produced?
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Here are the steps to solve this problem:
1) Balance the equation: Zn + I2 → ZnI2
2) Convert given mass of Zn to moles:
125.0 g Zn → 2.25 mol Zn (molar mass = 65.38 g/mol)
3) Use mole ratio from balanced equation:
2.25 mol Zn → 2.25 mol ZnI2
4) Convert moles of ZnI2 to grams:
2.25 mol ZnI2 x 143.32 g/mol = 322.5 g ZnI2
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1. The document contains 7 chemistry problems involving calculation of equilibrium constants (Keq and Kc) for various chemical reactions.
2. Problem 3 calculates Keq as 1.84x10-2 for the decomposition of HI gas into H2 and I2 gases at 425C, given the partial pressures of reactants and products at equilibrium.
3. Problem 7 involves a multiple step chemical equilibrium problem, calculating the initial and equilibrium partial pressures and concentrations of gases in a reaction between CO2, H2 and H2O, and determining the equilibrium constant Kp.
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3) Differential rate equations are described as relating the rates of change of concentrations of reactants and products based on stoichiometry. Examples of writing differential rate equations are provided.
4) Graphs illustrating concentration changes over time are shown and used to calculate average and instantaneous rates of reaction based on slope calculations.
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To determine rate exponents experimentally, concentrations of reactants must be changed one at a time to observe how the reaction rate changes. Rate exponents represent the order of the reaction with respect to each reactant and can be determined by running experiments where one concentration is doubled, tripled, or quadrupled and observing the corresponding change in reaction rate. The rate law and rate constant can then be used to predict reaction rates under different conditions.
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- The ideal gas equation and how it incorporates these variables
- Standard temperature and pressure conditions for reporting gas measurements
- Gas density calculations using molar mass and gas stoichiometry problems
- Dalton's law of partial pressures which states that in a mixture of gases, the total pressure is equal to the sum of the partial pressures of the individual gases.
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The document discusses making assumptions to simplify equilibrium calculations when the equilibrium constant (Keq) is extremely small or large. It provides an example of calculating equilibrium concentrations for the decomposition of CO2 at 2000°C assuming negligible decomposition due to the small Keq value of 6.40 x 10-7. The assumptions allow obtaining approximate equilibrium concentrations without solving the full cubic equation.
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This document provides instructions for completing stoichiometry calculations using mass-to-mass conversions. It includes a sample reaction and 5 practice problems using the reaction between sulfur, oxygen, hydrogen, and sulfuric acid. The instructions emphasize converting between grams and moles in 3 steps: 1) convert mass to moles, 2) use mole ratios from the balanced chemical equation, and 3) convert moles back to mass. Work must be shown for full credit.
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The document provides an explanation of stoichiometry and how to use mole ratios to solve stoichiometry problems. It defines mole ratios as conversion factors made from the coefficients of a balanced chemical equation. It outlines the steps to solve stoichiometry problems: 1) use a balanced equation, 2) convert quantities to moles if needed, 3) use mole ratios to convert between compounds, and 4) convert back to other units like grams if needed. An example problem is shown calculating moles of products from moles of reactants.
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This document discusses measurement units including the SI units, mass and weight, volume, and temperature scales. It provides information on the seven SI basic units, converting between units, derived units, common volume units like liters and milliliters, and temperature scales like Celsius, Kelvin and Fahrenheit. Examples are given for converting between units of length, mass, volume, density and temperature. The goal is for the reader to understand fundamental measurement units and conversions between unit systems.
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The document provides information about key concepts in general chemistry including the definitions of matter, homogeneous mixtures, and pressure. It also summarizes several gas laws including Boyle's law that pressure and volume are inversely proportional at constant temperature and amount, Charles' law that volume is directly proportional to temperature at constant pressure and amount, and Avogadro's law relating volume and amount of gas particles. Sample problems are provided to demonstrate how to use the gas laws and equations of state to calculate pressure, volume, temperature and amount in gas systems.
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Here are the steps to solve this problem:
1) Given: 12.04 X 1023 molecules of H2O
2) Conversion factors:
1 mol H2O / 6.02 X 1023 molecules H2O
18.02 g H2O / 1 mol H2O
3) Calculations:
12.04 X 1023 molecules H2O x (1 mol H2O / 6.02 X 1023 molecules H2O) x (18.02 g H2O / 1 mol H2O) = 36.04 g
The answer is d.
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- The teacher emphasizes studying purposefully and asks students to focus on following class rules around interfering with learning and leaving seats without permission.
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The document contains information from a chemistry class, including:
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12 14 Balancing Review
The document contains information from a chemistry class, including:
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12 14 Balancing Review1
The document contains information from a chemistry class, including:
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11 16 Is The Equation Balanced
The document contains a lesson on chemical equations and balancing chemical reactions. It provides examples of balanced and unbalanced chemical equations and instructs students to determine if equations are balanced by counting the number of atoms on each side. The lesson emphasizes the principle of conservation of matter and that the number of atoms must be equal on both sides of a balanced chemical equation.
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The document provides information about chemical equations including:
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2. Coefficients indicate the number of units of each element or compound and subscripts indicate the number of atoms in a molecule or compound.
3. Examples show how to use coefficients and subscripts to determine the number of each type of atom in a chemical formula or equation.
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Balancing_Chemical_Equations.pdf
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Chemistry - Chp 12 - Stoichiometry - Notes
When 20.0 g of FeS2 reacts with 16.0 g of O2 according to the balanced equation 4FeS2 + 15O2 → 2Fe2O3 + 8SO3:
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3.5 chemical equations
Chemical equations are used to represent chemical changes. They show the reactants and products of a reaction. A balanced chemical equation has the same number and type of atoms of each element on both sides. To balance an equation, the coefficients of the molecules are adjusted without changing subscripts so that atomic balance is achieved. Balanced equations indicate the mole ratios in which substances react, allowing for calculations relating amounts in moles of reactants and products.
Balancing chemical equation
This document discusses balancing chemical equations. It explains that a chemical equation describes a chemical reaction by showing the reactants and products. It also notes that balancing a chemical equation establishes the quantitative relationship between reactants and products by using coefficients. There are three main steps to balancing an equation: 1) writing the unbalanced equation, 2) balancing the equation by applying the law of conservation of mass so each element has the same number of atoms on both sides, and 3) indicating the states of matter of the reactants and products using abbreviations like (g) or (s). An example problem of balancing the equation for the reaction of tin oxide with hydrogen gas to form tin and water vapor is provided.
Chapter 8.1 : Describing Chemical Reactions
The document describes chemical equations and reactions, including:
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Balancing_Chemical_Equations.pptx
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2) Balancing equations involves adjusting coefficients so that the same number and type of atoms are on both sides of the reaction arrow.
3) Examples show balancing equations by changing coefficients like N2 + H2 → 2NH3 and KClO3 → 2KCl + 3/2O2.
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1 27 Unit #6 Jeopardy
The document provides a lesson plan and materials for a chemistry class reviewing solutions, acids, and bases. It includes:
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This document provides instructions and information for a chemistry class. It notes that the class scored 94% on a recent quiz and that today's objective is to learn about using the pH scale to characterize acids and bases. It includes sample practice questions, instructions to organize binders, and an exit slip quiz on pH concepts.
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The document outlines the agenda for a class on calculating molarity, including having students take their seats, get out homework from the previous day, and states the learning objective is to calculate the concentration of a solution in terms of molarity. It provides no data from the previous day's exit slips. Students are instructed to calculate the concentration of solutions in terms of molarity.
1 11 What Is Concentration
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1 7 What Is Solubility
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1 6 How Are Solutes Dissolved
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12 16 Mole Review Part Ii
I apologize, upon reviewing the document again I do not see enough context to summarize it in 3 sentences or less. The document provides some details about an upcoming exam and homework assignments but does not have an overall topic or theme that can be succinctly summarized. Please provide more context if you would like me to attempt a summary.
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11 17 How Are Equations Balanced
- 1. Launch: 11/17 Grab your binder and immediately take a seat! Place homework (practice questions) on desk. Mg + 2H2O Mg(OH)2 + 2H2 3. Is the above equation balanced? a. No, because the number of H atoms is not equal b. No, because the number of O atoms is not equal c. Yes, because the number of Mg atoms is not equal d. Yes, because there are equal numbers of atoms on both sides of the equation
- 2. Launch 11/17 4. Hydrogen gas (H2) and oxygen gas (O2) react to form water (H2O). Which of these represents a properly balanced equation for this reaction? a. H2 + O2 H2O b. 2H2 + 2O2 2H2O c. 2H2 + O2 2H2O d. H2 + 2O2 2H2O
- 3. Announcements Unit #5: Chemical Reactions & Stoichiometry Class points end next week Get your HW done Be respectful Quiz on Friday!
- 4. How are equations balanced? Mr. Heffner 11/17/09
- 5. Review Is the below equation balanced? Cl2 + H2 2HCl 2 X Cl 2XH 2XH 2 X Cl Multiply the coefficients and subscripts Yes, it is balanced!
- 6. How are equations balanced? There is a 3-step process for balancing equations: 1. Count the atoms on both sides 2. Change a coefficient & multiply out 3. Check tally (and repeat) Only one rule: you can’t change the subscripts
- 7. Example #1 Balance the equation, H2 + O2 H2O
- 8. Example #1 Step #1: Count the atoms on both sides H2 + O2 H2O 2XH 2XO 2XH 1XO Unbalanced!it underneath Put
- 9. Example #1 Step #2: Change coefficient & multiply out H2 + O2 H2O 2 2XH 2XO ( 2XH 2X 1 X O ) 4XH 2XO
- 10. Example #1 Step #3: Check tally (and repeat) 2H2 + O2 H2O 2 ( 2X 2 X H ) 2XO 4XH 2XO 4XH Still unbalanced!
- 11. Example #1 Step #3: Check tally (and repeat) 2H2 + O2 H2O 2 4XH 2XO 4XH 2XO Balanced!
- 12. Activity Make the equation on your whiteboard Balance the equation with the 3-step method 1. Count the atoms 2. Add a coefficient 3. Check tally (and repeat) Raise hand to get checked off Record reactants, products, and final balanced equation on worksheet
- 13. Practice Questions Practice question handouts
- 14. Exit Slip 1. In a balanced equation a. the reactant atoms are destroyed to make the products b. the number of atoms on the reactants side of the equation is equal to the number of atoms on the products c. the number of atoms on the reactants side of the equation is not equal to the number of atoms on the products d. all compounds have a coefficient of at least 2
- 15. Exit Slip H2 + Cl2 HCl 2. Is the above equation balanced? a. No, because H2 has a coefficient of 2. b. No, because the number of Cl and H atoms are not equal. c. Yes, because the number of H atoms is not equal on both sides. d. Yes, because there are equal numbers of atoms on both sides of the equation.
- 16. Exit Slip 4K + O2 2K2O 3. In the already balanced reaction above, what is the coefficient for O2? a. 0 b. 1 c. 2 d. 4
- 17. Exit Slip __C + __Br2 __CBr4 4. When the equation above is completely balanced, the coefficient for Br2 would be a. 0 b. 1 c. 2 d. 4
- 18. Exit Slip __Na + __H2O __NaOH + __H2 5. When the above equation is completely balanced, the coefficient for NaOH would be a. 1. b. 2. c. 4. d. 8.
- 19. Homework Finish practice question worksheet