Margaret E. Vorndam‚ M.S. Version 42-0038-00-01 Lab Report Assistant This document is not meant to be a substitute for a formal laboratory report. The Lab Report Assistant is simply a summary of the experiment’s questions‚ diagrams if needed‚ and data tables that should be addressed in a formal lab report. The intent is to facilitate students’ writing of lab reports by providing this information in an editable
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Chapter 3: Stoichiometry 3: Stoichiometry 5: Thermochemistry 8: Covalent Bonding and Molecular Structure 15: Chemical Equilibrium 16: Acids and Bases 3.2 Stoichiometry and Compound Formulas 3.1 The Mole and Molar Mass 3.2 Stoichiometry and Compound Formulas 3.3 Stoichiometry and Chemical Reactions 3.4 Stoichiometry and Limiting Reactants 3.5 Chemical Analysis Chapter Summary Chapter Summary Assignment Reference Tools Periodic Table Molarity Calculator Molar Mass Calculator Unit
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1 Measuring and calculating equilibrium constants Clearly‚ if the concentrations or pressures of all the components of a reaction are known‚ then the value of K can be found by simple substitution. Observing individual concentrations or partial pressures directly may be not always be practical‚ however. If one of the components is colored‚ the extent to which it absorbs light of an appropriate wavelength may serve as an index of its concentration. Pressure measurements are ordinarily able to measure
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Determining the Stoichiometry of Chemical Reactions Mrs. Farrales Nikita Pandya October 23‚ 2012 December 3‚ 2012 INRODUCTION In the method of continuous variations the total number of moles of reactants is kept constant for the series of measurements. Each measurement is made with a different mole ratio of reactants. A mole ratio
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Na2CO3(s)+CO2(g)+H2O(g). The was the only equation that matched up exactly with my data in terms of percentage. To start off with‚ when I balanced out the equation‚ I got 2 NaHCO3→ 1 Na2CO3(s)+ 1 CO2(g)+ 1 H2O(g). Therefore when I set up my stoichiometry problem I got 3.2 grams NaHCO3 over 1 x 1 mol NaHCO3 over 84.007g NaHCO3 x 1mol Na2CO3 over 2 mol NaHCO3 x 105.987g Na2CO3 x 1 mol Na2CO3. Hence‚ I multiplied 3.2 x 1 x 1 x 105.987 and got 339.1584. Afterwards‚ I divided 339.1584 by 84.007 and
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4‚5‚ and 6 Study Guide Chapter 4 - Chemical Quantities and Aqueous Reactions * Reactions Stoichiometry * mole-mole conversions * mass-mass conversions * Limiting Reactants * What is the Limiting Reagent * How do we find the L.R. * Solutions * Molarity - definition and how to calculate * Dilutions Calculations (M1V1 = M2V2‚ careful with M2) * Solution Stoichiometry * volume-volume conversions * volume-mass conversions
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Basic Stoichiometry PhET Lab rvsd 2/2011 Let’s make some sandviches! _ Introduction: When we bake/cook something‚ we use a specific amount of each ingredient. Imagine if you made a batch of cookies and used way too many eggs‚ or not enough sugar. YUCK! In chemistry‚ reactions proceed with very specific recipes. The study of these recipes is stoichiometry. When the reactants are present in the correct amounts‚ the reaction will produce products. What happens if there are more or less of
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Lab 3: Stoichiometry of a Precipitation Reaction NOTE: All photos should be taken so the reading on the electronic balance is readable. Photo 1: filter paper being weighed on electronic balance‚ along with your student information card. Photo 2: beaker with precipitate slurry in it (after step 5) ‚ along with your student information card Photo 3: dried precipitate/filter paper being weighed on electronic balance‚ along with your student information card Additional Question Guidelines:
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Lab Report for Experiment #10 Stoichiometry of a Precipitation Reaction Student’s Name ____________________ Date of Experiment ___________ Date Report Submitted _________________ Title: Purpose: Instructor Changes: Weigh out about 1.7 g of CaCl2·2H2O and record your mass to +/- 0.1 g (for example 1.6 g‚ 1.7 g‚ or 1.8 g). We have made this change so that you will have 2 sig figs in subsequent calculations. Have you made any changes to the procedure? Please explain: Data Tables and Observation:
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AP Chemistry Unit 2 Notes Stoichiometry You should understand all that is presented in chapter 3 of your text (Zumdahl: Chemistry‚ 8th edition). Some of the highlights are presented below. Atomic Masses (Section 3.1) Nearly every element is made up of atoms of more than one isotope for that element. A few‚ like Be‚ only have one isotope. Others can have a large number of isotopes. Tin (Sn) has ten isotopes. (No pun intended.) Isotopic abundance is determined by the use of
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