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Lab #4: Chemical Reactions
Lab #4: Chemical Reactions

... PART E-3 Mercury (II) Nitrate + Sodium Phosphate ...
File - Flipped Out Science with Mrs. Thomas!
File - Flipped Out Science with Mrs. Thomas!

... used to classify elements 2.2 Chemical Equations 5D recognize that chemical formulas are used to identify substances and determine the number of atoms of each element in chemical formulas containing substances 5F recognize whether a chemical equation containing coefficients is balanced or not and ho ...
File - Flipped Out Science with Mrs. Thomas!
File - Flipped Out Science with Mrs. Thomas!

... used to classify elements 2.2 Chemical Equations 5D recognize that chemical formulas are used to identify substances and determine the number of atoms of each element in chemical formulas containing substances 5F recognize whether a chemical equation containing coefficients is balanced or not and ho ...
C6_rev - boswellsrcd
C6_rev - boswellsrcd

... (eg could get too hot if exothermic; gas could be produced to quickly and pressure build up) If it is too slow, then product would be made too slowly, and yield low, so profit too low. (economic factors) ...
Spring 2001 Key
Spring 2001 Key

AP Chem Summer Assign Gen Chem Rev Problems
AP Chem Summer Assign Gen Chem Rev Problems

... For each of the following problems, identify the pH, pOH, AND tell if the solution is acidic, basic, or neutral: l. [OH-] = 7.6x10-10 M m. [H+] = 1.3x10-7 M n. [OH-] = 10-3 M o. [H+] = 10-12 M For a-o above, indicate the result of a litmus paper test. ...
CP-Chem Ch 3 PowerPoint(Atomic Theory
CP-Chem Ch 3 PowerPoint(Atomic Theory

... sample • Ex: Salt is NaCl, • it’s composition is always 39.4 % Na & 60.6% Cl ...
Reaction Systems Engineering II (part 1)
Reaction Systems Engineering II (part 1)

... rH° = (–230.0) – (–285.8) = 55.8 kJ mol–1. rS° = –10.8 – 69.9 = –80.7 J K–1 mol–1. rG° = rH° – TrS° = 55.8 – 298(–80.7) / 1000 = 79.85 kJ mol–1 Kw = exp(–rG° / RT) = exp[–79.851000 / (8.3145298)] = 1.0110–14 mol2 kg–2 2) rG° = rH° – TrS° = 55.8 – 348(–80.7) / 1000 = 83.88 kJ mol–1 Kw = ...
practice spring final exam
practice spring final exam

... 66. How many moles of iron, Fe, are produced with 25.0 grams of magnesium, Mg? 3 Mg + 1 Fe2O3  2 Fe + 3 MgO (A) 0.686 moles (B) 3 moles (C) 1.5 moles (D) 6 moles 67. Acetylene gas (C2H2) is produced as a result of the following reaction. CaC2(s) + 2H2O(l)  C2H2(g) + Ca(OH)2(aq) If 3.2 moles of CaC ...
handout 4
handout 4

Chemistry! - Duplin County Schools
Chemistry! - Duplin County Schools

... • A chemical equation shows what happens during a chemical reaction • There is a reactant, an arrow, and a product in every chemical equation (RAP) • It is important for you to know if chemical equations are balanced or not ...
Energetics II - Miller, Jonathan
Energetics II - Miller, Jonathan

All That Matters - Teach-n-Learn-Chem
All That Matters - Teach-n-Learn-Chem

... 3. Atoms can’t e created or destroyed by chemical reaction or changed into other kinds of atoms. 4. Compounds are formed by combining atoms of more than one kind of element, in the same ratio each time. ...
Chapter 10 Notes
Chapter 10 Notes

... 10.4 Gas Stoichiometry Summary of Gravimetric, Gas & Solution Stoichiometry STEP 1: Write a balanced chemical equation and list the measurements and conversion factors for the given substance and the one to be calculated. STEP 2: Convert the measurement to an amount in moles using the appropriate co ...
Writing Chemical Formulas and Chemical Reactions
Writing Chemical Formulas and Chemical Reactions

Balancing Chemical Equations Activity by Liz LaRosa www
Balancing Chemical Equations Activity by Liz LaRosa www

... few days. Once done, you can laminate them and have them forever! The materials account for one complete set which is good for 2-3 students to use. Print activity cards on card stock instead of making index cards for quicker set up. The color coding is very important for visualization. It is easier ...
Practice Final Exam, Chemistry 2220, Organic Chem II 1. Rank the
Practice Final Exam, Chemistry 2220, Organic Chem II 1. Rank the

The Mole - Solon City Schools
The Mole - Solon City Schools

... Ex – P4O6 C6H9 CH2OHCH2OH BrCl2 ...
Question Paper - Revision Science
Question Paper - Revision Science

Final Exam Review
Final Exam Review

... b. The molecules become separated to relatively great distances during evaporation. c. The molecules of vapor have a different chemical composition from those in the liquid. d. The density of the vapor is less than that of the liquid. e. The vapor is much more compressible than the liquid. Once vapo ...
Unit 2 Test Review - Liberty High School
Unit 2 Test Review - Liberty High School

1 mol H 2
1 mol H 2

... 2K(s) + 2H2O(l) 2KOH(aq) + H2(g)  Step 2: Identify the known/given and the unknown in the problem. 0.04 mol K is given; mol H2 is unknown. ...
The Cool Balancing Chemical Reactions Presentation
The Cool Balancing Chemical Reactions Presentation

... Yes! The law of conservation of mass was established in 1789 by French Chemist Antoine Lavoisier. The law states that matter cannot be destroyed or created in any ordinary chemical reaction. This simply means that the mass of the reactants must be equal to the mass of the product. This is the reaso ...
Organometallic Compounds and Catalysis: Synthesis
Organometallic Compounds and Catalysis: Synthesis

Calculations from Balanced Equations
Calculations from Balanced Equations

... You can use the relative numbers of moles of substances, as shown in balanced equations, to calculate the amounts of reactants needed or the amounts of products produced. A limiting reactant is the substance that is fully used up and thereby limits the possible extent of the reaction. Other reactant ...
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Stoichiometry



Stoichiometry /ˌstɔɪkiˈɒmɨtri/ is the calculation of relative quantities of reactants and products in chemical reactions.Stoichiometry is founded on the law of conservation of mass where the total mass of the reactants equals the total mass of the products leading to the insight that the relations among quantities of reactants and products typically form a ratio of positive integers. This means that if the amounts of the separate reactants are known, then the amount of the product can be calculated. Conversely, if one reactant has a known quantity and the quantity of product can be empirically determined, then the amount of the other reactants can also be calculated.As seen in the image to the right, where the balanced equation is:CH4 + 2 O2 → CO2 + 2 H2O.Here, one molecule of methane reacts with two molecules of oxygen gas to yield one molecule of carbon dioxide and two molecules of water. Stoichiometry measures these quantitative relationships, and is used to determine the amount of products/reactants that are produced/needed in a given reaction. Describing the quantitative relationships among substances as they participate in chemical reactions is known as reaction stoichiometry. In the example above, reaction stoichiometry measures the relationship between the methane and oxygen as they react to form carbon dioxide and water.Because of the well known relationship of moles to atomic weights, the ratios that are arrived at by stoichiometry can be used to determine quantities by weight in a reaction described by a balanced equation. This is called composition stoichiometry.Gas stoichiometry deals with reactions involving gases, where the gases are at a known temperature, pressure, and volume and can be assumed to be ideal gases. For gases, the volume ratio is ideally the same by the ideal gas law, but the mass ratio of a single reaction has to be calculated from the molecular masses of the reactants and products. In practice, due to the existence of isotopes, molar masses are used instead when calculating the mass ratio.
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