Exam Review Sheet #1 Balancing Equations and Simple Stoichiometry Answers are provided on the second sheet. Please try to do the worksheet without referring to them‚ because you’ll be expected to know this stuff the first day of school! Balance the following equations: 1) ___ N2 + ___ F2 ( ___ NF3 2) ___ C6H10 + ___ O2 ( ___ CO2 + ___ H2O 3) ___ HBr + ___ KHCO3 ( ___ H2O + ___ KBr + ___ CO2 4) ___ GaBr3 + ___ Na2SO3 ( ___ Ga2(SO3)3 + ___ NaBr 5) ___ SnO + ___ NF3 ( ___ SnF2
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Data Table: Record your observations in the following table Na3PO4 NaI Na2SO4 NaCl NaHCO3 Na2CO3 NaOH Co(NO3)2 Precipitated a blue color NR NR NR Turned a milky purple color Precipitated a light purple color Precipitated a light blue color Cu(NO3)2 Precipitated a light blue color Precipitated a dark amber color NR NR Precipitated a light blue color Precipitated a light green color Precipitated a light blue color Fe(NO3)3 Precipitated a light milky yellow color Turned a dark black
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Mass % Determination of empirical formula and molecular formula Combustion analysis Balancing equations Chapter 4 Limiting reactant Mole to mole conversion from reaction stoichiometry Theoretical yield‚ Percent yield‚ Actual yield Solution concentration (molarity) M = n/V ‚ V always in L M1V1=M2V2 ( dilution calculations) Stoichiometry of reactions in solutions M1V1=M2V2 ( dilution calculations) Ionic reactions (formula unit equation‚ complete ionic and net ionic equation). Solubility rule Types of
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in action – delicate silver crystals begin to grow on the wire surface and the colour of copper(II) ions gradually appears in the solution. Stoichiometry is the area of chemistry that deals with the numerical relationships and mathematical proportions of reactants and products in a chemical reaction. One of the most important lessons of stoichiometry is that the amounts of reactants and products in a chemical reaction are related to one another on a mole basis. Chemical reactions are normally
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Questions: B. Yes C. Carbonates‚ Phosphates‚ sulfides‚ oxides or hydroxides form precipitates. Alkali metal hydroxides are the exceptions. D. Most salts. E. Group one and two cations F. A1. Cobalt nitrate with Sodium phosphate. CO(NO3)2+6 H2O+Na3PO4+12 H2O Na3+PO2Co(NO3)2+18 H2OD5. Ba(N03)2 - NaHC03 A7. Cobalt nitrate with sodium hydroxide Co(N03)2 - 6 H20 – NaOH B6. Copper nitrate with sodium carbonate Cu(N03)2 - 3 H20 - Na2C03 C1. Iron nitrate with sodium phosphate Fe(N03)3 - 9
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-A solid seems to form 6 Copper(II) hydroxide à copper(II) oxide + water Cu(OH)2 à CuO + H2O Decomposition -The liquid turns black 10 Copper(II) oxide + hydrochloric acid à copper(II) chloride + water CuO + 2 HCl à CuCl2 + H2O Double Replacement -The liquid clears up and turns blue 11 Copper(II) chloride + aluminum à copper + aluminum 3CuCl2 + 2Al à 3 Cu + 2 AlCl3 Single Replacement -When the aluminum wire was added‚ the liquid began
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Pre-requisite: None 4. Co-requisite: None 5. Credit: 2 units 6. Course Description: A course designed for freshmen engineering students‚ it covers the structure of the atom‚ writing chemical formulas‚ naming of compounds and stoichiometry involving chemical formulas and reactions. 7. Student Outcomes and Relationship to Program Educational Objectives |Student Outcomes |Program Educational
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When phosphoric acid (a trihydrogen acid) is neutralized with a base‚ the hydrogens are neutralised in 3 steps: STEP 1: H3P04 + NaOH > NaH2PO4 + H2O gives X amount of heat STEP 2: NaH2PO4 + NaOH > Na2HPO4 + H2O gives Y amount of heat STEP 3: Na2PO4 + NaOH > Na3PO4 +H2O gives Z amount of heat Therefore‚ the total heat of reaction ( Hrxn) is equal to: There are two (2) ways to calculate the heat of reaction using Hess’ Law: 1 Equation Method (Algebraic Method) 2 Heat of Formation Method (Summation Method)
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Acid-Base Titration Background Information A titration is a controlled addition of one substance into another substance. In an acid-base titration‚ the experimenter will add a base of known concentration to an acid of unknown concentration (or vice-versa). The goal of the titration is usually to use the substance of known concentration to determine the concentration of the other substance. In order to run a titration‚ the following materials are needed: • A buret filled with the base (or acid) of
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The first flask was placed under a buret. Finally we titrated the KHP by adding the base until the end point was reached‚ which was when it turned pink completely. We repeated this twice and then cleaned up. The procedure does not include very difficult math‚ however the calculations did. The harder math calculations included finding moles of the acid‚ moles of the base that was used to neutralize‚ and the molarity of the base. After all calculations‚ below is what we concluded. The molarity you
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