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Chemistry Exam 2

by: Katerina Kushla

Chemistry Exam 2 CH 1213

Katerina Kushla
GPA 4.0

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Exam 2 is Oct. 4, 2016 Covers Chapter 4 Good Luck All!
Chemistry 1
Dr. Eric Van Dornshuld
Study Guide
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This 16 page Study Guide was uploaded by Katerina Kushla on Friday September 30, 2016. The Study Guide belongs to CH 1213 at Mississippi State University taught by Dr. Eric Van Dornshuld in Fall 2016. Since its upload, it has received 38 views. For similar materials see Chemistry 1 in Chemistry at Mississippi State University.


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Date Created: 09/30/16
Exam 2 Study Guide Oct. 4, 2016 Covers Chapter 4: Stoichiometry of Chemical Reactions: 16 Multiple Choice – 4pts apiece Show Work – 36 points total Section Number – 2 pts Balancing Equations - Balancing Equations with polyatomic ions Writing Equations - Molecular Equations - Ionic Equations - Net Ionic Equations - States of matter – Solubility Rules (MEMORIZE) Chemical Reactions (MEMORIZE) - Precipitation Reaction - Acid-Base Reactions o Neutralization Reaction o Strong/Weak Acids and Bases (MEMORIZE) - Redox o Disproportionation Reaction o Oxidation  Half-Reactions o Activity Series - Combination Reaction - Decomposition Reaction - Combustion Reaction Stoichiometry - Limiting and Excess Reactants - Theoretical, Actual, and Percent Yields Quantitative Chemical Analysis - Titration - Gravimetric Analysis Exam 2 Study Guide Oct. 4, 2016 Balancing Equations: Practice * Remember to treat polyatomic ions as a single unit if it appears on both sides of equation SnO +2H → S2 + H O 2 SnO +22H → Sn2+ 2H O 2 KOH + H PO → K PO + H O 3 4 3 4 2 3KOH + H PO →3K PO4+ 3H O3 4 2 (NH )4 3 + P4(NO ) → Pb 3 4 ) + NH 3O 4 4 4 3 4(NH ) 4 3+ 34b(NO ) → Pb (3 4) + 12N3 NO 4 4 4 3 Writing Equations: Chemical Equations Things to remember: - Naming Compounds (Chapter 2) - Solubility Rules MEMORIZE ) Exam 2 Study Guide Oct. 4, 2016 Soluble compounds are aqueous Insoluble compounds are solid; and will precipitate out of a solution Water is always liquid Practice: When lithium hydroxide pellets are added to a solution of sulfuric acid, lithium sulfate and water are formed. Chemical Equation: LiOH(s) + H SO 2aq)4 LiSO (aq) + H4O(l) 2 Balanced Chemical Equation: LiOH(s) + H SO 2aq)4 LiSO (aq) + 24 O(l) 2 In a precipitation reaction, sodium hydroxide solution is mixed with iron(II) chloride solution. Sodium Chloride solution and insoluble iron(II) hydroxide are produced. Write a balanced chemical equation including the state symbols. Chemical Equation: NaOH(aq) + FeCl (aq) 2 NaCl(aq) + Fe(OH) (s) 2 Balanced Chemical Equation: 2NaOH(aq) + FeCl (aq) 2 2NaCl(aq) + Fe(OH) (s) 2 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Ionic Equations: - Balancing Equations - Determining charge of Ions Molecular Equation – the standard chemical equation for a given reaction; this form does not explicitly represent the ionic species that are present in a solution CaCl (aq) + 2AgNO (aq)  Ca(NO ) (aq) + 2AgCl(s) 2 3 3 2 Ionic Equation aka “Complete Ionic Equation” – contains all species given in the molecular equation but explicitly represents all the dissolved ions Ca (aq) + 2Cl (aq) + 2Ag (aq) + 2NO (aq) 3 Ca (aq) + 2NO (aq) + 2AgC3(s) Exam 2 Study Guide Oct. 4, 2016 Net Ionic Equation – a chemical equation that omits spectator ions in an ionic equation Spectator Ions – ions whose presence is required to maintain charge neutrality; they are neither chemically nor physically changed by the reaction; they appear on both sides of an ionic equation 2Cl (aq) + 2Ag (aq)  2AgCl(s) Cl (aq) + Ag (aq)  AgCl(s) Types of Chemical Reactions 1. Precipitation Reaction One in which dissolved substances react to from on (or more) solid products (known as a “precipitate”) * sometimes called double displacement, double replacement, or metathesis reaction 2KI(aq) + Pb(NO ) (aq)  PBI (s) + 2KNO (aq) 3 2 2 3 2. Acid-base reactions Reaction in which a hydrogen ion (H ) is transferred from one chemical species to another MEMORIZE STRONG/WEAK ACIDS AND BASES Acid – in the context of aqueous solutions, is an aqueous species that donates a proton to water to yield hydronium ions (H O )3 + Strong Acid – acids that dissociate completely when dissolved in water Weak Acid – acids that slightly dissociates when dissolved in water Base – a substance that will dissolve in water to yield hydroxide ions (OH ) - Strong Base – bases that dissociate completely in water Exam 2 Study Guide Oct. 4, 2016 Weak Base – bases that slightly dissociate in water Neutralization Reaction – acid reacts with base and produces a salt and water (neither reactant is the water itself) acid + base  salt + H O 2 2HCl(aq) + Mg(OH) (s2  MgCl (aq)2+ 2H O(l)2 Exam 2 Study Guide Oct. 4, 2016 Exam 2 Study Guide Oct. 4, 2016 Exam 2 Study Guide Oct. 4, 2016 Redox Reactions Oxidation-Reduction Reactions Reaction involving a change in oxidation number for one or more reactant element 2Na(s) + Cl (2)  2NaCl(s) Half Reaction – an equation that shows whether each reactant gains or loses electrons 2Na(s) 2Na + 2e + - - - Cl 2g) + 2e  2Cl(s) Oxidation and Reduction Terminology Oxidation – “loss of electrons” Oxidized – the process of losing electrons Reduction – “gain of electrons” Reduced – the process of gaining electrons OIL RIG – “Oxidation Is Losing, Reduction Is Gaining” Oxidizing Agent – Species that is reduced (because it oxidizes something else) Reducing Agent – Species that is oxidized (because it reduces something else) Critical Info: for something to undergo oxidation, something else must undergo reduction (and vice-versa) Disproportionation Reaction – a reaction in which an element is both oxidized and reduced 2H O222H O + O2 2 Oxidation Number a.k.a Oxidation State – the charge each atom of an element would have in a compound if the compound were ionic Exam 2 Study Guide Oct. 4, 2016 MEMORIZE – Oxidation Numbers 1. The oxidation number of an atom in an elemental substance is zero 2. The oxidation number of a monatomic ion is equal to the ion’s charge 3. Rules higher in the table overrule those lower in the table 4. The sum of oxidation numbers for all atoms in a molecule or polyatomic ion equals the charge on the whole species Exam 2 Study Guide Oct. 4, 2016 Activity Series IMPORTANT TIP: In this type of equation, the solid metal reactant will always be the one undergoing oxidation, therefore, the dissolved metal would have to undergo reduction Zn(s) + CuCl (2q)  ZnCl (aq) 2 Cu(s) In this equation, Zinc was oxidized and Copper was oxidized The reverse reaction however, would not occur. Cu(s) + ZnCl (2q)  CuCl (aq)2+ Zn(s) We know this by referencing the activity series chart. * This chart does not have to be memorized, it will be provided on the exam The activity series is a list of metals (and hydrogen) arranged from bottom to top in order of increasing ease of oxidation So, if given a choice between Gold or Lithium being oxidized, Lithium would oxidize An element in the series will be oxidized by the ions of any element that appears below it Elements will not be oxidized by elements that appear above it. Exam 2 Study Guide Oct. 4, 2016 Chemical Reaction Names Combination Reaction – two reactants form one product NH (3) + HCl(g)  NH Cl(s) 4 Decomposition Reaction – a reactant decomposes into two or more products  CaCO (s3  CaO(s) + CO (g) 2 Combustion Reaction – a reaction where a substance containing C, H, and (sometimes) O is burned in the presence of oxygen to produce only C2 and H2O CH O2l) + O (g)2 CO (g) + H2O(l) 2 Stoichiometry Stoichiometry – relationships/ratios/conversion factors between amounts of reactant and products in a chemical reaction 1. Balance the chemical equation 2. Convert mass to moles 3. Implement appropriate stoichiometric relationship from equation 4. Convert moles to mass Exam 2 Study Guide Oct. 4, 2016 Limiting Reactant – the reactant that limits the maximum number of product you can make; it is the substance that is to be completely consumed first; once a reaction is completely consumed the reaction stops. Excess Reactant – any reactant that is not the limiting reactant; will be present after a reaction stops Theoretical Yield – the amount of product that may be produced by a reaction under specific conditions as calculated per the stoichiometry of a balanced chemical equation Actual Yield – the amount of product actually (less that the theoretical yield) Percent Yield – the extent to which a reaction’s theoretical yield is achieved; unit-less quantity if actual and theoretical yields are expressed using the same units Limiting Reactant Example 1. Convert mass to moles 2. Determine product yields; identify limiting reactant 3. Calculate product mass 4. Find how much excess reactant is consumed and leftover 5. Convert mass to moles Exam 2 Study Guide Oct. 4, 2016 If you had 2.00 g Fe and 1.55 g O , 2ow much product could you make (in g) according to the following chemical equation? How much excess reactant (in g) would be left over? Limiting Reactant (Fe) Excess Reactant (0.672g O ) 2 Fe2O 32.87g) Exam 2 Study Guide Oct. 4, 2016 Percent Yield Example 3.45 g Fe and excess O w2re reacted and produced 4.12 g of Fe O . Wha2 i3 the percent yield? Actual Yield (4.12 g) Theoretical Yield (4.95 g) Percent Yield (83.23%) Summary of Steps: 1. Convert mass to moles 2. Determine theoretical yield (in g); use stoichiometric coefficients 3. Find percent yield Quantitative Analysis Quantitative Analysis – the determination of the amount or concentration of a substance in a sample Titration Analysis – quantitative chemical analysis method that involves measuring the volume of a reactant solution required to completely react with the analyte in a sample Exam 2 Study Guide Oct. 4, 2016 Titration Example Titration of a 20.0 mL sample of acid rain required 1.7 mL of 0.0811 M NaOH to reach the end point. The acidity of the rain is due to the presence of sulfuric acid. What was the concentration of the sulfuric acid in this sample of rain? Summary of Steps: 1. Convert volume of NaOH to mmol NaOH 2. Determine mmol H SO f2om 4toichiometric relationship 3. Calculate concentration of H SO2(mo4arity equation) Exam 2 Study Guide Oct. 4, 2016 Gravimetric Analysis – a technique where a sample is subjected to some treatment that causes a change in the physical state of the analyte that permits its separation from other components in the sample Gravimetric Analysis Example A sample of gallium bromide weighing 0.165 was dissolved in water and treated with silver nitrate resulting in the precipitation of 0.299 g AgBr. Find the percent of Ga (by mass) in the GaBr 3ample. Summary of Steps: 1. Write out a balanced chemical equation 2. Convert mass of AgBr to moles 3. Find moles of GaBr f3om stoichiometric relationship 4. Convert moles to mass 5. Find mass percent


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