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Get Full Access to C of C - MATH 111 - Study Guide - Midterm
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1 Note Taker= Lacy Jones Professor/ institution=Dr. Lavrich-CofC Book= Atoms First 2nd edition by Julia Burdge and Jason Overby Chem 111Test III Study Guide Covers chapters 7 8 & 9 Key: * = This means that Dr. Lavrich Stressed this during class, or the book made note of it. OR *=excited state → This arrow means furthering information of the statement () This means that there is an AKA form or side note, or there is a source. My notes contain images from sources other than the Atoms First book, so I have to cite. This color means book topic Headlines This color is for definitions This color is for example problems This color with a high light is the answer2 This color means that the notes come from lecture and what day This color means Lecture Headlines This color emphasizes such as valence electrons This color is for the very end of the notes pages and covers key tables the professor wants us to memorize, constants and Equations, and also difficult concepts. = 1s^2 fully occupied = 1s^1 half occupied --------------------------------------------------------------------------------------------------------------------- Chapter 7 Bonding Theory and Molecular Geometry (3d Shape) 1. Valence Shell e- pair repulsion (VSEPR) AKA Lewis Theory~ -E- pairs repel one another - Non bonding (lone pair) -Shared pair (Bonds)  : H-O-H  : E- Domain: Pair of e Ex. CO2  : :  O=C=O  : : 2 e- Domains, no matter what the bond type is it counts as a single domain. : : :3 O=O-O: : : 3 Domains, → 2 bonds (single double) 1 lone pair Arrangement of e- domains~   For Trigonal Bipyramidal There is axial bonds by 90 degrees, and Equatorial bonds by 120 degrees4 Procedure for these arrangements~ 1. Draw Lewis Structure 2. Determine the number of e- domains on central atom 3. Determine the electron domain geometry with VSEPR → Lone pair be as far away as possible 4. Determine the molecular geometry → Based on the positions of the bonded atoms, not on lone pairs (Neglecting them) Ex: System Examples # of domains E- Domain Geometry Molecular Geometry AB3 3 Trigonal Planar Trigonal Planar AB2 1 lone pair 3 Trigonal Planar Bent (Because linear but another electron domain from the lone pair pushes the bonds down) AB4 4 Tetrahedral Tetrahedral (AB3) 3 bonds, 1 lone pair 4 Tetrahedral Trigonal Pyramid. Angles are Less than 109.5 (AB2) 2 bonds, 2 lone pairs 4 Tetrahedral Bent

60 moles of CO(g) reacts with excess of O2, how much can be produced?

5 AB5 all bonded 5 Trigonal Bipyramid Trigonal Bipyramid AB4 4 bonded, 1 lone pair 5 Trigonal Bipyramid See-saw AB3 3 Bonds and 2 lone Pairs 5 Trigonal Bipyramid T-Shape AB2 2 bonds 3 lone pairs 5 Trigonal Bipyramid Linear AB6 all bonded 6 Octahedral Octahedral AB5 5 bonds 1 lone pair 6 Octahedral Square Pyramid AB4 4 bonds 2 lone pairs 6 Octahedral Square Planar

NO3-/ ClO3-/ C2H3O2-

Halogens: cl-, Br-, I- Ag+/ Hg2 +2/ Pb+2 when paired with these SO4 -2 Ca+2/ Sr+2/ Ba+2 when paired with these

How many moles of H and N2 are used?

Don't forget about the age old question of fin 305

Insoluble Exceptions CO3 2-/ PO4 -3/ CrO4 -2/ S- 2 Group 1 cations/ NH4 +/ OH- Group 1 cations, NH4 +/ Ba +2

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Don't forget about the age old question of dna replication lecture notes

39 Steps Debrief: 1. Write and balance the molecular equation 2. Write the ionic → separate strong electrolytes into constituent ions 3. Cancel Spectators to produce net ionic equations ** which is what we want 9.22 a. Na2S(aq) + ZnCl2(aq) → NaCl(aq) +ZnS(s) = molecular equation 2Na+(aq) + S-2(aq) +Zn+2(aq)+2Cl-(aq) → Na+(aq)+Cl-(aq)+ ZnS(s)=ionic S-2(aq) +Zn+2(aq) → ZnS(s)= net ionic B. K3PO4(aq) +Sr(NO3)2(aq) → KNO3(aq) + Sr(PO4)2(s) =unbalanced 2K3PO4(aq) +3Sr(NO3)2(aq) → 6KNO3(aq) + Sr(PO4)2(s) =balanced molecular 6K+ (aq) +2PO4(aq) +3Sr+2 6NO3-(aq) → 6K+ (aq) + 6NO3-(aq) + Sr(PO4)2(s) =Ionic 2PO4(aq) +3Sr(aq) → Sr(PO4)2(s) =net Ionic C. Mg(NO3)2(aq) + NaOH(aq) → Mg(OH)2 (s) + NaNO3 (aq)= unbalanced Mg(NO3)2(aq) + 2NaOH(aq) → Mg(OH)2 (s) + 2NaNO3 (aq)= balanced molecular Mg+2(aq)+2NO3(aq) + 2Na+(aq)+2OH-(aq) → Mg(OH)2 (s) + 2Na+(aq)+2NO3- (aq)= Ionic Mg+2(aq)2OH-(aq) → Mg(OH)2 (s)= net Ionic Acid Base Reactions:40 Chapter 5 definition: The acid produces H3O+ when added to water Bronsted: Acid: Proton donor Base: Proton Acceptor HCl+H2O → H3O+ +Cl HCl= acid because becomes cl- loses proton H2O base because gains proton in H3O+ NH3+H2O → NH4+ + OH H2O= Proton donor/ base cause H2O → OH NH3= proton acceptor/ acid cause goes to NH4+ Monoprotic acid: Acid with 1 proton to donate→ 1 H to donate. Eg. HCl HNO3 Diprotic acid: Acid with 2 Protons to donate→ 2 H to donate. Eg. H2SO4 H2SO4 + H2O → H3O+ + HSO4- =strong acid HSO4- + H2O ←> H3O+ + SO4-2 = weak acid Triprotic acid: Acid with 3 Protons to donate→ 3 H to donate. Eg. H3PO4 H3PO4 + H2O ←> H2PO4- + H3O+ H2PO4- + H2O ←> HPO4-2 + H3O+ Acid Base Neutralization Acid + Base → Salt + H2O HCl(aq) +NaOH (aq) → NaCl(Aq) +H2O (l) H+(aq) + Cl-(aq_+Na+(aq) +OH-(aq0 → Na+(aq) +Cl-(aq) +H2O (l) H+(aq) +OH-(aq) → +H2O (l) HCN (aq) + NAOH (aq) → NACN (aq) +H2O(l) HCN (aq) + Na+ + OH- (aq) → Na+ + CN- (aq) +H2O(l) HCN (aq) + OH- (aq) → CN- (aq) +H2O(l)41 3. Oxidation Reduction -Transfer of e Oxidation→ Losing an e- (donates) Reduction → Gaining an e- (Receives) Oxidation Number: -Assigned to elements in reactants and products -Keeps track of e- *The increase in oxidation number, this process means the element is being oxidized *The decrease in oxidation number in a particular element, this element is reduced OIL RIG → Oxidation is loss Reduction is Gain H2(g) + Cl2(g) → 2HCl (g) H: 0 → +1 Cl: 0 → -1 Assigning Oxidation Numbers: 1. Free elements~ Oxidation number= 0 eg. Na, Be, H2, O2 2. Monatomic ions (single atom ions)~ Oxidation number= charge eg. Li+1=+1, Ba+2=+2, Fe+3=+3 3. Oxygen atoms in compounds~ Oxidation number= -2 eg. MgO= -2 H2O=-2 Exceptions: O2 -2(peroxide) =-1 O2-(superoxide)= -½ 4. Hydrogen atoms and compounds~ Oxidation number= +1 eg. HCl=+1 H2O=+1 Exception: Binary compounds with metals eg. LiH NaH CaH2 = oxidation number is -1 5. Fluorine atoms in compounds~ oxidation number=-1 ALWAYS because most electronegative (Cl, Br, I= ox#=-1 unless combined with O2) 6. Neutral Compounds~ The sum of oxidation numbers must=0 eg. HCl H=+1 and Cl=-1 7. Charged Compounds~ Sum of Oxidation numbers= charge. Eg. NH4+ (Ox # N)(1)+ (Ox #H)(4)= +1 → Ox # N=-3 8. Oxidation numbers can be non-integers~ eg. O2- (Ox#O)(2)=-1 → Ox # of O2=-1/2 Rule 6 example: HNO2 H: +1 O2: -2 (Ox # H)(1) + (Ox # N)(1)+(Ox#O)(2)= 0 (+1)(1)+(N)(1)+(-2)(2)=0 Ox # of N= +3 Rule 7 example SO4 -2 (Ox # S)(1)+(Ox#O)(4)=-242 (Ox #S)(1)+(-2)(4)=-2 Ox#S=+6 9.48 Assigning Oxidation Numbers a. Mg3N2 Mg+2/N-3 b. CsO2 Cs+ (+1)(1)+(ox#O2)(2)=0 c. CO3 -2 (Ox # C)(1)+ (Ox # O)(3)=-2 → (Ox#C)(1)+(2)(3)=-2 +4, -2 d. NaBH4 (1)(Ox#N)+(Ox # B)(1)+(Ox#H)(4)=0 → (Ox #N) +1, -5, +1 Concentration of Solutions: Molarity(M): The moles of solute/ L of solution => How much of a solute is in the solution. Dissolve 40.0 g NaCl in 1L of H2O, what is the concentration? 40.0g=1.00 mol NaCl → 1 mo lNaCl/ 1L = 1.00 M NaCl 9.62 How many moles of KOH are present in 35.0 mL of a 5.50 M in a KOH solution? 5.50 moles KOH/ 1 L Solution (0.035 L solution) (5.50 moles KOH/ 1 L Solution)= 0.193 moles of KOH 9.60 How to prepare 250 mL of 0.707M NaNO3 solution(This means 1L contains .707 moles NaNO3) But we’re not going to have that much (.707 mole NaNO3/ 1L solution)(.250 L solution)=/177 mol NaNO3= 15.05 g → Dissolved in 250 mL H20 9.64 Calculate the molarity of the following solutions: a. 6.57g CH3OH + 1.5X10^2 mL H2O 6.57 g= .205 mol CH3OH → 0.205 mol/ 0.15L solution= 1.37 M CH3OH B. 10.4 g CaCl2 in 2.20X10^2 mL H2O 0.34M CaCl2 Solution Dilution ↪Reducing the concentration ***M1V1=M2V243 Ex. What volume of a 12.0 M HCl solution is needed to make 250 mL of .125 M HCl solution M1= 12.0 M M2= .125 M V1= ? V2= 250mL V1= (.125M X 250 mL)/12.0 M V1= 2.60 mL 9.70 25.0 mL of a 0.866M KNO3 solution is diluted with water until final volume is 500 mL, what is the new concentration? M1= 0.866 M M2= ? V1= 25.0 mL V2= 500mL M2= M1V1/V2 → M2= 0.043 M  Consider the following: 1.0 M NaCl (means 1 L contains 1 mol NaCl, also 1 L will contain 1 mol Na+ and 1 L will contain 1 mol Cl-) and 1.0 M CaCl2 (this means 1 L contains 1 mol CaCl2, also 1 L will contain 1 mol Ca+ and 1 L will contain 2 mol Cl-) 9.78 What is the Na+ Concentration in the following: 3.25 M Na2SO4 (3.25 M Na2SO4)(2 mol/1 mol Na2SO4)= 6.5 mol/ 1L Na+ → Concentration: 6.5 M Na+ 9.81 Calculate the concentration of NO3- when the following 2 solutions are mixed: 95.0 mL of .992 M KNO3 is mixed with 155.5 mL of 1.570 M Ca(NO3)2 solution *the answer has to be mol NO3-/ L solution Vf (final volume)= 250.5 mL Solution 1: 0.095L (0.992 mol KNO3/ 1L)(1 mol NO3-/1 mol KNO3)= 0.094 mol NO3- Solution 2: (0.1555 L) (1.570 mol Ca(NO3)2/ 1L) (2 mol NO3-/ 1 mol Ca(NO3)2) = 0.44 mol NO3-/ 0.582 mol NO3- → 0.582 mol NO3/ 0.250L= 2.32 M NO3- Combine the Following: (aq) reactions with solution stoichiometry to get 2 different analysis techniques.44 1. Gravimetric analysis 2. Acid/ base titration 1. Gravimetric analysis → Precipitation reaction, compare actual(experimental) vs. theoretical yield Ex. AgNO3 (aq) + NaCl → AgCl (s) + NaNO3 (aq) Net Ionic: Ag+ (aq) + Cl- (aq) → AgCl (s) Use Gravimetric analysis→ Purity of NaCl sample 1. Dissolve 1.0 g of sample in 100 mL of H2O If it’s pure the concentration should be 0.172 M NaCl = .0172 mol NaCl/ L  2. Add excess of AgNO3 (aq) → 1st compare actual and theoretical yield of NaCl We expect 2.47 g AgCl (s) theoretical 2. Acid Base Titration → Going to be acid base duh, - Volume of a known concentration of acid/ base required to neutralize a known volume of an unknown concentration of base/ acid - Ex. 46.3 mL of a 0.203 M NaOh solution neutralizes 25.0 mL of a H2SO4 solution of an unknown concentration. What is the concentration of the H2SO4 Write the reaction 1st: H2SO4 + 2NaOH → Na2SO4 + 2H2O 1. Determine the moles of NaOH needed 2. Relate the moles of NaOH to the moles of H2SO4 3. Calculate the concentration of the molarity of H2SO4 1. (0.0463 L NaOH solution)(0.203 mol NaOH/ 1 L) = 0.009 mol NaOH 2. (0.009 mol NaOH)(1 mol H2SO4/ 2 mol NaOH)= 0.005 mol H2SO4 3. .005 mol H2SO4/ .250 mL= .188 M H2SO445
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