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Study Guide for Final

by: Nana Boateng Serebour

Study Guide for Final CHEM 111 - 02

Nana Boateng Serebour

General Chemistry I
Professor Ezra Wood

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This is the study guide for Chemistry 111
General Chemistry I
Professor Ezra Wood
Study Guide
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This 20 page Study Guide was uploaded by Nana Boateng Serebour on Saturday May 2, 2015. The Study Guide belongs to CHEM 111 - 02 at University of Massachusetts taught by Professor Ezra Wood in Spring 2015. Since its upload, it has received 156 views. For similar materials see General Chemistry I in Chemistry at University of Massachusetts.


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Date Created: 05/02/15
FINAL CHEMISTRY STUDY GUIDE Chapter 1 Chemistry The Study of Matter and the Changes It Undergoes 11 Types of Matter Matter Matter Mixtures Pure Substances Heterogeneous Homogeneous Element Compound Contains more than one element A compound is 2 or more different elements chemically bonded course you together such as can differentiate Just a pure NaCl Always different parts Solution simple element contain elements in visuallywith a can39t such as Gold or same mass microscope differentiate Carbon percentages 0 To separate mixtures Filtration to separate heterogeneous mixtures Distillation to separate homogeneous mixtures Chromatography to separate volatile gasses O O 12 Measurements 0 Metric Prefixes 106 Mega M 103 Kilo k 102 hecto h 101 deka 1 10391 Deci 1 10392 centi 0 10393 milli m 10396 micro it 10399 nano n 103912 pico p 0 Precision basically how many decimal places your measurement has eg 4125 is more precise than 41 0 Accuracy how well your measurement agrees with rightdefined answer 0 Signi cant Digits NonZero digits are significant Zeros 0 Leading Zeros not significant 000025 only 2 and 5 are significant 0 Captive Zeros are significant 1008 or 1105 in both all numb are sig 0 Trailing Zeros When number does not have decimal 150 not significant but when number does 10500 trailing zeros are significant Exact Numbers have no significant value meaning 0 O I Can be counted 5 pennies 0 If defined 12 inches in a foot 254 cm in an inch AdditionSubtraction 0 Result can only contain as many decimal places as number With least amount of decimal places MultiplicationDivision 0 Result can only have as many significant digits as number With least amount of significant digits 50125 12500 but 13x104 because 25 only has 2 sigs 0 Important Conversions To Remember T F 95T C 320 T C 59T F32 T KT C27315 254cm 1 inch 13 Properties of Substances 0 Chemical Properties 0 When substance takes part in a chemical reaction a change that converts it to a new substance I Decomposition Mecury Oxide turns into Mecury and Oxygen When heated 600 C 0 Physical Properties 0 Wo changing chemical identity or makeup I Melting points I Boling point I Color I Taste 0 E g Whether Nitrogen freezes or boils its still nitrogen 0 Density 0 D MassVolume 0 Doesn t matter What amount of something density is still the same I Solubility o ProcessExtent to Which solute dissolves in solvent 0 Ex 487g sugar dissolves completely in 100g water at 100 C and 247g dissolves in 100g water at 20 C how much water does it take to dissolve 100g of sugar at 100 C 0 100g sugar x 100g water 487g sugar 205 grams of water 0 If cooled to 20 C 205g water x 204g sugar100g water 418g of sugar is in the water and the remaining 582 crystalizes Saturated completely dissolved Unsaturated not enough solute Supersaturated too much solute Crystallization mixture tries to correct an imbalance 0 Ex a solution that is saturated at 30g sugar100g sugar is instead mixed With 35g sugar100g water So When shaken some of the mixture crystallizes To find the amount 35 100 30 100 5g sugar crystallizes 0000 There are 113 elements 91 of Which occur naturally Chapter 2 1 Atoms and Atomic Theory a Early Schools of Thought i Democrtitus indivisible particles atoms from atomos ii Aristotle matter is infinitely divisible 4 elements earth wind fire water iii Lavosisier the father of modern analytical chemistry 1 Law of the Conservation of Matter In chemical reactions mass in the beginning and mass after are same mass is neither lost nor gained Proust 1v 1 Law of Definite Properties Elements in compound are present in fixed proportion by weight E g Water is always HzO v Dalton 1 Law of Multiple Proportions Many elements combine in more than one ratio e g H20 water is 21 H202 is Hydrogen Peroxide 11 2 Dalton Atomic Theory On page 27 a Chemical elements are made of atoms b c Atoms of an element are the same mass except for isotopes Atoms of different elements have different masses except for isobars d Atoms combine in small number ratios 11 12 2 Components of the Atom a Electrons then protons then neutrons were discovered b A cathode ray is an electron beam 3 Ionic Compounds With Metal and Nonmetal a Ion Any charged particle b Ionic Bond Bond formed between 2 oppositely charged particles from transfer of electrons from one element to another i Charge is caused because atoms strive to be noble gasses ii Generally form between metals and nonmetals iii Element that loses electron becomes positivelv charged called CATION and one that gains electrons becomes negativelv charged and is called a ANION iv Metals Lose electrons positively charged CATION v Nonmetals Gain electrons negatively charged ANION c Important Notes i LEAVE NUCLEUS ALONE only manipulate electrons ii No such thing as an ionic molecule because the compound iii Ionic compounds do not exist independentlv iust as crvstals iv In chemical reaction reaction is transfer of electrons d Naming i For higher charge Cation use suffix ic ii For lower charge Cation use suffix ous 4 Empirical Formula Formula for Ionic Compounds a It provides information about the quantity or ratio or combined elements in a compound written in lowest terms 1 When writing empirical formula Cation is listed before Anion c In an ionic compound all electrons must be accounted for no leftovers d To solve the formula recognize What the charges on the elements to become noble gasses Rb and then use Crisscross Method i Examples 1 Ionic Compound of Rubidium and Oxygen 0 A compound of Rubidium Rb and Oxygen O are charged as ions as Rb1 and 0239 b 16 gt 16 Rb szO 2 Rubidiums for every oxygen c OR use Crisscross method and simply move integers from charges of opposite atom to the number of elements For example move the 2 from the O to the number of Rubidiums and move the 1 from the Rubidium to the number of oxygen Rb201 ampQ 2 Sr and S 0 Charges are SI 2 and 8239 b Use Crisscross and the compound is SrZSZ and reduce to smallest terms SrS e In Naming Ionic Compounds keep leading Cation the same name and change the suffix of the Anion to ide i Example 1 Na and C1 Na1 and C1139 NaCl Sodium Chloride 5 Ionic Compounds WPolyatomic Ion a Polyatomic Ion a group of atoms such as N03 1 With polyatomic ions the charge is over the Whole group of atoms The identity of the polyatomic ion must be preserved Use parenthesis if you need more than one and don t just multiply through i Ex CaN032 not CaN206 c To make compounds use same crisscross method as a regular ionic compound While preserving identity of groups of atoms d To name Ionic Compounds W Polyatomic Ions use the suf x ate i Examples 1 Ca and N02 Ca2 and N03139 CaN032 Calcium Nitrate 2 Ga and C2H302 Ga3 and C2H302139 GaC2H3023 Gallium Acetate 3 Ammonium NH4 and Phosphate PO43 NH43PO4 Ammonium Phosphate ii For groups of atoms with variable charges use appropriate roman numerals after name to clarify or use latin name if available iii 6 Molecule With Nonmetals a Molecule compound between two nonmetals i Joined by Covalent Bonds which mean that elements share electrons instead of exchanging them 7 Binary Molecular Compounds a Different than ionic compounds because they have no charge and empirical formula ratios don t apply i Ex N02 and N204 are different compounds 1 Naming i End in ide ii Prefixes 1 Mono 2 Di 3 Tri 4 Tetra 5 Pent 6 Hex 7 Hept 8 Oct 9 Non 10 Dec i Don t use mono for first occurring element in compound iv Examples 1 SF6 Sulfur Hex uoride 2 C1207 Dichloride Heptoxide 3 CF4 Carbon Tetra uoride 8 Acids a Acid A compound With the ability to produce H ions in a solution i Example 1 C1 H In H20 gtH or I 2 This is an example of Disassociation come off a molecule 1 The strength of an acid is its ability to produce H ions c To identify a compound as an acid look for a leading hydrogen d Acids are polar covalent so treat acid like ionic With H as a 1i e Naming i Suffix changes 1 ate goes to ic 0 Nitrate Nitric 2 ite goes to ous ii Hydrogen names If acid does not contain oxygen hydrogen is called hydro and acts as a prefix to the second element If there is an oxygen hydrogen is not included in the name iii Examples 1 HBr Hydrogen Bromine and no oxygen Hydrobromic Acid 2 HN03 Hydrogen and Nitrate Nitric Acid 3 H2SO4 Hydrogen and Sulfate Sulfuric Acid 4 HNOZ Hydrogen and Nitrite Nitrous Acid 5 HC2H302 Hydrogen and Acetate Acetic Acid iv Acetic Acid 1 Can be called multiple names 0 HC2H302 b HCH3COO 9 But most common is CH3COOH Chapter 3 Atomic Masses I Molar Mass Atomic weight of an element converted to Grams 0 Ex Molar Mass of 159940 15994 Grams I Calculating Atomic Mass from Isotopes w abundances 0 Multiply each of the Isotope s weights by their abundances and add I Ex Oxygen exists in 3 isotopes 1639000 9976 1739000 04 and 1839000 2 0 169976 170004 18002 16 amu I Calculating Isotope Abundance 0 Set up algebraic equation I Known mass of element isotopel weightx isotope weight 2x1 0 Ex Oxygen atomic mass 159994amu exists in 3 isotopes with masses 159949amu 169993amu and 179992amu occurring 204 0 159994 169993x 1599491x 17999200204 159582 169993x 159949 159949x 0367 10044x X 1699930 1599490 100 365 204 9976 0000 Converting From Grams Moles Units and So on I Grams gt Moles Divide by MM 0 Moles gt Units Multiply by Avogadro s Number 6022e23 I Units gt Moles Divide by Avogadro s Number 6022e23 I Molesgt Grams Multiply by MM 0 Example 0 Convert 1e8 g Silver Ag to moles then to Atoms of Ag I 1e8MM Ag 9271e11 mol Ag6022e23 558625e13 Atoms Ag Molar Mass and MoleGram Conversions To Find Molar Mass of Molecule Multiply molar mass of each element by number of atoms then add results 0 Examples I Molar mass of CF2C12 1MM C 2MM F 2MM C1 120914 I Convert 35g of CF2C12 to moles 35MM of CF2C12 35120914 289mol CF2C12 To Find number of atoms in a molecule 0 Convert into unitmolecules and multiply by corresponding number of atoms of element in the molecule Percent Composition From Analysis Take molar mass of element multiplied by its number of atoms in molecule and divide by molar mass of molecule 0 Example Percent composition of H in C6H1002S I 10mm of HMM of C6H1002S 69 Simplest Formula From Analysis Take percent of each element and use a 100g sample so each percent is converted to grams of an element 30 O 30g 0 Then divide by molar mass of element then divide by lowest common term to find simplest form 0 Example 459 C 275H 262 0 1750 S 765 N I 459mm C 275mm H 262mm O 175mm S 765mm N 39 C3822H272801638S546N546 divide all by 546 because it is the smallest 39 C7H503SN If after you have divided each by smallest term and you are left with nonwhole numbers multiply by an integer to get round numbers Ex if you had C267 multiply by 3 C8 Writing and Balancing Equations 0 Be extra sure to write molecules correctly charge balancing if between metal and nonmetal 0 Make sure number of each type of element is same on both sides MoleMass Relations in Reactions 0 Basic Procedure If you start with grams convert to moles by dividing by MM of element Then multiply moles by ratio of element you are looking for over the element you know Then multiply those moles of the element you were looking for back to moles if asked to 0 Example Phosphine gas reacts with oxygen according to equation a 4PH3 802 9 P4010 6H20 c Find mass of Tetraphosphorous decaoxide produced from 1243 mole of phosphine o 1243 mol Phosphine 1mol P40104mol PH3 31075mol P4010 o 31075mol P4010 mm of P4010 882184 g P4010 Theoretical Yield Problems 0 Work backwards from theo yield 0 Example Problem 66 Limiting Reagent Problems 0 Obvious because problem gives you masses you have of the reactants what combines to react 0 Work equation first starting with one reactant to find the amount needed of other reactant to react completely If you find you need a greater amount of the other reactant to react completely the other reactant is the limiting reagent Then work the problem with the limiting reagent to find whatever other amount of molecule you react with 0 Example 68 Combustion Reaction 0 Usually involves Carbon and Hydrogen and you burn it 0 How To Solve Find mass percentages of Hydrogen and Carbon in resulting molecules and then multiply each by corresponding masses of corresponding molecules Go through usual procedure of Simplest Form and you are left with empirical formula of Hydrocarbon Chapter 4 Molarity 0 M molvolume L 0 Ex How many moles of Potassium Chloride are there in 20 ml of KCL 39 15M mol KCL02 L I 15M02 003 M01 KCL 0 Dilutions When you have a concentrate with a certain molarity that you add water to to dilute it 0 Use M1V1 M2V2 o Calc Volume of 2M stock solution needed to produce 600ml of a desired 05M solution I 2V1 056 I V1 15 L then add water up to 600 ml Double Displacement ReactionPrecipitations I In double displacement reactions switch cations and anions and balance equations If a reaction occurs one becomes solid and the other is soluble The ions of the soluble one are eliminated from both sides of the equation Solubility Rules 0 Soluble o Nitrates all alkali Ammonium NH4 o Sulfates all except Ag Ca Sr Ba Hg Pb o Salts of F except Mg Ca Sr Ba Pb o Salts of Cl Br and I AS IN STRONG ACIDS except Ag Hg Pb 0 Insoluble NOTE SOULBLE RULES OVER INSOLUBLE o Carbonates o Phosphates o Sulfides o Hydroxides Acid Base Neutralization Reaction Switch cations and anions Form water when possible 0 Strong Acids HCl HBr HI Sulfiric Acid Sulfate Nitric Acid Nitrate Perchloric Acid HCLO4 Strong Bases All alkali metal hydroxides Calcium Hydroxide Strontium Hydroxide Barium Hydroxide O O In an acid base equation strong vs weak is based on the ability to dissociate Net Equations Strong acidStrong Base I End up with water Weak AcidStrong Base I End up with water and other part of acid Because base is strong it will become a spectator and will be crossed out Strong AcidWeak base I End up with base plus an H and the second part of the acid will be crossed out Titration 0 When you are given a solution of precisely known content and use it to find another solution 0 Steps Write a balanced acidbase equation Convert molarity to moles Compare mole ratios Convert moles back to molarity 0 Example 0 25ml of 02 Strontium Hydroxide is required to titrate 35ml of Nitric Acid to its endpoint What is the molarity of Nitric Acid SrOH2 2HN03 9 2HzO SI39NO32 Str base str acid SrOH2 mols 02M025L 0005 Mol SrOH2 2HN031SrOH2 001 M01 HNO3035L 029M Redox Reactions 0 A redox reaction is when there is a transfer of electrons from both sides to the other sides 0 When you reduce the charge of the left side by adding electrons reduction 0 When you increase the charge of the left side by adding electrons to the right side oxidation 0 When electrons show up on right Reduction 0 When electrons show up on left Oxidation Steps for doing a redox problem a Break the equation into 2 half reactions Balance elements except for oxygen and hydrogen Balance oxygen by adding water Balance hydrogen by adding H ions Balance charge by adding e39 Mult by each half reaction by the LCF of the two electron numbers The two electrons should be on opposite sides of the two half reactions Cancel them out Add the two sides of the half reactions together left left 9 right right Simplify water and hydrogen ions by adding and subtracting from sides IF IN BASE SOLUTION Add an 0H for every H ion Simplify by creating waters Simplify waters H and OH 00000000 Chapter 6 OVERALL EQUATIONS V7tv or CMI C Speed of Light 3x108 ms EhV EhCIK H Planks Constant 6626x103934 l E Rh1n2hi 1n210 Rh 2180X103918J Ad hmv 211an Waves 0 Frequency V How many waves pass by a point in a certain amount of time 0 Normally measured in Hertz Hz Ilsec 0 Wavelength 7 Distance between two consecutive identical points on a wave 0 Measured in nanometers nm 10399m 0 EQUATIONS o or Ckv Velocitywavelength 9 frequency v Speed of Light 3x108 ms wavelength Afrequency v 0 Wave Spectrum 0 Frequency decreasing 9 Gamma Visible Inna Radio Wavelength Increasing 9 o Visible Light 400nm V I B G Y O R 0 Photoelectric Effect 0 Frequency carries the energy of the wave 0 EQUATIONS Ehv 0 EnergyJ Planks Constant 6626x103934Frequency EhCv 0 EnergyJ Planks Constant 6626X103934Speed of lightwavelength a Light is a massless particle known as a photon 0 Threshold Frequencv Minimum frequency required to knock off an electron off metal Beyond the threshold frequency the intensity of the light amplitude determined how many electrons were ejected per unit of time o Amplitude refers to brightness Photontime o EQUATIONS quotE Rh1n2hi 1n210 0 R1 2180X103918J Ad hmv 0 DeBroglie Wavelength Planks constantmass of an electron 911X103931velocity 211an Quantum Mechanics 0 Quantum Numbers and Fields Psi I2 give the 90 probability region in 3D space Principle Quantum Number 11 determines energy level and average distance from nucleus Angular Momentum Quantum Number 1 determines shape of sublevel Possible Values 1 0 nl Field Letters 0S Sphere 1P3 dumbbells 2D 5 3F 7 4G Some People Don t Fart Gas Magnetic Quantum Number determines spatial orientation of sublevel orbital Namthl O 0 Possible Values 1 0 1 Spin Orbital Number ms determines spin orientation of electron in an orbital Possible Values il2 Electron Filling in An Atom 0 O pairing 0 numbers SSublevel s 23 3s 4s 53 63 7s DSublevel 1 Down from Period I Each orbital can contain a maximum of 2 electrons so S can have 2 P can have 6 D can have 10 Aufbau Principle Electrons fill energy positions from lowest to highest Hund s Rule When filling orbitals Within an energy sublevel electrons fill orbitals singly before Pauli Exclusion Rule No two electrons in a given atom can possess the same set of four quantum PSublevel 3d 4d 5d 6d FSublevel 2 Down From Period 4f 5f Spectroscopic Notation Ex H 1s1 1 11 Principal Quantum Number s sublevel Ex Br 0 O 1 electron in that sublevel 1s22s22p63 s23p64s23d1 4p5 Condensed Go back to the previous noble gas and add notations until you reach the element Ex Li 1s22s1 or He 2s1 Orbital Box Diagrams O gt Any atom With filled outer S and P sublevels is considered Valence Related to spectroscopic notation Fill boxes in according to electrons in energy levels Ex Fluorine ls22s22p5 0 0 1 0 1 Tl Tl Tl Tl T ls 2s 2p 2p 2p Exceptions To Aufbau s Principle Want to fill a Whole energy level or at least half Removes an electron from previous full electron level to supplement Element Chromium Cr Expected Ar 4s23d4 Ar 4s13d5 0 Actual Element Copper Cu Expected Ar 4s23d9 Ar 4s13d10 Actual Periodic Table Trends 0 0 Atomic Radius Radius of an atom from nucleus to edge of electron cloud Decreases from left to right along same energy level Due to increase in nuclear charge Increases top to bottom Due to electron being placed in increasingly large energy levels Ionic Radius Radius of an ion compared to its neutral counterpart Anions are always LARGER Due to added repulsion among electrons Cations are always SMALLER Due to fewer electrons attracted by the same of protons in the nucleus Ionization Energy Energy required to remove an outer electron from an atom in its gaseous state Increases from Bottom to Top Due to outermost electrons being closer to nucleus Increases Left to Right Due to increase in nuclear attraction as protons are added Electro Negativity Relative tendency for an atom in a bond to pull electrons toward it Increases Bottom to Top Due to bonding valence electrons being closer to the nucleus Increases Left to Right Due to increase in nuclear attractions as protons are added 0 0 Chapter 7 Covalent Bonding I Generally between nonmetals I Electrons are shared to achieve a stable octet I Usually takes place between valance electrons outermost s and p sublevels Lewis Dot Structure 0 Used to visualize electron structure 0 X Represents Element 0 AB represents electron pair AB AB 1AB 1 L AB 0 Writing Lewis Structures for Molecules 0 Determine the most likely arrangement of Atoms I Lease EN atom is usually in the center I Molecules are often symmetrical I Hydrogen is never central and can only bond to one thing duet 0 Determine the total number of valence electrons 0 Fill in the octets of the outer atoms place electrons in pairs around outer atoms 0 Check for an octet on the central atom I If the central atom has an octet and there are no leftover electrons then you are done I If the central atom doesn t contain an octet consider a double or triple bond I If there are electrons left over place them as long pairs on the central atom 0 Extra Misc 0 Sometimes a line is used to represent a pair of electrons Free Radicals 0 Molecules that have a lone electron not paired 0 Very reactive harmful to health cancer aging combated with antioxidants 0 For example N02 has a free radical on the N Resonance 0 Resonance occurs when there are more than one equivalent structure for the same molecule 0 For example on NO39 a double bond can be drawn from any one of the O s Bond Order 0 Bond order is the number of bonds shared between two atoms referring 0 For example in NO since the double bond can be from any three of the O the Bond Order is 1 and 13 This is because there is a single bond between each O and N and then there is one extra bond3 O atoms to make it 1 and 13 Formal Charge 0 Helps you determine the best structure for a given compound 0 Formal charge must be determined for the each element in the molecule 0 Ideally the formal charge would be zero If it is not zero it should still be a very small number eg i 12 with the more negative charge on the more electronegative element 0 Formal Charge Normal Valence Nonbonding Electrons 12 Bonding Electrons 0 For example a double bonded oxygen would have the formal charge of 6420 Electron Deficient Molecules 0 These are elements that are exceptions to the octet rule 0 Boron B Only needs three pairs of electrons to fulfill its valence shell 0 Beryllium Be Only needs two pairs of electrons to fulfill its valence shell Expanded Valence 0 This is when a central element has more electron pairs around it than it needs to fill its valance shell VSEPR 0 Stands for VALENCE SHELL ELECTRON PAIR REPULSION THEORY 0 Valence shells electron pairs will arrange themselves around a central atom to be as geometrically far apart as possible for the lease repulsion Major Classes For an atom The electronic The bond Wlth a LeWIs geometry Wlll angles Wlll Shape Dot Structure be be of Less than an Linear 180O octet XA X AXz 2S Trigonal 120O AX Planar AX3 X 0 Octet XAX Tetrahedral 10939s SE AX4 X Trigonal 120 900 X A OX Xo oX Blpyramldal AXS Greater than an octet 2g X X 900 A t h l X xxx 0c a edra AX6 X Electronic geometries refer to the arrangement of electron pairs around the central atom and will always be a major class The molecular geometries refer to the actual physical appearance of a molecule and may be a major class no nonbonding pairs on the central atom or a subclass one ore more nonbonding pairs on the central atom Subclass Geometries Subclass Geometries Those in which there are one or more nonbonding pairs of electrons on the central atom Trigonal Planar Lewis Structure Molecular Geometry Bond Angles 120O XzAX lt Bent AXZE Tetrahedral Lewis Structure Molecular Geometry Bond Angles xXX lt1095 0 Tr1 onal rarmd 1 x g N a AX3E o lt10950 X X Bent A AX2E2 Trigonal bipyramidal NOTE Nonbonding electron pairs MUST be on the equatorial region of the molecule Lewis Structure Molecular Geometry Bond Angles 22 X A o See Saw lt120 lt90o X39 o39X Distorted tetrahedron AX4E co lt900 39 39 Tsha ed XAX P AX3E2 X 39oX Linear 1800 4A AX2E3 Octahedral NOTE Remove electrons from the axial regions BEFORE the equatorial regions Lewis Structure Molecular Geometry Bond Angles lt90 lt180 A 39 X X Square Pyrarmdal AXSE X lt90 X 0X Square Planar AX4E2 Xoo lt900 A X quotoX T Shaped AX3E3 180 C A O X u X L1near AX2E4 Hybridization 0 The blending of existing orbitals to produce a new set of orbitals that are consistent With the VESPR theory Name Angle Initial Promoted Hybridized SP Linear 180 1313 wig D 3ij 1 120 SP2 Trigonal Planar SP3 Tetrahedral 1095 SP3D Trigonal Bipyrimidal 90 120 SPBLD2 Octahedral 90 180 1 l I Tquot Uj I TE I i F swat 0L 0 Rules a Only 1 bond of a multiple bond is hybridized The other bonds are unhybridized POrbitals o Nonbonding electron pairs on the central atom are hybridized a Sigma bonds D are bonds from hybridized orbitals Pi bonds D are unhybridized Porbitals Polarization 0 All polar molecules are considered dipole 0 Things to Look For 0 Nonbonding pairs on the central atom usually polar Symmetry about the central atom usually nonpolar o 0 Differences in electro negativity Bigger difference greater polarity o THINK ABOUT THE MOLECULAR STRUCTURE 0 After deciding if polar or not if polar draw charge arrow towards the most electronegative atom Chapter 1314 Acids Bases and Buffers Review 0 Arrhenius Definition 0 Acids that produce H ions in solution a Bases that produce OH39 ions in solution 0 BronsteadLowry Definition 0 Acid Proton Donor a Base Proton Acceptor 0 Ex NH3 H20 9 NH4 OH39 0 Water is called Amphiprotic o Amphiprotic Has a proton to donate or accept 0 Strength of an Acid or Base depends on its ability to dissociate 0 Strong Acids o HN03 Nitric Acid 0 HCl4 Perchloric Acid 0 H2SO4 Sulfiric Acid 0 HCl Hydrochloric Acid 0 HBr Hydrobromic Acid 0 HI Hydroiodic Acid 0 Strong Bases o LiOH 9CsOH o o o B aOH2 Neutralization Reactions Acid Base 9 H20 Salt 0 EX 9 H20 Conjugate AcidBase Pairs Differ by a hydrogen 0 EX NH3 H20 9 NH439 OH39 Base Acid Conj A Conj B AutoIonization of Water H20 H20 9 H30 OH39 EQUATION IONIZATION CONSTANT FOR WATER H3OOH39 10 X 103914 pH and pOH EQUATION pH LogH3O H3Oquot1039pH For pure water pH Log1xlO397 7 EQUATION pOH LogOH39 OH391039P H EQUATION pH pOH 14 Dissociation 0 EQUATION Dissociation XNM o X is the dissociation of H30 or 0H and NM is the nominal molarity o EQUATION Ka X2NMX o Ka is the dissociation constant Visual pH Tests 0 Litmus Paper 0 Blue Litmus I If Blue Litmus goes from Blue 9 Red the solution is acidic I If it doesn t change the solution is basic 0 Red Litmus I If Red Litmus goes from Red 9 Blue the solution is basic I If it doesn t change the solution is acidic I WideRange pH Paper 0 Red 9Blue o Acidic Basic 0 Colors displayed correspond directly to pH Acidic or Basic Anhydrides 0 Metal Oxides tend to produce Basic solutions 0 NazO H20 9 2NaOH I NonMetal Oxides tend to produce Acidic Solutions a 03 H20 9 H2804 Polyprotic Acid 0 These acids have more than one hydrogen capable of dissociating Generally each H on the acid has a different Ka value 0 Ka1gtKa2gtKa3 0 EX 0 H2304 0 But not CH3COOH Ka and Kb Values Both acids and bases have a KaKb value EQUATION KKb1X103914 When using a Kb value to calculate pH be aware that the molarity you find Will be of the base ions and not acid ions therefore find the pOH and then subtract it from 14 to get the pH Acid Base Neutralization Reactions 0 Acid Base 9 Salt Water 0 Different Cases 0 EX I Write Balanced Equation HCl NaOH 9 NaCl H20 I Check for hydrolysis Further dissociation of salt I Na H20 9 NaOH H It reforms the original base so it doesn t count 0 C139 H20 9 HCl OH39 It reforms the original acid so it doesn t count I If hydrolysis does occur additional creation of H ions turns the solution acidic Whereas OH turns the solution basic Buffer Systems 0 A buffer system uses a combination of elements to create a system that resists dramatic changes in pH when an additional acid or base is added 0 The principal behind a buffer is if an acid is added the base of the solution reacts with it consuming an amount of the base and adding that amount to the acid to balance it Visaversa with a added base 0 EQUATION KaXBase M XAcid M x 0 EQUATION pHpKa Log B39HB Chapter 18 Activity Series of Metals Oxidation Potential 0 Based on ability to lose electrons oxidize 0 The more potential an element has to oxidize the higher its activity 0 Activity Series In decreasing order 0 KgtCagtNagtMggtAlgtCrgtZngtFegtCdgtNigtSngtPbgtH2gtCugtAggtHggtAu 0 Electrons Flow from more active to less active electrode Standard Reduction Potentials STP 0 Standard Reduction Potential is the exact opposite of activityoxidation potential Therefore a highly active metal would have a low reduction potential and a low activity metal would have a high reduction potential 0 For example a metal with an activity of 25 would have an STP of 25 Standard Hydrogen Electrode SHE 000 V 0 Standard Conditions 0 Gases 1 ATM 1 bar 0 Solutions 1 M 0 Values measured at 298K 25C VoltaicGalvanic Cell 0 Two electrodes are connected by a wire and a salt bride 0 The negative terminal is the anode usually on the left The anode is the more active metal and is the place of oxidation Oxidization is the taking away electrons 0 The positive terminal is the cathode usually on the right The cathode is the less active metal with a higher reduction potential and is the place of reduction Reduction is adding electrons 0 Electrons always ow from the more active metal to the less active or else you have a negative voltage 0 Cell Diagram 0 Anodel Anode Solution Cathode Solution Cathode 0 Double bar represents a salt bridge 0 Measuring Voltages o Voltages are measured using standard reduction potentials o Etotal Ecathode 39 Eanode Etotal Ereduction onidation o For example if the anode of a cell has an STP of 2 and the cathode has an STP of 3 then the total voltage produced would be Etota1 32 5 Concentration Cell 0 A potential difference can be created with two halfcells having the same metals in solutions containing their ions if there is a molarity difference between the two solutions 0 La Chateliers Principal A chemical system will react in such a way as to reduce any stress placed on that system AKA A system will try to balance itself by moving electrons to relieve stress 0 In most concentration cells the anode is the place of the more dilute solution The cathode is the place of the concentrated solution The dilute solution is oxidized and ions are added to balance The concentrated solution is reduced and the ions are transferred to the dilute solution Batteries 0 LeClanche o A LeClanche battery is the average battery used in CD players stereos and ashlights o The anode is the zinc metal casing and the cathode is the NH4Cl and ZnClz paste 0 2Mn02 2NH4 2e39 9 Mn203 2NH3 2H20 o Cons Large currents cause the battery to produce a buildup of NH3 gas that lowers voltage 0 Alkaline Batteries 0 KOH is used instead of NH4Cl a Zn 2Mn02 9 ZnO Mn203 0 Button Battery Mercury HgO o The small at battery used in watches and calculators a Zn HgO H20 9 ZnOH2 Hg 0 If made with silver there is no toxic mercury 0 Lead Acid Car Battery 0 Produces a very large current but contains lead and sulfuric acid and it is very large and heavy 0 Good because it recharges very well 0 Pb P1302 2H 2HSO439 9 PbSO4 2H20 0 Fuel Cell 0 In a fuel cell oxygen and hydrogen combine to drive a reaction that creates pure water as a byproduct o 2H2 02 9 2H20 0 NiCd o NiCd batteries suffer from weight are toxic and have bad memory problems ability to recharge decreases over time Cathodic Protection 0 With cathodic protection an active metal in a place is prevented from oxidizing away A more active metal is connected to the original piece and is sacrificial and is oxidized instead of the original piece This saves the original from oxidizing away Electrolysis 0 Using electrical energy to drive a reaction in a nonspontaneous direction Instead of using chemistry to produce electricity electrolysis uses electricity to produce chemistry 0 Anodes and cathode polarity is switched anode is now positive and cathode is negative 0 In battery diagram long line is positive and short line I is negative 0 Follow Specific steps to determine aspects of electroplating problems 0 Amps ChargeTime Usually Amps Coulombssecond 0 With the given information solve the amps equation for the missing value a To continue to find moles or grams divide Coulombs by the Faraday constant96480 coulombsmole e39 0 Use the molecular coefficients to find mole ratios 0 Convert moles of a product to grams 0 Example How many grams of copper are deposited on the cathode of an electrolytic cell if an electric current of 200 A is run through a solution of CuSO4 for a period of 20 minutes I 200 A 20 min60 secmin 2400 Coulombs I 2400 C 96480 Cmol e39 5201 39 520112 0124 Copper is oxidized from Cu to Cu2 which means it needs two electrons for every molecule so a mole of electrons can only do half a mold of copper therefore we multiply by 12 I 0124mm of Cu 63546 79g Cu 0 If given moles or grams of a substance just work backwards to get back to amps charge or time 0 Example I How many hours would it take to produce 75g of metallic Chromium by the electrolytic reduction of Cr3 with a current of 225 Amps 0 Cr3 3e39 9 Cr 0 75gmm of Cr 519961 1442 0 14423 432 3 is the number of electrons needed for each molecule notice it is the reverse of the usually downward process 0 432 96480 417492 C 0 417492 C225 A 1855237 Seconds6O secmin 60 minhr 5154 hrs This document is the work of a student and is not reviewed by Mr Feebeck The information is not necessarily accurate


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