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Solved: An inductor and a capacitor are to be connected to

Physics | 4th Edition | ISBN: 9780321611116 | Authors: James S. Walker ISBN: 9780321611116 152

Solution for problem 44 Chapter 24

Physics | 4th Edition

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Physics | 4th Edition | ISBN: 9780321611116 | Authors: James S. Walker

Physics | 4th Edition

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Problem 44

An inductor and a capacitor are to be connected to a generator. Will the generator supply more current at low frequency if the inductor and capacitor are connected in series or in parallel? Explain.

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Organic Chemistry 2230 Chapter 1 Part B January 11, 2016 Amanda Biddlecome 1) Geometries -­‐Linear: draw Lewis Structure; then look at bond angles-­‐180 degrees; 2 regions with electrons -­‐Trigonal Planar: 3 regions with electrons; bond angles-­‐120 degrees; draw Lewis Structures with formal charges *Formal Charge=Group number-­‐(lone electron pairs + ½ bonding electrons) -­‐Tetrahedral: most common; 4 regions of electron density; Lewis Structure first; bond angle-­‐109.5 degrees *when you have lone pair electrons, bond angles change slightly and you have a different molecular geometry *trigonal pyramidal: 4 regions of electrons; 1 lone pair; bond angle-­‐ 107 degrees *bent: 4 regions of electrons; 2 lone pairs; bond angle-­‐104.5 degrees 2) Orbitals=where the electrons are located -­‐S Orbital *spherical; groups 1 and 2 -­‐P Orbital *dumbbell shaped; groups 13-­‐18 *3 possible orientations: 2px, 2py, or 2pz -­‐Bonding *start as atomic orbitals and when two of those combine, you get a molecular orbital *2 types: bonds with s orbitals=sigma bonds; bonds with p orbitals=pi bonds 3) Hybridization -­‐Carbon forms 4 single bonds by causing an s orbital electron to become excited and jump to the p orbital; now there is 1 electron in the s orbital and 3 electrons in the p orbital all unpaired; they go through hybridization and create a new orbital: sp3; tetrahedral; all bonds=sigma (4 sigma bonds) -­‐Carbon forms a double bond by creating a new, excited orbital: sp2; 3 regions of electron density; trigonal planar; bond angle=120 degrees; double bond=pi bond and others=sigma bonds (1 pi bond and 1 sigma bond) -­‐Carbon forms a triple bond by forming a new hybridized orbital: sp; triple bond has 1 sigma bond and 2 pi bonds; 2 regions of electron density; linear; 180 degree bond angles 4) Types of Bonds -­‐sigma is an overlap of s orbitals -­‐pi is an overlap of p orbitals Organic Chemistry 2230 Chapter 2: Acids and Bases January 13,2016 Amanda Biddlecome 1) Acids and Bases -­‐common acids: HClO4, H2SO4, HBr, HCl, HNO3, H3PO4 -­‐common bases: NaOH, KOH, Ba(OH)2 -­‐Bronsted-­‐Lowry Acids and Bases *Acid=proton donor: H^+=H3O^+=All 7 strong acids because they dissociate completely; always positively charged *Base=proton acceptor; always negatively charged *in reactions, negative always attacks positive (general rule) -­‐an acid + a base = conjugate acid + conjugate base *acid-­‐base reaction -­‐Keq=concentration of products/concentration of reactants -­‐Ka=acid-­‐dissociation constant *every acid has one *if it’s large, then the acid is strong *Ka values are very small values and hard to quantify, so we use the pKa to describe acid strength -­‐pKa= -­‐logKa *the smaller the pKa, the stronger the acid *used mostly in organic chemistry 2) Acid Strengths -­‐the smaller the pKa, the stronger the acid -­‐the larger the pKa, the stronger the base -­‐we don’t find the pKb in organic chemistry, so we have to determine base strength by looking at the pKa -­‐anytime you have an equilibrium arrow, it can shift to either the reactant side or the product side *look at pKa values to determine which sides have weak and strong acids and bases; the formation of the weaker acid and base is always favored in equilibrium, so the equation will always shift to the side with the weaker acid and base -­‐strong acids form weaker conjugate bases 3) Organic Acids -­‐R=Carbon or Hydrogen or anything really *place-­‐holder in Lewis structures -­‐Carboxylic Acid: pKa=5 -­‐Alcohol: pKa=15 -­‐Mines (Nitrogen based): pKa=10 -­‐Strong Acids: pKa<10 4) Acidity of Hydrogen Halides -­‐Group 7 -­‐atomic size increases going down the group -­‐electronegativity decreases going down the group -­‐acid strength increases going down the group *deals with bond length; the longer the bond, the easier it is to break -­‐base strength decreases going down the group -­‐the weaker the base, the more stable it is *this is why equilibrium favors weaker bases -­‐acid strength increase with electronegativity but competes with atomic size 5) Acidity of Carbon Compounds -­‐shorter bonds are stronger -­‐the more s character, the shorter and stronger the bond and the larger the bond angle -­‐single bond: pKa>60; double bond: pKa=44; triple bond: pka=25 -­‐electronegativity increases from single bond to triple bond because there’s more electrons involved in more bonds -­‐acid strength increases with electronegativity *atomic size doesn’t help you much 6) Atomic Radius versus Electronegativity -­‐when atoms are similar in size, the stronger acid has its proton attached to the more electronegative atom -­‐atomic size wins here -­‐atomic radius and electronegativity compete for control *increasing electronegativity causes increasing acidity *we use electronegativity to judge acidity because the differences in electronegativity between atoms is much more significant than most atomic radii 7) Resonance and Acidity -­‐resonance structures distribute charge across atoms to increase stability of compounds *a stable conjugate base means you had a strong acid which is displayed in the pKa value 8) Substituent Affects and Acidity -­‐substituents are things bonded to carbon that influence the molecules ability to lose electrons *when you put halides in, the greatest effect is noticed -­‐Inductive Electron Withdrawal-­‐when you put something very electronegative on a carbon, it pulls on all of the electrons in the molecule, stabilizes it, and increases its acidity -­‐Inductive Effect diminishes as you move farther away from the central atom, or the acidic site 9) pH Effects on Organic Compounds -­‐if you put a molecule into a solution with a pH equal to its pKa value, it will be equally acidic and basic (neutral) *buffers -­‐Henderson-­‐Hasselbalch Equation pKa=pH+log([HA]/[A]) -­‐when pH=pKa, the compound is neutral -­‐when pHpKa, the compound is basic -­‐the magnitude of the deviation from neutral doesn’t matter -­‐you can have more than one acidic sites on a single molecule that can have different pKa values *if placed in a solution with a pH equal to one of the pKas, the equal pKa would remain the same (neutral) and the other would either become acidic or basic and would behave appropriately *a pH value between the two pKa values would result in both groups being utilized -­‐pH of human mouth=6.8; stomach=1.5-­‐3.5; intestine=8.5 *these values determine which foods get broken down where in the digestive tract 10) Lewis Acids and Bases -­‐acid=non-­‐proton-­‐donating; will accept two electrons -­‐base=will donate two electrons

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Chapter 24, Problem 44 is Solved
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Textbook: Physics
Edition: 4
Author: James S. Walker
ISBN: 9780321611116

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Solved: An inductor and a capacitor are to be connected to