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Chemistry 111 - Final Exam Review St. Thomas CHEM 111
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  • Title: Chemistry 111 - Final Exam Review
  • Type: Lab
  • School: St. Thomas
  • Course: CHEM 111
  • Term: Fall

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111 CHEM From Chapter 9 you should: Exam 4 Review Sheet - Draw Lewis Structures (molecules, polyatomic ions, exceptions to octet rule, and those with more than one central atom). Ex. Ethane. Formal Charge: Hypothetical charge you obtain by assuming that bonding electrons are equally shared between bonded atoms and that electrons of each lone pair belong completely to one atom. o o o Equation: (Periodic electrons) (number of bonds) (number of non-bonded) = FORMAL CHARGE RULE A Write the Lewis Dot formula that has the lowest magnitudes of formal charges. RULE B When two Lewis formulas for a molecule have the same magnitude choose the negative formal charge or more electronegative atom. - - Delocalized Electrons: An electron that is shared over the entire region of the molecule; seen most often in resonance. Bond Energies--Generally, the enthalpy of reaction is approx. equal to the sum of the bond energies; if you subtract the sum of the bond energies from the bonds broken. o Example: - H H H H H C C Br H H C = C + (H-Br) H H Bonds Broken -> (C=C), 4(C-H), (H-Br) Bonds Formed -> 5(C-H), (C-Br), (C-C) (Cancel out the 4(C-H) = left with 1(C-H) in formed (Add bonding energies found in table 9.5 [given]) Subtract: (Bonds broken) (Bonds formed) = Change in Energies From Chapter 10 you should: - VSEPR Theory: 1) Write the electron dot structure. 2) Determine the number of electron pairs. 3) Determine arrangement. 4) Obtain molecular geometry from the bonding pairs. The ideal bond angle(s) within a molecule and the directions of any deviations. Axial: (Trigonal bipyramidal) there are 2 axial atoms found. One coming out the top and one coming out the bottom. Equatorial: (Trigonal bipyramidal) there are 3 equatorial atoms found in various positions coming from the center of the model. Dipole Moment: Quantitative measure of the degree of charge separation in a molecule; the arrows drawn to note polarity creating a net force whether that be neutral or charged. - - o o o - Measurements are based on the fact that polar molecules can be oriented by an electric field. Polar molecules tend to align themselves. Has a dipole moment = is polar, and follows the direction of the non-canceling charge. Valence Bond Theory: An approx. theory to explain the electron pair or covalent bond in terms of quantum mechanics. o 1- an orbital on one atom comes to occupy a portion of the same region of space as an orbital on the other atom. The two orbitals overlap. o 1- the total number of electrons in both orbitals is no more than two. Hybrid Orbitals: Orbitals used to describe bonding and are obtained by combining atomic orbitals of the isolated atoms. Hybrid Orbitals Sp Sp2 Sp3 Sp3d Sp3d2 Number of Orbitals Geometric Arrangement 3 2 4 5 6 Linear Trigonal Planar Tetrahedral Trigonal bipyramidal Octahedral - Sigma Bonds: Bond that has a cylindrical shape about the bond axis (Hot dog Model). Pi Bonds: Bond that has an electron distribution above and below the bond axis (Corn dog Model). Bond rotation = Cis and Trans isomers. - For Chapter 11 you should: - Gas: Molecules in random motion through a largely empty space >> Weak Negligible Forces. Liquid: Molecules in random motion but packed more tightly >> Weak Present Forces. Solid: (Atoms, Ions, and molecules) are closely packed and vibrate about fixed sites >> Strong Forces. Sublimation Melting Solid Freezing Liquid Condensation Vaporization Gas Deposition Intramolecular: The force found between atoms in bonding. Intermolecular: The force found between bonded molecules. Dipole-Dipole Forces: Found in polar bonding. Hydrogen Forces: Found in the bonding of H to F, O, or N. (H-FON). London Forces: The weakest force found in all bonding. o The stronger the forces the higher the boiling point is going to be. Type of Solid Molecular Metallic Ionic Covalent o o Structural Units Atoms or molecules Pos. core surrounded by sea of atoms Ions Atoms Attractive Forces b/n Structural Unis Intermolecular Forces Metallic bonding (extreme delocalized bond) Ionic bonding Covalent bonding Examples Ne, H2O, CO2 Fe, Cu, Age CsCl, NaCl, ZnS Diamond and graphite Melting Point: Stronger the bonding the higher the melting point; Covalent and Ionic = high melting points. Molecular = Low melting point. Hardness: The weaker the intermolecular force, the softer the object; Diamond and Silicon are the hardest known substances. o Conductivity: Easy to move delocalized valence electrons; covalent and ionic = nonconductors and Ionic is more conductive in the liquid state. - Crystalline Solid: Composed of one or more crystals; each crystal has a well defined ordered structure in 3-dimensions. Ex NaCl and Sucrose (table sugar). Simple Cubic Unit Cell: lattice points only at the corners. (1/8) slices x 8 corners = 1 atom Body-Centered Cubic Unit Cell: lattice points at all corners and 1 atom in the center. (1/8) slices x 8 corners + 1center atom = 2 atoms Face-Centered Cubic Unit Cell: lattice points at the centers and half of an atom on each face. (1/8) slices x 8 corners + (1/2) x 6 faces = 4 atoms Covalent Network Solid: Solid that consists of atoms held together in large networks or chains by covalent bonds (majority are molecular) >>Upper Right Hand Corner of Periodic Table. Water is unique because ----------o High CHANGE in heat of VAPORIZATION o High BOILING POINT o High MELTING POINT o High SPECIFIC HEAT o SOLVEN (POLAR) o Its DENSITY is less as a SOLID then when it's a LIQUID (S < L) - -

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Path: St. Thomas >> CHEM >> 111 Fall, 2006

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Path: BU >> HUMANITIES >> HU 101-202 Spring, 2008
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Path: BU >> HUMANITIES >> HU 101-202 Spring, 2008
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Path: BU >> HUMANITIES >> HU 101-202 Spring, 2008
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Path: Texas State >> COMM >> 1310 Spring, 2008
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Path: SUNY Oswego >> BRC >> 350 Spring, 2008
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Path: UC Riverside >> ENGL >> 1A Spring, 2008
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Path: Vanderbilt >> POLYSCI >> 100 Spring, 2008
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Path: Vanderbilt >> POLYSCI >> 100 Spring, 2008
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Path: Vanderbilt >> POLYSCI >> 100 Spring, 2008
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AFP Midterm 2
Path: Vanderbilt >> POLYSCI >> 100 Spring, 2008
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Path: Vanderbilt >> POLYSCI >> 100 Spring, 2008
Description: 1. Is continued progress of the EU toward becoming more like a federal union in the interest of the United States? Should the US still continue to support ever higher levels of integration for the EU? a. No, the United States should limit the EU beca...
BedSt2_143
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.143 For the vectors U D 6i C 2j 4k, V D 2i C 7j, and W D 3i C 2k, evaluate the following mixed triple products: (a) U V W ; (b) W V U ; (c) V W U . Solution: Use Eq. (2.36). 6 2 (a) U V W D 2 7 3 0 D 6 14 3 (b) W V U D 2 6 D3 2 (c...
BedSt2_144
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.144 Use the mixed triple product to calculate the volume of the parallelepiped. y (140, 90, 30) mm (200, 0, 0) mm x z (160, 0, 100) mm Solution: We are given the coordinates of point D. From the geometry, we need to locate points A and ...
BedSt2_145
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.145 that By using Eqs. (2.23) and (2.34), show Uy Vy Wy Uz Vz Wz Ux U V W D Vx Wx . Solution: One strategy is to expand the determinant in terms of its components, take the dot product, and then collapse the expansion. Eq. (2.23) is an...
BedSt2_146
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.146 The vectors U D i C UY j C 4k, V D 2i C j 2k, and W D 3i C j 2k are coplanar (they lie in the same plane). What is the component Uy ? Solution: Since the non-zero vectors are coplanar, the cross product of any two will produce a vector ...
BedSt2_157
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.157 (a) Write the position vector rAB from point A to point B in terms of scalar components. (b) The vector F has magnitude jFj D 200 N and is parallel to the line from A to B. Write F in terms of scalar components. Solution: (a) rAB D xB D...
BedSt2_158
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.158 The rope exerts a force of magnitude jFj D 200 lb on the top of the pole at B. (a) (b) Determine the position vector Determine the position vector vector rAB F, where rAB is the from A to B. vector rAC F, where rAC is the from A to C....
BedSt2_159
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.159 The magnitude of FB is 400 N and jFA C FB j D 900 N. Determine the components of FA . FB y FA 60 30 50 40 x z Solution: jFB j D 400 N We need to write each vector in terms of its known or unknown components. From the diagram FAx D ...
BedSt2_160
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.160 Suppose that the forces FA and FB shown in Problem 2.159 have the same magnitude and FA FB D 600 N2 . What are FA and FB ? Solution: From Problem 2.159, the forces are: FA D jFA j i cos 40 sin 50 C j sin 40 C k cos 40 cos 50 D jFA j 0....
BedSt2_161
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.161 The magnitude of the force vector FB is 2 kN. Express it in terms of scalar components. y F D (4, 3, 1) m FC FB A z FA C x (6, 0, 0) m B (5, 0, 3) m Solution: The strategy is to determine the unit vector collinear with FB and then e...
BedSt2_137
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.137 The force vector F points along the straight line from point A to point B. Its magnitude is jFj D 20 N. The coordinates of points A and B are xA D 6 m, yA D 8 m, zA D 4 m and xB D 8 m, yB D 1 m, zB D 2 m. (a) (b) Express the vector F in...
BedSt2_138
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.138 The rope AB exerts a 50-N force T on the collar at A. Let rCA be the position vector from point C to point A. Determine the cross product rCA T. B y 0.15 m 0.4 m C T Solution: The vector from C to D is rCD D xD yC j C zD jrCD j D zC...
BedSt2_135
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.135 For the points O, A, and B in Problem 2.134, use the cross product to determine the length of the shortest straight line from point B to the straight line that passes through points O and A. Solution: rOA D 6i 2j C 3k (m) 4k m (The magn...
BedSt2_136
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.136 The cable BC exerts a 1000-lb force F on the hook at B. Determine rAB F. Solution: The coordinates of points A, B, and C are A (16, 0, 12), B (4, 6, 0), C (4, 0, 8). The position vectors are 6 ft rOA D 16i C 0j C 12k, rOB D 4i C 6j C 0...
BedSt2_133
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.133 Use the cross product to determine the components of a unit vector e that is normal to both of the vectors U D 8i 6j C 4k and V D 3i C 7j C 9k. (Notice that there are two answers.) Solution: First, find U V D R i RDUVD 8 3 RD RD eR D s...
BedSt2_134
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.134 (a) What is the cross product rOA rOB ? (b) Determine a unit vector e that is perpendicular to rOA and rOB . Solution: The two radius vectors are rOB D 4i C 4j (a) 4k, rOA D 6i 2j C 3k y B ( 4, 4, 4) m rOB O The cross product is i r...
BedSt2_132
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.132 By evaluating the cross product U V, prove the identity sin 1 2 D sin 1 cos 2 cos 1 sin 2 . y U V 1 2 x Solution: Assume that both U and V lie in the x-y plane. The strategy is to use the definition of the cross product (Eq. 2.28)...
BedSt2_131
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.131 The force F D 10i mine the cross product rAB F. 4j (N). Deter- y (6, 3, 0) m A rAB z x (6, 0, 4) m B F Solution: The position vector is rAB D 6 6 iC 0 3 jC 4 0 k D 0i 3j C 4k y A (6, 3, 0) The cross product: i 0 10 j k 3 4 D i 16 ...
BedSt2_129
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.129 The cross product of two vectors U and V is U V D 30i C 40k. The vector V D 4i 2j C 3k. Determine the components of U. Solution: We know i U V D Ux 4 j k Uy Uz 2 3 3Ux j C 2Ux 4Uy k (1) Equating components of (1) and (2), we get 3Uy C...
BedSt2_130
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.130 20. (a) (b) (c) (d) The magnitudes jUj D 10 and jVj D y V U Use the definition of the cross product to determine U V. Use the definition of the cross product to determine V U. Use Eq. (2.34) to determine U V. Use Eq. (2.34) to det...
BedSt2_128
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.128 Suppose that the cross product of two vectors U and V is U V D 0. If jUj 6D 0, what do you know about the vector V? Solution: Either V D 0 or VjjU c 2005 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This mater...
BedSt2_126
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.126 What is the cross product r F of the position vector r D 2i C 2j C 2k (m) and the force F D 20i 40k (N)? Solution: Use Eq. (2.34) and expand into 2 by 2 determinants. rFD i 2 20 j 2 0 k 2 Di 2 40 40 0 2 j 2 40 20 2 C k 2 0 rFD 80i C ...
BedSt2_127
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.127 Determine the cross product r F of 12j C 3k (m) and the force Solution: rFD i 4 16 j 12 22 k 3 10 40 j the position vector r D 4i F D 16i 22j 10k N . r F D 120 C 88 66 i C 48 192 k N-m r F D 186i C 88j C 104k N-m c 2005 Pearso...
BedSt2_124
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.124 (a) Determine the cross product U V of the vectors U D 4i C 6j 10k and V D 8i C 12j C 2k. (b) Use the dot product to prove that the vector U V is perpendicular to U and perpendicular to V. Strategy: The vectors are expressed in terms ...
BedSt2_123
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.123 Point P is at longitude 30 W and latitude 45 N on the Atlantic Ocean between Nova Scotia and France. (See Problem 2.86.) Point Q is at longitude 60 E and latitude 20 N in the Arabian Sea. Use the dot product to determine the shortest di...
BedSt2_139
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.139 The straight line L is collinear with the force vector F. Let D be the perpendicular distance from an arbitrary point P to L. Prove that DjFj D jr Fj, where r is a position vector from point P to any point on L. Solution: By definition...
BedSt2_140
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.140 The bar AB is 6 m long and is perpendicular to the bars AC and AD. Use the cross product to determine the coordinates xB , yB , zB of point B. Solution: The strategy is to determine the unit vector perpendicular to both AC and AD, and t...
BedSt2_141
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.141* Determine the minimum distance from point P to the plane defined by the three points A, B, and C. Solution: The strategy is to find the unit vector perpendicular to the plane. The projection of this unit vector on the vector OP: rOP e...
BedSt2_142
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.142* The force vector F points along the straight line from point A to point B. Use Eqs. (2.28)(2.31) to prove that rB F D rA F. y A F rA B rB x Strategy: Let rAB be the position vector from point A to point B. Express rB in terms of of...
Thermodynamics 6th Ed. - Chapter 12
Path: Washington >> CHEM E >> 260 Spring, 2008
Description: 12-1 Chapter 12 THERMODYNAMIC PROPERTY RELATIONS Partial Derivatives and Associated Relations 12-1C z dz (z)y (z)x x dx y dy dz = (z ) x + (z ) y y x dx x+dx dy y + dy y x 12-2C For functions that depend on one variable, they are identical. ...
Thermodynamics 6th Ed. - Chapter 7
Path: Washington >> CHEM E >> 260 Spring, 2008
Description: 7-1 Chapter 7 ENTROPY Entropy and the Increase of Entropy Principle 7-1C No. The Q represents the net heat transfer during a cycle, which could be positive. 7-2C No. A system may produce more (or less) work than it receives during a cycle. A st...
ch4jh
Path: Washington >> MATH >> 308 Spring, 2008
Description: Linear Algebra 1 2 3 1 x1 x3 2 1 6 2 8 1 Jim Hefferon Notation R N C {. . . . . . } . V, W, U v, w 0, 0V B, D En = e1 , . . . , en , RepB (v) Pn Mnm [S] M N V W = h, g H, G t, s T, S RepB,D (h) hi,j |T | R(h), N (h) R (h), N (h) real numbers n...
BedSt2_147
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.147 The magnitude of F is 8 kN. Express F in terms of scalar components. y (3, 7) m Solution: The unit vector collinear with the force F is developed as follows: The collinear vector is r D 7 3 i C 2 7 j D 4i 5j The magnitude: jrj D eD r D...
BedSt2_148
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.148 The magnitude of the vertical force W is 600 lb, and the magnitude of the force B is 1500 lb. Given that A C B C W D 0, determine the magnitude of the force A and the angle . B 50 W A Solution: The strategy is to use the condition o...
BedSt2_149
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.149 The magnitude of the vertical force vector A is 200 lb. If A C B C C D 0, what are the magnitudes of the force vectors B and C? 50 in. 70 in. 100 in. C E B A D F Solution: The strategy is to express the forces in terms of scalar ...
BedSt2_150
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.150 The magnitude of the horizontal force vector D in Problem 2.149 is 280 lb. If D C E C F D 0, what are the magnitudes of the force vectors E and F? Solution: The strategy is to express the force vectors in terms of scalar components, and...
BedSt2_151
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.151 What are the direction cosines of F? y F = 20i + 10j 10k (lb) A (4, 4, 2) ft Refer to this diagram when solving Problems 2.1512.157. B (8, 1, 2) ft x z Solution: Use the definition of the direction cosines and the ensuing discus...
BedSt2_152
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.152 Determine the scalar components of a unit vector parallel to line AB that points from A toward B. Solution: Use the definition of the unit vector, we get The position vectors are: rA D 4i C 4j C 2k, rB D 8i C 1j 2k. The 4 vector from A...
BedSt2_153
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.153 What is the angle between the line AB and the force F? Solution: Use the definition of the dot product Eq. (2.18), and Eq. (2.24): cos D rAB F . jrAB jjFj From the solution to Problem 2.130, the vector parallel to AB is rAB D 4i 3j...
BedSt2_154
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.154 Determine the vector component of F that is parallel to the line AB. Solution: Use the definition in Eq. (2.26): UP D e U e, where e is parallel to a line L. From Problem 2.152 the unit vector parallel to line AB is eAB D 0.6247i 0.46...
BedSt2_155
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.155 Determine the vector component of F that is normal to the line AB. Solution: Use the Eq. (2.27) and the solution to Problem 2.154. FN D F FP D 20 8.78 i C 10 C 6.59 j C 1.22k (lb) 10 C 8.78 k D 11.22i C 16.59j c 2005 Pearson Education...
BedSt2_156
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.156 Determine the vector rBA F, where rBA is the position vector from B to A. Solution: Use the definition in Eq. (2.34). Noting rBA D rAB , from Problem 2.155 rBA D 4i C 3j C 4k. The cross product is rBA F D j 4 3 20 10 i k 4 D 10 C D 30...
BedSt2_125
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.125 C 4j. (a) (b) Two vectors U D 3i C 2j and V D 2i Solution: Use Eq. (2.34) and expand into 2 by 2 determinants. i j k UVD 3 2 0 Di 2 0 2 4 0 Ck 3 4 i j k VUD 2 4 0 Di 4 0 3 2 0 Ck 2 2 4 0 j 3 0 2 0 What is the cross product U V? What...
BedSt2_119
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.119 The disk A is at the midpoint of the sloped surface. The string from A to B exerts a 0.2lb force F on the disk. If you resolve F into vector components parallel and normal to the sloped surface, what is the component normal to the surfa...
BedSt2_84
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.84 The pole supporting the sign is parallel to the x axis and is 2 m long. Point A is contained in the y-z plane. Express the position vector r from the origin to the end of the pole in terms of components. y A Bedford Falls Solution: Th...
BedSt2_79
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.79 Consider the structure described in Problem 2.78. After returning to the United States, an archaeologist discovers that a graduate student has erased the only data file containing the dimension b. But from recorded GPS data he is able to...
BedSt2_80
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.80 Observers at A and B use theodolites to measure the direction from their positions to a rocket in flight. If the coordinates of the rocket\'s position at a given instant are (4, 4, 2) km, determine the direction cosines of the vectors rAR...
BedSt2_72
Path: Washington >> A A >> 210 Spring, 2008
Description: Problem 2.72 Determine the components of the position vector rBD from point B to point D. Use your result to determine the distance from B to D. y D (4, 3, 1) m Solution: We have the following coordinates: A 0, 0, 0 , B 5, 0, 3 m, C 6, 0, 0 m, D 4, ...

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