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Course: CHEM 2000, Fall 2006
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2000 Chemistry Lecture 9: Intermolecular forces Marc R. Roussel Bond polarity When the two atoms at each end of a bond are dierent, they will not share electrons equally. This results in bond polarity, i.e. a separation of charge between the atoms at either end of the bond. The bond polarity will increase the greater the dierence in electronegativity between the atoms. Example: N2 is nonpolar. Example: CO should...

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2000 Chemistry Lecture 9: Intermolecular forces Marc R. Roussel Bond polarity When the two atoms at each end of a bond are dierent, they will not share electrons equally. This results in bond polarity, i.e. a separation of charge between the atoms at either end of the bond. The bond polarity will increase the greater the dierence in electronegativity between the atoms. Example: N2 is nonpolar. Example: CO should be more polar than NO. Molecular polarity in diatomic molecules In diatomic molecules, if the bond is polar, then the molecule is automatically polar. Molecular polarity can be calculated or measured. Dipole moment: vector measure of molecular polarity = qi ri atoms qi = partial charge on atom i ri = position vector of atom i For a neutral diatomic molecule, the size of the dipole moment vector (or just the dipole moment) is || = = |q |R |q | = absolute value of the partial charge carried by each atom R = bond length The dipole moment vector calculated from the formula points toward the positive end of the molecule. However, chemists draw their dipole moment vectors pointing the other way, using an arrow with a cross bar at the positive end of the dipole: + . :N .. O .. Measurement of dipole moments Dipole moments can be measured in two ways: Temperature dependence of the relative permittivity: The force between two charges in any medium is F= q1 q2 4 r 0 r 2 where r is the relative permittivity of the medium. For solids and liquids, the relative permittivity is easily measured by measuring the capacitance of a pair of parallel plates separated rst by vacuum and then by the material under study. r depends on temperature in a way which allows us to recover both the dipole moment and the polarizability of the medium. Stark eect in rotational spectroscopy: Molecules in the gas phase have discrete rotational energy levels. Molecules with a permanent dipole moment can undergo pure rotational transitions. These are studied by microwave spectroscopy. In the presence of an added electric eld, the rotational levels shift (Stark eect). The size of the shift tells us the size of the molecular dipole moment. A mess of units... The SI unit for the dipole moment is the C m. Dipole moments in C m are small ( 1030 C m). Example: The dipole moment of HCl is 3.60 1030 C m. Although its use is discouraged, many people like to use the debye (D), a non-SI unit, as a unit for dipole moments because these values are of order unity. 1 D = 3.335 64 1030 C m Example: The dipole moment of HCl is 1.08 D. A more modern approach which gives nice values is to use the dipole length = /e . is the distance we would have to separate two elementary charges (+e and e ) to get the dipole moment . Example: The dipole length of HCl is 22.5 pm. (Compare: HCl bond length = 127 pm) Polarity in polyatomic molecules We can get molecular dipole moments in polyatomic molecules by vector addition of bond dipoles. For our purposes, it will suce to carry out this procedure graphically (and qualitatively). Example: CO2 bond dipoles: .. O .. C .. O .. Because of the symmetry, the bond dipoles cancel, so CO2 is nonpolar ( = 0). Example: OCS bond dipoles: .. O .. C .. S .. The molecular dipole moment is nonzero, and the corresponding dipole moment vector points toward O. Example: BF3 bond dipoles: F B F F To add vectors, place them tip to tail: The bond dipoles add to zero, so BF3 has a zero dipole moment. Example: CH4 . For simplicity, Ill only draw the vectors without their cross bars: Adding the bottom two vectors gives a vector that points straight down: Adding the top two vectors gives a vector that points up with the same magnitude as the one shown above. Adding all four vectors therefore gives zero. Conclusion: Methane is nonpolar. Example: CH2 Cl2 . Cl Cl C H H Red = C-Cl bond dipoles. Green = C-H bond dipoles. Blue = net dipole moment (not to scale). Intermolecular forces The properties of chemical systems are strongly aected by the forces which act between molecules in these systems. There are dierent kinds of intermolecular forces. Some are stronger than others. We will survey the dierent types of intermolecular forces, from strongest to weakest. Ion-ion forces The Coulomb force is the strongest one which operates on the length scales of interest to chemists. Because of the strength of this force, ionic compounds typically have high boiling points. When we dissolve an ionic compound in a solvent, forces between the ions result in a relatively organized state where, on average, anions are closer to cations than to other anions. Ion-dipole forces There are also fairly strong forces between ions and dipoles. When we dissolve an ionic compound in a polar solvent, this results in substantial organization of the solvent around the ions. Mg 2+ Dipole-dipole forces Dipoles exert forces on each other. Dipoles tend to align themselves so that the positive end of one is close to the negative end of another. F F F F N N F F Dipole-induced dipole forces If we place a molecule next to a polar molecule, the electric eld of the dipole exerts a force on the electrons of the other molecule, resulting in an induced dipole. The polarizability is a measure of how large an induced dipole is produced for a given electric eld. The more loosely held the electrons are, the larger the polarizability. = atomic polarizability increases as we move down a group. All other things being equal, the polarizability will be larger for long molecules because a small charge separation over a large distance can result in a large dipole moment ( = |q |d ). The induced dipoles orientation will always generate an attractive force to the polar molecule. H O + H + N N Fluctuation-induced dipole Due to the random essentially motion of electrons around a molecule, from time to time, molecules will acquire a dipole moment when there happen to be more electrons on one side of the molecule than the other. This short-term uctuation will in turn induce dipoles in adjacent molecules. Fluctuation-induced dipole forces are also known as London dispersion forces. The strength of dispersion forces depends on the polarizability. Dispersion forces can occur even in molecules which have a permanent dipole moment. Induced dipoles can be produced in directions orthogonal to . Example: In H2 O, there is a permanent dipole moment parallel to the symmetry axis of the molecule. H O H Dipole moment components can be induced perpendicular to this axis (perpendicular to molecular plane, or parallel to the H-H axis). Boiling points All other things being equal, boiling points will be higher if the intermolecular forces are larger because it takes more energy to remove a molecule from the liquid. Example: Group 14 boiling points 200 180 Tb /K 160 140 120 100 (Why?) CH4 SiH4 GeH4 400 Group 15 Group 16 Group 17 H2O 350 Tb /K 300 250 Normal trend HF NH3 H2S 200 PH3 HCl H2Se AsH3 HBr H2Te SbH3 HI 150 2 3 4 Period Why do NH3 , H2 O and HF break the trend? 5 Hydrogen bonding In molecules containing O-H and N-H bonds and in HF, we observe dramatically increased intermolecular forces. We call this additional type of intermolecular force hydrogen bonding. Molecules involved in hydrogen bonding are much closer together than molecules experiencing (e.g.) dipole-dipole interactions, but not as close as covalent bonding distances. (Known from X-ray and neutron scattering experiments) Hydrogen bonding (continued) Hydrogen bonding is due to a combination of factors: Large electronegativity of N, O and F No core orbitals in H Availability of lone pairs on N, O and F The rst two factors result in a very small electron density on the side of H away from a covalent bond to N, O or F. This allows H to interact with a lone pair. This interaction is partly a dipole-dipole interaction, but also has some covalent character. Hierarchy of intermolecular forces 1. Ion-ion 2. Ion-dipole 3. Hydrogen bonding 4. Dipole-dipole 5. Dipole-induced dipole 6. Induced dipole-induced dipole (London dispersion) Higher members of the hierarchy are stronger. Higher members are usually accompanied by some or all of the lower members. Example: Water is subject to all the forces from hydrogen bonding to London dispersion. Some manifestations of intermolecular forces in liquids Other than boiling points, there are other macroscopic manifestations of the varying strengths of intermolecular forces in liquids. surface tension viscosity Surface tension Bubbles and drops tend to be round. Why? A sphere is the shape of minimum area-to-volume ratio. Increasing the surface area takes work. Well see why in a minute. surface tension = work increase in surface area On average, the intermolecular forces on a molecule in the bulk (away from the surface) sum to zero. A molecule at the surface experiences a net force directed inward. Adding molecules to the surface (increasing the surface area) requires that we push them to the surface against this unbalanced force. This is the origin of surface tension. Surfactants Suppose that we add a solute to a solvent. Suppose also that the forces between the solute and solvent are weaker than the forces between the solute molecules. As a rule in a situation like this, the solute-solute forces would also be weaker than solvent-solvent forces. For this kind of solute-solvent system, the solute molecules would tend to congregate on the surface because the energy required to put a solute on the surface is less than the energy required to put a solvent molecule on the surface. Accordingly, such a solute would tend to lower the surface tension. Solutes with these properties are called surfactants (surface active agents). Surfactants Surfactants vary greatly in chemical and physical properties. Detergents are a particularly important class of surfactants. They have a polar (or charged) head and a nonpolar tail. Example: the dodecylsulfate ion, CH3 (CH2 )11 OSO 3 Surface wetting When we put a drop of liquid on a solid surface, we can observe one of two qualitative situations: Wetting: the drop spreads out on the surface. Non-wetting: the drop assumes a spheroidal form. Wetting occurs when molecules in the liquid are more strongly attracted to the solid than to each other. In this case, part of the liquid surface, which requires energy to create, is replaced by an energetically favorable contact with the solid. Wetting is responsible for the familiar meniscus seen when water is placed in a glass container. Non-wetting is seen when the interactions between molecules of the liquid are stronger than between the liquid and solid. In this case, we get an inverted meniscus, for instance when we put mercury in a glass container. Capillary rise Wetting is also responsible for capillary rise. A capillary is a narrow tube. The attractive force between the tube and liquid causes liquid to rise within the capillary. The solid-liquid attraction has to support the weight of the column of water, so narrower capillaries lead to higher capillary rise. In non-wetting situations, we get capillary depression. Viscosity Viscosity is a liquids resistance to uid ow. It is analogous to friction as one part of the liquid moves relative to another. Viscosity is directly related to intermolecular forces: stronger intermolecular forces, higher viscosity. Some of the highest viscosities are seen in ionic liquids. Example: The viscosity of (bmim)(BF4 ) H 3C + N CH 3 N F B F is more than 100 times that of water. F F
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Lethbridge College - CHEM - 2000
Chemistry 2000 Lecture 10: The kineticmolecular theory of matterMarc R. RousselThe kinetic molecular theory of gasesMatter is in constant movement and, as we have seen, subjectto a variety of intramolecular (bonding) and intermolecularforces.Can we
Lethbridge College - CHEM - 2000
Chemistry 2000 Lecture 11: Entropy and thesecond law of thermodynamicsMarc R. RousselThe thermodynamic description of matterIn classical thermodynamics, we describe the state of a systemby macroscopic variables which can be measured usingordinary la
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Chemistry 2000 Lecture 13: Chemical equilibriumMarc R. RousselThermodynamic criterion for equilibriumRecall that under given reaction conditions (concentrations ofreactants and products),r Gm = r Gm + RT ln QIf r Gm < 0, the reaction is thermodynami
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2001 Section B Assignment 1Due: Thursday, Jan. 18, 10:50 a.m.Note: Im not terribly fussy about signicant gures, but I will ask the marker to penalize answerswhich give either a ridiculously small or a ridiculously large number of
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Chemistry 2000B Spring 2002 Assignment 1Due: Thursday, Jan. 24, noonNotes on signicant gures: In this and all subsequent assignments and tests in this course, you shouldreport your answers to a reasonable number of signicant gures. Slavish adherence to
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Chemistry 2000 Spring 2001 Section B Assignment 1Solutions1.4nH2 O0 0036 (a) Suppose that we take 1 L of this solution. This liter of solution contains 1.697 mol ofacetone and weighs 0.9849 kg (from the concentration and density, respectively). Th
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Chemistry 2000B Spring 2002 Assignment 1 Solutions(a) 1 L of solution contains 1.875 mol of sodium sulfate (Na2 SO4 ) and weighs 1.2106 kg.The mass of sodium sulfate is266 3 g1 875 mol 142 043 g molmNa2 SO41.The mass of water is therefore0 2663 kg
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Chemistry 2000 Spring 2001 Section B Assignment 2Due: Thursday, Feb. 15, 10:50 a.m.1. The liquid compound 1,3-di-t-butylcyclohexane exists in two forms which are known as thechair and boat conformations:chairboat(Only the carbon skeletons are shown.
Lethbridge College - CHEM - 2000
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Chemistry 2000B Spring 2002 Assignment 2 Solutionst t11t1 22ln x x0t1 2ln 1 2ln 0 055 3sln 1 2ln 2 t1 2 . The results at the(a) We can calculate the rate constant from the half-life by ktwo temperatures are as follows:23 slnt2T ( C) T (K
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Chemistry 2000 Spring 2001 Section B Assignment 3Due: Thursday, March 22, 10:50 a.m.Note: You will have to look up some data from your textbook to answer most of these questions.1. Suppose that you want to make a buffer with a pH of 3. Select an acid/c
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Chemistry 2000 Spring 2001 Section B Assignment 3Solutions1. A buffer is most effective when the target pH is close to the pKa of the acid. The pKa ofhydrouoric acid is 3.14 so an HF/F buffer would do nicely in at least some applications.(HF and uorid
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Chemistry 2000 Spring 2001 Section B Assignment 4Due: Thursday, April 5, 10:50 a.m.Note: You may have to hunt around your textbook for relevant data. Most of the data you need can be foundin the appendices.1. Calculate the solubility of magnesium hydr
Lethbridge College - CHEM - 2000
Chemistry 2000B Spring 2002 Assignment 4Due: Thursday, March 21, noonMost of the questions in this assignment will require data from the textbook. I will not tell youwhere to look. Think of it as a scavenger hunt. Unless otherwise stated, assume that t
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2001 Section B Assignment 4Solutions(a) The solubility equilibrium isPb2aq L.4 mol(b) In a pH 4 buffer, aH1010 10 . Since aMg2Using Kw , we nd that aOHKsp a2 , we nd a solubility of 1 5 109 mol L. This absurdly high number
Lethbridge College - CHEM - 2000
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Lethbridge College - CHEM - 2000
Chemistry 2000B Spring 2002 Assignment 5Due: Thursday, April 4, noonAs in the last assignment, you will need to look up several pieces of data in your textbook.(a) How much heat is liberated when 1.0043 g of benzoic acid (C6 H5 COOH) burns in anexcess
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Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2006 problem solutions:Biochemistry1. C4 H8 O4 + C5 H10 O5 C9 H16 O8 + H2 O2. A lipid is a biochemical compound which is soluble in nonpolar solvents.3.CH 3( CH 2) 10CHOOHCCCOHOHOHCC( CH 2) 10( CH 2) 10CH 3C
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2006 problem solutions:Biochemistry1. C4 H8 O4 + C5 H10 O5 C9 H16 O8 + H2 O2. A lipid is a biochemical compound which is soluble in nonpolar solvents.3.CH 3( CH 2) 10CHOOHCCCOHOHOHCC( CH 2) 10( CH 2) 10CH 3C
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2006 problems:Biochemistry1. Write a balanced reaction for the condensation of the sugars erythrose (C4 H8 O4 )and xylose (C5 H10 O5 ).2. What is a lipid?3. Draw the structure of the triglyceride formed by condensation of glycer
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2006 problems:Biochemistry1. Write a balanced reaction for the condensation of the sugars erythrose (C4 H8 O4 )and xylose (C5 H10 O5 ).2. What is a lipid?3. Draw the structure of the triglyceride formed by condensation of glycer
Lethbridge College - CHEM - 2000
Solutions to the Practice Problems on the Relationshipbetween Kinetics and Equilibrium1. The rate of production of C (for instance) is1k1012k18s111k 1Ck1Kk1k1 1 4 10 3 L mol 1 sK1 2 1015 L molC(a) A and B are reactants. C is an interm
Lethbridge College - CHEM - 2000
Solutions to the Practice Problems on the Relationshipbetween Kinetics and Equilibrium1. The rate of production of C (for instance) is1k1012k18s111k 1Ck1Kk1k1 1 4 10 3 L mol 1 sK1 2 1015 L molC(a) A and B are reactants. C is an interm
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Practice Problems on the Relationship between Kineticsand Equilibrium1. Suppose that, for the elementary reactionk1and the equilibrium constant K123 L mol 1 s 11011015 . What is k 1 ?14kk1C BANotes: Watch your units. For reasons which we
Lethbridge College - CHEM - 2000
Practice Problems on the Relationship between Kineticsand Equilibrium1. Suppose that, for the elementary reactionk1and the equilibrium constant K123 L mol 1 s 11011015 . What is k 1 ?14kk1C BANotes: Watch your units. For reasons which we
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Lethbridge College - CHEM - 2000
Practice Problems on EquilibriumAll of the problems given here assume the use of proper thermodynamic equilibrium constants(phrased in terms of activities).1. Write equilibrium constant expressions for the following reactions both in terms of activitie
Lethbridge College - CHEM - 2000
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Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2002 Final ExaminationTime: 3 hoursMarks: 100Aids allowed: Calculator. No printed materials (e.g. periodic tables, calculator manuals) are allowed.Formulas and data are given on page 4.Instructions: Answer all questions in the b
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2001 Section BFinal ExaminationTime: 3 hoursAids allowed: Calculator. In particular, periodic tables and other written or printed materials areexcluded.Additional data is given at the end of this paper.Write all your answers in
Lethbridge College - CHEM - 2000
Up: Back to the Chemistry 2000 test indexChemistry 2000, Fall 1996 Final examinationYou have three hours to complete this exam. You may not leave the examination hall in the first hour orin the last 10 minutes. The aggregate value of all questions on t
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2006 Final ExaminationName:Student number:Time: 3 hoursAids allowed: CalculatorInstructions: Answer all questions in the spaces provided. You can use the backs of pages forscratch work or for extra answer space. If you do need
Lethbridge College - CHEM - 2000
Solutions to the Practice Problems onRequired Mathematical Skills1. Simplify each of the following expressions as much as possible:(a) pH(b) log10 (10x2 ) = log10 10 + log10 x2 = 1 + 2 log10 x(c) x(x + 1)2doesnt simplify.(d)x2 + 12. Solve the fo
Lethbridge College - CHEM - 2000
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Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2002 Midterm ExaminationSolutions(a) The concentrations in experiments 1 and 2 are in a 3:1 ratio, as are the rates. This istherefore a rst-order reaction with rate law2k CH3 Nv1.The rate constant can be determined from eithe
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2002 Midterm ExaminationName:Student number:Aids allowed: Calculator. No printed materials (e.g. periodic tables, calculator manuals) are allowed.Formulas and data are given on page 7.Instructions: Answer all questions in the sp
Lethbridge College - CHEM - 2000
Chemistry 2000: MO and VB descriptions ofbondingMarc R. RousselMO descriptions of bondingTo describe the bonding in a molecule using MO theory, proceedas follows:Look at the valence AOs and, if available, at the AO energies.Separate the AOs into si
Lethbridge College - CHEM - 2000
Up: Back to the Chemistry 2000 assignment indexChemistry 2000, Fall 1996, ThermodynamicsPractice Problem Solutions1. This is a heat balance problem. Since there is no indication to the contrary, we must assume thatno heat is either gained or lost by t
Lethbridge College - CHEM - 2000
Up: Back to the Chemistry 2000 assignment indexChemistry 2000, Fall 1996, ThermodynamicsPractice Problems1. A 10kg piece of wood (specific heat capacitywater (specific heat capacitywood and water?2. Oxygen boils at) at) atis thrown into 100kg of
Lethbridge College - CHEM - 2000
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Lethbridge College - CHEM - 2000
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Lethbridge College - CHEM - 2000
Next: Useful data Up: Back to the Chemistry 2000 test indexChemistry 2000, Fall 96, Test 11. Suppose that a chemist measures the initial rate of disappearance of a reactant A as a function ofconcentration for the reaction:What is the experimental rat
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2001 Section B Test 1Name:Student number:Aids allowed: Calculator. In particular, periodic tables and other written or printed materials areexcluded.Instructions: Answer all questions in the spaces provided. Use the backs of the
Lethbridge College - CHEM - 2000
Next: Useful information Up: Back to the Chemistry 2000 test indexChemistry 2000, Section ASpring 1996 Test 1All questions are equally weighted. Bonus mark policy: Bonuses are applied to this test and to the lastassignment, to a cumulative maximum of
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2006 Test 1Name:Student number:Time: 50 minutesAids allowed: CalculatorInstructions: Answer all questions in the spaces provided. You can use the backs of pagesfor scratch work or for extra answer space. If you do need to conti
Lethbridge College - CHEM - 2000
Test 112111098765432100-9.9%10-19.9%20-29.9%30-39.9%40-49.9%50-59.9%60-69.9%70-79.9%80-89.9%N = 761 quartile: 12/45 = 27%Median: 20/45 = 44%rd3 quartile: 28.5/45 = 63%st1003rd quartilemedian1st quartilePercentage806040
Lethbridge College - CHEM - 2000
Up: Back to the Chemistry 2000 test indexChemistry 2000, Section ASpring 1996 Test 1 Solutions1. Using the first two lines of the table, we see that the rate goes up by a factor of 5 when [A] goesup by a factor of 5. Using the last two lines, we see t
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2001 Section B Test 1 Solutions(a) We rst compare experiments 1 and 2. At constant O2 , the concentration of NO isdoubled, resulting in a (roughly) fourfold increase in the rate. Thus, the order withrespect to [NO] is 2. Experimen
Lethbridge College - CHEM - 2000
Up: Back to the Chemistry 2000 test indexChemistry 2000, Fall 96, Test 1 Solutions1. From the last two lines of the table, we see that increasing [B] has no effect on the rate. Therefore,the reaction is zero-order with respect to B so we can ignore thi
Lethbridge College - CHEM - 2000
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Lethbridge College - CHEM - 2000
Next: Useful information Up: Back to the Chemistry 2000 test indexChemistry 2000, Section ASpring 1996 Test 2This exam will be marked out of 100. The detailed breakdown for each question is given in each question.The bonus questions are worth one mark
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2001 Section B Test 2Aids allowed: Calculator. In particular, periodic tables and other written or printed materials areexcluded.Useful data is given at the end of this paper.Answer all questions in the booklet provided.1. Here
Lethbridge College - CHEM - 2000
Next: Useful data Up: Back to the Chemistry 2000 test indexChemistry 2000, Fall 96, Test 2There are six questions.1. Starch is a glucose polymer, i.e. a very large molecule made by stringing many glucose moleculestogether. The osmotic pressure of a st
Lethbridge College - CHEM - 2000
Chemistry 2000 Spring 2006 Test 2Time: 50 minutesTotal marks: 49Aids allowed: CalculatorInstructions: Answer all questions in the booklets provided. You can answer the questions in anyorder, but make sure that your answers are clearly marked with the
Lethbridge College - CHEM - 2000
Mark distributions for tests 1 and 218test 1test 216Percentage of class141210864Test 1:1st quartile:Median:3rd quartile:30%46%65%90-100%80-89.9%70-79.9%60-69.9%50-59.9%40-49.9%30-39.9%20-29.9%10-19.9%00-9.9%2Test 2:1st quar
Lethbridge College - CHEM - 2000
Up: Back to the Chemistry 2000 test indexChemistry 2000, Section ASpring 1996 Test 2 Solutions1. It sublimes.2. On the solid/liquid phase boundary, K=1. (Solid and liquid are in equilibrium and since theactivities of pure solids and liquids are both