CHEMISTRY 202 Hour Exam III. Dr. D. DeCoste T.A (60 pts.) 31 (20 pts.) 32 (20 pts.) 33 (20 pts.) Total (120 pts)
|
|
- Maude Shepherd
- 5 years ago
- Views:
Transcription
1 CHEMISTRY 202 Hour Exam III December 3, 2015 Dr. D. DeCoste Name Signature T.A. This exam contains 33 questions on 13 numbered pages. Check now to make sure you have a complete exam. You have two hours to complete the exam. Determine the best answer to the first 30 questions and enter these on the special answer sheet. Also, circle your responses in this exam booklet. Show all of your work and provide complete answers to questions 31, 32, and (60 pts.) 31 (20 pts.) 32 (20 pts.) 33 (20 pts.) Total (120 pts) Useful Information: R = J/Kmol = Latm/molK k = Ae -Ea/RT k 2 E ln( ) = a 1 [ k1 R T 1 1 T ] 2
2 Hour Exam III Page No Which of the following best completes the following sentence concerning trends on the periodic table? While there can be exceptions, in general a) smaller atoms have larger ionization energies, and smaller electronegativity values. b) smaller atoms have smaller ionization energies, and smaller electronegativity values. c) smaller atoms have smaller ionization energies, and larger electronegativity values. d) smaller atoms have larger ionization energies, and larger electronegativity values. e) There are no general trends among atomic radius, ionization energy, and electronegativity. 2. Given that the first two ionization energy values for Rb(g) = 403 kj/mol and 2633 kj/mol, respectively, determine the electron affinity of Rb(g). a) 403 kj/mol b) 2230 kj/mol c) 2633 kj/mol d) 3036 kj/mol e) The electron affinity of Rb(g) cannot be determined with these data. 3. The following graph plots the first, second, and third ionization energies for Be, B, and C. #1 #2 #3 Which of the following corrects matches the plot to the element? a) #1 = Be #2 = B #3 = C b) #1 = Be #2 = C #3 = B c) #1 = C #2 = B #3 = Be d) #1 = C #2 = Be #3 = B e) #1 = B #2 = C #3 = Be
3 Hour Exam III Page No Consider a compound formed from a Group 2 element (such as Be or Mg and designated as M below) and a Group 6 element (such as O or S and designated as X below). Which of the following is the most reasonable statement? a) An ionic compound with the ions M 2+ and X 2- is the only stable form of the compound. b) An ionic compound with the ions M + and X - could be thermodynamically favorable, but an ionic compound with the ions M 2+ and X 2- is more thermodynamically favorable. c) The compound between M and X will be in a 1:1 ratio but it is best to think of the bond as polar covalent since atoms do not want to lose electrons (thermodynamically speaking). d) The bond between M and X is polar covalent as explained in c above, but we cannot determine the molecular formula without a molar mass. e) An ionic compound will form but we cannot predict the charges with any degree of accuracy. 5. Consider the following reaction: C 2 H 4 (g) + X 2 (g) CH 2 XCH 2 X(g) ΔH = 549 kj Estimate the C X bond energy given that the C H bond energy is 413 kj/mol, the C C bond energy is 347 kj/mol, the C=C bond energy is 614 kj/mol, and the X X bond energy is 154 kj/mol. Note: you will need to draw Lewis structures for the reactants and products to determine the possible presence of multiple bonds. a) 198 kj b) 352 kj c) 485 kj d) 704 kj e) 970 kj 6. Consider the following molecules/ions: O 3 NO 3 PO 4 3 CH 4 SO 3 How many total Lewis structures (resonance structures) can be drawn for the molecules/ions above that obey the octet rule? a) 5 b) 8 c) 10 d) 12 e) Consider the following molecules: SF 4 ICl 5 XeF 4 NF 3 For how many of the molecules above are the geometry (electron-pair arrangement) and shape (molecular structure) the same? a) 0 b) 1 c) 2 d) 3 e) 4
4 Hour Exam III Page No How many of the following statements is/are false? I. When the difference in electronegativity between two atoms is very large, the bond most likely to form is an ionic bond. II. Covalent bonding results from the sharing of valence electrons between two atoms. III. The valence electrons in a polar bond are found nearer (on the average) to the more electronegative atom in the bond. IV. VSEPR theory states that the central atom in a molecule has the bonded atoms and lone pairs arranged so to minimize electron-electron repulsions. V. If a molecule has polar bonds it is a polar molecule. VI. It is possible for a molecule with polar bonds to have no overall dipole moment. a) 0 b) 1 c) 2 d) 3 e) 4 9. What type(s) of intermolecular force(s) is/are exhibited by propane (C 3 H 8 )? a) hydrogen bonding and London dispersion forces b) hydrogen bonding only c) London dispersion forces only d) dipole-dipole and London dispersion forces e) dipole-dipole only 10. For which of the following mixtures is ΔH soln expected to be the most positive? a) C 6 H 14 and C 7 H 16 b) H 2 O and CH 3 CH 2 OH c) (CH 3 ) 2 CO and H 2 O d) CH 3 CH 2 OH and CH 3 OH e) C 7 H 16 and H 2 O 11. When table salt (solid NaCl) is dissolved in water, the overall process is slightly endothermic (the change in enthalpy is 4 kj/mol). The lattice energy for NaCl is 787 kj/mol. Determine the enthalpy of hydration. a) 783 kj/mol b) 791 kj/mol c) 783 kj/mol d) 791 kj/mol e) Consider two pure gaseous substances A and B each made of molecules of approximately the same size. Substance A consists of molecules which are more polar than those of substance B. How many of the following statements is/are true? I. Substance A has a higher vapor pressure than substance B. II. Substance A has a higher boiling point than substance B. III. Substance A is a more ideal gas than substance B IV. The bonds in molecule A must be more polar than the bonds in molecule B. a) 0 b) 1 c) 2 d) 3 e) 4
5 Hour Exam III Page No For each of the following molecules, choose the correct molecular geometry, shape, and polarity. 13. Sulfur tetrafluoride (sulfur is the central atom) GEOMETRY SHAPE POLARITY a) trigonal bipyramid see-saw polar b) tetrahedral tetrahedral polar c) tetrahedral tetrahedral non-polar d) octahedral square planar non-polar e) trigonal bipyramid trigonal pyramid polar 14. Sulfur dioxide (sulfur is the central atom) GEOMETRY SHAPE POLARITY a) trigonal planar bent non-polar b) linear linear non-polar c) tetrahedral bent polar d) linear linear polar e) trigonal planar bent polar 15. Nitrogen triiodide (nitrogen is the central atom) GEOMETRY SHAPE POLARITY a) trigonal planar trigonal planar non-polar b) tetrahedral trigonal planar non-polar c) tetrahedral tetrahedral non-polar d) tetrahedral trigonal pyramid polar e) trigonal planar bent polar Radioactive nuclei decay according to first-order kinetics. If a radioactive sample decays from 1.00 x 10 4 nuclei to 625 nuclei in 10.0 minutes, determine the half-life of this radioactive species. a) min b) 1.50 min c) 2.00 min d) 2.50 min e) 5.33 min 17. Consider the reaction type aa products which is zero-order in A. How many of the following statements is/are true? I. [A] remains constant with time. II. The half-life is equal to the rate constant, k. III. The rate of the reaction remains constant over time. IV. An increase in temperature will not affect the rate of the reaction. a) 0 b) 1 c) 2 d) 3 e) For a reaction: aa Products, [A] 0 = 4.00 M, and the first two half-lives are 48 and 24 minutes, respectively. Calculate [A] at t= 81 minutes. a) 0.48 M b) 0.62 M c) 1.24 M d) 1.49 M e) 2.81 M
6 Hour Exam III Page No Consider the following data concerning the equation: H 2 O 2 (aq) + 3I (aq) + 2H + (aq) I 3 (aq) + 2H 2 O [H 2 O 2 ] [I ] [H + ] rate I M 5.00 x 10-4 M 1.00 x 10-2 M M/sec II M 1.00 x 10-3 M 1.00 x 10-2 M M/sec III M 1.00 x 10-3 M 1.00 x 10-2 M M/sec IV M 1.00 x 10-3 M 2.00 x 10-2 M M/sec 19. The rate law for this reaction is a) rate = k[h 2 O 2 ][I ][H + ] b) rate = k[h 2 O 2 ] 2 [I ] 2 [H + ] 2 c) rate = k[i ][H + ] d) rate = k[h 2 O 2 ][H + ] e) rate = k[h 2 O 2 ][I ] 20. The average value for the rate constant (units with M and sec) is a) 108 b) 137 c) 2.71 x 10 3 d) 2.74 x 10 4 e) 3.14 x Two mechanisms are proposed: I. H 2 O 2 + I H 2 O + OI II. H 2 O 2 + I + H + H 2 O + HOI OI + H + HOI HOI + I + H + I 2 + H 2 O HOI + I + H + I 2 + H 2 O I 2 + I I 3 I 2 + I - I 3 Which of the following describes a potentially correct mechanism? a) Mechanism I with the first step the rate determining step. b) Mechanism I with the second step the rate determining step. c) Mechanism II with the first step rate determining. d) Mechanism II with the second step rate determining. e) Neither mechanism supports the correct rate law How many of the following are true concerning catalysts? I. The catalyzed reaction has a different rate constant from the uncatalyzed reaction. II. The catalyzed reaction has a different value of ΔH rxn from the uncatalyzed reaction. III. The catalyzed reaction has a different equilibrium constant from the uncatalyzed reaction. IV. The catalyzed reaction has a different activation energy from the uncatalyzed reaction. a) 0 b) 1 c) 2 d) 3 e) 4
7 Hour Exam III Page No Consider the following reaction: The following mechanism is proposed: 2NO(g) + H 2 (g) N 2 O(g) + H 2 O(g) 1. 2NO(g) N 2 O 2 (g) 2. N 2 O 2 (g) + H 2 (g) N 2 O(g) + H 2 O(g) Answer the following questions. Note: the k terms in the rate laws takes into account the rate law constants for the elementary steps and are not necessarily equivalent in the choices below. 23. Assuming the first step is a fast equilibrium step and the second step is rate-determining, which of the following best represents the rate law? a) rate = k[h 2 ] b) rate = k[no][h 2 ] c) rate = k[no][h 2 ] 2 d) rate = k[no] 2 [H 2 ] e) rate = k[no] 2 [H 2 ] 2 d [H ] 24. Using rate = 2, use the steady-state approximation to determine the rate law for the dt proposed mechanism and choose the correct statement below. a) If [H 2 ] is very low, the rate law appears to be the same as the rate law you determined in #23. If [H 2 ] is very high, the rate law appears to be rate = k[no] 2. b) If [H 2 ] is very high, the rate law appears to be the same as the rate law you determined in #23. If [H 2 ] is very low, the rate law appears to be rate = k[no] 2. c) If [NO] is very low, the rate law appears to be the same as the rate law you determined in #23. If [NO] is very high, the rate law appears to be rate = k[h 2 ]. d) If [NO] is very high, the rate law appears to be the same as the rate law you determined in #23. If [NO] is very low, the rate law appears to be rate = k[h 2 ]. e) The rate law appears the same as what you determined in #23 regardless of [NO] or [H 2 ] Recall the hydrogen chloride cannon demo from lecture in which a mixture of hydrogen gas and chlorine gas was initiated by a burning magnesium strip: H 2 (g) + Cl 2 (g) 2HCl(g) Suppose the activation energy for the reaction is changed from 239 kj/mol to kj/mol at 298K by the introduction of a catalyst. Calculate the ratio of rate(catalyzed):rate(uncatalyzed). Assume the pre-exponential factor, A, is the same for the catalyzed and uncatalyzed reactions. a) b) 1.03 c) 1.54 d) 2.30 x 10 7 e) 5.30 x 10 14
8 Hour Exam III Page No Recall the activated complex demonstration performed in lecture (the one which reacted to form a green transition state, and then turned pink to reveal the original catalyst). Suppose we run two trials, one at 55 C and one at 65 C. The times for the reaction are 85 seconds and 41 seconds, respectively. Determine the activation energy for this reaction. Assume the frequency factor, A, to be constant. a) 261 J/mol b) 663 J/mol c) 48.5 kj/mol d) 67.2 kj/mol e) 98.3 kj/mol 27. A first-order reaction is 42% complete at the end of 17 minutes. Determine the value of the rate constant (units are min -1 ). a) 3.2 x 10-2 b) 5.1 x 10-2 c) 1.8 d) 20 e) Choose the correct graph for the plots described below. A graph can be chosen once, more than once, or not at all. a) b) c) d) e) 28. A plot of [A] vs. time for reaction type aa products which is zero-order in A. [C] 29. A plot of t ½ vs. [A] for reaction type aa products which is first-order in A. [E] 30. t ½ vs k at constant temperature for a reaction type aa products which is first-order in A. [D]
9 Hour Exam III Page No We discussed in lecture how compounds can have the same chemical formula but different structures, and therefore different properties. Four different compounds exist with the chemical formula C 3 H 9 N. One such compound is isopropyl amine which is used in chemical weapons. Isopropyl amine has the following Lewis structure: a. Draw the three other distinct Lewis structures with the chemical formula C 3 H 9 N. For each of these structures you should minimize formal charge. Can you choose a best Lewis structure based on formal charge? If so, rank the structures. If not, explain why not. [8 pts.] All have 0 formal charge on all atoms; each equally good. b. Of the three compounds in part a, one is a gas at room conditions and the other two are liquids. Label which is the gas and which are liquids, and rank the two liquids in terms of boiling point (which boils at the higher temperature?). Support your answer with a discussion of VSEPR theory, explain the nature of any relevant intermolecular force, and explain the relationship between intermolecular forces and boiling points. [12 pts.] Trimethylamine is a gas and n-propylamine has a higher boiling point than N-methylethylamine. All same size, thus need to look at types of intermolecular forces. All tetrahedral/trigonal planar around the nitrogen, thus all polar (dipole-dipole interactions). Trimethylamine exhibits dipole-dipole interactions but not hydrogen bonding. N-methylethylamine contains an N-H bond which gives rise to hydrogen bonding. And n-propylamine contains two N-H bonds rather than one. H-bonding strong form of dipole-dipole interactions because N-H bond is particularly polar and the hydrogen atom is small. Strong IMF the more attraction between the molecules, thus the more energy required to separate the molecules to form a gas (higher boiling point).
10 Hour Exam III Page No In the future, you are visiting friends at their 10 th College Reunion (not UIUC). They are looking through an old chemistry notebook when they come across the following chemical equation: H 2 (g) + Cl 2 (g) 2HCl You all notice that there is no phase for HCl. I think it is a solid, states one of your friends. It is made up of H + and Cl ions. Another friend replies, I think it is a gas. The bond between hydrogen and chlorine is covalent not ionic. They are both ready to call off their lifelong friendship when you smile, take your Chemistry 202 textbook from your backpack, and tell them that thermodynamics can solve this problem. So, which friend is correct?; that is, which is the correct chemical equation for the reaction of hydrogen and chlorine gases? Your choices are: H 2 (g) + Cl 2 (g) 2HCl(s) or H 2 (g) + Cl 2 (g) 2HCl(g) Use the data below to provide quantitative support to your answer; that is, provide numbers and calculations along with your discussion. A complete answer will include a discussion of ΔH, ΔS, ΔS surr, and ΔS univ along with calculating any of these values you can while estimating the sign (with explanation) for those you cannot calculate. NOTE: Assume the lattice energy of HCl(s) = 1125 kj/mol
11 Hour Exam III Page No Provide your answer in the space below. See the previous page for what is required. [20 pts.] ===================================================================== Consider H 2 (g) + Cl 2 (g) 2HCl(g) We need ΔS univ to be positive for the process to be spontaneous. In this case, ΔS 0 because 2 moles of gas are becoming 1 mole of gas. ΔS surr due to enthalpy change of reaction, and we can estimate this with bond energies: Form 2HCl(g): 2(427 kj/mol) Break H 2 (g): +432 kj/mol Break Cl 2 (g): +239 kj/mol Overall: 183 kj Thus ΔH rxn is negative (exothermic), so ΔS surr is positive (heat released to surroundings) Thus, ΔS univ = ΔS surr + ΔS = (positive) + ( 0) = positive; should be spontaneous Consider H 2 (g) + Cl 2 (g) 2HCl(s) We need ΔS univ to be positive for the process to be spontaneous. In this case, ΔS is negative because 2 moles of gas are becoming 1 mole of solid. Thus, ΔS surr must be positive for the reaction to occur spontaneously. ΔS surr due to enthalpy change of reaction, and we can determine this as follows: H 2 2H (bond energy = +432 kj) 2H 2H + + 2e (2 x ionization energy = 2(~ kj/mol) = ~ 2400 kj Cl 2 2Cl (bond energy = +239 kj) 2Cl +2e 2Cl (2 x electron affinity = 2(~ 350 kj/mol) = ~ 700 kj 2H + + 2Cl 2HCl(s) (2 x lattice energy) = 2( 1125 kj/mol) = 2250 kj Overall = +121 kj Thus ΔH rxn is positive (endothermic), so ΔS surr is negative (heat released to surroundings) Thus, ΔS univ = ΔS surr + ΔS = (negative) + (negative) = negative; not spontaneous. Thus, reaction is H 2 (g) + Cl 2 (g) 2HCl(g)
12 Hour Exam III Page No The following data were collected in two studies of the reaction 2H 2 (g) + 2NO(g) N 2 (g) + 2H 2 O(g) EXPERIMENT #1 EXPERIMENT #2 TIME (sec) [NO] [NO] x 10-2 M 1.00 x 10-2 M x 10-3 M 5.00 x 10-3 M x 10-3 M 3.33 x 10-3 M x 10-3 M 2.50 x 10-3 M x 10-3 M 2.00 x 10-3 M In experiment #1, [H 2 ] 0 = 10.0 M In experiment #2, [H 2 ] 0 = 20.0 M The following three mechanisms are proposed: I H 2 + NO H 2 O + N (slow) N + NO N 2 + O O + H 2 H 2 O II 2NO N 2 O 2 (fast equil.) N 2 O 2 + H 2 N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O III H 2 + 2NO N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O Your overall goal is to evaluate these mechanisms and choose the best mechanism. Provide a complete defense of your answer, which includes an explanation of which mechanism(s) you exclude and why. If you believe two (or even all three) mechanisms are equally good, explain why Along the way answer the questions on the following pages.
13 Hour Exam III Page No (con t). a. Use the concentration vs. time data to determine the rate law for the reaction. Show all work/explain your thinking (support your answer). Full credit is given for answers not relying on graphs. [4 pts.] rate = k[h 2 ][NO] 2 Data from Expt #2 show that the half-life increases by a factor of two as [NO] changes. Thus the reaction is second order with respect to NO. Between Expt #1 and Expt #2, [H 2 ] was doubled, and the half-life was cut by a factor of 2 (from 20 seconds to 10 seconds). Thus, doubling [H 2 ] doubled the rate. This means that the reaction is first order with respect to H 2. b. Solve for the rate constant (k) for the reaction. Include units. Show all work/explain your thinking (support your answer). [4 pts.] Using t=0 and t =10 from Expt #2 (for example): 1/(5.00 x 10-3 ) = k (10) + 1/(1.00 x 10-2 ); k = 10 = k[h 2 ]; 10 = k[20] k = M -2 s -1 Using t=0 and t =30 from Expt #1 (for example): 1/(4.00 x 10-3 ) = k (30) + 1/(1.00 x 10-2 ); k = 5 = k[h 2 ]; 5 = k[10] k = M -2 s -1 k = M -2 s -1 c. Calculate the concentration of NO(g) in experiment #2 at t=90.0 seconds. Show all work. [2 pts.] 1/[NO] = (10)(90) + 1/(1.00 x 10-2 ) [NO] = 1.00 x 10-3 M
14 Hour Exam III Page No (con t) d. Choose the best mechanism. Provide a complete defense of your answer, which includes an explanation of which mechanism(s) you exclude and why. If you believe two (or even all three) mechanisms are equally good, explain why. [10 pts.] I H 2 + NO H 2 O + N (slow) N + NO N 2 + O O + H 2 H 2 O Adds up to correct stoichiometry Seems chemically reasonable Gives wrong rate law REJECTED rate = k 1 [NO][H 2 ] II 2NO N 2 O 2 (fast equil.) N 2 O 2 + H 2 N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O Adds up to correct stoichiometry Seems chemically reasonable Gives correct rate law POSSIBLE rate = k 2 [N 2 O 2 ][H 2 ] k 1 [NO] 2 = k -1 [N 2 O 2 ] [N 2 O 2 ] = k 1 /k -1 [NO] 2 rate = k 2 k 1 /k -1 [NO] 2 [H 2 ] III H 2 + 2NO N 2 O + H 2 O (slow) N 2 O + H 2 N 2 + H 2 O Adds up to correct stoichiometry Seems chemically reasonable Gives correct rate law POSSIBLE rate = k 1 [NO] 2 [H 2 ] Thus, mechanisms II and III are possible. However, mechanism II is a slightly better mechanism because it does not rely on a termolecular elementary step (which is possible, but unlikely).
CHEMISTRY 202 Practice Hour Exam III. Dr. D. DeCoste T.A (60 pts.) 21 (15 pts.) 22 (20 pts.) 23 (25 pts.) Total (120 pts)
CHEMISTRY 202 Practice Hour Exam III Fall 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 23 questions on 10 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationCHEMISTRY 202 Hour Exam III. Dr. D. DeCoste T.A. 21 (16 pts.) 22 (21 pts.) 23 (23 pts.) Total (120 pts)
CHEMISTRY 202 Hour Exam III December 1, 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 23 questions on 12 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationCHEMISTRY 202 Hour Exam III. Dr. D. DeCoste T.A. 21 (16 pts.) 22 (21 pts.) 23 (23 pts.) Total (120 pts)
CHEMISTRY 202 Hour Exam III December 1, 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 23 questions on 12 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationDr. D. DeCoste T.A (80 pts.) 21 (30 pts.) 22 (25 pts.) 23 (25 pts.) Total (160 pts)
CHEMISTRY 202 Hour Exam III December 6, 2018 Dr. D. DeCoste Name KEY Signature T.A. This exam contains 23 questions on 13 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationCHEMISTRY 102B Hour Exam III. Dr. D. DeCoste T.A. Show all of your work and provide complete answers to questions 16 and (45 pts.
CHEMISTRY 102B Hour Exam III April 28, 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 17 questions on 8 numbered pages. Check now to make sure you have a complete exam. You have one hour and
More informationCHEMISTRY 101 Hour Exam III. Dr. D. DeCoste T.A (30 pts.) 16 (12 pts.) 17 (18 pts.) Total (60 pts)
CHEMISTRY 101 Hour Exam III April 27, 2017 Dr. D. DeCoste Name Signature T.A. This exam contains 17 questions on 6 numbered pages. Check now to make sure you have a complete exam. You have one hour and
More informationCHEMISTRY 202 Hour Exam III (Multiple Choice Section) Dr. D. DeCoste T.A.
CHEMISTRY 0 Hour Exam III (Multiple Choice Section) December 7, 07 Dr. D. DeCoste Name Signature T.A. This exam contains 0 questions on 4 numbered pages. Chec now to mae sure you have a complete exam.
More informationCHEMISTRY 101 Hour Exam III. Dr. D. DeCoste T.A (30 pts.) 16 (12 pts.) 17 (18 pts.) Total (60 pts)
CHEMISTRY 101 Hour Exam III April 27, 2017 Dr. D. DeCoste Name Signature T.A. This exam contains 17 questions on 6 numbered pages. Check now to make sure you have a complete exam. You have one hour and
More informationCHEMISTRY 202 Hour Exam III. Dr. D. DeCoste T.A (60 pts.) 31 (10 pts.) 32 (20 pts.) 33 (30 pts.) Total (120 pts)
CHEMISTRY 202 Hour Exam III December 4, 2014 Dr. D. DeCoste Name Signature T.A. This exam contains 33 questions on 12 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationBonding and IMF practice test MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name Bonding and IMF practice test MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) There are paired and unpaired electrons in the Lewis symbol
More informationLevel 3 Chemistry Demonstrate understanding of thermochemical principles and the properties of particles and substances
1 ANSWERS Level 3 Chemistry 91390 Demonstrate understanding of thermochemical principles and the properties of particles and substances Credits: Five Achievement Achievement with Merit Achievement with
More informationCHEM 101 Fall 09 Final Exam (a)
CHEM 101 Fall 09 Final Exam (a) On the answer sheet (scantron) write your name, student ID number, and recitation section number. Choose the best (most correct) answer for each question and enter it on
More informationQuestions 1-2 Consider the atoms of the following elements. Assume that the atoms are in the ground state. a. S b. Ca c. Ga d. Sb e.
AP Chemistry Fall Semester Practice Exam 5 MULTIPLE CHOICE PORTION: Write the letter for the correct answer to the following questions on the provided answer sheet. Each multiple choice question is worth
More informationPLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET. c) SeF 4
Chem 130 EXAM 4 Fall 99 PLEASE DO NOT MARK ON THE EXAM. ALL ANSWERS SHOULD BE INDICATED ON THE ANSWER SHEET QUESTIONS 1-5 MAY HAVE MORE THAN ONE POSSIBLE ANSWER CIRCLE ALL CORRECT RESPONSES TO EACH QUESTION
More informationChem 1210 Final Spring points Dr. Luther Giddings
Chem 1210 Final Spring 2002 150 points Dr. Luther Giddings Name Instructions: This is a closed book, closed notebook test. You may not discuss this exam with anyone, either during or after the exam, until
More informationChapter Eight. p328. Bonding: General Concepts
Chapter Eight p328 Bonding: General Concepts 1 Contents 8-1 Types of Chemical Bonds p330 Coulomb s law The energy of interaction between a pair of ions can be calculated using Coulomb s law: E 19 Q1Q 2
More informationSuccess means having the courage, the determination, and the will to become the person you believe you were meant to be.
CHEMISTRY 101 Hour Exam III December 7, 2006 Adams/Le Name KEY Signature T.A./Section Success means having the courage, the determination, and the will to become the person you believe you were meant to
More informationChapter 6 Chemistry Review
Chapter 6 Chemistry Review Multiple Choice Identify the choice that best completes the statement or answers the question. Put the LETTER of the correct answer in the blank. 1. The electrons involved in
More informationBROOKLYN COLLEGE. FINAL EXAMINATION IN CHEMISTRY 1. Fall, Paul Cohen, instructor.
Name BROOKLYN COLLEGE. FINAL EXAMINATION IN CHEMISTRY 1. Fall, 2008. Paul Cohen, instructor. Formulas: Write the chemical formula for each of the following: Al 2 (S0 4 ) 3 1. Aluminum sulfate Ni(OH) 2
More informationChemical Bonding AP Chemistry Ms. Grobsky
Chemical Bonding AP Chemistry Ms. Grobsky What Determines the Type of Bonding in Any Substance? Why do Atoms Bond? The key to answering the first question are found in the electronic structure of the atoms
More informationChemical Reactions and Equations
Chemical Reactions and Equations 1991 B The molecular formula of a hydrocarbon is to be determined by analyzing its combustion products and investigating its colligative properties. (a) The hydrocarbon
More informationCHEMISTRY 110 EXAM 2 Feb 25, 2013 FORM A
EMISTRY 110 EXAM 2 Feb 25, 2013 FORM A 1. ow many valence electrons and lone pairs are in the structure of the ammonium ion? # valence electrons # lone pairs A. 8 0 B. 10 1. 8 1 D. 10 2 E. 12 3 2. Which
More informationChap 10 Part 4Ta.notebook December 08, 2017
Chapter 10 Section 1 Intermolecular Forces the forces between molecules or between ions and molecules in the liquid or solid state Stronger Intermolecular forces cause higher melting points and boiling
More informationGas Laws. Bonding. Solutions M= moles solute Mass %= mass solute x 100. Acids and Bases. Thermochemistry q = mc T
Name Period Teacher Practice Test: OTHS Academic Chemistry Spring Semester 2017 The exam will have 100 multiple choice questions (1 point each) Formula sheet (see below) and Periodic table will be provided
More informationWhat is a Bond? Chapter 8. Ionic Bonding. Coulomb's Law. What about covalent compounds?
Chapter 8 What is a Bond? A force that holds atoms together. Why? We will look at it in terms of energy. Bond energy- the energy required to break a bond. Why are compounds formed? Because it gives the
More informationChapter 8. Basic Concepts of Chemical Bonding
Chapter 8. Basic Concepts of Chemical Bonding 8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule 8.2 Ionic Bonding Consider the reaction between sodium and chlorine: Na(s) + ½ Cl 2 (g) NaCl(s) H o f
More informationChapter 8. Basic Concepts of Chemical Bonding
Chapter 8. Basic Concepts of Chemical Bonding 8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule 8.2 Ionic Bonding positive and negative ions form an ionic lattice, in which each cation is surrounded
More informationCh 10 Chemical Bonding, Lewis Structures for Ionic & Covalent Compounds, and Predicting Shapes of Molecules
Fructose Water Ch 10 Chemical Bonding, Lewis Structures for Ionic & Covalent Compounds, and Predicting Shapes of Molecules Carbon Dioxide Ammonia Title and Highlight TN Ch 10.1 Topic: EQ: Right Side NOTES
More informationKWANTLEN UNIVERSITY COLLEGE DEPARTMENT OF CHEMISTRY
KWANTLEN UNIVERSITY COLLEGE DEPARTMENT OF CHEMISTRY Final Examination: CHEM 1110 Name: Student Number: December 17, 2001 Time: 3 hours INSTRUCTIONS: 1. All calculations must be shown in order to receive
More informationMore Chemical Bonding
More Chemical Bonding Reading: Ch 10: section 1-8 Ch 9: section 4, 6, 10 Homework: Chapter 10:.31, 33, 35*, 39*, 43, 47, 49* Chapter 9: 43, 45, 55*, 57, 75*, 77, 79 * = important homework question Molecular
More informationCP Covalent Bonds Ch. 8 &
CP Covalent Bonds Ch. 8 & 9 2015-2016 Why do atoms bond? Atoms want stability- to achieve a noble gas configuration ( ) For bonds there is a transfer of electrons to get an octet of electrons For covalent
More informationCh 6 Chemical Bonding
Ch 6 Chemical Bonding What you should learn in this section (objectives): Define chemical bond Explain why most atoms form chemical bonds Describe ionic and covalent bonding Explain why most chemical bonding
More informationForm J. Test #4 Last Name First Name Zumdahl, Chapters 8 and 9 November 23, 2004
Form J Chemistry 1441-023 Name (please print) Test #4 Last Name First Name Zumdahl, Chapters 8 and 9 November 23, 2004 Instructions: 1. This exam consists of 27 questions. 2. No scratch paper is allowed.
More informationBonding. Honors Chemistry 412 Chapter 6
Bonding Honors Chemistry 412 Chapter 6 Chemical Bond Mutual attraction between the nuclei and valence electrons of different atoms that binds them together. Types of Bonds Ionic Bonds Force of attraction
More information1) Based on the octet rule, magnesium most likely forms a ion. A) Mg- B) Mg6+ C) Mg2+ D) Mg6- E) Mg2-
1) ased on the octet rule, magnesium most likely forms a ion. ) Mg- ) Mg6+ ) Mg2+ ) Mg6- E) Mg2-2) ased on the octet rule, phosphorus most likely forms a ion. ) P5- ) P5+ ) P+ ) P3- E) P3+ 3) Which ion
More informationAP Chemistry. Unit #7. Chemical Bonding & Molecular Shape. Zumdahl Chapters 8 & 9 TYPES OF BONDING BONDING. Discrete molecules formed
AP Chemistry Unit #7 Chemical Bonding & Molecular Shape Zumdahl Chapters 8 & 9 TYPES OF BONDING BONDING INTRA (Within (inside) compounds) STRONG INTER (Interactions between the molecules of a compound)
More informationChapter 8. Basic Concepts of Chemical Bonding
Chapter 8. Basic Concepts of Chemical Bonding 8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule 8.2 Ionic Bonding positive and negative ions form an ionic lattice, in which each cation is surrounded
More informationSuccess means having the courage, the determination, and the will to become the person you believe you were meant to be.
CHEMISTRY 101 Hour Exam III December 1, 2016 Adams/ Huynh Name Signature Section Success means having the courage, the determination, and the will to become the person you believe you were meant to be.
More informationChemical Bonding Chapter 8
Chemical Bonding Chapter 8 Get your Clicker, 2 magnets, goggles and your handouts Nov 15 6:15 PM Recall that: Ionic-Involves the transfer of electrons - forms between a metal and a nonmetal Covalent-Involves
More informationSTD-XI-Science-Chemistry Chemical Bonding & Molecular structure
STD-XI-Science-Chemistry Chemical Bonding & Molecular structure Chemical Bonding Question 1 What is meant by the term chemical bond? How does Kessel-Lewis approach of bonding differ from the modern views?
More informationChemical bonding is the combining of elements to form new substances.
Name Covalent Bonding and Nomenclature: Unit Objective Study Guide Class Period Date Due 1. Define chemical bonding. What is chemical bonding? Chemical bonding is the combining of elements to form new
More informationUnit 7: Basic Concepts of Chemical Bonding. Chemical Bonds. Lewis Symbols. The Octet Rule. Transition Metal Ions. Ionic Bonding 11/17/15
Unit 7: Basic Concepts of Chemical Bonding Topics Covered Chemical bonds Ionic bonds Covalent bonds Bond polarity and electronegativity Lewis structures Exceptions to the octet rule Strength of covalent
More informationSL Score. HL Score ! /30 ! /48. Practice Exam: Paper 1 Topic 4: Bonding. Name
Name Practice Exam: Paper 1 Topic 4: Bonding SL SL Score! /30 HL Score! /48 1. What is the correct Lewis structure for hypochlorous acid, a compound containing chlorine, hydrogen and oxygen? A. B. C. D.
More information2) C 2 H 2 (g) + 2 H 2 (g) ---> C 2 H 6 (g) Information about the substances
Thermochemistry 1) 2 C 4 H 10 (g) + 13 O 2 (g) ------> 8 CO 2 (g) + 10 H 2 O(l) The reaction represented above is spontaneous at 25 C. Assume that all reactants and products are in their standard states.
More informationNOTES #28 Bonds & Thermochemistry AP Chemistry
NOTES #28 Bonds & Thermochemistry AP Chemistry - When studying thermochemistry, we determined ΔH or ΔH rxn of a reaction by using ΔH f values. For practice s sake, determine ΔH rxn for the formation of
More informationChemical Bonding I: Basic Concepts
Chemical Bonding I: Basic Concepts Chapter 9 Valence electrons are the outer shell electrons of an atom. The valence electrons are the electrons that participate in chemical bonding. Group e - configuration
More informationChapter 8. Bonding: General Concepts
Chapter 8 Bonding: General Concepts Chapter 8 Table of Contents 8.1 Types of Chemical Bonds 8.2 Electronegativity 8.3 Bond Polarity and Dipole Moments 8.4 Ions: Electron Configurations and Sizes 8.5 Energy
More informationCHEMISTRY 107 Section 501 Final Exam Version A December 12, 2016 Dr. Larry Brown
NAME: (print) UIN #: CHEMISTRY 107 Section 501 Final Exam Version A December 12, 2016 Dr. Larry Brown This is a 2-hour exam, and contains 11 problems. There should be 14 numbered pages, including this
More informationWe study bonding since it plays a central role in the understanding of chemical reactions and understanding the chemical & physical properties.
AP Chapter 8 Notes Bonding We study bonding since it plays a central role in the understanding of chemical reactions and understanding the chemical & physical properties. Chemical Bond: holding atoms together
More informationAP Chemistry A. Allan Chapter 8 Notes - Bonding: General Concepts
AP Chemistry A. Allan Chapter 8 Notes - Bonding: General Concepts 8.1 Types of Chemical Bonds A. Ionic Bonding 1. Electrons are transferred 2. Metals react with nonmetals 3. Ions paired have lower energy
More informationChapter 8. Basic Concepts of Chemical Bonding
Chapter 8. Basic Concepts of Chemical Bonding 8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule 8.2 Ionic Bonding Consider the reaction between sodium and chlorine: Na(s) + ½ Cl 2 (g) NaCl(s) H o f
More informationChapter 8. Bonding: General Concepts. Copyright 2017 Cengage Learning. All Rights Reserved.
Chapter 8 Bonding: General Concepts Chapter 8 Table of Contents (8.1) (8.2) (8.3) (8.4) (8.5) (8.6) (8.7) (8.8) Types of chemical bonds Electronegativity Bond polarity and dipole moments Ions: Electron
More informationChapter 8. Bonding: General Concepts
Chapter 8 Bonding: General Concepts Chapter 8 Table of Contents 8.1 Types of Chemical Bonds 8.3 Bond Polarity and Dipole Moments 8.5 Energy Effects in Binary Ionic Compounds 8.6 Partial Ionic Character
More informationChem 105 Final Exam. Here is the summary of the total 225 points plus 10 bonus points. Carefully read the questions. Good luck!
May 3 rd, 2012 Name: CLID: Score: Chem 105 Final Exam There are 50 multiple choices that are worth 3 points each. There are 4 problems and 1 bonus problem. Try to answer the questions, which you know first,
More informationChapter 8. Basic Concepts of Chemical Bonding
Chapter 8. Basic Concepts of Chemical Bonding 8.1 Lewis Symbols and the Octet Rule When atoms or ions are strongly attracted to one another, we say that there is a chemical bond between them. In chemical
More informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (30 pts.) 22 (30 pts.)
CHEMISTRY 202 Hour Exam II October 25, 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 10 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationCHEMISTRY HONORS LEOCE Study Guide
BENCHMARK: N.1.1-1.2, N.1.6, N.3.1, N.3.3, N.3.4 CHEMISTRY HONORS CHEMISTRY AND SCIENTIFIC MEASUREMENT TEXTBOOK: Glencoe, Chemistry: Matter and Change, Chapters 1-3 ESSENTIAL QUESTION: How is measurement
More informationChemistry 3.4 AS WORKBOOK. Working to Excellence Working to Excellence
Chemistry 3.4 AS 91390 Demonstrate understanding of thermochemical principles and the properties of particles and substances WORKBOOK Working to Excellence Working to Excellence CONTENTS 1. Writing Excellence
More informationNa Cl Wants to lose ONE electron! Na Cl Ionic Bond TRANSFER of electrons between atoms. Ionic Bonding. Ionic Bonding.
BONDING Chemical Bond Attraction that holds atoms together Types include IONIC, METALLIC, or COVALENT Differences in electronegativity determine the bond type Ionic Bond TRANSFER of electrons between atoms
More informationName Unit Three MC Practice March 15, 2017
Unit Three: Bonding & Molecular Geometry Name Unit Three MC Practice March 15, 2017 1. What is the hybridization of the oxygen atom in water? a) sp b) sp 2 c) sp 3 d) It is not hybridized 2. When a double
More informationName AP CHEM / / Chapter 8 Outline Bonding: General Concepts
Name AP CHEM / / Chapter 8 Outline Bonding: General Concepts Types of Chemical Bonds Information about the strength of a bonding interaction is obtained by measuring the bond energy, which is the energy
More informationName AP CHEM / / Collected AP Exam Essay Answers for Chapter 16
Name AP CHEM / / Collected AP Exam Essay Answers for Chapter 16 1980 - #7 (a) State the physical significance of entropy. Entropy (S) is a measure of randomness or disorder in a system. (b) From each of
More informationChapter 8. Bonding: General Concepts
Chapter 8 Bonding: General Concepts Chapter 8 Questions to Consider What is meant by the term chemical bond? Why do atoms bond with each other to form compounds? How do atoms bond with each other to form
More information"Let me tell you the secret that has led me to my goal. My strength lies solely in my tenacity." -Louis Pasteur-
CHEMISTRY 101 Hour Exam III December 4, 2014 Adams/Esbenshade Name Signature Section "Let me tell you the secret that has led me to my goal. My strength lies solely in my tenacity." -Louis Pasteur- This
More informationMULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. 1) The F-B-F bond angle in the BF3 molecule is. A) 109.5e B) 120e C) 180e D) 90e E) 60e
More informationChapter 8: Concepts of Chemical Bonding
Chapter 8: Concepts of Chemical Bonding Learning Outcomes: Write Lewis symbols for atoms and ions. Define lattice energy and be able to arrange compounds in order of increasing lattice energy based on
More informationChapter 8 Covalent Boding
Chapter 8 Covalent Boding Molecules & Molecular Compounds In nature, matter takes many forms. The noble gases exist as atoms. They are monatomic; monatomic they consist of single atoms. Hydrogen chloride
More informationMULTIPLE CHOICE PORTION:
AP Chemistry Fall Semester Practice Exam 4 MULTIPLE CHOICE PORTION: Write the letter for the correct answer to the following questions on the provided answer sheet. Each multiple choice question is worth
More informationbond energy- energy required to break a chemical bond -We can measure bond energy to determine strength of interaction
bond energy- energy required to break a chemical bond -We can measure bond energy to determine strength of interaction ionic compound- a metal reacts with a nonmetal Ionic bonds form when an atom that
More informationIonic Bond TRANSFER of electrons between atoms. Ionic Bonding. Ionic Bonding. Ionic Bonding. Attraction that holds atoms together
BONDING Chemical Bond Attraction that holds atoms together Types include IONIC, METALLIC, or COVALENT Differences in electronegativity determine the bond type Ionic Bond TRANSFER of electrons between atoms
More informationCHEM PRACTICE EXAM IV CLASS - SPRING 2017 ANSWER KEY
CHEM 1031 - PRACTICE EXAM IV CLASS - SPRING 2017 ANSWER KEY 1. When Group 1A (except for H) and Group 17 (7A) elements react with each other, they are most likely to form: A. Covalent or ionic bonds B.
More informationExample 9.1 Using Lewis Symbols to Predict the Chemical Formula of an Ionic Compound
Example 9.1 Using Lewis Symbols to Predict the Chemical Formula of an Ionic Compound For Practice 9.1 Use Lewis symbols to predict the formula for the compound that forms between magnesium and nitrogen.
More informationAtoms have the ability to do two things in order to become isoelectronic with a Noble Gas.
CHEMICAL BONDING Atoms have the ability to do two things in order to become isoelectronic with a Noble Gas. 1.Electrons can be from one atom to another forming. Positive ions (cations) are formed when
More information1. When two pure substances are mixed to form a solution, then always
Name: Date: 1. When two pure substances are mixed to form a solution, then always A) there is an increase in entropy. B) there is a decrease in entropy. C) entropy is conserved. D) heat is released. E)
More informationCHEMICAL BONDING. Chemical Bonds. Ionic Bonding. Lewis Symbols
CHEMICAL BONDING Chemical Bonds Lewis Symbols Octet Rule whenever possible, valence electrons in covalent compounds distribute so that each main-group element is surrounded by 8 electrons (except hydrogen
More informationBonding. Polar Vs. Nonpolar Covalent Bonds. Ionic or Covalent? Identifying Bond Types. Solutions + -
Chemical Bond Mutual attraction between the nuclei and valence electrons of different atoms that binds them together. Bonding onors Chemistry 412 Chapter 6 Types of Bonds Ionic Bonds Force of attraction
More informationCHEMISTRY 202 Practice Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (40 pts.) 22 (20 pts.)
CHEMISTRY 202 Practice Hour Exam II Fall 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 7 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationCartoon courtesy of NearingZero.net. Chemical Bonding and Molecular Structure
Cartoon courtesy of NearingZero.net Chemical Bonding and Molecular Structure Chemical Bonds Forces that hold groups of atoms together and make them function as a unit. 3 Major Types: Ionic bonds transfer
More information(03) WMP/Jun10/CHEM4
Thermodynamics 3 Section A Answer all questions in the spaces provided. 1 A reaction mechanism is a series of steps by which an overall reaction may proceed. The reactions occurring in these steps may
More informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 31 (20 pts.) 32 (40 pts.)
CHEMISTRY 202 Hour Exam II October 27, 2015 Dr. D. DeCoste Name Signature T.A. This exam contains 32 questions on 11 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationChapter 7. Ionic & Covalent Bonds
Chapter 7 Ionic & Covalent Bonds Ionic Compounds Covalent Compounds 7.1 EN difference and bond character >1.7 = ionic 0.4 1.7 = polar covalent 1.7 Electrons not shared at
More informationTopics to Expect: Periodic Table: s, p, d, f blocks Metal, Metalloid, Non metal, etc. Periodic Trends, Family names Electron Configuration: Orbitals a
Chemistry Final Exam Review and Practice Chapters Covered ESSENTIALLY CUMMULATIVE List of Chapters: Ch: 6, 7, 8, 9, 10, 13, 14, 15, 16, 19, 20 Topics to Expect: Periodic Table: s, p, d, f blocks Metal,
More informationThe energy associated with electrostatic interactions is governed by Coulomb s law:
Chapter 8 Concepts of Chemical Bonding Chemical Bonds Three basic types of bonds: Ionic Electrostatic attraction between ions Covalent Sharing of electrons Metallic Metal atoms bonded to several other
More informationChapter 8. Basic Concepts of Chemical Bonding
Chapter 8 Basic Concepts of Chemical Bonding Chemical Bonds An attractive force that holds two atoms together in a more complex unit Three basic types of bonds Ionic Electrons are transferred from one
More informationChapter 7. Chemical Bonding I: Basic Concepts
Chapter 7. Chemical Bonding I: Basic Concepts Chemical bond: is an attractive force that holds 2 atoms together and forms as a result of interactions between electrons found in combining atoms We rarely
More informationChem 401 Unit 1 (Kinetics & Thermo) Review
KINETICS 1. For the equation 2 H 2(g) + O 2(g) 2 H 2 O (g) How is the rate of formation of H 2 O mathematically related to the rate of disappearance of O 2? 2. Determine the relative reaction rates of
More informationChem Hughbanks Final Exam, May 11, 2011
Chem 107 - Hughbanks Final Exam, May 11, 2011 Name (Print) UIN # Section 503 Exam 3, Version # A On the last page of this exam, you ve been given a periodic table and some physical constants. You ll probably
More informationName: Hr: 8 Basic Concepts of Chemical Bonding
8.1-8.2 8.3-8.5 8.5-8.7 8.8 Name: Hr: 8 Basic Concepts of Chemical Bonding 8.1 Chemical Bonds, Lewis Symbols, and the Octet Rule State the type of bond (ionic, covalent, or metallic) formed between any
More informationCHEMISTRY 202 Hour Exam II. Dr. D. DeCoste T.A (60 pts.) 21 (30 pts.) 22 (30 pts.)
CHEMISTRY 202 Hour Exam II October 25, 2016 Dr. D. DeCoste Name Signature T.A. This exam contains 22 questions on 10 numbered pages. Check now to make sure you have a complete exam. You have two hours
More informationAP Chemistry- Practice Bonding Questions for Exam
AP Chemistry- Practice Bonding Questions for Exam Multiple Choice Identify the choice that best completes the statement or answers the question. 1. Which of the following is a correct Lewis structure for
More information12A Entropy. Entropy change ( S) N Goalby chemrevise.org 1. System and Surroundings
12A Entropy Entropy change ( S) A SPONTANEOUS PROCESS (e.g. diffusion) will proceed on its own without any external influence. A problem with H A reaction that is exothermic will result in products that
More informationLewis Structure. Lewis Structures & VSEPR. Octet & Duet Rules. Steps for drawing Lewis Structures
Lewis Structure Lewis Structures & VSEPR Lewis Structures shows how the are arranged among the atoms of a molecule There are rules for Lewis Structures that are based on the formation of a Atoms want to
More informationGroup 1 Group 2 Group 3 Group 4 Group 5 Group 6 Group 7 Group 8. Na Mg Al Si P S Cl Ar
CHM 111 Chapters 7 and 8 Worksheet and Study Guide Purpose: This is a guide for your as you work through the chapter. The major topics are provided so that you can write notes on each topic and work the
More informationChemistry 112 Spring 2007 Prof. Metz Exam 1 KEY
Chemistry 112 Spring 27 Prof. Metz Exam 1 KEY 1. Ammonia, NH 3, has a much higher boiling point than phosphine, PH 3. This is because: (A) NH 3 has a lower molecular weight than PH 3. (B) NH 3 is extensively
More informationChemistry 112 Spring 2007 Prof. Metz Exam 1 KEY
Chemistry 112 Spring 27 Prof. Metz Exam 1 KEY 1. The predominant intermolecular attractive force in solid sodium is: (A) ionic (B) covalent (C) metallic (D) dipole-dipole (E) induced dipole-induced dipole
More informationChemistry 112 Spring 2007 Prof. Metz Exam 1 KEY
Chemistry 112 Spring 27 Prof. Metz Exam 1 KEY 1. The predominant intermolecular attractive force in solid sodium is: (A) covalent (B) metallic (C) ionic (D) dipole-dipole (E) induced dipole-induced dipole
More informationMultiple Choice Identify the letter of the choice that best completes the statement or answers the question.
CHAPTER 4 Multiple Choice Identify the letter of the choice that best completes the statement or answers the question. 1. A substance is a brittle crystal that conducts electricity in molten liquid state
More informationName: Class: Date: Multiple Choice Identify the choice that best completes the statement or answers the question.
Name: Class: Date: Practice test 2013 Multiple Choice Identify the choice that best completes the statement or answers the question. Important constants For Water H fus = 6.01 kj/mol or 334 J/g H vap =
More informationCovalent Bonding. In nature, only the noble gas elements exist as uncombined atoms. All other elements need to lose or gain electrons
In nature, only the noble gas elements exist as uncombined atoms. They are monatomic - consist of single atoms. All other elements need to lose or gain electrons To form ionic compounds Some elements share
More informationg of CO 2 gas is at a temperature of 45 o C and a pressure of 125 kpa. What is the volume of the container? 11 L
Name period AP Chemistry Unit 5 answers 1. A fixed quantity of gas at 23⁰C exhibits a pressure of 748 torr and occupies a volume of 10.3 L. Calculate the volume the gas will occupy if the temperature is
More information