Chem 1A Dr. White Fall 2013 1 Chem istry W orkbook 1: Problem s For Exam 1 Section 1-1: Significant Figures How many significant figures do the following numbers have? 1) 1234 2) 0.023 3) 890 4) 91010 5) 9010.0 6) 1090.0010 4-3 7) 0.00120 8) 3.4 x 10 9) 9.0 x 10-2 10) 9.010 x 10 11) 0.00030 12) 1020010 13) 780. 14) 1000 15) 918.010 Solve the following mathematical problems such that the answers have the correct number of significant figures: 16) 334.54 grams + 198 grams = 17) 34.1 grams / 1.1 ml = 3 18) 2.11 x 10 joules / 34 seconds = 19) 0.110 meters 0.10 m = 20) 349 cm + 1.10 cm + 100 cm = 21) 450 meters / 114 seconds = 22) 298.01 kilograms + 34.112 kilograms = 23) 84 m/s x 31.221 s = 24) (5.159 + 82.3) x (0.024 + 3.00) = 25) 2.34 x 4.4391 / 3.465 = 26) (0.00015 x 54.6) + 1.0020 = 27) Read the following measurements Section 1-2: Dimensional Analysis and Significant Figures Round numerical answer to contain the correct number of significant digits. If the answer is smaller than 0.01 or larger than 1000, use scientific notation 1. Convert 0.0456 nm to cm. 2. Convert 0.03030 kg to ng. 3. Convert 72.6 C to F. 4. Convert 14.5 F to Kelvin. 5. A piece of metal alloy weighs 12.0000g and contains iron, cobalt, and nickel. What is the weight of the nickel is the iron and cobalt are determined to be 11.53g and 0.1233g respectively? 6. If 18.57 g of water is combined with 15.3 ml of water, what is the total volume? The density of water is 0.998 g/ml. 3 3 3 7. The density of iron is 7.20 g/cm. Calculate the mass, in kilograms, of 7.6 x 10 cm or iron? 3 3 8. What is the volume, in liters, of 544 g of iron? The density of iron is 7.20 g/cm. (Recall that one cm is exactly 1 ml).
Chem 1A Dr. White Fall 2013 2 9. Light travels at 186,000 miles per second (3 sig figs). How many kilometers does light travel in one day? 10. The speed of sound in water is 4.7 x 10 3 feet per second. How long, in minutes, does it take for sound to travel 19,000 km? 11. A solution contains 4.52 g of sugar per liter. What volume of the solution, in liters is needed to supply 2.0 kg of sugar? 12. A solution contains 144.1 mg of NaCl per liter. What volume of the solution, in milliliters, is needed to supply 27.8µg of NaCl? 13. A cylindrical redwood tree is 255 ft tall and 16 ft in diameter. The density of the redwood is 0.9 g/cm 3. Calculate the mass of the tree in pounds. (V cylinder = πr 2 h) 14. The density of Saran Wrap is 0.80 g /cm 3. Calculate the mass, in grams, of a roll that is 2.00 x 10 2 ft long, 12.00 inches in diameter. (assume it is a cylinder). 15. What is the volume, in cubic centimeters, of a rectangular block whose dimensions are 3.88 in by 5.0 in by 36 in? 16. What is the diameter, in millimeters, of a red blood cell whose volume is 9.0 x 10 1 µm 3? Assume that the cell is spherical (Recall that the volume of a sphere is V=4/3πr 3 ). 17. A piece of metal with mass 34.65 is placed into a 50-mL graduated cylinder partially filled with water, raising the water level from 22.5 ml to 30.7 ml. What is the density of the metal? 18. The mass of a vial is 18.41 g. After a 10.00 ml sample of methanol is pipetted into it, the vial and sample have a mass of 26.34 g. Calculate the density of methanol. Remember that the volume of a pipet is accurate to 0.01 ml. 19. The mass of a vial is 28.20 g. After five milliliters of benzene is pipetted into the vial, the vial and its contents have a mass of 32.55 g. What is the density of benzene? 20. If 50 ml of water is pipetted into a flask weighing 90.45 g, what will be the total weight of the flask with the water? The density of water is 0.998g/mL. Section 1-3: Atomic Structure Intro 1. In the early 1900 s a common model for the structure of the atom was the plum pudding model. In this model, negatively charged electrons reside in the atom surrounded by a diffuse, continuous medium of positive charge (like plums in a pudding, or for a more modern analogy, like chocolate chips in a cookie). (a) In 1911 Earnest Rutherford carried out an experiment that changed the way we view the atom. Briefly describe this gold foil experiment. (b) How is the plum pudding model inconsistent with Rutherford s experimental findings? Approach this by considering the plum pudding hypothesis, and then predict what you would have expected to happen in Rutherford s experiment if the plum pudding model was true. (c) Draw a picture of Rutherford s model of the atom. Use your picture to describe why most of the alpha particles pass through the gold foil in Rutherford s experiment.
Chem 1A Dr. White Fall 2013 3 2. Atoms X, Y, Z, and R have the following nuclear compositions: Which two are isotopes? Explain. 3. Fill in the following table for neutral atoms of the given isotopes. 4. Silicon, which makes up about 25% of Earth's crust by mass, is used widely in the modern electronics industry. It has three naturally occurring isotopes, 28 Si, 29 Si, and 30 Si. Calculate the atomic mass of silicon. (28.09 amu) Isotope Isotopic Mass (amu) Abundance % 28 Si 27.976927 92.23 29 Si 28.976495 4.67 Si 29.973770 3.10 Section 1-4: Wave and Particle Behavior 1. Consider the following types of electromagnetic radiation: i) microwave ii) ultraviolet iii) radio waves iv) infrared a) arrange them in order of increasing wavelength b) arrange them in order of increasing frequency c) arrange them in order of decreasing energy d) 2. In the diagram below, draw in each of the following. Make sure you label each new wave! a) A wave with shorter wavelength b) A wave with smaller frequency c) A wave with the same wavelength and frequency but less amplitude. v) x-ray 3. Some of the new cordless phones are said to operate at 900.MHz. Calculate the wavelength and energy of these waves.
Chem 1A Dr. White Fall 2013 4 4. Cobalt-60 is a radioactive isotope used to treat cancers of the brain and other tissues. A gamma ray emitted by an atom of this isotope has a wavelength of 9.32 x 10-13 m. a. What is this wavelength in nm? b. What is the frequency of this light in Hz c. Does radiation with λ = 242 nm have greater or lesser energy than the radiation emitted by the gamma ray above? What about radiation with λ = 2200 Å (1 Å = 1 x 10-10 m)? (You shouldn t need to do calculations to answer these questions) 5. Molybdenum metal will eject an electron if it absorbs a photon with an energy of at least 7.22 x 10 19 J. If Mo is irradiated with UV light with a wavelength of 120. nm, could an electron be ejected? 6. Calculate the de Broglie wavelengths for (a) an electron with a speed of 3.0 x 10 6 m/s (mass of an electron is 9.109 x 10 31 kg) and (b) a 12 g bullet whose speed is 200 m/s. 7. Two objects are moving at the same speed. Which (if any) of the following statements about them are true? a. The wavelength of the heavier object is longer. b. If one object has twice as much mass as the other, its wavelength is one-half the wavelength of the other. c. Doubling the mass of one of the objects will have the same effect on its wavelength as does doubling its speed. 8. The speed of the electron in the ground state of the hydrogen atom is 2.2 x 10 6 m/s. What is the wavelength of the electron? 9. At what speed must a human weighing 150. lb. be traveling in order to have a wavelength in the visible region, say at 650 nm? Does it seem likely that a human would attain such a speed? (1 kg = 2.20 lb) 10. Identify the following transitions as either absorption or emission: a) n = 3 n = 1 b) n = 6 n = 4 c) n = 1 n = 5 d) n = 1 n = 2 11. Which of these transitions correspond to absorption and which to emission of radiation (or energy in general)? Illustrate each transition on the provided energy level diagram. a) n = 2 to n = 4 b) n = 3 to n = 1 c) n = 5 to n = 2 n=5 n=4 n=3 d) n = 3 to n = 4 E n=2 n=1 (note that as n increases, the levels get closer and closer together) e) Of the transitions listed above, which corresponds to emission of radiation with the longest wavelength? Explain. You should not need to use your calculator to answer this! f) Of the transitions listed above, which corresponds to absorption of radiation with the highest frequency? Explain. You should not need to use your calculator to answer this!
Chem 1A Dr. White Fall 2013 5 12. Calculate the wavelength (in nm) of the photon emitted when a hydrogen atom undergoes a transition from n = 5 to n = 2. Is this visible light? 13. An electron in a hydrogen atom undergoes a transition from the n = 3 level to the n = 6 level. To accomplish this, energy, in the form of light, must be absorbed by the hydrogen atom. a) Calculate the energy of the light (in kj/photon) associated with this transition. Watch the sign of your answer! b) In order to move to the n = 5 level, would this same electron need to absorb more or less energy? Explain. 14. An electron in a hydrogen atom in the n = 5 level emits a photon with wavelength 1281 nm. a) To what energy level does this electron move? b) In order to move to the n = 1 level, will this electron emit a photon with longer or shorter wavelength than 1281 nm? Explain. Again, you should not need to use your calculator here. 15. An electron in Hydrogen in an unknown energy level emits a photon with a wavelength of 1005 nm. The final state of the electron is the 2 nd excited state. What was the initial state of the electron? 16. What is the minimum energy needed to ionize a hydrogen atom with the electron in the first excited state? Report your answer is J and kj/mol. Section 1-5: Atomic Structure and Electron Configurations 1. Fill in the following table: Quantum Number Orbital or Electron Property Allowed Values m l m s energy level and size 0 to n-1 2. a) If the principal quantum number n of an atomic orbital is 4, what are the possible values of l? b) If the angular momentum quantum number l is 3, what are the possible values of m l? 3. There is something wrong with each set of quantum numbers given below. Make a correction so that each set is valid. a) n = 1, l = 1, m l = 0, m s = 1/2 b) n = 2, l = 1, m l = -1, m s = 1 c) n = 4, l = 2, m l = -3, m s = -1/2 4. Identify the pictured orbital as s, p or d (write your answer next to the picture) and then answer the following questions: a. What is the l value for this type of orbital? b. Can this type of orbital be found in the n = 2 energy level? If not, explain why. c. List one complete and valid set of allowed quantum numbers for an electron occupying this type of orbital in the n = 4 level.
Chem 1A Dr. White Fall 2013 6 5. Write the electron configuration of Osmium. 6. Name the element with the electron configuration [Kr]5s 2 4d 10 5p 2. 7. Write a set of quantum numbers that could describe a 5d electron. 8. Which one of the following equations correctly represents the process relating to the ionization energy of X? A. X(s) X + (g) + e - B. X 2 (g) X + (g) + X - (g) C. X(g) + e - X - (g) D. X - (g) X(g) + e - E. X(g) X + (g) + e - 9. Give the full electronic configurations for the following. Which are paramagnetic? (a) Be (b) N (c) Ar (d) Sc (e) Cr 10. Write a set of quantum numbers (n, l, m l, m s ) for one of the highest energy electrons in europium (Eu) (i.e., one of the last electron to fill in) 11. Arrange strontium, cesium, iodine and tin in order of increasing atomic size (radii) 12. Use the figure below to answer the following questions. a. Given the representation of a chlorine atom, which circle might represent an atom of sulfur? b. Given the representation of a chlorine atom, which circle might represent an atom of argon? c. Given the representation of a chlorine atom, which circle might represent an atom of fluorine? 13. The energy required to remove an electron from an atom is known as 14. Which would you predict to be more reactive, potassium or rubidium? Explain your answer. 15. State the relationship between the group number in the chart and the number of valence or outermost electrons in an atom. 16. Name the element with the outermost electron configuration 4s 2 4p 4. 17. a. Name the element with the electron configuration [Kr]5s 2 4d 7. b. How many valence electrons does it have? c. Draw an orbital box diagram for the valence electrons. 18. Choose the element or ion in each pair that has the greater electron affinity. a. Iodine or Chlorine b. Al or P
Chem 1A Dr. White Fall 2013 7 19. Choose the element or ion in each pair that has the larger radius. a. Oxygen or Selenium b. Sc or Zn 20. Choose the element or ion in each pair that has the smaller ionization energy. a. Boron or Neon b. K or Ge 21. Which of the following atoms has the smallest atomic radius? a) Se b) S c) Te d) O 22. Put the following elements (N, O, F, Be) in order of decreasing ionization energy. Circle the correct answer. a. N > O > Be > F b. O > N > F > Be c. Be > N > O > F d. F > O > N > Be e. F > Be > N > O f. 23. Which of the following atoms has the smallest electron affinity? Br, Se, Ca, K 24. Fill in the following orbital box diagram for the 2p subshell with 4 electrons total: (use arrows to indicate electrons arrows can point up or down) a) b) How many unpaired electrons are there? c) Which atom is represented here? (only one correct answer) 25. Would you expect the ionization energy for Li + be less than or greater than the ionization energy for Li? Briefly explain your answer. 26. Fill in the following table for each of the following atoms and ions. Full Electron Configuration Abbreviated Electron Configuration # core electron s # valence electrons Paramagnetic or Diamagnetic? N, Cr, Li, Ga, Zn, Ag, Rb, Fe 2+, K +, P 3-, The two common ions of Pb* 27. Use your electron configurations above to answer the following: a. A set of quantum numbers (not all possible values) for the first electron removed in the formation of Zn 2+ b. A set of quantum numbers (not all possible values) for the first electron removed in the formation of one of the lead ions.
Chem 1A Dr. White Fall 2013 8 Answers: Section 1-1: Significant Figures 1. 4 2. 2 3. 2 4. 4 5. 5 6. 8 7. 3 8. 2 9. 2 10. 4 11. 2 12. 6 13. 3 14. 1 15. 6 16. 533 g 17. 31 g/ml 18. 62 J/s 19. 0.01 m 20. 500 cm 21. 3.95 m/s 22. 332.12 kg 23. 2.6 x 10 3 m 24. 264 25. 3.00 26. 1.0102 27. Left figure 16.8 C Center Figure 22.21 ml Right Figure 11.4 ml Section 1-2: Dimensional Analysis and Significant Figures 21. 4.56 x 10-9 cm 22. 3.030 x 10 10 ng 23. 163 F 24. 263.43K 25. 0.35 g 26. 33.9 ml 27. 55 kg 28. 7.56 x 10-2 L 29. 2.59 x 10 10 km/day 30. 2.2 x 10 2 min or 220 min 31. 4.4 x 10 2 L 32. 0.193 ml 33. 3 x 10 6 lb 34. 3.6 x 10 6 g 35. 1.1 x 10 4 cm 3 36. 5.6 x 10-3 mm 37. 4.2 g/ml 38. 0.793 g/ml 39. 0.870 g/ml 40. 140.4 g Section 1-3: Atomic Structure Intro 1. (a) Rutherford used positive alpha particles. These alpha particles were directed at the gold foli. The paths of these particles were examined. (b) In the plum pudding model, negatively charged electrons reside in the atom surrounded by a continuous medium of positive charge. At the time, it was believed that this continuous medium of positive charge was very diffuse both in terms of charge and mass. The pudding was viewed as having the same amount of charge as the electrons, but that charge was spread out over the whole atom, as was the mass of the atom. Thus, Rutherford s hypothesis was that the alpha particles would pass right through the foil. As they did not all do this, but many were deflected or bounced back, he had to revise the initial hypothesis to account for this. He did this by proposing that the atom has a positively charged nucleus that contained most of the mass of the atom. (c) Since most of the atom is empty space, the particles will pass through the atom. Those that hit the dense nucleus will bounce off the nucleus and are deflected. 2. Isotopes have different mass numbers due to a difference in the number of neutrons. They have the same number of protons. Thus, X and Z are isotopes. They have the same number of protons (186) but different mass numbers (412 and 410). 3. Symbol Atomic Number Mass Number Number of protons Number of neutrons 18 38 18 20 18 Number of electrons
Chem 1A Dr. White Fall 2013 9 15 7N 7 15 7 8 7 1 2 1 1 1 16 34 16 18 16 3 7 3 4 3 38 84 38 46 38 4. 28.09 amu Section 1-4: Wave and Particle Behavior 1. a. x-ray < ultraviolet < infrared < microwave < radio waves b. radio waves < microwaves < infrared < ultraviolet < x-ray c. x-ray > ultraviolet > infrared > microwave > radio waves 2.. 3. 0.333m, 5.96 x 10 25 J 4. a. 9.32 x 10-4 nm b. 3.22 x 10 20 Hz c. Energy is inversely proportional to wavelength, so a larger wavelength means a smaller Energy and vice versa. A wavelength of 242 nm is larger than 9.32 x 10-4 nm, so it would have a smaller energy. A wavelength of 2200 Å is equal to 220 nm, so it also has a smaller energy. 5. yes, an electron can be ejected E = hc/λ = (6.626 x 10-34 Js)(2.998 x 10 8 m/s)/120 x 10-9 m = 1.66 x 10-18 J>7.22 x 10 19 J, so an electron could be ejected. 6. 2.4 x 10-10 m, 2.8 x 10-34 m 7. a. FALSE, wavelength of heavier object is shorter b. TRUE c. TRUE 8. 3.31 x 10-12 m 9. 1.49 x 10-29 m/s - It doesn t seem likely a human could be at this speed, we would have to be moving VERY slowly, but not stopped. 10. a. emission b. emission c. absorption d. absorption 11. a. absorption b. emission c. emission d. absorption e. Longest wavelength corresponds to lowest change in energy. So, the EMISSION with the longest wavelength is n=5 to n=2. f. Highest frequency corresponds to highest change in energy, so the ABSORPTION with the highest frequency is n=2 to n=4. 12. 433.9 nm 13. a. 1.815 x 10-22 kj b. The electron would need to absorb less energy since it is moving to a lower energy level. 14. a. n = 3 b. To move to the n=1 level, the electron would emit a photon with more energy. Energy is inversely proportional to wavelength, so it would emit a photon with a shorter wavelenth 15. n f = 7 16. 5.445 x 10-19 J and 327.9 kj/mol
Chem 1A Dr. White Fall 2013 10 1. Section 1-5: Atomic Structure and Electron Configurations Quantum Number Orbital or Electron Property Allowed Values ml Orientation of orbital in 3-D space Can range from l to + l, including 0 n Positive Integers from energy level and size n = 1 to l orbital shape 0 to n-1 m s Electron spin +1/2 or 1/2 2. a) If n = 4, l = 3, 2, 1, 0 b) If l = 3, ml = 3, 2, 1, 0, -1, -2, -3 3. a. n = 1, l = 1, ml = 0, m s = 1/2 if n = 1, l = 0 b. n = 2, l = 1, ml = -1, m s = 1 m s = 1/2 or 1/2 only c. n = 4, l = 2, ml = -3, m s = -1/2 if l = 2, ml = -2, -1, 0, 1 and 2, NOT -3!! 4. a. this is a d orbital, L = 2 b. a d orbital cannot be found in the n = 2 level, when n = 2, l can only be either 1 or 0, for l to equal 2, must be in n = 3 or higher c. n = 4, l = 2 (we re talking about d orbitals), m l = 1, m s = 1/2 (these last two, just pick an allowed value.) 5. [Xe]6s 2 4f 14 5d 6 6. Tin (Sn) 7. n=5, l=2, m l =-2, m s =1/2 (m l =-1, 0, 1, 2 also valid and m s =-1/2 also valid) 8. E 9. Give the full electronic configurations for the following. Which are paramagnetic? a. Be 1s 2 2s 2 b. 1s 2 2s 2 2p 3 paramagnetic c. Ar 1s 2 2s 2 2p 6 3s 2 3p 6 d. Sc 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 1 paramagnetic e. Cr 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 paramagnetic 10. n=4, l=3, m l =+3 (could also be -3, -2, -1, 0, 1, or 2), m s =1/2 11. I<Sn<Sr<Cs 12. a. D b. B c. B 13. ionization energy 14. Which would you predict to be more reactive, potassium or rubidium? Explain your answer. Rb would be more reactive/ It is larger, so it has a smaller ionization energy and hence is more reactive. 15. The group number tells the number of valence electrons for main group atoms. 16. There is no such atoms it is missing the 3d electrons. 17. a. Rh b. 7 valence electrons c. 5s 18. a. Chlorine b. P 19. a. Selenium b. Sc 20. a. Boron b. K 21. d 22. F > O > N > Be 4d
Chem 1A Dr. White Fall 2013 11 23. K 24. a) b. How many unpaired electrons are there? 2 c. Which atom is represented here? (only one correct answer) Oxygen 25. I expect the ionization energy for Li+ to be greater. Li+ has a full octet in its outer shell (it has the same electron configuration as a noble gas), therefore it is very stable. It would be much harder to remove the 2 nd electron from the very stable octet.
Chem 1A Dr. White Fall 2013 12 26. Full Electron Configuration Abbreviated Electron Configuration # core electrons # valence electrons P or D N 1s 2 2s 2 2p 3 [He]2s 2 2p 3 2 5 P Cr 1s 2 2s 2 2p 6 3s 2 3p 6 4s 1 3d 5 [Ar]4s 1 3d 5 18 6 P Li 1s 2 2s 1 [He]2s 1 2 1 P Ga 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 1 [Ar]4s 2 3d 10 4p 1 28 3 P Zn 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 [Ar]4s 2 3d 10 28 2 D Ag 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 4d 10 (like Cu, full d subshell is very stable!) [Kr]5s 1 4d 10 46 1 P Rb 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 1 [Kr]5s 1 36 1 P Fe 2+ 1s 2 2s 2 2p 6 3s 2 3p 6 3d 6 [Ar]3d 6 18 6 (we had 8 valence and removed 2) P K + 1s 2 2s 2 2p 6 3s 2 3p 6 [Ar] 18 0 (we removed the 2 valence electrons!) D P 3-1s 2 2s 2 2p 6 3s 2 3p 6 [Ar] 10 Pb 2+ Pb 2+ : 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 6s 2 4f 14 5d 10 [Xe]6s 2 4f 14 5d 10 78 8 (we had 5 valence and added 3) 2 (we had 4 valence and removed 2) D D Pb 4+ 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 4f 14 5d 10 [Xe]4f 14 5d 10 78 0 (we had 4 valence and removed 4) D 27. a. Since the 4s electron is removed: n= 4; l = 0; m l = 0, and m s = ½ [ or another possible answer is n= 4; l = 0; m l = 0, and m s =- ½] b. Since the 6p electron is removed first: n = 6 ; l = 1; m l = -1; m s = ½ [other answers possible for m l and m s ; m l can be -1, 0, or 1 and m s can be + or ½)