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MT1 Solutions - Chemistry 1A, Spring 2010 Midterm 1...

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Unformatted text preview: Chemistry 1A, Spring 2010 Midterm 1 Solutions Feb 8, 2010 Problems 1. Of the reactions below, which one is not a combination reaction? E) 2 CH4 + 4 O2 → 2 CO2 + 4 H2 O, because it has two products 2. Vanadium has two naturally occurring isotopes, 50 V, with an atomic mass of 49.9472 amu and 51 V, with an atomic mass of 50.9440 amu. The atomic weight of vanadium is 50.9415. The percent abundances of the vanadium isotopes are 50 V and 51 V. x ∗ m50 V + y ∗ m51 V x ∗ m50 V + (1 − x) ∗ m51 V = 50.9415 = 50.9415 x+y =1⇒y = 1−x 49.9472x + 50.9440 − 50.9440x = 50.9415 0.9968x = 0.0025 y = 0.9975 (A) Vanadium is 0.25% 50 V x = 0.002508 and 99.75% 51 V 3. One moles of ? contains the largest number of atoms. B) C10 H8 has 18 moles of atoms 4. When the following equations is balanced, the coefficients are: (B) 4 NH3 (g) + 7 O2 (g) → 4 NO2 + 6 H2 O 5. The combustion of propane (C3 H8 ) produces CO2 and H2 O. The reaction of 2.5 mole of C3 H8 in air will produce ? moles of H2 O theoretically. 4 mole H2 O ✭ 2.5✭✭✭✭3 H8 × mole C ✭ ✭ = 10 moles H2 O 1✭✭✭✭3 H8 mole C ✭ (D) 10 moles H2 O 1 C3 H8 + 5 O2 ⇒ 3 CO2 + 4 H2 O 1 6. A 4.000g mixture of NaCl and KCl is dissolved in water and reacted with AgNO3 to make 8.5904 g AgCl. What is the mass percent of NaCl in the original mixture? 8.5904✘✘✘✘ × g AgCl ✘ ✭ 1✭✭✭✭✭ mol AgCl 1 mol Cl ✘× ✭ = 0.05993 mol Cl atoms 143.35✘✘✘✘ g AgCl 1✭✭✭✭✭ mol AgCl We know that the total number of Cl atoms must be conserved in the reactants and the products, which we know to be 0.05993 mol Cl atoms. Therefore let’s define the following: x= 0.05993 − x = ✭ ✭✭✭✭ number of moles Cl from NaCl number of moles of Cl from KCl We can now relate this to the weight of the mixture (4.000 g) via the individual molecular weights. 1 mol NaCl ✭ ✭✭ = x mol NaCl ✭ 1✭✭✭✭✭ mol Cl atoms ✭✭ 1 mol KCl ✭✭✭ 0.05993 mol Cl ✭✭✭ from KCl × ✭ atoms ✭✭ = 0.05993 − x mol KCl ✭✭✭ 1✭✭✭✭✭ mol Cl atoms x mol Cl ✭✭✭✭ ✭✭✭ atoms from NaCl × ✭ ✭ x✭✭✭✭ mol NaCl ￿ 58.44 g NaCl ✭ ✭ 1✭✭✭✭ mol NaCl ￿ ✭ + (0.05993 − x) ✭✭✭✭ mol KCl ￿ 74.55 g KCl = 4.000 g ✭ 1✭✭✭✭ mol KCl 58.44x + 4.4675 − 74.55x = 4.000 ￿ 16.11x = 0.46749 x = 0.02902 = mol NaCl ￿ 58.44 g NaCl = 1.696 g NaCl ✭ ✭ 1✭✭✭✭ mol NaCl ￿ ￿ ✭ 74.55 g KCl ✭✭✭ 0.05993 − 0.02902 = 0.03091✭ KCl mol = 2.304 g KCl ✭ 1✭✭✭✭ mol KCl ✭ ✭ 0.02902✭✭✭✭ mol NaCl ￿ %NaCl = %KCl = (D) 42.40% NaCl 7. ? and ? reside in the atomic nucleus? (C) protons, neutrons 1.696g ✁ × 100 = 42.40% 4.000g ✁ 2.304g ✁ × 100 = 57.60% 4.000g ✁ 8. An ion has 8 protons, 9 neutrons, and 10 electrons. The symbol for this ion is ?. (A) 17 O2− . 8 protons makes the element oxygen, 10 electrons gives an overall -2 charge, and 9 neutrons + 8 protons gives a weight of 17. 2 9. In the Rutherford nuclear-atomic model, ?. (A) The heavy subatomic particles, protons and neutrons, reside in the nucleus 10. Which atom has the smallest number of neutrons? (B) Nitrogen-14 11. The empirical formula of a compound with molecules containing 12 carbon atoms, 14 hydrogen atoms, and 6 oxygen atoms is ? . (D) C6 H7 O3 . The smallest common factor is 2, so divide all the coefficients by two to get the empirical formula. 12. The combustion products of a hydrocarbon with which empirical formula yield the following mass spectrum? The height of the 18 peak is twice that of the 44 peak. Therefore, we know that there is twice as much H2 O as CO2 . That means for every 4 mol of H, there is only 1 mol C in the original hydrocarbon, giving (E), CH4 13. A compound that is composed of carbon, hydrogen, and oxygen contains 70.6% C, 5.9% H, and 23.5% O by mass. The molecular weight of the compound is 136 amu. What is the molecular formula? Since 136 g of the compound will be one mole, we use this as the mass. By determining the moles of C, H, and O are in one mole of compound, we obtain the molecular formula 1 mol C = 7.994 mol C 12.011✟✟ g✟ C 1 mol H 136 g compound × 5.9% H = 8.024✟✟ × g✟ H = 7.96 mol H 1.00794✟✟ g✟ H 1 mol O 136 g compound × 23.5% O = 31.96✟✟ × g✟ O = 1.998 mol O 15.9994✟✟ g✟ O 136 g compound × 70.6% C = 96.016✟✟ × g✟ C (A) C8 H8 O2 14. Which is the best description of the color of an object with the following spectrum? (E) Blue. Since most of the transmitted light is in the blue end, and most of the rest of the spectrum is absorbed, the color is blue. 15. Electromagnetic radiation with a wavelength of 531 nm appears as green light to the human eye. Thus, a laser that emits 1.3×102 J of energy in a pulse of light at this wavelength produces ? photons in each pulse. 3 E531 1 photon 3.74 × 10−19 J ✁ 1.3 × 10−2 J × ✁ ￿￿ ￿ 6.626 × 10−34 J✁ 2.998 × 108 ✟✟ s m/s = ✟ 5.31 × 10−7 ✟ m −19 = 3.74 × 10 J = 3.475 × 1016 photons (D) ￿ 16. The energy of a photon that has a wavelength of 9.0 m is ? J. E 9 .0 = ￿ ￿￿ ￿ 6.626 × 10−34 J✁ 2.998 × 10−8 ✟✟ s m/s = 2.21 × 10−26 J (A) ✟ 9.0✟ m 17. The photoelectric effect is ? . (C) The ejection of electrons by a metal when struck with light of sufficient energy. 18. It takes 254 kJ/mol to eject electrons for a certain metal surface. What is the longest wavelength of light (nm) that can be used to eject electrons from the surface of this metal via the photoelectric effect? ✘ ✚ kJ 1000￿ J 1✘✘ mol ￿ ✚ 254 ✟ × × ✟ mol 1✟ kJ 6.02 × 1023 atoms ✟ = 4.22 × 10−19 J λ= hc E ￿ ￿￿ ￿ 6.626 × 10−34 J✁ 2.9948 × 108 ✟✟ s m/s = 4.22 × 10−19 J ✁ 1 × 109 nm = 4.707 × 10−7 ✚ × m = 471 nm (A) ✚ ✟ 1✟ m 19. The de Broglie wavelength of an electron is 8.7×10−11 m. The mass of an electron is 9.1×10−31 kg. The velocity of this electron is ? m/s. λ= p= p= 4 h p h λ 6.626 × 10−34 Js 8.7 × 10−11 m ￿ 9.109 × 10 −31 p = mv ￿ kg v = 7.62 × 10−24 kg m/s v= p = 7.62 × 10−24 kg m/s ✚ 7.62 × 10−24 ✚ m/s kg 9.109 × 10−31 ￿ kg ￿ 6 v = 8.36 × 10 m/s (D) 20. A spectrum containing only specific wavelengths is called a ? spectrum. (A) Line spectrum 21. According to the Heisenberg Uncertainty Principle, it is impossible to know both the position and ? of an electron? (C) momentum. The uncertainty relation is δ xδ p ≥ ￿ , so as the uncertainty in position decreases 2 (or, the more accurately you know position), the higher the uncertainty in the momentum must be. 22. A photon of wavelength 400 nm is split into two photons, one at 800 nm wavelength. What is the wavelength of the other photon? Energy must be conserved when the photon is split, so we have: E∝ E400 E? 1 λ? 1 λ? λ 1 λ = E800 + E? = E400 − E800 1 1 = − 400 800 1 = 800 = 800 nm (D) 23. Consider a two-slit experiment using monochromatic light. Which statement is NOT true? (B) Constructive interference results in dimmed or no intensity. This is false, as constructive interference leads to maxima (bright spots), not minima (dark spots) in the interference pattern. 24. Of the following, ? radiation has the shortest wavelength. (A) X-Ray. It has the highest energy, which gives it the highest frequency (E = hν ), and the shortest wavelength (c = λν ). 5 25. The n = 8 to n = 2 transition in the Bohr hydrogen atom is in the ? region of the electromagnetic spectrum. ∆E = −R∞ ￿ 1 1 −2 n2 ni f −18 ￿ J ∆E = −2.198 × 10 ∆E = 5.152 × 10−19 J hc λ= E ￿ ￿￿ ￿ ✟ 6.626 × 10−34 ✟ Js 2.998 × 108 m/✁ s = 5.152 × 10−20 ￿ J ￿ −7 λ = 3.86 × 10 m = 386 nm (E) 386 nm is in the ultraviolet region of the electromagnetic spectrum 26. Blue, red, yellow, and green light eject electrons from the surface of potassium. In which case do the ejected electrons have the higher average kinetic energy? (A) Blue. It has the highest energy photons, so for a constant work function, there will be more energy left for the electron kinetic energy (Ekin = Ehν − Φ). 27. Of the following transitions in the Bohr hydrogen atom, the ? transition results in the emission of the highest energy photon. (B) n = 6 → n = 1. For emission, the￿ electron must move from ￿ high n to low n, and these two levels 1 1 35 have the highest difference in energy E ∝ 2 − 2 ⇒ E ∝ 6 1 36 28. Using the Bohr’s equation for the energy levels of the electron in the hydrogen atom, determine the energy (J) of an electron in the n = 4 level. ￿ 1 1 − 4 64 ￿ E4 = −R∞ 1 42 1 = −2.198 × 10−18 J × 16 = −1.37 × 10−19 J (A) = −2.198 × 10−18 J × Z2 n2 29. All of the ? have a valence shell electron configuration ns1 . 6 (C) Alkali Metals. They all have 1 electron in the outer most valence orbital All students were given credit for this question, as the material had not be reached in lecture. 30. OMIT For the next three questions, consider the following set of five orbitals 31. How many nodes are displayed in orbital D? (B) 1. This is the angular node for the p orbital 32. ha tis the best label for orbital A? (C) 2p. It is a p orbital, and since no radial nodes are shown, n − ￿ − 1 = 0, which gives n = 2 for ￿ = 1. 33. Which orbital has the highest energy? (C) The d orbital will be the highest in energy, as it must have a minimum n = 3, whereas s and p can have smaller n. 34. There are ? unpaired electrons in a ground state phosphorus atom. (D) 3, Phosphorus has 3 p electrons, which will each go in a different p orbital. All students were given credit for this question, as the material had not be reached in lecture. 35. In a hydrogen atom, and electron in a ? orbital can absorb a photon, but cannot emit a photon. (D) 1s, this is the lowest energy level, so it cannot lose any energy in order to emit a photon. 36. Which one of the following is an incorrect subshell notation? (B) 2d. The d orbital has ￿ = 2, therefore the minimum value for n is 3. 37. In the px orbital, the subscript x denotes (E) axis along which the orbital is aligned 38. Which set of three quantum numbers (n, ￿, m￿ ) corresponds to a 3d orbital? (A) 3,2,2. For a 3d, we must have n = 3 and ￿ = 2. The allowed values for m￿ = 2, 1, 0, −1, −2. 39. The requirements for the airbag that you developed in the first lab were that it could be deployed at a given time and that it inflates to the largest possible volume. The constraints on the airbag design 7 were that it could not weigh more than 3.5 g and you could only use the materials available. The chemical reaction that you used to produce the gas (CO2 ) NaHCO3 (s) + CH3 COOH (aq) → CO2 (g) + CH3 COONa (aq) + H2 O (l) If you are given an apparatus design that weighs 1.95 g (without the addition of reaction chemicals); what is the maximum amount of CO2 that you can produce? 3.5 g total - 1.95 g apparatus = 1.55 g chemicals 1 mol CH3 COOH 60 g CH3 COOH x g CH3 COOH = = 1 mol NaHCO3 84 g NaHCO3 (1.55 − x) g NaHCO3 .714 (1.55 − x) = x 1.107 = 1.714x x = 0.6458 g CH3 COOH 1.55 − x = 0.9402 g NaHCO3 ✘ ✭✭ ✚ 1 mol CH✭✭ 1✘✘✘✘ 2 mol CO 22.4✚ L 1000 mL ✭✭ 3 COOH ✭✭✭✭ × ✭✭ 0.6458✭✭✭3 COOH g CH × × × ✭✭✭✭ ✭✭✭✭ ✘✘✘ ✚ g CH 60✭✭✭3 COOH 1✚ L mol 1 mol CH3 COOH 1✘✘ CO2 ✭✭✭ ✭✭ ✭ ✘ ✘✘ ✭✭✭ ✭✭✭✭ 3 1 mol NaHCO 1 mol 22.4 1000 mL ✘L ✭ ✭✭ CO2 ✭ 0.9042✭✭✭✭✭ 3 × ✭✭ g NaHCO ✭× ✭× ✘× ✘✘ ✭✭✭✭3 ✭✭✭✭ 3 84 g 1 mol 1✟ 1 mol ✟L ✭✭ NaHCO ✘✘ gas ✭✭✭ NaHCO = 240 mL = 240 mL So, 240 mL is the maximum amount of CO2 gas that can be produced. 40. When experimentally determining the value of Avogadro’s number, many students obtain a result that is much lower than the actual value of 6.02 × 1023 molecules per mole. There are a number of reasons for the underestimate. Which of the following is not a valid reason for this? (C) Water was in the pipette because it was not thoroughly rinsed with stearic acid. or (D) The water on the watch glass was not completely full with water. For (C), if there is water in the pipet, then there will be fewer stearic acid molecules on the surface then you think, so your calculated answer will be larger than expected. For (D), There will be fewer drops required to make the monolayer than if the watch glass were completely full. This will lead to fewer stearic acid molecules being present than for a full watch glass, leading to an overestimate of Avogadro’s number. 8 ...
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