Unformatted text preview: S117 Homework 1 Out of 89 points Chapter 2 (not to turn in) 38. Symbol # Protons 92 20 23 39 35 15 # Neutrons 146 20 28 50 44 16 # Electrons 92 18 20 39 36 18 Net Charge 0 +2 +3 0 1 3 45. a. sulfur difluoride b. dinitrogen tetroxide d. tetraphosphorus hexoxide c. iodine trichloride 49. a. SO2 g. Cr(C2H3O2)2 m. HBrO b. SO3 h.SnF4 n. HBr c. Na2SO3 i. NH4HSO4 d. KHSO3 j. (NH4)2HPO4 e. Li3N k. KClO4 f. Cr2(CO3)3 l. NaH Chapter 12 (not to turn in) 21. The frequency of a photon of length 1.0cm is energy is energy of a mole of photons is . , so its . Therefore, the 23. a. Wavelength of light is given by b. This is in the infrared region (figure on page 524). c. , so the energy per mole is . . d. It is less energetic since the energy of a photon is proportional to its frequency. 25. The maximum wavelength which can remove an electron from lithium metal corresponds to the photon of minimum energy which can remove an electron. The energy of the photon must at least equal the energy required to eject an electron, which is given by , where is the work function of lithium and NA is Avogadro's number, so Now , so . . 27. To remove a single electron from the surface of rubidium metal requires . The energy of 254nm light is . The kinetic energy of the ejected electron is the difference between the energy of the photon and the energy required to eject the electron, . 31. a. Assuming that the neutron's mass is its rest mass, . b. Rearranging the above formula, . 36. The energy required to change the energy state is so hydrogen atom, essentially making , so , so , so , so . Of a photon, . For a , . Removing the electron completely is . Finally, the negative of RH will be used so that increasing the energy state of an electron requires positive energy to be put into the system. To remove an electron from requires a wavelength of no greater than . To remove an electron from requires a wavelength of no greater than . Chapter 2 (to turn in) 46. (12 points) a. sodium perchlorate b. magnesium phosphate c. aluminum sulfate d. sulfur difluoride e. sulfur hexafluoride f. sodium hydrogen phosphate g. sodium dihydrogen phosphate h. lithium nitride i. sodium hydroxide j. magnesium hydroxide k. aluminum hydroxide l. silver chromate 50. (14 points) a. Na2O g. PbO2 m. HNO2 b. Na2O2 h. CuCl n. P2O5 c. KCN i. GaAs d. Cu(NO3)2 j. CdSe e. SiCl4 k. ZnS f. PbO l. Hg2Cl2 56. (10 points) The reaction is , and , so 1mol Cr2O3 reacts with 2mol Al. . Therefore, , so neither reagent is limiting. (Note that calculating the limiting reagent was unnecessary in this case since the amount of product was given. I just wanted to demonstrate a limiting reagent calculation.) Now 1mol Al2O3 is produced for every 2mol Cr; therefore, 0.224mol Al2O3 were produced. . Alternatively, since the mass is given of all species but one, conservation of mass can be invoked. , so . , and 58. (8 points) Ion Fe+2 Fe+3 Ba+2 Cs+ S2 P3 BrN3 # Protons 26 26 56 55 16 15 35 7 # Electrons 24 23 54 54 18 18 36 10 with O2 or Al+3 FeO Fe2O3 BaO Cs2O Al2S3 AlP AlBr3 AlN Chapter 12 (to turn in) 22. (10 points) Clearly, wave a has the longer wavelength. There are a total of four periods of the wave within a length of , so the wavelength is . Clearly, wave b has the higher frequency and, therefore, the higher energy. Frequency of light is given by , and energy of light is given by . Both waves have the same velocity since all light has the same velocity within the same medium. Wave a has a wavelength of , so it is infrared. Wave b has a wavelength of , so it is also infrared. 24. (5 points) Energy of a photon is given by then, has energy . An entire mole of those photons, . 30. (5 points) a. The de Broglie wavelength is given by of light, b. Of the tennis ball: . . For an electron moving at 10.% the speed . 38. (15 points) The energy required to change the energy state of a oneelectron species is . For a hydrogen atom, , so . The negative of RH will be used so that increasing the energy state of an electron requires positive energy to be put into the system. The energy required to excite an electron from the n = 1 to the n = 5 energy state is, then, . The highest energy photon of visible light has energy , so a visible photon does not have enough energy to excite a ground state electron in the hydrogen atom to the n = 5 state. The energy required to excite an electron from the n = 2 to the n = 6 energy state is , so a visible photon could excite the electron between those two states. Extra (10 points) 1. red star power density blue star Power density corresponds to the amount of light emitted. If the red star is brighter, then it is giving off significantly more radiation. The blue star is shifted to lower wavelengths since blue is higher in energy than red, and its spectrum is still the general blackbody radiation spectrum, but the spectrum is much smaller since the blue star is not as bright. wavelength ...
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This note was uploaded on 01/17/2012 for the course S 117 taught by Professor Stephenjacobson during the Fall '11 term at Indiana.
 Fall '11
 StephenJacobson

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