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Chem 1B - CLAS – Ch 12 - Key1. Calculate the energy of 2.9 moles of yellow photons with a wavelength of 580 nm.for one photon ? E = hcλ ? E = (6.626x10-34Js)(3x108m/s)(580 nm)(1 m109nm) = 3.43x10-19J/photon(3.43x10-19J/photon)(6.022x1023 photons/mol)(2.9 mol) = 5.99x105 J2. Calculate the wavelength of a beryllium atom traveling at 15% the speed of light.for particles ? λ = hmv mass of a beryllium atom = 1 atom 1 mol6.022 x 1023atoms x9.012 gmolx1 kg1000g = 1.5 x 10-26 kgλ = hmv= 6.626 x 10-34Js1.5 x 10-26 kg0.15 x 3 x 108 ms-1= 9.84 x 10-16 m3. It takes 208.4 kJ of energy to remove 1 mol of electrons from the atoms on the surface of rubidium metal. If rubidium metal is irradiated with 254-nm light, what is the maximum kinetic energy (kJ/mol) the released electrons can have? KEe- = Ephoton - EbindingKEe- = hcλ- EbindingKEe- = 6.626 x 10-34Js3 x 108 ms-12.54 x 10-7mxkJ103Jx6.022 x 1023mol – 208.4 KJ/mol = 262.9kJ/mol 4. Consider the following transitions. Which will emit light with a longer wavelength?a. n = 4 → n = 2 or n = 3 → n = 2b. n = 3 → n = 1 or n = 1 → n = 3c. n = 5 → n = 3 or n = 3 → n = 15. Calculate the wavelength (nm) emitted for the electronic transition n = 5 to n = 3 in a Li2+ ion.ΔE = -2.178 x 10–18JZ2nf2-Z2ni2 and E photon = | ΔE | = hcλ hcλ = |2.178 x 10–18JZ2nf2-Z2ni2|6.626 x 10-34Js3 x 108 ms-1λ109nmm = |2.178 x 10–18J3232- 3252 ? λ =142 nm6. An excited hydrogen atom with an electron in n = 5 state emits light having a frequency of 6.9 x 1014 s-1. Determine the principal quantum level (n) for the final state in this electronic transition. hv = |2.178 x 10–18JZ2nf2-Z2ni2|(6.626 x 10-34Js)( 6.9 x 1014 s-1) = |2.178 x 10–18J12x2- 1252 ? nf = 27. The ground-state ionization energy of the one-electron ion Xm+ is 1.18x104 kJ/mol. What is the value of “m” in Xm+?Xm+ is an one electron species ? ΔE = -2.178 x 10–18JZ2nf2-Z2ni2ground state ? ni = 1 and ionization ? nf = ∞ ? ? ΔE = (2.178 x 10–18J)(Z2) ? 1.18x104 kJ/mol = (2.178 x 10–18J)(Z2)1 kJ1000J6.022 x 1023mol ? Z = 3 ? Li2+8. How many electrons in any one atom can have the following quantum numbers?a. n = 5 ? 50b. n = 6, l = 0 ? 2c. n = 4, l = 2 ? 10d. n = 4, l = 3, ml = -2 ? 2e. n = 2, l = 0, ml = 0, ms = -1/2 ? 19. Fill in the following table: a. Cl b. Ni c. Cr d. Ag e. Te2– f. Ba2+ Cl ? 1s22s22p63s23p5 ? 1 unpairedNi ? [Ar]4s23d8 ? 2 unpairedCr ? [Ar]4s13d5 ? 6 unpairedAg ? [Kr]5s14d10 ? 1 unpairedTe2– ? [Kr] 5s24d105p6 ? 0 unpairedBa2+? [Kr] 5s24d105p6 ? 0 unpaired10. Determine if each of the following corresponds with an excited state or ground state electron configuration.a. [Ar]4s24p5 ? excitedb. [Kr]6s1? excitedc. [Ne]3s23p4? ground11. Which of the following has the largest radius?a. Al or Sib. F or Clc. S or S2-d. K or K+12. Which of the following has the greatest ionization energy?a. K or Cab. P or Asc. Sr or Sr2+13. Which of the following has the most negative electron affinity?a. Br or Krb. C or Si14. The successive ionization energies for an unknown element are:I1 = 896 kJ/molI2 = 1752 kJ/molI3 = 14,807 kJ/molI4 = 17,948 kJ/molWhich family does the unknown element most likely belong? Group 2 - When you notice a jump in the order of magnitude going from 103 to 104 this implies that all of the valence electrons have been removed and the 104 magnitude would be referring to the core – the first and second ionization energies are within reason however the third ionization energy is much higher – therefore there must have been only 2 electrons in the valences shell ................
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