Exam1 - ctt253 – Exam 1a – Sparks –(53535 1 This print-out should have 35 questions Multiple-choice questions may continue on the next column

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Unformatted text preview: ctt253 – Exam 1a – Sparks – (53535) 1 This print-out should have 35 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. It is extremely important that you bubble your EID, name, and version number cor- rectly! Double check these three things! 001 14.3 points The ions K + , Ca 2+ , and Cl − each have eigh- teen electrons. The order of sizes of these three ions would be 1. Ca 2+ is smaller than Cl − which is smaller than K + . 2. Cl − is smaller than K + which is smaller than Ca 2+ . 3. Cl − is smaller than Ca 2+ which is smaller than K + . 4. K + is smaller than Ca 2+ which is smaller than Cl − . 5. Ca 2+ is smaller than K + which is smaller than Cl − . correct Explanation: These ions are isoelectronic. In an isoelec- tronic series, ionic size decreases with increas- ing atomic number due to increased nuclear charge. 002 14.3 points For which would the de Broglie wavelength be significant: an electron or a golf ball? 1. a golf ball 2. an electron correct Explanation: 003 14.3 points Cobalt-60 ( 60 27 Co) is an artificial radioisotope that is produced in a nuclear reactor for use as a gamma-ray source in the treatment of certain types of cancer. If the wavelength of the gamma radiation from a cobalt-60 source is 1 . 00 × 10 − 3 nm, calculate the energy of a photon of this radiation. 1. 2 . × 10 − 13 J correct 2. None of these 3. 2 . × 10 − 19 J 4. 6 . 4 × 10 − 10 J 5. 6 . 63 × 10 − 34 J 6. 6 . 4 × 10 − 14 J 7. 2 . × 10 − 16 J Explanation: λ = 1 . 00 × 10 − 3 nm E photon = ? E photon = h ν = h c λ = (6 . 626 × 10 − 34 J · s)(3 . × 10 8 m/s) (1 . 00 × 10 − 3 nm)(10 − 9 m/nm) = 2 . × 10 − 13 J 004 14.3 points In the formula for copper(II) phosphate, how many phosphorus atoms are there? 1. one 2. four 3. two correct 4. three Explanation: The copper(II) ion is Cu 2+ ; the phosphate ion is PO 3 − 4 . Two PO 3 − 4 are needed to balance the charge on every three Cu 2+ . (This gives a total anion charge of- 6 and a total cation charge of +6.) The formula is Cu 3 (PO 4 ) 2 . 005 14.3 points The first ionization potential of the elements B, C, and N (atomic numbers 5, 6, and 7) steadily increases, but that of O is less than that of N. The best interpretation of the lower value for O is that ctt253 – Exam 1a – Sparks – (53535) 2 1. there is more shielding of the nuclear charge in O than in B, C, or N. 2. the ionization potential of N is a maxi- mum and the values decrease steadily for the elements O, F, and Ne. 3. the electron removed from O is farther from the nucleus and therefore less tightly bound than that in N. 4. the half-filled set of p orbitals in N makes it more difficult to remove an electron from N than from O. correct 5. the electron removed from O corresponds to a different value of the quantum number ℓ than that of the electron removed from B, C, or N....
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This note was uploaded on 03/19/2008 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas at Austin.

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Exam1 - ctt253 – Exam 1a – Sparks –(53535 1 This print-out should have 35 questions Multiple-choice questions may continue on the next column

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