HW7 - momin(rrm497 Homework 7 Sutclie(53120 This print-out...

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momin (rrm497) – Homework 7 – Sutclife – (53120) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – Fnd all choices be±ore answering. NOTE: Read the questions care±ully. I± it asks ±or a CHANGE (Delta) in a quantity or the VALUE o± that quantity, then you must be sure to include the correct sign. Remem- ber that the quantities are ±or the SYSTEM, unless stated diferently. This covers most but not all o± Ch. 9 and the rest will be on HW8, the last HW. PLEASE NOTE the due date by checking the main page on Quest. As always, the deadline is 11pm. 001 10.0 points In the reaction o± permanganate anion with iodide in acidic solution 2 MnO 4 + 16 H + + 10 I 2 Mn +2 + 8 H 2 O + 5 I 2 , what has been oxidized and what is its change in oxidation number? 1. iodine; ±rom - 10 to - 5 2. manganese; ±rom +7 to +5 3. oxygen; ±rom - 1 to - 2 4. hydrogen; ±rom +1 to 0 5. iodine; ±rom - 1 to 0 correct Explanation: Oxidation is the loss o± electrons and is indicated by the algebraic increase o± the ox- idation number. Oxidation numbers are per atom. The oxidation number o± an element in a monatomic ion is simply the charge on the ion, so I in I is - 1, H in H + is +1, and Mn in Mn 2+ is +2. The oxidation number o± ±ree, uncombined elements is zero, includ- ing polyatomic elements such as I 2 . In com- pounds, the oxidation number o± O is - 2 and o± H is +1. The sum o± oxidation numbers in a polyatomic ion is the charge on the ion. We calculate the oxidation number o± Mn in MnO 1 4 : x + 4( - 2) = - 1 x = +7 The oxidation number o± I increases ±rom - 1 to 0, indicating that it is being oxidized. 002 10.0 points The oxidation numbers exhibited by iron in the red-brown oxide ²e 2 O 3 (iron rust), the greenish-black oxide ²eO (unstable and al- most impossible to get pure) and the very unstable reddish-purple ±errate ion ²eO 2 4 are 1. +3, +2, and +6, respectively. correct 2. +3, +2, and +8, respectively. 3. - 3, - 2, and - 4, respectively. 4. +2, +1, and +4, respectively. 5. +2, +1, and +6, respectively. 6. - 3, - 2, and - 6, respectively. 7. - 2, - 1, and - 4, respectively. 8. - 3, - 2, and - 8, respectively. 9. +3, +2, and +4, respectively. 10. - 2, - 1, and - 6, respectively. Explanation: Oxidation numbers are per atom. The sum o± oxidation numbers in a neutral molecule is zero and the sum o± oxidation numbers in a polyatomic ion is equal to the charge on the ion. Oxygen has an oxidation number o± - 2. We set the sum o± the oxidation numbers in each compound equal to the appropriate value and solve ±or the oxidation number o± ²e (symbolized by x ). 2 x + 3( - 2) = 0 ²e 2 O 3 : x = +3 x + ( - 2) = 0 ²eO : x = +2
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momin (rrm497) – Homework 7 – Sutclife – (53120) 2 x + 4( - 2) = - 2 FeO 2 4 : x = +6 003 10.0 points The combustion o± a certain amount o± methane gives of 100 kJ o± energy. Dur- ing this process the amount o± energy in the universe 1. decreases by 100 kJ.
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This note was uploaded on 12/11/2009 for the course CH 301 taught by Professor Fakhreddine/lyon during the Spring '07 term at University of Texas.

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HW7 - momin(rrm497 Homework 7 Sutclie(53120 This print-out...

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