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

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momin (rrm497) – Homework 7 – Sutcliffe – (53120) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. NOTE: Read the questions carefully. If it asks for a CHANGE (Delta) in a quantity or the VALUE of that quantity, then you must be sure to include the correct sign. Remem- ber that the quantities are for the SYSTEM, unless stated differently. This covers most but not all of 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 of 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; from - 10 to - 5 2. manganese; from +7 to +5 3. oxygen; from - 1 to - 2 4. hydrogen; from +1 to 0 5. iodine; from - 1 to 0 correct Explanation: Oxidation is the loss of electrons and is indicated by the algebraic increase of the ox- idation number. Oxidation numbers are per atom. The oxidation number of 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 of free, uncombined elements is zero, includ- ing polyatomic elements such as I 2 . In com- pounds, the oxidation number of O is - 2 and of H is +1. The sum of oxidation numbers in a polyatomic ion is the charge on the ion. We calculate the oxidation number of Mn in MnO 1 4 : x + 4( - 2) = - 1 x = +7 The oxidation number of I increases from - 1 to 0, indicating that it is being oxidized. 002 10.0 points The oxidation numbers exhibited by iron in the red-brown oxide Fe 2 O 3 (iron rust), the greenish-black oxide FeO (unstable and al- most impossible to get pure) and the very unstable reddish-purple ferrate ion FeO 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 of oxidation numbers in a neutral molecule is zero and the sum of oxidation numbers in a polyatomic ion is equal to the charge on the ion. Oxygen has an oxidation number of - 2. We set the sum of the oxidation numbers in each compound equal to the appropriate value and solve for the oxidation number of Fe (symbolized by x ). 2 x + 3( - 2) = 0 Fe 2 O 3 : x = +3 x + ( - 2) = 0 FeO : x = +2

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momin (rrm497) – Homework 7 – Sutcliffe – (53120) 2 x + 4( - 2) = - 2 FeO 2 4 : x = +6 003 10.0 points The combustion of a certain amount of methane gives off 100 kJ of energy. Dur- ing this process the amount of energy in the universe 1. decreases by 100 kJ.
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