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hw 4 - Garg Pallavi Homework 4 Due Nov 7 2007 2:00 am Inst...

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Garg, Pallavi – Homework 4 – Due: Nov 7 2007, 2:00 am – Inst: Donna C Lyon 1 This print-out should have 51 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. This HW assignment is due Wednesday, November 7 by 2:00AM. DO NOT WAIT UNTIL THE LAST HOUR TO TURN IN YOUR HOMEWORK ANSWERS. KEEP UP WITH THE MATERIAL BY ANSWER- ING QUESTIONS AS THEY ARE COV- ERED IN CLASS. 001 (part 1 of 1) 10 points When the ions CO 2 - 3 , Cl - , NO - 3 , Na + , NH + 4 , Fe 2+ are mixed in aqueous solution, what sub- stance is likely to form an insoluble precip- itate? 1. NH 4 Cl 2. FeCO 3 correct 3. ClNO 3 4. Na 2 CO 3 5. NH 4 NO 3 6. NaNO 3 Explanation: The common carbonates ( CO 2 - 3 ) are insol- uble except for those of the alkali metals and NH + 4 . 002 (part 1 of 1) 10 points In four of the following groups, the three substances are either all strong electrolytes, all weak electrolytes, or all nonelectrolytes. Which group has one member which is NOT like the others in the group? 1. NH 3 ; CH 3 COOH; HCN 2. H 2 SO 4 ; NaOH; HCl 3. BaCl 2 ; NaNO 3 ; HF correct 4. O 2 ; CO(NH 2 ) 2 ; CH 3 OH 5. NaCl; Ca(OH) 2 ; HNO 3 Explanation: 003 (part 1 of 1) 10 points Assign oxidation numbers to each atom in MnO - 4 . 1. Mn = +1, O = - 2 2. Mn = - 2, O = +2 3. Mn = +3, O = - 1 4. Mn = +7, O = - 2 correct 5. Mn = +5, O = - 2 Explanation: oxidation numbers = ? According to rule 4, O has an oxidation number of - 2 : Mn - 2 O - 4 The total oxidation number of the oxygen atoms is - 2 × 4 = - 8 : Mn - 2 O - 4 -→ Mn - 2 O - 4 - 8 Balance the oxidation numbers: Mn +7 - 2 O - 4 - 8 -→ +7 Mn +7 - 2 O - 4 - 8 Mn therefore has an oxidation number of +7: Mn = +7 , O = - 2 004 (part 1 of 1) 10 points Depending on conditions, vanadium forms any of 4 different cations: VO 2 + (light yel- low), V 2+ (violet), VO 2+ (blue), or V 3+ (green) in acid solution. The oxidation num- bers exhibited by vanadium in these ions are 1. +1, +2, +2, and, +3, respectively. 2. - 5, - 2, - 4, and - 3, respectively.
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Garg, Pallavi – Homework 4 – Due: Nov 7 2007, 2:00 am – Inst: Donna C Lyon 2 3. +5, +2, +4, and +3, respectively. cor- rect 4. - 1, - 2, - 2, and - 3, respectively. Explanation: Oxidation numbers are per atom. The ox- idation number of a monoatomic ion is equal to the charge on the ion. Vanadium in V 2+ has an oxidation of +2. 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 oxida- tion numbers in each compound equal to the appropriate value and solve for the oxidation number of V (symbolized by x ): x + 2( - 2) = +1 VO + 2 : x = +5 x + ( - 2) = +2 VO 2+ : x = +4 005 (part 1 of 2) 10 points An excess of aqueous AgNO 3 reacts with 48 . 5 mL of 5 M K 2 CrO 4 (aq) to form a precip- itate. What is the precipitate? 1. AgNO 3 2. K 2 CrO 4 3. Ag 2 CrO 4 correct 4. KNO 3 Explanation: 2 AgNO 3 (aq) + K 2 CrO 4 (aq) Ag 2 CrO 4 (s) + 2 KNO 3 (aq) 006 (part 2 of 2) 10 points What mass of precipitate is formed?
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