Spring 09 Recitation Quiz 7

Spring 09 Recitation Quiz 7 - CHEM 0120 Golde SPRING 2009...

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CHEM 0120 SPRING 2009 Monday 1:00 PM Golde QUIZ 7 WEEK 9 March 2, 2009 Name_______________________________ 1 Coulomb = 1 Amp x 1 sec ; F = 96500 Coul/mol e - , ΔE = Δm.c 2 , c = 3.00 x 10 8 m/s 1. Given the following half reactions: 2 H 2 O(l) + 2 e - H 2 (g) + 2 OH - (aq) E o red = -0.83 V 2 H + (aq) + 2 e - H 2 (g) E o red = 0 V 2 H 2 O(l) O 2 (g) + 4 H + (aq) + 4 e - E o ox = -1.23 V A certain mystery metal, M, has the following half reaction: M 2+ (aq) + 2 e - M(s) E o red = -0.25V i.e. M(s) M 2+ (aq) + 2e - E o ox = +0.25V (a) Is elemental M(s) able to displace H 2 from 1.0M acids? (b) Is elemental M(s) able to displace H 2 from water? Give your reasoning, making use of the E o data. 2. Complete the following nuclear reactions by inserting the missing symbol. Give the name of each of these missing symbols. a) Br ? Kr 74 35 74 36 + b) 7 14 N + ? O + H 8 17 1 1 ANSWERS 1. (a) Yes. As M is to be oxidized, we require H + to be reduced, i.e. the second given half reaction. For this process, E o cell = 0 + (+0.25)V = 0.25V: positive, therefore reaction occurs. (b) No. Here, water is to be reduced, so we use the first given half reaction. For this process, E o cell = -0.83 + (+0.25) V = -0.58V. The negative value indicates that the reaction does not occur. 2. a) Br e Kr 74 35 0 1 74 36 + → Positron b) H + O He + N 1 1 17 8 4 2 14 7 Alpha particle (or He nucleus) CHEM 0120 SPRING 2009 Monday 6:00 PM For Make-up Only Regular recitation Day: Time: instructor: Quiz Score: Group score: Scores recorded by:
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Golde QUIZ 7 WEEK 9 March 2, 2009 Name_______________________________ 1 Coulomb = 1 Amp x 1 sec ; F = 96500 Coul/mol e - , ΔE = Δm.c 2 , c = 3.00 x 10 8 m/s 1. A certain battery contains 46.3g of metallic Zn. (a) How many Coulombs of charge can this battery deliver, if all of the Zn is oxidized to Zn 2+ ? (b) If the average current drawn is 0.65Amp, for how many hours will this battery run? 2. a) Write a complete nuclear equation for alpha emission by 228 88 Ra b) Write a complete nuclear equation for electron capture by Fe 52 26 . c) Explain why electron capture is a favored mode of decay of this isotope of Fe. ANSWERS 1. (a) Moles of Zn = 46.3g / 65.4g/mol = 0.708 mol Zn. Mole of e - = 0.708 mol Zn x 2 mol e - /1 mol Zn = 1.42 mol e - . Charge delivered = 1.42 mol e - x 96500 Coul/mol e - = 1.37 x 10 5 Coul (b) Time = (1.37 x 10 5 Coul / 0.65 Amp) x 1 hr / 3600 sec = 58 hr. 2. a) 228 4 224 88 2 86 Ra He + Rn b) Mn e Fe 52 25 0 1 - 52 26 + c) The molar mass of Fe is 55.8g/mol. Thus 52 Fe has fewer neutrons than are needed for stability. Such isotopes prefer to decay by electron capture or positron emission (conversion of a
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This note was uploaded on 04/06/2009 for the course CHEM 0120 taught by Professor Golde during the Spring '07 term at Pittsburgh.

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Spring 09 Recitation Quiz 7 - CHEM 0120 Golde SPRING 2009...

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