Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393615159
Author: Stacey Lowery Bretz, Geoffrey Davies, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss
Publisher: W. W. Norton & Company
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Chapter 19, Problem 19.89AP

(a)

Interpretation Introduction

Interpretation: The use of 241Am in smoke detectors and the ionization of Oxygen and Nitrogen by the alpha particles that are emitted by this isotope that helps in creating the current are given. Various questions based on the radioactivity of this isotope have to be answered.

Concept introduction: The physical phenomenon that involves the spontaneous emission of radiations is known as radioactivity.

To determine: The device to which smoke detector shows close resemblance.

(b)

Interpretation Introduction

To determine: The time taken for the radioactivity decay of sample of 241Am to decay to 1% of its original radioactivity.

(c)

Interpretation Introduction

To determine: The reason behind the safe handling of smoke detectors containing Americium 241 without any protective equipment.

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MISSED THIS? Read Section 19.9 (Pages 878-881); Watch IWE 19.10 Consider the following reaction: CH3OH(g) CO(g) + 2H2(g) (Note that AG,CH3OH(g) = -162.3 kJ/mol and AG,co(g)=-137.2 kJ/mol.) Part A Calculate AG for this reaction at 25 °C under the following conditions: PCH₂OH Pco PH2 0.815 atm = 0.140 atm 0.170 atm Express your answer in kilojoules to three significant figures. Ο ΑΣΦ AG = -150 Submit Previous Answers Request Answer □? kJ × Incorrect; Try Again; 2 attempts remaining Calculate the free energy change under nonstandard conditions (AGrxn) by using the following relationship: AGrxn = AGrxn + RTInQ, AGxn+RTInQ, where AGxn is the standard free energy change, R is the ideal gas constant, T is the temperature in kelvins, a is the reaction quotient. Provide Feedback Next >
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Chapter 19 Solutions

Chemistry: The Science in Context (Fifth Edition)

Ch. 19 - Prob. 19.5VPCh. 19 - Prob. 19.6VPCh. 19 - Prob. 19.7VPCh. 19 - Prob. 19.8VPCh. 19 - Prob. 19.9VPCh. 19 - Prob. 19.10VPCh. 19 - Prob. 19.11QPCh. 19 - Prob. 19.12QPCh. 19 - Prob. 19.13QPCh. 19 - Prob. 19.14QPCh. 19 - Prob. 19.15QPCh. 19 - Prob. 19.16QPCh. 19 - Prob. 19.17QPCh. 19 - Prob. 19.18QPCh. 19 - Prob. 19.19QPCh. 19 - Prob. 19.20QPCh. 19 - Prob. 19.21QPCh. 19 - Prob. 19.22QPCh. 19 - Prob. 19.23QPCh. 19 - Prob. 19.24QPCh. 19 - Prob. 19.25QPCh. 19 - Prob. 19.26QPCh. 19 - Prob. 19.27QPCh. 19 - Prob. 19.28QPCh. 19 - Prob. 19.29QPCh. 19 - Prob. 19.30QPCh. 19 - Prob. 19.31QPCh. 19 - Prob. 19.32QPCh. 19 - Prob. 19.33QPCh. 19 - Prob. 19.34QPCh. 19 - Prob. 19.35QPCh. 19 - Prob. 19.36QPCh. 19 - Prob. 19.37QPCh. 19 - Prob. 19.38QPCh. 19 - Prob. 19.39QPCh. 19 - Prob. 19.40QPCh. 19 - Prob. 19.41QPCh. 19 - Prob. 19.42QPCh. 19 - Prob. 19.43QPCh. 19 - Prob. 19.44QPCh. 19 - Prob. 19.45QPCh. 19 - Prob. 19.46QPCh. 19 - Prob. 19.47QPCh. 19 - Prob. 19.48QPCh. 19 - Prob. 19.49QPCh. 19 - Prob. 19.50QPCh. 19 - Prob. 19.51QPCh. 19 - Prob. 19.52QPCh. 19 - Prob. 19.53QPCh. 19 - Prob. 19.54QPCh. 19 - Prob. 19.55QPCh. 19 - Prob. 19.56QPCh. 19 - Prob. 19.57QPCh. 19 - Prob. 19.58QPCh. 19 - Prob. 19.59QPCh. 19 - Prob. 19.60QPCh. 19 - Prob. 19.61QPCh. 19 - Prob. 19.62QPCh. 19 - Prob. 19.63QPCh. 19 - Prob. 19.64QPCh. 19 - Prob. 19.65QPCh. 19 - Prob. 19.66QPCh. 19 - Prob. 19.67QPCh. 19 - Prob. 19.68QPCh. 19 - Prob. 19.69QPCh. 19 - Prob. 19.70QPCh. 19 - Prob. 19.71QPCh. 19 - Prob. 19.72QPCh. 19 - Prob. 19.73QPCh. 19 - Prob. 19.74QPCh. 19 - Prob. 19.75QPCh. 19 - Prob. 19.76QPCh. 19 - Prob. 19.77QPCh. 19 - Prob. 19.78QPCh. 19 - Prob. 19.79QPCh. 19 - Prob. 19.80QPCh. 19 - Prob. 19.81QPCh. 19 - Prob. 19.82QPCh. 19 - Prob. 19.83QPCh. 19 - Prob. 19.84QPCh. 19 - Prob. 19.85APCh. 19 - Prob. 19.86APCh. 19 - Prob. 19.87APCh. 19 - Prob. 19.88APCh. 19 - Prob. 19.89APCh. 19 - Prob. 19.90APCh. 19 - Prob. 19.91APCh. 19 - Prob. 19.92APCh. 19 - Prob. 19.93APCh. 19 - Prob. 19.94APCh. 19 - Prob. 19.95APCh. 19 - Prob. 19.96APCh. 19 - Prob. 19.97APCh. 19 - Prob. 19.98APCh. 19 - Prob. 19.99APCh. 19 - Prob. 19.100AP
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