General Chemistry - Standalone book (MindTap Course List)
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN: 9781305580343
Author: Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher: Cengage Learning
Question
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Chapter 20, Problem 20.101QP

(a)

Interpretation Introduction

Interpretation:

Bromine-82 decays by emission of beta particle. The half-life of Bromine-82 is 1.471d.

The balanced equation for decay of Bromine-82 and decomposition of H82Br has to be written.

(a)

Expert Solution
Check Mark

Answer to Problem 20.101QP

The balanced equation for decay of Bromine-82 is,

3582Br3682Kr +-1 0β

The balanced equation for decomposition of H82Br

2H3582Br(g)H2+ 23682Kr +-1 0β

Explanation of Solution

When Bromine-82 undergoes beta ( -10β ) decay it gives Krypton-82. This balanced equation is given as,

3582Br3682Kr +-1 0β

Already given that Hydrogen bromide made with Bromine-82 isotope. Above equation shows that Br82 decays to gives Kr-82 and beta particle. Hence Decomposition of H82Br is given by,

2H3582Br(g)H2+ 23682Kr +-1 0β

(b)

Interpretation Introduction

Interpretation:

Bromine-82 decays by emission of beta particle. The half-life of Bromine-82 is 1.471d.

The quantity of HBr remains after 10h has to be calculated and the pressure in the flask at 22°C should be identified.

(b)

Expert Solution
Check Mark

Answer to Problem 20.101QP

The quantity of HBr remains after 10h is 0.0123 mol .

At 22°C the pressure in the flask is 0.3957 atm .

Explanation of Solution

Use the rate law to identify the fraction after 10.0h. The half-life is 1.471d.

lnNtNo= -kt =-0.693tt12=-0.693(0.4167d)1.471 d= -0.1963

Taking antilog on both sides gives,

NtNo=AtAo=e-0.1963= 0.8218

So, the number of moles of H82Br remains after 10h is

At = 0.0150 mol × 0.8218= 0.0123 mol

The reaction can be summarized as follows,

2HBr           H2 + 2Kr0.0150-2yy2y

From this,we can calculate y,0.0150mol-2y = 0.01233 moly = 0.00134mol

To determine the pressure in the flask, we required the total moles of gas. This is,

Total mass = mol H3582Br + mol Kr= (0.0150-2y) + y + 2y= 0.0150 + y = 0.0150 + 0.00134 = 0.01634mol

Therefore, the pressure in the flask is,

P =nRTV=0.01634 mol×0.0821L•atm/K•mol×295K1.00L=0.396 atm

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Chapter 20 Solutions

General Chemistry - Standalone book (MindTap Course List)

Ch. 20.4 - Prob. 20.9ECh. 20.4 - Prob. 20.10ECh. 20.4 - Prob. 20.11ECh. 20.4 - Prob. 20.3CCCh. 20.6 - Prob. 20.12ECh. 20 - Prob. 20.1QPCh. 20 - Prob. 20.2QPCh. 20 - Prob. 20.3QPCh. 20 - Prob. 20.4QPCh. 20 - Prob. 20.5QPCh. 20 - Prob. 20.6QPCh. 20 - Prob. 20.7QPCh. 20 - Prob. 20.8QPCh. 20 - Prob. 20.9QPCh. 20 - Prob. 20.10QPCh. 20 - Prob. 20.11QPCh. 20 - Prob. 20.12QPCh. 20 - Prob. 20.13QPCh. 20 - Prob. 20.14QPCh. 20 - Prob. 20.15QPCh. 20 - Prob. 20.16QPCh. 20 - Prob. 20.17QPCh. 20 - Prob. 20.18QPCh. 20 - Prob. 20.19QPCh. 20 - Prob. 20.20QPCh. 20 - Prob. 20.21QPCh. 20 - Prob. 20.22QPCh. 20 - Prob. 20.23QPCh. 20 - Prob. 20.24QPCh. 20 - Prob. 20.25QPCh. 20 - Prob. 20.26QPCh. 20 - Prob. 20.27QPCh. 20 - Prob. 20.28QPCh. 20 - Prob. 20.29QPCh. 20 - Prob. 20.30QPCh. 20 - Prob. 20.31QPCh. 20 - Prob. 20.32QPCh. 20 - Prob. 20.33QPCh. 20 - Prob. 20.34QPCh. 20 - Prob. 20.35QPCh. 20 - Prob. 20.36QPCh. 20 - Prob. 20.37QPCh. 20 - Prob. 20.38QPCh. 20 - Prob. 20.39QPCh. 20 - Prob. 20.40QPCh. 20 - Prob. 20.41QPCh. 20 - Prob. 20.42QPCh. 20 - Prob. 20.43QPCh. 20 - Prob. 20.44QPCh. 20 - Prob. 20.45QPCh. 20 - Prob. 20.46QPCh. 20 - Prob. 20.47QPCh. 20 - Prob. 20.48QPCh. 20 - Prob. 20.49QPCh. 20 - Prob. 20.50QPCh. 20 - Prob. 20.51QPCh. 20 - Prob. 20.52QPCh. 20 - Fill in the missing parts of the following...Ch. 20 - Fill in the missing parts of the following...Ch. 20 - Prob. 20.55QPCh. 20 - Prob. 20.56QPCh. 20 - Prob. 20.57QPCh. 20 - Prob. 20.58QPCh. 20 - Prob. 20.59QPCh. 20 - Prob. 20.60QPCh. 20 - Prob. 20.61QPCh. 20 - Prob. 20.62QPCh. 20 - Prob. 20.63QPCh. 20 - Prob. 20.64QPCh. 20 - Prob. 20.65QPCh. 20 - Prob. 20.66QPCh. 20 - Prob. 20.67QPCh. 20 - Prob. 20.68QPCh. 20 - Prob. 20.69QPCh. 20 - Prob. 20.70QPCh. 20 - Prob. 20.71QPCh. 20 - Prob. 20.72QPCh. 20 - Prob. 20.73QPCh. 20 - Prob. 20.74QPCh. 20 - Prob. 20.75QPCh. 20 - Prob. 20.76QPCh. 20 - Prob. 20.77QPCh. 20 - Prob. 20.78QPCh. 20 - Find the change of mass (in grams) resulting from...Ch. 20 - Find the change of mass (in grams) resulting from...Ch. 20 - Prob. 20.81QPCh. 20 - Prob. 20.82QPCh. 20 - Prob. 20.83QPCh. 20 - Prob. 20.84QPCh. 20 - Prob. 20.85QPCh. 20 - Prob. 20.86QPCh. 20 - Prob. 20.87QPCh. 20 - Prob. 20.88QPCh. 20 - Prob. 20.89QPCh. 20 - Prob. 20.90QPCh. 20 - Prob. 20.91QPCh. 20 - Prob. 20.92QPCh. 20 - Prob. 20.93QPCh. 20 - Prob. 20.94QPCh. 20 - Prob. 20.95QPCh. 20 - Prob. 20.96QPCh. 20 - Prob. 20.97QPCh. 20 - Prob. 20.98QPCh. 20 - Prob. 20.99QPCh. 20 - Prob. 20.100QPCh. 20 - Prob. 20.101QPCh. 20 - Prob. 20.102QPCh. 20 - Prob. 20.103QPCh. 20 - Prob. 20.104QPCh. 20 - Prob. 20.105QPCh. 20 - Prob. 20.106QPCh. 20 - Prob. 20.107QPCh. 20 - Prob. 20.108QPCh. 20 - Prob. 20.109QPCh. 20 - Prob. 20.110QPCh. 20 - Prob. 20.111QPCh. 20 - Prob. 20.112QPCh. 20 - Prob. 20.113QPCh. 20 - Prob. 20.114QPCh. 20 - Prob. 20.115QPCh. 20 - Prob. 20.116QPCh. 20 - Prob. 20.117QPCh. 20 - Prob. 20.118QPCh. 20 - Prob. 20.119QPCh. 20 - Prob. 20.120QPCh. 20 - Prob. 20.121QPCh. 20 - Prob. 20.122QPCh. 20 - Prob. 20.123QPCh. 20 - Prob. 20.124QPCh. 20 - Prob. 20.125QPCh. 20 - Prob. 20.126QPCh. 20 - Prob. 20.127QPCh. 20 - Prob. 20.128QP
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