General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 21, Problem 21.73QP
Interpretation Introduction

Interpretation:

The pressure present inside the flask after 78.8 min has to be calculated using the given data.

Concept Introduction:

Radiocarbon dating: The dynamic equilibrium exists in all living organism by exhaling or inhaling, maintain the same ratio of 12C and 14C that takes in. When the living organism dies, the intake of C stops and it’s the ratio no longer exhibits equilibrium since 14C undergoes radioactive decay.

Half-life period: The time required to reduce to half of its initial value.

Formula used to calculate half-life:

t1/2=0.693kwhere,kis rate constant.(or)ln[N]t- ln[N]0= -kt

Ideal gas equation:

An ideal gas equation is,

  PV = n RTwhere, P , pressure           V is volume,           n is mole of He           R is gas constant           T is temperature in Kelvin

Expert Solution & Answer
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Answer to Problem 21.73QP

The value of pressure exist in the flask is 22.42mmHg

Explanation of Solution

Given: Half-life of Bismuth 214 is t1/2=19.7 min by emitting alpha particle (Helium nuclei). Half-life of Bismuth 214 after 78.8 min is,

  t1/2= 19.7 mink = 0.693t1/2=0.69319.7 min=3.52×102min1

The decay constant is 3.52×102min1

Calculate the number of Bismuth atoms in the flask:

Mass of 214Bi in the flask is 5.26mg=5.26×103g

Number of Bismuth atoms in the flask is,

No.of atoms=(Moles)(Avogadro'snumber)=(5.26×103g214g/mol)(6.023×1023atoms/mol)=1.48×1019atoms

The number of Bismuth atoms [N]0 present initially, is 1.48×1019atoms

Predict the equation from first order decay process:

First order process equation:  ln[N]t ln[N]0 ktAs known, k =0.693t1/2substituteinaboveequationln[N]t ln[N]0=(3.52×102min1)(78.8 min)ln([N]t[N]0)2.77([N]t[N]0)=e2.77=6.27×102

By comparing the values,

[N]t=(6.27×102)(1.48×1019atoms)=9.28×1017atoms

The number of Bismuth atoms [N]t present after decay, is 9.28×1017atoms

Atoms of Helium is same as Atoms of Bismuth decayed

Atoms of He emitted is given as:

  [N]0[N]t=(1.48×1019atoms)(9.28×1017atoms)=1.3872×1019atoms

Conversion of Helium atom into mole:

Mole of Helium 1.3872×1019atoms6.023×1023atoms/mol=2.30×105molHe

The mole of Helium emitted is 2.30×105molHe

Calculate the pressure in the flask:

  T= 40oC= 313KV=20.0mL=0.0200L

An ideal gas equation is,

  PV = n RTwhere, P , pressure           V is volume,           n is mole of He           R is gas constant           T is temperature in Kelvin

By substituting the known values into the above equation is,

  P = n RTV=(2.30×105molHe)(0.08206L.atomK.mol)(313K)0.0200L=2.95×102atmP = 2.95×102atm×760mmHg1atm=22.42mmHg

The value of pressure exist in the flask is 22.42mmHg

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Chapter 21 Solutions

General Chemistry

Ch. 21 - Prob. 21.2QPCh. 21 - Prob. 21.3QPCh. 21 - Prob. 21.4QPCh. 21 - Prob. 21.5QPCh. 21 - Prob. 21.6QPCh. 21 - Prob. 21.7QPCh. 21 - Prob. 21.8QPCh. 21 - Prob. 21.9QPCh. 21 - Prob. 21.10QPCh. 21 - Prob. 21.11QPCh. 21 - Prob. 21.12QPCh. 21 - Prob. 21.13QPCh. 21 - Prob. 21.14QPCh. 21 - Prob. 21.15QPCh. 21 - Prob. 21.16QPCh. 21 - Prob. 21.17QPCh. 21 - Prob. 21.18QPCh. 21 - Prob. 21.19QPCh. 21 - Prob. 21.20QPCh. 21 - Prob. 21.21QPCh. 21 - Prob. 21.22QPCh. 21 - Prob. 21.23QPCh. 21 - Prob. 21.24QPCh. 21 - Prob. 21.25QPCh. 21 - Prob. 21.26QPCh. 21 - Prob. 21.27QPCh. 21 - Prob. 21.28QPCh. 21 - Prob. 21.29QPCh. 21 - Prob. 21.30QPCh. 21 - Prob. 21.31QPCh. 21 - Prob. 21.32QPCh. 21 - Prob. 21.33QPCh. 21 - Prob. 21.34QPCh. 21 - Prob. 21.35QPCh. 21 - Prob. 21.36QPCh. 21 - Prob. 21.37QPCh. 21 - Prob. 21.38QPCh. 21 - Prob. 21.39QPCh. 21 - Prob. 21.40QPCh. 21 - Prob. 21.41QPCh. 21 - Prob. 21.42QPCh. 21 - Prob. 21.43QPCh. 21 - Prob. 21.44QPCh. 21 - Prob. 21.45QPCh. 21 - Prob. 21.46QPCh. 21 - Prob. 21.47QPCh. 21 - Prob. 21.48QPCh. 21 - Prob. 21.49QPCh. 21 - Prob. 21.50QPCh. 21 - Prob. 21.51QPCh. 21 - Prob. 21.52QPCh. 21 - Prob. 21.53QPCh. 21 - Prob. 21.54QPCh. 21 - Prob. 21.55QPCh. 21 - Prob. 21.56QPCh. 21 - Prob. 21.57QPCh. 21 - Prob. 21.58QPCh. 21 - Prob. 21.59QPCh. 21 - Prob. 21.60QPCh. 21 - Prob. 21.61QPCh. 21 - Prob. 21.62QPCh. 21 - Prob. 21.63QPCh. 21 - Prob. 21.64QPCh. 21 - Prob. 21.65QPCh. 21 - Prob. 21.66QPCh. 21 - Prob. 21.67QPCh. 21 - Prob. 21.68QPCh. 21 - Prob. 21.69QPCh. 21 - Prob. 21.70QPCh. 21 - Prob. 21.71QPCh. 21 - Prob. 21.72QPCh. 21 - Prob. 21.73QPCh. 21 - Prob. 21.74QPCh. 21 - Prob. 21.75QPCh. 21 - Prob. 21.76QPCh. 21 - Prob. 21.77QPCh. 21 - Prob. 21.78QPCh. 21 - Prob. 21.79QPCh. 21 - Prob. 21.80QPCh. 21 - Prob. 21.81QPCh. 21 - Prob. 21.82QPCh. 21 - Prob. 21.83QPCh. 21 - Prob. 21.84QPCh. 21 - Prob. 21.85QPCh. 21 - Prob. 21.86QPCh. 21 - Prob. 21.87SPCh. 21 - Prob. 21.88SPCh. 21 - Prob. 21.89SPCh. 21 - Prob. 21.90SPCh. 21 - Prob. 21.91SPCh. 21 - Prob. 21.92SPCh. 21 - Prob. 21.93SPCh. 21 - Prob. 21.94SPCh. 21 - Prob. 21.95SPCh. 21 - Prob. 21.96SPCh. 21 - Prob. 21.97SPCh. 21 - Prob. 21.98SPCh. 21 - Prob. 21.99SP
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