EBK CHEMICAL PRINCIPLES
EBK CHEMICAL PRINCIPLES
8th Edition
ISBN: 8220101425812
Author: DECOSTE
Publisher: Cengage Learning US
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Chapter 20, Problem 16E

(a)

Interpretation Introduction

Interpretation:

The unknown element for equation for ?+H24eB97243k+n01 needs to be determined.

Concept Introduction:

A radioactive decay is unstable nuclei which undergoes decay as it dissociates to smaller fragments releasing alpha, beta or gamma particles. This is because the ratio of number of protons to number of neutrons is not equal. If the ratio of number of protons to number of neutrons is less or equal to 1, then positron is released and if the ratio is more than 1 then beta particle is released.

(a)

Expert Solution
Check Mark

Answer to Problem 16E

The unknown element for equation ?+H24eB97243k+n01 is A95240m .

Explanation of Solution

The equation given is ?+H24eB97243k+n01

To determine the mass number and atomic number of unknown elements, the mass numbers and atomic numbers are equated on both sides.

  Mass number : ? + 4 = 243 + 1  x=2444=240Atomic number : x + 2 = 97 + 0  x=972=95

Looking through the periodic table, the element with mass number 240 and atomic number 95 is A95240m .

The equation is written as A95240m+H24eB97243k+n01 .

(b)

Interpretation Introduction

Interpretation:

The unknown element for equation for U92238 + C612? + 6n01 needs to be determined.

Concept Introduction:

An unstable nuclei which undergo decay to dissociate to smaller fragments releasing alpha, beta or gamma particles as there is an imbalance in the ratio of number of protons to number of neutrons. If the ratio of number of protons to number of neutrons is less or equal to 1, then positron is released and if the ratio is more than 1 then beta particle is released.

(b)

Expert Solution
Check Mark

Answer to Problem 16E

The unknown element for equation U92238 + C612? + 6n01 is C98244f .

Explanation of Solution

The equation given is U92238 + C612? + 6n01

To determine the mass number and atomic number of unknown elements, the mass numbers and atomic numbers are equated on both sides.

  Mass number : 238 + 12 = ? + 6  x=2506=244Atomic number : 92 + 6 = ?+ 0  x=980=98

Looking through the periodic table, the element with mass number 244 and atomic number 98 is C98244f .

The equation is written as U92238 + C612C98244f + 6n01 .

(c)

Interpretation Introduction

Interpretation:

The unknown element for equation for C98249f + ?D105260b + 4n01 needs to be determined.

Concept Introduction:

A radioactive decay undergoes when an unstable nuclei dissociates to smaller fragments releasing alpha, beta or gamma particles due to imbalance in the number of protons to number of neutrons If the ratio of protons to number of neutrons is less or equal to 1, a positron is released and if the ratio is more than 1 then there is release of beta particle.

(c)

Expert Solution
Check Mark

Answer to Problem 16E

The unknown element for equation C98249f + ?D105260b + 4n01 is N715 .

Explanation of Solution

The equation given is C98249f + ?D105260b + 4n01

To determine the mass number and atomic number of unknown elements, the mass numbers and atomic numbers are equated on both sides.

  Mass number : 249 + ? = 260 + 4  x=264249=15Atomic number : 98 + ? = 105+ 0  x=10598=7

Looking through the periodic table, the element with mass number 15 and atomic number 7 is N715 .

The equation is written as C98249f + N715D105260b + 4n01 .

(d)

Interpretation Introduction

Interpretation:

The unknown element for equation for C98249f + B510L103257r + ? needs to be determined.

Concept Introduction:

If nucleus is unstable, it will undergo dissociation such that it dissociates to smaller fragments releasing alpha, beta or gamma particles. The instability is due to an imbalance in the ratio of number of protons to number of neutrons. If the ratio of number of protons to number of neutrons is less or equal to 1, then positron is released and if the ratio is more than 1 then beta particle is released.

(d)

Expert Solution
Check Mark

Answer to Problem 16E

The unknown element for equation C98249f + B510L103257r + ? is 2n01 .

Explanation of Solution

The equation given is C98249f + B510L103257r + ?

To determine the mass number and atomic number of unknown elements, the mass numbers and atomic numbers are equated on both sides.

  Mass number : 249 + 10 = 257 + ?  x=259257=2Atomic number : 98 + 5 = 103+ ?  x=103103=0

Looking through the periodic table, the element with mass number 2 and atomic number 0 is 2n01 .

The equation is written as C98249f + B510L103257r + 2n01 .

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Chapter 20 Solutions

EBK CHEMICAL PRINCIPLES

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