If three neutrons are produced when a neutron bombards a U- 235 atom, determine the second fission product isotope when the first fission fragment is (a) xenon-133, (b) barium-144, (c) cesium-143, (d) tellurium-137, and (e) lanthanum-146.

Explanation:

235 U92 + 1n0 --> A + B + 3 1n0

when A = ;

(a) xenon-133

Mass number of B = 235 + 1 - 133 -3 = 100

Atomic number = 92 - 54 = 38

Element = Sr - Strontium

(b) barium-144

Mass number of B = 235 + 1 - 144 -3 = 89

Atomic number = 92 - 56 = 36

Element = Kr - Krypton

(c) cesium-143

Mass number of B = 235 + 1 - 143 -3 = 90

Atomic number = 92 - 55 = 37

Element = Rb - Rubidium

(d) tellurium-137

Mass number of B = 235 + 1 - 137 -3 = 96

Atomic number = 92 - 52 = 40

Element = Zr - Zirconium

(e) lanthanum-146

Mass number of B = 235 + 1 - 146 -3 = 87

Atomic number = 92 - 57 = 35

Element = Br - Bromine

substituting a hydrogen atom with a halogen in a hydrocarbon:

- the single bond remains single. a is wrong.

- the bond with hydrogen is as strong as with a halogen. c is wrong

- a hydrocarbon is saturated if there is no double bond. we do not know that so d is wrong.

- carbon will always have four valence electrons. e is wrong.

by elimination, the correct answer is b. the boiling point of the new compound increases.