(a) Consider the following nuclear process, in which a proton is removed from an
ID: 1394656 • Letter: #
Question
Explanation / Answer
Here,
16O8 = 15.99491 amu
15O8 = 15.00306 amu
15N7 = 15.00010 amu
neutron = 1.00893 amu
proton = 1.00814 amu
a) The reaction is
16O8 + E -> 15N7 + 1H1
Sum of the masses of reactants = 15.99491 amu
Sum of the masses of products =15.00010 amu + 1.00814 amu = 16.00824 a.m.u.
Therefore, ?M = Sum of the masses of products
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.