Asked by Alexus
How do you write balanced equations
A. Neutron emission by 88Br
B. Electron absorption by 116Sb
C. Positron emission by 184Hg
D. Alpha emission by 229Th
E. Neutron capture by 200Hg
A. Neutron emission by 88Br
B. Electron absorption by 116Sb
C. Positron emission by 184Hg
D. Alpha emission by 229Th
E. Neutron capture by 200Hg
Answers
Answered by
DrBob222
It's SUPPOSED to be atomic number as a lower left subscript and mass number as upper left superscript BUT that's difficult to do on this board so what I do is write the atomic number before the symbol as in 35Br and mass number after the symbol as in 35Br80 etc.
The secret to writing these is to make the subscripts add up and the superscripts add up.
35Br88 ==> 0n1 + yXz
Note: the neutron has a zero charge and a mass of 1. We are looking for element X with atomic number y and mass number z.
35 = 0 + y; therefore, y must be 35
88 = 1+z; therefore 87 = z
Since atomic number is 35 it still is Br and the full equation is
35Br88 ==> 0n1 + 35Br87
51Sb116 + -1e0 = yXz
51-1 = y; therefore, y = 50
116 + 0 = z; therefore, z = 116
element #50 is Sn so the full equation is
51Sb116 + -1e0 = 50Sn116
I'll leave the others for you.
The secret to writing these is to make the subscripts add up and the superscripts add up.
35Br88 ==> 0n1 + yXz
Note: the neutron has a zero charge and a mass of 1. We are looking for element X with atomic number y and mass number z.
35 = 0 + y; therefore, y must be 35
88 = 1+z; therefore 87 = z
Since atomic number is 35 it still is Br and the full equation is
35Br88 ==> 0n1 + 35Br87
51Sb116 + -1e0 = yXz
51-1 = y; therefore, y = 50
116 + 0 = z; therefore, z = 116
element #50 is Sn so the full equation is
51Sb116 + -1e0 = 50Sn116
I'll leave the others for you.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.