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
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.
A. Well, to write a balanced equation for neutron emission by 88Br, just imagine the 88Br feeling spontaneous and saying "Bye, bye, neutron!" So, the equation would be: 88Br -> 1n + something else. Don't worry, that "something else" is just the leftover product.
B. Now, for electron absorption by 116Sb, it's like 116Sb saying, "Hey, electron, come on over here, join the party!" So, the equation would be: 116Sb + 1e- -> something else.
C. Ah, positron emission by 184Hg, huh? Well, it's like 184Hg feeling a bit positive and saying, "I'm feeling so positive, I'll release a positron!" So, the equation would be: 184Hg -> 1e+ + something else.
D. Next, we have alpha emission by 229Th. It's like 229Th thinking, "You know what? I'll send out an alpha particle, just to lighten the mood!" So, the equation would be: 229Th -> 4He + something else.
E. Lastly, for neutron capture by 200Hg. It's like 200Hg saying, "Hey, neutron, wanna be a part of me?" So, the equation would be: 1n + 200Hg -> something else.
To write balanced nuclear equations, we need to ensure that the total number of protons (atomic number) and total number of nucleons (mass number) remain balanced on both sides of the equation. Here are the step-by-step balanced equations for the given processes:
A. Neutron emission by 88Br:
1. Identify the initial element: 88Br
2. Determine the product after neutron emission: 87Kr (one less neutron)
3. Write the balanced equation:
88Br → 87Kr + 1n
B. Electron absorption by 116Sb:
1. Identify the initial element: 116Sb
2. Determine the product after electron absorption: 116Te (one additional proton)
3. Write the balanced equation:
116Sb + 1e → 116Te
C. Positron emission by 184Hg:
1. Identify the initial element: 184Hg
2. Determine the product after positron emission: 184Au (one less proton)
3. Write the balanced equation:
184Hg → 184Au + 0e
D. Alpha emission by 229Th:
1. Identify the initial element: 229Th
2. Determine the product after alpha emission: 225Ra (four less protons and four less neutrons)
3. Write the balanced equation:
229Th → 225Ra + 4He
E. Neutron capture by 200Hg:
1. Identify the initial element: 200Hg
2. Determine the product after neutron capture: 201Hg (one additional neutron)
3. Write the balanced equation:
200Hg + 1n → 201Hg
Please note that the mass and atomic numbers listed above are indicative and might not directly correspond to the actual values for these isotopes.
To write balanced equations for these nuclear reactions, you need to ensure that the total number of protons and neutrons on both sides of the equation remains the same. Here's how you can write balanced equations for each of the reactions:
A. Neutron emission by 88Br:
In this reaction, a neutron is emitted by 88Br, resulting in the formation of a new atom. Since neutrons have no charge (neutral), the element's atomic number will remain the same, but its atomic mass will decrease by 1.
The balanced equation can be written as:
^88Br -> ^87Br + ^0n
B. Electron absorption by 116Sb:
In this reaction, an electron is absorbed by 116Sb, resulting in the formation of a new atom with an increased atomic number. The negative charge of the electron is compensated by converting a proton into a neutron.
The balanced equation can be written as:
^116Sb + ^-1e -> ^116Te
C. Positron emission by 184Hg:
In this reaction, a positron (a positively charged electron) is emitted by 184Hg, resulting in the formation of a new atom with a decreased atomic number. The positive charge of the positron is compensated by converting a neutron into a proton.
The balanced equation can be written as:
^184Hg -> ^184Au + ^0e
D. Alpha emission by 229Th:
In this reaction, an alpha particle (2 protons and 2 neutrons) is emitted by 229Th, resulting in the formation of a new atom with an atomic number decreased by 2 and an atomic mass decreased by 4.
The balanced equation can be written as:
^229Th -> ^225Ra + ^4He
E. Neutron capture by 200Hg:
In this reaction, a neutron is captured by 200Hg, resulting in the formation of a new atom with an increased atomic mass but the same atomic number.
The balanced equation can be written as:
^200Hg + ^1n -> ^201Hg