fill in the missing symnbols in the following nuclear bombardment reactions
a. 96 4 100
42Mo + 2He = 43 Tc+ ?
b. 59 1 56
27Co+ 0n = 25+ ?
c. 23 23 1
11Na+ ? = 12Mg+ 0n
d. 209 210 1
83Bi+ ? = 84Po+ 0n
e. 238 16 1
92U + 8O= ? + 5 0n
TEACHER DID NOT EXPLAIN THIS AT ALL SO CONFUSED
There is no way you can make sense on this board because it doesn't allow spacing. So here is what you should do and I'll do the first for you.
Place the atomic number just before the symbol, as in 42Mo.
Placd the mass number as an exponent to the right of the symbol, as in
42Mo^96. (I know the 96 goes as a superscript on the left side BUT there is no room on this board for that so we do the best we can. When you translate this to your paper, you can put the number where it belongs.. So here is the way the equation looks.
I'm having trouble reading your post but I think I can do the d.
83Bi^209 + ? ==>84Po^210 + oN^1
So we make the subscripts = and superscripts equal.
83 on the left + ? = 84+0 so the number for ? must be 1. Now it looks like this.
83Bi^209 + 1? ==> 84Po^210 + 0N^1
Now for the top.
On the right we have 210 + 1 = 211. on the left we have 209; therefore, ? must be 2 so it reads this now.
83Bi^209 + 1?^2 ==> 84Po^210 + 0N^1
The only thing left to do is to identify ?. How do we do that? It has an atomic number of 1 so you look at the periodic table and find the the element with atomic number 1 which is H and place an H for ?
83Bi^209 + 1H^2 ==>84Po^210 + 0N^1
Done. By the way, 1H^2 is deuterium an isotope of hydrogen.
a. 42Mo^96 + 2He^4= 43Tc100 + ?
b. 27Co^59 + 0n^1 = 25Mn^56 + ?
c. 11Na^23 + ? = 12Mg^23 + 0n^1
e. 92U^238 + 8O^16 = ? + 5(0n^1)
I did one for you. Now that you see how to do, try your hand at the others.
In order to fill in the missing symbols in the nuclear bombardment reactions, you need to know the principles of nuclear reactions and nuclear notation.
In these reactions, the numbers on the left side represent the atomic mass or mass number, while the numbers on the bottom-left represent the atomic number or proton number. The atomic mass is the sum of protons and neutrons, while the atomic number is the number of protons in the nucleus.
a. 96 4 100
42Mo + 2He = 43 Tc + x
To balance the equation, we need to consider the number of protons and neutrons on each side of the reaction. The sum of the atomic numbers on the left side should be equal to the sum on the right side, and the sum of the atomic masses on the left side should be equal to the sum on the right side.
For this equation, the sum of the atomic numbers is 42 + 2 = 44 on the left side, and 43 + x on the right side. To balance the equation, x should be equal to 44 - 43 = 1.
So the balanced equation is:
42Mo + 2He = 43Tc + 1n
b. 59 1 56
27Co + 0n = 25x + y
Similarly, for this equation, the sum of the atomic numbers is 27 + 0 = 27 on the left side, and x + y on the right side. To balance the equation, x should be equal to 27 and y should be equal to 1.
So the balanced equation is:
27Co + 0n = 25Mn + 1H
c. 23 23 1
11Na + x = 12Mg + 0n
Again, for this equation, the sum of the atomic numbers is 11 + x on the left side, and 12 + 0 = 12 on the right side. To balance the equation, x should be equal to 12 - 11 = 1.
So the balanced equation is:
11Na + 1H = 12Mg + 0n
d. 209 210 1
83Bi + x = 84Po + 0n
In this equation, the sum of atomic numbers is 83 + x on the left side, and 84 + 0 = 84 on the right side. To balance the equation, x should be equal to 84 - 83 = 1.
So the balanced equation is:
83Bi + 1H = 84Po + 0n
e. 238 16 1
92U + 8O = x + 5 0n
In this equation, the sum of atomic numbers is 92 + 8 = 100 on the left side, and x + 0 = x on the right side. To balance the equation, x should be equal to 100 - 5 = 95.
So the balanced equation is:
92U + 8O = 95Am + 5 0n
Remember to always check if the balanced equation satisfies the conservation of mass and charge.