# chemistry

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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

• chemistry -

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.

• chemistry -

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)

• chemistry -

I did one for you. Now that you see how to do, try your hand at the others.

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