I need help completing nuclear equations:

a. ? + 59/27Co -> 56/25Mn + 4/2He

b. ? -> 14/7N + 0/-1e
(I think this one is 14/6C, but I'm not sure.)

c. 0/-1e + 76/36Kr -> ?

d. 4/2He + 241/95Am -> ? + 2 1/0 n

5 answers

These are easier than they look. The upper numbers must add up on both sides and the lower numbers must add up on both sides.
a. ? + 59/27Co -> 56/25Mn + 4/2He
upper numbers: On the right you have 56 + 4 = 60 so the left must be 59 + ? = 60. ? must be 1.
lower numbers: On the right 25 + 2 = 27 so th left must be 27 + ? = 27. ? must be 0. What element is 27? Must be Co. So what has a mass of 1 and a charge of 0? must be a neutron.
1/0n + 59/27Co -> 56/25Mn + 4/2He Things can't get much easier than that. Your solution to b is correct. I'll leave the others for you. Shout if you have trouble.
Thank you so much for your help!

So would C be 76/37Rb?

I'm still having trouble with d though. 4/2He would make 241/27Co into 237/25 Mn, right? But I'm confused by the 2 1/0 n
I think you're confused by more than the 2 1/0 n. Where did the Co come from? That isn't in the problem. Co is in another problem. Where did the Mn come from? That Am is americium and you've ignored that.
"I'm still having trouble with d though. 4/2He would make 241/27Co into 237/25 Mn, right? But I'm confused by the 2 1/0 n"
Here is what I would do.

4/2He + 241/95Am -> ? + 2 1/0 n .
If that 2 1/0 n is confusing you let's write it this way.
4/2He + 241/95Am -> ? + 1/0 n + 1/0 n
upper numbers: on the left we have 4 + 241 = 245. On the right we have ? + 1 + 1 so ? must be 243.
lower numbers: on the left we have 2 + 95 = 97. On the right we have ? + 0 + 0 so ? must be 97. What's atomic number 97 on the periodic table. That's Bk, Berkelium or 243/97 Bk. Final looks this way.
4/2He + 241/95Am -> 243/97Bk + 2 1/0 n

No, c isn't correct. Here is what you have.
0/-1e + 76/36Kr -> 76/37Rb but you didn''t add correctly. -1 + 36 = 35 and not 37 so it must be Br and not Rb
Ah, thank you so much! I didn't realize I mixed up problems c and d when I was writing the question.

For c, I thought that for beta decay, we would add 1 to the atomic number.. if it's on the left side of the equation, I should subtract 1?
For beta DECAY you DO add one because you're keeping the total on the left = to total on the right as in this equation.
238/92 U ==> 238/93 Np + 0/-1e but when you're ADDING a beta particle to something on the left such as
0/-1e + 76/36Kr ==> 76/35Br you must START with a higher number to account for the negative charge on the electrons. Your rule of thumb is good for DECAY and it make it easy to pronounce quickly what the product is but just remember it's reversed when you reverse the reaction. To be safe always remember the numbers on both sides must add up. That gets it right EVERY time.