use the periodic table to determine what element cesium-135 () becomes after beta decay.

D.

Explanation:

Your unbalanced nuclear equation is

It is convenient to replace the question mark by an atomic symbol, , where x = the atomic number, y = the mass number, and Z = the symbol of the element .

Then your equation becomes

The main point to remember in balancing nuclear equations is that the sums of the superscripts and the subscripts must be the same on each side of the equation.  

Then

55 = x - 1, so x = 55 + 1 = 56

135 = y + 0, so y = 135

Element 56 is barium, so the nuclear equation becomes

This question is much trickier than it looks. There are two possible answers: D and E and both can happen in nature. The only thing you know for certain is that you can only create something that is one step on either side of the given atom. It's atomic number can go up one or down one, but it can only move 1 step in either direction.

If the cesium gives up an electron in from nucleus, then a proton is left behind and the Cs goes from 55 to 56. That means it becomes a Barium atom.

If the Cesium gives up a positron from the nucleus, then the atomic number goes down one and the atomic number goes down 1 from Cs to Xe. I cannot refine this any further without a hint about what kind of reaction took place.

All I can suggest is that you look at your notes to see what kind of example you were taught. When I was teaching this material we gave out only 1 example and that was the first one. It would be easier to detect the emission of an electron and the process has a longer time frame. Things may have changed since then.

So the olden days answer woud have been it becomes Ba and that's D.

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.