239U+1n=?+140Xe and 8Be+4He=?+y?

Question

please complete the following reaction? thanks

Answer 1

239U(92) + 1n(0) ===> 240U(92) ===> 140Xe(54) + 100Sr(38)

8Be(4) + 4He(2) ===> 12C(6)

In both cases, the numbers of neutrons and protons (in parentheses) are conserved.

Answer 2

Start by counting the protons and neutrons on each side of the reactions. I'll solve the first one to show you how it's done:

239U + 1n = ? + 140Xe

Uranium has 92 protons (use your periodic table. remember that the atomic number of an element is equal to the number of protons on that element's atom's nucleus), so 239U has 239 - 92 = 147 neutrons. In total, our reactives have 92 protons and 148 neutrons.
Xenon has 54 protons, so 140Xe has 140 - 54 = 86 neutrons. We had 92 protons on our reactives, and Xe accounts for 54 of them, meaning the remaining product must account for the other 92 - 54 = 38 protons, which makes it a strontium atom. We also had 148 neutrons, 86 of which are on the Xe atom. The 148 - 86 = 62 neutrons left must be on the Sr atom, which now has a total of 38 + 62 = 100 nucleons, making the final result:

239U + 1n = 100Sr + 140Xe