Complete each of the following nuclear equations:
a. 239/92U - ? + o/-1e
b. 11/5B - 7/3Li + ?
thank you!
No arrow makes the equations difficult to read. I hope I interpreted the question correctly.
92U239 ==> X + -1e0
The subscripts must add up on both sides and the superscripts must add up on both sides. On the left, we have 92 protons (+ charge) for U and on the right we have -1; therefore, X must be an element with 93 [because +93 +(-1) = +92)]. For the superscripts, we have 239 on the left and 0 on the right; therefore, the element X must have a mass number of 239 (239+0=239). What is element X? Look on the periodic chart for element number 93. That is Np. Therefore, we write that as
93Np239
I know the mass number is supposed to go in the upper left corner but since it is difficult to write both a subscript and a superscript in the same spot with a computer, I placed the atomic number as a subscript in the correct place and used the upper right corner for the mass number since we don't need to worry about the charge on the U or Np.
To complete nuclear equations, it's important to consider the conservation of both mass number (A) and atomic number (Z). The mass number refers to the total number of protons and neutrons in the nucleus, while the atomic number represents the number of protons.
a.
Given: 239/92U - ? + o/-1e
The mass number of the uranium-239 nucleus (A) is 239, and the atomic number (Z) is 92. The subscript represents the atomic number, while the superscript indicates the mass number.
Since there is a negative charge (-1e) involved, it means an electron is produced. Electrons can be represented as o/-1e.
To find the missing element, we need to calculate the sum of the mass numbers and atomic numbers on each side of the equation:
On the left side: mass number (239) + atomic number (92) = 331
On the right side: ?? + 0 + 1
To balance the equation, the sum of mass numbers and atomic numbers should be the same on both sides. Therefore, the missing element on the right side will have:
Mass number: 331 - 1 = 330
Atomic number: 330 - 0 = 330
Hence, the missing element is 330/130. This element can be represented as 130/330X.
Therefore, the completed nuclear equation is:
239/92U → 330/130X + o/-1e
b.
Given: 11/5B - 7/3Li + ?
Similarly, let's consider the conservation of mass number (A) and atomic number (Z).
On the left side: mass number (11) + atomic number (5) = 16
On the right side: mass number of lithium (7) + atomic number of lithium (3) + mass number of missing element + atomic number of missing element
To balance the equation, we need to find the missing element that results in the sum of mass numbers and atomic numbers equal on both sides:
On the left side: 16
On the right side: 7 + 3 + ?? + ??
To make them equal, the missing element should have:
Mass number: 16 - 7 - 3 = 6
Atomic number: 6 - 3 = 3
Hence, the missing element is 6/3C. This element can be represented as 3/6C.
Therefore, the completed nuclear equation is:
11/5B → 7/3Li + 3/6C