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November 29, 2014

November 29, 2014

Posted by **Pernell** on Wednesday, May 25, 2011 at 1:45pm.

a. 239/92U - ? + o/-1e

b. 11/5B - 7/3Li + ?

thank you!

- Chemistry -
**DrBob222**, Wednesday, May 25, 2011 at 4:46pmNo arrow makes the equations difficult to read. I hope I interpreted the question correctly.

_{92}U^{239}==> X +_{-1}e^{0}

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_{93}Np^{239}

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.

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