Ka is known as the acid dissociation constant. It is a measure of an acid strength of a solution. It is an equilibrium constant for the dissociation of an acid or base in a solvent. A very large value of Ka represents a strong acid. For instance, for HI, Ka is 3.2 x 10^9 and for HBr, Ka is 1.0 *10^9.
1. An atom has an equel number of protons and electrons. Since protons are positive and electrons are negative, they are balanced out and there is no charge.
Answer:
![\boxed{_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} + \,_{2}^{4}\text{He}}](https://tex.z-dn.net/?f=%5Cboxed%7B_%7B84%7D%5E%7B210%7D%5Ctext%7BPo%7D%20%5Clongrightarrow%20%5C%2C%20_%7B82%7D%5E%7B206%7D%5Ctext%7BPb%7D%20%2B%20%5C%2C_%7B2%7D%5E%7B4%7D%5Ctext%7BHe%7D%7D)
Explanation:
The unbalanced nuclear equation is
![_{84}^{210}\text{Po} \longrightarrow \, ? + \,_{2}^{4}\text{He}](https://tex.z-dn.net/?f=_%7B84%7D%5E%7B210%7D%5Ctext%7BPo%7D%20%5Clongrightarrow%20%5C%2C%20%3F%20%2B%20%5C%2C_%7B2%7D%5E%7B4%7D%5Ctext%7BHe%7D)
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
![_{84}^{210}\text{Po} \longrightarrow \, _{x}^{y}\text{Z} + \,_{2}^{4}\text{He}](https://tex.z-dn.net/?f=_%7B84%7D%5E%7B210%7D%5Ctext%7BPo%7D%20%5Clongrightarrow%20%5C%2C%20_%7Bx%7D%5E%7By%7D%5Ctext%7BZ%7D%20%2B%20%5C%2C_%7B2%7D%5E%7B4%7D%5Ctext%7BHe%7D)
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
84 = x + 2, so x = 84 - 2 = 82
210 = y + 4, so y = 206
Element 82 is lead, so the nuclear equation becomes
![\boxed{_{84}^{210}\text{Po} \longrightarrow \, _{82}^{206}\text{Pb} + \,_{2}^{4}\text{He}}](https://tex.z-dn.net/?f=%5Cboxed%7B_%7B84%7D%5E%7B210%7D%5Ctext%7BPo%7D%20%5Clongrightarrow%20%5C%2C%20_%7B82%7D%5E%7B206%7D%5Ctext%7BPb%7D%20%2B%20%5C%2C_%7B2%7D%5E%7B4%7D%5Ctext%7BHe%7D%7D)