<span>During neutralization reaction
Milliequivalent of acid=milliequivalent of base.
Milliequivalent of HCl=0.100*12*1
Milliequivalent of NaOH=M*10*1
So 0.100*12*1=M*10*1
Solve for M.</span>
Average atomic mass of an element is the sum of atomic mass of its isotopes multiplied by their respective percentage abundance.
There are four isotopes of element X:
1. 4.350 % with mass 49.94605 amu
2. 83.79% with mass 51.94051 amu
3. 9.5% with mass 52.94065 amu
4. 2.360% with mass 53.93888 amu
Thus, average atomic mass will be:

Or,

Therefore, average atomic mass of element X will be 51.99 amu (four significant figures)
Given
Mass of K2O2 = 1.0 g
Molar mass of K2O2 = 110.0 g/mol
Calculate the # moles of K2O2
# moles = mass of K2O2/molar mass
= 1.0 g/110.0 g.mol-1 = 0.0091 moles
1 mole of K2O2 contains 2*6.023*10^23 atoms of oxygen
Therefore, 0.0091 moles of K2O2 will correspond to:
= 0.0091 moles * 2* 6.023*10^23 atoms/1 mole
= 1.096 * 10^24 atoms of oxygen