.50 M KCl because 5% is the same as .05, which makes the .50M more concentrated.
The 2nd one I think but I need some points
The number of atoms present, on average, will be the natural abundance of the isotope times the number of atoms in the sample => number of C-13 atoms = C-13 abundance * number of atoms in the sample = 1.07% * 30,000 = 1.07 * 30,000 / 100 = 321 atoms.<span> Answer: 321 atoms.</span>
