Multiply the mass by the abundance and add each. Make sure to convert the percentage into a decimal. (49.946 * .043) + (51.941 * .838) + (52.941 * .095) + (53.939 * .024) = 51.99 round up using sig figs and the answer is c. 52.00 amu
None of the alpha particles fired at the foil are being repelled back, like they were in the Rutherford atom simulation.I hope this correct.