identify the following element if it has three naturally occurring isotopes 50% of which have an atomic mass of 127 amu, 40% at 128 amu, and 10% at 129 amu
The element with three naturally occurring isotopes, where 50% have an atomic mass of 127 amu, 40% have an atomic mass of 128 amu, and 10% have an atomic mass of 129 amu is iodine (I).
The element with three naturally occurring isotopes (50% at 127 amu, 40% at 128 amu, and 10% at 129 amu) is Tellurium (Te).
To identify the element with three naturally occurring isotopes, we can start by calculating the weighted average atomic mass using the information provided.
Step 1: Multiply the atomic mass of each isotope by its percentage abundance:
Isotope 1: 127 amu * 50% = 63.5
Isotope 2: 128 amu * 40% = 51.2
Isotope 3: 129 amu * 10% = 12.9
Step 2: Add up the results from step 1:
63.5 + 51.2 + 12.9 = 127.6 amu
Step 3: Round the result to the nearest whole number:
Rounding 127.6 to the nearest whole number gives 128.
Therefore, the element with three naturally occurring isotopes, where 50% have an atomic mass of 127 amu, 40% have an atomic mass of 128 amu, and 10% have an atomic mass of 129 amu, is element 128 on the periodic table.