show all calculation setups. adjust for significant figures and give units.\n- protium is the most prevalent…

show all calculation setups. adjust for significant figures and give units.\n- protium is the most prevalent hydrogen isotope, with an abundance of 99.98%. it consists of one proton and one electron.\n- deuterium is a hydrogen isotope consisting of one proton, one neutron and one electron.\n- tritium is a hydrogen isotope consisting of one proton, two neutrons and one electron.\n1. the term \average atomic mass\ is a ______________ average and so is calculated differently from a \normal\ average.\n2. the element copper has naturally - occurred isotopes with mass numbers of 63 and 65. the relative abundance and atomic masses are 69.2% for a mass of 62.93amu and 30.8% for a mass of 64.93amu. calculate the average atomic mass of copper.

show all calculation setups. adjust for significant figures and give units.\n- protium is the most prevalent hydrogen isotope, with an abundance of 99.98%. it consists of one proton and one electron.\n- deuterium is a hydrogen isotope consisting of one proton, one neutron and one electron.\n- tritium is a hydrogen isotope consisting of one proton, two neutrons and one electron.\n1. the term \average atomic mass\ is a ______________ average and so is calculated differently from a \normal\ average.\n2. the element copper has naturally - occurred isotopes with mass numbers of 63 and 65. the relative abundance and atomic masses are 69.2% for a mass of 62.93amu and 30.8% for a mass of 64.93amu. calculate the average atomic mass of copper.

Answer

Explanation:

Step1: Recall the formula for average atomic mass

The formula for average atomic mass ($A_{avg}$) of an element with isotopes is $A_{avg}=\sum_{i}(x_i\times A_i)$, where $x_i$ is the relative - abundance (as a decimal) of the $i$-th isotope and $A_i$ is the atomic mass of the $i$-th isotope.

Step2: Convert the relative - abundances to decimals

For copper, the relative - abundance of the first isotope $x_1 = 0.692$ and its atomic mass $A_1=62.93$ amu, and the relative - abundance of the second isotope $x_2 = 0.308$ and its atomic mass $A_2 = 64.93$ amu.

Step3: Calculate the average atomic mass

$A_{avg}=x_1\times A_1 + x_2\times A_2=(0.692\times62.93)+(0.308\times64.93)$ $=0.692\times62.93+0.308\times64.93$ $=43.54756+20.09844$ $=63.646$ amu

Answer:

$63.65$ amu (adjusted to four significant figures)