6. calculate the atomic mass of the unknown element. then identify the element.\nisotope mass (amu) percent…

6. calculate the atomic mass of the unknown element. then identify the element.\nisotope mass (amu) percent abundance operations element:\n¹⁸⁴x 184.953 37.40%\n¹⁸⁶x 186.956 62.60%\n7. calculate the atomic mass of the unknown element. then identify the element.\nisotope mass (amu) percent abundance operations element:\n¹¹²x 112.904 4.30%\n¹¹⁴x 114.904 95.70%\nshow 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.\nthe relative abundances and atomic masses are 69.2% for a mass of 62.93 amu and 30.8% for a mass of 64.93 amu.

6. calculate the atomic mass of the unknown element. then identify the element.\nisotope mass (amu) percent abundance operations element:\n¹⁸⁴x 184.953 37.40%\n¹⁸⁶x 186.956 62.60%\n7. calculate the atomic mass of the unknown element. then identify the element.\nisotope mass (amu) percent abundance operations element:\n¹¹²x 112.904 4.30%\n¹¹⁴x 114.904 95.70%\nshow 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.\nthe relative abundances and atomic masses are 69.2% for a mass of 62.93 amu and 30.8% for a mass of 64.93 amu.

Answer

Explanation:

Step1: Recall atomic mass formula

The average atomic mass ($A_{avg}$) of an element with isotopes is calculated using the formula $A_{avg}=\sum_{i = 1}^{n}(m_i\times p_i)$, where $m_i$ is the mass of the $i$-th isotope and $p_i$ is the percent - abundance of the $i$-th isotope (expressed as a decimal).

Step2: Calculate atomic mass for problem 6

For the first isotope of problem 6 with mass $m_1 = 184.953$ amu and percent - abundance $p_1=0.3740$, and the second isotope with mass $m_2 = 186.956$ amu and percent - abundance $p_2 = 0.6260$. $A_{avg}=(184.953\times0.3740)+(186.956\times0.6260)$ $A_{avg}=184.953\times0.3740 = 69.172422$ $A_{avg}=186.956\times0.6260=117.034456$ $A_{avg}=69.172422 + 117.034456=186.206878\approx186.21$ amu. Looking up this atomic mass in the periodic table, the element is Rhenium (Re).

Step3: Calculate atomic mass for problem 7

For the first isotope of problem 7 with mass $m_1 = 112.904$ amu and percent - abundance $p_1 = 0.0430$, and the second isotope with mass $m_2=114.904$ amu and percent - abundance $p_2 = 0.9570$. $A_{avg}=(112.904\times0.0430)+(114.904\times0.9570)$ $A_{avg}=112.904\times0.0430 = 4.854872$ $A_{avg}=114.904\times0.9570 = 110.953128$ $A_{avg}=4.854872+110.953128 = 115.808\approx115.81$ amu. Looking up this atomic mass in the periodic table, the element is Indium (In).

Answer:

Problem 6: Atomic mass is approximately 186.21 amu, element is Rhenium (Re). Problem 7: Atomic mass is approximately 115.81 amu, element is Indium (In).