isotopes and relative atomic mass challenge practice direction: complete the table by filling in the missing…

isotopes and relative atomic mass challenge practice direction: complete the table by filling in the missing information in the tables below. direction: solve each question and show all your steps in the space provided. 1. given the percentage abundance of (_{10}^{20}ne) is 90.48%, (_{10}^{21}ne) is 0.27%, and (_{10}^{22}ne) is 9.25%, calculate the relative atomic mass of neon. relative atomic mass : ________ 2. antimony has two isotopes, (_{51}^{121}sb) and (_{51}^{123}sb), with an average atomic mass of 121.76 amu. determine the percentage abundances of both isotopes.

isotopes and relative atomic mass challenge practice direction: complete the table by filling in the missing information in the tables below. direction: solve each question and show all your steps in the space provided. 1. given the percentage abundance of (_{10}^{20}ne) is 90.48%, (_{10}^{21}ne) is 0.27%, and (_{10}^{22}ne) is 9.25%, calculate the relative atomic mass of neon. relative atomic mass : ________ 2. antimony has two isotopes, (_{51}^{121}sb) and (_{51}^{123}sb), with an average atomic mass of 121.76 amu. determine the percentage abundances of both isotopes.

Answer

Explanation:

Step1: Recall relative - atomic - mass formula

The relative atomic mass ($A_r$) of an element with isotopes is calculated using the formula $A_r=\sum_{i = 1}^{n}(m_i\times x_i)$, where $m_i$ is the mass number of the $i$-th isotope and $x_i$ is the percentage abundance of the $i$-th isotope (expressed as a decimal).

Question 1

Step1: Convert percentage abundances to decimals

For $^{20}{10}Ne$: $x_1 = 90.48%=0.9048$, for $^{21}{10}Ne$: $x_2 = 0.27% = 0.0027$, for $^{22}_{10}Ne$: $x_3=9.25% = 0.0925$. The mass numbers of the isotopes are $m_1 = 20$, $m_2=21$, $m_3 = 22$.

Step2: Calculate the relative atomic mass

$A_r=(20\times0.9048)+(21\times0.0027)+(22\times0.0925)$ $A_r = 18.096+0.0567 + 2.035$ $A_r=20.1877\approx20.19$

Question 2

Let the percentage abundance of $^{121}Sb$ be $x$, then the percentage abundance of $^{123}Sb$ is $(1 - x)$.

Step1: Set up the equation

The average atomic mass formula gives us $121.76=121x + 123(1 - x)$.

Step2: Expand the equation

$121.76=121x+123 - 123x$.

Step3: Combine like - terms

$121.76=123 - 2x$.

Step4: Solve for $x$

$2x=123 - 121.76$ $2x = 1.24$ $x=\frac{1.24}{2}=0.62$

The percentage abundance of $^{121}Sb$ is $62%$ and the percentage abundance of $^{123}Sb$ is $1 - 0.62 = 0.38$ or $38%$.

Answer:

  1. $20.19$
  2. Percentage abundance of $^{121}Sb$: $62%$, Percentage abundance of $^{123}Sb$: $38%$