Question

part c
what is the identity of element x from part b?
express your answer as a chemical symbol.

part d
the two isotopes of chlorine are $\ce{_{17}^{35}cl}$ and $\ce{_{17}^{37}cl}$. which isotope is the most abundant?
view available hint(s)
$\circ$ $\ce{_{17}^{35}cl}$
$\circ$ $\ce{_{17}^{37}cl}$

Explanation:

Part C

To determine the identity of element \( X \), we typically use the atomic number (the subscript in the isotope notation \( _Z^A X \), where \( Z \) is the atomic number and \( A \) is the mass number). The atomic number determines the element's identity because it equals the number of protons in the nucleus. For example, if in Part B the isotope of \( X \) had an atomic number (subscript) of 17, the element with atomic number 17 is chlorine, with the chemical symbol \( \text{Cl} \).

Part D

The average atomic mass of chlorine is approximately 35.45 amu. The average atomic mass is a weighted average of the masses of its isotopes (\( ^{35}\text{Cl} \) with mass ~35 amu and \( ^{37}\text{Cl} \) with mass ~37 amu). Since 35.45 is closer to 35 than to 37, the isotope \( ^{35}\text{Cl} \) must be more abundant (because the weighted average is pulled closer to the mass of the more abundant isotope).

Part C Answer:

\( \text{Cl} \) (assuming the atomic number from Part B was 17; if Part B had a different atomic number, adjust accordingly, but typically for chlorine-related problems, this is the case)

Part D Answer:

A. \( \ce{^{35}_{17}Cl} \)

Answer:

Part C

To determine the identity of element \( X \), we typically use the atomic number (the subscript in the isotope notation \( _Z^A X \), where \( Z \) is the atomic number and \( A \) is the mass number). The atomic number determines the element's identity because it equals the number of protons in the nucleus. For example, if in Part B the isotope of \( X \) had an atomic number (subscript) of 17, the element with atomic number 17 is chlorine, with the chemical symbol \( \text{Cl} \).

Part D

The average atomic mass of chlorine is approximately 35.45 amu. The average atomic mass is a weighted average of the masses of its isotopes (\( ^{35}\text{Cl} \) with mass ~35 amu and \( ^{37}\text{Cl} \) with mass ~37 amu). Since 35.45 is closer to 35 than to 37, the isotope \( ^{35}\text{Cl} \) must be more abundant (because the weighted average is pulled closer to the mass of the more abundant isotope).

Part C Answer:

\( \text{Cl} \) (assuming the atomic number from Part B was 17; if Part B had a different atomic number, adjust accordingly, but typically for chlorine-related problems, this is the case)

Part D Answer:

A. \( \ce{^{35}_{17}Cl} \)