Question

327 chemistry prep b cycle 4 homework atoms and isotopes integrity statement: i affirm that this assignment is my own original work. i have not given or received any unauthorized assistance to/from other students, and i have not copied content from internet sources or used ai tools irresponsibly to obtain answers to questions. initials part i: label the subatomic particles of the atom. word bank: proton electron neutron electron cloud nucleus part ii: fill in the chart with the missing information about the given isotopes. name notation notation mass number atomic # protons neutrons electrons boron - 14 5 5 135 55 134 80 part iii: calculate the average atomic mass for the following isotopes. 1) neon has two isotopes: ne - 20 with an abundance of 91% and ne - 22 with an abundance of 9% 2) an unknown element “x” has two isotopes: x - 56 with an abundance of 60% and x - 58 with an abundance of 40%

Explanation:

Part I:

In an atom, the nucleus is at the center. Protons and neutrons are in the nucleus. Electrons are in the electron - cloud outside the nucleus.

Part II:

For Boron - 14:

The atomic number is equal to the number of protons. Given atomic number = 5, so protons = 5. For a neutral atom, number of electrons = number of protons = 5. Mass number = protons+neutrons. So neutrons=Mass number - protons = 14 - 5=9. The notation for Boron - 14 is $^{14}_{5}B$.

For the second row (assuming it's a neutral atom):

If electrons = 55, then protons = 55 (for neutral atom), atomic number = 55. Mass number = 135, neutrons=Mass number - protons = 135 - 55 = 80.

For the third row:

Mass number = 134, neutrons = 80. Protons=Mass number - neutrons = 134 - 80 = 54. For a neutral atom, electrons = protons = 54, atomic number = 54.

Part III:

1) For Neon:

The formula for average atomic mass ($A_{avg}$) is $A_{avg}=\sum_{i} (A_i\times \%_{i})$ where $A_i$ is the mass of isotope $i$ and $\%_{i}$ is the percentage abundance of isotope $i$ in decimal form.
$A_{Ne - 20}=20$, $\%_{Ne - 20}=0.91$, $A_{Ne - 22}=22$, $\%_{Ne - 22}=0.09$.
$A_{avg}=(20\times0.91)+(22\times0.09)=18.2 + 1.98=20.18$.

2) For element X:

$A_{X - 56}=56$, $\%_{X - 56}=0.6$, $A_{X - 58}=58$, $\%_{X - 58}=0.4$.
$A_{avg}=(56\times0.6)+(58\times0.4)=33.6+23.2 = 56.8$.

Answer:

Part I:

(Labeling the atom diagram): Nucleus at the center, Protons and Neutrons in the nucleus, Electron Cloud outside the nucleus, Electrons in the electron - cloud.

Part II:

Name Notation	Notation	Mass Number	Atomic #	Protons	Neutrons	Electrons
(Unknown element)	$^{135}_{55}X$ (assuming X is the element symbol)	135	55	55	80	55
(Unknown element)	$^{134}_{54}X$ (assuming X is the element symbol)	134	54	54	80	54

Part III:

1. Average atomic mass of Neon: 20.18
2. Average atomic mass of element X: 56.8