Question

1. practice: turn off show isotope notation. use what you have learned to fill in the table. when you are finished, check your answers in the gizmo.

protons neutrons mass number isotope notation symbol
17 35 cl
1 1 h i
30 56 fe
95 148 am
31 14 si
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Explanation:

Step1: Recall neutron - proton - mass number relationship

The mass number ($A$) is the sum of the number of protons ($p$) and neutrons ($n$), i.e., $A=p + n$.

Step2: First row calculation

Given $p = 17$ and $A=35$. Then $n=A - p=35 - 17 = 18$. The isotope notation for an element $X$ with atomic number $Z$ (equal to the number of protons) and mass number $A$ is $_{Z}^{A}X$. For chlorine ($Cl$) with $Z = 17$ and $A = 35$, the isotope notation is $_{17}^{35}Cl$.

Step3: Second row analysis

For the isotope $_{1}^{1}H$, the number of protons $p=1$, the number of neutrons $n=1 - 1=0$ (since $A = 1$ and $Z = 1$), and the symbol is $H$.

Step4: Third row calculation

Given $n = 30$ and $A = 56$ for iron ($Fe$). Then $p=A - n=56 - 30 = 26$. The isotope notation is $_{26}^{56}Fe$.

Step5: Fourth row analysis

For the isotope $_{14}^{31}Si$, the number of protons $p = 14$, the number of neutrons $n=31 - 14 = 17$, and the symbol is $Si$.

Step6: Fifth row calculation

Given $p = 95$ and $n = 148$. The mass number $A=p + n=95+148 = 243$. The isotope notation for americium ($Am$) with $Z = 95$ and $A = 243$ is $_{95}^{243}Am$.

Answer:

Protons	Neutrons	Mass number	Isotope notation	Symbol
1	0	1	$_{1}^{1}H$	$H$
26	30	56	$_{26}^{56}Fe$	$Fe$
14	17	31	$_{14}^{31}Si$	$Si$
95	148	243	$_{95}^{243}Am$	$Am$