Question

1. what subatomic particles are in a zirconium (iv) ion if it has a mass number of 105?
2. what is the isotope name for $_{13}^{23}al$?
3. what subatomic particles are in an uncharged isotope with atomic number 19 and 40 nucleons?
4. what is the element symbol for magnesium?
5. what is the name of the element with atomic #55?
6. what is the atomic number for germanium?

Explanation:

Step1: Determine zirconium's atomic number

Zirconium (Zr) has an atomic number of 40. In a $Zr^{4 +}$ ion, the number of protons remains the same as the atomic number, so # $p^{+}$ = 40. Mass number (A)=105, and since $A = p^{+}+n^{0}$, then $n^{0}=A - p^{+}=105 - 40 = 65$. A $Zr^{4+}$ ion has lost 4 electrons, so # $e^{-}=40 - 4=36$.

Step2: Identify isotope name for $_{13}^{27}Al$

The element is aluminum (Al), and the mass - number is 27, so it is called aluminum - 27.

Step3: Analyze uncharged isotope

Given atomic number (Z) = 19, so # $p^{+}=19$. Since the isotope is uncharged, # $e^{-}=19$. Nucleons are protons + neutrons, and given nucleons (mass number A) = 40, then $n^{0}=A - p^{+}=40 - 19 = 21$.

Step4: Recall magnesium symbol

The element symbol for magnesium is Mg from the periodic table.

Step5: Find element with atomic number 55

Looking at the periodic table, the element with atomic number 55 is cesium.

Step6: Determine germanium's atomic number

From the periodic table, the atomic number of germanium (Ge) is 32.

Answer:

1.

• # $p^{+}$: 40
• # $n^{0}$: 65
• # $e^{-}$: 36

1. Aluminum - 27

3.

• # $p^{+}$: 19
• # $n^{0}$: 21
• # $e^{-}$: 19

1. Mg
2. Cesium
3. 32