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8. how many electrons, protons, and neutrons are found in each of the f…

Question

  1. how many electrons, protons, and neutrons are found in each of the following? a) $^{24}mg$ b) $^{23}na^{+}$ c) $^{35}cl$ d) $^{35}cl^{-}$ e) $^{56}fe^{3 + }$ f) $^{15}n$ g) $^{16}o^{2 - }$ h) $^{27}al^{3 + }$

Explanation:

Step1: Recall atomic number and mass number concepts

Atomic number ($Z$) = number of protons. Mass number ($A$)= number of protons + number of neutrons. For neutral atoms, number of electrons = number of protons. For ions, adjust electron number based on charge.

Step2: Analyze $^{24}Mg$

Magnesium ($Mg$) has atomic number $Z = 12$. So, number of protons = 12. Mass number $A=24$, number of neutrons $n=A - Z=24 - 12 = 12$. For neutral $Mg$, number of electrons = 12.

Step3: Analyze $^{23}Na^{+}$

Sodium ($Na$) has atomic number $Z = 11$. So, number of protons = 11. Mass number $A = 23$, number of neutrons $n=A - Z=23 - 11 = 12$. Since it is a $+ 1$ ion, number of electrons = $11-1 = 10$.

Step4: Analyze $^{35}Cl$

Chlorine ($Cl$) has atomic number $Z = 17$. So, number of protons = 17. Mass number $A = 35$, number of neutrons $n=A - Z=35 - 17 = 18$. For neutral $Cl$, number of electrons = 17.

Step5: Analyze $^{35}Cl^{-}$

Chlorine has atomic number $Z = 17$, so number of protons = 17. Mass number $A = 35$, number of neutrons $n=A - Z=35 - 17 = 18$. Since it is a $-1$ ion, number of electrons = $17 + 1=18$.

Step6: Analyze $^{56}Fe^{3 + }$

Iron ($Fe$) has atomic number $Z = 26$. So, number of protons = 26. Mass number $A = 56$, number of neutrons $n=A - Z=56 - 26 = 30$. Since it is a $+3$ ion, number of electrons = $26-3 = 23$.

Step7: Analyze $^{15}N$

Nitrogen ($N$) has atomic number $Z = 7$. So, number of protons = 7. Mass number $A = 15$, number of neutrons $n=A - Z=15 - 7 = 8$. For neutral $N$, number of electrons = 7.

Step8: Analyze $^{16}O^{2 - }$

Oxygen ($O$) has atomic number $Z = 8$. So, number of protons = 8. Mass number $A = 16$, number of neutrons $n=A - Z=16 - 8 = 8$. Since it is a $- 2$ ion, number of electrons = $8 + 2=10$.

Step9: Analyze $^{27}Al^{3+}$

Aluminum ($Al$) has atomic number $Z = 13$. So, number of protons = 13. Mass number $A = 27$, number of neutrons $n=A - Z=27 - 13 = 14$. Since it is a $+3$ ion, number of electrons = $13-3 = 10$.

Answer:

a) Electrons: 12, Protons: 12, Neutrons: 12
b) Electrons: 10, Protons: 11, Neutrons: 12
c) Electrons: 17, Protons: 17, Neutrons: 18
d) Electrons: 18, Protons: 17, Neutrons: 18
e) Electrons: 23, Protons: 26, Neutrons: 30
f) Electrons: 7, Protons: 7, Neutrons: 8
g) Electrons: 10, Protons: 8, Neutrons: 8
h) Electrons: 10, Protons: 13, Neutrons: 14