Question

1. which species is represented by the following information?

p^+ = 12 n^0 = 12 e^- = 10
a. ne^+2
b. mg^+2
c. mg^-2
d. ne^-2

1. how many protons are in ^{32}_{15}p?
2. calculate the average atomic mass of the following isotope,
3. circle the ions which have the same number of valence electrons? (for each atom, find the original number of valence electrons, then add or subtract the ion number to find the total number of valence elections in the ion)

a. li^+1
b. h^+1
c. mg^+2
d. mg^+1

1. calculate the average atomic mass of the following ion:

a. 31.548 amu
b. 93 amu
c. 31.9 amu
d. 30.962 amu

Explanation:

Step1: Determine the element

The number of protons (\(p^+\)) is 12. Looking at the periodic table, magnesium (\(Mg\)) has an atomic number of 12.

Step2: Calculate the charge

The charge of an ion is calculated as \(p^+ - e^-\). Given \(p^+ = 12\) and \(e^- = 10\), the charge is \(12 - 10=+ 2\).

The notation \(\ce{^{32}_{16}P}\) is incorrect. If we assume it's \(\ce{^{32}_{15}P}\) (phosphorus), the subscript in the nuclear notation represents the atomic number. The atomic number is equal to the number of protons.

Step1: Use the formula for average atomic mass

The formula for average atomic mass (\(M\)) is \(M=\sum (m_i\times a_i)\), where \(m_i\) is the mass of the isotope and \(a_i\) is the percent abundance (in decimal form).
For \(X - 26\) (\(m_1 = 26\), \(a_1=0.7\)), for \(X - 27\) (\(m_2 = 27\), \(a_2 = 0.15\)), for \(X - 28\) (\(m_3=28\), \(a_3 = 0.15\))

Step2: Calculate each contribution

\(m_1\times a_1=26\times0.7 = 18.2\)
\(m_2\times a_2=27\times0.15=4.05\)
\(m_3\times a_3=28\times0.15 = 4.2\)

Step3: Sum the contributions

\(M=18.2 + 4.05+4.2=26.45\)

Step1: Determine valence electrons for neutral atoms

• For \(Li\) (atomic number 3), electron configuration is \(1s^22s^1\), so neutral \(Li\) has 1 valence electron.
• For \(H\) (atomic number 1), electron configuration is \(1s^1\), so neutral \(H\) has 1 valence electron.
• For \(Mg\) (atomic number 12), electron configuration is \(1s^22s^22p^63s^2\), so neutral \(Mg\) has 2 valence electrons.

Step2: Calculate valence electrons in ions

• For \(Li^{+1}\): \(1 - 1=0\)
• For \(H^{+1}\): \(1 - 1=0\)
• For \(Mg^{+2}\): \(2- 2=0\)
• For \(Mg^{+1}\): \(2-1 = 1\)

Answer:

b. \(Mg^{+2}\)