Solved ISOTOPES AND NUCLEAR CHEMISTRY WORKSHEET Chemistry 2A ... - Free Printable

Here is the step-by-step solution to the problems on your worksheet.

Part 1: Writing Atomic Symbols

To write atomic symbols, we need three numbers based on the key in the top right corner:
* Z (Atomic Number): This is equal to the number of protons. It goes in the bottom left box. You can find the element symbol by looking up this number on a periodic table.
* A (Mass Number): This is the total mass. It equals Protons + Neutrons. It goes in the top left box.
* Charge: This goes in the top right box. Positive (+) for cations, negative (-) for anions.

Problem A: Two isotopes of gallium
* Isotope 1: Gallium always has 31 protons (Atomic Number = 31).
* Given: 37 neutrons.
* Mass Number ($A$) = $31 \text{ (protons)} + 37 \text{ (neutrons)} = 68$.
* Symbol: $^{68}_{31}\text{Ga}$
* Isotope 2: Still Gallium (31 protons).
* Given: 39 neutrons.
* Mass Number ($A$) = $31 \text{ (protons)} + 39 \text{ (neutrons)} = 70$.
* Symbol: $^{70}_{31}\text{Ga}$

Problem B: A cation with charge +3, 13 protons, 14 neutrons
* Protons (Z): 13. Looking at the periodic table, element 13 is Aluminum (Al).
* Mass Number (A): $13 \text{ (protons)} + 14 \text{ (neutrons)} = 27$.
* Charge: +3.
* Symbol: $^{27}_{13}\text{Al}^{+3}$

Problem C: An anion with charge -1, 53 protons, 74 neutrons
* Protons (Z): 53. Looking at the periodic table, element 53 is Iodine (I).
* Mass Number (A): $53 \text{ (protons)} + 74 \text{ (neutrons)} = 127$.
* Charge: -1.
* Symbol: $^{127}_{53}\text{I}^{-1}$

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Part 2: Completing the Table

Let's break down each row using the rules:
* $\text{Mass Number} = \text{Protons} + \text{Neutrons}$
* $\text{Charge} = \text{Protons} - \text{Electrons}$

Row 1: Symbol is $^{22}_{11}\text{Na}^+$
* Protons: The bottom number is 11.
* Electrons: Charge is +1. So, $11 - \text{electrons} = 1$. Electrons = 10.
* Neutrons: Top number (22) - Bottom number (11) = 11.
* Mass Number: 22.
* Charge: +1.

Row 2: Protons = 8, Electrons = 10, Mass = 16
* Element: Atomic number 8 is Oxygen (O).
* Neutrons: $16 (\text{Mass}) - 8 (\text{Protons}) = 8$.
* Charge: $8 (\text{Protons}) - 10 (\text{Electrons}) = -2$.
* Symbol: $^{16}_{8}\text{O}^{-2}$

Row 3: Electrons = 10, Neutrons = 11, Charge = +2
* Protons: Since the charge is +2 and there are 10 electrons, the atom must have 2 more protons than electrons. Protons = $10 + 2 = 12$.
* Element: Atomic number 12 is Magnesium (Mg).
* Mass Number: $12 (\text{Protons}) + 11 (\text{Neutrons}) = 23$.
* Symbol: $^{23}_{12}\text{Mg}^{+2}$

Row 4: Protons = 6, Neutrons = 7, Charge = 0
* Element: Atomic number 6 is Carbon (C).
* Electrons: Charge is 0, so electrons = protons = 6.
* Mass Number: $6 + 7 = 13$.
* Symbol: $^{13}_{6}\text{C}$

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Part 3: Average Atomic Mass Calculation

To find the average atomic mass, we multiply the mass of each isotope by its percentage abundance (converted to a decimal), then add the results together.

Data:
* Isotope 1 ($^{10}\text{B}$): Mass = 10.0129, Abundance = 19.9% (0.199)
* Isotope 2 ($^{11}\text{B}$): Mass = 11.009, Abundance = 80.1% (0.801)

Calculation:
1. Contribution from Boron-10: $10.0129 \times 0.199 = 1.9925671$
2. Contribution from Boron-11: $11.009 \times 0.801 = 8.818209$
3. Total Average Mass: $1.9925671 + 8.818209 = 10.8107761$

Rounding to two decimal places (standard for atomic masses), we get 10.81 amu.

Final Answer:
1. Symbols: $^{68}_{31}\text{Ga}$, $^{70}_{31}\text{Ga}$, $^{27}_{13}\text{Al}^{+3}$, $^{127}_{53}\text{I}^{-1}$
2. Table:
* Row 1: 11, 10, 11, 22, +1
* Row 2: $^{16}_{8}\text{O}^{-2}$, -, 8, -, -2
* Row 3: $^{23}_{12}\text{Mg}^{+2}$, 12, -, 23, -
* Row 4: $^{13}_{6}\text{C}$, -, 6, 13, -
3. Average Atomic Mass: 10.81 amu