Question

1. why are noble metal atoms so inactive chemically?
2. can you change the atomic number of an atom and still have the same atom or does it become an isotope?
3. in an isotope, why is the \abundance\ of an atom important when calculating the atomic mass of an atom on the periodic table of elements?
4. calculate the isotope abundance of lithium 6.015122 that is 7.59% and li 7.016004 that is 92.4%.
5. what part of an atom makes up the majority of its atomic mass?
6. which element has more ionization energy b or ba?

a. boron b. barium

1. which number is used to organize the periodic table of elements?

a. atomic mass b. atomic number

1. which temperature scale is important in the calculations?

a. celsius b. fahrenheit c. kelvin

1. in the flame test lab, what was absorbing energy and releasing energy?

a. protons b. neutrons c. electrons

1. in the american system of measurement a gallon is the standard of measurement at a gas station, what unit is used in the metric system?

a. liter b. meter c. second d. gram

1. in american cars the speedometer measures speed as miles/hr. how is it measured in the metric system?
2. what do coefficient numbers represent in an equation for chemical reactions?
3. what is the significance of a subscript in an equation?
4. what is the difference between a covalent bond and an ionic bond in a compound?

Explanation:

Question 18 Solution (Chemistry, a subfield of Natural Science)

Step 1: Recall the formula for atomic mass

The atomic mass ($A$) of an element with isotopes is calculated using the formula:
$$A = (m_1 \times f_1) + (m_2 \times f_2) + \dots$$
where $m_i$ is the mass of isotope $i$, and $f_i$ is the fractional abundance of isotope $i$ (converted from percentage by dividing by 100).

Step 2: Convert percentages to fractions

For Lithium - 6 (mass $m_1 = 6.015122$):
Fractional abundance $f_1 = \frac{7.59}{100} = 0.0759$

For Lithium - 7 (mass $m_2 = 7.016004$):
Fractional abundance $f_2 = \frac{92.41}{100} = 0.9241$ (Note: The problem states 92.4%, but $7.59 + 92.41 = 100$, so we use 92.41% for accuracy.)

Step 3: Calculate the atomic mass

Substitute into the formula:
$$A = (6.015122 \times 0.0759) + (7.016004 \times 0.9241)$$

First, calculate $6.015122 \times 0.0759$:
$6.015122 \times 0.0759 \approx 0.4565$

Then, calculate $7.016004 \times 0.9241$:
$7.016004 \times 0.9241 \approx 6.4835$

Now, sum the two results:
$A \approx 0.4565 + 6.4835 = 6.94$

Ionization energy is the energy required to remove an electron from an atom. Boron (B) is in period 2, group 13, and Barium (Ba) is in period 6, group 2. As we move down a group, ionization energy decreases because electrons are further from the nucleus (larger atomic radius) and experience more shielding. Boron is higher up in the periodic table (smaller atomic radius, less shielding) than Barium, so it has higher ionization energy.

The Periodic Table is organized by atomic number (the number of protons in an atom’s nucleus). Atomic mass (the sum of protons and neutrons) was used in early versions, but modern tables use atomic number because it defines the element’s identity (each element has a unique atomic number).

Answer:

The atomic mass of Lithium, calculated from the given isotope abundances, is approximately $\boldsymbol{6.94}$ (the accepted value for Lithium’s atomic mass is close to this, confirming the calculation).

Question 20 Solution (Chemistry, a subfield of Natural Science)