Question
The Periodic Table Online Practice
Complete this assessment to review what you’ve learned. It will not count toward your grade.
6 of 156 of 15 Items
Question
Suppose that hypothetical element Jovian occurs as three isotopes with mass numbers 44, 46, and 48 respectively. If, in nature, Jovian occurs as Jovian-44 approximately 22.4% of the time, and as Jovian-46 approximately 44.0% of the time, what is the best value to use as the atomic mass of Jovian?(1 point)
Responses
46.0 amu
46.0 amu
30.1 amu
30.1 amu
39.0 amu
39.0 amu
46.2 amu
Complete this assessment to review what you’ve learned. It will not count toward your grade.
6 of 156 of 15 Items
Question
Suppose that hypothetical element Jovian occurs as three isotopes with mass numbers 44, 46, and 48 respectively. If, in nature, Jovian occurs as Jovian-44 approximately 22.4% of the time, and as Jovian-46 approximately 44.0% of the time, what is the best value to use as the atomic mass of Jovian?(1 point)
Responses
46.0 amu
46.0 amu
30.1 amu
30.1 amu
39.0 amu
39.0 amu
46.2 amu
Answers
GPT-4o mini
To find the atomic mass of the hypothetical element Jovian, we need to take a weighted average of its isotopes based on their natural abundances.
The information provided is as follows:
- Jovian-44: 22.4% abundance
- Jovian-46: 44.0% abundance
- Jovian-48: The remaining percentage, which we can calculate as:
\[
100\% - (22.4\% + 44.0\%) = 100\% - 66.4\% = 33.6\%
\]
Now we have:
- Jovian-44: 22.4% or 0.224
- Jovian-46: 44.0% or 0.440
- Jovian-48: 33.6% or 0.336
Next, we calculate the weighted average using the formula:
\[
\text{Atomic mass} = (0.224 \times 44) + (0.440 \times 46) + (0.336 \times 48)
\]
Calculating each term:
1. \(0.224 \times 44 = 9.856\)
2. \(0.440 \times 46 = 20.240\)
3. \(0.336 \times 48 = 16.128\)
Now, summing these values:
\[
\text{Atomic mass} = 9.856 + 20.240 + 16.128 = 46.224 \, \text{amu}
\]
Rounding this value, the best estimation for the atomic mass of Jovian is approximately \(46.2 \, \text{amu}\).
So, the correct answer is:
**46.2 amu**
The information provided is as follows:
- Jovian-44: 22.4% abundance
- Jovian-46: 44.0% abundance
- Jovian-48: The remaining percentage, which we can calculate as:
\[
100\% - (22.4\% + 44.0\%) = 100\% - 66.4\% = 33.6\%
\]
Now we have:
- Jovian-44: 22.4% or 0.224
- Jovian-46: 44.0% or 0.440
- Jovian-48: 33.6% or 0.336
Next, we calculate the weighted average using the formula:
\[
\text{Atomic mass} = (0.224 \times 44) + (0.440 \times 46) + (0.336 \times 48)
\]
Calculating each term:
1. \(0.224 \times 44 = 9.856\)
2. \(0.440 \times 46 = 20.240\)
3. \(0.336 \times 48 = 16.128\)
Now, summing these values:
\[
\text{Atomic mass} = 9.856 + 20.240 + 16.128 = 46.224 \, \text{amu}
\]
Rounding this value, the best estimation for the atomic mass of Jovian is approximately \(46.2 \, \text{amu}\).
So, the correct answer is:
**46.2 amu**