naturally occurring boron is 80.20% boron 11 (atomic mass= 11.01amu) and 19.80% of some other isoptopic form of boron. what must the atmoci mass be be in order to account for the 10.81 amu averafe atomic mass of boron? round to two decimal places
i got 9.999 and rounded to 10.00
10.00amu
(0.8020*11.01)+(0.1980*x) = 10.81
Solve for x
To find the atomic mass of the other isotopic form of boron, we can set up the following equation:
(0.8020 * 11.01amu) + (0.1980 * x) = 10.81amu
Simplifying the equation:
8.82402 + 0.1980x = 10.81
Subtracting 8.82402 from both sides:
0.1980x = 1.98598
Dividing both sides by 0.1980:
x = 10.02
Therefore, the atomic mass of the other isotopic form of boron must be 10.02 amu in order to account for the average atomic mass of boron (10.81 amu).
To find the atomic mass of the other isotopic form of boron, you can use the following equation:
(percentage of isotopic form A * atomic mass of isotopic form A) + (percentage of isotopic form B * atomic mass of isotopic form B) = average atomic mass of boron
Let's solve the equation step by step:
First, convert the percentages to decimals:
Percentage of boron 11 = 80.20% = 0.8020
Percentage of other isotopic form = 19.80% = 0.1980
Now, plug in the known values into the equation:
(0.8020 * 11.01) + (0.1980 * x) = 10.81
Where x represents the atomic mass of the other isotopic form.
Next, solve for x:
(0.8020 * 11.01) + (0.1980 * x) = 10.81
Simplify the equation:
8.8242 + 0.1980x = 10.81
Subtract 8.8242 from both sides:
0.1980x = 1.9858
Divide both sides by 0.1980:
x = 1.9858 / 0.1980
x ≈ 10.03
Therefore, the atomic mass of the other isotopic form of boron must be approximately 10.03 amu.