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The fractional abundance is calculated by dividing the abundance of the isotope of interest by the abundance of all the isotopes of the element. For chlorine-37, the percent abundance is 0.2434, or 24.34%.
Boron-11 (80%) and boron-10 (20%).
Boron-10: 19,9 % Boron-11: 80,1 %
There are no radioactive isotopes of boron that are ordinarily found in nature. All elements have synthetic radioactive isotopes, however.
Boron is abundantEstimated Crustal Abundance: 1.0×101 milligrams per kilogramEstimated Oceanic Abundance: 4.44 milligrams per liter
The fractional abundance is calculated by dividing the abundance of the isotope of interest by the abundance of all the isotopes of the element. For chlorine-37, the percent abundance is 0.2434, or 24.34%.
There is 13 isotopes in the element boron. xD
Zero. There are NO boron atoms with a mass of 10.81 amu. The value of 10.81 is an average of the masses of the isotopes of boron.There are two stable isotopes of boron: boron-10 and boron-11, with masses of 10.012 amu and 11.009 amu. B-10 has a relative abundance of 19.9% and B-11 has a relative abundance of 80.1%.Do the math:10.012 x 0.199 + 11.009 x 0.801 = 10.81 amu
Boron-11 (80%) and boron-10 (20%).
Boron-10: 19,9 % Boron-11: 80,1 %
Europium 150.9196 has relative abundance of 51.99%, while Europium 152.9209 has a relative abundance of 48.04% (Assuming that these are the only 2 isotopes of Europium
Two stable ones, 10 & 11. Several unstable.
There are no radioactive isotopes of boron that are ordinarily found in nature. All elements have synthetic radioactive isotopes, however.
Boron is abundantEstimated Crustal Abundance: 1.0×101 milligrams per kilogramEstimated Oceanic Abundance: 4.44 milligrams per liter
To calculate average atomic mass from different isotopes of an element, we take into account the relative atomic masses of isotopes and their relative abundance on Earth. The following formula is used to calculate the needful : atomic mass = mass of isotope x percent abundance + mass of isotope x percent abundance / 100 (whole expression divided by 100)
There are several isotopes of Boron, which have different levels of abundance. I don't know the accurate numbers, but say like 25% of Zu-isotopes are 1 amu, and 75% of Zu-isotopes is 2, it would be 1.75 amu for the Atomic mass. Relating Zu to B of course, what I mean is Boron's different isotopes appear in different abundances and have different masses. 14% of boron may be about 6.882 amu, but just think about it like the above analogy of Zu, where Zu either weighs 1 or 2, but never 1.75. By the way, to find amu for an element: (%1*Iso1)+(%2*Iso2)+[...]
How do you calculate percent abundance of an isotope?You find the isotope number and then you calculate that into a fraction and then turn the fraction into a percentage and divide it by the atomic number then times it by the mass and turn that answer into a percent and voila, there you have it.