Anaturally occurring sample of an element contains only two isotopes. the first isotope has a mass of 68.9255 amu and a natural abundance of 60.11%. the second isotope has a mass of 70.9247 amu. find the atomic mass of the element.

- The average atomic mass or the atomic mass of the element is 69.7230 amu.

Solution:- The atomic mass or also known as average atomic mass of an element is calculated from it's isotopes by using the formula:

average atomic mass = mass of First isotope(abundance) + mass of second isotope(abundance)

From given data, the mass of first isotope is 68.9255 amu and mass of second isotope is 70.9247 amu. percent abundance for the first isotope is 60.11%. The sum of percent abundance of all the isotopes of an element is always 100. So, the percent abundance of second isotope = 100 - 60.11 = 39.89%

We convert the percent abundances to the decimals and then plug in the values in the formula to calculate the average atomic mass.

First isotope abundance = 0.6011

second isotope abundance = 0.3989

average atomic mass = 68.9255(0.6011) + 70.9247(0.3989)

average atomic mass = 41.4311 + 28.2919

average atomic mass = 69.7230 amu

So, the average atomic mass or the atomic mass of the element is 69.7230 amu.

its going to be mutualism cuz they both benefit from the relationship.

~batmans wife dun dun

"standard molar enthalpy:

h2o ( liquid water ) : - 285.8 kj/mole

h2o ( water vapor ) : - 241.8 kj / mole.

δhrxn = - 241. 8 - ( - 285.8 )= -241.8 + 285.8 = 44.0

answer: 44.0 kj"