Anaturally occurring element consists of three isotopes. the data for the isotopes are: isotope 1: 46.972 u, 69.472% isotope 2: 48.961 u, 21.667% isotope 3: 49.954 u, 8.8610% what is the average atomic mass of this naturally occurring element?i keep getting an answer that is not one of the multiple choices

The atomic mass of the element has been found to be 87.5048 amu.

The atomic mass of the element has been given as the average atomic mass of the atom with respect to the isotopes in the nature and abundance.

The average atomic mass of a compound has been given by:

Computation for atomic mass

The mass of isotopes of a given element has been given as:

Isotope 1 = 46.972, abundance = 0.69472

Isotope 2= 48.961 abundance = 0.21667

Isotope 3= 49.954 abundance = 0.88610

Substituting the values for atomic mass:

The atomic mass of the element has been found to be 87.5048 amu.

For more information about atomic mass., refer to the link:

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(46.972* 69.472% + 48.961*21.667% + 49.954*8.8610%)/100% =

= 46.972* 0.69472 + 48.961*0.21667 + 49.954*0.088610 =47.667 u

Idon't know hope someone you

  remember that hf is only the energy of one photon. not two. not three. one.  1) 5.12 * 10^3 photons  find the energy of one photon, then determine how many photons are needed to reach the sensitivity.  e = hf  = (6.626 * 10^-34 js)(5.95 * 10^19 hz)  = 3.94247 * 10^-14 j per photon  (2.02 * 10^-10 j) / (3.94247 * 10^-14 j/photon) = 5123.69149 photons  round to three sig figs.  2) 5.77 * 10^-17 j  energy of pulse = energy per photon * number of photons  = hfn  = (6.626 * 10^-34 js)(1.66 * 10^13 hz)(5250)  = 5.774559 * 10^-17 j  round to three sig figs.  3) 4.07 * 10^8 kj/mol  10^12 pm = 1 m  c = fλ  f = c/λ  e = hf  = hc/λ  = [(6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.94 * 10^-12 m)](1 kj/1000 j)(6.02 * 10^23 photons/mol)  = 40702571.4 kj/mol  round to three sig figs.  4) 9.94 * 10^-1 m  e = hc/λ  λ = hc / e  = (6.626 * 10^-34 js)(3.00 * 10^8 m/s) / (2.00 * 10^-25 j)  = 0.9939 m  round to three sig figs.  5) 9.06 * 10^3 photons  energy of pulse = energy per photon * number of photons  number of photons = energy of pulse / energy per photon  = energy of pulse / (hf)  = (3.87 * 10^-13 j) / [(6.626 * 10^-34 js)(6.45 * 10^16 hz)]  = 9055.23695 photons  round to three sig figs.  6) 2.78 * 10^-15 j  this is problem 1, except in reverse.  sensitivity = minumum number of photons * energy per photon  = (5930)(6.626 * 10^-34 js)(7.077 * 10^14 hz)  = 2.78070758 × 10^-15 j  round to three sig figs.