Please help asap!! Need help for problem #2. Naturally occuring iron, contains 5.82% ^54Fe, 91.66% ^56Fe, 2.19% ^57Fe, and 0.33% ^58Fe. The reslective atomic masses are 53.940 amu, 55.935 amu, 56.935 amu and 57.933 amu. Calculate the average atmoic mass of iron. (show work)

55.56 amu

Explanation:

Let A, B, C and D represent the four isotopes of iron.

The following data were obtained from the question:

Isotope A (Fe-54):

Mass of A = 53.940 amu

Abundance (A%) = 5.82%

Isotope B (Fe-56):

Mass of B = 55.935 amu

Abundance (B%) = 91.66%

Isotope C (Fe-57):

Mass of C = 56.935 amu

Abundance (C%) = 2.19%

Isotope D (Fe-58):

Mass of D = 57.933 amu

Abundance (D%) = 0.33%

Average atomic mass =.?

The average atomic mass of the iron can obtained as follow:

Average atomic mass = [(mass of A × A%)/100] + [(mass of B × B%)/100] + [(mass of C × C%)/100] + [(mass of D × D%)/100]

Average atomic mass = [(53.940 × 5.82)/100] + [(55.935 × 91.66 )/100] + [(56.935 × 2.19)/100] + [(57.933 × 0.33)/100]

= 2.848 + 51.270 + 1.247 + 0.191

= 55.56 amu

Therefore, the average atomic mass of the iron is 55.56 amu

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the formula for the conversion of temperature from degree celsius to kelvin is:

the sample of water at is converted to kelvin temperature by using the above formula as:

on comparing the two samples of water at or we can say that at and at respectively. the difference between the temperatures of two samples of water is:

hence, one can say that the sample of water at is greater than the sample of water at by .