7. oxygen is composed of three isotopes: one has a a mass of 16.999 amu. this isotope
makes up 0.037% of oxygen. of the other two, one has a mass of 15.995 amu, and the
other has a mass of 17.999 amu. calculate the abundance of the other two isotopes, using
the average atomic mass of 15.9994 amu.

Answer : The abundance in % for other two isotopes of the oxygen is 99.762% and 0.201%

Explanation :

The average atomic mass of an element is the total of isotopic mass times the percent abundance for all the isotopes.

Oxygen has 3 isotope, ¹⁶O, ¹⁷O and ¹⁸O.

The average atomic mass of oxygen can be calculated as follows.

Average Atomic Mass = (Isotopic mass of ¹⁶O x % abundance of ¹⁶O) + (isotopic mass of ¹⁷O x % abundance of ¹⁷O) + (Isotopic mass of ¹⁸O x % abundance of ¹⁸O)

The mass of ¹⁷O = 16.999 amu and its % abundance is 0.037% which is 0.00037

The mass of ¹⁶O = 15.995 amu. Let us assume its % abundance as a

The mass of ¹⁸O = 17.999 amu

The total % abundance for all the isotopes is 100% or 1 .

Therefore % abundance for ¹⁸O would be 1 - ( a + 0.00037) = 0.99963 - a

Let us plug in the above values in the average atomic mass formula .

Average atomic mass = ( 15.995)(a) + (16.999)(0.00037) + (17.999)(0.99963 - a)

The average atomic mass of element O is 15.9994

Therefore we have,

15.9994 = 15.995a + 0.0062896 + 17.99234 - 17.999a

15.9994 = -2.004a + 17.99863

-1.99923 = -2.004a

a = 0.99762

a = 99.762%

The percent abundance of ¹⁶O is 99.762%

The percent abundance of ¹⁸O is 0.99963 - 0.99762 = 0.00201 or 0.201%

Hope this helps

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f(2x−x2)=−3x2+6x−1

step-by-step explanation:

set up the composite function and evaluate.

f(2x−x2)=−3x2+6x−1