Antimony consists of two major isotopes with the following atomic masses: 121sb (120.904 amu) and 123sb (122.904 amu). the average atomic mass of antimony is 121.760 amu. calculate the percent of each of the isotopes. report your answer to 3 significant figures.

121Sb=57.2%

123Sb=42.8%

Explanation:

We are given that

Atomic mass of 121Sb=120.904 amu

Atomic mass of 123Sb=122.904 amu

Average atomic mass of antimony=121.760 amu

We have to find the percent of each of the isotope.

Let x be the percent of 121Sb and 1-x be the percent of 123Sb.

Using formula of average atomic weight

Average atomic weight=atomic weight of 121Sb percentage abundance of isotope 121Sb+atomic weight of 123Sbpercentage abundance of isotope 123Sb

Substitute the values

Percentage of 121Sb=57.2%

Abundance of isotope 123Sb=1-0.572=0.428

Percentage of isotope 123Sb=42.8%

because it says use -2,0,and 2 for x, we can plug those numbers into the equation. y/4=3/4(-2)+2   y/4= -1 1/2 +2 now to get x by itself, we can multiply each side by 4. y=-6+8 which equals y=2 when x=-2. now you can do the same thing with 0 and 2 as x. you get y=8 when x=0 and y=14 when x=2.

x   |   y

-2 |   2

0   |   8

2   |   14

explanation:

could i give you a exaplo that could you?