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Serhud [2]
2 years ago
7

silver has an atomic mass of 107.868 amu. silver has two common isotopes. one of the isotopes has a mass of 106.906 amu and a re

lative abundace of 51.881%. a. what is the % abundance of the other isotope. b. what is the mass if the other isotope.​
Chemistry
1 answer:
Sunny_sXe [5.5K]2 years ago
6 0

A. The abundance of the 2nd isotope is 48.119%

B. The mass of the 2nd isotope is 108.905 amu

Let the 1st isotope be A

Let the 2nd isotope be B

A. Determination of the abundance of the 2nd isotope

Abundance of isotope A = 51.881%.

<h3>Abundance of isotope B =? </h3>

Abundance of B = 100 – A

Abundance of B = 100 – 51.881

<h3>Abundance of B = 48.119%</h3>

B. Determination of the mass of the 2nd isotope

Atomic mass of silver = 107.868 amu.

Mass of 1st isotope (A) = 106.906 amu

Abundance of isotope A (A%) = 51.881%.

Abundance of isotope B (B%) = 48.119%

<h3>Mass of 2nd isotope (B) =? </h3>

atomic \: mass =  \frac{mass \: of \:A \times \:A\%}{100}  + \frac{mass \: of \:B \times \:B\%}{100} \\  \\ 107.868 = \frac{106.906\times \ \: 51.881}{100}  + \frac{mass \: of \:B \times \:48.119}{100} \\  \\ 107.868 = \: 55.464 + 0.48119 \times mass \: of \:B  \\  \\  collect \: like \: terms \\  \\ 0.48119 \times mass \: of \:B  = 107.868  - 55.464  \\  \\ divide \: both \: side \: by \: 0.48119 \\  \\ mass \: of \:B  = \frac{107.868  - 55.464 }{0.48119}  \\  \\ mass \: of \:B  =108.905 \: amu \\  \\

Therefore, the mass of the 2nd isotope is 108.905 amu

Learn more: brainly.com/question/7955048

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