PLEASE HELP!!! CHEMISTRY

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Hydrogen bonds are typically stronger than Van der Waals forces bc they are based on permanent dipoles, that form when hydrogen comes in vicinity of a highly electronegative atom (like F, N, or O). These bonds are long-lasting and pretty strong.

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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

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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>

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