Explain the effect of change in concentration of a system in Equilibrium with at least one example.

Chemistry

1 answer:

Luda [366]3 years ago

3 0

Explanation:

If the concentration of a substance is changed, the equilibrium will shift to minimise the effect of that change. If the concentration of a reactant is increased the equilibrium will shift in the direction of the reaction that uses the reactants, so that the reactant concentration decreases. For example, decreased volume and therefore increased concentration of both reactants and products for the following reaction at equilibrium will shift the system toward more products.

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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>

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