Chemistry, 16.09.2019 08:30 fangkronos

What are the factors that affect the average atomic mass of a mixture of isotopes?

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Relative abundance of each different isotopes. So when you calculate, you get the weighed average of the atomic mass of the element.

Answer from: buiratsamah

The factors that affect the average atomic mass of a mixture of isotopes is the relative abundance of each different isotope.

Explanation:

Each chemical element is characterized by the number of protons in its nucleus, which is called the atomic number Z. But in the nucleus of each element it is also possible to find neutrons, the number of which can vary. The atomic mass (A) is obtained by adding the number of protons and neutrons of a given nucleus.

The same chemical element may consist of different atoms, that is, their atomic numbers are the same, but the number of neutrons is different. These atoms are called element isotopes. Isotopes means "same place," that is, since all isotopes of an element have the same atomic number, they occupy the same place in the Periodic Table. Then, if a neutron is added to an atom, it becomes an isotope of that chemical element

The isotopes of an element have the same chemical properties, but differ somewhat in their physical properties.

On the other hand, the atomic mass of an element is the weighted average mass of its natural isotopes. Therefore, the atomic mass of an element is not an integer. The weighted average means that not all isotopes have the same percentage. In other words, the atomic masses of the chemical elements are usually calculated as the weighted average of the masses of the different isotopes of each element taking into account the relative abundance of each of them.

atomic mass of the element=Σ(relative abundance of isotope*atomic mass of the isotope)

Then, the factors that affect the average atomic mass of a mixture of isotopes is the relative abundance of each different isotope, as can be seen in the definitions set forth above.

Answer from: Quest

explanation:

Answer from: Quest

the dissociation of nickel hydroxide is follows:

$ni(oh)_2 > ni^2^+ + 2 oh^-$

$ksp for no(oh)2 = 5.5 * 10^-^1^6$

$5.5 * 10^-^1^6 = s * (2 s)^2 = 4 s^3$

$s^3 = 137.5 * 10^-^1^8$

$s = 5.14 * 10^-^6 = [oh^-]$

the concentration of hydroxide ion is

$5.14 * 10^-^6 = [oh^-]$

the equilibrium constant value of nickel hydroxide is = 6.1 * 10^-^1^6 and thus equilibrium equation can be written as:

$k = [ ni^2^+] * [oh^-]^2$

$6.1 * 10^-^1^6 = [ ni^2] * [5.14 * 10^-^6]^2$

$[ ni^2^+] = 0.23 * 10^-^4 mol /l$

thus, concentration of ni is 0.23 * 10^-^4 mol /l

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