Chemistry, 10.11.2020 20:00 FlyingUnicorn123

How many electrons are there in the atom with a mass number of 66 and n = 35?

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Answers

Answer from: 21hendlill

Explanation:

Given parameters:

Mass number = 66

Number of electrons = 35

Unknown:

Number of electrons

Solution:

In an atom, the number of protons and electrons are the same. In ions, there is a difference between the two numbers.

For this problem, once we can find the number of protons, it is therefore the number of electrons.

Mass number = number of protons + number of neutrons

Insert the parameters and solve;

66 = number of protons + 35

Number of protons = 66 - 35 = 31

So the number of electrons is 31

Answer from: Quest

assuming that your periodic table has eighteen columns, you should highlight the four elements in the third column from the left of the table.

explanation

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the $3d$ have energy higher than those in the $4 s$ orbital while lower than those of the $4 p$ orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the $4s$ orbital is therefore filled right after $3p$ orbitals and before $3 d$ orbitals.

potassium has electron configuration $[\text{ar}] 4 s^{1}$ and calcium $[\text{ar}] 4 s^{2}$ . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration $[\text{ar}] 4 s^{2} \; 3d^{1}$ and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the $ns^{2} \; (n-1) p^{1}$ and should be highlighted as well.

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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