When an alpha particle is emitted from an unstable nucleus, the atomic mass numberof the nucleus

Answer:

9.88

Explanation:

The higher the Ksp, the more soluble a compound is. Hence, Co(OH)₂ ranks as the least soluble hydroxide.

Its maximum concentration must be 1x10⁻⁶ M, and the reaction can be represented as:

Co(OH)₂(s) ⇄ Co⁺²(aq) + 2OH⁻(aq)

Thus, [Co⁺²] = 1x10⁻⁶M

Ksp =  [Co⁺²] *[OH⁻]²

[[OH⁻]² = 5.9x10⁻¹⁵/1x10⁻⁶

[[OH⁻] = √(5.9x10⁻⁹)

[[OH⁻] = 7.6811x10⁻⁵

pOH = -log[OH⁻]

pOH = -log(7.6811x10⁻⁵)

pOH = 4.11

Knowing that pH + pOH = 14

pH = 14 - 4.11

The equal mass indicates that both atoms have the same number of protons and neutrons.

A positive charge signifies a difference in electron count.

Assuming the atomic number is A,

the mass number equals M.

In a neutral atom, there are A electrons.

A negatively charged atom would have A + 1 electrons [while the count of protons and mass number remains unchanged].

A positively charged atom contains A - 1 electrons [with consistent protons and mass number].

For instance: Cl- and Cl+.

Answer:

For A: The change in free energy for the reaction is -5339.76 J/mol

For B: Free energy change is expressed in kJ/mol

For C: The forward reaction favors progression, while the reverse reaction does not.

Explanation:

Regarding the specified chemical reaction:

• For A:

The relationship between standard Gibbs free energy and equilibrium constant is as follows:

The free energy change can be calculated using the following equation:

Or,

where,

= Change in free energy

R = Gas constant =

= standard temperature =

T = temperature of the cell =

= equilibrium constant =

Q = reaction quotient =

= 0.06 M

[DHAP] = 0.125 M

Substituting the values into the equation yields:

Thus, the change in free energy for the reaction is -5339.76 J/mol

• For B:

To convert the free energy change to kilojoules, we apply the conversion factor:

1 kJ = 1000 J

So,

Consequently, the free energy change's units are kJ/mol

• For C:

For spontaneity in the reaction, the Gibbs free energy must be negative. However, the calculations indicate a positive Gibbs free energy, leading to the conclusion that the reaction is not spontaneous.

The free energy change of the reaction is negative.

Consequently, the forward reaction is favored and the reverse reaction is not favored.