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LiRa [457]
3 years ago
14

As a Ca atom undergoes oxidation to Ca2+, the number of neutrons in its nucleus

Chemistry
2 answers:
maria [59]3 years ago
6 0
Oxidation refers to a change in electrons (In this case, loss of electrons). It has no impact on the number of neutrons, or for that matter on any sub-atomic particles in the nucleus. So the number of neutrons (3) remains the same
Lemur [1.5K]3 years ago
3 0

Answer:

Hence the correct answer is remains the same

Explanation:

As a Ca atom undergoes oxidation to Ca2+,as follows:

Ca --- Ca 2+ + 2e-

In the process of oxidation calcium loss 2 electron only there is no change in the number of protons as well neutrons therefore there is no change in the number of neutrons in its nucleus

Hence the correct answer is remains the same

Oxidation:

Oxidation is a process in which either 1 or all following changes occurs:

1. Gaining of oxygen atoms

2. Loss of electrons

3. Loss of hydrogen atom.

4. Increasing oxidation number.

Reduction:

Reduction is a process in which either 1 or all following changes occurs:

1. Loss of oxygen atoms

2. Gaining of electrons

3. Gaining of hydrogen atom.

4. Decreasing oxidation number.

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Arrange the following compounds in order of decreasing basicity: MgO, Cs2O, Cl2O7, SnO2, P4O10
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Genrally as the metalic character decreases basicity character decreases while acidic character increases .

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Can someone please please help
Llana [10]

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oxidizer

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7 0
3 years ago
Ca(OH)₂(s) precipitates when a 1.0 g sample of CaC₂(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g sa
Alex

Answer:

D) Ca(OH)₂ will not precipitate because Q <  Ksp

Explanation:

This is a question in which we will employ the reaction quotient Q to determine whether a precipitate will form.

Here we have first a chemical reaction in which Ca(OH)₂  is produced:

CaC₂(s)  + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂  is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product  constant for Ca(OH)₂  is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

Thus the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M  = 0.002 M  ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q=  [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is (D)

6 0
3 years ago
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