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3 years ago

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

2 answers:

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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What is the progression of a space rock as it from space to Earth’s surface?

Bezzdna [24]

I think that's called "Trajectory"

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3 years ago

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The partial pressure of CH4 is 0.175 atm and that of O2 is 0.250 atm in a mixture of the two gases. What is the mole fraction of

tresset_1 [31]

Answer:

The total number of moles of gas in the mixture is 0.16939.

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Explanation:

Total volume of the mixture = V = 10.5 L

Temperature of the mixture = T = 35°C = 308.15K

Pressure of the mixture = P

Total moles of mixture = n = $n_1+n_2$

Using an ideal gas equation :

$PV=nRT$

$P\times 10.0 L=n\times 0.0821 atm L/mol K\times 308.15 K$

$P=2.509 n$

Partial pressure of the methane= $p_1=0.175 atm$

Moles of the methane= $n_1$

Partial pressure of the oxygen gas= $p_2=0.250 atm$

Moles of the methane= $n_2$

Mole fraction of the methane= $\chi_1=\frac{n_1}{n_1+n_2}$

Mole fraction of the oxygen gas= $\chi_2=\frac{n_2}{n_1+n_2}$

$p_1=p\times \chi_1$ (Dalton's law)

$0.175 atm=P\times \frac{n_1}{n_1+n_2}$

$0.175 atm=(2.509 n)\times \frac{n_1}{n}$

$n_1=0.06975 mol$

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

$p_2=p\times \chi_2$ (Dalton's law)

$0.250 atm=P\times \frac{n_2}{n_1+n_2}$

$0.250 atm=(2.509 n)\times \frac{n_2}{n}$

$n_2=0.09964 mol$

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n = $n_1+n_2$

[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]

7 0

3 years ago

What would happen if N2 were added to N2(g) + O2(g) = 2NO(9) at

DerKrebs [107]

Balanced chemical equation is :

$N_2(g)+O_2(g)-->2NO(g)$

It is given that the equation is in equilibrium.

We need to find what will happen if we add more $N_2$ is added .

By Le Chatelier's principle :

Changing the concentration of a chemical will shift the equilibrium to the side that would counter that change in concentration.

It means production of the side where content is added will decrease and concentration on other side will increase .

So , more NO would form .

Therefore, option B. is correct.

Hence, this is the required solution.

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3 years ago

The half-life of a radioactive isotope is 20.0 minutes. What is the total amount of a 1.00-gram sample of this sample of this is

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NISA [10]

Answer:

i think its fission

Explanation:

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