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3 years ago

What are the exceptions to the periodic trends in ionization energy?

Chemistry

1 answer:

Verdich [7]3 years ago

7 0

Answer:

The exceptions to the periodic trends in ionization energy are the first ionization energy of beryllium is higher than that of boron and the first ionization energy of nitrogen is also higher than that of oxygen.

Explanation:

Taking a close look at the figure of first ionization energies, it clearly shows that the first ionization energy of beryllium is higher than that of boron and the first ionization energy of nitrogen is also higher than that of oxygen.

This is as a result of Hund's rule and electron configuration. For example, the first ionization potential electron of beryllium is obtained from a 2s orbital while that of boron comes from a 2p electron. However, for oxygen and nitrogen, their electrons are obtained from 2p orbitals. While spin is uniform for all 2p electrons of nitrogen, it is different for oxygen.

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Which of the following atoms will form an ion with the most negative charge?

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Answer:

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3 years ago

Which of the following radioactive emissions is the least penetrating?

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Answer:

alpha particles

Explanation:

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2 years ago

My professor gave me two questions to solve using the Van Der Waals Equation. She told us to solve for P and the second one we h

Fed [463]

Answer:

P=atm

$b=\frac{L}{mol}$

Explanation:

The problem give you the Van Der Waals equation:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

First we are going to solve for P:

$(P+\frac{n^{2}a}{V^{2}})=\frac{nRT}{(V-nb)}$

$P=\frac{nRT}{(V-nb)}-\frac{n^{2}a}{v^{2}}$

Then you should know all the units of each term of the equation, that is:

$P=atm$

$n=mol$

$R=\frac{L.atm}{mol.K}$

$a=atm\frac{L^{2}}{mol^{2}}$

$b=\frac{L}{mol}$

$T=K$

$V=L$

where atm=atmosphere, L=litters, K=kelvin

Now, you should replace the units in the equation for each value:

$P=\frac{(mol)(\frac{L.atm}{mol.K})(K)}{L-(mol)(\frac{L}{mol})}-\frac{(mol^{2})(\frac{atm.L^{2}}{mol^{2}})}{L^{2}}$

Then you should multiply and eliminate the same units which they are dividing each other (Please see the photo below), so you have:

$P=\frac{L.atm}{L-L}-atm$

Then operate the fraction subtraction:

P= $P=\frac{L.atm-L.atm}{L}$

$P=\frac{L.atm}{L}$

And finally you can find the answer:

P=atm

Now solving for b:

$(P+\frac{n^{2}a}{V^{2}})(V-nb)=nRT$

$(V-nb)=\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$nb=V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}$

$b=\frac{V-\frac{nRT}{(P+\frac{n^{2}a}{V^{2}})}}{n}$

Replacing units:

$b=\frac{L-\frac{(mol).(\frac{L.atm}{mol.K}).K}{(atm+\frac{mol^{2}.\frac{atm.L^{2}}{mol^{2}}}{L^{2}})}}{mol}$

Multiplying and dividing units,(please see the second photo below), we have:

$b=\frac{L-\frac{L.atm}{atm}}{mol}$

$b=\frac{L-L}{mol}$

$b=\frac{L}{mol}$

7 0

3 years ago

Thorium is an element that occurs naturally in the earth's crust. The only naturally occurring isotope of thorium is 232Th and i

Murrr4er [49]

A, radium-228. Alpha radiation is essentially just two protons and two neutrons leaving the atom, causing the atomic number to drop by two (since the protons left) and the mass number to drop by four (since two neutrons and protons left).

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3 years ago

Read 2 more answers

Explain three factors that can affect the size of a line in a spectrum.

Sunny_sXe [5.5K]

Answer:For atoms and molecules, the width of spectral lines is governed mainly by the broadening of the energy levels of the atoms or molecules during interactions with surrounding particles and by the broadening of the spectral lines as a result of the Doppler effect.

Explanation:

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3 years ago