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4 years ago

15

How many of the elements in the second row of the periodic table (li through ne) will have at least one unpaired electron in the

ir electron configurations?

1 answer:

nexus9112 [7]4 years ago

4 0

The second row of the periodic table consists of the elements:
Lithium (3)
Beryllium (4)
Boron (5)
Carbon (6)
Nitrogen (7)
Oxygen (8)
Fluorine (9)
Neon (10)

The number in the bracket shows the atomic number of the elements, which also indicates the number of electrons they have. We see that the odd numbered elements, Li, B, N and F will all have unpaired electrons. Therefore, 4 elements will have unpaired electrons in their electronic configuration.

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3 years ago

The coefficient of thermal expansion α = (1/V)(∂V/∂T)p. Using the equation of state, compute the value of α for an ideal gas. Th

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Answer:

The coefficient of thermal expansion α is

$\alpha = \frac{1}{T}$

The coefficient of compressibility

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

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So

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

Explanation:

From the question we are told that

The coefficient of thermal expansion is $\alpha = \frac{1}{V} * (\frac{\delta V}{ \delta P}) P$

The coefficient of compressibility is $\beta = - (\frac{1}{V} ) * (\frac{\delta V}{ \delta P} ) T$

Generally the ideal gas is mathematically represented as

$PV = nRT$

=> $V = \frac{nRT}{P} --- (1)$

differentiating both side with respect to T at constant P

$\frac{\delta V}{\delta T } = \frac{ n R }{P}$

substituting the equation above into $\alpha$

$\alpha = \frac{1}{V} * ( \frac{ n R }{P}) P$

$\alpha = \frac{nR}{PV}$

Recall from ideal gas equation $T = \frac{PV}{nR}$

So

$\alpha = \frac{1}{T}$

Now differentiate equation (1) above with respect to P at constant T

$\frac{\delta V}{ \delta P} = -\frac{nRT}{P^2}$

substituting the above equation into equation of $\beta$

$\beta = - (\frac{1}{V} ) * (-\frac{nRT}{P^2} ) T$

$\beta =\frac{ (\frac{n RT}{PV} )}{P}$

Recall from ideal gas equation that

$\frac{PV}{nRT} = 1$

So

$\beta = \frac{1}{P}$

Now considering $(\frac{ \delta P }{\delta T} )V$

From equation (1) we have that

$\frac{ \delta P}{\delta T} = \frac{n R }{V}$

From ideal equation

$nR = \frac{PV}{T}$

So

$\frac{\delta P}{\delta T} = \frac{PV}{TV}$

=> $\frac{\delta P}{\delta T} = \frac{P}{T}$

=> $\frac{\delta P}{\delta T} = \frac{\alpha }{\beta}$

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