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3 years ago

Which of the following is a reasonable ground-state electron configuration?A) 1s21p62d2B) 1s22s42p6C) 1s22s22p5D) 1s22s22d6

Chemistry

1 answer:

Semenov [28]3 years ago

4 0

Answer:

Option C) 1s² 2s² 2p⁵

Explanation:

Ground-state is the configuration of the electrons when every electron of the atom occupies the lowest possible energy level.

The rank of energy of the orbitals may be remembered using Aufbau rules, and it is:

1s² < 2s² < 2p⁶ < 3s² < 3p⁶ < 4s² < 3d¹⁰ < 4p⁶ < 5s² < 4d¹⁰ < 5p⁶ < 6s² < 4f¹⁴ < 5d¹⁰ < 6p⁶ < 7s² < 5f¹⁴ < 6d¹⁰ < 7p⁶

Then, to find whether an electron configuration corresponds to a ground-state one, you must check that the previous order is preserved.

Option A). 1s² 1p⁶ 2d²

The orbital 1p does not exist. So this one is discarded.

Option B) 1s² 2s⁴ 2p⁶

The maximum number of electrons in an s orbital is 2, so 2s⁴ is impossible, and this option is discarded.

Option C) 1s² 2s² 2p⁵

This is correct because the orbitals are filled in the correct increasing energy order, without jumping any one.

Option D) 1s² 2s² 2d⁶

Orbital 2d does not exist, so this option is also discarded.

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3 years ago

Consider the following reaction: A(g)⇌2B(g). Find the equilibrium partial pressures of A and B for each of the following differe

Illusion [34]

Answer:

a. Kp=1.4

$P_{A}=0.2215 atm$

$P_{B}=0.556 atm$

b.Kp=2.0 * 10^-4

$P_{A}=0.495atm$

$P_{B}=0.00995 atm$

c.Kp=2.0 * 10^5

$P_{A}=5*10^{-6}atm$

$P_{B}=0.9999 atm$

Explanation:

For the reaction

A(g)⇌2B(g)

Kp is defined as:

$Kp=\frac{(P_{B})^{2}}{P_{A}}$

The conditions in the system are:

A B

initial 0 1 atm

equilibrium x 1atm-2x

At the beginning, we don’t have any A in the system, so B starts to react to produce A until the system reaches the equilibrium producing x amount of A. From the stoichiometric relationship in the reaction we get that to produce x amount of A we need to 2x amount of B so in the equilibrium we will have 1 atm – 2x of B, as it is showed in the table.

Replacing these values in the expression for Kp we get:

$Kp=\frac{(1-2x)^{2}}{x}$

Working with this equation:

$x*Kp=(1-2x)^{2} - -> x*Kp=4x^{2}-4x+1- - >4x^{2}-(4+Kp)*x+1=0$

This last expression is quadratic expression with a=4, b=-(4+Kp) and c=1

The general expression to solve these kinds of equations is:

$x=\frac{-b(+-)*\sqrt{(b^{2}-4ac)}}{2a}$ (equation 1)

We just take the positive values from the solution since negative partial pressures don´t make physical sense.

Kp = 1.4

$x_{1}=\frac{(1.4+4)+\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=1.128$

$x_{1}=\frac{(1.4+4)-\sqrt{(-(1.4+4)^{2}-4*4*1)}}{2*4}=0.2215$

With x1 we get a partial pressure of:

$P_{A}=1.128 atm$

$P_{B}=1-2*1.128 = -1.256 atm$

Since negative partial pressure don´t make physical sense x1 is not the solution for the system.

With x2 we get:

$P_{A}=0.2215 atm$

$P_{B}=1-2*0.2215 = 0.556 atm$

These partial pressures make sense so x2 is the solution for the equation.

We follow the same analysis for the other values of Kp.

Kp=2*10^-4

X1=0.505

X2=0.495

With x1

$P_{A}=0.505atm$

$P_{B}=1-2*0.505 = -0.01005 atm$

Not sense.

With x2

$P_{A}=0.495atm$

$P_{B}=1-2*0.495 = 0.00995 atm$

X2 is the solution for this equation.

Kp=2*10^5

X1=50001

$X2=5*10^{-6}$

With x1

$P_{A}=50001atm$

$P_{B}=1-2*50001=-100001atm$

Not sense.

With x2

$P_{A}= 5*10^{-6}atm$

$P_{B}=1-2*5*10^{-6}= 0.9999 atm$

X2 is the solution for this equation.

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3 years ago

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3 years ago

Robert has pure samples of both D-ribose and D-arabinose, but he forgot to label them. He only has some nitric acid, the reagent

Sauron [17]

Answer:

Following are the responses to the given question:

Explanation:

Since HN03 is an oxidation substance D-ribose u.ith oxidized to form in rubric acid Ribose is chiral, but rubric acid is achiral because of its symmetry mirror level, Hence no infrared roster in the sample holder is observed.

Please find the attached file.

D-Arabinose, on either hand, gives optical aldaric acid with such a net optical rotation observed inside the polarimeter for diagnosis with HN03.

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3 years ago