The atmospheric gases in the thermosphere are heated ____

The quantum numbers for the last electron placed in three elements are listed below. Which of these is(are) NOT correct? Er (4 3

UNO [17]

Answer:

The three elements Erbium, Thallium and Osmium have incorrect quantum numbers for the last electron placed.

Explanation:

The 4 quantum numbers are (n,l,ml,ms):

n (Principal quantum number): it is the number of the shell (level) where the electron is placed.
l (Angular momentum quantum number or Secondary): it represents the sublevel where the electron is placed. There are 4 subleves: s, p d and f so secondary quantum number can take the number 0 (s), 1 (p), 2 (d) or 3 (f) depending on which sublevel the electron is placed.
ml (Magnetic quantum number): it represents the spatial orientation of the electron in respect of the sublevel the electron is placed. For example: if the electron occupies the s sublevel the magnetic number will be 0, if the electron occupies the p sublevel the magnetic number could be -1,0,1, if the electron occupies the d sublevel the magnetic number could be -2,-1,0,1,2 and if the electron occupies the f sublevel the magnetic number could be -3,-2,-1,0,1,2,3. You can see this in the attachment related to the correct sublevel for the example.

ms (Spin quantum number): this number represents the possible rotation of the electron so it could be 1/2 (which is represented by an up arrow) or -1/2 (represented by an down arrow).

Let's analyze the last electron of each element. You can see the attachment for better understanding. The last electron it is represented with orange color.

- Erbium:

This element has 68 electrons so following the Moeller's Diagram to fill the the electronic configuration, we found that the last electron of Erbium it is in the 4th level (shell), in the f sublevel. As Erbium has 12 electrons in the f sublevel, it is necessary to follow the Hund's rule (electrons must be placed singly in every sublevel before place a parallel electron) to placed correctly all of them. Finally, the last electron of Erbium stays in the middle of the sublevel and it is represented by a down arrow so the correct quantum numbers in the Erbium element are (4,3,1,-1/2).

- Thallium:

This element has 81 electrons and following the Moeller's Diagram, we found that it last electron it is in the 6th level, in the p sublevel. As Thallium has 1 electron in the p sublevel, it is placed singly in the sublevel. So the last electron of Thallium it is represented by an up arrow so the correct quantum numbers in the Thallium element are (6,1,-1,1/2).

- Osmium:

Osmium has 76 electrons and following the steps that we did with we the other elements, we noticed that its last electron it is in the 5th level, in the d sublevel. Following the Hund's rule the last electron of Osmium has a magnetic quantum number of -2 and its spin quantum number is -1/2, so the quantum numbers in the Osmium element are (5,2,-2,-1/2).

Note:

- Remember that the s sublevel has place for 2 electrons, the p sublevel has place for 6 electrons, the d sublevel has place for 10 electrons and the f sublevel has place for 14 electrons.

3 0

3 years ago

As human populations grow, it becomes more important to use renewable sources of energy Which of the following objects uses ener

deff fn [24]

Explanation:

Moreover, increasing use of fossil fuels also causes serious environmental problems. Hence, there is a pressing need to use renewable energy sources like solar energy, wind, tide, biomass and energy from waste material. (i) it reduces the dependence on fossil fuels for energy. (ii) they donot produces pollution.

4 0

3 years ago

Read 2 more answers

The rate constant for the second-order reaction: 2NOBr(g) → 2NO(g) + Br2(g) is 0.80/(M · s) at 10°C. Starting with a concentrati

a_sh-v [17]

Answer : The concentration of NOBr after 95 s is, 0.013 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = $0.80M^{-1}s^{-1}$

t = time taken = 95 s

[A] = concentration of substance after time 't' = ?

$[A]_o$ = Initial concentration = 0.86 M

Now put all the given values in above equation, we get:

$0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)$

[A] = 0.013 M

Hence, the concentration of NOBr after 95 s is, 0.013 M

4 0

3 years ago

100.mL of a .795 M solution of KBr is diluted to 500.mL. what is the new concentration of the solution?

kap26 [50]

Answer:

0.159 M

Explanation:

convert from mL to L then use the equation:

M1V1 = M2V2

rearrange to find M2

$\frac{M1V1}{V2} = M2$

$\frac{(0.795 M)(0.100 L)}{0.500 L} = 0.159 M$

4 0

3 years ago

The formula K₂S indicates that _______.

Pachacha [2.7K]

The subscriot 2 means that in the formula there are two parts of K, and the subscript 1 (implicit) for S, indicates that there is one part of S.

This is, the formula gives the ratio of the elements K and S in the compound, which is:

2 atoms of K : 1 atom of S.

Answer: there are 2 atoms of K and 1 atom of S in a molecule of K2S.

5 0

3 years ago

The atmospheric gases in the thermosphere are heated _____. Pick the best one.