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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

Vinegar is a solution of acetic acid (the solute) in water (the solvent) with a solution density of 1010 g/L. If vinegar is 0.80

Damm [24]

Answer:

4.8 %

Explanation:

We are asked the concentration in % by mass, given the molarity of the solution and its density.

0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:

MW acetic acid = 60.0 g/mol

mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g

mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)

= 1010 g

% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %

6 0

3 years ago

The stomach and intestines are organs of the body system which _______.

yulyashka [42]

Answer:

A. Digests food

Explanation:

Look about diggestive process in Google

Best regards

3 0

3 years ago

Read 2 more answers

What is the maximum mass of P2I4 that can be prepared from 8.80g of P4O6 and 12.37g of iodine according to the reaction:

Dvinal [7]

The balanced chemical reaction would be as follows:

5P4O6 +8I2 ---> 4P2I4 +3P4O10

We are given the amount of reactants used for the reaction. We first need to determine the limiting reactant from the given amounts. We do as follows:

8.80 g P4O6 (1 mol / 219.88 g) = 0.04 mol P4O6
12.37 g I2 ( 1 mol / 253.809 g ) = 0.05 mol I2

Therefore, the limiting reactant is iodine since less it is being consumed completely in the reaction. We calculate the amount of P2I4 prepared as follows:

0.05 mol I2 ( 4 mol P2I4 / 8 mol I2 ) (569.57 g / 1 mol) = 14.24 g P2I4

8 0

3 years ago

Read 2 more answers

The burning of a sample of propane

r-ruslan [8.4K]

Answer:

70.0°C

Explanation:

We are given;

Amount of heat generated by propane as 104.6 kJ or 104600 Joules
Mass of water is 500 g
Initial temperature as 20.0 ° C

We are required to determine the final temperature of water;

Taking the initial temperature is x°C

We know that the specific heat of water is 4.18 J/g°C

Quantity of heat = Mass × specific heat × change in temperature

In this case;

Change in temp =(x-20)° C

Therefore;

104600 J = 500 g × 4.18 J/g°C × (x-20)

104600 J = 2090x -41800

146400 = 2090 x

x = 70.0479

=70.0 °C

Thus, the final temperature of water is 70.0°C

7 0

3 years ago