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Dmitriy789 [7]
3 years ago
6

State the exclusion principle. What does it imply about the number and spin of electrons in an atomic orbital?

Chemistry
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

The Pauli exclusion principle was developed by Austrian physicist Ernst Pauli in 1925.  This principle of quantum says that two electrons in an atom cannot have all four equal quantum numbers .

Explanation:

This fact would explain that electrons are dispersed in layers or levels around the nucleus of the atom and therefore, atoms that have more electrons occupy more space, because the number of layers of which the atom consists increases. The maximum number of electrons that a layer or level can have is 2n ^ 2.

In order to fully describe the electron within the hydrogen atom, we need to enter a fourth quantum number to those already known. Said fourth quantum number is represented by the letters ms, and is known as the quantum number of spin, which is closely related to the magnetic properties of electrons. The quantum number ms can only have two different values, +1/2 or -1/2. To electrons whose values ​​of ms are equal, it is said that they have what is known as parallel spins, however, if the values ​​that present more are different it is said that they have opposite spins or also called antiparallels.

In order to describe an orbital, three quantum numbers (the numbers n, l and ml) are needed, at the same time that an electron that is in an atom is given by a combination of four quantum numbers, the main three plus the number ms . Pauli's exclusion principle tells us that in an atom it is impossible for two electrons to coexist with the four identical quantum numbers. According to this principle, in an atomic type orbital, which is determined by the quantum numbers n, l, and ml, there can only be two electrons: one of them with a positive spin +1/2 and another with its opposite spin negative -1/2.

Then we say that each of the types of orbitals can only contain 2 electrons at most, which must necessarily have opposite spins. These electrons will have all their equal quantum numbers, and will only differ in the quantum number ms (spin).

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a compound is composed of only c h and O. Combustion of a 519 gram sample of the compound yields 1.24 grams of CO2 and?
densk [106]

Answer:

C3H3O

Explanation:

Question incomplete needs to be rewritten:

A compound is composed of only C, H, and O. The combustion of a .519-g sample of the compound yields 1.24g of CO_2 and 0.255 g of H_2 O. What is the empirical formula of the compound

We can get the answer through calculations as follows.

From the mass of carbon iv oxide produced, we can get the number of moles of carbon produced. We first divide the mass by the molar mass of carbon iv oxide. The molar mass of carbon iv oxide is 44g/mol

The number of moles of carbon iv oxide is 1.24/44= 0.0282

Since there is only one carbon atom in CO2, the number of moles of carbon is same as above

The mass of carbon in the compound is simply the number of moles multiplied by the atomic mass unit. The atomic mass unit of carbon is 12. The mass of carbon in the compound is thus 12 * 0.0282= 0.338

From the number of moles of water, we can get the number of moles of hydrogen. To get the number of moles of water, we need to divide the mass of water by its molar mass. Its molar mass is 18g/mol. The number of moles here is thus 0.255/18= 0.0142 moles

But there are 2 atoms of hydrogen in 1 mole of water and thus, the number of moles of hydrogen is 2 * 0.0142= 0.0283

The mass of hydrogen is thus 0.0283 * 1 = 0.0283g

The mass of oxygen equals the mass of the compound minus that of hydrogen and that of carbon.

= 0.519 - 0.338 - 0.0283= 0.1527g

The number of moles of oxygen is the mass of oxygen divided by its atomic mass unit.

That equals 0.1527/16= 0.00954375 moles

The empirical formula can be obtained by dividing the number of moles of each by the smallest which is that oxygen 0.00954375 moles

H = 0.0284/0.00954375 = 2.97 = 3

O = 0.00954375/0.00954375= 1

C = 0.0282/0.00954375 = 2.95 = 3

The empirical formula is thus C3H3O

7 0
4 years ago
The specific heat of aluminum is 0.902 J/g C. If a one gram sample of aluminum absorbs 50 Joules of energy, how much with its te
PtichkaEL [24]
Answer:
55.4324℃
Explanation:
Q=cm△t
50J = 0.902J/(g·℃) x 1g x △t
△t =55.4324℃
7 0
3 years ago
How many molecules are in 13.2 g NO2?
DochEvi [55]

Answer:

No. of molecules= mass/molar mass ×avogadro no.

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  No. of molecules =1.59×10°23

Explanation:

5 0
3 years ago
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