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omeli [17]
3 years ago
11

How many electron shells/orbitals will be around the nucleus of an atom of aluminum?

Chemistry
2 answers:
AnnyKZ [126]3 years ago
7 0
C.
There is 13 electrons in a aluminum atom. Looking at the picture, you can see there are two electrons in shell one, eight in shell two, and three in shell three.
Romashka-Z-Leto [24]3 years ago
6 0
C would be the correct answer I believe
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Will vote brainliest and 5 star
svetoff [14.1K]

Answer:

it is B

Explanation:

hope it helps you

4 0
3 years ago
Which characteristic of a substance is considered a chemical property?
slavikrds [6]
B. It’s reactivity.
8 0
1 year ago
What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weigh
Natasha_Volkova [10]

Answer:

1088.89 Pa

Explanation:

According to the Newton's second law of motion:-

Force=Mass\times Acceleration

Mass = 50.0 kg

Acceleration = g = 9.81 m/s²

So,  

Force=50.0\times 9.8\ kgm/s^2

Force = 490 N

Area of the base = length\times breath = 0.500\times 0.900 m² = 0.45 m²

<u>Pressure = Force/Area = \frac{490\ N}{0.45\ m^2} = 1088.89 Pa</u>

8 0
3 years ago
Which of the following would you except to see in the death of a star that is less than 0.5 solar mass
Ket [755]

B. White Dwarf.

<h3>Explanation</h3>

The star would eventually run out of hydrogen fuel in the core. The core would shrink and heats up. As the temperature in the core increases, some of the helium in the core will undergo the triple-alpha process to produce elements such as Be, C, and O. The triple-alpha process will heat the outer layers of the star and blow them away from the core. This process will take a long time. Meanwhile, a planetary nebula will form.

As the outer layers of gas leave the core and cool down, they become no longer visible. The only thing left is the core of the star. Consider the Chandrasekhar Limit:

Chandrasekhar Limit: 1.4 \;M_\odot.

A star with core mass smaller than the Chandrasekhar Limit will not overcome electron degeneracy and end up as a white dwarf. Most of the outer layer of the star in question here will be blown away already. The core mass of this star will be only a fraction of its 0.5 \;M_\odot, which is much smaller than the Chandrasekhar Limit.

As the star completes the triple alpha process, its core continues to get smaller. Eventually, atoms will get so close that electrons from two nearby atoms will almost run into each other. By Pauli Exclusion Principle, that's not going to happen. Electron degeneracy will exert a strong outward force on the core. It would balance the inward gravitational pull and prevent the star from collapsing any further. The star will not go any smaller. Still, it will gain in temperature and glow on the blue end of the spectrum. It will end up as a white dwarf.

7 0
2 years ago
Match the atoms to their correct electron configuration. <br> (Answer choices in photo)
zysi [14]

Answer:

The answer to your questions are given below

Explanation:

The answer to the questions given above can simply be obtained by writing the electronic configuration of each atom.

The electronic configuration of each atoms can be written as follow:

Fluorine, F (9) => 1s² 2s²2p⁵

Sodium, Na (11) => [Ne] 3s¹

Helium, He (2) => 1s²

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Nitrogen, N (7) => 1s² 2s²2p³

Sulphur, S (16) => [Ne] 3s²3p⁴

6 0
3 years ago
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