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Gnesinka [82]
3 years ago
6

How is pressure affected as depth increases?

Chemistry
1 answer:
Romashka [77]3 years ago
5 0

pressure increases as depth increases.

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Can someone please help me
mrs_skeptik [129]
The answer will be D
8 0
3 years ago
A molecular orbital that decreases the electron density between two nuclei is said to be ____.
nikklg [1K]

A molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>

The bonding orbital, which would be more stable and encourages the bonding of the two H atoms into H_{2}, is the orbital that is located in a less energetic state than just the electron shells of the separate atoms. The antibonding orbital, which has higher energy but is less stable, resists bonding when it is occupied.

An asterisk (sigma*) is placed next to the corresponding kind of molecular orbital to indicate an antibonding orbital. The antibonding orbital known as * would be connected to sigma orbitals, as well as antibonding pi orbitals are known as \pi* orbitals.

Therefore,  molecular orbital that decreases the electron density between two nuclei is said to be <u>antibonding.</u>

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Hence, the correct answer will be option (b)

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To know more about molecular orbital

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6 0
2 years ago
Which statement best describes the building?
Kamila [148]

Answer:

the answer is A

Explanation:

because abiotic things are non-living things

7 0
3 years ago
Read 2 more answers
Which of the following types of radiation can be shielded by piece of paper or fabric
Keith_Richards [23]
There are 3 major types of radiation. The Alpha, Beta and Gamma radiation. It is the Alpha radiation that can be stopped by a piece of paper or fabric. Beta on the other hand can be stopped by plastic or a thin sheet of metal and Gamma by a piece of aluminum foil or lead. 
8 0
2 years ago
What is the energy released in this β − β − nuclear reaction 40 19 K → 40 20 C a + 0 − 1 e 19 40 K → 20 40 C a + − 1 0 e ? (The
Effectus [21]

<u>Answer:</u> The energy released in the given nuclear reaction is 1.3106 MeV.

<u>Explanation:</u>

For the given nuclear reaction:

_{19}^{40}\textrm{K}\rightarrow _{20}^{40}\textrm{Ca}+_{-1}^{0}\textrm{e}

We are given:

Mass of _{19}^{40}\textrm{K} = 39.963998 u

Mass of _{20}^{40}\textrm{Ca} = 39.962591 u

To calculate the mass defect, we use the equation:

\Delta m=\text{Mass of reactants}-\text{Mass of products}

Putting values in above equation, we get:

\Delta m=(39.963998-39.962591)=0.001407u

To calculate the energy released, we use the equation:

E=\Delta mc^2\\E=(0.001407u)\times c^2

E=(0.001407u)\times (931.5MeV)    (Conversion factor:  1u=931.5MeV/c^2  )

E=1.3106MeV

Hence, the energy released in the given nuclear reaction is 1.3106 MeV.

6 0
3 years ago
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