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Sphinxa [80]
3 years ago
5

If electrons are collectively compressed into a very small volume (e.g., within the core of a dying white dwarf star) where quan

tum mechanical considerations become important in preventing one electron from occupying space near to a second electron (Pauli exclusion principle), what is the result?
Chemistry
1 answer:
m_a_m_a [10]3 years ago
7 0

Answer:  A degenerate pressure will generate a large force to repel further  compression.

Explanation: In the production of new stars from the core of old dying white dwarf stars, the inner parts of the star will experience contraction with the release of  heat , as they contract, their atoms will be  squeezed such that their electrons start to overlap, and because of the Pauli's exclusive principle which states that no two electrons can occupy same space, the electrons will begin to repel each other and an opposing pressure called degenerate pressure will create a force so that the electrons  cannot continually be crushed or overlap. With the limit of contraction, the outer parts of the star will expand and be repelled releasing the old stars called nebula  and creating space for the inner new stars to form.

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How many liters of 1.0 M HCl do you need to neutralize 2.0 L of 3.0 M NaOH? How many liters of 1.0 M HCl do you need to neutrali
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<u>Answer: </u>

<u>For 1:</u> The volume of HCl required is 6 L.

<u>For 2:</u> The volume of HCl required is 9 L.

<u>For 3:</u> The volume of sulfuric acid required is 4.5 L.

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2        ......(1)

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base

  • <u>For 1:</u>

We are given:

n_1=1\\M_1=1M\\V_1=?L\\n_2=1\\M_2=3.0M\\V_2=2.0L

Putting values in equation 1, we get:

1\times 1\times V_1=1\times 3\times 2\\\\V_1=6L

Hence, the volume of HCl required is 6 L.

  • <u>For 2:</u>

We are given:

n_1=1\\M_1=1M\\V_1=?L\\n_2=2\\M_2=3.0M\\V_2=1.5L

Putting values in equation 1, we get:

1\times 1\times V_1=2\times 3.0\times 1.5\\\\V_1=9L

Hence, the volume of HCl required is 9 L.

  • <u>For 3:</u>

To calculate the volume of acid, we use the equation:

N_1V_1=N_2V_2

where,

N_1\text{ and }V_1 are the normality and volume of acid

N_2\text{ and }V_2 are the normality and volume of base

We are given:

N_1=1.0N\\V_1=?L\\N_2=3.0N\\V_2=1.5L

Putting values in above equation, we get:

1.0\times V_1=3.0\times 1.5\\\\V_1=4.5L

Hence, the volume of sulfuric acid required is 4.5 L.

8 0
3 years ago
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