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2 years ago

Please help!!!!

Chemistry

2 answers:

Nat2105 [25]2 years ago

8 0

Answer:

When an atom gains or loses energy, the energy of an electron can change.

An electron in an atom can move from one energy level to another when the atom gains or loses energy.

Explanation:

The possible energies that electrons in an atom

can have are called energy levels.

• An electron cannot exist between energy levels.

https://1.cdn.edl.io/tTlW7xRtvD62xSe7RcZlJr7kSR7XsL93akcgJkbGJBNNcpwY.pdf

*this link can also help* brainly ist plz

Ivan2 years ago

7 0

Answer: Option (c) is the correct answer.

Explanation:

When energy is supplied to an electron then upon absorption of energy it tends to move from a lower energy level to higher energy level.

And, when this electron moves from higher energy level to lower energy level then excess of energy absorbed by it is released in the form of visible light or radiation.

This emission of energy is also responsible for imparting color to a compound.

Thus, we can conclude that when an electron loses its energy then it may drop to a lower energy level.

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Aclosed system contains an equimolar mixture of n-pentane and isopentane. Suppose the system is initially all liquid at 120°C an

garri49 [273]

Explanation:

The given data is as follows.

T = $120^{o}C$ = (120 + 273.15)K = 393.15 K,

As it is given that it is an equimolar mixture of n-pentane and isopentane.

So, $x_{1}$ = 0.5 and $x_{2}$ = 0.5

According to the Antoine data, vapor pressure of two components at 393.15 K is as follows.

$p^{sat}_{1}$ (393.15 K) = 9.2 bar

$p^{sat}_{1}$ (393.15 K) = 10.5 bar

Hence, we will calculate the partial pressure of each component as follows.

$p_{1} = x_{1} \times p^{sat}_{1}$

= $0.5 \times 9.2 bar$

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and, $p_{2} = x_{2} \times p^{sat}_{2}$

= $0.5 \times 10.5 bar$

= 5.25 bar

Therefore, the bubble pressure will be as follows.

P = $p_{1} + p_{2}$

= 4.6 bar + 5.25 bar

= 9.85 bar

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$y_{1} = \frac{p_{1}}{p}$

= $\frac{4.6}{9.85}$

= 0.467

and, $y_{2} = \frac{p_{2}}{p}$

= $\frac{5.25}{9.85}$

= 0.527

Calculate the dew point as follows.

$y_{1}$ = 0.5, $y_{2}$ = 0.5

$\frac{1}{P} = \sum \frac{y_{1}}{p^{sat}_{1}}$

$\frac{1}{P} = \frac{0.5}{9.2} + \frac{0.5}{10.2}$

$\frac{1}{P}$ = 0.101966 $bar^{-1}$

P = 9.807

Composition of the liquid phase is $x_{i}$ and its formula is as follows.

$x_{i} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{9.2}$

= 0.5329

$x_{z} = \frac{y_{i} \times P}{p^{sat}_{1}}$

= $\frac{0.5 \times 9.807}{10.5}$

= 0.467

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