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Whitepunk [10]
1 year ago
8

The following do not represent valid ground-state electron configurations for an atom either because they violate the Pauli excl

usion principle or because orbitals are not filled in order of increasing energy. Indicate which of these two principles is violated in each example or whether both or neither are violated.
Part 1
1s22s23s2
A) The orbitals are not filled in order of increasing energy.
B) The Pauli exclusion principle is violated.
C) Orbitals are not filled in order of increasing energy and the Pauli exclusion principle is violated.
D) The ground-state electron configuration is valid.
Part 2
[Rn]7s26d4
A) The orbitals are not filled in order of increasing energy.
B) The Pauli exclusion principle is violated.
C) Orbitals are not filled in order of increasing energy and the Pauli exclusion principle is violated.
D) The ground-state electron configuration is valid.
Part 3
[Ne]3s23d5
A) The orbitals are not filled in order of increasing energy.
B) The Pauli exclusion principle is violated.
C) Orbitals are not filled in order of increasing energy and the Pauli exclusion principle is violated.
D) The ground-state electron configuration is valid.
Chemistry
2 answers:
svetoff [14.1K]1 year ago
7 0

Answer:

<em>For both cases the answer is C</em>

Explanation:  

We can see that the orbitals are not filled in the order of increasing energy and the Pauli exclusion principle is violated because it does not follow the correct order of the electron configuration; In the first exercise after the 2s2 orbital, the 2p2 orbital follows.

For the second exercise, you must start in order with level 1 and correctly filling each of the sublevels corresponding to each level until reaching level 7 and thus completing the desired number of electrons.

Kipish [7]1 year ago
7 0

Answer:

for both parts the ans is c

hope helps you

have a great day

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Vlad1618 [11]

Answer:

The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of FeNCS^{2+} as indicated on the spectrum. The reaction between the Fe^{3+} and the SCN^{-} are both clear liquids that form the orange liquid product  FeNCS^{2+} which creates the absorbance spectrum. Because the color of the solution is orange, it reflects this and similar colors while absorbing blueish hues. We can find the absorption of only the FeNCS^{2+} by pre-rinsing the cuvette with each solution we intend to measure before placing it in the spectrometer. Also, wipe each cuvette with a kimwipe to remove all fingerprints that could effect the data collection.

Explanation:

The cuvette was blank with the solution so that the spectrometer will only read the solute absorbance. This also ensures that the spectrometer will ignore other absorbance fluctuations that normally occur due to the chemical make-up of water. The spectrometer only considered the absorbance of FeNCS^{2+} as indicated on the spectrum.

3 0
1 year ago
The Lewis dot model of a molecule is shown. A molecule is shown with atoms arranged in the order HCCH. There are three horizonta
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Answer:

B)The octet of carbon atom remains incomplete in the molecule.

Explanation:

8 0
2 years ago
In this experiment, 0.170 g of caffeine is dissolved in 10.0 ml of water. the caffeine is extracted from the aqueous solution th
zmey [24]

solution:

Weight of caffeine is W = 0.170 gm.

Volume of water is V= 10 ml

Volume of methylene chloride which extracted caffeine is v= 5ml

No of portions n=3

Distribution co-efficient= 4.6

Total amount of caffeine that can be unextracted is given by

w_{n}=w\times[\frac{k_{Dx}v}{k_{Dx}v+v}]^n\\w_{3}=0.170[\frac{4.6\times10}{(4.6\times10+5)}]^3\\=0.170[\frac{46}{46+5}]^3\\=0.170[\frac{46}{51}]^3\\=0.170[\frac{97336}{132651}]\\=0.170\times0.734=0.125gms

amount of caffeine un extracted is 0.125gms

amount of caffeine extracted=0.170-0.125

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6 0
2 years ago
The graph shows the distribution of energy in the particles of two gas samples at different temperatures, T1 and T2. A, B, and C
coldgirl [10]

Answer:

  • <u><em>The average speed of gas particles at T</em></u><em><u>₂</u></em><u><em> is lower than the average speed of gas particles at T</em></u><em><u>₁</u></em><u><em>.</em></u>

Explanation:

<em>Particles A and C</em> are shown as if they are on the same vertical line, which means with the same kinetic energy. Both particle A and C are to the lett of <em>particle B</em>, which means that the formers have a lower kinetic energy than the latter.

Since the likelyhood of a particle to participate in the reaction increases with the kinetic energy, particle B is more likely to participate in the reaction than particles A and C. Hence, the first choice is incorrect.

The graph, although not perfectly symmetrical, does show a bell shape, hence there are many particles will low kinetic energy and many particles with high kinetic energy. You cannot assert that most of the particles of the two gases have high high speeds. Hence, second statement is incorrect, too.

At high values of kinetic energy (toward the right of the curve), the line labeled T₁ is higher than the line labeled T₂, meaning that at T₁ more particles have an elevated kinetic energy than the number of particles that have an elevated kinetic energy at T₂.

On the other hand, at low values of kinetic energy (toward the left of the curve) the line T₂ is higher than the line T₁, meaning that at T₂ more particles have a low kinetic energy than the number of particles that have low kinetic energy at T₁.

Hence, the last two paragraphs are telling that the average kinetic energy of gas particles at T₂ is is lower than the average kinetic energy of gas particles at T₁.

Since the average speed is proportional the the square root of the temperature, the same trend for the average kinetic energy is true for the average speed, and you conclude that the last statement is true: "The average speed of gas particles at T₂ is lower than the average speed of gas particles at T₁".

Since more particles at T₁ have high kinetic energy than the number of particles at T₂ that have a high kinetic energy, more particles of gas at T₁ are likely to participate in the reaction  than the gas at T₂, and the third statement is incorrect.

7 0
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scoray [572]

Answer:

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6 0
2 years ago
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