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3 years ago

Classify each orbital diagram for ground-state electron configurations by the rule or principle it violates.

1 answer:

6 0

Thank you for posting your question here at brainly. I hope the answer will help. Please see below:

Aufbau violation:
4s ↑↓ 3d ↑↓ ↑↓ ↑↓ ↑↓ ↑↓ 5s ↑
3s ↑ 3p ↑ ↑ ↑ 
Hund violation: 
4s ↑↓ 3d ↑↓ ↑ ↓ ↑ ↑ 
3s ↑↓ 3p ↑↓ ↑↓
Pauli violation
2s ↑↑ 2p ↑ ↑ ↑

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For the reaction: C(s) + 2H2(g) → CH4(g), how many moles of H2 are required to

Sliva [168]

You would need 20 moles of H2 to get 10 moles of CH4 because the ration of H2 to CH4 is 2:1

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2 years ago

Which is NOT an example of mechanical weathering?

coldgirl [10]

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2 years ago

Which of the following cannot be separated by physical means?

tatuchka [14]

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2 years ago

A sample of neon gas at a pressure of 1.08 atm fills a flask with a volume of 250 mL at a temperature of 24.0 °C. If the gas is

musickatia [10]

Answer:

124.91mL

Explanation:

Given parameters:

P₁ = 1.08atm

V₁ = 250mL

T₁ = 24°C

P₂ = 2.25atm

T₂ = 37.2°C

V₂ = ?

Solution:

To solve this problem, we are going to apply the combined gas law;

$\frac{P_{1} V_{1} }{T_{1} } = \frac{P_{2} V_{2} }{T_{2} }$

P, V and T represents pressure, volume and temperature

1 and 2 delineates initial and final states

Convert the temperature to kelvin;

T₁ = 24°C, T₁ = 24 + 273 = 297K

T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K

Input the variables and solve for V₂

$\frac{1.08 x 250}{298} = \frac{2.25 x V_{2} }{310.2}$

V₂ = 124.91mL

6 0

3 years ago

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

2 years ago