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uysha [10]
2 years ago
13

Determine the ground-state electron configuration for each of the following elements: A. chlorine ______________________________

________________________________________ B. cesium ______________________________________________________________________ C. vanadium______________________________________________________________________ D. rhenium______________________________________________________________________ 2. Predict the number of valence electrons present in each of the following atoms (include the outermost d- electrons): A. Sn ___________________________ B. La___________________________ C. Mn__________________________ D. Zn __________________________ 3. Determine the ground-state electron configuration for each of the following ions:
Chemistry
1 answer:
ki77a [65]2 years ago
8 0

In ground-state electron configuration, the<u> principal quantum number</u> (n), <u>the orbital</u> (s, p, d, or f), and the total number of <u>electrons</u> are used to represent electron configurations.

Case 1 Ground-state electron configuration of elements.

A) Chlorine: 1s²2s²2p^{6}3s²3p^{5}

B) Cesium: 1s² 2s²2p⁶ 3s²3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s¹

C) Vanadium: 1s²2s²2p⁶3s²3p⁶3d³4s²

D) Rhenium: 1s² 2s² 2p⁶ 3s² 3p⁶3d¹⁰ 4s²4p⁶ 4d¹⁰ 5s² 5p⁶ 4f¹⁴ 5d⁵ 6s²

Case 2 : Number of valence electron in atoms

A) Sn -Tin - Four valence electrons

B) La - Lanthanum - Three valence electrons

C) Mn - Manganese - Seven valence electrons

D) Zn- Zinc - Two valence electrons

Case 3 Ground-state electronic configuration of ions

A) Co⁺³ : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹3d⁵

B) Mo⁺² : 1s² 2s² 2p⁶3s²3p⁶ 4s² 3d¹⁰ 4p⁶5s² 4d²

C) Ra⁺² : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 4f¹⁴ 5s² 5p⁶ 5d¹⁰ 6s² 6p⁶

D) I⁻ : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶

Learn more about Quantum numbers here brainly.com/question/2292596

#SPJ4

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Assume that you have a cylinder with a movable piston. What would happen to the gas pressure inside the cylinder if you do the f
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Answer:

A. The pressure will increase 4 times. P₂ = 4 P₁

B. The pressure will decrease to half its value. P₂ = 0.5 P₁

C. The pressure will decrease to half its value. P₂ = 0.5 P₁

Explanation:

Initially, we have n₁ moles of a gas that occupy a volume V₁ at temperature T₁ and pressure P₁.

<em>What would happen to the gas pressure inside the cylinder if you do the following?</em>

<em />

<em>Part A: Decrease the volume to one-fourth the original volume while holding the temperature constant. Express your answer in terms of the variable P initial.</em>

V₂ = 0.25 V₁. According to Boyle's law,

P₁ . V₁ = P₂ . V₂

P₁ . V₁ = P₂ . 0.25 V₁

P₁ = P₂ . 0.25

P₂ = 4 P₁

<em>Part B: Reduce the Kelvin temperature to half its original value while holding the volume constant. Express your answer in terms of the variable P initial.</em>

T₂ = 0.5 T₁. According to Gay-Lussac's law,

\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}\\\frac{P_{1}}{T_{1}} =\frac{P_{2}}{0.5T_{1}}\\\\P_{2}=0.5P_{1}

<em>Part C: Reduce the amount of gas to half while keeping the volume and temperature constant. Express your answer in terms of the variable P initial.</em>

n₂ = 0.5 n₁.

P₁ in terms of the ideal gas equation is:

P_{1}=\frac{n_{1}.R.T_{1}}{V_{1}}

P₂ in terms of the ideal gas equation is:

P_{2}=\frac{n_{2}.R.T_{1}}{V_{1}}=\frac{0.5n_{1}.R.T_{1}}{V_{1}}=0.5P_{1}

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