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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: 1s²2s²23s²3
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
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Answer:
Empirical formula is C3H3O
Molecular formula C9H9O3
Explanation:
From the question given, we obtained the following data:
Carbon = 63.15%
Hydrogen = 5.30%
Oxygen = 31.55%
We can obtain the empirical and molecular formula by doing the following as illustrated in the attached file. Please see attachment for explanation.
