File:///Users/user/Downloads/Chemistry%20Exam%20Study%20Guide%20Fall%202012.pdf
Idk if this will help you, cuz it says what question are going to be in ur final
Answer:
(a) 1s² 2s² 2p⁶ 3s² 3p⁴
(b) 1s² 2s² 2p⁶ 3s² 3p⁵
(c) sp³
(d) No valence orbital remains unhybridized.
Explanation:
<em>Consider the SCl₂ molecule. </em>
<em>(a) What is the electron configuration of an isolated S atom? </em>
S has 16 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁴.
<em>(b) What is the electron configuration of an isolated Cl atom? </em>
Cl has 17 electrons. Its electron configuration is 1s² 2s² 2p⁶ 3s² 3p⁵.
<em>(c) What hybrid orbitals should be constructed on the S atom to make the S-Cl bonds in SCl₂? </em>
SCl₂ has a tetrahedral electronic geometry. Therefore, the orbital 3s hybridizes with the 3 orbitals 3 p to form 4 hybrid orbital sp³.
<em>(d) What valence orbitals, if any, remain unhybridized on the S atom in SCl₂?</em>
No valence orbital remains unhybridized.
Answer:
272.31× 10²³ atom of strontium
Explanation:
Given data:
Number of moles of strontium = 45.22 mol
Number of atoms = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance. The number 6.022 × 10²³ is called Avogadro number.
1 mole = 6.022 × 10²³ atoms
45.22 mol × 6.022 × 10²³ atom / 1 mol
272.31× 10²³ atom
Answer:
2.1 × 10⁻¹ M
2.0 × 10⁻¹ m
Explanation:
Molarity
The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:
3.9 g × (1 mol/93.13 g) = 0.042 mol
The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:
M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M
Molality
The moles of solute are 0.042 mol.
The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:
200 mL × 1.05 g/mL = 210 g = 0.210 kg
The molality of aniline is:
m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m
