Answer:
1. Orbital diagram
2p⁴ ║ ↑↓ ║ "↑" ║ ↑
2s² ║ ↑↓ ║
1s² ║ ↑↓ ║
2. Quantum numbers
- <em>n </em>= 2,
- <em>l</em> = 1,
= 0,
= +1/2
Explanation:
The fill in rule is:
- Follow shell number: from the inner most shell to the outer most shell, our case from shell 1 to 2
- Follow the The Aufbau principle, 1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s<5f<6d<7p
- Hunds' rule: Every orbital in a sublevel is singly occupied before any orbital is doubly occupied. All of the electrons in singly occupied orbitals have the same spin (to maximize total spin).
So, the orbital diagram of given element is as below and the sixth electron is marked between " "
2p⁴ ║ ↑↓ ║ "↑" ║ ↑
2s² ║ ↑↓ ║
1s² ║ ↑↓ ║
The quantum number of an electron consists of four number:
- <em>n </em>(shell number, - 1, 2, 3...)
- <em>l</em> (subshell number or orbital number, 0 - orbital <em>s</em>, 1 - orbital <em>p</em>, 2 - orbital <em>d...</em>)
(orbital energy, or "which box the electron is in"). For example, orbital <em>p </em>(<em>l</em> = 1) has 3 "boxes", it was number from -1, 0, 1. Orbital <em>d</em> (<em>l </em>= 2) has 5 "boxes", numbered -2, -1, 0, 1, 2
(spin of electron), either -1/2 or +1/2
In our case, the electron marked with " " has quantum number
- <em>n </em>= 2, shell number 2,
- <em>l</em> = 1, subshell or orbital <em>p,</em>
= 0, 2nd "box" in the range -1, 0, 1
= +1/2, single electron always has +1/2
Answer:
C. 1.17 grams
Explanation:
- The molarity is the no. of moles of solute in a 1.0 L of the solution.
<em>M = (mass/molar mass)solute x (1000/ V)</em>
M = 0.1 M, mass = ??? g, molar mass of NaCl = 58.44 g/mol, V = 200.0 mL.
∴ mass of NaCl = (M)(molar mass)(V)/1000 = (0.1 M)(58.44 g/mol)(200.0 mL)/1000 = 1.168 g ≅ 1.17 g.
Answer:
0.37atm
Explanation:
Given parameters:
Initial pressure = 0.25atm
Initial temperature = 0°C = 273K
Final temperature = 125°C = 125 + 273 = 398K
Unknown:
Final pressure = ?
Solution:
To solve this problem, we use a derivative of the combined gas law;
=
P and T are pressure and temperature
1 and 2 are initial and final values
=
P2 = 0.37atm
The answer is A. Permeability.