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
Option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
<h3>
Reaction between oxygen and ethene</h3>
Ethene (C2H4) burns in the presence of oxygen (O2) to form carbon dioxide (CO2) and water (H2O) along with the evolution of heat and light.
C₂H₄ + 3O₂ ----- > 2CO₂ + 2H₂O
from the equation above;
3 moles of O₂ ---------> 2(18 g) of water
3.5 moles of O₂ ----------> x
![x = 3.2 \times [\frac{2 \ moles \ H_2O}{3 \ moles \ O_2} ] \times[ \frac{18.02 \ g \ H_2O}{1 \ mole \ H_2O} ]](https://tex.z-dn.net/?f=x%20%3D%203.2%20%5Ctimes%20%5B%5Cfrac%7B2%20%5C%20moles%20%5C%20H_2O%7D%7B3%20%5C%20moles%20%5C%20O_2%7D%20%20%5D%20%5Ctimes%5B%20%5Cfrac%7B18.02%20%5C%20g%20%5C%20H_2O%7D%7B1%20%5C%20mole%20%5C%20H_2O%7D%20%5D)
Thus, option C is the correct set of the problem for mass of water produced by 3.2 moles of oxygen and an excess ethene.
Learn more about reaction of ethene here: brainly.com/question/4282233
#SPJ1
Answer:
a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+
b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -
c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.
d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base
e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt
f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt
g) NaCl -> Na+ + Cl- ionization of the salt, Sodium Chloride
Explanation:
Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.
Answer:
25 mL
Explanation:
Step 1: Given data
- Concentration of the concentrated solution (C₁): 2 M
- Volume of the concentrated solution (V₁): ?
- Concentration of the diluted solution (C₂): 0.1 M
- Volume of the diluted solution (V₂): 0.500 L
Step 2: Calculate the volume of the concentrated NaCl solution
We will use the dilution rule.
C₁ × V₁ = C₂ × V₂
V₁ = C₂ × V₂ / C₁
V₁ = 0.1 M × 0.500 L / 2 M
V₁ = 0.025 L = 25 mL
