Answer:
[He]: 2s² 2p⁵.
[Ne]: 3s².
[Ar]: 4s² 3d¹⁰ 4p².
[Kr]: 5s² 4d¹⁰ 5p⁵.
[Xe]: 6s² 4f¹⁴ 5d¹⁰ 6p².
Explanation:
- Noble elements are used as blocks in writing the electronic configuration of other elements as they are stable elements.
He contains 2 electrons fill 1s (1s²).
So, [He] can be written before the electronic configuration of 2s² 2p⁵.
Ne contains 10 electrons fill (1s² 2s² 2p⁶).
So, [Ne] can be written before the electronic configuration of 3s².
Ar contains 18 electrons is configured as ([Ne] 3s² 3p⁶).
So, [Ar] can be written before the electronic configuration of 4s² 3d¹⁰ 4p².
Kr contains 36 electrons is configured as ([Ar] 4s² 3d¹⁰ 4p⁶).
So, [Kr] can be written before the electronic configuration of 5s² 4d¹⁰ 5p⁵.
Xe contains 54 electrons is configured as ([Kr] 5s² 4d¹⁰ 5p⁶).
So, [Xe] can be written before the electronic configuration of 6s² 4f¹⁴ 5d¹⁰ 6p².
Answer: The two main types of chemical bonds are ionic and covalent bonds. An ionic bond essentially donates an electron to the other atom participating in the bond, while electrons in a covalent bond are shared equally between the atoms. The only pure covalent bonds occur between identical atoms.Jan 23, 2020
Explanation: Hope this helped!!!!
Answer:
The answer to your question is V = 0.108 L or 108 ml
Explanation:
Data
Volume = ?
mass = 0.405 g
Temperature = 273°K
Pressure = 1 atm
Process
1.- Convert mass of Kr to moles
83.8 g of Kr -------------------- 1 mol
0.405 g ------------------- x
x = (0.405 x 1) / 83.8
x = 0.0048 moles
2.- Use the Ideal gas law to solve this problem
PV = nRT
- Solve for V
V = nRT / P
- Substitution
V = (0.0048)(0.082)(273) / 1
- Simplification
V = 0.108 / 1
- Result
V = 0.108 L
