Answer:
the emission of visible light by a body, caused by its high temperature.Compare luminescence.
the light produced by such an emission.
the quality of being incandescent.
Answer:materials that impede the free flow of electrons from atom to atom and molecule to molecule.
In rubidium oxide - Rb₂O , the ions are Rb⁺ and O²⁻
Rb is a group one element with one valence electron. To become stable it loses its outer electron to gain a complete outer shell.
Electronic configuration of Rb is - 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 5s¹
Once it loses its valence electron the configuration is;
- 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶
The noble gas with this configuration is Krypton - Kr
Oxygen electron configuration is 1s² 2s² 2p⁴
Once it gains 2 electrons the configuration is - 1s² 2s² 3p⁶
The noble gas with this configuration is Neon - Ne
Answer:
This question is incomplete
Explanation:
This question is incomplete, however, the element that has 52 electrons only is Tellurium (Te) and when the electronic configuration of elements with more than 52 electrons are written, the 52nd electron is indicated/paired the same way the 52nd electron of Te is indicated/paired. Hence, while writing the electronic configuration of Te, it is written as
[Kr] 4d¹⁰ 5s² 5p⁴ where [Kr] is the electronic configuration of krypton. Based on this, we can deduce that the 52nd electron will be in the first orbital of the P subshell (as attached in the picture). This is because when indicating the electrons in the subshell, one electron will be spread across each orbital and if any electron is still remaining, it will be added starting from to the first orbital of the subshell, however no two electrons in an orbital in a subshell can have the same spin and hence must face opposite direction based on pauli's exclusion principle (as seen in attached); thus for the 5p-orbital of elements with 52 or more electrons, when one electron each is represented in each box (3 boxes in total) in the 5p-orbital, the remaining electron is paired with the the first electron in the first box of the 5p-orbital
Answer:
The total mass of D-Glucose dissolved in a 2μL aliquot is 1 E-4 g
Explanation:
providing a solution to 5% weight-volume as found in commerce:
⇒ % 5 = (5g d-glucose/ 100 mL sln)×100
⇒ 0.05 = g C6H12O6/mL sln
⇒ g C6H12O6 = (2 μL sln)×(0.001 mL/μL)×(0.05 g C6H12O6/mL sln)
⇒ g C6H12O6 = 1 E-4 g C6H12O6