Answer: The last electron will be filled in first orbital of 3p sub-shell.
Explanation: Filling of electrons in orbitals is done by using Hund's Rule.
Hund's rule states that the electron will be singly occupied in the orbital of the sub-shell before any orbital is doubly occupied.
For filling up of the electrons in Sulfur atom having 16 electrons. First 10 electrons will completely fill according to Aufbau's Rule in 1s, 2s and 2p sub-shells and last 6 electrons are the valence electrons which will be filled in the order of 3s and then 3p.
3s sub-shell will be fully filled and the orbitals of 3p sub-shell will be first singly occupied and then pairing will take place. Hence, the last electron will be filled in the first orbital of 3p-sub-shell.
Answer: Atoms of carbon and hydrogen
Explanation:
I believe there would still be 3 protons in Lithium-8 because its usually the neutrons that are affected.
Answer:
2H⁺(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2H₂O(ℓ)
Explanation:
You aren't dumb. You just need more time to learn the concepts.
There are three steps you must follow. You must write the:
- Molecular equation
- Ionic equation
- Net ionic equation
1. Molecular equation
2HBr + Sr(OH)₂ ⟶ SrBr₂ + 2H₂O
To predict the states of the substances, we must remember some solubility rules:
- HBr is a strong acid. It dissociates completely in water.
- Most hydroxides are only slightly soluble. Unless the solution is quite dilute, I would write their states in water as "(s)", i.e., a suspension of the solid in water.
- Salts containing Br⁻ are generally soluble.
Acids and bases react to give salts and water.
Thus, the molecular equation is
2HBr(aq) + Sr(OH)₂(s) ⟶ SrBr₂(aq) + 2H₂O(ℓ)
B. Ionic equation
You write all the soluble substances as ions.
2H⁺(aq)+ 2Br⁻(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2Br⁻(aq) + 2H₂O(ℓ)
C. Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
2H⁺(aq) + <u>2Br⁻(aq)</u> + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + <u>2Br⁻(aq)</u> + 2H₂O(ℓ)
The net ionic equation is
2H⁺(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2H₂O(ℓ)
