Answer:
C. The lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, all of which have the same spin, in degenerate orbitals.
Explanation:
The Hund's rule is used to place the electrons in the orbitals is it states that:
1. Every orbital in a sublevel is singly occupied before any orbital is doubly occupied;
2. All of the electrons in singly occupied orbitals have the same spin.
So, the electrons first seek to fill the orbitals with the same energy (degenerate orbitals) before paring with electrons in a half-filled orbital. Orbitals doubly occupied have greater energy, so the lowest-energy electron configuration of an atom has the maximum number of unpaired electrons, and for the second statement, they have the same spin.
The other alternatives are correct, but they're not observed by the Hund's rule.
Answer:
Grey precipitate implies the presence of silver ions
Yellow precipitate implies the presence of lead II ions
Explanation:
Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.
In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.
The empirical formula : MnO₂.
<h3>Further explanation</h3>
Given
632mg of manganese(Mn) = 0.632 g
368mg of oxygen(O) = 0.368 g
M Mn = 55
M O = 16
Required
The empirical formula
Solution
You didn't include the pictures, but the steps for finding the empirical formula are generally the same
- Find mol(mass : atomic mass)
Mn : 0.632 : 55 = 0.0115
O : 0.368 : 16 =0.023
- Divide by the smallest mol(Mn=0.0115)
Mn : O =
The empirical formula : MnO₂
Answer: 10
Explanation:
The detailed solution is contained in the image attached. The molar mass of hydrated and anhydrous salts are obtained and the number of moles of hydrated and hydrated salts are equated. The masses of hydrated and anhydrous salts are gives in the question and are simply substituted accordingly. This can now be used to obtain the number of molecules of water of crystallization as required in the question.
Answer:
i think that the answer is B
Explanation:
I DON
