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.
If it’s hydraulic turbine then it’s potential and kinetic energy and if it’s a thermal process then heat energy from the fuel burnt runs the turbine
Answer : The mass defect required to release energy is 6111.111 kg
Explanation :
To calculate the mass defect for given energy released, we use Einstein's equation:
E = Energy released = 
= mass change = ?
c = speed of light = 
Now put all the given values in above equation, we get:
Therefore, the mass defect required to release energy is 6111.111 kg
<h3>Answer:</h3>
Rb = + 1
S = + 4
O = - 2
<h3>Explanation:</h3>
Oxidation states of the elements were calculated keeping in mind the basic rules of assigning oxidation states which included assignment of +1 charge to first group elements i.e. Rubidium (Rb) and assignment of -2 charge to Oxygen atom. Then the oxidation state of Sulfur was calculated as follow,
Rb₂ + S + O₃ = 0
Above zero (0) means that the overall molecule is neutral.
Putting values of Rb and O,
(+1)₂ + S + (-2)₃ = 0
(+2) + S + (-6) = 0
+2 + S - 6 = 0
S - 6 = -2
S = -2 + 6
S = + 4
Answer:
Explanation:
In theory, not much of anything. The vast majority of nitrates are water soluble. Aside, not sure what chemistry level you are at but you will probably be asked to know or memorize some solubility rules. This, for lack of a better phrase, Nitrate rule, is near spot on. With one exception—a rare one—all metal cationic nitrates are soluble in water. All of them. So, assuming you are talking about aqueous, water-based solutions of these salts and mixing them together, I expect nothing to occur. Both solutions, I believe are colorless in water and will thus remain so. If you had say a solution of Iron (III) nitrate and copper (II) nitrate, slightly different story. Both are colorful solutions and I would think you might see blending of colors but no reaction; no precipitate will form. You will probably learn about markers of a chemical reaction. One of these is a color change. Note, you should read this as a change of color from what you previously had. Going from red to blue or colorless to colored (or vice versa) is a strong indication of a reaction (e. g. evidence of bond-breaking and bond-formation). The mere mixing of colors does not constitute a chemical reaction.
