Hey there!
No of hybrid orbitals , H = ( V +S - C + A ) / 2
Where H = no . of hybrid orbitals
V = Valence of the central atom = 5
S = No . of single valency atoms = 4
C = No . of cations = 1
A = No . of anions = 0
For PCl4 +
Plug the values we get H = ( 5+4-1+0) / 2
H = 4 ---> sp3 hybridization
sp3 hybrid orbitals are used by phosphorous in the PCl4+ cations
Answer C
Hope that helps!
Reduction <span>always results in a lowering of the oxidation number. The reaction of the system above is written as:
</span><span>Cu2+(aq) + Fe(s) --> Cu(s) + Fe2+(aq)
</span>
From the reaction, we see that copper goes from the +2 to a neutral charge. Lowering of the oxidation number happens so this is the element that is being reduced.
Answer:
for the balanced equation
The concentration of the solution reduces and the number of moles of solute isn't affected.
Data;
- V1 = 50mL
- C1 = 12.0M
- V2 = 200mL
- C2 = ?
<h3>Facts about the diluted solution</h3>
1. When the solution is diluted, the concentration changes and this time, the concentration reduces.
Using dilution formula
The concentration of the solution reduces.
2. The number of moles remains the same.
When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.
Learn more on concentration of a solution here;
brainly.com/question/2201903
