Tetrahedral arrangement is resulted upon mixing one s and three p atomic orbitals, resulting in 4 hybridized 
orbitals → hybridization.
<h3>What is
orbital hybridization?</h3>
In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.
For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.
Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.
Learn more about Hybridization
brainly.com/question/22765530
#SPJ4
The rate constant of a reaction : 8.3 x 10⁻⁴
<h3>Further explanation</h3>
Given
rate = 1 x 10⁻² (mol/L)/s, [A] is 2 M, [B] is 3 M, m = 2, and n = 1
Required
the rate constant
Solution
For aA + bB ⇒ C + D
Reaction rate can be formulated:
![\large{\boxed{\boxed{\bold{r~=~k.[A]^a[B]^b}}}](https://tex.z-dn.net/?f=%5Clarge%7B%5Cboxed%7B%5Cboxed%7B%5Cbold%7Br~%3D~k.%5BA%5D%5Ea%5BB%5D%5Eb%7D%7D%7D)
the rate constant : k =
![\tt k=\dfrac{rate}{[A]^m[B]^n}\\\\k=\dfrac{1.10^{-2}}{2^2\times 3^1}\\\\k=8.3\times 10^{-4}](https://tex.z-dn.net/?f=%5Ctt%20k%3D%5Cdfrac%7Brate%7D%7B%5BA%5D%5Em%5BB%5D%5En%7D%5C%5C%5C%5Ck%3D%5Cdfrac%7B1.10%5E%7B-2%7D%7D%7B2%5E2%5Ctimes%203%5E1%7D%5C%5C%5C%5Ck%3D8.3%5Ctimes%2010%5E%7B-4%7D)
Answer : B) In step 2, there was a chemical change which was observed in sugar.
Explanation : In A step 2 there was a physical change that was seen. By just boiling the dissolved salt solution salt was obtained. Therefore, it is a physical change. In B step 2 there was a chemical change seen as sugar solution was thickened and turned brown. It was not obtained in its original form; there was a chemical reaction that took place during sugar evaporation. As chemical change is the one where the reaction is irreversible.
Therefore only in B step 2 there was a chemical change that was observed.
The Answer is A
Hope this helped ;)
