is the orbital hybridization of a central atom that has one lone pair and bonds to three other atoms.
<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
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The oxidation state is zero for Fe (s)
The solution would be like
this for this specific problem:
<span>(78.6 kJ) / (92.0 g /
(46.0684 g C2H5OH/mol)) = 39.4 kJ/mol </span>
<span>39.3 </span>
So the approximate molar
heat of vaporization of ethanol in kJ/mol is 39.3.
I hope this answers your question.
Answer:
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
Explanation:
Step 1: Data given
The temperature of a gas = 25.0°C
AT 25 °C the gas occupies a volume of 10.0L and a pressure of 667 torr.
The volume reduces to 7.88 L but the temperature stays constant.
Step 2: Boyle's law
(P1*V1)/T1 = (P2*V2)/T2
⇒ Since the temperature stays constant, we can simplify to:
P1*V1 = P2*V2
⇒ with P1 = the initial pressure 667 torr
⇒ with V1 = the initial volume = 10.0 L
⇒ with P2 = the final pressure = TO BE DETERMINED
⇒ with V2 = the final volume = 7.88L
P2 = (P1*V1)/V2
P2 = (667*10.0)/7.88
P2 = 846 torr
The pressure, when the volume is reduced to 7.88L, is 846 torr (option A)
