is the type of orbital hybridization of a central atom that has one lone pair and bonds to four 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.
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Answer:
Mole fraction for solute = 0.1, or 10%
Molality = 6.24 mol/kg
Explanation:
22.3% by mass → In 100 g of solution, we have 22.3 g of HCOOH
Mass of solution = 100 g
Mass of solute = 22.3 g
Mass of solvent = 100 g - 22.3g = 77.7 g
Let's convert the mass to moles
22.3 g . 1mol/ 46 g = 0.485 moles
77.7 g. 1mol / 18 g = 4.32 moles
Total moles = 4.32 moles + 0.485 moles = 4.805 moles
Xm for solute = 0.485 / 4.805 = 0.100 → 10%
Molality → mol/ kg → we convert the mass of solvent to kg
77.7 g. 1 kg / 1000g = 0.0777 kg
0.485 mol / 0.0777 kg = 6.24 m
Answer: orientation , energy , frequency
Explanation:
According to the collision theory , the number of collisions that take place per unit volume of the reaction mixture is called collision frequency. The effective collisions are ones which result into the formation of products.
Effective collisions depends on the following two factors:-
1. Orientation factor: The colliding molecules must have proper orientation at the time of collision to result into formation of products.
2. Energy factor: For collision to be effective, the colliding molecules must have energy more than a particular value called as threshold energy.
Answer:
Question Clear
Explanation:
Make sure your question is clear.
In chemical reactions, the actual yield is not the same as the expected yield . Actual yield is lower than the theoretical yield . Then we have to find the yield percentage. To see what percentage of the theoretical yield is the actual yield.
Percent yield = actual yield / theoretical yield x 100%
Percent yield = 24.6/55.9 x100%
Percent yield = 44%