Atomic or hybrid orbital on the central br atom makes up the sigma bond between this br and an outer f atom in bromine trifluoride, brf3 is sp2 hybridization
Trigonal hybridization is another name for sp2 hybridization. It entails combining one's' orbital with two 'p' orbitals of equal energy to create a new hybrid orbital known as sp2. A trigonal symmetry combination of s and p orbitals that is kept at 120
One of the hybrid orbitals formed when one s orbital and two p orbitals are mathematically merged to form three new equivalent orbitals orientated toward the corners of a triangle is sp2 hybridization.
The only feasible molecule geometry for sp2 hybridized center atoms is trigonal planar. When all of the bonds are in place, the shape is trigonal planar as well.
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Answer:
Explanation:
The given reaction equation is:
2A + 4B → C + 3D
We know the mass of compound A in the reaction above. We are to find the mass of compound D.
We simply work from the known mass to calculate the mass of the unkown compound D
Using the mole concept, we can find the unknown mass.
Procedures
- We first find the molar mass of the compound A from the atomic units of the constituent elements.
- We then use the molar mass of A to calculate its number of moles using the expression below:
Number of moles of A = 
- Using the known number of moles of A, we can work out the number of moles of D.
From the balanced equation of the reaction, it is shown that:
2 moles of compound A was used up to produced 3 moles of D
Then 
x number of moles of A would give the number of moles of D
- Now that we know the number of moles of D, we can find its mass using the expression below:
Mass of D = number of moles of D x molar mass of D
Answer:
D
Explanation:
Meteorologists use this symbol to show that the front is a stationary front.
Answer:
We can use heat = mcΔT to determine the amount of heat, but first we need to determine ΔT. Because the final temperature of the water is 55°C and the initial temperature is 20.0°C, ΔT is as follows:
ΔT = Tfinal − Tinitial = 55.0°C − 20.0°C = 35.0°C
given the specific heat of water as 1 cal/g·°C. Substitute the known values into heat = mcΔT and solve for amount of heat:
= heat=(75.0 g)(1 cal/ g· °C )(35.0°C) =
= 75x1x35=2625 cal
