first find the atomic weight of CH3 which would be
atomic weight: 12.011 (3×1.008) = 36.32 g/mol
then find the moles in the given mass
36.32 ÷ 45.7 = 0.794
I HOPE I'M NOT WRONG I HAVENT DONE CHEM IN SO LONG
For the given molecule, we are asked to give-
- The electron configuration of an isolated B atom
- The electron configuration of an isolated F atom
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride
- valence orbitals, if any, remain unhybridized on the B atom.
- The electron configuration of an isolated B atom:
as atomic number of B is 5
electronic configuration will be [He] 2s² 2p¹
- The electron configuration of an isolated F atom:
as atomic number of F is 9
electronic configuration will be [He] 2s² 2p5
- Hybrid orbitals should be constructed on the B atom to make the B–F bonds in Boron tri flouride will be sp2.
as the one s and two of p orbital from the valance shell will hybridised to make 3 hybrid orbital of B resulting in 3 B-F bonds.
- valence orbitals, if any, remain unhybridized on the B atom will be 1
To know more about hybrisisation:
brainly.com/question/23038117
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<span>Force = total mass * acceleration = 330 * 4 = 1320 N so D is correct !!
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Answer:
That is extremely confusing. Try contacting your prof.
Explanation:
