the force between the electron and the proton.
a) Use F = k * q1 * q2 / d²
where k = 8.99e9 N·m²/C²
and q1 = -1.602e-19 C (electron)
and q2 = 1.602e-19 C (proton)
and d = distance between point charges = 0.53e-10 m
The negative result indicates "attraction".
the radial acceleration of the electron.
b) Here, just use F = ma
where F was found above, and
m = mass of electron = 9.11e-31kg, if memory serves
a = radial acceleration
the speed of the electron.
c) Now use a = v² / r
where a was found above
and r was given
<span> the period of the circular motion.</span>
d) period T = 2π / ω = 2πr / v
where v was found above
and r was given
Although it is omitted, the reaction equation for the decomposition of phosphorus pentachloride is:
PCl₅ → PCl₃ + Cl₂
The equilibrium constant's equation then becomes:
Kc = [PCl₃]*[Cl₂] / [PCl₅]
Kc = (0.02 * 0.02) / 0.0095
Kc = 0.042
The equilibrium constant is 0.042.
4 l ------ 3,2 g
x l ------ 8 g
x = 8 g × 4 l / 3,2 g = 10 l
Answer: 10 l of sulfuric acid is needed to produce 8 g of product.
:-) ;-)
Yes i think so
hope i helped
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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