There will be four unpaired electrons
The metal complex is [FeX₆]³⁻
X being the halogen ligand
X = F, CL, Br, and I
The oxidation of metal state is +3
The ground state configuration is
₂₆Fe =Is² 2s²2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Metal, Fe(III) ion electron configures
₂₆Fe³⁺ = Is2 2s² 2p⁶ 3s² 3p⁶ 3d⁵
Technically yes you are i think
Answer:
The minumum speed the pail must have at its highest point if no water is to spill from it
= 2.64 m/s
Explanation:
Working with the forces acting on the water in the pail at any point.
The weight of water is always directed downwards.
The normal force exerted on the water by the pail is always directed towards the centre of the circle of the circular motion.
And the centripetal force, which keeps the system in its circular motion, is the net force as a result of those two previously mentioned force.
At the highest point of the motion, the top of the vertical circle, the weight and the normal force on the water are both directed downwards.
Net force = W + (normal force)
But the speed of this motion can be lowered enough to a point where the normal force becomes zero at the moment the pail reaches the highest point of its motion. Any speed lower than this value would result in the water spilling out of the pail. The water would not be able to resist the force of gravity.
At this point of minimum velocity,
Normal force = 0
Net force = W
Net force = centripetal force = (mv²/r)
W = mg
(mv²/r) = mg
r = 0.710 m
g = 9.8 m/s²
v² = gr = 9.8 × 0.71 = 6.958
v = √(6.958) = 2.64 m/s
Hope this Helps!!!
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of
from the center of the pattern. In the formula, m is the order of the minimum,
the wavelenght,
the distance of the screen from the slit and
the width of the slit.
In our problem, the distance of the first-order band (m=1) is
. The distance of the screen is D=86 cm while the wavelength is
. Using these data and re-arranging the formula, we can find a, the width of the slit:
