The CO ligand can easily back-bond, accepting electron density from the metal centre through pi bonds.
This is because of the empty anti-bonding orbitals.
1. metal to ligand charge transfer transition 2. ligand to metal charge transfer transition 3. spin allowed and Laporte forbidden d-d transition 4. spin allowed and Laporte allowed d-d transition
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Co
Epinephrine or also known as adrenaline contains 6 pi bonds. Epinephrine is a naturally occurring hormone that is also commercially manufactured for use as heart stimulant.
d=m/v v= (4/3)(pi)(r^3) set up equation for aluminum and lead 11.3*10^3=m/( (4/3)(pi)(ri^3)) 2.7*10^3=m/( (4/3)(pi)(ra^3)) solve for m. then set two equations equal to each other, since they have the same masses. finally, solve your new equations for ra/ri 11.3*10^3( (4/3)(pi)(ri^3)) =m 2.7*10^3( (4/3)(pi)(ra^3)) =m 11.3*10^3( (4/3)(pi)(ri^3)) = 2.7*10^3( (4/3)(pi)(ra^3)) 11.3*10^3/2.7*10^3 = (ra^3)/(ri^3) (11.3*10^3/2.7*10^3)^(1/3)=ra/ri
pi-4 is the opposite and 1/4-pi is the reciprocal
11pi/12 = pi - pi/12 cos(11pi/12) = cos(pi - pi/12) cos(a-b) = cos(a)cos(b)+sin(a)sin(b) cos(pi -pi/12) = cos(pi)cos(pi/12) + sin(pi)sin(pi/12) sin(pi)=0 cos(pi)=-1 Therefore, cos(pi -pi/12) = -cos(pi/12) pi/12=pi/3 -pi/4 cos(pi/12) = cos(pi/3 - pi/4) = cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) cos(pi/3)=1/2 sin(pi/3)=sqrt(3)/2 cos(pi/4)= sqrt(2)/2 sin(pi/4) = sqrt(2)/2 cos(pi/3)cos(pi/4)+sin(pi/3) sin(pi/4) = (1/2)(sqrt(2)/2 ) + (sqrt(3)/2)( sqrt(2)/2) = sqrt(2)/4 + sqrt(6) /4 = [sqrt(2)+sqrt(6)] /4 Therefore, cos(pi/12) = (sqrt(2)+sqrt(6))/4 -cos(pi/12) = -(sqrt(2)+sqrt(6))/4 cos(11pi/12) = -(sqrt(2)+sqrt(6))/4
Entire surface area = (2*pi*4)+(4*pi*height) = 312 (4*pi*height) = 312-(2*pi*4) (4*pi*height) = 286.8672588 Divide both sides by 4*pi to find the height:- height = 22,82817112 inches Check: (2*pi*4)+(4*pi*22.82817112) = 312 square inches Formula: (2*pi*radius2)+(diameter*pi*height) = area
cos(a)cos(b)-sin(a)sin(b)=cos(a+b) a=7pi/12 and b=pi/6 a+b = 7pi/12 + pi/6 = 7pi/12 + 2pi/12 = 9pi/12 We want to find cos(9pi/12) cos(9pi/12) = cos(3pi/4) cos(3pi/4)= cos(pi-pi/4) cos(pi)cos(pi/4)-sin(pi)sin(pi/4) cos(pi)=-1 sin(pi)=0 cos(pi/4) = √2/2 sin(pi/4) =√2/2 cos(pi)cos(pi/4)-sin(pi)sin(pi/4) = - cos(pi/4) = -√2/2
Area = (Pi/4)*(dia2) = (Pi/4)*22 = (Pi/4)*4 = Pi sq.in = 3.1416 square inches.
The four roots are cos(theta)+i*sin(theta) where theta = pi/4, 3*pi/4, 5*pi/4 and 7*pi/4.
4-(4/3)+(4/5)-(4/7)+(4/9)-(4/11)....=pi There are no non-infinite serieses known for finding pi
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tangent of pi/4 = 1
Consider pi and 4 - pi. 4 - pi + pi = 4, which is clearly rational. However, both pi and 4 - pi are irrational, as you can verify. plz to be lerning numburs Then consider pi + pi = 2pi, which is clearly irrational. The sum of two irrational numbers, therefore, may or may not be rational.
The surface area of a sphere with radius 'R' is 4(pi)R2 The volume of the same sphere is (4/3)(pi)R3 . Their ratio is (4 pi R2)/(4/3 pi R3) = (12 pi R2)/(4 pi R3) = 3/R
4 πr2π is pi... Also it is 4(pi(r^2))