Answer:
Explanation:
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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:
4.75 is the equilibrium constant for the reaction.
Explanation:

Equilibrium concentration of reactants :
![[CO]=0.0590 M,[H_2O]=0.00600 M](https://tex.z-dn.net/?f=%5BCO%5D%3D0.0590%20M%2C%5BH_2O%5D%3D0.00600%20M)
Equilibrium concentration of products:
![[CO_2]=0.0410 M,[H_2]=0.0410 M](https://tex.z-dn.net/?f=%5BCO_2%5D%3D0.0410%20M%2C%5BH_2%5D%3D0.0410%20M)
The expression of an equilibrium constant is given by :
![K_c=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)


4.75 is the equilibrium constant for the reaction.
Answer:
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Explanation:
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It is given that the person weighs 62 kg = 62,000 g
Natural abundances in mass percent are:
O = 65%
C = 18%
H = 10%
N = 3.0%
Ca = 1.6%
P = 1.2%
Corresponding weights of the elements are:
O = 65/100 * 62000 g = 40.30 * 10^3 g
C = 18/100 * 62000 g = 11.16 * 10^3 g
H = 10/100 * 62000 g = 62.00 * 10^2 g
N = 3.0/100 * 62000 g = 18.60 * 10^2 g
Ca = 1.6/100 * 62000 g = 9.92 * 10^2 g
P = 1.2/100 * 62000 g = 7.44 * 10^2 g