What orbitals are used to form the 10 sigma bonds in propane (ch3ch2ch3)? label each atom with the appropriate hybridization. drag the appropriate labels to their respective targets. note: not all labels will be used?

Mixing of pure orbitals having nearly equal energy to form equal number of completely new orbitals is said to be hybridization.

For the compound, the electronic configuration of the atoms, carbon and hydrogen are:

Carbon (atomic number=6): In ground state=

In excited state:

Hydrogen (atomic number=1):

All the bonds in the compound is single bond(-bond) that is they are formed by head on collision of the orbitals.

The structure of the compound is shown in the image.

The Carbon-Hydrogen bond is formed by overlapping of s-orbital of hydrogen to p-orbital of carbon.

In order to complete the octet the required number of electrons for carbon is 4 and for hydrogen is 1. So, the electron in of hydrogen will overlap to the 2p^{3}-orbital of carbon.

Thus, the hybridization of Hydrogen is -hybridization and the hybridization of Carbon is -hybridization.

The hybridization of each atom is shown in the image.

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.