Answer: 50%
Explanation:
The number of electron pairs are 2 for hybridization to be
and the electronic geometry of the molecule will be linear.
1. percentage of s character in sp hybrid orbital =![\frac{\text {number of s orbitals}}{\text {total number of orbitals}}=\frac{1}{2}\times 100=50\%](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bnumber%20of%20s%20orbitals%7D%7D%7B%5Ctext%20%7Btotal%20number%20of%20orbitals%7D%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20100%3D50%5C%25)
2. percentage of s character in
hybrid orbital =![\frac{\text {number of s orbitals}}{\text {total number of orbitals}}=\frac{1}{3}\times 100=33.3\%](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bnumber%20of%20s%20orbitals%7D%7D%7B%5Ctext%20%7Btotal%20number%20of%20orbitals%7D%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Ctimes%20100%3D33.3%5C%25)
3. percentage of s character in
hybrid orbital =![\frac{\text {number of s orbitals}}{\text {total number of orbitals}}=\frac{1}{4}\times 100=25\%](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7Bnumber%20of%20s%20orbitals%7D%7D%7B%5Ctext%20%7Btotal%20number%20of%20orbitals%7D%7D%3D%5Cfrac%7B1%7D%7B4%7D%5Ctimes%20100%3D25%5C%25)
Thus percentage of s-character in an sp hybrid is 50%.
Answer:
A.) 1
Explanation:
Propane only exists in one conformation. It does not have enough carbons to form branches, and there are only hydrogens attached to each carbon. Furthermore, there is no way to twist the carbon or change its orientation (ex. cis- and trans-) to result in a different structure of propane. There is no other way to represent the molecule without drawing a different molecule.
Answer:
the answer is b Li + Cl2 .....
Average atomic mass is the weighted average atomic masses with regard to the relative abundance of the isotopes
average atomic mass of Li = relative abundance of Li-6 x mass of Li-6 + relative abundance of Li-7 x mass of Li-7
average atomic mass of Li = (7.42% x 6.0151 a.m.u) + (92.58% x 7.0160 a.m.u)
= 0.446 + 6.495
= 6.941 amu
average atomic mass of Li is 6.941 amu
The balanced reaction
is:
4NH3 + 3O2 --> 2N2 + 6H2O
<span>We
are given the amount of reactants to be used for the reaction. This
will be the starting point of our calculation.</span>
83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2
2.81 moles of NH3
From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.
<span>2.62 mol O2</span><span> (6 mol H2O / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>