Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.
<span> I got (a). The answer was 7.53. But when I try to solve (d), I keep getting the wrong answer. I subtracted the moles of NaOH from the acid and added the moles to the base. Then I did Ka = (x*([NaClO]+x))/([HClO - x) and then I found the pH</span>
<u>Two hybrid </u>orbitals of <u>sp </u>type forms when each set of atomic orbitals mixes one p and one s
<h3>What is
orbital hybridization?</h3>
In the context of valence bond theory, orbital hybridization (or hybridisation) refers to the idea of combining atomic orbitals to create new hybrid orbitals (with energies, forms, etc., distinct from the component atomic orbitals) suited for the pairing of electrons to form chemical bonds.
For instance, the valence-shell s orbital joins with three valence-shell p orbitals to generate four equivalent sp3 mixes that are arranged in a tetrahedral configuration around the carbon atom to connect to four distinct atoms.
Hybrid orbitals are symmetrically arranged in space and are helpful in the explanation of molecular geometry and atomic bonding characteristics. Usually, atomic orbitals with similar energies are combined to form hybrid orbitals.
Learn more about Hybridization
brainly.com/question/22765530
#SPJ4
Answer:
The solutions of hydrobromic acid and of chloric acid are the most electrically conductive.
Explanation:
The electrical conductivity of a solution is determined by the concentration of ions in the solution. The acids listed react with water to form ions. In the case of hydrobromic acid, for example, the reaction is as follows:
HBr + H2O >> H3O(+) + Br(-)
Hydrobromic acid and chloric acid are strong acids, meaning the reaction is considered to proceed at 100%. Acetic acid, however, is a weak acid and the reaction with water does not go to completion, so there are less ionic species in the solution of acetic acid, which makes it less electrically conductive.
Answer:
0 kJ/mol.
Explanation:
Hello,
In this case, since the chemical potential can be represented in terms of the Gibbs free energy of formation:
Thus, since the Gibbs free energy of formation of an element is zero, the chemical potential is also zero, or just 0 kJ/mol.
Best regards.
