Answer:
bonding molecular orbital is lower in energy
antibonding molecular orbital is higher in energy
Explanation:
Electrons in bonding molecular orbitals help to hold the positively charged nuclei together, and they are always lower in energy than the original atomic orbitals.
Electrons in antibonding molecular orbitals are primarily located outside the internuclear region, leading to increased repulsions between the positively charged nuclei. They are always higher in energy than the parent atomic orbitals.
The reactants are oxygen and nitrogen monoxide.
<h3>What is nitrogen monoxide?</h3>
Nitrogen oxide, also known as nitrogen monoxide or nitric oxide, is an inert gas with the chemical formula NO. It is one of the main nitrogen oxides. Free radical nitric oxide (•N=O or •NO) possesses an unpaired electron, which is commonly indicated by a dot in its chemical formula. As a heteronuclear diatomic molecule, nitric oxide also contributed to the development of early modern theories of chemical bonding.
Nitric oxide is a chemical compound that occurs in combustion systems and can be produced by lightning during thunderstorms. It is a crucial intermediate in industrial chemistry. In many physiological and pathological processes in animals, including humans, nitric oxide serves as a signaling molecule.
To learn more about nitrogen monoxide from the given link:
brainly.com/question/13428103
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The answer is D. Fertilizer and vinegar
Answer:
One extraction: 50%
Two extractions: 75%
Three extractions: 87.5%
Four extractions: 93.75%
Explanation:
The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.
qⁿ = (V₁/(V₁ + KV₂))ⁿ
We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:
qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ = (V₁/(2V₁))ⁿ = (1/2)ⁿ
When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.
When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.
When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.
When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.
Solving part-1 only
#1
KMnO_4
- Transition metal is Manganese (Mn)
#2
Actually it's the oxidation number of Mn
Let's find how?
- x is the oxidation number
#3
- Purple as per the color of potassium permanganate
#4
