Answer:
This is the typical route of alcohol metabolism, where in the liver it is first transformed to acetaldehyde and then to acetate.
Explanation:
Ethanol is not digested but absorbed and follows its metabolic pathway in the liver, producing in the first instance acetaldehyde, which is the main substance that causes the hangover and then this compound is transformed into the final product, which is acetate Later acetate is metabolized to Acetyl-CoA. The enzyme responsible for the metabolism of ethanol is alcohol dehydrogenase in the liver, and for the cytochrome P-450 dependent system and for catalase in the liver.
Answer:
The value of the missing equilibrium constant ( of the first equation) is 1.72
Explanation:
First equation: 2A + B ↔ A2B Kc = TO BE DETERMINED
⇒ The equilibrium expression for this equation is written as: [A2B]/[A]²[B]
Second equation: A2B + B ↔ A2B2 Kc= 16.4
⇒ The equilibrium expression is written as: [A2B2]/[A2B][B]
Third equation: 2A + 2B ↔ A2B2 Kc = 28.2
⇒ The equilibrium expression is written as: [A2B2]/ [A]²[B]²
If we add the first to the second equation
2A + B + B ↔ A2B2 the equilibrium constant Kc will be X(16.4)
But the sum of these 2 equations, is the same as the third equation ( 2A + 2B ↔ A2B2) with Kc = 28.2
So this means: 28.2 = X(16.4)
or X = 28.2/16.4
X = 1.72
with X = Kc of the first equation
The value of the missing equilibrium constant ( of the first equation) is 1.72
If the person had clogged arteries and high blood pressure, then went on a healthy diet, the person would start losing weight. They would have a more stable blood pressure (since they are eating healthy foods, with less saturated fat) and the clogged arteries would look significantly better. He/she might lose her/his risk of developing diabetes, heart disease, cancer, etc. Most of her/his health issues would start to balance out if they started eating a diet low in saturated fat.
I hope this helps!
~kaikers
A machine called a barometer is what is used to measure atmospheric pressure :D
Answer:
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺
Explanation:
This is an easy redox reaction:
Br₂ + 2KI → I₂ + 2KBr
We determine the oxidation states.
0 for the elements at ground state.
K does not change the oxidation state during the reaction.
Bromine is reduced to bromide (oxidation state decreases)
and iodide is oxidized to Iodine (oxidation state increases)
Br₂ + 2e⁻ ⇄ 2Br⁻ Half reaction of reduction
2I⁻ ⇄ 2e⁻ + I₂ Half reaction of oxidation
In oxidation, electrons are released while in reduction, the electrons are gained. To make the ionic equation, we just add K⁺
So we sum both reactions
Br₂ + 2e⁻ + 2I⁻ + 2K⁺ ⇄ 2e⁻ + I₂ + 2Br⁻ + 2K⁺
We cancel the electrons on both sides of the equation:
Br₂ + 2I⁻ + 2K⁺ ⇄ I₂ + 2Br⁻ + 2K⁺