Answer:
Long answer
Proteins are the building blocks of the body, each and every organ muscle and skin and cells are made of proteins. They can also be used for providing energy to the body in a state where other energy sources aren't available.
Fats are stored energy blocks which can be used by the body at its own convenience whenever there is a deficit of energy in the body. They can also be directly burnt to give energy in cases where energy needs of the body are not met by dietary intake
Carbohydrates are the most convenient and preferable source of energy in the body and are easily converted to give out immidiate energy to the body, excess carbohydrates can be converted into fats and stored for later usage as and when required by the body.
Proteins, fats and carbohydrates have many secondary functions and roles in the human body. If you are keen to know anything more specific feel free to ask.
<h3>
Short answer :</h3>
Fats are used for energy after they are broken into fatty acids. Protein can also be used for energy, but the first job is to help with making hormones, muscle, and other proteins.
A catalyst reduces H°rnx in most reactions. The answer is false
<h3>Do catalysts reduce delta H?</h3>
By reducing the activation energy required for the reaction to occur, a catalyst just modifies the route used to go from reactants to products. However, because it doesn't alter the state of the products or reactants, delta H is unaffected.
A catalyst reduces a reaction's activation energy, enabling a chemical reaction to occur. The number of reactant particles with energy above the activation energy increases as the temperature of a reaction rises.
learn more about catalyst refer
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PH scale is used to determine how acidic or basic a solution is.
pH can be calculated as follows;
by knowing the ph we can calculate pOH
pH + pOH = 14
pOH = 14 - 8.1
pOH = 5.9
pOH is used to calculate the hydroxide ion concentration
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 1.26 x 10⁻⁶ M
therefore hydroxide ion concentration is 1.26 x 10⁻⁶ M
Answer:
E₁ ≅ 28.96 kJ/mol
Explanation:
Given that:
The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,
Let the activation energy for a catalyzed biochemical reaction = E₁
E₁ = ??? (unknown)
Let the activation energy for an uncatalyzed biochemical reaction = E₂
E₂ = 50.0 kJ/mol
= 50,000 J/mol
Temperature (T) = 37°C
= (37+273.15)K
= 310.15K
Rate constant (R) = 8.314 J/mol/k
Also, let the constant rate for the catalyzed biochemical reaction = K₁
let the constant rate for the uncatalyzed biochemical reaction = K₂
If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:
K₁ = 3.50 × 10³
K₂ = 1
Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;
we can use the formula for Arrhenius equation;
If 
&
E₁ ≅ 28.96 kJ/mol
∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol
