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The coefficient for Lithium Acetate in the balanced chemical equation is 6
Why?
This is the reaction:
(CH₃COO)₂Ca(aq) + Li₃PO₄(aq) → Ca₃(PO₄)₂(s) + CH₃COOLi(aq)
For balancing the equation, we're going to apply the Law of Conservation of Mass, which states that matter is not created nor destroyed, and the number of atoms of each element should be the same on both sides of the equation, from that, the balanced equation is:
3(CH₃COO)₂Ca(aq) + 2Li₃PO₄(aq) → Ca₃(PO₄)₂(s) + 6CH₃COOLi(aq)
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Taking into account the definition of density, the density of the metal is 16.33
.
<h3>Definition of density</h3>
Density is a quantity that allows us to measure the amount of mass in a certain volume of a substance. Then, the expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

From this expression it can be deduced that density is inversely proportional to volume: the smaller the volume occupied by a given mass, the higher the density.
<h3>Density of the metal in this case</h3>
In this case, you know that:
- Mass= 34.3 g
- Volume= 2.1 mL
Replacing in the definition of density:

Solving:
<u><em>density= 16.33 </em></u>
In summary, the density of the metal is 16.33
.
Learn more about density:
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Answer:
The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Explanation:
From the given information:
The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.
In this same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;
Rate factor in the absence of catalyst:

Rate factor in the presence of catalyst:

Assuming the catalyzed reaction and the uncatalyzed reaction are taking place at the same temperature :
Then;
the ratio of the rate factors can be expressed as:

![\dfrac{k_2}{k_1}={ \dfrac {e^{[ Ea_1 - Ea_2 ] }}{RT} }}](https://tex.z-dn.net/?f=%5Cdfrac%7Bk_2%7D%7Bk_1%7D%3D%7B%20%20%5Cdfrac%20%7Be%5E%7B%5B%20%20Ea_1%20-%20Ea_2%20%5D%20%7D%7D%7BRT%7D%20%7D%7D)
Thus;

Let say the assumed temperature = 25° C
= (25+ 273)K
= 298 K
Then ;



The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme
Answer:
Complex II
Explanation:
Complex II does not directly pump protons but rather sends two protons on to Complex III in the form of the reduced UQ known as ubiquinol.
Thus, the correct answer is Complex II