Answer:
used to complete the glicolisis process
Explanation:
A glucose molecule will generate the necessary energy to the body to compete their own metabolic processes.
At the begging of the process, the glucose change into 2 molecules of gliceraldehyde 3 - phosphate. In this first process the glucose molecule consume 2 molecules of ATP (the energetic molecule).
After this first process, the aldehyde turns into piruvate, during this process, there is produced 2 molecules of ATP, so it is producing chemical energy.
If during this process the cell need to enter into another metabolic process (pentose ways) then the gain of ATP deplets.
The phenomenon known as "salting-out" occurs at very high ionic strengths, when protein solubility declines as ionic strength rises. As a result, salting out may be used to segregate proteins according to how soluble they are in salt solutions.
Because large levels of sodium chloride disturb the bonds and structure of the active site, the rate of enzyme activity will gradually decrease as the concentration of sodium chloride rises. As a result, some of the active sites get denaturized and the starch loses its ability to attach to them. As more enzymes get denatured and eventually cease to function, enzyme activity will steadily wane.
Answer:
The equilibrium value of [CO] is 1.04 M
Explanation:
Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.
Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:
aA + bB ⇔ cC + dD
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
In this case:
![Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7DOH%20%5D%7D%7B%5BCO%5D%2A%5BH_%7B2%7D%20%5D%5E%7B2%7D%20%7D)
You know:
- Kc= 14.5
- [H₂]= 0.322 M
- [CH₃OH] =1.56 M
Replacing:
![14.5=\frac{1.56}{[CO]*0.322^{2} }](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B1.56%7D%7B%5BCO%5D%2A0.322%5E%7B2%7D%20%7D)
Solving:
![[CO]=\frac{1.56}{14.5*0.322^{2} }](https://tex.z-dn.net/?f=%5BCO%5D%3D%5Cfrac%7B1.56%7D%7B14.5%2A0.322%5E%7B2%7D%20%7D)
[CO]= 1.04 M
The equilibrium value of [CO] is 1.04 M
Answer:
C
Explanation:
All thing in option c is flammable
D = M/V
D = 0.10
V = 1000 cm^3
0.10 = M / 1000
M = 100
Porque la densidad es 10%, se puede usar 0.10 en este formulario para calcular el peso. No sé lo que es la unidad para el peso pero es 100.