There are 0.000076 moles in 4.6 x 10^19 atoms.
Explanation:
Zn + Sn3(PO4)4 = Zn3(PO4)2 + Sn - Balanced Chemical Equation
4.3 is limiting
is the answer
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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:
ACTIVATION OF PHOSPHORYLASE KINASE AND ALLOSTERICALLY ACTIVATION OF PHOSPHORYLASE KINASE B
Explanation:
In a contracting skeletal muscle, there is a rapid need of ATP by the muscle cell. The energy need is obtained by the degradation of glycogen into glucose which then enters glycolysis. Muscle contraction causes muscle depolarization in which there is the rapid influx of calcium ions from the sarcoplasmic reticulum into the sacroplasm of the myocytes. This in turn causes the binding of calcium ion with calmodulin which thenb activates phosphorylase kinase from which it allosterically activate the b form of the enzyme needed for the conversion of glycogen to glucose. The other options do not occur as a result of increase in cytolic calcium concentration.
Answer:
ΔG <0 , ΔH > 0 , ΔS > 0 .
Explanation:
From the data given in question , the reaction is a spontaneous process , hence , the value of change in gibbs free energy would be negative , ΔG <0
And , on dissolution process , the temperature of the water decreases , i.e. , it is an endothermic process , i.e. , the change in enthalphy value is positive , ΔH > 0
And , during the process of dissolution , the ammonia salt break does to ions , i.e. , the randomness increases , hence the ΔS > 0
