Answer:
The Phosphorylated glucose(glucose +inorganic phosphate), with the energy supplied from ATP hydrolysis formed glucose 6- phosphate, which is later converted to 2 molecules of fructose 6-phosphate- this is phosphorylation.And represented the fate of glucose -6-phosphate.
The fructose 6-phosphate are converted to triose phosphate- which is a 2-molecules of 3C compound. The latter is oxidized by NAD→ NADH+ to form intermediates in the glycolytic pathways .
These intermediates are converted to ribose 5-phosphates in the presence of transketolase and transaldolase enzymes.And they are finally converted to pyruvate in the glycolytic pathway with the production of 2ATPs per molecule of glucose.
Basically the phosphate pathway reaction is very slow due to enzyme catalysis.
Answer:
20ppm
Explanation:
parts per million are defined as the mass of solute in mg (In this case, mass of DDT) per kg of sample.
To solve this question we must find the mass of DDT in mg and the mass of sample in kg:
<em>Mass DDT:</em>
0.10g * (1000mg / 1g) = 100mg
<em>Mass sample:</em>
5000g * (1kg / 1000g) = 5kg
Parts per Million:
100mg / 5kg =
<h3>20ppm</h3>
Answer: D. Slow down the chain reaction by absorbing free neutrons
Explanation: just got it right on the quiz A P E X
Answer:
At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.
Explanation:
The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.
The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.
This problem is providing us with the molality of a solution of calcium iodide as 0.01 m. So the most likely van't Hoff factor is required and theoretically found to be 3 due to the following:
<h3>Van't Hoff factor:</h3>
In chemistry, the correct characterization of solutions also imply the identification of the ions it will release in aqueous solution. For that reason, the van't Hoff factor gives us an idea of this number, according to the formula the solute has got.
In such a way, for calcium iodide, we write its ionization equation as shown below:
Assuming it is able to ionize due to the low molality, because if it was higher, then it won't ionize. Hence, since we have three moles of ion products, one Ca²⁺ and two I⁻, we can conclude the van't Hoff factor would be 3, although calculations may lead to a different, yet close result.
Learn more about the van't Hoff factor: brainly.com/question/23764376
