Answer:
Present in both catabolic and anabolic pathways
Explanation:
Glyceraldehyde-3-phosphate abbreviated as G3P occurs as intermediate in glycolysis and gluconeogenesis.
In photosynthesis, it is produced by the light independent reaction and acts as carrier for returning ADP, phosphate ions Pi, and NADP+ to the light independent pathway. Photosynthesis is a anbolic pathway.
In glycolysis, Glyceraldehyde-3-phosphate is produced by breakdown of fructose-1,6 -bisphosphate. Further Glyceraldehyde-3-phosphate converted to pyruvate and pyruvate is further used in citric acid cycle for energy production. Therefore, it is used in catabolic pathway too.
Glyceraldehyde-3-phosphate is an important intermediate molecule in the cell's metabolic pathways because it is present in both catabolic and anabolic pathways.
Answer :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and the unpaired electrons present in the molecule.
The given molecule is, 
Beryllium has '2' valence electrons and fluorine has '7' valence electrons. Beryllium is the central atom and fluorine is the terminal atom.
Total number valence electrons in
= 2 + 2(7) = 16
According to the Lewis-dot structure, there are '4' number of bonding electrons and 12 number of non-bonding electrons (lone-pair).
The Lewis-dot structure is shown below.
The question is incomplete, complete question is :
The first two steps in the industrial synthesis of nitric acid produce nitrogen dioxide from ammonia:
Step 1 : 
Step 2 : 
The net reaction is:

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants
and
. If you need to include any physical constants, be sure you use their standard symbols
Answer:
Equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:

Explanation:
Step 1 : 
Expression of an equilibrium constant can be written as:
![K_1=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^5}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E5%7D)
Step 2 : 
Expression of an equilibrium constant can be written as:
![K_2=\frac{[NO_2]^2}{[NO]^2[O_2]}](https://tex.z-dn.net/?f=K_2%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BNO%5D%5E2%5BO_2%5D%7D)
The net reaction is:

Expression of an equilibrium constant can be written as:
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D)
Multiply and divide
;
![K=\frac{[NO_2]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO_2%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=\frac{[NO]^4[H_2O]^6}{[NH_3]^4[O_2]^7}\times \frac{[NO_2]^4}{[NO]^4}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BNO%5D%5E4%5BH_2O%5D%5E6%7D%7B%5BNH_3%5D%5E4%5BO_2%5D%5E7%7D%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BNO%5D%5E4%7D)
![K=K_1\times \frac{[NO_2]^4}{[O_2]^2[NO]^4}](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%5Cfrac%7B%5BNO_2%5D%5E4%7D%7B%5BO_2%5D%5E2%5BNO%5D%5E4%7D)
![K=K_1\times (\frac{[NO_2]^2}{[O_2]^1[NO]^2})^2](https://tex.z-dn.net/?f=K%3DK_1%5Ctimes%20%28%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BO_2%5D%5E1%5BNO%5D%5E2%7D%29%5E2)

So , the equation that gives the overall equilibrium constant K in terms of the equilibrium constants
:
