The chemical element of atomic number 7, a colorless, odorless unreactive gas that forms
about 78 percent of the earth's atmosphere.
Answer:
They are synthesized from raw materials
Explanation:
All synthesized products goes through lots of processes. This processes involves the conversion of substances known as raw materials into a more useful and modified form.
These synthesized products are usually more useful and made in order to solve a given problem or offer a particular service. Raw materials are cheaper than the synthesized products and are common in all industries and fields.
Answer:
No.
Explanation:
If the temperature changes a substance from it's original form, going from solid to liquid, then the temperature required to change it from a liquid to a solid are the opposite. For example, Ice is Water (H2O) in it's solid form. In order to freeze water into ice, the temperature must be very low. In order to melt the ice back into water, the temperature must be higher than the freezing point, but not too high or it will boil, making it a gaseous substance instead.
Answer:
As temperature increases, the molecules and atoms move faster.
Explanation:
Molecules and atoms move according to how hot or cold the temperature is. As the temperature rises, the molecules and atoms will begin to move faster. As the temperature cools, the molecules and atoms will begin to move slower.
Hope this helps! :)
Answer:
The concentration of fructose-6-phosphate F6P ≅ 1.35 mM
Explanation:
Given that:
ΔG°′ is the conversion of glucose-6-phosphate to fructose-6-phosphate (F6P) = +1.67 kJ/mol = 1670 J/mol
concentration of glucose-6-phosphate at equilibrium = 2.65 mM
Assuming temperature = 25.0°C
=( 25 + 273)K
= 298 K
We are to find the concentration of fructose-6-phosphate
Using the relation;
ΔG' = -RT In K_c
where;
R = 8.314 J/K/mol
1670 = - (8.314 × 298 ) In K_c
1670 = -2477.572 × In K_c
1670/ 2477.572 = In K_c
0.67 = In K_c
0.511
Now using the equilibrium constant 
![K_c = \dfrac{[F6P]}{[G6P]}](https://tex.z-dn.net/?f=K_c%20%3D%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5BG6P%5D%7D)
![0.511 = \dfrac{[F6P]}{[2.65]}](https://tex.z-dn.net/?f=0.511%20%3D%20%20%5Cdfrac%7B%5BF6P%5D%7D%7B%5B2.65%5D%7D)
F6P = 0.511 × 2.65
F6P = 1.35415
F6P ≅ 1.35 mM
