In chemistry, yes...yes it is.
Answer:
When you freeze the already cracked glow stick the chemical reactions are slowed down. When you take it out of the freezer and it reaches room temperature the chemical reaction will go on as 'normal'. XD hope this helps
Explanation:
This problem is asking for the equilibrium constant at two different temperatures by describing the chemical equilibrium between gaseous nitrogen, oxygen and nitrogen monoxide at 25 °C and 1496 °C as the room temperature and the typical temperature inside the cylinders of a car's engine respectively:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Thus, the calculated equilibrium constants turned out to be 6.19x10⁻³¹ and 9.87x10⁻⁵ at the aforementioned temperatures, respectively, according to the following work:
There is a relationship between the Gibbs free energy, enthalpy and entropy of the reaction, which leads to the equilibrium constant as shown below:

Which means we can calculate the enthalpy and entropy of reaction and subsequently the Gibbs free energy and equilibrium constant. In such a way, we calculate these two as follows, according to the enthalpies of formation and standard entropies of N₂(g), O₂(g) and NO(g) since these are assumed constant along the temperature range:

Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:

And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:
![K=exp(-\frac{\Delta _rG}{RT} )\\\\K_{25\°C}=exp[-\frac{172420 J/mol}{(8.3145\frac{J}{mol*K})(298.15K)} ]=6.19x10^{-31}\\\\K_{1496\°C}=exp[-\frac{135650J/mol}{(8.3145\frac{J}{mol*K})(1769K)} ]=9.87x10^{-5}](https://tex.z-dn.net/?f=K%3Dexp%28-%5Cfrac%7B%5CDelta%20_rG%7D%7BRT%7D%20%29%5C%5C%5C%5CK_%7B25%5C%C2%B0C%7D%3Dexp%5B-%5Cfrac%7B172420%20J%2Fmol%7D%7B%288.3145%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%28298.15K%29%7D%20%5D%3D6.19x10%5E%7B-31%7D%5C%5C%5C%5CK_%7B1496%5C%C2%B0C%7D%3Dexp%5B-%5Cfrac%7B135650J%2Fmol%7D%7B%288.3145%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%281769K%29%7D%20%5D%3D9.87x10%5E%7B-5%7D)
Learn more:
Answer is: <span>edge of the length of the box measured is 1280,57 kilometers.
</span>V(raindrop) = 3,48 cm³.
V(cubic box) = 3,48 cm³ · 6,023·10²³.
V(cubic box) = 2,1·10²⁴ cm³.
V(cubic box) = 2,1·10²⁴ cm³ ÷ 10¹⁵.
V(cubic box) = 2,1·10⁹ km³.
r(cubic box) = ∛2,1·10⁹ km³.
r(cubic box) = 1280,57 km.