<span>Their orbitals are completely filled</span>
We could (a) stir faster and (b) warm the mixture.
<em>Stirring faster</em> moves freshly-dissolved sugar away from the solid and allows new water molecules to contact with the surface,
<em>Warming the mixture</em> gives the water molecules more kinetic energy, so their collisions with the surface of the sugar will be more effective in removing the sugar molecules.
The question is incomplete, here is the complete question:
At a certain temperature this reaction follows second-order kinetics with a rate constant of 14.1 M⁻¹s⁻¹

Suppose a vessel contains SO₃ at a concentration of 1.44 M. Calculate the concentration of SO₃ in the vessel 0.240 seconds later. You may assume no other reaction is important. Round your answer to 2 significant digits.
<u>Answer:</u> The concentration of
in the vessel after 0.240 seconds is 0.24 M
<u>Explanation:</u>
For the given chemical equation:

The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken= 0.240 second
[A] = concentration of substance after time 't' = ?
= Initial concentration = 1.44 M
Putting values in above equation, we get:
![14.1=\frac{1}{0.240}\left (\frac{1}{[A]}-\frac{1}{1.44}\right)](https://tex.z-dn.net/?f=14.1%3D%5Cfrac%7B1%7D%7B0.240%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B1.44%7D%5Cright%29)
![[A]=0.245M](https://tex.z-dn.net/?f=%5BA%5D%3D0.245M)
Hence, the concentration of
in the vessel after 0.240 seconds is 0.24 M
Answer:

Explanation:
Hello there!
In this case, for such unit conversion we need to realize that 1 kg is equal to 1000 g, 1 g is equal to 1x10⁹ and 1 mL equals 1 cm³, therefore we apply:

Best regards!