Nope. The definition talks about the RATES in each direction, but there's no reason that it must happen at any particular STAGE during the whole thing. Choice 'E' even talks about a reaction that doesn't reach equilibrium until it's almost done. The one you want is 'A'.
Answer:
The concentration of the most dilute solution is 0.016M.
Explanation:
First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:
![[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M](https://tex.z-dn.net/?f=%5BNa_%7B2%7DSO_%7B4%7D%5D%3D%5Cfrac%7Bmoles%28Na_%7B2%7DSO_%7B4%7D%29%7D%7Bliters%28solution%29%7D%20%3D%5Cfrac%7Bmass%28%28Na_%7B2%7DSO_%7B4%7D%29%29%7D%7Bmolarmass%28moles%28Na_%7B2%7DSO_%7B4%7D%29%20%5Ctimes%200.100L%29%7D%20%3D%5Cfrac%7B2.5316g%7D%7B142g%2Fmol%5Ctimes%200.100L%20%7D%20%3D0.178M)
<u>First dilution</u>
We can use the dilution rule:
C₁ x V₁ = C₂ x V₂
where
Ci are the concentrations
Vi are the volumes
1 and 2 refer to initial and final state, respectively.
In the first dilution,
C₁ = 0.178 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,

<u>Second dilution</u>
C₁ = 0.053 M
V₁ = 15 mL
C₂ = unknown
V₂ = 50 mL
Then,

This was because the experiment showed that a substance could emit radiation even while it was not exposed to light.
The critical point is said to be the end point of a phase diagram. The most known example is the liquid-vapor critical point, the point in a PT curve that signifies that at this point both vapor and liquid coexist. The critical point is characterized by a critical pressure and a critical temperature.
Basically it means carbon dioxide is removed from the atmosphere and is trapped in e.g. calcium carbonate (like shells), rocks, oceans etc.