Wood, Water, and Neon Gas, because the wood is more of a solid its particles are tightly arranged and water next because it takes the shape of the container it is in and then the neon gas because with a gas the particles are not at all arranged
Explanation:
Reaction equation is as follows.

Here, 1 mole of
produces 2 moles of cations.
![[Na^{+}] = 2[Na_{2}SO_{3}] = 2 \times 0.58](https://tex.z-dn.net/?f=%5BNa%5E%7B%2B%7D%5D%20%3D%202%5BNa_%7B2%7DSO_%7B3%7D%5D%20%3D%202%20%5Ctimes%200.58)
= 1.16 M
= 0.58 M
The sulphite anion will act as a base and react with
to form
and
.
As, 
= 
=
According to the ICE table for the given reaction,

Initial: 0.58 0 0
Change: -x +x +x
Equilibrium: 0.58 - x x x
So,
![K_{b} = \frac{[HSO^{-}_{3}][OH^{-}]}{[SO^{2-}_{3}]}](https://tex.z-dn.net/?f=K_%7Bb%7D%20%3D%20%5Cfrac%7B%5BHSO%5E%7B-%7D_%7B3%7D%5D%5BOH%5E%7B-%7D%5D%7D%7B%5BSO%5E%7B2-%7D_%7B3%7D%5D%7D)


x = 0.0003 M
So, x =
= 0.0003 M
= 0.58 - 0.0003
= 0.579 M
Now, we will use
= 0.0003 M
The reaction will be as follows.

Initial: 0.0003
Equilibrium: 0.0003 - x x x


= 
= 
Therefore, 
As, x <<<< 0.0003. So, we can neglect x.
Therefore, 
= 
x = 
x =
= 
![[H^{+}] = \frac{10^{-14}}{[OH^{-}]}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B%5BOH%5E%7B-%7D%5D%7D)
= 
=
M
Thus, we can conclude that the concentration of spectator ion is
M.
Out of the following given choice:
A. The increase in
the airspace occupied by vinegar molecules
B. The chemical reaction with nerves, which is slower than
other sensory processes
C. Attractive forces between the air and vinegar molecules
D. Random collisions between the air and vinegar molecules.
<span>The answer is
D. While the particles may be
moving at high velocities even at room
temperatures, the delay is due to the numerous collisions
between the vinegar molecules and the air molecules.This changes the vinegar’s
molecules directions from straight lines to random unpredictable paths</span>
Answer:
I do not know the Answer I'm just trying to get my point
Explanation:
Thank you