Answer:
answer is a since solute dissolve a solvent to give a solution
Ethanoic (Acetic) acid is a weak acid and do not dissociate fully. Therefore its equilibrium state has to be considered here.

In this case pH value of the solution is necessary to calculate the concentration but it's not given here so pH = 2.88 (looked it up)
pH = 2.88 ==>
![[H^{+}]](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D)
=

= 0.001

The change in Concentration Δ
![[CH_{3}COOH]](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D)
= 0.001

CH3COOH H+ CH3COOH
Initial

0 0
Change

-0.001 +0.001 +0.001
Equilibrium

- 0.001 0.001 0.001
Since the

value is so small, the assumption
![[CH_{3}COOH]_{initial} = [CH_{3}COOH]_{equilibrium}](https://tex.z-dn.net/?f=%5BCH_%7B3%7DCOOH%5D_%7Binitial%7D%20%3D%20%5BCH_%7B3%7DCOOH%5D_%7Bequilibrium%7D)
can be made.
![k_{a} = [tex]= 1.8*10^{-5} = \frac{[H^{+}][CH_{3}COO^{-}]}{[CH_{3}COOH]} = \frac{0.001^{2}}{x}](https://tex.z-dn.net/?f=%20k_%7Ba%7D%20%3D%20%5Btex%5D%3D%201.8%2A10%5E%7B-5%7D%20%20%3D%20%20%5Cfrac%7B%5BH%5E%7B%2B%7D%5D%5BCH_%7B3%7DCOO%5E%7B-%7D%5D%7D%7B%5BCH_%7B3%7DCOOH%5D%7D%20%3D%20%20%5Cfrac%7B0.001%5E%7B2%7D%7D%7Bx%7D%20)
Solve for x to get the required concentration.
note: 1.)Since you need the answer in 2SF don&t round up values in the middle of the calculation like I've done here.
2.) The ICE (Initial, Change, Equilibrium) table may come in handy if you are new to problems of this kind
Hope this helps!
According to ideal gas equation, we know for 1 mole of gas: PV=RT
where P = pressure, T = temperature, R = gas constant, V= volume
If '1' and '2' indicates initial and final experimental conditions, we have

Given that: V1 = 100.0 kPa, T1 = 100.0 K, V1 = 2.0 m3, T2 = 400 K, P2 = 200.0 kPa
∴ on rearranging above eq., we get V2 =

∴ V2 = 4 m3
<span>Elements are matter made of one kind of atom.</span>
Answer:
The pH is equal to 4.41
Explanation:
Since HClO is a weak acid, its dissociation in aqueous medium is:
HClO ⇄ ClO- + H+
start: 0.05 0 0
change -x +x +x
balance 0.05-x x x
As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.
the acidity constant when equilibrium is reached is equal to:
![Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BClO-%5D%2A%5BH%2B%5D%7D%7B%5BHClO%5D%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.05-x%7D%3D3x10%5E%7B-8%7D)
The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:

clearing the x and calculating its value we have:
![x=3.87x10^{-5}=[H+]=[ClO-]](https://tex.z-dn.net/?f=x%3D3.87x10%5E%7B-5%7D%3D%5BH%2B%5D%3D%5BClO-%5D)
the pH can be calculated by:
![pH=-log[H+]=-log[3.87x10^{-5}]=4.41](https://tex.z-dn.net/?f=pH%3D-log%5BH%2B%5D%3D-log%5B3.87x10%5E%7B-5%7D%5D%3D4.41)