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!
Answer: The Process is Deposition
Explanation:
Answer:
The Gravitation Force between two objects depend upon :
1).The<u> product of masses</u> of the objects
2) <u>Distance </u>Between the Objects
Explanation:
Universal Law of Gravitation : It states that Force of gravitation between two object is directly proportional to the product of masses and inversely proportional to the square of distance between them.

Here,
G = Universel Gravitational Constant
m1 = mass of 1st object
m2 = mass of 2nd object
R = distance between the objects
Answer:
There are 2,04x10 e-6 moles of water in 1.23x10e18 molecules
Explanation:
We use Avogadro's number, according to which in 1 mole of any substance there are 6,023 x 10 e23 particles (example: atoms, molecules). We calculate the number of moles according to the following simple three rule:
6,023 x 10 e23 molecules-----1 mol of water
1.23x10e18 molecules----x=
x=(1.23x10e18 molecules x1 mol of water)/6,023 x 10 e23 molecules
<em>x=2,04x10 e-6 moles of water</em>
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