Potassium hydroxide (KOH) is formed when Potassium forms ionic bonds with OH-ions while Potassium Oxide (K2O) is formed when potassium forms ionic bonds with the Oxide (O2-) ions. i.e. This reaction is a neutralization reaction and occurs when an acid (HCl) reacts with a base (KOH).
Prescribed to you by your doctor.
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Answer:
If the zero is between fwo nonzeros
Explanation:
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:
26.0 g/mol is the molar mass of the gas
Explanation:
We have to combine density data with the Ideal Gases Law equation to solve this:
P . V = n . R .T
Let's convert the pressure mmHg to atm by a rule of three:
760 mmHg ____ 1 atm
752 mmHg ____ (752 . 1)/760 = 0.989 atm
In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.
Moles = Mass / molar mass.
We can replace density data as this in the equation:
0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K
(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x
0.0405 mol = 1.053 g / x
x = 1.053 g / 0.0405 mol = 26 g/mol