The kind of reaction that occurs when you mix aqueous solutions of barium sulfide and sulfuric acid is a precipitation reaction.
<h3>Further Explanation</h3>
- The chemical reaction between Ba(OH)2(aq) and H2SO4(aq) is given by;
Ba(OH)₂(aq) + H₂SO4(aq) --> BaSO₄(aq) + 2H₂O(l)
- This is a type of precipitation reaction, where a precipitate is formed after the reaction, that is Barium sulfate.
<h3>Other types of reaction</h3><h3>Neutralization reactions </h3>
- These are reactions that involve reacting acids and bases or alkali to form salt and water as the only products.
- For example a reaction between sodium hydroxide and sulfuric acid.
NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l)
<h3>Displacement reactions</h3>
- These are reactions in which a more reactive atom or ion displaces a less reactive ion from its salt.
Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)
<h3>Redox reactions </h3>
- These are reactions that involve both reduction and oxidation occuring simultaneously durin a chemical reaction.
- For example,
Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)
- Magnesium atom undergoes oxidation while copper ions undergoes reduction.
<h3>Decomposition reactions</h3>
- These are type of reactions that involves breakdown of a compound into its constituents elements.
- For example decomposition of lead nitrate.
Pb(NO3)2(S) → PbO(s) + O2(g) + NO2(g)
Keywords: Precipitation
<h3>Learn more about: </h3>
Level: High school
Subject: Chemistry
Topic: Chemical reactions
Sub-topic: Precipitation reactions
<span>Express the answer in scientific notation and with the correct number of significant figures:
(6.32 x 10-4) ÷ 12.64
5.00 x 10^-5</span>
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,
pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
![\frac{[Deprotonated]}{[Protonated]}=0.01](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D%3D0.01)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
![\frac{[Protonated]}{[Deprotonated]}=100](https://tex.z-dn.net/?f=%5Cfrac%7B%5BProtonated%5D%7D%7B%5BDeprotonated%5D%7D%3D100)
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100
4.5 moles! I hope my answer helps
2 atome nitrogen , 1 is correct
