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victus00 [196]
3 years ago
14

What kind of reaction occurs when you mix aqueous solutions of barium sulfide and sulfuric acid? what kind of reaction occurs wh

en you mix aqueous solutions of barium sulfide and sulfuric acid? neutralization gas evolution precipitation precipitation and gas evolution?
Chemistry
2 answers:
My name is Ann [436]3 years ago
3 0

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>
  • Precipitation reaction: brainly.com/question/11194650
  • Examples of precipitation reactions: brainly.com/question/11194650
  • Neutralization reactions brainly.com/question/3243813

Level: High school

Subject: Chemistry

Topic: Chemical reactions

Sub-topic: Precipitation reactions

lapo4ka [179]3 years ago
3 0

The reactions that occur by mixing aqueous solutions of barium sulfide and sulfuric acid are

\boxed{{\text{Precipitation and gas evolution reactions}}}

Further Explanation:

<u>Precipitation reaction:</u>

It is the type of reaction in which an<em> insoluble salt </em>is formed by the combination of two solutions containing soluble salts. That insoluble salt is known as<em> precipitate</em> and therefore such reactions are named precipitation reactions. An example of precipitation reaction is,

{\text{AgN}}{{\text{O}}_3}\left( {aq}\right)+{\text{KBr}}\left( {aq}\right)\to {\text{AgBr}}\left( s \right)+{\text{KN}}{{\text{O}}_3}\left( {aq}\right)

Here, AgBr is a precipitate.

<u>Neutralization reaction:</u>

It is the reaction that occurs between an acid and a base in order to form salt and water. It is named so as it neutralizes the excess amount of hydrogen or hydroxide ions present in the solution. It is used to decrease the acidity in the stomach, wastewater treatment, antacid tablets and to control the pH of soil. An example of neutralization reaction is,

{\text{HCl}}+{\text{NaOH}}\to{\text{NaCl}}+{{\text{H}}_2}{\text{O}}

<u>Gas evolution reaction:</u>

It is the type of chemical reaction in which one of the products is a gas. These reactions are often carried out in a fume chamber if poisonous gases are produced. An example of gas evolution reaction is,

{\text{Zn}}+2{\text{HCl}}\to{\text{ZnC}}{{\text{l}}_2}+{{\text{H}}_2}

Here, {{\text{H}}_2} is evolved so it is a gas evolution reaction.

The solubility rules to determine the solubility of the compound are as follows:

1. The common compounds of group 1A are soluble.

2. All the common compounds of ammonium ion and all acetates, chlorides, nitrates, bromides, iodides, and perchlorates are soluble in nature. Only the chlorides, bromides, and iodides of  {\text{A}}{{\text{g}}^ + } , {\text{P}}{{\text{b}}^{2 + }} , {\text{C}}{{\text{u}}^ + } and {\text{Hg}}_2^{2 + } are not soluble.

3. All common fluorides, except for {\text{Pb}}{{\text{F}}_{\text{2}}} and group 2A fluorides, are soluble. Moreover, sulfates except {\text{CaS}}{{\text{O}}_{\text{4}}} , {\text{SrS}}{{\text{O}}_{\text{4}}} , {\text{BaS}}{{\text{O}}_{\text{4}}} , {\text{A}}{{\text{g}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}} and {\text{PbS}}{{\text{O}}_{\text{4}}} are soluble.

4. All common metal hydroxides except {\text{Ca}}{\left( {{\text{OH}}}\right)_{\text{2}}} , {\text{Sr}}{\left( {{\text{OH}}} \right)_{\text{2}}} , {\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}  and hydroxides of group 1A, are insoluble.

5. All carbonates and phosphates, except those formed by group 1A and ammonium ion, are insoluble.

6. All sulfides, except those formed by group 1A, 2A, and ammonium ion are insoluble.

7. Salts that contain {\text{C}}{{\text{l}}^ - } , {\text{B}}{{\text{r}}^ - } or {{\text{I}}^ - } are usually soluble except for the halide salts of {\text{A}}{{\text{g}}^ + } , {\text{P}}{{\text{b}}^{2 + }} and {\left({{\text{H}}{{\text{g}}_2}}\right)^{{\text{2 + }}}}.

8. The chlorides, bromides, and iodides of all the metals are soluble in water, except for silver, lead, and mercury (II). Mercury (II) iodide is water insoluble. Lead halides are soluble in hot water.

9. The perchlorates of group 1A and group 2A are soluble in nature.

10. Almost all the sulfides of transition metals are highly insoluble. These include CdS, FeS, ZnS, and {\text{A}}{{\text{g}}_2}{\text{S}}. The sulfides of arsenic, antimony, bismuth, and lead are also insoluble.

11. All the acetates and chlorates are soluble in nature.

The reaction between barium sulfide and sulfuric acid is as follows:

{\text{BaS}}\left({aq} \right)+{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\left( {aq}\right) \to {\text{BaS}}{{\text{O}}_4}\left( s \right)+{{\text{H}}_2}{\text{S}}\left( g \right)

According to the solubility rules, {\text{BaS}}{{\text{O}}_4} is an insoluble salt. So {\text{BaS}}{{\text{O}}_4} will form precipitate in the above reaction and therefore this is a precipitation reaction.

Also, {{\text{H}}_2}{\text{S}} is evolved during this reaction, so it is also a gas evolution reaction.

Learn more:

1. Balanced chemical equation brainly.com/question/1405182

2. The main purpose of conducting experiments: brainly.com/question/5096428

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: precipitate, soluble, insoluble, solubility rules, solubility, precipitation reaction, BaSO4, H2S, BaS, H2SO4, neutralization reaction, gas evolution reaction.

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Answer: True.

Explanation.

The group 7A is actually named group 17.

That group is the halogens: F, Cl, Br, I, At, and Ts (Ts is one of the last elements discovered).

Those elements have 7 valence electrons (notice that it is the same number as the second digit in 17).

The atoms with 7 valence electrons will "easily" gain one electron to get the configuration of the next noble gas (8 valence electrons). That is why these elements gain electrons to form ions.

When atoms gain electrons form anions (negative ions). For example: F(-), Cl(-), Br(-), I(-).
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2 years ago
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A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 64.0 mL of 0.0600 M EDTA . Titration of the excess unreacted EDTA req
tigry1 [53]

Answer:

the concentration of Cd^{2+}  in the original solution= 0.0088 M

the concentration of Mn^{2+} in the original solution = 0.058 M

Explanation:

Given that:

The volume of the sample  containing Cd2+ and Mn2+ =  50.0 mL; &

was treated with 64.0 mL of 0.0600 M EDTA

Titration of the excess unreacted EDTA required 16.1 mL of 0.0310 M Ca2+

i.e the strength of the Ca2+ = 0.0310 M

Titration of the newly freed EDTA required 14.2 mL of 0.0310 M Ca2+

To determine the concentrations of Cd2+ and Mn2+ in the original solution; we have the following :

Volume of newly freed EDTA = \frac{Volume\ of \ Ca^{2+}* Sample \ of \ strength }{Strength \ of EDTA}

= \frac{14.2*0.0310}{0.0600}

= 7.3367 mL

concentration of  Cd^{2+} = \frac{volume \ of \  newly  \ freed \ EDTA * strength \ of \ EDTA }{volume \ of \ sample}

= \frac{7.3367*0.0600}{50}

= 0.0088 M

Thus the concentration of Cd^{2+} in the original solution = 0.0088 M

Volume of excess unreacted EDTA = \frac{volume \ of \ Ca^{2+} \ * strength \ of Ca^{2+} }{Strength \ of \ EDTA}

= \frac{16.1*0.0310}{0.0600}

= 8.318 mL

Volume of EDTA required for sample containing Cd^{2+}   and  Mn^{2+}  = (64.0 - 8.318) mL

= 55.682 mL

Volume of EDTA required for Mn^{2+}  = Volume of EDTA required for

                                                                sample containing  Cd^{2+}   and  

                                                             Mn^{2+} --  Volume of newly freed EDTA

Volume of EDTA required for Mn^{2+}  = 55.682 - 7.3367

= 48.3453 mL

Concentration  of Mn^{2+} = \frac{Volume \ of EDTA \ required \ for Mn^{2+} * strength \ of \ EDTA}{volume \ of \ sample}

Concentration  of Mn^{2+} =  \frac{48.3453*0.0600}{50}

Concentration  of Mn^{2+}  in the original solution=   0.058 M

Thus the concentration of Mn^{2+} = 0.058 M

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According to Le Chatelier’s principle, a system in chemical equilibrium responds to stress by shifting the equilibrium in a dire
shepuryov [24]

Answer:

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Answer:

  The rate of the forward reaction is equal to the rate of the reverse reaction

Explanation:

Equilibrium is a condition in which the net force for change is zero.

In chemistry, equilibrium exists when reactant concentrations in a reaction are not changing. This condition means the reaction proceeding left to right in the reaction equation is producing products at the same rate those products are being used up by the reaction proceeding right to left. In short, the rate of the forward reaction is equal to the rate of the reverse reaction.

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