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77julia77 [94]
3 years ago
7

What happens when sodium and sulfur combine

Chemistry
1 answer:
Eduardwww [97]3 years ago
7 0

Answer:

Sodium sulfide is the chemical compound with the formula Na2S, or more commonly its hydrate Na2S·9H2O. Both the anhydrous and the hydrated salts are colorless solids. They are water-soluble, giving strongly alkaline solutions. When exposed to moist air, Na2S and its hydrates emit hydrogen sulfide, which smells like rotten eggs. Some commercial samples are specified as Na2S·xH2O, where a weight percentage of Na2S is specified. Commonly available grades have around 60% Na2S by weight, which means that x is around 3. Such technical grades of sodium sulfide have a yellow appearance owing to the presence of polysulfides. These grades of sodium sulfide are marketed as 'sodium sulfide flakes'.

Contents

1 Structure

2 Production

3 Reactions with inorganic reagents

4 Uses

4.1 Reagent in organic chemistry

5 Safety

6 References

Structure

Na2S adopts the antifluorite structure,[2][3] which means that the Na+ centers occupy sites of the fluoride in the CaF2 framework, and the larger S2− occupy the sites for Ca2+.

Production

Industrially Na2S is produced by carbothermic reduction of sodium sulfate often using coal:[4]

Na2SO4 + 2 C → Na2S + 2 CO2

In the laboratory, the salt can be prepared by reduction of sulfur with sodium in anhydrous ammonia, or by sodium in dry THF with a catalytic amount of naphthalene (forming sodium naphthalenide):[5]

2 Na + S → Na2S

Reactions with inorganic reagents

The sulfide ion in sulfide salts such as sodium sulfide can incorporate a proton into the salt by protonation:

S2−

+  H+ → SH−

Because of this capture of the proton ( H+), sodium sulfide has basic character. Sodium sulfide is strongly basic, able to absorb two protons. Its conjugate acid is sodium hydrosulfide (SH−

). An aqueous solution contains a significant portion of sulfide ions that are singly protonated.

S2−

+ H

2O {\displaystyle {\ce {<=>>}}}{\displaystyle {\ce {<=>>}}} SH−

+  OH−

 

 

 

 

(1)

SH−

+ H

2O {\displaystyle {\ce {<<=>}}}{\displaystyle {\ce {<<=>}}} H

2S +  OH−

 

 

 

 

(2)

Sodium sulfide is unstable in the presence of water due to the gradual loss of hydrogen sulfide into the atmosphere.

When heated with oxygen and carbon dioxide, sodium sulfide can oxidize to sodium carbonate and sulfur dioxide:

2 Na2S + 3 O2 + 2 CO

2 → 2 Na2CO3 + 2 SO2

Oxidation with hydrogen peroxide gives sodium sulfate:[6]

Na2S + 4 H2O2 → 4 H

2O + Na2SO4

Upon treatment with sulfur, polysulfides are formed:

2 Na2S + S8 → 2 Na2S5

Uses

Sodium sulfide is primarily used in the kraft process in the pulp and paper industry.

It is used in water treatment as an oxygen scavenger agent and also as a metals precipitant; in chemical photography for toning black and white photographs; in the textile industry as a bleaching agent, for desulfurising and as a dechlorinating agent; and in the leather trade for the sulfitisation of tanning extracts. It is used in chemical manufacturing as a sulfonation and sulfomethylation agent. It is used in the production of rubber chemicals, sulfur dyes and other chemical compounds. It is used in other applications including ore flotation, oil recovery, making dyes, and detergent. It is also used during leather processing, as an unhairing agent in the liming operation.

Reagent in organic chemistry

Alkylation of sodium sulfide give thioethers:

Na2S + 2 RX → R2S + 2 NaX

Even aryl halides participate in this reaction.[7] By a broadly similar process sodium sulfide can react with alkenes in the thiol-ene reaction to give thioethers. Sodium sulfide can be used as nucleophile in Sandmeyer type reactions.[8] Sodium sulfide reduces1,3-dinitrobenzene derivatives to the 3-nitroanilines.[9] Aqueous solution of sodium sulfide can be refluxed with nitro carrying azo dyes dissolved in dioxane and ethanol to selectively reduce the nitro groups to amine; while other reducible groups, e.g. azo group, remain intact.[10] Sulfide has also been employed in photocatalytic applications.[11]

Explanation:there you go

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PLEASE ANSWER + BRAINLIEST !!!
Sunny_sXe [5.5K]
Hope this would help you

3 0
3 years ago
Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium
dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

8 0
3 years ago
What is the volume at stp of 720.0 ml of a gas collected at 20.0 °c and 3.00 atm pressure?
MArishka [77]
Combined gas law is 
       PV/T = K (constant)

P = Pressure
V = Volume
T = Temperature in Kelvin

For two situations, the combined gas law can be applied as,
     P₁V₁ / T₁ = P₂V₂ / T₂

P₁ = 3.00 atm                                     P₂ = standard pressure = 1 atm
V₁ = 720.0 mL                                    T₂ = standard temperature = 273 K
T₁ = (273 + 20) K = 293 K
  
By substituting,
 3.00 atm x 720.0 mL / 293 K = 1 atm x V₂ / 273 K
                                          V₂ = 2012.6 mL

hence the volume of gas at stp is 2012.6 mL
5 0
3 years ago
Use calc to determine whether it is possible to remove 99.99% Cu2 by converting it to Cu(s) in a solution mixture containing 0.1
inessss [21]

Answer:

it is possible to remove 99.99% Cu2 by converting it to Cu(s)

Explanation:

So, from the question/problem above we are given the following ionic or REDOX equations of reactions;

Cu2+ + 2e- <--------------------------------------------------------------> Cu (s) Eo= 0.339 V

Sn2+ + 2e- <---------------------------------------------------------------> Sn (s) Eo= -0.141 V

In order to convert 99.99% Cu2 into Cu(s), the equation of reaction given below is needed:

Cu²⁺ + Sn ----------------------------------------------------------------------------> Cu + Sn²⁺.

Therefore, E°[overall] = 0.339 - [-0.141] = 0.48 V.

Therefore, the change in Gibbs' free energy, ΔG° = - nFE°. Where E° = O.48V, n= 2 and F = 96500 C.

Thus, ΔG° = - 92640.

This is less than zero[0]. Therefore,  it is possible to remove 99.99% Cu2 by converting it to Cu(s) because the reaction is a spontaneous reaction.  

7 0
3 years ago
What role does heat have in plate tectonics?
kap26 [50]
The answer to your question is the option D, it moves crust plates around
5 0
3 years ago
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