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Norma-Jean [14]
3 years ago
7

Identify the precipitate (if any) that forms when Na2S and CdSO4 are mixed.

Chemistry
2 answers:
andrezito [222]3 years ago
6 0

Answer : Cadmium sulfide (CdS) precipitate is formed.

Explanation :

When the sodium sulfide, (Na_2S) is mixed with cadmium sulfate, (CdSO_4) then cadmium sulfide precipitate and sodium sulfate, (Na_2SO_4) is formed. The cadmium sulfide is insoluble in hot and cold water but sodium sulfate is soluble in water.

The balanced chemical reaction will be,

Na_2S+CdSO_4\rightarrow Na_2SO_4+CdS

This reaction is an example of double displacement reaction in which the cation and anion of the two reactants exchange their places to give two new products.

This is a chemical reaction where the two clear solutions mixed to produce a bright yellow colored precipitate of cadmium sulfide.

Flauer [41]3 years ago
4 0

The precipitate that is formed when {\text{N}}{{\text{a}}_2}{\text{S}} reacts with {\text{CdS}}{{\text{O}}_4}  is  \boxed{{\text{CdS}}}

The given reaction occurs as follows:

 {\text{N}}{{\text{a}}_2}{\text{S}}+{\text{CdS}}{{\text{O}}_4}\to{\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}\left( {aq}\right)+{\text{CdS}}\left( s \right)

Further Explanation:

<u>Precipitation reaction:</u>

It is the type of reaction in which an insoluble salt is formed by the combination of two solutions containing soluble salts. That insoluble salt is known as precipitate, 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.

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 and that of transition metals are insoluble in nature.

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. All sulfates of metals are soluble, except for lead, mercury (I), barium, and calcium sulfates.

The given reaction occurs as follows:

 {\text{N}}{{\text{a}}_2}{\text{S}}+{\text{CdS}}{{\text{O}}_4}\to{\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}+{\text{CdS}}

This is an example of a double displacement reaction in which two ionic compounds are exchanged with each other and two new compounds are formed. {\text{N}}{{\text{a}}_2}{\text{S}}{{\text{O}}_4}  is a soluble salt according to the solubility rules. So CdS will form precipitate in this reaction.

Learn more:

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

2. The net ionic equation for the reaction of {\text{MgS}}{{\text{O}}_{\text{4}}} with {\text{Sr}}{\left({{\text{N}}{{\text{O}}_3}}\right)_2} : brainly.com/question/4357519

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Chemical reaction and equation

Keywords: precipitation reaction, precipitate, insoluble, soluble, CdS, Na2SO4, Na2S, CdSO4, double displacement, salt, solubility rules.

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A rigid container of O 2 has a pressure of 340 kPa at a temperature of 713 K. What is the pressure at 273°C ?
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Answer:

260 kPa

Explanation:

To answer this problem we can use <em>Gay-Lussac's law</em>, which states that at a constant volume (such as is the case with a rigid container):

  • P₁T₂=P₂T₁

Where in this case:

  • P₁ = 340 kPa
  • T₂ = 273 °C ⇒ 273 + 273.16 = 546 K
  • P₂ = ?
  • T₁ = 713 K

We <u>input the data</u>:

  • 340 kPa * 546 K = P₂ * 713 K

And <u>solve for P₂</u>:

  • P₂ =  260 kPa
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Elaborate on how the isotopes and their relative abundances affect the average atomic mass of an element.
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ACTIVITY 1

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A solution is prepared by dissolving 15.0 g of nh3 in 250.0 g of water. the density of the resulting solution is 0.974 g/ml. the
iris [78.8K]

To solve this problem, first we assume the volume is purely additive. The density of the mixture can then be calculated by the summation of mass fraction of each component divided by its individual density:

1 / ρ mixture = (x NH3 / ρ NH3) + (x H2O / ρ<span> H2O)                        ---> 1</span>

Calculating for mass fraction of NH3:

x NH3 = 15 g / (15 g + 250 g)

x NH3 = 0.0566

Therefore the mass fraction of water is:

x H2O = 1 – x NH3 = 1 – 0.0566

x H2O = 0.9434

Assuming that the density of water is 1 g / mL and substituting the known values back to equation 1:

1 / 0.974 g / mL = [0.0566 / (ρ NH3)] + [0.9434 / (1 g / mL)]

ρ NH3 = 0.680 g / mL

Given the density of NH3, now we can calculate for the volume of NH3:

V NH3 = 15 g / 0.680 g / mL

V NH3 = 22.07 mL

The number of moles NH3 is: (molar mass NH3 is 17.03 g / mol)

n NH3 = 15 g / 17.03 g / mol

n NH3 = 0.881 mol

Therefore the molarity of NH3 in the solution is:

<span>Molarity = 0.881 mol / [(22.07 mL  + 250 mL) * (1L / 1000 mL)</span>

<span>M = 3.238 mol/L = 3.24 M</span>

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