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guapka [62]
3 years ago
15

Describe and explain the stages of making copper chloride crystals and acid.

Chemistry
2 answers:
bulgar [2K]3 years ago
6 0

Answer:

Add excess copper carbonate to hydrochloric acid in a beaker, stirring until there is no further reaction. Filter the mixture to remove the remaining copper carbonate, then heat the remaining solution to the point of crystallisation. Leave the solution to cool and collect the crystals formed by filtration.

Explanation:

yKpoI14uk [10]3 years ago
6 0

Answer:

CHICKEN

Explanation:

EGG

LEG

PIC

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How many formula units are contained in 23.5 g of Sb2S3
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The molar mass of Sb2S3 is approximately equal to 339.7 g/mol. We calculate the number of moles of Sb2S3 by dividing the given mass by the molar mass. 
                                     n = 23.5 g / (339.7 g/mol)
                                        n = 0.0692 mols
To calculate for the number of formula units, we multiply the number of mols by the Avogadro's number, 
                            number of formula units = (0.0692 mols)(6.022 x 10^3)
                                                   = 4.167 x 10^22 formula units
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Sodium sulfate is slowly added to a solution containing 0.0500 M Ca 2 + ( aq ) and 0.0390 M Ag + ( aq ) . What will be the conce
levacccp [35]

The given question is incomplete. The complete question is as follows.

Sodium sulfate is slowly added to a solution containing 0.0500 M Ca^{2+}(aq) and 0.0390 M Ag^{+}(aq). What will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate? What percentage of the Ca^{2+}(aq) can be separated from the Ag(aq) by selective precipitation?

Explanation:

The given reaction is as follows.

      Ag_{2}SO_{4} \rightleftharpoons 2Ag^{+} + SO^{2-}_{4}

[Ag^{+}] = 0.0390 M

When Ag_{2}SO_{4} precipitates then expression for K_{sp} will be as follows.

         K_{sp} = [Ag^{+}]^{2}[SO^{2-}_{4}]

        1.20 \times 10^{-5} = (0.0390)^{2} \times [SO^{2-}_{4}]

       [SO^{2-}_{4}] = 0.00788 M

Now, equation for dissociation of calcium sulfate is as follows.

         CaSO_{4} \rightleftharpoons Ca^{2+} + SO^{2-}_{4}

      K_{sp} = [Ca^{2+}][SO^{2-}_{4}]

     4.93 \times 10^{-5} = [Ca^{2+}] \times 0.00788

           [Ca^{2+}] = 0.00625 M

Now, we will calculate the percentage of Ca^{2+} remaining in the solution as follows.

               \frac{0.00625}{0.05} \times 100

                 = 12.5%

And, the percentage of Ca^{2+} that can be separated is as follows.

                     100 - 12.5

                     = 87.5%

Thus, we can conclude that 87.5% will be the concentration of Ca^{2+}(aq) when Ag_{2}SO_{4}(s) begins to precipitate.

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