Use the sequence E (NaCl, Na2SO4, then Na2S). Silver is insoluble as a chloride, so it would be removed first, the others (Pb and Ni) are soluble as chlorides(Note; lead chloride is soluble as a hot solution but will ppt when cold), next, PbSO4 is insoluble but NiSO4 is soluble so use Na2SO4 to separate lead from nickel. Lastly, nickel sulfide is insoluble and can be separated and collected.

Hope I helped :)

6 0

2 years ago

Read 2 more answers

Other questions:

vlabodo [156]

Answer:

The functional group present in Cinnamon are : Aromatic , alkene , aldehyde

The functional group present in Almonds : Aromatic , aldehyde

The functional group present in DHA : Ketone , alcohol

Explanation:

Almond :

Chemical formula : $C_{7}H_{6}O$

It contain : A benzene ring (Aromatic) , a -CHO group (aldehyde)

Cinnamon Sticks :

Chemical Formula : $C_{9}H_{8}O$

It contain : A benzene ring (Aromatic) ,an alkene group (C=C) , a -CHO group (aldehyde)

DHA:

Chemical Formula : $C_{3}H_{6}O_{3}$

It contain : Alcohol group (-OH) , ketone group (C=O)

8 0

3 years ago

Why is the formation of ionic compounds exothermic?

bazaltina [42]

TBH i dont know but can you give advice to get more points on this site

6 0

3 years ago

Ca(OH)₂(s) precipitates when a 1.0 g sample of CaC₂(s) is added to 1.0 L of distilled water at room temperature. If a 0.064 g sa

Alex

Answer:

D) Ca(OH)₂ will not precipitate because Q < Ksp

Explanation:

This is a question in which we will employ the reaction quotient Q to determine whether a precipitate will form.

Here we have first a chemical reaction in which Ca(OH)₂ is produced:

CaC₂(s) + H₂O ⇒ Ca(OH)₂ + C₂H₂

Ca(OH)₂ is slightly soluble, and depending on its concentration it may precipitate out of solution.

The solubility product constant for Ca(OH)₂ is:

Ca(OH)₂(s) ⇆ Ca²⁺(aq) + 2OH⁻(aq)

Ksp = [Ca²⁺][OH⁻]²

and the reaction quotient Q:

Q = [Ca²⁺][OH⁻]²

So by comparing Q with Ksp we will be able to determine if a precipitate will form.

From the stoichiometry of the reaction we know the number of moles of hydroxide produced, and since the volume is 1 L the molarity will also be known.

mol Ca(OH)₂ = mol CaC₂( reacted = 0.064 g / 64 g/mol = 0.001 mol Ca(OH)₂

Thus the concentration of ions will be:

[Ca²⁺ ] = 0.001 mol / L 0.001 M

[OH⁻] = 2 x 0.001 M = 0.002 M ( From the coefficient 2 in the equilibrium)

Now we can calculate the reaction quotient.

Q= [Ca²⁺][OH⁻]² = 0.001 x (0.002)² = 4.0 x 10⁻⁹

Q < Ksp since 4.0 x 10⁻⁹ < 8.0 x 10⁻⁸

Therefore no precipitate will form.

The answer that matches is (D)

6 0

2 years ago