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schepotkina [342]
3 years ago
11

Which statement best describes the properties of ionic compounds? They are brittle, solid at room temperature, and have a high m

elting point. They are always composed of an equal proportion of cations to anions. They mostly have low melting and boiling points. They are good conductors of heat and electricity and are ductile and malleable.
Chemistry
2 answers:
djverab [1.8K]3 years ago
6 0
They are good conductors of heat and electricity.

They are solid at room tempature

They have a high melting point
Aleks04 [339]3 years ago
4 0
The answer is the first one! 

I hope this helps. :)
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A solution contains 0.036 M Cu2+ and 0.044 M Fe2+. A solution containing sulfide ions is added to selectively precipitate one of
Ratling [72]

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The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

Explanation:

<u>Step 1: </u>Data given

The solution contains 0.036 M Cu2+ and 0.044 M Fe2+

Ksp (CuS) = 1.3 × 10-36

Ksp (FeS) = 6.3 × 10-18

Step 2:  Calculate precipitate

CuS → Cu^2+ + S^2-         Ksp= 1.3*10^-36

FeS → Fe^2+ + S^2-      Ksp= 6.3*10^-18

Calculate the minimum of amount needed to form precipitates:

Q=Ksp

<u>For copper</u>  we have:  Ksp=[Cu2+]*[S2-]

Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]

[S2-]= 3.61*10^-35 M

<u>For Iron</u>  we have: Ksp=[Fe2+]*[S2-]

Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]

[S2-]= 1.43*10^-16 M

CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.

The precipitate is CuS.

Sulfide will precipitate at  [S2-]= 3.61*10^-35 M

3 0
3 years ago
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